Dr. GWF Drake's Research Group - User contributions [en]

In The News

2026-03-03T22:08:36Z

Drake: /* 2024 */

==2025==
* Evan Petrimoulx completed his 4th year undergraduate thesis project entitled "[[Media:Petrimoulx Thesis.pdf|Investigation of the Higher-order Zeeman Effect]]."

==2019==
*Congratulations to M.Sc. student Aaron Bondy for the award of a CGS Scholarship by NSERC!

==2017==
*Dr. Drake retires from his regular faculty position, but remains active in graduate studies and research as a Professor Emeritus.
==2015==
*Dr. Drake awarded the CAP Peter Kirkby Memorial Medal for outstanding service to Canadian Physics
**[https://services.cap.ca/medal/publicity/press.php?year=2015&medal_id=8 CAP Press Release]

==2010==
[[Image:Sharcnet.png|x45px|right]]
*Dr. Drake named SHARCnet site leader
**[[Media:UW_Sharcnet_head.pdf|U of W Press Release]] [http://web4.uwindsor.ca/units/physics/PhysicsM.nsf/SubCategoryFlyOut/AD4B5DFB74DAD9BD8525740C006ABB75]

==2009==
*Dr. Drake to head Canterbury College as Principal
**[[Media:UW_Canterbury.pdf|U of W Press Release]] [http://web4.uwindsor.ca/units/physics/PhysicsM.nsf/SubCategoryFlyOut/AD4B5DFB74DAD9BD8525740C006ABB75]
**[[Media:UW_Canterbury2.pdf|U of W Physics Dept.]] [http://web4.uwindsor.ca/units/physics/PhysicsM.nsf/SubCategoryFlyOut/AD4B5DFB74DAD9BD8525740C006ABB75]

==2008==
[[Image:PSAS2008.jpg|x120px|right]]
*International conference on the Precision Physics of Simple Atomic Systems [http://web2.uwindsor.ca/psas/ (PSAS)] hosted by the Windsor Physics Department (Dr. Drake, Chair)
**[[Media:Star_PSAS.pdf|Windsor Star]] [http://www.canada.com/windsorstar/news/story.html?id=862e2548-bfb5-4ddc-a7be-6e4eb6d14ed0]
**[[Media:UW_PSAS.pdf|U of W Press Release]] [http://web4.uwindsor.ca/units/physics/PhysicsM.nsf/SubCategoryFlyOut/AD4B5DFB74DAD9BD8525740C006ABB75]
*Physicists Assert No Conflict Between Religion and Science
**[[Media:Star_Religion.pdf|Windsor Star]] [http://www.canada.com/windsorstar/news/local/story.html?id=ac4f5f43-a89e-4715-b3fc-718ea820662b]

==2006==
*Dr. Drake named as Editor of the journal ''Physical Review A'', published by the American Physical Society
**[https://journals.aps.org/pra/edannounce/PRAv73i2_2.html Announcement]

In The News

2026-03-03T21:56:19Z

Drake: /* 2024 */

==2024==
Evan Petrimoulx completed his undergraduate thesis project entitled "[[Media:Petrimoulx Thesis.pdf|Investigation of the Higher-order Zeeman Effect]]."

==2019==
*Congratulations to M.Sc. student Aaron Bondy for the award of a CGS Scholarship by NSERC!

==2017==
*Dr. Drake retires from his regular faculty position, but remains active in graduate studies and research as a Professor Emeritus.
==2015==
*Dr. Drake awarded the CAP Peter Kirkby Memorial Medal for outstanding service to Canadian Physics
**[https://services.cap.ca/medal/publicity/press.php?year=2015&medal_id=8 CAP Press Release]

==2010==
[[Image:Sharcnet.png|x45px|right]]
*Dr. Drake named SHARCnet site leader
**[[Media:UW_Sharcnet_head.pdf|U of W Press Release]] [http://web4.uwindsor.ca/units/physics/PhysicsM.nsf/SubCategoryFlyOut/AD4B5DFB74DAD9BD8525740C006ABB75]

==2009==
*Dr. Drake to head Canterbury College as Principal
**[[Media:UW_Canterbury.pdf|U of W Press Release]] [http://web4.uwindsor.ca/units/physics/PhysicsM.nsf/SubCategoryFlyOut/AD4B5DFB74DAD9BD8525740C006ABB75]
**[[Media:UW_Canterbury2.pdf|U of W Physics Dept.]] [http://web4.uwindsor.ca/units/physics/PhysicsM.nsf/SubCategoryFlyOut/AD4B5DFB74DAD9BD8525740C006ABB75]

==2008==
[[Image:PSAS2008.jpg|x120px|right]]
*International conference on the Precision Physics of Simple Atomic Systems [http://web2.uwindsor.ca/psas/ (PSAS)] hosted by the Windsor Physics Department (Dr. Drake, Chair)
**[[Media:Star_PSAS.pdf|Windsor Star]] [http://www.canada.com/windsorstar/news/story.html?id=862e2548-bfb5-4ddc-a7be-6e4eb6d14ed0]
**[[Media:UW_PSAS.pdf|U of W Press Release]] [http://web4.uwindsor.ca/units/physics/PhysicsM.nsf/SubCategoryFlyOut/AD4B5DFB74DAD9BD8525740C006ABB75]
*Physicists Assert No Conflict Between Religion and Science
**[[Media:Star_Religion.pdf|Windsor Star]] [http://www.canada.com/windsorstar/news/local/story.html?id=ac4f5f43-a89e-4715-b3fc-718ea820662b]

==2006==
*Dr. Drake named as Editor of the journal ''Physical Review A'', published by the American Physical Society
**[https://journals.aps.org/pra/edannounce/PRAv73i2_2.html Announcement]

In The News

2026-03-03T21:54:13Z

Drake: /* 2024 */

==2024==
Evan Petrimoulx completed his undergraduate thesis project entitled "Investigation of the Higher-order Zeeman Effect."
[[Media:Petrimoulx Thesis.pdf|Click here]]

==2019==
*Congratulations to M.Sc. student Aaron Bondy for the award of a CGS Scholarship by NSERC!

==2017==
*Dr. Drake retires from his regular faculty position, but remains active in graduate studies and research as a Professor Emeritus.
==2015==
*Dr. Drake awarded the CAP Peter Kirkby Memorial Medal for outstanding service to Canadian Physics
**[https://services.cap.ca/medal/publicity/press.php?year=2015&medal_id=8 CAP Press Release]

==2010==
[[Image:Sharcnet.png|x45px|right]]
*Dr. Drake named SHARCnet site leader
**[[Media:UW_Sharcnet_head.pdf|U of W Press Release]] [http://web4.uwindsor.ca/units/physics/PhysicsM.nsf/SubCategoryFlyOut/AD4B5DFB74DAD9BD8525740C006ABB75]

==2009==
*Dr. Drake to head Canterbury College as Principal
**[[Media:UW_Canterbury.pdf|U of W Press Release]] [http://web4.uwindsor.ca/units/physics/PhysicsM.nsf/SubCategoryFlyOut/AD4B5DFB74DAD9BD8525740C006ABB75]
**[[Media:UW_Canterbury2.pdf|U of W Physics Dept.]] [http://web4.uwindsor.ca/units/physics/PhysicsM.nsf/SubCategoryFlyOut/AD4B5DFB74DAD9BD8525740C006ABB75]

==2008==
[[Image:PSAS2008.jpg|x120px|right]]
*International conference on the Precision Physics of Simple Atomic Systems [http://web2.uwindsor.ca/psas/ (PSAS)] hosted by the Windsor Physics Department (Dr. Drake, Chair)
**[[Media:Star_PSAS.pdf|Windsor Star]] [http://www.canada.com/windsorstar/news/story.html?id=862e2548-bfb5-4ddc-a7be-6e4eb6d14ed0]
**[[Media:UW_PSAS.pdf|U of W Press Release]] [http://web4.uwindsor.ca/units/physics/PhysicsM.nsf/SubCategoryFlyOut/AD4B5DFB74DAD9BD8525740C006ABB75]
*Physicists Assert No Conflict Between Religion and Science
**[[Media:Star_Religion.pdf|Windsor Star]] [http://www.canada.com/windsorstar/news/local/story.html?id=ac4f5f43-a89e-4715-b3fc-718ea820662b]

==2006==
*Dr. Drake named as Editor of the journal ''Physical Review A'', published by the American Physical Society
**[https://journals.aps.org/pra/edannounce/PRAv73i2_2.html Announcement]

In The News

2026-03-03T21:52:13Z

Drake: /* 2024 */

==2024==
Evan Petrimoulx completed his undergraduate thesis project entitled "Investigation of the Higher-order Zeeman Effect."
[[Media:Petrimoulx_thesis.pdf|Click here]]

==2019==
*Congratulations to M.Sc. student Aaron Bondy for the award of a CGS Scholarship by NSERC!

==2017==
*Dr. Drake retires from his regular faculty position, but remains active in graduate studies and research as a Professor Emeritus.
==2015==
*Dr. Drake awarded the CAP Peter Kirkby Memorial Medal for outstanding service to Canadian Physics
**[https://services.cap.ca/medal/publicity/press.php?year=2015&medal_id=8 CAP Press Release]

==2010==
[[Image:Sharcnet.png|x45px|right]]
*Dr. Drake named SHARCnet site leader
**[[Media:UW_Sharcnet_head.pdf|U of W Press Release]] [http://web4.uwindsor.ca/units/physics/PhysicsM.nsf/SubCategoryFlyOut/AD4B5DFB74DAD9BD8525740C006ABB75]

==2009==
*Dr. Drake to head Canterbury College as Principal
**[[Media:UW_Canterbury.pdf|U of W Press Release]] [http://web4.uwindsor.ca/units/physics/PhysicsM.nsf/SubCategoryFlyOut/AD4B5DFB74DAD9BD8525740C006ABB75]
**[[Media:UW_Canterbury2.pdf|U of W Physics Dept.]] [http://web4.uwindsor.ca/units/physics/PhysicsM.nsf/SubCategoryFlyOut/AD4B5DFB74DAD9BD8525740C006ABB75]

==2008==
[[Image:PSAS2008.jpg|x120px|right]]
*International conference on the Precision Physics of Simple Atomic Systems [http://web2.uwindsor.ca/psas/ (PSAS)] hosted by the Windsor Physics Department (Dr. Drake, Chair)
**[[Media:Star_PSAS.pdf|Windsor Star]] [http://www.canada.com/windsorstar/news/story.html?id=862e2548-bfb5-4ddc-a7be-6e4eb6d14ed0]
**[[Media:UW_PSAS.pdf|U of W Press Release]] [http://web4.uwindsor.ca/units/physics/PhysicsM.nsf/SubCategoryFlyOut/AD4B5DFB74DAD9BD8525740C006ABB75]
*Physicists Assert No Conflict Between Religion and Science
**[[Media:Star_Religion.pdf|Windsor Star]] [http://www.canada.com/windsorstar/news/local/story.html?id=ac4f5f43-a89e-4715-b3fc-718ea820662b]

==2006==
*Dr. Drake named as Editor of the journal ''Physical Review A'', published by the American Physical Society
**[https://journals.aps.org/pra/edannounce/PRAv73i2_2.html Announcement]

In The News

2026-03-03T21:48:25Z

Drake: /* 2024 */

==2024==
Evan Petrimoulx completed his undergraduate thesis project entitled "Investigation of the Higher-order Zeeman Effect."
[[Media:Petrimoulx_thesis.pdf]]

==2019==
*Congratulations to M.Sc. student Aaron Bondy for the award of a CGS Scholarship by NSERC!

==2017==
*Dr. Drake retires from his regular faculty position, but remains active in graduate studies and research as a Professor Emeritus.
==2015==
*Dr. Drake awarded the CAP Peter Kirkby Memorial Medal for outstanding service to Canadian Physics
**[https://services.cap.ca/medal/publicity/press.php?year=2015&medal_id=8 CAP Press Release]

==2010==
[[Image:Sharcnet.png|x45px|right]]
*Dr. Drake named SHARCnet site leader
**[[Media:UW_Sharcnet_head.pdf|U of W Press Release]] [http://web4.uwindsor.ca/units/physics/PhysicsM.nsf/SubCategoryFlyOut/AD4B5DFB74DAD9BD8525740C006ABB75]

==2009==
*Dr. Drake to head Canterbury College as Principal
**[[Media:UW_Canterbury.pdf|U of W Press Release]] [http://web4.uwindsor.ca/units/physics/PhysicsM.nsf/SubCategoryFlyOut/AD4B5DFB74DAD9BD8525740C006ABB75]
**[[Media:UW_Canterbury2.pdf|U of W Physics Dept.]] [http://web4.uwindsor.ca/units/physics/PhysicsM.nsf/SubCategoryFlyOut/AD4B5DFB74DAD9BD8525740C006ABB75]

==2008==
[[Image:PSAS2008.jpg|x120px|right]]
*International conference on the Precision Physics of Simple Atomic Systems [http://web2.uwindsor.ca/psas/ (PSAS)] hosted by the Windsor Physics Department (Dr. Drake, Chair)
**[[Media:Star_PSAS.pdf|Windsor Star]] [http://www.canada.com/windsorstar/news/story.html?id=862e2548-bfb5-4ddc-a7be-6e4eb6d14ed0]
**[[Media:UW_PSAS.pdf|U of W Press Release]] [http://web4.uwindsor.ca/units/physics/PhysicsM.nsf/SubCategoryFlyOut/AD4B5DFB74DAD9BD8525740C006ABB75]
*Physicists Assert No Conflict Between Religion and Science
**[[Media:Star_Religion.pdf|Windsor Star]] [http://www.canada.com/windsorstar/news/local/story.html?id=ac4f5f43-a89e-4715-b3fc-718ea820662b]

==2006==
*Dr. Drake named as Editor of the journal ''Physical Review A'', published by the American Physical Society
**[https://journals.aps.org/pra/edannounce/PRAv73i2_2.html Announcement]

In The News

2026-03-03T21:46:36Z

Drake: /* 2024 */

==2024==
Evan Petrimoulx completed his undergraduate thesis project entitled "Investigation of the Higher-order Zeeman Effect."
[[Media:Petrimoulx thesis.pdf]]

==2019==
*Congratulations to M.Sc. student Aaron Bondy for the award of a CGS Scholarship by NSERC!

==2017==
*Dr. Drake retires from his regular faculty position, but remains active in graduate studies and research as a Professor Emeritus.
==2015==
*Dr. Drake awarded the CAP Peter Kirkby Memorial Medal for outstanding service to Canadian Physics
**[https://services.cap.ca/medal/publicity/press.php?year=2015&medal_id=8 CAP Press Release]

==2010==
[[Image:Sharcnet.png|x45px|right]]
*Dr. Drake named SHARCnet site leader
**[[Media:UW_Sharcnet_head.pdf|U of W Press Release]] [http://web4.uwindsor.ca/units/physics/PhysicsM.nsf/SubCategoryFlyOut/AD4B5DFB74DAD9BD8525740C006ABB75]

==2009==
*Dr. Drake to head Canterbury College as Principal
**[[Media:UW_Canterbury.pdf|U of W Press Release]] [http://web4.uwindsor.ca/units/physics/PhysicsM.nsf/SubCategoryFlyOut/AD4B5DFB74DAD9BD8525740C006ABB75]
**[[Media:UW_Canterbury2.pdf|U of W Physics Dept.]] [http://web4.uwindsor.ca/units/physics/PhysicsM.nsf/SubCategoryFlyOut/AD4B5DFB74DAD9BD8525740C006ABB75]

==2008==
[[Image:PSAS2008.jpg|x120px|right]]
*International conference on the Precision Physics of Simple Atomic Systems [http://web2.uwindsor.ca/psas/ (PSAS)] hosted by the Windsor Physics Department (Dr. Drake, Chair)
**[[Media:Star_PSAS.pdf|Windsor Star]] [http://www.canada.com/windsorstar/news/story.html?id=862e2548-bfb5-4ddc-a7be-6e4eb6d14ed0]
**[[Media:UW_PSAS.pdf|U of W Press Release]] [http://web4.uwindsor.ca/units/physics/PhysicsM.nsf/SubCategoryFlyOut/AD4B5DFB74DAD9BD8525740C006ABB75]
*Physicists Assert No Conflict Between Religion and Science
**[[Media:Star_Religion.pdf|Windsor Star]] [http://www.canada.com/windsorstar/news/local/story.html?id=ac4f5f43-a89e-4715-b3fc-718ea820662b]

==2006==
*Dr. Drake named as Editor of the journal ''Physical Review A'', published by the American Physical Society
**[https://journals.aps.org/pra/edannounce/PRAv73i2_2.html Announcement]

In The News

2026-03-03T21:44:39Z

Drake: /* 2024 */

==2024==
Evan Petrimoulx completed his undergraduate thesis project entitled "Investigation of the Higher-order Zeeman Effect."
[[File:Petrimoulx thesis.pdf]]

==2019==
*Congratulations to M.Sc. student Aaron Bondy for the award of a CGS Scholarship by NSERC!

==2017==
*Dr. Drake retires from his regular faculty position, but remains active in graduate studies and research as a Professor Emeritus.
==2015==
*Dr. Drake awarded the CAP Peter Kirkby Memorial Medal for outstanding service to Canadian Physics
**[https://services.cap.ca/medal/publicity/press.php?year=2015&medal_id=8 CAP Press Release]

==2010==
[[Image:Sharcnet.png|x45px|right]]
*Dr. Drake named SHARCnet site leader
**[[Media:UW_Sharcnet_head.pdf|U of W Press Release]] [http://web4.uwindsor.ca/units/physics/PhysicsM.nsf/SubCategoryFlyOut/AD4B5DFB74DAD9BD8525740C006ABB75]

==2009==
*Dr. Drake to head Canterbury College as Principal
**[[Media:UW_Canterbury.pdf|U of W Press Release]] [http://web4.uwindsor.ca/units/physics/PhysicsM.nsf/SubCategoryFlyOut/AD4B5DFB74DAD9BD8525740C006ABB75]
**[[Media:UW_Canterbury2.pdf|U of W Physics Dept.]] [http://web4.uwindsor.ca/units/physics/PhysicsM.nsf/SubCategoryFlyOut/AD4B5DFB74DAD9BD8525740C006ABB75]

==2008==
[[Image:PSAS2008.jpg|x120px|right]]
*International conference on the Precision Physics of Simple Atomic Systems [http://web2.uwindsor.ca/psas/ (PSAS)] hosted by the Windsor Physics Department (Dr. Drake, Chair)
**[[Media:Star_PSAS.pdf|Windsor Star]] [http://www.canada.com/windsorstar/news/story.html?id=862e2548-bfb5-4ddc-a7be-6e4eb6d14ed0]
**[[Media:UW_PSAS.pdf|U of W Press Release]] [http://web4.uwindsor.ca/units/physics/PhysicsM.nsf/SubCategoryFlyOut/AD4B5DFB74DAD9BD8525740C006ABB75]
*Physicists Assert No Conflict Between Religion and Science
**[[Media:Star_Religion.pdf|Windsor Star]] [http://www.canada.com/windsorstar/news/local/story.html?id=ac4f5f43-a89e-4715-b3fc-718ea820662b]

==2006==
*Dr. Drake named as Editor of the journal ''Physical Review A'', published by the American Physical Society
**[https://journals.aps.org/pra/edannounce/PRAv73i2_2.html Announcement]

In The News

2026-03-03T21:41:17Z

Drake: /* 2024 */

==2024==
Evan Petrimoulx completed his undergraduate thesis project entitled "Investigation of the Higher-order Zeeman Effect."
[[File:Petrimoulx_thesis.pdf]]

==2019==
*Congratulations to M.Sc. student Aaron Bondy for the award of a CGS Scholarship by NSERC!

==2017==
*Dr. Drake retires from his regular faculty position, but remains active in graduate studies and research as a Professor Emeritus.
==2015==
*Dr. Drake awarded the CAP Peter Kirkby Memorial Medal for outstanding service to Canadian Physics
**[https://services.cap.ca/medal/publicity/press.php?year=2015&medal_id=8 CAP Press Release]

==2010==
[[Image:Sharcnet.png|x45px|right]]
*Dr. Drake named SHARCnet site leader
**[[Media:UW_Sharcnet_head.pdf|U of W Press Release]] [http://web4.uwindsor.ca/units/physics/PhysicsM.nsf/SubCategoryFlyOut/AD4B5DFB74DAD9BD8525740C006ABB75]

==2009==
*Dr. Drake to head Canterbury College as Principal
**[[Media:UW_Canterbury.pdf|U of W Press Release]] [http://web4.uwindsor.ca/units/physics/PhysicsM.nsf/SubCategoryFlyOut/AD4B5DFB74DAD9BD8525740C006ABB75]
**[[Media:UW_Canterbury2.pdf|U of W Physics Dept.]] [http://web4.uwindsor.ca/units/physics/PhysicsM.nsf/SubCategoryFlyOut/AD4B5DFB74DAD9BD8525740C006ABB75]

==2008==
[[Image:PSAS2008.jpg|x120px|right]]
*International conference on the Precision Physics of Simple Atomic Systems [http://web2.uwindsor.ca/psas/ (PSAS)] hosted by the Windsor Physics Department (Dr. Drake, Chair)
**[[Media:Star_PSAS.pdf|Windsor Star]] [http://www.canada.com/windsorstar/news/story.html?id=862e2548-bfb5-4ddc-a7be-6e4eb6d14ed0]
**[[Media:UW_PSAS.pdf|U of W Press Release]] [http://web4.uwindsor.ca/units/physics/PhysicsM.nsf/SubCategoryFlyOut/AD4B5DFB74DAD9BD8525740C006ABB75]
*Physicists Assert No Conflict Between Religion and Science
**[[Media:Star_Religion.pdf|Windsor Star]] [http://www.canada.com/windsorstar/news/local/story.html?id=ac4f5f43-a89e-4715-b3fc-718ea820662b]

==2006==
*Dr. Drake named as Editor of the journal ''Physical Review A'', published by the American Physical Society
**[https://journals.aps.org/pra/edannounce/PRAv73i2_2.html Announcement]

In The News

2026-03-03T21:36:39Z

Drake: /* 2024 */

==2024==
Evan Petrimoulx completed his undergraduate thesis project.
[[File:Petrimoulx_thesis.pdf]]

==2019==
*Congratulations to M.Sc. student Aaron Bondy for the award of a CGS Scholarship by NSERC!

==2017==
*Dr. Drake retires from his regular faculty position, but remains active in graduate studies and research as a Professor Emeritus.
==2015==
*Dr. Drake awarded the CAP Peter Kirkby Memorial Medal for outstanding service to Canadian Physics
**[https://services.cap.ca/medal/publicity/press.php?year=2015&medal_id=8 CAP Press Release]

==2010==
[[Image:Sharcnet.png|x45px|right]]
*Dr. Drake named SHARCnet site leader
**[[Media:UW_Sharcnet_head.pdf|U of W Press Release]] [http://web4.uwindsor.ca/units/physics/PhysicsM.nsf/SubCategoryFlyOut/AD4B5DFB74DAD9BD8525740C006ABB75]

==2009==
*Dr. Drake to head Canterbury College as Principal
**[[Media:UW_Canterbury.pdf|U of W Press Release]] [http://web4.uwindsor.ca/units/physics/PhysicsM.nsf/SubCategoryFlyOut/AD4B5DFB74DAD9BD8525740C006ABB75]
**[[Media:UW_Canterbury2.pdf|U of W Physics Dept.]] [http://web4.uwindsor.ca/units/physics/PhysicsM.nsf/SubCategoryFlyOut/AD4B5DFB74DAD9BD8525740C006ABB75]

==2008==
[[Image:PSAS2008.jpg|x120px|right]]
*International conference on the Precision Physics of Simple Atomic Systems [http://web2.uwindsor.ca/psas/ (PSAS)] hosted by the Windsor Physics Department (Dr. Drake, Chair)
**[[Media:Star_PSAS.pdf|Windsor Star]] [http://www.canada.com/windsorstar/news/story.html?id=862e2548-bfb5-4ddc-a7be-6e4eb6d14ed0]
**[[Media:UW_PSAS.pdf|U of W Press Release]] [http://web4.uwindsor.ca/units/physics/PhysicsM.nsf/SubCategoryFlyOut/AD4B5DFB74DAD9BD8525740C006ABB75]
*Physicists Assert No Conflict Between Religion and Science
**[[Media:Star_Religion.pdf|Windsor Star]] [http://www.canada.com/windsorstar/news/local/story.html?id=ac4f5f43-a89e-4715-b3fc-718ea820662b]

==2006==
*Dr. Drake named as Editor of the journal ''Physical Review A'', published by the American Physical Society
**[https://journals.aps.org/pra/edannounce/PRAv73i2_2.html Announcement]

File:Petrimoulx Thesis.pdf

2026-03-03T21:35:01Z

Drake: Evan Petrimoulx's 4th year project thesis entitled Investigation of the Higher Order Zeeman Effect

== Summary ==
Evan Petrimoulx's 4th year project thesis entitled Investigation of the Higher Order Zeeman Effect

In The News

2026-03-03T21:29:31Z

Drake:

==2024==
Evan Petrimoulx completed his undergraduate thesis project.

==2019==
*Congratulations to M.Sc. student Aaron Bondy for the award of a CGS Scholarship by NSERC!

==2017==
*Dr. Drake retires from his regular faculty position, but remains active in graduate studies and research as a Professor Emeritus.
==2015==
*Dr. Drake awarded the CAP Peter Kirkby Memorial Medal for outstanding service to Canadian Physics
**[https://services.cap.ca/medal/publicity/press.php?year=2015&medal_id=8 CAP Press Release]

==2010==
[[Image:Sharcnet.png|x45px|right]]
*Dr. Drake named SHARCnet site leader
**[[Media:UW_Sharcnet_head.pdf|U of W Press Release]] [http://web4.uwindsor.ca/units/physics/PhysicsM.nsf/SubCategoryFlyOut/AD4B5DFB74DAD9BD8525740C006ABB75]

==2009==
*Dr. Drake to head Canterbury College as Principal
**[[Media:UW_Canterbury.pdf|U of W Press Release]] [http://web4.uwindsor.ca/units/physics/PhysicsM.nsf/SubCategoryFlyOut/AD4B5DFB74DAD9BD8525740C006ABB75]
**[[Media:UW_Canterbury2.pdf|U of W Physics Dept.]] [http://web4.uwindsor.ca/units/physics/PhysicsM.nsf/SubCategoryFlyOut/AD4B5DFB74DAD9BD8525740C006ABB75]

==2008==
[[Image:PSAS2008.jpg|x120px|right]]
*International conference on the Precision Physics of Simple Atomic Systems [http://web2.uwindsor.ca/psas/ (PSAS)] hosted by the Windsor Physics Department (Dr. Drake, Chair)
**[[Media:Star_PSAS.pdf|Windsor Star]] [http://www.canada.com/windsorstar/news/story.html?id=862e2548-bfb5-4ddc-a7be-6e4eb6d14ed0]
**[[Media:UW_PSAS.pdf|U of W Press Release]] [http://web4.uwindsor.ca/units/physics/PhysicsM.nsf/SubCategoryFlyOut/AD4B5DFB74DAD9BD8525740C006ABB75]
*Physicists Assert No Conflict Between Religion and Science
**[[Media:Star_Religion.pdf|Windsor Star]] [http://www.canada.com/windsorstar/news/local/story.html?id=ac4f5f43-a89e-4715-b3fc-718ea820662b]

==2006==
*Dr. Drake named as Editor of the journal ''Physical Review A'', published by the American Physical Society
**[https://journals.aps.org/pra/edannounce/PRAv73i2_2.html Announcement]

Theory Notes

2024-09-23T00:51:35Z

Drake: /* Matrix Elements of H */

<center>
<H1>Notes on solving the Schrödinger equation in Hylleraas coordinates for heliumlike atoms
Gordon W.F. Drake
Department of Physics, University of Windsor
Windsor, Ontario, Canada N9B 3P4
(Transcribed from hand-written notes and edited by Lauren Moffatt. Last revised November 22, 2017 to correct a missing subscript in the expression for $B_1$ in the Hamiltonian in Sect. 11 General Hermitian Property.)
</H1>
</center>

==Introduction==
These notes describe some of the technical details involved in solving the Schrödinger equation for a heliumlike atom of nuclear charge Z in correlated Hylleraas coordinates. The standard for computational purposes is to express all quantities in a form that is valid for electronic states of arbitrary angular momentum, and to express matrix elements of the Hamiltonian in a manifestly Hermitian symmetrized form. Further information and derivations of the mathematical techniques for angular integrals and radial recursion relations can be found in the paper
G.W.F. Drake, Phys. Rev. A <math>{\bf 18}</math>, 820 (1978).

The starting point is the two-electron Schrödinger equation for infinite nuclear mass
\begin{equation} \left[-\frac{\hbar^2}{2m}(\nabla^2_1 +\nabla^2_2) - \frac{Ze^2}{r_1} -
\frac{Ze^2}{r_2}+\frac{e^2}{r_{12}} \right]\psi = E\psi\nonumber
\end{equation}
where <math>m</math> is the electron mass, and <math>r_{12} = |{\bf r}_1 - {\bf r}_2|</math> (see diagram below).

[[File:Pic1.jpg|250px]]

Begin by rescaling distances and energies so that the Schrödinger equation can be expressed in a dimensionless form. The dimensionless Z-scaled distance is defined by <math>\begin{eqnarray} \rho = \frac{Zr}{a_0} \end{eqnarray} </math> where <math> \begin{eqnarray} a_0 = \frac{\hbar^2}{me^2} \end{eqnarray} </math> is the atomic unit (a.u.) of distance equal to the Bohr radius <math>0.529\, 177\, 210\, 92(17) \times 10^{-10}</math> m. Then

\begin{equation}[-\frac{\hbar^2}{2m}Z^2(\frac{me^2}{\hbar^2})^2(\nabla^2_{\rho_1}+\nabla^2_{\rho_2})
- Z^2\frac{e^2}{a_0}\rho^{-1}_1 - Z^2\frac{e^2}{a_0}\rho^{-1}_2 +
\frac{e^2}{a_0}Z\rho^{-1}_{12}]\psi= E\psi\nonumber\end{equation}

But <math>\begin{eqnarray} \frac{\hbar^2}{m}(\frac{me^2}{\hbar^2})^2 = \frac{e^2}{a_0} \end{eqnarray} </math> is the hartree atomic unit of of energy <math>(E_h = 27.211\,385\,05(60)</math> eV, or equivalently <math>E_h/(hc) = 2194.746\,313\,708(11)</math> m<math>^{-1})</math>. Therefore, after multiplying through by <math>a_0/(Ze)^2</math>, the problem to be solved in Z-scaled dimensionless units becomes

\begin{equation} \left[-\frac{1}{2}(\nabla^2_{\rho_1}+\nabla^2_{\rho_2}) -
\frac{1}{\rho_1} - \frac{1}{\rho_2} + \frac{Z^{-1}}{\rho_{12}}\right]\psi =
\varepsilon\psi\nonumber \end{equation}

where <math>\begin{eqnarray}\varepsilon = \frac{Ea_0}{(Ze)^2} \end{eqnarray}</math> is the energy in Z-scaled atomic units. For convenience, rewrite this in the conventional form
\begin{equation}
H\psi = \varepsilon\psi\nonumber \end{equation}
where (using <math>r</math> in place of <math>\rho</math> for the Z-scaled distance)
\begin{equation}
H = -\frac{1}{2}(\nabla^2_1+\nabla^2_2) - \frac{1}{r_1}-\frac{1}{r_2} + \frac{Z^{-1}}{r_{12}}
\nonumber\end{equation}
is the atomic Hamiltonian for infinite nuclear mass.

==The Hartree Fock Method==

For purposes of comparison, and to define the correlation energy, assume that <math>\psi({\bf r}_1,{\bf r}_2)</math> can be written in the separable product form
\begin{equation}
\psi({\bf r}_1,{\bf r}_2) = \frac{1}{\sqrt{2}} \left[ u_1(r_1) u_2(r_2) \pm u_2(r_1) u_1(r_2) \right] \nonumber
\end{equation}
for the <math>1s^2\ ^1\!S</math> ground state. Because of the <math>\frac{1}{r_{12}}</math> term in the Schrödinger equation, the exact solution connot be expressed in this form as a separable product. However, the Hartree-Fock approximation corresponds to finding the best possible solution to the Schrödinger equation
\begin{equation}
H\psi({\bf r}_1,{\bf r}_2) = E\psi({\bf r}_1,{\bf r}_2)\nonumber
\end{equation}
that can nevertheless be expressed in this separable product form, where as before
\begin{equation}
H = -\frac{1}{2}(\nabla^2_1+\nabla^2_2) - \frac{1}{r_1}- \frac{1}{r_2} + \frac{Z^{-1}}{r_{12}}\nonumber
\end{equation}
is the full two-electron Hamiltonian. To find the best solution, substitute into <math>\left\langle\psi \lvert H-E \rvert \psi\right\rangle</math> and require this expression to be stationary
with respect to arbitrary infinitesimal variations <math>\delta u_1</math> and <math>\delta
u_2</math> in <math>u_1</math> and <math>u_2</math> respectively; i.e.

\begin{equation}
\frac{1}{2} \left\langle \delta u_1(r_1)u_2(r_2) \pm u_2(r_1)\delta
u_1(r_2) \lvert H-E \rvert u_1(r_1) u_2(r_2)\pm u_2(r_1) u_1(r_2)\right\rangle\nonumber
\end{equation}

\begin{equation}
=\int\delta u_1(r_1) d{\bf r}_1 \left\{ \int d{\bf r}_2 u_2(r_2) \left( H - E \right) \left[ u_1(r_1)u_2(r_2) \pm u_2(r_1) u_1(r_2) \right] \right\}\nonumber
\end{equation}

\begin{equation}
= 0 \nonumber
\end{equation}

for arbitrary <math> \delta u_1(r_1).\nonumber </math> Therefore <math>\{\int d{\bf r}_2 \ldots \} = 0</math>.

Similarly, the coefficient of <math>\delta u_2</math> would give

\begin{equation}
\int d{\bf r}_1 u_1(r_1) \left( H-E \right) \left[ u_1(r_1) u_2(r_2) \pm
u_2(r_1) u_1(r_2) \right] = 0\nonumber
\end{equation}

Define

\begin{equation}
I_{12} = I_{21} = \int d{\bf r}\, u_1(r)u_2(r), \nonumber
\end{equation}

\begin{equation}
H_{ij} = \int d{\bf r}\, u_i(r)(-\frac{1}{2}\nabla - \frac{1}{r})u_j(r), \nonumber
\end{equation}

\begin{equation}
G_{ij}(r) = \int d{\bf r}^\prime u_i(r^\prime)\frac{1}{|{\bf r} - {\bf r}\prime|}u_j(r^\prime)\nonumber
\end{equation}

Then the above equations become the pair of integro-differential equations

\begin{equation}
[ H_0 - E + H_{22}+G_{22}(r)]u_1(r) = \mp [ I_{12}(H_0-E) + H_{12}+G_{12}(r)]u_2(r)\nonumber
\end{equation}

\begin{equation}
[H_0-E+H_{11}+G_{11}(r)]u_2(r) = \mp [I_{12}(H_0-E) + H_{12}+G_{12}(r)]u_1(r)\nonumber
\end{equation}

These must be solved self-consistently for the "constants" <math>I_{12}</math> and
<math>H_{ij}</math> and the function <math>G_{ij}(r)</math>.

The Hartree Fock energy is <math>E \simeq -2.87\ldots </math>a.u. while the exact energy is <math>E = -2.903724\ldots </math>a.u.

The difference is called the "correlation energy" because it arises from the
way in which the motion of one electron is correlated to the other. The
Hartree Fock equations only describe how one electron moves in the average field
provided by the other (mean-field theory).

==Configuration Interaction==
Expand
\begin{equation}
\psi({\bf r}_1,{\bf r}_2)= C_0u^{(s)}_1(r_1)u^{(s)}_1(r_2) + C_1u^{(P)}_1({\bf r}_1)u^{(P)}_1({\bf r}_2){\cal Y}^0_{1,1,0}(\hat{\bf r}_1, \hat{\bf r}_2)+C_2u^{(d)}_1({\bf r}_1)u^{(d)}_2({\bf r}_2){\cal Y}^0_{2,2,0}(\hat{\bf r}_1, \hat{\bf r}_2)+... \pm {\rm\ exchange}
\end{equation}
where
\begin{equation}
{\cal Y}^M_{l_1,l_2,L}(\hat{\bf r}_1, \hat{\bf r}_2)=\sum_{m_1,m_2}Y^{m_1}_{l_1}({\bf r}_1)Y^{m_2}_{l_2}({\bf r}_2)\times <l_1l_2m_1m_2\mid LM>
\end{equation}

This works, but is slowly convergent, and very laborious. The best CI calculations are accurate to <math> ~10^{-7}</math> a.u.

==Hylleraas Coordinates==
[E.A. Hylleraas, Z. Phys. <math>{\bf 48}, 469(1928)</math> and <math>{\bf 54}, 347(1929)</math>]
suggested using the co-ordinates <math>r_1</math>, <math>r_2</math> and <math>r_{12}</math> or equivalently

\begin{eqnarray}
s &=& r_1 + r_2, \nonumber\\
t &=& r_1-r_2, \nonumber\\
u &=& r_{12}\nonumber
\end{eqnarray}

and writing the trial functions in the form

\begin{equation}
\Psi({\bf r}_1,{\bf r}_2) = \sum^{i+j+k\leq
N}_{i,j,k}c_{i,j,k}r_1^{i+l_1}r_2^{j+l_2}r_{12}^ke^{-\alpha r_1 - \beta r_2}
\mathcal{Y}^M_{l_1,l_2,L}(\hat{r}_1,\hat{r}_2)\pm \text{exchange}\nonumber
\end{equation}

Diagonalizing H in this non-orthogonal basis set is equivalent to solving
<math> \begin{eqnarray}
\frac{\partial E}{\partial c_{i,j,k}} = 0\nonumber
\end{eqnarray} </math>
for fixed <math>\alpha</math> and <math>\beta</math>.

The diagonalization must be repeated for different values of <math>\alpha</math> and
<math>\beta</math> in order to optimize the non-linear parameters.

==Completeness==
The completeness of the above basis set can be shown by first writing
\begin{equation}
r_{12}^2 = r_1^2 + r_2^2 - 2r_1r_2\cos\theta_{12}
\end{equation}
and, from the spherical harmonic addition theorem,
\begin{equation}
\cos(\theta_{12})=\frac{4\pi}{3}\sum^1_{m=-1}Y^{m*}_1(\theta_1,\varphi_1)Y^m_1(\theta_2,\varphi_2)
\end{equation}
Consider first the S-states. The <math>r_{12}^0</math> terms are like the ss terms in a CI calculation. The
<math>r_{12}^2</math> terms bring in p-p type contributions, and the higher powers bring
in d-d, f-f etc. type terms. In general

\begin{equation}
P_l(\cos\theta_{12}) = \frac{4\pi}{2l+1}\sum^l_{m=-l}{Y^{m}_l}^*(\theta_1,\varphi_1)Y^m_l(\theta_2,\varphi_2)\nonumber
\end{equation}

For P-states, one would have similarly

\begin{array}{lr}
r_{12}^0 & (sp)P\nonumber\\
r_{12}^2 & (pd)P\nonumber\\
r_{12}^4 & (df)P\nonumber\\
\vdots & \vdots\nonumber
\end{array}

For D-states

\begin{array}{lr}
r_{12}^0 & (sd)D & (pp^\prime)D\nonumber\\
r_{12}^2 & (pf)D & (dd^\prime)D\nonumber\\
r_{12}^4 & (dg)D & (ff^\prime)D\nonumber\\
\vdots & \vdots & \vdots\nonumber
\end{array}

In this case, since there are two "lowest-order" couplings to form a
D-state, both must be present in the basis set, i.e.

\begin{equation}
\Psi(r_2,r_2) = \sum c_{ijk}r_1^ir_2^{j+2}r_{12}^ke^{-\alpha r_1-\beta
r_2}\mathcal{Y}^M_{022}(\hat{r}_1,\hat{r}_2)
+\sum d_{ijk}r_1^{i+1}r_2^{j+1}r_{12^k}e^{-\alpha^\prime r_1 - \beta^\prime
r_2}\mathcal{Y}^M_{112}(\hat{r}_1,\hat{r}_2)\nonumber
\end{equation}

For F-states, one would need <math>(sf)F</math> and <math>(pd)F</math> terms.

For G-states, one would need <math>(sg)G</math>, <math>(pf)G</math> and <math>(dd^\prime)G</math> terms.

Completeness of the radial functions can be proven by considering the
Sturm-Liouville problem

\begin{equation}
\left(-\frac{1}{2}\nabla^2-\frac{\lambda}{r_s}-E\right)\psi({\bf r}) = 0\nonumber
\end{equation}

or

\begin{equation}
\left(-\frac{1}{2}\frac{1}{r^2}\frac{\partial}{\partial r}\left({r^2}\frac{\partial}{\partial r}\right) -
\frac{l\left(l+1\right)}{2r^2} - \frac{\lambda}{r} - E\right)u(r) = 0.\nonumber
\end{equation}

For fixed E and variable <math>\lambda</math> (nuclear charge).

The eigenvalues are <math>\lambda_n = (E/E_n)^{1/2}</math>, where <math>E_n =- \frac{1}{2n^2}</math>

[[File:Pic2.jpg|500px]]

and the eigenfunctions are

\begin{equation}
u_{nl}(r) =
\frac{1}{(2l+1)!}\left(\frac{(n+l)!}{(n-l-1)2!}\right)^{1/2}(2\alpha)^{3/2}e^{-\alpha
r}\nonumber
\times \ \left( 2 \alpha r \right)^l\,_{1}F_{1}\left(-n+l+1,2l+2; 2 \alpha r \right)
\end{equation}

with <math>\alpha = (-2E)^{1/2}</math> and <math>n\geq l+1</math>. The confluent hypergeometric function <math> _1F_1(a,b,;z)</math> then denotes a finite polynomial since <math>a=-n+l+1</math> is a negative integer or zero.

Unlike the hydrogen spectrum, which has both a discrete part for <math>E<0</math> and a
continuous part for <math>E>0</math>, this forms an entirely discrete set of finite
polynomials, called Sturmian functions. They are orthogonal with respect to
the potential, i.e.

\begin{equation}
\int^\infty_0 r^2dr \left( u_{n^\prime l}(r)\frac{1}{r}u_{nl}(r) \right) = \delta_{n,n^\prime}\nonumber
\end{equation}

Since they become complete in the limit <math>n\rightarrow\infty</math>, this assures the
completeness of the variational basis set.

[See also B Klahn and W.A. Bingel Theo. Chim. Acta (Berlin) 44, 9 and 27
(1977)].

==Solutions of the Eigenvalue Problem==

For convenience, write

\[
\Psi({\bf r}_1,{\bf r}_2) = \sum^N_{m=1}c_m\varphi_m\nonumber
\]

where <math>m</math> represents the <math>m</math>'th combination of <math>i,j,k</math> and

\begin{equation}
\varphi_{ijk}=r_1^ir_2^jr_{12}^ke^{-\alpha r_1 - \beta
r_2}\mathcal{Y}^M_{l_1,l_2,L}(\hat{r}_1,\hat{r}_2) \pm \mbox{exchange}\nonumber
\end{equation}

\begin{align}

&
\left( \begin{array}{lr}
\cos(\theta) & \sin(\theta)\\
-\sin(\theta) & \cos(\theta)
\end{array} \right)
\left( \begin{array}{lr}
H_{11} & H_{12} \\
H_{12} & H_{22}
\end{array}\right)
\left( \begin{array}{lr}
\cos(\theta) & -\sin(\theta)\\
\sin(\theta) & \cos(\theta)
\end{array} \right)
\\[5pt]
= & \left(\begin{array}{lr}
cH_{11}+sH_{12} & cH_{12} + sH_{22}\\
-sH_{11} + cH_{12} & -sH_{12} + cH_{22}
\end{array}\right)
\left( \begin{array}{cc}
c & -s \\
s & c
\end{array}\right)
\\[5pt]
= & \left(\begin{array}{lr}
c^2H_{11}+s^2H_{22} + 2csH_{12} & (c^2-s^2)H_{12}+cs(H_{22}-H_{11})\\
(c^2-s^2)H_{12}+cs(H_{22}-H_{11}) & s^2H_{11}+c^2H_{22}-2csH_{12}
\end{array}\right)
\end{align}

Therefore

\begin{equation}
\left[\cos^2(\theta)-\sin^2(\theta)\right]H_{12} =
\cos(\theta)\sin(\theta)(H_{11}-H_{22})\nonumber
\end{equation}

and

\begin{equation}
\tan(2\theta) = \frac{2H_{12}}{H_{11}-H_{22}}\nonumber
\end{equation}

i.e.

\begin{equation}
\cos(\theta)=\left(\frac{r+\omega}{2r}\right)^{1/2}\nonumber
\end{equation}

\begin{equation}
\sin(\theta)=-\mbox{sgn}(H_{12})\left(\frac{r-\omega}{2r}\right)^{1/2}\nonumber
\end{equation}

where

\begin{equation}
\omega = H_{22}-H_{11}\nonumber
\end{equation}

\begin{equation}
r=\left(\omega^2+4H_{12}^2\right)^{1/2}\nonumber
\end{equation}

\begin{equation}
E_1=\frac{1}{2}\left(H_{11}+H_{22}-r\right)\nonumber
\end{equation}

\begin{equation}
E_2=\frac{1}{2}\left(H_{11}+H_{22}+r\right)\nonumber
\end{equation}

===Brute Force Method===

- Gives all the eigenvalues and eigenvectors, but it is slow

- First orthonormalize the <math>N</math>-dimensional basis set; i.e. form linear combinations

\begin{equation}
\Phi_m = \sum_{n=1}^N\varphi_nR_{nm}\nonumber
\end{equation}

such that
\begin{equation}
\langle \Phi_m | \Phi_n \rangle = \delta_{m,n}
\end{equation}

This can be done by finding an orthogonal tranformation T such that

\begin{equation}
T^TOT = I = \left(
\begin{array}{ccccc}
I_1 & 0 & 0 & \ldots & 0 \\
0 & I_2 & 0 & \ldots & 0 \\
0 & 0 & I_3 & \ldots & 0 \\
\vdots & \vdots & \vdots & \ddots & \vdots \\
0 & 0 & 0 & \ldots & I_N
\end{array}\right);
\end{equation}

\begin{equation}
O_{mn} = \langle \varphi_m | \varphi_n \rangle \nonumber
\end{equation}

and then applying a scale change matrix

\begin{equation}
S = \left(\begin{array}{ccccc}
\frac{1}{I_1^{1/2}} & 0 & 0 & \ldots & 0\\
0 & \frac{1}{I_2^{1/2}} & 0 & \ldots & 0 \\
0 & 0 & \frac{1}{I_3^{1/2}} & \ldots & 0 \\
\vdots & \vdots & \vdots & \ddots & \vdots \\
0 & 0 & 0 & \ldots & I_N^{1/2}
\end{array}\right)= S^T\nonumber
\end{equation}

Then

\begin{equation}
S^TT^TOTS = 1
\end{equation}

i.e.

\begin{equation}
R^TOR = 1
\end{equation}

with <math>R=TS</math>.

If H is the matrix with elements <math>H_{mn} = \langle \varphi_m | \varphi_n \rangle </math>, then H
expressed in the <math>\Phi_m</math> basis set is

\begin{equation}
H^\prime = R^THR.\nonumber
\end{equation}

We next diagonalize <math>H^\prime</math> by finding an orthogonal transformation W such
that

\begin{equation}
W^T H^\prime W = \lambda = \left(
\begin{array}{cccc}
\lambda_1 & 0 & \ldots & 0 \\
0 & \lambda_2 & \ldots & 0 \\
\vdots & \vdots & \ddots & \vdots \\
0 & 0 & \ldots & \lambda_N
\end{array}\right)\nonumber
\end{equation}

The q'th eigenvector is

\begin{equation}
\Psi^{(q)} = \sum_{n=1}^N\Phi_n W_{n,q} = \sum_{n,n^\prime} \varphi_{n^\prime} R_{n^\prime ,n} W_{n,q} \nonumber
\end{equation}

i.e.

\begin{equation}
c_{n^\prime}^{(q)} = \sum_{n=1}^N R_{n^\prime n} W_{n,q}
\end{equation}

===The Power Method===

- Based on the observation that if H has one eigenvalue, <math>\lambda_M</math>, much bigger than all the rest, and <math>\chi = \left(
\begin{array}{c}a_1\\a_2\\\vdots\end{array}\right)</math>
is an arbitrary starting vector, then <math>\chi = \sum_{q=1}^N x_q\Psi^{(q)}</math>.

\begin{equation}
(H)^n\chi = \sum_{q=1}^N x_q \lambda^n_q\Psi^{(q)}\nonumber\rightarrow x_M\lambda_M^n\Psi^{(M)}\nonumber
\end{equation}

provided <math>x_M\neq 0</math>.

To pick out the eigenvector corresponding to any eigenvalue, write the original problem in the form

\begin{align}
H\Psi & = \lambda O \Psi \nonumber \\
\left( H - \lambda_q O \right) \Psi & = \left( \lambda - \lambda_q \right) O \Psi \nonumber
\end{align}

Therefore,

\begin{equation}
G \Psi = \frac{1}{\lambda - \lambda_q } \Psi \nonumber \\
\end{equation}

where <math>G=(H-\lambda_qO)^{-1}O</math> with eigenvalues <math>\frac{1}{\lambda_n - \lambda_q}</math>.

By picking <math>\lambda_q</math> close to any one of the <math>\lambda_n</math>, say
<math>\lambda_{n^\prime}</math>, then <math>\frac{1}{\lambda_n-\lambda_q}</math> is much larger for
<math>n=n^\prime</math> than for any other value. The sequence is then

\begin{array}{c}
\chi_1=G\chi\nonumber\\
\chi_2=G\chi_1\nonumber\\
\chi_3=G\chi_2\nonumber\\
\vdots\nonumber
\end{array}

until the ratios of components in <math>\chi_n</math> stop changing.

- To avoid matrix inversion and multiplication, note that the sequence is equivalent to
\begin{equation}
F \chi_n = \left( \lambda - \lambda_q \right) O \chi_{n-1} \nonumber \\
\end{equation}
where <math>F = H-\lambda_qO</math>. The factor of <math>\left( \lambda - \lambda_q \right)</math> can be
dropped because this only affects the normalization of <math>\chi_n</math>. To find <math>\chi_n</math>, solve
\begin{equation}
F \chi_n = O \chi_{n-1} \nonumber
\end{equation}
(N equations in N unknowns). Then
\begin{equation}
\lambda = \frac{ \langle \chi_n | H | \chi_n \rangle}{\langle \chi_n | \chi_n \rangle} \nonumber
\end{equation}

==Matrix Elements of H==

\begin{equation}
H=-\frac{1}{2}\nabla^2_1 -\frac{1}{2}\nabla^2_2 - \frac{1}{r_1} - \frac{1}{r_2} +\frac{Z^{-1}}{r_{12}}\nonumber
\end{equation}

Taking <math>r_1, r_2</math> and <math>r_{12}</math> as independent variables,

\begin{equation}
\nabla_1^2 = \frac{1}{r_1^2}\frac{\partial}{\partial r_1}\left(r_1^2\frac{\partial}{\partial r_1}\right)+ \frac{1}{r_{12}^2}\frac{\partial}{\partial r_{12}}
\left(r_{12}^2\frac{\partial}{\partial r_{12}}\right)\nonumber -\frac{l_1(l_1+1)}{r_1^2}+2(r_1 - r_2\cos(\theta_{12}))\frac{1}{r_{12}}\frac{\partial^2}{\partial r_1 \partial r_{12}} \nonumber - 2(\nabla_1^Y \cdot {\bf r}_2)\frac{1}{r_{12}}\frac{\partial}{\partial r_{12}}\nonumber
\end{equation}

where <math>\nabla_1^Y</math> acts only on the spherical harmonic part of the wave function and the diagram

[[File:Pic3.jpg|400px]]

defines the complete set of 6 independent variables is <math>r_1</math>, <math>r_2</math>, <math>r_{12}</math>, <math>\theta_1</math>, <math>\varphi_1</math>, <math>\chi</math>.

If <math>r_{12}</math> were not an independent variable, then one could take the volume element to be

\begin{equation}
d\tau = r_1^2dr_1\sin(\theta_1)d\theta_1d\varphi_1r_2^2dr_2\sin(\theta_2)d\theta_2d\varphi_2.\nonumber
\end{equation}

However, <math>\theta_2</math> and <math>\varphi_2</math> are no longer independent variables. To eliminate them, take the point <math>{\bf r}_1</math> as the
origin of a new polar co-ordinate system, and write

\begin{equation}
d\tau=-r_1^2dr_1\sin(\theta_1)d\theta_1d\varphi_1r_{12}^2dr_{12}\sin(\psi)d\psi d\chi\nonumber
\end{equation}

and use

\begin{equation}
r_2^2=r_1^2+r_{12}^2 +2r_1r_{12}\cos(\psi)
\end{equation}

Then for fixed <math>r_1</math> and <math>r_{12}</math>,

\begin{equation}
2r_2dr_2 = -2r_1r_{12}\sin(\psi)d\psi
\end{equation}

Thus

\begin{equation}
d\tau= r_1dr_1r_2dr_2r_{12}dr_{12}\sin(\theta_1)d\theta_1d\varphi_1d\chi
\end{equation}

The basic type of integral to be calculated is

\begin{align}
I \left( l_1, m_1, l_2, m_2; R \right) & = \int \sin \left( \theta_1 \right) d\theta_1 d\varphi_1 d\chi Y^{m_1}_{l_1} \left( \theta_1, \varphi_1 \right)^{*} Y^{m_2}_{l_2} \left( \theta_2, \varphi_2 \right) \\
& \times \int r_1dr_1r_2dr_2r_{12}dr_{12}R \left( r_1, r_2, r_{12} \right)
\end{align}

Consider first the angular integral. <math>Y^{m_2}_{l_2} \left( \theta_2, \varphi_2 \right)</math> can be expressed in terms of the independent variables <math>\theta_1, \varphi_1, \chi</math> by use of the rotation matrix relation
\begin{equation}
Y^{m_2}_{l_2}( \theta_2, \varphi_2) = \sum_m \mathcal{D}^{(l_2)}_{m_2,m}(\varphi_1,\theta_1,\chi)^*Y^m_{l_2}( \theta_{12}, \varphi)
\end{equation}
where <math>\theta, \varphi</math> are the polar angles of <math>{\bf r}_2</math> relative to <math>{\bf r}_1</math>. The angular integral is then

\begin{align}
I_{ang} & = \int^{2 \pi}_{0} d\chi \int^{2\pi}_{0} d\varphi_1 \int^{\pi}_{0} \sin \left( \theta_1 \right) d\theta_{1} Y^{m_1}_{l_1} \left( \theta_1, \varphi_1 \right)^{*} \\
& \times \sum_m \mathcal{D}^{(l_2)}_{m_2,m} \left( \varphi_1,\theta_1,\chi \right)^{*} Y^{m}_{l_2} \left( \theta_{12},\varphi \right) \\
\end{align}

Use

\begin{equation}
Y^{m_1}_{l_1} \left( \theta_1, \varphi_1 \right)^{*} = \sqrt{\frac{2l_1+1}{4\pi}}\mathcal{D}^{(l_1)}_{m_1,0} \left( \varphi_1, \theta_1, \chi \right)
\end{equation}

together with the orthogonality property of the rotation matrices (Brink and Satchler, p 147)

\begin{align}
& \int \mathcal{D}^{(j)*}_{m,m^\prime}\mathcal{D}^{(J)}_{M,M^\prime}\sin(\theta_1)d\theta_1d\varphi_1d\chi \\
& = \frac{8\pi^2}{2j+1}\delta_{jJ}\delta_{mM}\delta_{m^\prime M^\prime}\nonumber
\end{align}

to obtain

\begin{align}
I_{ang} & = \sqrt{ \frac{2l_1+1}{4\pi}} \frac{8\pi^2}{2l_1+1} \delta_{l_1,l_2} \delta_{m_1,m_2} Y^{0}_{l_2} \left( \theta_{12}, \varphi \right) \\
& = 2 \pi \delta_{l_1,l_2} \delta_{m_1,m_2} P_{l_2} \left( \cos \theta_{12} \right)
\end{align}

since

\begin{equation}
Y^0_{l_2} \left( \theta_{12}, \varphi \right) = \sqrt{\frac{2l_1+1}{4\pi}} P_{l_2} \left( \cos \left( \theta_{12} \right) \right)
\end{equation}

Note that <math>P_{l_2} \left( \cos \theta \right)</math> is just a short hand expression for a radial function because

\begin{equation}
\cos \theta_{12} = \frac{r_1^2 + r_2^2 - r_{12}^2}{2r_1r_2}
\end{equation}

The original integral is thus

\begin{equation}
I \left( l_1, m_1, l_2, m_2; R \right) = 2 \pi \delta_{l_1,l_2} \delta_{m_1,m_2} \int^{\infty}_{0} r_{1} dr_{1} \int^{\infty}_{0} r_{2} dr_{2} \int^{r_{1} + r_{2}}_{|r_{1} - r_{2}|} r_{12} dr_{12} R \left( r_{1}, r_{2}, r_{12} \right) P_{l_2} \left( \cos \theta_{12} \right)
\end{equation}

where again

\begin{equation}
\cos \theta_{12} = \frac{r_{1}^{2} +r^{2}_{2} - r_{12}^{2}}{2 r_{1} r_{2}}
\end{equation}

is a purely radial function.

The above would become quite complicated for large <math>l_2</math> because <math>P_{l_2}(\cos \theta_{12})</math> contains terms up to <math>\left( \cos \theta_{12} \right)^{l_2}</math>. However, recursion relations exist which allow any integral containing <math>P_l \left( \cos \theta_{12} \right)</math> in terms of those containing just <math>P_0 \left( \cos \theta_{12} \right) = 1</math> and <math>P_1 \left( \cos \theta_{12} \right) = \cos \theta_{12}</math>.

==Radial Integrals and Recursion Relations==

The basic radial integral is [see G.W.F. Drake, Phys. Rev. A 18, 820 (1978)]

\begin{align}
I_0 \left( a, b, c \right) & =
\int^{\infty}_{0} r_{1} dr_{1} \int^{\infty}_{r_{1}} r_{2} dr_{2} \int^{r_{1}+r_{2}}_{r_{2} - r_{1}} r_{12} dr_{12} r_{1}^{a} r_{2}^{b} r_{12}^{c} e^{-\alpha r_{1} - \beta r_{2}} + \int^{\infty}_{0} r_{2} dr_{2} \int^{\infty}_{r_{2}} r_{1} dr_{1} \int^{r_{1} + r_{2}}_{r_{1} - r_{2}} r_{12} dr_{12} r_{1}^{a} r_{2}^{b} r_{12}^{c} e^{- \alpha r_{1} - \beta r_{2}} \\
& = \frac{2}{c+2} \sum^{[(c+1)/2]}_{i=0}
\left(\begin{array}{c}
c+2 \\
2i+1
\end{array} \right)
\left\{ \frac{q!}{\beta^{q+1} \left( \alpha + \beta \right)^{p+1}} \sum_{j=0}^{q} \frac{\left( p + j \right)!}{j!} \left( \frac{\beta}{\alpha + \beta} \right)^j + \frac{q^{\prime}!}{\alpha^{q^{\prime}+1} \left( \alpha + \beta \right)^{p^{\prime}+1}} \sum^{q^{\prime}}_{j=0} \frac{\left( p^{\prime} + j \right)!}{j!} \left( \frac{\alpha}{\alpha + \beta} \right)^j \right\}
\end{align}

where

\begin{array}{ll}
p = a + 2i + 2 & p^{\prime} = b + 2i + 2 \\
q = b + c - 2i + 2 & q^{\prime} = a + c - 2i + 2
\end{array}

The above applies for <math> a, b \geq -2, c\geq -1</math>.
<math>[x]</math> means "greatest integer in" <math>x</math>.

Then

\begin{align}
I_1 \left( a, b, c \right) & = \int d\tau_{r} r_{1}^{a} r_{2}^{b} r_{12}^{c} e^{ -\alpha r_{1} - \beta r_{2}} P_{1}\left( \cos \theta \right) \\
& = \frac{1}{2} \left( I_{0} \left( a + 1, b - 1, c \right) + I_{0} \left( a - 1, b + 1, c \right) - I_{0} \left( a - 1, b - 1, c + 2 \right) \right)
\end{align}

===The Radial Recursion Relation===

Recall that the full integral is

\begin{equation}
I(l_1m_1,l_2m_2;R)=2\pi\delta_{l_1,l_2}\delta_{m_1,m_2}I_{l_2}(R)
\end{equation}

where

\begin{equation}
I_{l_2}(R) = \int d\tau_r R(r_1,r_2,r_{12})P_{l_2}(\cos \theta_{12})
\end{equation}

To obtain the recursion relation, use

\begin{equation}
P_l(x) = \frac{[P^{\prime}_{l+1}(x)-P^{\prime}_{l-1}(x)]}{2l+1}
\end{equation}

with

\begin{equation}
P_{l+1}^\prime (x) = \frac{d}{dx}P_{l+1}(x)
\end{equation}

Here <math>x=\cos\theta_{12}</math> and

\begin{align}
\frac{d}{d \cos \theta_{12}} & = \frac{dr_{12}}{d \cos \theta_{12}} \frac{d}{dr_{12}} \\
& = -\frac{r_{1} r_{2}}{r_{12}} \frac{d}{dr_{12}}
\end{align}

Then,

\begin{equation}
I_{l}\left( R \right) = - \int d\tau_r R \frac{r_{1} r_{2}}{r_{12}} \frac{d}{dr_{12}} \frac{ \left[ P_{l+1} \left( \cos \theta_{12} \right) - P_{l-1} \left( \cos \theta_{12} \right) \right]}{2l+1}
\end{equation}

The <math>r_{12}</math> part of the integral is

\begin{align}
& \int^{r_{1} + r_{2}}_{|r_{1} - r_{2}|} r_{12} dr_{12} R \frac{r_{1} r_{2}}{r_{12}} \frac{d}{dr_{12}} \left[P_{l+1}-P_{l-1} \right] \\
& = R r_{1} r_{2} \left[ P_{l+1} - P_{l-1} \right]^{r_{1} + r_{2}}_{|r_{1} - r_{2}|} - \int^{r_{1} + r_{2}}_{|r_{1} - r_{2}|} r_{12} dr_{12} \left( \frac{d}{dr_{12}} R \right) \frac{r_{1} r_{2}}{r_{12}} \frac{ \left[ P_{l+1} - P_{l-1} \right]}{2l+1}
\end{align}

The integrated term vanishes because

\begin{equation}
\cos \theta_{12} = \frac{r_{1}^{2} + r_{2}^{2} - r_{12}^{2}}{2 r_{1} r_{2}}
\begin{array}{lll}
& = -1 & {\rm when} & r_{12}^{2} = \left( r_{1} + r_{2} \right)^{2} \\
& = 1 & {\rm when} & r_{12}^{2} = \left( r_{1} - r_{2} \right)^{2}
\end{array}
\end{equation}

and <math>P_l(1)=1</math>, <math>P_{l}(-1)=(-1)^{l}</math>.

Thus,

\begin{equation}
I_{l+1}\left(\frac{r_1r_2}{r_{12}}R^\prime\right) = \left( 2l+1 \right) I_{l}(R) - I_{l-1} \left( \frac{r_{1} r_{2}}{r_{12}} R^{\prime} \right)
\end{equation}

If

\begin{equation}
R = r_{1}^{a-1} r_{2}^{b-1} r_{12}^{c+2} e^{-\alpha r_{1} - \beta r_{2}}
\end{equation}

then [G.W.F. Drake, Phys. Rev. A 18, 820 (1978)]

\begin{equation}
I_{l+1}(r_1^ar_2^br_{12}^c) =
\frac{2l+1}{c+2} I_{l} \left( r_{1}^{a-1} r_{2}^{b-2} r_{12}^{c+2} \right)+I_{l-1} \left( r_{1}^{a} r_{2}^{b} r_{12}^{c} \right)
\end{equation}

For the case <math>c=-2</math>, take

\begin{equation}
R = r_{1}^{a-1} r_{2}^{b-1} \ln r_{12} e^{ -\alpha r_{1} -\beta r_{2}}
\end{equation}

Then

\begin{equation}
I_{l+1} \left( r_{1}^{a} r_{2}^{b} r_{12}^{c} \right) = I_{l} \left( r_{1}^{a-1} r_{2}^{b-1} \ln r_{12} \right) \left( 2l + 1 \right) - I_{l-1} \left( r_{1}^{a} r_{2}^{b} r_{12}^{-2} \right)
\end{equation}

This allows all <math>I_{l}</math> integrals to be calculated from tables of <math>I_0</math> and
<math>I_1</math> integrals.

===The General Integral===

The above results for the angular and radial integrals can now be combined into a general formula for integrals of the type

\begin{equation}
I = \int \int d{\bf r}_{1} d{\bf r}_{2} R_{1} \mathcal{Y}^{M^{\prime}}_{l^{\prime}_{1} l^{\prime}_{2} L^{\prime}} \left( \hat{r}_{1}, \hat{r}_{2} \right) T_{k_{1} k_{2} K}^{Q} \left( {\bf r}_{1},{\bf r}_{2} \right) R_{2} \mathcal{Y}_{l_{1} l_{2} L}^{M} \left( \hat{r}_{1}, \hat{r}_{2} \right)
\end{equation}

where

\begin{equation}
\mathcal{Y}^{M}_{l_{1} l_{2} L} \left( \hat{r}_{1}, \hat{r}_{2} \right) = \sum_{m_1,m_2} \langle l_{1} l_{2} m_{1} m_{2} | LM \rangle Y^{m_{1}}_{l_{1}} \left( \hat{r}_{1} \right) Y^{m_{2}}_{l_{2}} \left( \hat{r}_{2} \right)
\end{equation}

and

\begin{equation}
T^Q_{k_{1} k_{2} K} \left( {\bf r}_{1}, {\bf r}_{2} \right) = \sum_{q_{1}, q_{2}} \langle k_{1} k_{2} q_{1} q_{2} | KQ \rangle Y_{k_{1}}^{q_{1}} \left( \hat{r}_{1} \right) Y^{q_{2}}_{k_{2}} \left( \hat{r}_{2} \right)
\end{equation}

The basic idea is to make repeated use of the formula

\begin{equation}
Y^{m_{1}}_{l_{1}} \left( \hat{r}_{1} \right) Y^{m_{2}}_{l_{2}} \left( \hat{r}_{1} \right) = \sum_{l m} \left( \frac{\left( 2l_{1}+1 \right) \left(2l_{2}+1 \right) \left(2l+1 \right)}{4 \pi} \right)^{1/2}
\times \left( \begin{array}{ccc}
l_1 & l_2 & l \\
m_1 & m_2 & m
\end{array} \right) \left(\begin{array}{ccc}
l_1 & l_2 & l \\
0 & 0 & 0
\end{array} \right)
Y^{m*}_{l} \left( \hat{r}_{1} \right)
\end{equation}

where

\begin{equation}
Y^{m*}(\hat{r}) = (-1)^mY_l^{-m}(\hat{r})
\end{equation}

and

\begin{equation}
\left(\begin{array}{ccc}
l_1 & l_2 & l \\
m_1 & m_2 & m
\end{array} \right) = \frac{\left( -1 \right)^{l_{1} - l_{2} - m}}{\left( 2l + 1 \right)^{1/2}} \left( l_{1} l_{2} m_{1} m_{2} | l, -m \right)
\end{equation}

is a 3-<math>j</math> symbol. In particular, write

\begin{array}{cc}
Y^{m_1^\prime}_{l_1^\prime}(\hat{r}_1)^{*} \underbrace{Y_{k_{1}}^{q_{1}} \left( \hat{r}_{1} \right) Y^{m_{1}}_{l_{1}} \left( \hat{r}_{1} \right)} & = & \sum_{\Lambda M}(\ldots)Y^{M}_\Lambda(\hat{r}_1)\\
\ \ \ \ \ \ \ \sum_{\lambda_1\mu_1}Y_{\lambda_1}^{\mu_1}(\hat{r}_1) & & \\
& & \\
Y^{m_{2}^{\prime}}_{l_{2}^{\prime}} \left( \hat{r}_{2} \right) \underbrace{Y^{q_{2}}_{k_{2}} \left( \hat{r}_{2} \right) Y_{l_{2}}^{m_{2}} \left( \hat{r}_{2} \right)} & = & \sum_{\Lambda^{\prime} M^{\prime}} \left( \ldots \right)Y_{\Lambda^{\prime}}^{M^{\prime}*} \left( \hat{r}_{2} \right) \\
\ \ \ \ \ \ \ \ \ \sum_{\lambda_{2} \mu_{2}}Y_{\lambda_{2}}^{\mu_{2}} \left( \hat{r}_{2} \right) & &
\end{array}

The angular integral then gives a factor of <math>2 \pi \delta_{\Lambda, \Lambda^{\prime}} \delta_{M, M^{\prime}} P_{\Lambda} \left( \cos \theta_{12} \right)</math>. The total integral therefore reduces to the form

\begin{equation}
I = \sum_{\Lambda} C_{\Lambda} I_{\Lambda} \left( R_{1} R_{2} \right)
\end{equation}

where <math>C_{\Lambda} = \sum_{\lambda_{1}, \lambda_{2}} C_{ \lambda_{1}, \lambda_{2}, \Lambda}</math>. For further details and derivations, including graphical representations, see G.W.F. Drake, Phys. Rev. A <math>{\bf 18}</math>, 820 (1978).

==Graphical Representation==

[[File:GraphRep1.jpg|500px]]

==Matrix Elements of H==
Recall that

\begin{equation}
H = -\frac{1}{2} \nabla^{2}_{1} - \frac{1}{2} \nabla^{2}_{2} - \frac{1}{r_1} - \frac{1}{r_2} + \frac{Z^{-1}}{r_{12}}
\end{equation}

Consider matrix elements of

\begin{equation}
\nabla^2_1 = \frac{1}{r_1^2} \frac{\partial}{\partial r_1}\left(r_1^2\frac{\partial}{\partial r_1} \right) + \frac{1}{r^2}\frac{\partial}{\partial r}\left(r^2\frac{\partial}{\partial r}\right) - \frac{(\vec{l}_1^Y)^2}{r_1^2} + \frac{2 \left( r_1 - r_2 \cos \theta \right)}{r} \frac{\partial^2}{\partial r_1 \partial r} - 2 \left( \nabla^Y_1 \cdot {\bf r}_2 \right) \frac{1}{r} \frac{\partial}{\partial r}
\end{equation}

where <math>r \equiv r_{12}</math> and <math>\cos\theta \equiv \cos\theta_{12}</math>. Also in what follows, define <math>\hat{\nabla}^Y_1 \equiv r_1\nabla^Y_1</math> where <math>\nabla^Y</math> operates only on the spherical harmonic part of the wave function, and similarly for <math>\vec{l}_1^Y</math>.

A general matrix element is

\begin{equation}
\langle r_1^{a^\prime} r_2^{b^\prime} r_{12}^{c^\prime} e^{-\alpha^\prime r_1 - \beta^\prime r_2} \mathcal{Y}^{M^\prime}_{l^\prime_1 l^\prime_2 L^\prime} \left( \hat{r}_1, \hat{r}_2 \right) \left| \nabla^2_1 \right| r_1^a r_2^b r_{12}^c e^{-\alpha r_1 - \beta r_2} \mathcal{Y}^M_{l_1 l_2 L} \left( \hat{r}_1, \hat{r}_2 \right) \rangle
\end{equation}

Since <math>\nabla^2_1</math> is rotationally invariant, this vanishes unless <math>L=L^\prime</math>, <math>M=M^\prime</math>. Also <math>\nabla_1^2</math> is Hermitian, so that the result must be the same whether it operates to the right or left, even though the results look very different. In fact, requiring the results to be the same yields some interesting and useful integral identities as follows:

Put

\begin{equation}
\mathcal{F} = F \mathcal{Y}^M_{l_1 l_2 L} \left( \hat{r}_1, \hat{r_2} \right)
\end{equation}

and

\begin{equation}
F = r_1^a r_2^b r_{12}^c e^{-\alpha r_1 - \beta r_2}
\end{equation}

Then

\begin{equation}
\nabla^2_1 \mathcal{F} = \left\{ \frac{1}{r_1^2} \left[ a \left( a + 1 \right) - l_1 \left( l_1 + 1 \right) \right] + \frac{c \left( c + 1 \right)}{r^2} + \alpha^2 + \frac{2\alpha \left( a + 1 \right)}{r_1} + \frac{2 \left(r_1 - r_2 \cos \theta \right) }{r_1r^2}c \left( a - \alpha r_1 \right) - \frac{2c}{r^2} \left( \hat{\nabla}^Y_1 \cdot \hat{r}_2 \right) \frac{r_2}{r_1} \right\} \mathcal{F}
\end{equation}

and

\begin{align}
\langle \mathcal{F}^\prime \left| \nabla^2_1 \right| \mathcal{F} \rangle & = \sum_\Lambda \int d\tau_r F^\prime C_\Lambda \left( 1 \right) P_\Lambda \left( \cos \theta \right) \times \left\{ \frac{1}{r_1^2} \left[ a \left( a + 1 \right) - l_1 \left( l_1 + 1 \right) \right] -\frac{2 \alpha \left( a + 1 \right)}{r_1} + \frac{c \left( c + 1 \right)}{r^2} +\alpha^2 + \frac{2 \left( r_2 - r_2 \cos \theta \right)}{r_1r^2}c \left( a - \alpha r_1 \right) \right\} F \\
& + \sum_\Lambda \int d\tau_r F^\prime C_\Lambda \left( \hat{\nabla}^Y_1 \cdot \hat{r}_2 \right) P_\Lambda \left( \cos \theta \right) \left( \frac{-2c r_2}{r_1 r^2} \right) F
\end{align}

where

\begin{equation}
\int d\tau_r = \int^\infty_0 r_1 dr_1 \int^\infty_0 r_2 dr_2 \int^{r_1+r_2}_{\left| r_1 - r_2 \right|} r dr
\end{equation}

For brevity, let the sum over <math>\Lambda</math> and the radial integrations be understood, and let <math>\left( \nabla_1^2 \right)</math> stand for the terms which appear in the integrand. Then operating to the right gives

\begin{equation}
(\nabla_1^2)_R = \frac{1}{r_1^2} \left[ a \left( a + 1 \right) - l_1 \left( l_1 + 1 \right) \right] + \frac{c \left( c + 1 \right)}{r^2} + \alpha^2 - \frac{2 \alpha \left( a + 1 \right)}{r_1}
+ \frac{2 \left( r_1 - r_2 \cos \theta \right)}{r_1 r^2}c \left( a - \alpha r_1 \right) - \frac{2c r_2}{r_1 r^2} \left( \hat{\nabla}^Y_1 \cdot \hat{r}_2 \right)
\end{equation}

Operating to the left gives

\begin{equation}
(\nabla_1^2)_L = \frac{1}{r_1^2} \left[a^\prime \left( a^\prime + 1 \right) - l_1^\prime \left( l_1^\prime + 1 \right) \right] + \frac{c^\prime \left( c^\prime + 1 \right)}{r^2} + \alpha^{\prime 2} -\frac{2 \alpha^\prime \left( a^\prime + 1 \right)}{r_1}
+ \frac{2 \left( r_1 - r_2 \cos \theta \right)}{r_1 r^2} c^\prime \left( a^\prime - \alpha^\prime r_1 \right) - \frac{2 c^\prime r_2}{r_1 r^2} \left( \hat{\nabla}^{Y^\prime}_1 \cdot \hat{r}_2 \right)
\end{equation}

Now put

\begin{array}{cc}
a_+ = a+a^\prime, & \hat{\nabla}_1^+ = \hat{\nabla}_1^Y + \hat{\nabla}_1^{Y\prime} \\
a_{-} = a-a^\prime, & \hat{\nabla}_1^{-} = \hat{\nabla}_1^Y - \hat{\nabla}_1^{Y\prime}
\end{array}

etc., and substitute <math>a^\prime = a_+ - a</math>, <math>c^\prime = c_+ - c</math>, and <math>\alpha^\prime = \alpha_+ - \alpha</math> in <math>(\nabla^2_1)_L</math>. If <math>a_+</math>, <math>c_+</math> and <math>\alpha_+</math> are held fixed, then the equation

\begin{equation}
(\nabla^2_1)_R = (\nabla^2_1)_L
\end{equation}

must be true for arbitrary <math>a</math>, <math>c</math>, and <math>\alpha</math>. Their coefficients must thus vanish.

This yields the integral relations

\begin{equation}
\label{I}
\frac{\left( r_1 - r_2 \cos \theta \right)}{r_1 r^2} = \frac{1}{c_+} \left( \frac{- \left( a_+ + 1 \right)}{r_1^2} + \frac{\alpha_+}{r_1} \right) \tag{I}
\end{equation}

from the coefficient of <math>a</math> and

\begin{equation}
\label{II}
\frac{ \left( r_1 - r_2 \cos \theta \right) \left( a_+ - \alpha_+ r_1 \right)}{r_1 r^2} = \frac{r_2}{r_1 r^2}\left( \hat{r}_2 \cdot \hat{\nabla}_1^+ \right) - \frac{ \left( c_+ + 1 \right)}{r^2} \tag{II}
\end{equation}

from the coefficient of <math>c</math>. The coefficient of <math>\alpha</math> gives an equation equivalent to (I).

Furthermore, if can be show that (see problem)

\begin{align}
& \sum_\Lambda \int d\tau_r \frac{r^c}{r^2} C_\Lambda \left( \hat{r}_2 \cdot \hat{\nabla}_1^Y \right) P_\Lambda \left( \cos \theta \right) \\
= & \sum_\Lambda \int d\tau_r \frac{r^c}{c r_1 r_2}C_\Lambda \left( 1 \right)P_\Lambda \left( \cos \theta \right) \times \left( \frac{l_1^\prime \left( l_1^\prime + 1 \right) - l_1 \left( l_1 + 1 \right) -\Lambda \left( \Lambda + 1 \right)}{2} \right)
\end{align}

and similarly for <math>\left( \hat{r}_2 \cdot \hat{\nabla}^{Y\prime}_1 \right)</math> with <math>l_1</math> and <math>l_1^\prime</math> interchangeable, then it follows that

\begin{equation}
\frac{r^{c_+}}{r^2} \left( \hat{r}_2 \cdot \hat{\nabla}_1^+ \right) = - \frac{r^{c_+}}{c_+ r_1 r_2} \Lambda \left( \Lambda + 1 \right)
\end{equation}

and

\begin{equation}
\frac{r^{c_+}}{r^2} \left( \hat{r}_2 \cdot \hat{\nabla}_1^- \right) = \frac{r^{c_+}}{c_+ r_1 r_2} \left[ l_1^\prime \left( l_1^\prime + 1 \right) - l_1 \left( l_1 + 1 \right) \right]
\end{equation}

where equality applies after integration and summation over <math>\Lambda</math>.

Thus (II) becomes

\begin{equation}\label{IIa}
\frac{ \left( r_1 - r_2 \cos \theta \right) \left( a_+ - \alpha_+ r_1 \right)}{r_1 r^2} = - \frac{\Lambda \left( \Lambda + 1 \right)}{c_+ r^2_1} - \frac{ \left( c_+ + 1 \right)}{r^2} \tag{IIa}
\end{equation}

===Problem===

Prove the integral relation

\begin{align}
& \sum_\Lambda \int d\tau_r f \left( r_1, r_2 \right) \frac{1}{r} \left( \frac{d}{dr} q(r) \right) C_\Lambda \left( \hat{r}_2 \cdot \hat{\nabla}^Y_1 \right) P_\Lambda \left( \cos \theta \right) \\
= & \sum_\Lambda \int d\tau_r \frac{ f \left( r_1, r_2 \right)}{r_1, r_2} q(r) C_\Lambda \left( 1 \right) P_\Lambda \left( \cos \theta \right) \left(\frac{l_1^\prime \left( l_1^\prime + 1 \right) - l_1 \left( l_1 + 1 \right) - \Lambda \left( \Lambda + 1 \right)}{2}\right)
\end{align}

where the coefficients <math>C_\Lambda \left( 1 \right)</math> are the angular coefficients from the overlap integral

\begin{align}
\int & d\Omega\, \mathcal{Y}^{M*}_{l_1^\prime l_2^\prime L}(\hat{r}_1,\hat{r}_2)\, \mathcal{Y}^{M}_{l_1 l_2 L}(\hat{r}_1,\hat{r}_2)\\
= & \sum_\Lambda C_\Lambda(1)P_\Lambda(\cos\theta_{12}).
\end{align}

Hint: Use the fact that <math>l_1^2</math> is Hermitian so that
<math>
\int d\tau (l_1^2\mathcal{Y}^\prime)^*q(r)\mathcal{Y} = \int d\tau
\mathcal{Y}^{\prime *}l_1^2(q(r)\mathcal{Y})
</math>
with <math>\ \vec{l}_1= \frac{1}{i}\vec{r}_1\times\nabla_1</math>.

It is also useful to use
<math>\begin{eqnarray}\ \
\cos\theta\, P_L(\cos\theta) &=& \frac{1}{2L+1}[L\,P_{L-1}(\cos\theta) + (L+1)P_{L+1}(\cos\theta)]
\\
(\cos^2\theta-1)P_L(\cos\theta) &=&
\frac{L(L-1)}{(2L-1)(2L+1)}P_{L-2}(\cos\theta)\nonumber
-\frac{2(L^2+L-1)}{(2L-1)(2L+3)}P_L(\cos\theta) + \frac{(L+1)(L+2)}{(2L+1)(2L+3)}P_{L+2}(\cos\theta)\nonumber\\
&=& \frac{L(L-1)}{(2L-1)(2L+1)}[P_{L-2}(\cos\theta) - P_L(\cos\theta)]
+ \frac{(L+1)(L+2)}{(2L+1)(2L+3)}[P_{L+2}(\cos\theta) - P_L(\cos\theta)]
\end{eqnarray}</math>

together with a double application of the integral recursion relation
<math>
I_{L+1}\left(\frac{1}{r}\frac{d}{dr}q(r)\right) =
(2L+1)I_L\left(\frac{1}{r_1r_2}q(r)\right)+I_{L-1}\left(\frac{1}{r}\frac{d}{dr}q(r)\right)
</math>

Of course <math>l_1^2\mathcal{Y}^M_{l_1l_2L}(\hat{r}_1,\hat{r}_2)=l_1(l_1+1)\mathcal{Y}^M_{l_1l_2L}(\hat{r}_1,\hat{r}_2)</math>. Begin by expanding
<math>
l_1^2(g\mathcal{Y}) = \mathcal{Y}l_1^2g + g\,l_1^2\mathcal{Y} + 2(\vec{l}_1g)\cdot(\vec{l}_1\mathcal{Y})
</math>, and show that

<math>
\begin{eqnarray}\ \
(\vec{l}_1g)\cdot(\vec{l}_1\mathcal{Y}) & = & \frac{r_1^2}{r}\frac{dg}{dr}\vec{r}_2\cdot \nabla_1\mathcal{Y}\,\,\,\,\mbox{ and}\\
l_1^2g(r) &=& 2r_1r_2\cos\theta\frac{1}{r}\frac{dg}{dr} + r_1^2r_2^2(\cos^2\theta - 1)
\frac{1}{r}\frac{d}{dr}\left(\frac{1}{r}\frac{dg}{dr}\right).
\end{eqnarray}
</math>

The proof amounts to showing that the term <math>l_1^2g(r)</math> can be replaced by
<math>\Lambda(\Lambda+1)g(r)</math> after multiplying by <math>P_\Lambda(\cos\theta)</math>
and integrating by parts with respect to the radial integrations over <math>r_1</math>, <math>r_2</math> and
<math>r \equiv |\vec{r}_1 - \vec{r}_2|</math>. Remember that <math>\cos\theta</math> is just a short-hand notation for the radial function <math>(r_1^2 + r_2^2 - r^2)/(2r_1r_2)</math>.

==General Hermitian Property==

Each combination of terms of the form
\begin{equation}
<f>_R = a^2f_1+af_2+abf_3+b^2f_4+bf_5+af_6\nabla_1^Y+bf_7\nabla^Y_2+f_8(Y)
\end{equation}
can be rewritten
\begin{equation}
<f>_L= (a_+-a)^2f_1+(a_+-a)f_2 + (a_+-a)(b_+-b)f_3+(b_+-b)^2f_4
+(b_+-b)f_5+(a_+-a)f_6\nabla^{Y\prime}_1+(b_+-b)f_7\nabla^{Y\prime}_2+f_8)Y^\prime
\end{equation}
where as usual <math>\nabla_1^Y</math> acts only on the spherical harmonic part of the wave function denoted for short by <math>Y</math>.
Since these must be equal for arbitrary <math>a</math> and <math>b</math>,
\begin{equation}
a_+^2f_1+a_+f_2+_+b_+f_3+b_+f_4+b_+f_5+a_+f_6\nabla^{Y\prime}_1 +
b_+f_7\nabla^{Y\prime}_2+f_8(Y^\prime)-f_8(Y)=0
\end{equation}

Adding the corresponding expression with <math>Y</math> and <math>Y^\prime</math> interchanged
yields
<math>\label{III}
a_+^2f_1+a_+f_2+a_+b_+f_3+b_+^2f_4+b_+f_5+\frac{1}{2}a_+f_6\nabla^+_1+\frac{1}{2}b_+f_7\nabla^+_2=0\tag{III}
</math>

Subtracting gives
\begin{eqnarray*}
f_8(Y)-f_8(Y^\prime)&=&-\frac{1}{2}[a_+f_6\nabla^-_1+b_+f_7\nabla_2^-]\tag{IV}\\
a[-2a_+f_1-2f_2-b_+f_3-f_6\nabla^+_1]&=&0\tag{V}\\
b[-2b_+f_4-2f_5-a_+f_3-f_7\nabla^+_2]&=&0\tag{VI}\\
\end{eqnarray*}

Adding the two forms gives

\begin{eqnarray}
<f>_R+<f>_L=\frac{1}{2}(a_+^2+a_-^2)f_1+a_+f_2+\frac{1}{2}(a_+b_++a_-b_-)f_3\nonumber\\
\ +\frac{1}{2}(b_+^2+b_-^2)f_4+b_+f_5+\frac{1}{2}f_6(a_+\nabla_1^++a_-\nabla^-_1)\nonumber\\
\ +\frac{1}{2}(b_+\nabla_2^++b_-\nabla^-_2)+f_8(Y)+f_8(Y^\prime)\nonumber
\end{eqnarray}

Subtracting <math>\frac{x}{2}\times({\rm I})</math> gives

\begin{eqnarray}\ \
<f>_R+<f>_L = \frac{1}{2}[(1-x)a_+^2 + a_-^2]f_1 +
(1-\frac{x}{2})a_+f_2+\frac{1}{2}[(1-x)a_+b_++a_-b_-]f_3\nonumber\\
+\frac{1}{2}[(1-x)b_+^2+b_-^2]f_4+(1-\frac{x}{2})f_5b_++\frac{1}{2}f_6[(1-\frac{x}{2})a_+\nabla_1^++a_-\nabla^-_1]\nonumber\\
+\frac{1}{2}[(1-\frac{x}{2})b_+\nabla^+_2+b_-\nabla^-_2]+f_8(Y)+f_8(Y^\prime)\nonumber
\end{eqnarray}

If <math>x=1</math>,
\begin{eqnarray}
<f>_R+<f>_L &=&
\frac{1}{2}[a_-^2f_1+a_+f_2+a_-b_-f_3+b_-^2f_4+b_+f_5\nonumber\\
&+&\left(\frac{1}{2}a_+\nabla^+_1+a_-\nabla^-_1\right)f_6+\left(\frac{1}{2}b_+\nabla^+_2+b_-\nabla^-_2\right)f_7]+f_8(Y)+f_8(Y^\prime)\nonumber
\end{eqnarray}

The General Hermitian Property for arbitrary <math>x</math> gives
\begin{eqnarray}\ \
\nabla_1^2=
\frac{1}{4}\{\frac{1}{r_1^2}[(1-x)a_+^2+a_-^2+2(1-\frac{x}{2})a_+-2[l_1(l_1+1)+l_1^\prime(l_1^\prime+1)]\nonumber\\
-\frac{2}{r_1}[(1-x)\alpha_+a_++\alpha_-a_-+2(1-\frac{x}{2})\alpha_+]+(1-x)\alpha_+^2
+ \alpha^2_-]\nonumber\\
+\frac{2(r_1-r_2\cos\theta)}{r_1r^2}[(1-x)(a_+-\alpha_+r_1)c_++(a_--\alpha_-r_2)c_-]\nonumber\\
-\frac{2r_2}{r_1r^2}[(1-\frac{x}{2})c_+\hat{r}_2\cdot\hat{\nabla}_1^++c_-\hat{r}_2\hat{\nabla}_1^-]+\frac{(1-x)c_+^2+c_-^2+2(1-\frac{x}{2})c_+}{r^2}\}\nonumber
\end{eqnarray}.

Use
<math>\begin{eqnarray}\ \
\frac{-2r_2}{r_1r^2}[(1-\frac{x}{2})c_+\hat{r}_2\hat{\nabla}_1^++c_-\hat{r}_2\hat{\nabla}_1^-]=\frac{2}{r_1^2}\left[(1-\frac{x}{2})\Lambda(\Lambda+1)-\frac{c_-}{c_+}[l_1^\prime(l_1^\prime+1)+l_1(l_1+1)]\right]\end{eqnarray}</math>,

<math>\begin{eqnarray}
\frac{2(r_1-r_2\cos\theta)}{r_1r^2}(a_--\alpha_-r_1)c_- = 2\frac{c_-}{c_+}\left[\frac{-1}{r_1^2}[a_-(a_++1)]+\frac{1}{r_1}[a_-\alpha_++\alpha_-(a_++2)]-\alpha_-\alpha_+\right]\nonumber
\end{eqnarray}</math>,

and
<math>\begin{eqnarray}\ \
\frac{2(r_1-r_2\cos\theta)}{r_1r^2}(a_+-\alpha_+r_1)c_+ = 2\left[-\frac{1}{r_1^2}[a_+(a_++1)]+\frac{1}{r_1}[a_+\alpha_++\alpha_+(a_++2)]-\alpha_+^2\right]
\end{eqnarray}</math>.

Substituting into <math>\nabla_1^2</math> gives

\begin{eqnarray}
\nabla^2_1 = \frac{1}{4}\{\frac{1}{r_1^2}[-(1-x)a_+^2 + a_-^2 +xa_+ +
2(1-\frac{x}{2})\Lambda(\Lambda+1)\nonumber\\
-2l_1(l_1+1)(1-\frac{c_-}{c_+}) -
2l_1^\prime(l_1^\prime+1)(1+\frac{c_-}{c_+})-\frac{2c_-a_-}{c_+}(a_++1)]\nonumber\\
- \frac{2}{r_1}\left[-(1-x)\alpha_+(a_++2)+\alpha_-a_-+2(1-\frac{x}{2})\alpha_+
-\frac{c_-}{c_+}[a_-\alpha_++\alpha_-(a_++2)]\right]\nonumber\\
-(1-x)\alpha_+^2+\alpha_-^2-2\frac{c_-}{c_+}\alpha_-\alpha_+ +\frac{1}{r^2}\left[(1-x)c_+^2+c_-^2+2(1-\frac{x}{2})c_+\right]\}\nonumber
\end{eqnarray}

This has the form
<math>
\nabla_1^2 =
\frac{1}{4}\left[\frac{A_1}{r_1^2}+\frac{B_1}{r_1}+\frac{C_1}{r^2}+D_1\right]
</math>
with

<math>\begin{array}{ccc}
A_1=-(1-x)a_+^2+a_-^2
+xa_++2(1-\frac{x}{2})\Lambda(\Lambda+1)-2l_1(l_1+1)(1-\frac{c_-}{c_+})\
-2l_1^\prime(l_1^\prime+1)(1+\frac{c_-}{c_+})-\frac{2c_-a_-}{c_+}(a_++1) \ \
\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \
\end{array}
</math>

<math>
B_1 =
2\left[(1-x)\alpha_+(a_++2)-\alpha_-a_--2(1-\frac{x}{2})\alpha_++\frac{c_-}{c_+}[a_-\alpha_++\alpha_-(a_++2)]\right]
</math>

<math>
C_1 = (1-x)c_+^2+ c_-^2 +2(1-\frac{x}{2})c_+
</math>

<math>
D_1 = -(1-x)\alpha_+^2 +\alpha_-^2 - 2\frac{c_-}{c_+}\alpha_-\alpha_+
</math>

The complete Hamiltonian is then (in Z-scaled a.u.)

<math>\begin{eqnarray}
H =-\frac{1}{2}\nabla^2_1 -\frac{1}{2}\nabla^2_2 -\frac{1}{r_1} -\frac{1}{r_2} +\frac{Z^{-1}}{r}\nonumber\\
= -\frac{1}{2}\nabla^2_1 -\frac{1}{2}\nabla^2_2 -\frac{1}{r_1} -\frac{Z-1}{Zr_2}+Z^{-1}\left(\frac{1}{r}-\frac{1}{r_2}\right)\nonumber\\
\ =-\frac{1}{8}\left[\frac{A_1}{r_1^2}+\frac{B_1+8}{r_1}+\frac{C_1}{r^2}+D_1+D_2 +\frac{A_2}{r_2^2} + \frac{B_2 + 8(Z-1)/Z}{r_2}+\frac{C_2}{r^2}\right]\nonumber
+ Z^{-1}\left(\frac{1}{r}-\frac{1}{r_2}\right)\nonumber
\end{eqnarray}</math>

The screened hydrogenic energy is
<math>\begin{eqnarray}
E_{SH}=-\frac{1}{2}\left[\frac{1}{n_1^2}+ \left(\frac{Z-1}{Z}\right)^2\frac{1}{n_2^2}\right]
\end{eqnarray}</math>
so that

<math>\begin{eqnarray}
H-E_{SH} &=& -\frac{1}{8}[
\frac{A_1}{r_1^2}+\frac{B_1+8}{r_1}+\frac{C_1}{r^2} + D_1 + D_2
-\frac{4}{n_1^2} -\left(\frac{Z-1}{Z}\right)^2\frac{4}{n_2^2}\nonumber
+\frac{A_2}{r_2}+\frac{B_2+8(Z-1)/Z}{r_2}+\frac{C_2}{r^2}] +
Z^{-1}\left(\frac{1}{r} -\frac{1}{r_2}\right)\nonumber
\end{eqnarray}</math>

==Optimization of Nonlinear Parameters==
- The traditional method of performing Hylleraas calculations is to write the
basis set in the form

<math>
\Psi = \sum_{i,j,k}c_{i,j,k}\varphi_{i,j,k}(\alpha,\beta)\pm exchange
</math>

with

<math>
\varphi_{i,j,k}(\alpha,\beta) = r_1^ir_2^jr_{12}^ke^{-\alpha r_1 - \beta
r_2}\mathcal{Y}^M_{l_1l_2L}.
</math>

The usual procedure is to set <math>\alpha = Z</math> so that it represents the inner
<math>1s</math> electron, and then to vary <math>\beta</math> so as to minimize the energy.

[[File:OptNon.jpg|500px]]

Since <math>\beta</math> appears in <math>\Psi</math> as a non-linear parameter, the entire
calculation must be repeated for each value of <math>\beta</math>. However, the minimum
becomes progressively smaller as the basis set is enlarged.

[[File:OptNon02.jpg|500px]]

*Difficulties

1. If the basis set is constructed so that <math> i +j +j \leq N</math>, the the
number of terms is
<math>
(N+1)(N+2)(N+3)/6
</math>
N=14 already gives 680 terms and an accuracy of about 1 part in <math>10^{10}</math> for
low-lying states. A substantial improvement in accuracy would require much
larger basis sets, together with multiple precision arithmetic to avoid loss
of significant figures when high powers are included.

2. The accuracy rapidly deteriorates as one goes to more highly excited
states - about 1 significant figure is lost each time the principle quantum
number is increased.

*Cure

We have found that writing basis sets in the form
\begin{eqnarray}
\Psi &=& \sum_{i,j,k}\left(c^{(1)}_{i,j,k}\varphi_{ijk}(\alpha_1\beta_1) +
c_{ijk}^{(2)}\varphi_{ijk}(\alpha_2\beta_2)\right) \pm exchange \nonumber\\
&=& \psi({\bf r}_1,{\bf r}_2) \pm \psi({\bf r}_2,{\bf r}_1)\nonumber
\end{eqnarray}
so that each combination of powers is included twice with different nonlinear
parameters gives a dramatic improvement in accuracy for basis sets of about
the same total size.

However, the optimization of the nonlinear parameters is now much more
difficult, and an automated procedure is needed.

If
<math>
E = \frac{<\Psi|H|\Psi>}{<\Psi|\Psi>}</math> and we assume <math><\Psi|\Psi> = 1</math>, then

<math>
\frac{\partial E}{\partial\alpha_t} = -2 <\Psi|H-E|r_1\psi({\bf r}_1,{\bf
r}_2;\alpha_t)\pm r_21\psi({\bf r}_2,{\bf
r}_1;\alpha_t)>
</math>
where
<math>
\psi({\bf r}_1,{\bf r}_2;\alpha_t) =
\sum_{i,j,k}c_{ijk}^{(t)}\varphi_{ijk}(\alpha_t\beta_t)
</math>

*Problem

1. Prove the above.

2. Prove that there is no contribution from the implicit dependence of
<math>c_{ijk}^{(t)}</math> on <math>\alpha_t</math> if the linear parameters have been optimized.

==The Screened Hydrogenic Term==

If the Hamiltonian is written in the form (in Z-scaled a.u.)
<math>\begin{eqnarray}\
H = H_0({\bf r}_1,Z) + H_0({\bf r}_2,Z-1) +
Z^{-1}\left(\frac{1}{r}-\frac{1}{r_2}\right)
\end{eqnarray}</math>
with
<math>\
H_0({\bf r}_1,Z) = -\frac{1}{2}\nabla_1^2 - \frac{1}{r_1}\nonumber
</math> and
<math>
H_0({\bf r}_2,Z-1) = -\frac{1}{2}\nabla_2^2 - \frac{Z-1}Z{r_2}\nonumber
</math>,
then the eigenvectors of <math>H_0({\bf r},Z) +H_0({\bf r}_2,Z-1)</math> are products of
hydrogenic orbitals
<math>
\Psi_0=\psi_0(1s,Z)\psi_0(nl,Z-1)
</math>
and the eigenvalue is
<math>\begin{eqnarray}
E_{SH} = \left[-\frac{1}{2} - \left(\frac{Z-1}{Z}\right)^2\frac{1}{2n^2}\right]Z^2 a.u.
\end{eqnarray}</math>
called the screened hydrogenic eigenvalue. From highly excited states,
<math>E_{SH}</math> and <math>\Psi_0</math>are already excellent approximations.

For example, for the 1s8d states, the energies are

<math>\begin{eqnarray}\\
E(1s8d^1D)&=& -2.007\,816\,512\,563\,81 {\rm\ a.u.}\nonumber\\
E(1s8d^3D)&=& - 2.007\,817\,934\,711\,71 {\rm\ a.u.}\nonumber\\
E_{SH} &=& -2.007\,812\,5 {\rm\ a.u.} \nonumber
\end{eqnarray}</math>

It is therefore advantageous to include the screened hydrogenic terms in the
basis set so that the complete trial function becomes
<math>
\Psi = c_0\Psi_0 +
\sum_{ijk}\left[c_{ijk}^{(1)}\varphi_{ijk}(\alpha_1,\beta_1)+c_{ijk}^{(2)}\varphi(\alpha_2,\beta_2)\right]\pm
{\rm exchange}
</math>

Also, the variational principal can be re-expressed in the form
<math>\begin{eqnarray}
E= E_{SH}+\frac{<\Psi|H-E_{SH}|\Psi>}{<\Psi|\Psi>}
\end{eqnarray}</math>
so that the <math>E_{SH}</math> term can be cancelled analytically from the matrix
elements.

For example
<math>
<\Psi_0|H-E_{SH}|\Psi_0> = 0
</math>

and
<math>
<\varphi_{ijk}|H-E_{SH}|\Psi_0> =
Z^{-1}<\varphi_{ijk}|\frac{1}{r}-\frac{1}{r_2}|\Psi_0>
</math>

Recall that
<math>\begin{eqnarray}\ \
I_0(a,b,c) = \frac{2}{c+2}\sum^{[c+1]/2}_{i=0}\left(\begin{array}{c}
c+2\\
2i+1
\end{array}\right) \left[f(p,q;\beta) + f(p^\prime,q^\prime;\alpha)\right]
\end{eqnarray}</math>
where

<math>\begin{eqnarray}
f(p,q;x) =
\frac{q!}{x^{q+1}(\alpha+\beta)^{p+1}}\sum^p_{j=0}\frac{(p+j)!}{j!}\left(\frac{x}{\alpha+\beta}\right)^j
\end{eqnarray}</math>

<math>
\begin{array}{cc}
p=a+2i+2 & p^\prime = b+2i+2\\
q= b+c-2i+2 & q^\prime =a+c-2i+2
\end{array}
</math>

If <math>\beta\ll\alpha</math>, the <math>f(p,q;\beta)\gg f(p^\prime,q^\prime;\alpha)</math> and the
<math>i=0</math> term is the dominant contribution to <math>f(p,q;\beta)</math>. However, since
this term depends only on the sum of powers <math>b+c</math> for <math>r_2</math>, it cancels exactly
from the matrix element of
<math>
\frac{1}{r_{12}} - \frac{1}{r_2}
</math>
and can therefore
be omitted in the calculations of integrands, thereby saving many significant
figures. This is especially valuable when <math>b+c</math> is large and <math>a</math> is small.

==Small Corrections==
*Mass Polarization
If the nuclear mass is not taken to be infinite, then the Hamiltonian is
<math>\begin{eqnarray}\ \
H = \frac{P_N^2}{2M} + \sum_{i=1}\frac{p_i^2}{2m} + V
\end{eqnarray}</math>

[[File:SmallCor.jpg|500px]]

where <math>M =M_A-nm</math> is the nuclear mass and <math>m</math> is the electron mass.

Change variables to the C of M
<math>\begin{eqnarray}
{\bf R} = \frac{1}{M+nm}\left[M{\bf r}_N + m({\bf r}_1 + {\bf r}_2 +
\ldots)\right]
\end{eqnarray}</math>

and relative variables

<math>
{\bf s}_i = {\bf r}_i - {\bf r}_N.
</math>

Then

<math>\begin{eqnarray}\ \ \ \ \
H = \frac{1}{2(M+nm)}{\bf p}_R^2+\frac{1}{2m}\sum_{i=1}^n{\bf p}_{s_i}^2
+\frac{1}{2M}\sum_{i,k}{\bf p}_{s_i}\cdot {\bf p}_{s_k} + V({\bf s}_1, {\bf
s}_2, \ldots)\nonumber
= \frac{1}{M+nm}{\bf p}_R^2+\frac{1}{2\mu}\sum_{i=1}^n{\bf
p}_{s_i}^2+\frac{1}{2M}\sum_{i\neq k}{\bf p}_{s_i}\cdot {\bf p}_{s_k}+V({\bf s}_1, {\bf
s}_2, \ldots)\nonumber
\end{eqnarray}</math>

where
<math>\begin{eqnarray}
\mu = \left(\frac{1}{M} + \frac{1}{m}\right)^{-1}
\end{eqnarray}</math>

If <math>n=2</math>, the Schrödinger equation is
<math>\begin{eqnarray}\ \ \
\left[ -\frac{\hbar^2}{2\mu}(\nabla^2_{s_1}+ \nabla^2_{s_2}) -
\frac{\hbar^2}{M}\nabla_{s_1}\nabla_{s_2}-\frac{Ze^2}{s_1}-\frac{Ze^2}{s_2}+\frac{e^2}{s_{12}}\right]\psi
= E\psi
\end{eqnarray}</math>.

Define the reduced mass Bohr radius
<math>\begin{eqnarray}
a_\mu = \frac{\hbar^2}{\mu e^2} = \frac{m}{\mu}a_0
\end{eqnarray}</math>
and

<math>\begin{eqnarray}
\rho_i = \frac{Zs_i}{a_\mu}\nonumber\\
\varepsilon = \frac{E}{Z^2(e^2/a_\mu)}\nonumber
\end{eqnarray}</math>

The Schrödinger equation then becomes
<math>\begin{eqnarray}\ \ \
\left[
-\frac{1}{2}(\nabla^2_{\rho_1}+\nabla^2_{\rho_2})-\frac{\mu}{M}\vec{\nabla}_{\rho_1}\cdot\vec{\nabla}_{\rho_2} - \frac{1}{\rho_1}-\frac{1}{\rho_2}+\frac{Z^{-1}}{\rho_{12}}\right]\psi = \varepsilon \psi
\end{eqnarray}</math>

The effects of finite mass are thus

1. A reduced mass correction to all energy levels of <math>(e^2/a_\mu)/(e^2/a_0)
= \mu/m</math>, i.e.
<math>\begin{eqnarray}
E_M=\frac{\mu}{m}E_\infty
\end{eqnarray}</math>

2. A specific mass shift given in first order perturbation theory by
<math>\begin{eqnarray}
\Delta\varepsilon =
-\frac{\mu}{M}<\psi|\nabla_{\rho_1}\cdot\nabla_{\rho_2}|\psi>
\end{eqnarray}</math>
or

<math>\begin{eqnarray}
\Delta\varepsilon =
-\frac{\mu}{M}<\psi|\nabla_{\rho_1}\cdot\nabla_{\rho_2}|\psi>\frac{e^2}{a_\mu}\nonumber\\
= -\frac{\mu^2}{mM}<\psi|\nabla_{\rho_1}\cdot\nabla_{\rho_2}|\psi>\frac{e^2}{a_0}\nonumber
\end{eqnarray}</math>

Since <math>\nabla_{\rho_1}\cdot\nabla_{\rho_2}</math> has the same angular properties as
<math>{\bf\rho}_1 \cdot {\bf\rho}_2</math>, the operator is like the product of two
dipole operators. For product type wave functions of the form
$
\psi = \psi(1s)\psi(nl) \pm \mbox{exchange},
$
the matrix element vanishes for all but P-states.

=Radiative Transitions=

==Transition Integrals and the Line Strength==

For an electric dipole transition <math>\gamma'\,L'\,S'\,J'\,M' \rightarrow \gamma\,L\,S\,J\,M</math>, the line strength is defined by
\begin{equation}
S = \sum_{M',\mu,M}|\langle\gamma'\,L'\,S'\,J'\,M'|r_\mu|\gamma\,L\,S\,J\,M\rangle|^2
\end{equation}
Using the Wigner-Eckart theorem, this is related to the reduced matrix element by
\begin{eqnarray}
S(J',J) = |\langle\gamma'\,L'\,S'\,J'||r||\gamma\,L\,S\,J\,M\rangle|^2
\sum_{M',\mu,M}\left|\left(\matrix{J' & 1 & J \cr
-M'& \mu &M \cr}\right)\right|^2
\end{eqnarray}
Using the sum rule for the 3-<math>j</math> symbols
\begin{equation}
\sum_{M',\mu,M}\left|\left(\matrix{J' & 1 & J \cr
-M'& \mu &M \cr}\right)\right|^2 = 1
\end{equation}
we see that the line strength is identical to the reduced matrix element
\begin{equation}
S(J',J) = |\langle\gamma'\,L'\,S'\,J'||r||\gamma\,L\,S\,J\,M\rangle|^2
\end{equation}
For a nonrelativistic calculation in <math>LS</math> coupling, <math>S = S'</math> and we can next strip out the dependence on <math>J</math> and <math>J'</math> by use of Eq.\ (7.1.7) of Edmonds { Angular Momentum in Quantum Mechanics} with <math>k = 1</math> to obtain
\begin{equation}
S(J',J) = (-1)^{L'+S+J+1}[(2J+1)(2J'+1)]^{1/2}\left\{
\matrix{L' & J'& S\cr
J & L & 1\cr}\right\}\langle\gamma' L'|| r || \gamma L\rangle
\end{equation}
where <math>\langle\gamma' L'|| r || \gamma L\rangle</math> is the reduced matrix element if spin is neglected. Note that if spin is neglected, then the line-strength is simply
\begin{equation}
S(L',L) = |\langle\gamma' L'|| r || \gamma L\rangle|^2
\end{equation}
To calculate the reduced matrix element, use a particular example along with the Wigner-Eckart theorem
such as
\begin{equation}
\langle\gamma'\, L'|| r || \gamma\, L\rangle = (-1)^{L'}\langle\gamma'\, L'\, 0| z | \gamma\, L\, 0\rangle{\Large /}\left(\matrix{L' & 1 & L \cr
0 & 0 & 0 \cr}\right)
\end{equation}

===Correlated Two-electron Integrals===
For the case of two-electron integrals, <math>z = z_1 + z_2</math> and <math>r = r_1 + r_2</math>. As discussed in G.W.F. Drake, Phys. Rev. A 18, 820 (1978), the general integral in Hylleraas coordinates for the integral over the <math>z_1</math> term can be written in the form
\begin{equation}
\label{int1}
\langle\gamma'\, L'\, 0| z_1 | \gamma\, L\, 0\rangle = \sum_\Lambda C_\Lambda I_\Lambda(\gamma',\gamma)
\end{equation}
where <math>I_\Lambda(\gamma',\gamma)</math> stands for the purely radial integral
\begin{equation}
\label{radial}
I_\Lambda(\gamma',\gamma) = \int_0^\infty r_1\,dr_1\int_0^\infty r_2\,dr_2\int_{|r_1-r_2|}^{r_1+r_2}r_{12}\, dr_{12}
R_{\gamma'}^*\,r_1\, R_\gamma P_\Lambda(\cos\theta_{12})
\end{equation}
and <math>\cos\theta_{12} = (r_1^2 + r_2^2 - r_{12}^2)/(2r_1r_2)</math> is a short-hand notation for a radial function of <math>r_1</math>, <math>r_2</math> and <math>r_{12}</math>. Also, <math>P_\Lambda(\cos\theta_{12})</math> is a Legendre polynomial,
and, <math>R_\gamma = R_\gamma(r_1,r_2,r_{12})</math> is the radial part of the Hylleraas wave function for state <math>\gamma</math> such
that the total wave function is <math>\psi({\bf r}_1,{\bf r}_2) = R_\gamma(r_1,r_2,r_{12}){\cal Y}_{l_1,l_2,L}^M(\hat{\bf r}_1,\hat{\bf r}_2)\pm {\rm exchange}</math>, normalized to unity, and <math>{\cal Y}_{l_1,l_2,L}^M(\hat{\bf r}_1,\hat{\bf r}_2)</math> is a vector coupled product of spherical harmonics. The coefficents <math>C_\Lambda</math> in Eq.\ (\ref{int1}) can be calculated from Eq.\ (32) of the above 1978 paper. They need to be calculated separately for he direct-direct and the direct-exchange contributions. For the reduced matrix element of <math>z_1</math>, the leading 3-<math>j</math> symbol can be dropped, leaving (for the case <math>k = 1</math>)
\begin{eqnarray}
C_\Lambda^{\rm reduced} &=& \frac{(-1)^{l_2'+L'+L+1}}{2} (l_1,l_1',l_2,l_2',L,L')^{1/2}
\sum_{\lambda={\rm max}(l_1-1,0)}^{l_1+1} (\lambda,\Lambda,1)
\left(\matrix{l_1 & 1 & \lambda \cr
0 & 0 & 0 \cr}\right)
\left(\matrix{l_1'& \lambda & \Lambda \cr
0 & 0 & 0 \cr}\right)\nonumber\\
&\times&\left(\matrix{l_2' & 1_2 & \Lambda \cr
0 & 0 & 0 \cr}\right)
\left\{\matrix{L' & \lambda & l_2 \cr
\Lambda& l_2' & l_1' \cr}\right\}
\left\{\matrix{L' & \lambda & l_2 \cr
l_1 & L & 1 \cr}\right\}
\end{eqnarray}
where the notation <math>(a,b,c,\cdots)</math> means <math>(2a+1)(2b+1)(2c+1)\cdots</math>.
A similar expression can be obtained for the angular coefficients for the <math>z_2</math> term by interchanging the subscripts 1 and 2 throughout and multiplying by <math>(-1)^{L+L'+1}</math>.

===Example===
As an example, for the <math>1s^2\;^1S - 1s2p\;^1P</math> transition of helium, the only nonvanishing <math>C_\Lambda</math> angular coefficients are <math>C_1 = 1/2</math> for the direct-direct term (i.e. <math>P_1(\cos\theta_{12}) = \cos\theta_{12}</math> is included in the radial integral (\ref{radial})) and <math>C_0 = 1/2</math> for the direct-exchange term (i.e. <math>P_0(\cos\theta_{12}) = 1</math> is included in the radial integral (\ref{radial})). The results for infinite nuclear mass are
\begin{equation}
S(1\;^1S-2\;^1P) = (0.781\,127\,158\,667)^2
\end{equation}
\begin{equation}f(1\;^1S-2\;^1P) = -0.276\,164\,703\,736
\end{equation}
where
\begin{equation}
f(1\;^1S-2\;^1P) = \frac{2}{3}[E(1\;^1S) - E(2\;^1P)]S(1\;^1S-2\;^1P)
\end{equation}
is the absorption oscillator strength.

Dr. Gordon Drake's Research Group

2024-07-26T11:33:16Z

Drake:

University of Windsor - Canada

[[Image:IMGP8659.JPG|650px]]

== About Us ==
Dr. G.W.F. Drake's research group is a team of graduate students, undergraduate students and postdoctoral fellows based at the University of Windsor in Canada's southern-most border city.

We develop new tools to probe the properties of atoms and nuclei by combining the high precision theoretical techniques developed in-house with the high precision measurements made possible by the modern methods of laser spectroscopy. We also work in close collaboration with Dr. Zong-Chao Yan and his colleagues at the University of New Brunswick, and Wuhan University in China.

The results calculated using techniques developed at the University of Windsor help fuel an industry of research at Argonne National Laboratory, TRIUMF, GSI (Darmstadt), and other laboratories and universities around the world.

The theoretical techniques employed include variational calculations with large correlated basis sets to solve the Schroedinger equation, and the evaluation of relativistic and quantum electrodynamic contributions to atomic processes.

== Research Summary ==
The unifying theme of our research is the development of new
measurement tools through the combined application of both high precision theory and experiment to atoms. Two examples are a
significant new value for the fine structure constant, which is one
of the fundamental constants of nature, and table-top measurements
of the size of the atomic nucleus. Recent successes at Argonne
(U.S.), GSI (Germany) and TRIUMF (Canada) have focused on the
so-called neutron-rich "halo" nuclei. These exotic structures have
extra planetary neutrons surrounding a tightly bound core of
neutrons and protons. The significance of the results is that they
enable one to distinguish amongst the various theoretical models
proposed for the effective forces holding the nucleus together. By
studying structures that fall apart easily, we learn about the
forces holding them together. A fascinating new project for the
future involves a search for new physics beyond the standard model
of elementary particles, as revealed by angular correlations
following beta decay of the halo nucleus.

Two recent advances have made these initiatives possible. First,
our work has yielded unprecedented levels of accuracy in calculating
the properties of atoms from first principles. Some of this
requires the high performance computing facilities of SHARCNET,
together with fundamental advances in methods to calculate essential
corrections due to relativity and quantum electrodynamics. I have
formed an international collaboration of other researchers in Canada
and China to carry out this work, with the help of dedicated
resources from SHARCNET.

Second, we work in close collaboration with the top experimental
groups at particle accelerators around the world to exploit the
techniques of modern laser resonance spectroscopy to perform
measurements of exquisite accuracy and sensitivity. The ability to
trap and study just a single atom plays a key role. A long sequence
of joint papers published in Physical Review Letters has resulted
from these collaborations. Our theoretical results dominate the
literature in this area, and they are providing the motivation for
experimental groups to advance the state-of-the-art for high
precision laser metrology. Measurements of the nuclear charge
radius are now available all the way from helium-3 to helium-8,
lithium-6 to lithium-11, and beryllium-8 to beryllium-12.
__NOTOC__

File:IMGP8659.JPG

2024-07-26T11:27:04Z

Drake: Group picture 2024

== Summary ==
Group picture 2024

Contact

2024-07-25T19:00:54Z

Drake: removed hard-coded URL from link

<html><iframe src='/contact/contact-raw.php' width=540 height=650 frameborder=0 scrolling=no> </iframe> </html>

Mailing address: 
Department of Physics, University of Windsor 
401 Sunset street 
Windsor, Ontario, N9B 3P4 
Canada.

Downloadable Resources

2024-07-25T18:43:00Z

Drake: removed hard-coded URL from link

==Energy & Wave functions for Heliumlike Atoms==

Please refer to [[Program Notes]] for an explanation of how the information is stored in these files and a sample program to compute expectation values. For expectation values of various operators,open and scroll through the .MAT file for the particular state selected.

<html><iframe src="/download/download-raw.php" width=780 height=460 frameborder=0 scrolling=no> </iframe></html>

Lecture Notes

2024-04-17T01:15:45Z

Drake:

This page contains Gordon Drake's hand written lecture notes and other information for the University of Windsor Physics course 64-544 ''Theory of Atomic Structure and Atomic Spectra''.

----
==Theory of Atomic Structure and Atomic spectra==

Course outline: [[media:64-544.txt|64-544]]

Pages 1-4 [[media:Spectroscopy1a.pdf|Spectroscopy 1a]]

Pages 5-10 [[media:Spectroscopy1b.pdf|Spectroscopy 1b]]

Pages 11-22 [[media:Spectroscopy2.pdf|Spectroscopy 2]]

Pages 23-27b [[media:Spectroscopy3.pdf|Spectroscopy 3]]

544 Problems & Section from "Angular Momentum Theory for Diatomic Molecules" [[Media:Spectroscopy4.pdf|Spectroscopy 4]]

Pages 29-32.5 [[Media:Spectroscopy5.pdf|Spectroscopy 5]]

Pages 33-40 [[Media:Spectroscopy6.pdf|Spectroscopy 6]]

Pages 40.1-46 [[Media:Spectroscopy7.pdf|Spectroscopy 7]]

Pages 47-49.2 [[Media:Spectroscopy8.pdf|Spectroscopy 8]]

Pages 49.3-54 [[Media:Spectroscopy9.pdf|Spectroscopy 9]]

Pages 52-58 [[Media:Spectroscopy10.pdf|Spectroscopy 10]]

Pages 59-68 [[Media:Spectroscopy11(1).pdf|Spectroscopy 11]]

Pages 69-76 [[Media:Spectroscopy12.pdf|Spectroscopy 12]]

Pages 77-86 [[Media:Spectroscopy13.pdf|Spectroscopy 13]]

Pages 77-86 [[Media:Spectroscopy14(1).pdf|Spectroscopy 14]]

Pages 87-102 [[Media:Spectroscopy15(2).pdf|Spectroscopy 15]]

Pages 103-118 [[Media:Spectroscopy16(1).pdf|Spectroscopy 16]]

Pages 119-132 [[Media:Spectroscopy17.pdf|Spectroscopy 17]]

Pages 133-145 [[Media:Spectroscopy18.pdf|Spectroscopy 18]]

Group Theory [[Media:Spectroscopy19.pdf|Part 1]]

Group Theory [[Media:Spectroscopy20.pdf|Part 2]]

==Relativistic Effects in Atomic Physics==

Relativistic 0-12 [[Media:Relativistic 1.pdf|Relativistic 1]]

Relativistic 13-21 [[Media:Relativistic 2.pdf|Relativistic 2]]

Relativistic 22-30 [[Media:Relativistic 3.1-3.2.pdf|Relativistic 3.1-3.2]]

Relativistic 31-44 [[Media:Relativistic 3.3-3.4.pdf|Relativistic 3.3-3.4]]

Dirac Equations 23-30 [[Media:Dirac Equations 1.pdf|Dirac Equations 1]]

Dirac Equations 31-42 [[Media:Dirac Equations 2.pdf|Dirac Equations 2]]

Dirac Equations Addendum [[Media:Dirac Equations 3.pdf|Dirac Equations 3]]

==WIPM Lecture Notes==

[[Media:lectures.pdf|Lecture Notes 1 Oct. 9/19]]

[[Media:lecture1a.pdf|Lecture Notes 1 Appendix Oct. 11/19]]

[[Media:Lecture2.pdf|Lecture Notes 2 Oct. 14/19]]

[[Media:lecture3.pdf|Lecture Notes 3 Oct. 16/19]]

[[Media:Lecture4.pdf|Lecture Notes 4 Oct. 25/19]]

[[Media:lecture4a.pdf|Lecture Notes 4 Appendix Oct. 28/19]]

[[Media:lecture5.pdf|Lecture Notes 5 Oct. 30/19]]

[[Media:lecture6.pdf|Lecture Notes 6 Oct. 31/19]]

[[Media:lecture7.pdf|Lecture Notes 7 Nov. 1/19]]

[[Media:lecture8.pdf|Lecture Notes 8 Oct. 4/19]]

[[Media:lecture9.pdf|Lecture Notes 9 Nov. 5/19]]

[[Media:lecture10.pdf|Lecture Notes 10 Nov. 6/19]]

[[Media:lecture11.pdf|Lecture Notes 11 Nov. 6/19]]

[[Media:lecture12_GD.pdf|Lecture Notes 12 Nov. 11/19]]

==WIPM Recorded Lectures==
The above lectures were recorded at the Wuhan Institute of Physics and Mathematics (WIPM) during the fall of 2019. Click [https://www.uwindsor.ca/science/physics/466/high-precision-atomic-theory here] to view the recorded lectures.

Lecture Notes

2024-04-17T00:56:35Z

Drake: /* WIPM RECORDED LECTURES */

This page contains Gordon Drake's hand written lecture notes and other information for the University of Windsor Physics course 64-544 ''Theory of Atomic Structure and Atomic Spectra''.

----
==Theory of Atomic Structure and Atomic spectra==

Course outline: [[media:64-544.txt|64-544]]

Pages 1-4 [[media:Spectroscopy1a.pdf|Spectroscopy 1a]]

Pages 5-10 [[media:Spectroscopy1b.pdf|Spectroscopy 1b]]

Pages 11-22 [[media:Spectroscopy2.pdf|Spectroscopy 2]]

Pages 23-27b [[media:Spectroscopy3.pdf|Spectroscopy 3]]

544 Problems & Section from "Angular Momentum Theory for Diatomic Molecules" [[Media:Spectroscopy4.pdf|Spectroscopy 4]]

Pages 29-32.5 [[Media:Spectroscopy5.pdf|Spectroscopy 5]]

Pages 33-40 [[Media:Spectroscopy6.pdf|Spectroscopy 6]]

Pages 40.1-46 [[Media:Spectroscopy7.pdf|Spectroscopy 7]]

Pages 47-49.2 [[Media:Spectroscopy8.pdf|Spectroscopy 8]]

Pages 49.3-54 [[Media:Spectroscopy9.pdf|Spectroscopy 9]]

Pages 52-58 [[Media:Spectroscopy10.pdf|Spectroscopy 10]]

Pages 59-68 [[Media:Spectroscopy11(1).pdf|Spectroscopy 11]]

Pages 69-76 [[Media:Spectroscopy12.pdf|Spectroscopy 12]]

Pages 77-86 [[Media:Spectroscopy13.pdf|Spectroscopy 13]]

Pages 77-86 [[Media:Spectroscopy14(1).pdf|Spectroscopy 14]]

Pages 87-102 [[Media:Spectroscopy15(2).pdf|Spectroscopy 15]]

Pages 103-118 [[Media:Spectroscopy16(1).pdf|Spectroscopy 16]]

Pages 119-132 [[Media:Spectroscopy17.pdf|Spectroscopy 17]]

Pages 133-145 [[Media:Spectroscopy18.pdf|Spectroscopy 18]]

Group Theory [[Media:Spectroscopy19.pdf|Part 1]]

Group Theory [[Media:Spectroscopy20.pdf|Part 2]]

==Relativistic Effects in Atomic Physics==

Relativistic 0-12 [[Media:Relativistic 1.pdf|Relativistic 1]]

Relativistic 13-21 [[Media:Relativistic 2.pdf|Relativistic 2]]

Relativistic 22-30 [[Media:Relativistic 3.1-3.2.pdf|Relativistic 3.1-3.2]]

Relativistic 31-44 [[Media:Relativistic 3.3-3.4.pdf|Relativistic 3.3-3.4]]

Dirac Equations 23-30 [[Media:Dirac Equations 1.pdf|Dirac Equations 1]]

Dirac Equations 31-42 [[Media:Dirac Equations 2.pdf|Dirac Equations 2]]

Dirac Equations Addendum [[Media:Dirac Equations 3.pdf|Dirac Equations 3]]

==WIPM Lecture Notes==

[[Media:lectures.pdf|Lecture Notes 1 Oct. 9/19]]

[[Media:lecture1a.pdf|Lecture Notes 1 Appendix Oct. 11/19]]

[[Media:Lecture2.pdf|Lecture Notes 2 Oct. 14/19]]

[[Media:lecture3.pdf|Lecture Notes 3 Oct. 16/19]]

[[Media:Lecture4.pdf|Lecture Notes 4 Oct. 25/19]]

[[Media:lecture4a.pdf|Lecture Notes 4 Appendix Oct. 28/19]]

[[Media:lecture5.pdf|Lecture Notes 5 Oct. 30/19]]

[[Media:lecture6.pdf|Lecture Notes 6 Oct. 31/19]]

[[Media:lecture7.pdf|Lecture Notes 7 Nov. 1/19]]

[[Media:lecture8.pdf|Lecture Notes 8 Oct. 4/19]]

[[Media:lecture9.pdf|Lecture Notes 9 Nov. 5/19]]

[[Media:lecture10.pdf|Lecture Notes 10 Nov. 6/19]]

[[Media:lecture11.pdf|Lecture Notes 11 Nov. 6/19]]

[[Media:lecture12_GD.pdf|Lecture Notes 12 Nov. 11/19]]

==WIPM RECORDED LECTURES==
<a href="https://www.uwindsor.ca/science/physics/466/high-precision-atomic-theory/"> Click here to view the recorded lectures</a>

Lecture Notes

2024-04-17T00:51:51Z

Drake: /* WIPM Lecture Notes */

This page contains Gordon Drake's hand written lecture notes and other information for the University of Windsor Physics course 64-544 ''Theory of Atomic Structure and Atomic Spectra''.

----
==Theory of Atomic Structure and Atomic spectra==

Course outline: [[media:64-544.txt|64-544]]

Pages 1-4 [[media:Spectroscopy1a.pdf|Spectroscopy 1a]]

Pages 5-10 [[media:Spectroscopy1b.pdf|Spectroscopy 1b]]

Pages 11-22 [[media:Spectroscopy2.pdf|Spectroscopy 2]]

Pages 23-27b [[media:Spectroscopy3.pdf|Spectroscopy 3]]

544 Problems & Section from "Angular Momentum Theory for Diatomic Molecules" [[Media:Spectroscopy4.pdf|Spectroscopy 4]]

Pages 29-32.5 [[Media:Spectroscopy5.pdf|Spectroscopy 5]]

Pages 33-40 [[Media:Spectroscopy6.pdf|Spectroscopy 6]]

Pages 40.1-46 [[Media:Spectroscopy7.pdf|Spectroscopy 7]]

Pages 47-49.2 [[Media:Spectroscopy8.pdf|Spectroscopy 8]]

Pages 49.3-54 [[Media:Spectroscopy9.pdf|Spectroscopy 9]]

Pages 52-58 [[Media:Spectroscopy10.pdf|Spectroscopy 10]]

Pages 59-68 [[Media:Spectroscopy11(1).pdf|Spectroscopy 11]]

Pages 69-76 [[Media:Spectroscopy12.pdf|Spectroscopy 12]]

Pages 77-86 [[Media:Spectroscopy13.pdf|Spectroscopy 13]]

Pages 77-86 [[Media:Spectroscopy14(1).pdf|Spectroscopy 14]]

Pages 87-102 [[Media:Spectroscopy15(2).pdf|Spectroscopy 15]]

Pages 103-118 [[Media:Spectroscopy16(1).pdf|Spectroscopy 16]]

Pages 119-132 [[Media:Spectroscopy17.pdf|Spectroscopy 17]]

Pages 133-145 [[Media:Spectroscopy18.pdf|Spectroscopy 18]]

Group Theory [[Media:Spectroscopy19.pdf|Part 1]]

Group Theory [[Media:Spectroscopy20.pdf|Part 2]]

==Relativistic Effects in Atomic Physics==

Relativistic 0-12 [[Media:Relativistic 1.pdf|Relativistic 1]]

Relativistic 13-21 [[Media:Relativistic 2.pdf|Relativistic 2]]

Relativistic 22-30 [[Media:Relativistic 3.1-3.2.pdf|Relativistic 3.1-3.2]]

Relativistic 31-44 [[Media:Relativistic 3.3-3.4.pdf|Relativistic 3.3-3.4]]

Dirac Equations 23-30 [[Media:Dirac Equations 1.pdf|Dirac Equations 1]]

Dirac Equations 31-42 [[Media:Dirac Equations 2.pdf|Dirac Equations 2]]

Dirac Equations Addendum [[Media:Dirac Equations 3.pdf|Dirac Equations 3]]

==WIPM Lecture Notes==

[[Media:lectures.pdf|Lecture Notes 1 Oct. 9/19]]

[[Media:lecture1a.pdf|Lecture Notes 1 Appendix Oct. 11/19]]

[[Media:Lecture2.pdf|Lecture Notes 2 Oct. 14/19]]

[[Media:lecture3.pdf|Lecture Notes 3 Oct. 16/19]]

[[Media:Lecture4.pdf|Lecture Notes 4 Oct. 25/19]]

[[Media:lecture4a.pdf|Lecture Notes 4 Appendix Oct. 28/19]]

[[Media:lecture5.pdf|Lecture Notes 5 Oct. 30/19]]

[[Media:lecture6.pdf|Lecture Notes 6 Oct. 31/19]]

[[Media:lecture7.pdf|Lecture Notes 7 Nov. 1/19]]

[[Media:lecture8.pdf|Lecture Notes 8 Oct. 4/19]]

[[Media:lecture9.pdf|Lecture Notes 9 Nov. 5/19]]

[[Media:lecture10.pdf|Lecture Notes 10 Nov. 6/19]]

[[Media:lecture11.pdf|Lecture Notes 11 Nov. 6/19]]

[[Media:lecture12_GD.pdf|Lecture Notes 12 Nov. 11/19]]

==WIPM RECORDED LECTURES==
<a href="https://www.uwindsor.ca/science/physics/466/high-precision-atomic-theory> Click here to view the recorded lectures</a>

Publications

2022-10-02T01:15:56Z

Drake:

Recent journal publications

H. S. Dhindsa, V. J. Marton and G. W. F. Drake, "Search for light bosons with King and second-King plots optimized for lithium ions," Phys. Part. Nuclei, 53, 800 (2022), 5 pp.

B. M. Henson, J. A Ross, K. F. Thomas, C. N. Kuhn, D. K. Shin, S. S. Hodgman, Y. H. L. Y. Zhang, G. W. F. Drake, A. T. Bondy, A. G. Truscott, and K. G. H. Baldwin, "Measurement of a helium tune-out frequency: an independent test of quantum electrodynamics," Science, 376, 199 (2022).

G.W.F. Drake, Harvir S. Dhindsa and Victor J. Marton, "King and second-King plots with optimized sensitivity for lithium ions," Phys. Rev. A, 104, L060801 (2021), 6 pages.

A.T. Bondy, D.C. Morton, and G.W.F. Drake, "Two-photon decay rates in heliumlike ions: Finite-nuclear-mass effects," Phys. Rev. A, 102, 052807 (2020), 10 pages.

G.W.F. Drake and E. Tiesinga, "Simplify Your Life," Nature Physics 16, 1242 (2020): https://rdcu.be/cbBQS

G.W.F. Drake, "Accuracy in Atomic and Molecular Data," J. Phys. B: At. Mol. Opt. Phys. 53, 223001 (2020).

A.T. Bondy, D.C. Morton, and G.W.F. Drake, "Two-photon decay rates in heliumlike ions: Finite-nuclear-mass effects," Phys. Rev. A 102, 052807 (2020)

X.-Q Qi, P.-P. Zhang, Z.-C. Yan, G.W.F. Drake, Z.-X. Zhong, T.-Y. Shi, S.-L. Chen, Y. Huang, H. Guan, and K.-L. Gao, "Precision Calculation of Hyperfine Structure and the Zemach Radii of Li-6.7(+) Ions
Phys. Rev. Lett. 125, 183002 (2020).

H. Guan, S.L. Chen, X.-Q Qi, S.Y. Liang, W. Sun, P.P. Zhou, Y. Huang, P.P. Zhang, Z.-X. Zhong, Z.-C. Yan, G.W.F. Drake, T.-Y. Shi, and K.L. Gao, "Probing atomic and nuclear properties with precision spectroscopy of fine and hyperfine structures in the Li-7(+) ion," Phys. Rev. A 102, 030801 (2020).

G.W.F. Drake, J.G. Manalo, P.-P. Zhang, and K.G.H. Baldwin, "Helium tune-out wavelength: Gauge invariance and retardation corrections," Hyperfine inter. 240, 31 (2019), 8 pp.

Gordon W.F. Drake, Jung-Sik Yoon, Daiji Kato, and Grzegorz Karwasz,"Atomic and Molecular Data and their Applications," Euro. Phys. J. D 72, R49 (2018), 3 pp.

Donald C. Morton and G.W.F. Drake, "Oscillator strengths for spin-changing P–D transitions in He I including the effect of a finite nuclear mass and intermediate coupling," Can. J. Phys. 95, 828(2017), 4 pp.

L.M. Wang, Chun Li, Z.-C. Yan, and G.W.F. Drake, "Isotope shifts and transition frequencies for the S and P states of lithium: Bethe logarithms and second-order relativistic recoil," Phys. Rev. A 95 R032504 (2017) 10 pp.

D. C. Morton and G. W. F. Drake, "Oscillator strengths for 1s^2 ^1S_0 - 1s2p ^3P_{1,2} transitions in helium-like carbon, nitrogen and oxygen including the effects of a finite nuclear mass," J. Phys. B -- At. Mol. Opt. Phys. 49 234002 (2016), 7 pp.

H.-K. Chung, B. J. Braams, K. Bartschat, A. G. Csaszar, G. W. F. Drake, T. Kirchner, V. Kokoouline and J. Tennyson, "Uncertainty Estimates for Theoretical Atomic and Molecular Data”, J. Phys. D: Applied Physics, 49 (2016); arXiv:1603.05923v2 [physics.atom-ph].

D.C. Morton, E. Schulhoff and G.W.F. Drake, "Oscillator strengths and radiative decay rates for spin-changing S-P transition in helium: finite nuclear mass effects,"
J. Phys. B. 48, 235001 (2015).

E. Schulhoff and G.W.F. Drake, "Electron emission and recoil effects following the beta-decay of helium-6 (6He)," Phys. Rev. A. 92, 050701 (2015).

L.M. Wang, C. Li, Z.-C. Yan, and G.W.F. Drake, "Fine structure and ionization energy of the 1s2s2p (4)P state of the helium negative ion He(-),"
Phys. Rev. Lett. 113, 263007 (2014) (4 pages).

C. Estienne, M. Busuttil, A. Moini, and G.W.F. Drake, "Critical nuclear charge for two-electron atoms," Phys. Rev. Lett. 112, 173001 (2014) (4 pages).

Z.-T. Lu, P. Mueller, G.W.F. Drake et. al.,
"Colloquium: Laser probing of neutron-rich nuclei in light atoms," Rev. Mod. Phys. 85, 1383-1400 (2013).

L.M. Wang, Z.-C. Yan, H.X. Qiao et al., "Variational energies and the Fermi contact term for the low-lying states of lithium: Basis-set completeness," Phys. Rev. A 85, 052513 (2012).

A.S. Titi, and G.W.F. Drake, "Quantum theory of longitudinal momentum transfer in above-threshold ionization," Phys. Rev. A 85, 041404 (2012).

W. ElMaraghy, H. ElMaraghy, T. Tomiyama, L. Monostori, Complexity in engineering design and
manufacturing, CIRP Annals - Manufacturing Technology, 61/2: 2012

A. Djuric, R. Al Saidi, W. ElMaraghy, Dynamics solution of n-DOF global machinery model, Robotics and
Computer-Integrated Manufacturing, 28: 621-630 pp, 2012

M. Brodeur, T. Brunner, C. Champagne, et al.,
"First Direct Mass Measurement of the Two-Neutron Halo Nucleus He-6 and Improved Mass for the Four-Neutron Halo He-8," Phys. Rev. Lett. 108, 052504 (2012).

D. Morton and G.W.F. Drake, 2011. "Spin-forbidden radiative decay rates from the 3 (3)P(1,2) and 3 (1)P(1) states of helium," Phys. Rev. A 83, 042503 (2011)

L. M. Wang, Z. -C. Yan, H.X. Qiao, et al., "Variational upper bounds for low-lying states of lithium," Phys. Rev. A 83, 034503 (2011) (4 pages)

W. Noertershaeuser, W., R. Sanchez, R. G. Ewald, G., et al. " Isotope-shift measurements of stable and short-lived lithium isotopes for nuclear-charge-radii determination," Phys. Rev. A, 83, 012516, (2011)

Donald C. Morton, Paul Moffatt, and G. W. F. Drake, "Relativistic corrections to He I transition rates," Can. J. Phys. 89, 1, (2011) (5 pages)
Conference: International Conference on Precision Physics of Simple Atomic Systems Location: Ecole de Physique, Les Houches, FRANCE Date: MAY 30-JUN 04, 2010

M. Zakova, Z. Andjelkovic, M.L. Bissell, K. Blaum, G.W.F. Drake, C Geppert, M. Kowalska, J. Kramer, A. Krieger, M. Lochmann, T. Neff, R. Neugart, W. N\"ortersh\"auser, R. Sanchez, F. Schmidt-Kaler, D. Tiedemann, Z.-C. Yan, D.T. Yordanov, And C. Zimmermann, "Isotope shift measurements in the 2s(1/2) --> 2p(3/2) transition of Be+ and extraction
of the nuclear charge radii for Be-7, Be-10, Be-11," J. Phys. G--Nucl. and Particle Phys., 37, 055107 (2010) (14 pages).

R. El-Wazni and G.W.F. Drake, "Energies for the high-L Rydberg states of helium: Asymptotic analysis," Phys. Rev. A 80, 064501 (2009) (4 pages).

R. Ringle, M. Brodeur, T. Brunner, S. Ettenauer, M. Smith, A. Lapierre, V.L. Ryjkov, P. Delheij, G.W.F. Drake, J. Lassen, D. Lunney, and J. Dilling, "High-Precision Penning-Trap Mass Measurements of 9,10Be and the One-Neutron Halo Nuclide 11Be," Phys. Lett.\ B 695, 170-174 (2009).

W. Nortershauser, D. Tiedemann, M. Zakova, Z. Andjelkovic, K. Blaum, M.L. Bissell, R. Cazan, G.W.F. Drake, C. Geppert, M. Kowalska, J. Kramer, A. Krieger, R. Neugart, R. Sanchez, F. Schmidt-Kaler, Z-C. Yan, D.T. Yordanov, C. Zimmermann, "Nuclear Charge Radii of Be-7, Be-9, and One-Neutron Halo Nucleus Be-11", Phys. Rev. Lett., 102, 062503 (2009).

M. Smith, M. Brodeur. T. Brunner, S. Ettenauer, A. Lapierre, R. Ringle, V. L. Ryjkov, F. Ames, P. Bricault, G. W. F. Drake, P. Delheij, D. Lunney, F. Sarazin, and J. Dilling, First Penning-Trap Mass Measurements of the Exotic Halo Nucleus, Phys. Rev. Lett., 101, 202501 (2008).

I.A. Sulai, Wu, Qixue, Bishof, M., Drake, G. W. F., Lu, Z. -T., Mueller, P., Santra, R., Hyperfine Suppression
of 2(3)(S(1)-3 (3)P(J)Transitions in 3He, Phys. Rev. Lett. 101, 173001 (2008).

Publications

2022-10-02T01:06:53Z

Drake:

Recent journal publications

H. S. Dhindsa, V. J. Marton and G. W. F. Drake, "Search for light bosons with King and second-King plots optimized for lithium ions," Phys. Part. Nuclei, 53</strong}, 800 (2022), 5 pp.

B. M. Henson, J. A Ross, K. F. Thomas, C. N. Kuhn, D. K. Shin, S. S. Hodgman, Y. H. L. Y. Zhang, G. W. F. Drake, A. T. Bondy, A. G. Truscott, and K. G. H. Baldwin, "Measurement of a helium tune-out frequency: an independent test of quantum electrodynamics," Science, 376, 199 (2022).

G.W.F. Drake, Harvir S. Dhindsa and Victor J. Marton, "King and second-King plots with optimized sensitivity for lithium ions," Phys. Rev. A, 104, L060801 (2021), 6 pages.

A.T. Bondy, D.C. Morton, and G.W.F. Drake, "Two-photon decay rates in heliumlike ions: Finite-nuclear-mass effects," Phys. Rev. A, 102, 052807 (2020), 10 pages.

G.W.F. Drake and E. Tiesinga, "Simplify Your Life," Nature Physics 16, 1242 (2020): https://rdcu.be/cbBQS

G.W.F. Drake, "Accuracy in Atomic and Molecular Data," J. Phys. B: At. Mol. Opt. Phys. 53, 223001 (2020).

A.T. Bondy, D.C. Morton, and G.W.F. Drake, "Two-photon decay rates in heliumlike ions: Finite-nuclear-mass effects," Phys. Rev. A 102, 052807 (2020)

X.-Q Qi, P.-P. Zhang, Z.-C. Yan, G.W.F. Drake, Z.-X. Zhong, T.-Y. Shi, S.-L. Chen, Y. Huang, H. Guan, and K.-L. Gao, "Precision Calculation of Hyperfine Structure and the Zemach Radii of Li-6.7(+) Ions
Phys. Rev. Lett. 125, 183002 (2020).

H. Guan, S.L. Chen, X.-Q Qi, S.Y. Liang, W. Sun, P.P. Zhou, Y. Huang, P.P. Zhang, Z.-X. Zhong, Z.-C. Yan, G.W.F. Drake, T.-Y. Shi, and K.L. Gao, "Probing atomic and nuclear properties with precision spectroscopy of fine and hyperfine structures in the Li-7(+) ion," Phys. Rev. A 102, 030801 (2020).

G.W.F. Drake, J.G. Manalo, P.-P. Zhang, and K.G.H. Baldwin, "Helium tune-out wavelength: Gauge invariance and retardation corrections," Hyperfine inter. 240, 31 (2019), 8 pp.

Gordon W.F. Drake, Jung-Sik Yoon, Daiji Kato, and Grzegorz Karwasz,"Atomic and Molecular Data and their Applications," Euro. Phys. J. D 72, R49 (2018), 3 pp.

Donald C. Morton and G.W.F. Drake, "Oscillator strengths for spin-changing P–D transitions in He I including the effect of a finite nuclear mass and intermediate coupling," Can. J. Phys. 95, 828(2017), 4 pp.

L.M. Wang, Chun Li, Z.-C. Yan, and G.W.F. Drake, "Isotope shifts and transition frequencies for the S and P states of lithium: Bethe logarithms and second-order relativistic recoil," Phys. Rev. A 95 R032504 (2017) 10 pp.

D. C. Morton and G. W. F. Drake, "Oscillator strengths for 1s^2 ^1S_0 - 1s2p ^3P_{1,2} transitions in helium-like carbon, nitrogen and oxygen including the effects of a finite nuclear mass," J. Phys. B -- At. Mol. Opt. Phys. 49 234002 (2016), 7 pp.

H.-K. Chung, B. J. Braams, K. Bartschat, A. G. Csaszar, G. W. F. Drake, T. Kirchner, V. Kokoouline and J. Tennyson, "Uncertainty Estimates for Theoretical Atomic and Molecular Data”, J. Phys. D: Applied Physics, 49 (2016); arXiv:1603.05923v2 [physics.atom-ph].

D.C. Morton, E. Schulhoff and G.W.F. Drake, "Oscillator strengths and radiative decay rates for spin-changing S-P transition in helium: finite nuclear mass effects,"
J. Phys. B. 48, 235001 (2015).

E. Schulhoff and G.W.F. Drake, "Electron emission and recoil effects following the beta-decay of helium-6 (6He)," Phys. Rev. A. 92, 050701 (2015).

L.M. Wang, C. Li, Z.-C. Yan, and G.W.F. Drake, "Fine structure and ionization energy of the 1s2s2p (4)P state of the helium negative ion He(-),"
Phys. Rev. Lett. 113, 263007 (2014) (4 pages).

C. Estienne, M. Busuttil, A. Moini, and G.W.F. Drake, "Critical nuclear charge for two-electron atoms," Phys. Rev. Lett. 112, 173001 (2014) (4 pages).

Z.-T. Lu, P. Mueller, G.W.F. Drake et. al.,
"Colloquium: Laser probing of neutron-rich nuclei in light atoms," Rev. Mod. Phys. 85, 1383-1400 (2013).

L.M. Wang, Z.-C. Yan, H.X. Qiao et al., "Variational energies and the Fermi contact term for the low-lying states of lithium: Basis-set completeness," Phys. Rev. A 85, 052513 (2012).

A.S. Titi, and G.W.F. Drake, "Quantum theory of longitudinal momentum transfer in above-threshold ionization," Phys. Rev. A 85, 041404 (2012).

W. ElMaraghy, H. ElMaraghy, T. Tomiyama, L. Monostori, Complexity in engineering design and
manufacturing, CIRP Annals - Manufacturing Technology, 61/2: 2012

A. Djuric, R. Al Saidi, W. ElMaraghy, Dynamics solution of n-DOF global machinery model, Robotics and
Computer-Integrated Manufacturing, 28: 621-630 pp, 2012

M. Brodeur, T. Brunner, C. Champagne, et al.,
"First Direct Mass Measurement of the Two-Neutron Halo Nucleus He-6 and Improved Mass for the Four-Neutron Halo He-8," Phys. Rev. Lett. 108, 052504 (2012).

D. Morton and G.W.F. Drake, 2011. "Spin-forbidden radiative decay rates from the 3 (3)P(1,2) and 3 (1)P(1) states of helium," Phys. Rev. A 83, 042503 (2011)

L. M. Wang, Z. -C. Yan, H.X. Qiao, et al., "Variational upper bounds for low-lying states of lithium," Phys. Rev. A 83, 034503 (2011) (4 pages)

W. Noertershaeuser, W., R. Sanchez, R. G. Ewald, G., et al. " Isotope-shift measurements of stable and short-lived lithium isotopes for nuclear-charge-radii determination," Phys. Rev. A, 83, 012516, (2011)

Donald C. Morton, Paul Moffatt, and G. W. F. Drake, "Relativistic corrections to He I transition rates," Can. J. Phys. 89, 1, (2011) (5 pages)
Conference: International Conference on Precision Physics of Simple Atomic Systems Location: Ecole de Physique, Les Houches, FRANCE Date: MAY 30-JUN 04, 2010

M. Zakova, Z. Andjelkovic, M.L. Bissell, K. Blaum, G.W.F. Drake, C Geppert, M. Kowalska, J. Kramer, A. Krieger, M. Lochmann, T. Neff, R. Neugart, W. N\"ortersh\"auser, R. Sanchez, F. Schmidt-Kaler, D. Tiedemann, Z.-C. Yan, D.T. Yordanov, And C. Zimmermann, "Isotope shift measurements in the 2s(1/2) --> 2p(3/2) transition of Be+ and extraction
of the nuclear charge radii for Be-7, Be-10, Be-11," J. Phys. G--Nucl. and Particle Phys., 37, 055107 (2010) (14 pages).

R. El-Wazni and G.W.F. Drake, "Energies for the high-L Rydberg states of helium: Asymptotic analysis," Phys. Rev. A 80, 064501 (2009) (4 pages).

R. Ringle, M. Brodeur, T. Brunner, S. Ettenauer, M. Smith, A. Lapierre, V.L. Ryjkov, P. Delheij, G.W.F. Drake, J. Lassen, D. Lunney, and J. Dilling, "High-Precision Penning-Trap Mass Measurements of 9,10Be and the One-Neutron Halo Nuclide 11Be," Phys. Lett.\ B 695, 170-174 (2009).

W. Nortershauser, D. Tiedemann, M. Zakova, Z. Andjelkovic, K. Blaum, M.L. Bissell, R. Cazan, G.W.F. Drake, C. Geppert, M. Kowalska, J. Kramer, A. Krieger, R. Neugart, R. Sanchez, F. Schmidt-Kaler, Z-C. Yan, D.T. Yordanov, C. Zimmermann, "Nuclear Charge Radii of Be-7, Be-9, and One-Neutron Halo Nucleus Be-11", Physical Review Letters, 102: (2009)

M. Smith, M. Brodeur. T. Brunner, S. Ettenauer, A. Lapierre, R. Ringle, V. L. Ryjkov, F. Ames, P. Bricault, G. W. F. Drake, P. Delheij, D. Lunney, F. Sarazin, and J. Dilling, First Penning-Trap Mass Measurements of the Exotic Halo Nucleus, Physical Review Letters, 101: 2008

I.A. Sulai, Wu, Qixue, Bishof, M., Drake, G. W. F., Lu, Z. -T., Mueller, P., Santra, R., Hyperfine Suppression
of 2(3)(S(1)-3 (3)P(J)Transitions in 3He, Physical Review Letters,<stong> 101</strong: 2008

Publications

2022-10-02T00:55:00Z

Drake:

Recent journal publications

H. S. Dhindsa, V. J. Marton and G. W. F. Drake, ``Search for light bosons with King and second-King
plots optimized for lithium ions," Phys.\ Part.\ Nuclei, {\bf 53}, 800 (2022), 5 pp.

\item[205]
B. M. Henson, J. A Ross, K. F. Thomas, C. N. Kuhn, D. K. Shin, S. S. Hodgman, Y. H. L. Y. Zhang, G. W.
F. Drake, A. T. Bondy, A. G. Truscott, and K. G. H. Baldwin, ``Measurement of a helium tune-out
frequency: an independent test of quantum electrodynamics," Science, {\bf 376} 199 (2022).

\item[204]
G.W.F. Drake, Harvir S. Dhindsa and Victor J. Marton, ``King and second-King plots with optimized
sensitivity for lithium ions," Phys.\ Rev.\ A, {\bf 104} L060801 (2021), 6 pages.

\item[203]
. A.T. Bondy, D.C. Morton, and G.W.F. Drake,``Two-photon decay rates in heliumlike ions: Finite-
nuclear-mass effects," Phys.\ Rev.\ A, {\bf 102} 052807 (2020), 10 pages.

G.W.F. Drake and E. Tiesinga, "Simplify Your Life," Nature Physics 16, 1242 (2020): https://rdcu.be/cbBQS

G.W.F. Drake, "Accuracy in Atomic and Molecular Data," J. Phys. B: At. Mol. Opt. Phys. 53, 223001 (2020).

A.T. Bondy, D.C. Morton, and G.W.F. Drake, "Two-photon decay rates in heliumlike ions: Finite-nuclear-mass effects," Phys. Rev. A 102, 052807 (2020)

X.-Q Qi, P.-P. Zhang, Z.-C. Yan, G.W.F. Drake, Z.-X. Zhong, T.-Y. Shi, S.-L. Chen, Y. Huang, H. Guan, and K.-L. Gao, "Precision Calculation of Hyperfine Structure and the Zemach Radii of Li-6.7(+) Ions
Phys. Rev. Lett. 125, 183002 (2020).

H. Guan, S.L. Chen, X.-Q Qi, S.Y. Liang, W. Sun, P.P. Zhou, Y. Huang, P.P. Zhang, Z.-X. Zhong, Z.-C. Yan, G.W.F. Drake, T.-Y. Shi, and K.L. Gao, "Probing atomic and nuclear properties with precision spectroscopy of fine and hyperfine structures in the Li-7(+) ion," Phys. Rev. A 102, 030801 (2020).

G.W.F. Drake, J.G. Manalo, P.-P. Zhang, and K.G.H. Baldwin, "Helium tune-out wavelength: Gauge invariance and retardation corrections," Hyperfine inter. 240, 31 (2019), 8 pp.

Gordon W.F. Drake, Jung-Sik Yoon, Daiji Kato, and Grzegorz Karwasz,"Atomic and Molecular Data and their Applications," Euro. Phys. J. D 72, R49 (2018), 3 pp.

Donald C. Morton and G.W.F. Drake, "Oscillator strengths for spin-changing P–D transitions in He I including the effect of a finite nuclear mass and intermediate coupling," Can. J. Phys. 95, 828(2017), 4 pp.

L.M. Wang, Chun Li, Z.-C. Yan, and G.W.F. Drake, "Isotope shifts and transition frequencies for the S and P states of lithium: Bethe logarithms and second-order relativistic recoil," Phys. Rev. A 95 R032504 (2017) 10 pp.

D. C. Morton and G. W. F. Drake, "Oscillator strengths for 1s^2 ^1S_0 - 1s2p ^3P_{1,2} transitions in helium-like carbon, nitrogen and oxygen including the effects of a finite nuclear mass," J. Phys. B -- At. Mol. Opt. Phys. 49 234002 (2016), 7 pp.

H.-K. Chung, B. J. Braams, K. Bartschat, A. G. Csaszar, G. W. F. Drake, T. Kirchner, V. Kokoouline and J. Tennyson, "Uncertainty Estimates for Theoretical Atomic and Molecular Data”, J. Phys. D: Applied Physics, 49 (2016); arXiv:1603.05923v2 [physics.atom-ph].

D.C. Morton, E. Schulhoff and G.W.F. Drake, "Oscillator strengths and radiative decay rates for spin-changing S-P transition in helium: finite nuclear mass effects,"
J. Phys. B. 48, 235001 (2015).

E. Schulhoff and G.W.F. Drake, "Electron emission and recoil effects following the beta-decay of helium-6 (6He)," Phys. Rev. A. 92, 050701 (2015).

L.M. Wang, C. Li, Z.-C. Yan, and G.W.F. Drake, "Fine structure and ionization energy of the 1s2s2p (4)P state of the helium negative ion He(-),"
Phys. Rev. Lett. 113, 263007 (2014) (4 pages).

C. Estienne, M. Busuttil, A. Moini, and G.W.F. Drake, "Critical nuclear charge for two-electron atoms," Phys. Rev. Lett. 112, 173001 (2014) (4 pages).

Z.-T. Lu, P. Mueller, G.W.F. Drake et. al.,
"Colloquium: Laser probing of neutron-rich nuclei in light atoms," Rev. Mod. Phys. 85, 1383-1400 (2013).

L.M. Wang, Z.-C. Yan, H.X. Qiao et al., "Variational energies and the Fermi contact term for the low-lying states of lithium: Basis-set completeness," Phys. Rev. A 85, 052513 (2012).

A.S. Titi, and G.W.F. Drake, "Quantum theory of longitudinal momentum transfer in above-threshold ionization," Phys. Rev. A 85, 041404 (2012).

W. ElMaraghy, H. ElMaraghy, T. Tomiyama, L. Monostori, Complexity in engineering design and
manufacturing, CIRP Annals - Manufacturing Technology, 61/2: 2012

A. Djuric, R. Al Saidi, W. ElMaraghy, Dynamics solution of n-DOF global machinery model, Robotics and
Computer-Integrated Manufacturing, 28: 621-630 pp, 2012

M. Brodeur, T. Brunner, C. Champagne, et al.,
"First Direct Mass Measurement of the Two-Neutron Halo Nucleus He-6 and Improved Mass for the Four-Neutron Halo He-8," Phys. Rev. Lett. 108, 052504 (2012).

D. Morton and G.W.F. Drake, 2011. "Spin-forbidden radiative decay rates from the
3 (3)P(1,2) and 3 (1)P(1) states of helium," Phys. Rev. A 83, 042503 (2011)

L. M. Wang, Z. -C. Yan, H.X. Qiao, et al.
"Variational upper bounds for low-lying states of lithium," Phys. Rev. A 83, 034503 (2011) (4 pages)

W. Noertershaeuser, W., R. Sanchez, R. G. Ewald, G., et al.
" Isotope-shift measurements of stable and short-lived lithium isotopes for nuclear-charge-radii determination," Phys. Rev. A, 83, 012516, (2011)

Donald C. Morton, Paul Moffatt, and G. W. F. Drake.
"Relativistic corrections to He I transition rates," Can. J. Phys. 89, 1, (2011) (5 pages)
Conference: International Conference on Precision Physics of Simple Atomic Systems Location: Ecole de Physique, Les Houches, FRANCE Date: MAY 30-JUN 04, 2010

M. Zakova, Z. Andjelkovic, M.L. Bissell, K. Blaum, G.W.F. Drake, C Geppert, M. Kowalska,
J. Kramer, A. Krieger, M. Lochmann, T. Neff, R. Neugart, W. N\"ortersh\"auser, R. Sanchez,
F. Schmidt-Kaler, D. Tiedemann, Z.-C. Yan, D.T. Yordanov, And C. Zimmermann,
"Isotope shift measurements in the 2s(1/2) --> 2p(3/2) transition of Be+ and extraction
of the nuclear charge radii for Be-7, Be-10, Be-11," J. Phys. G--Nucl. and Particle Phys.,
37, 055107 (2010) (14 pages).

R. El-Wazni and G.W.F. Drake, "Energies for the high-L Rydberg states of helium:
Asymptotic analysis," Phys. Rev. A 80, 064501 (2009) (4 pages).

R. Ringle, M. Brodeur, T. Brunner, S. Ettenauer, M. Smith, A. Lapierre, V.L. Ryjkov, P. Delheij, G.W.F. Drake, J. Lassen,
D. Lunney, and J. Dilling, "High-Precision Penning-Trap Mass Measurements of
9,10Be and the One-Neutron Halo Nuclide 11Be," Phys. Lett.\ B 695, 170-174 (2009).

W. Nortershauser, D. Tiedemann, M. Zakova, Z. Andjelkovic, K. Blaum, M.L. Bissell, R. Cazan, G.W.F. Drake,
C. Geppert, M. Kowalska, J. Kramer, A. Krieger, R. Neugart, R. Sanchez, F. Schmidt-Kaler, Z-C. Yan, D.T. Yordanov,
C. Zimmermann, "Nuclear Charge Radii of Be-7, Be-9, and One-Neutron Halo Nucleus Be-11", Physical
Review Letters, ,102: 2009

M. Smith, M. Brodeur. T. Brunner, S. Ettenauer, A. Lapierre, R. Ringle, V. L. Ryjkov, F. Ames, P. Bricault, G.
W. F. Drake, P. Delheij, D. Lunney, F. Sarazin, and J. Dilling, First Penning-Trap Mass Measurements of the
Exotic Halo Nucleus, Physical Review Letters, 101: 2008

I.A. Sulai, Wu, Qixue, Bishof, M., Drake, G. W. F., Lu, Z. -T., Mueller, P., Santra, R., Hyperfine Suppression
of 2(3)(S(1)-3 (3)P(J)Transitions in 3He, Physical Review Letters,<stong> 101</strong: 2008

Program Notes

2021-09-14T04:29:23Z

Drake:

==Energy & Wave Function Files==
There are 3 types of files in which the information is stored: 2 are Wave function files (.POW for infinite mass, .POL for finite mass) while .MAT files store energy information and the expectation values for a large variety of operators.

In order to decipher the information stored in these files refer to Busuttil & Drake 2008 ([[File:Busuttil_&_Drake_2008_-_PSAS.pdf]]) as presented at the International Conference on Precision Physics of Simple Atomic Systems (PSAS) in 2008.

An example of how these wave functions can be used is in calculating expectation values. Included below is a program written in FORTRAN 77 which calculates expectation values for various operators called MATL. The quantities calculated are <1/r12>, <1/r1>, <r1>, <r12>, <1/r122>, <1/r12>, <r12>, <r122>, <r1•r2>, <1/r1r12>, and <1/r1r2>.

Download MATL package: [[File:Matl.zip]]

Downloadable Resources

2021-09-14T04:26:07Z

Drake: /* Energy & Wave functions for Heliumlike Atoms */

==Energy & Wave functions for Heliumlike Atoms==

Please refer to [[Program Notes]] for an explanation of how the information is stored in these files and a sample program to compute expectation values. For expectation values of various operators,open and scroll through the .MAT file for the particular state selected.

<html><iframe src="http://drake.sharcnet.ca/download/download-raw.php" width=780 height=460 frameborder=0 scrolling=no> </iframe></html>

Theory Notes

2021-03-12T01:48:19Z

Drake: /* General Hermitian Property */

<center>
<H1>Notes on solving the Schrödinger equation in Hylleraas coordinates for heliumlike atoms
Gordon W.F. Drake
Department of Physics, University of Windsor
Windsor, Ontario, Canada N9B 3P4
(Transcribed from hand-written notes and edited by Lauren Moffatt. Last revised November 22, 2017 to correct a missing subscript in the expression for $B_1$ in the Hamiltonian in Sect. 11 General Hermitian Property.)
</H1>
</center>

==Introduction==
These notes describe some of the technical details involved in solving the Schrödinger equation for a heliumlike atom of nuclear charge Z in correlated Hylleraas coordinates. The standard for computational purposes is to express all quantities in a form that is valid for electronic states of arbitrary angular momentum, and to express matrix elements of the Hamiltonian in a manifestly Hermitian symmetrized form. Further information and derivations of the mathematical techniques for angular integrals and radial recursion relations can be found in the paper
G.W.F. Drake, Phys. Rev. A <math>{\bf 18}</math>, 820 (1978).

The starting point is the two-electron Schrödinger equation for infinite nuclear mass
\begin{equation} \left[-\frac{\hbar^2}{2m}(\nabla^2_1 +\nabla^2_2) - \frac{Ze^2}{r_1} -
\frac{Ze^2}{r_2}+\frac{e^2}{r_{12}} \right]\psi = E\psi\nonumber
\end{equation}
where <math>m</math> is the electron mass, and <math>r_{12} = |{\bf r}_1 - {\bf r}_2|</math> (see diagram below).

[[File:Pic1.jpg|250px]]

Begin by rescaling distances and energies so that the Schrödinger equation can be expressed in a dimensionless form. The dimensionless Z-scaled distance is defined by <math>\begin{eqnarray} \rho = \frac{Zr}{a_0} \end{eqnarray} </math> where <math> \begin{eqnarray} a_0 = \frac{\hbar^2}{me^2} \end{eqnarray} </math> is the atomic unit (a.u.) of distance equal to the Bohr radius <math>0.529\, 177\, 210\, 92(17) \times 10^{-10}</math> m. Then

\begin{equation}[-\frac{\hbar^2}{2m}Z^2(\frac{me^2}{\hbar^2})^2(\nabla^2_{\rho_1}+\nabla^2_{\rho_2})
- Z^2\frac{e^2}{a_0}\rho^{-1}_1 - Z^2\frac{e^2}{a_0}\rho^{-1}_2 +
\frac{e^2}{a_0}Z\rho^{-1}_{12}]\psi= E\psi\nonumber\end{equation}

But <math>\begin{eqnarray} \frac{\hbar^2}{m}(\frac{me^2}{\hbar^2})^2 = \frac{e^2}{a_0} \end{eqnarray} </math> is the hartree atomic unit of of energy <math>(E_h = 27.211\,385\,05(60)</math> eV, or equivalently <math>E_h/(hc) = 2194.746\,313\,708(11)</math> m<math>^{-1})</math>. Therefore, after multiplying through by <math>a_0/(Ze)^2</math>, the problem to be solved in Z-scaled dimensionless units becomes

\begin{equation} \left[-\frac{1}{2}(\nabla^2_{\rho_1}+\nabla^2_{\rho_2}) -
\frac{1}{\rho_1} - \frac{1}{\rho_2} + \frac{Z^{-1}}{\rho_{12}}\right]\psi =
\varepsilon\psi\nonumber \end{equation}

where <math>\begin{eqnarray}\varepsilon = \frac{Ea_0}{(Ze)^2} \end{eqnarray}</math> is the energy in Z-scaled atomic units. For convenience, rewrite this in the conventional form
\begin{equation}
H\psi = \varepsilon\psi\nonumber \end{equation}
where (using <math>r</math> in place of <math>\rho</math> for the Z-scaled distance)
\begin{equation}
H = -\frac{1}{2}(\nabla^2_1+\nabla^2_2) - \frac{1}{r_1}-\frac{1}{r_2} + \frac{Z^{-1}}{r_{12}}
\nonumber\end{equation}
is the atomic Hamiltonian for infinite nuclear mass.

==The Hartree Fock Method==

For purposes of comparison, and to define the correlation energy, assume that <math>\psi({\bf r}_1,{\bf r}_2)</math> can be written in the separable product form
\begin{equation}
\psi({\bf r}_1,{\bf r}_2) = \frac{1}{\sqrt{2}} \left[ u_1(r_1) u_2(r_2) \pm u_2(r_1) u_1(r_2) \right] \nonumber
\end{equation}
for the <math>1s^2\ ^1\!S</math> ground state. Because of the <math>\frac{1}{r_{12}}</math> term in the Schrödinger equation, the exact solution connot be expressed in this form as a separable product. However, the Hartree-Fock approximation corresponds to finding the best possible solution to the Schrödinger equation
\begin{equation}
H\psi({\bf r}_1,{\bf r}_2) = E\psi({\bf r}_1,{\bf r}_2)\nonumber
\end{equation}
that can nevertheless be expressed in this separable product form, where as before
\begin{equation}
H = -\frac{1}{2}(\nabla^2_1+\nabla^2_2) - \frac{1}{r_1}- \frac{1}{r_2} + \frac{Z^{-1}}{r_{12}}\nonumber
\end{equation}
is the full two-electron Hamiltonian. To find the best solution, substitute into <math>\left\langle\psi \lvert H-E \rvert \psi\right\rangle</math> and require this expression to be stationary
with respect to arbitrary infinitesimal variations <math>\delta u_1</math> and <math>\delta
u_2</math> in <math>u_1</math> and <math>u_2</math> respectively; i.e.

\begin{equation}
\frac{1}{2} \left\langle \delta u_1(r_1)u_2(r_2) \pm u_2(r_1)\delta
u_1(r_2) \lvert H-E \rvert u_1(r_1) u_2(r_2)\pm u_2(r_1) u_1(r_2)\right\rangle\nonumber
\end{equation}

\begin{equation}
=\int\delta u_1(r_1) d{\bf r}_1 \left\{ \int d{\bf r}_2 u_2(r_2) \left( H - E \right) \left[ u_1(r_1)u_2(r_2) \pm u_2(r_1) u_1(r_2) \right] \right\}\nonumber
\end{equation}

\begin{equation}
= 0 \nonumber
\end{equation}

for arbitrary <math> \delta u_1(r_1).\nonumber </math> Therefore <math>\{\int d{\bf r}_2 \ldots \} = 0</math>.

Similarly, the coefficient of <math>\delta u_2</math> would give

\begin{equation}
\int d{\bf r}_1 u_1(r_1) \left( H-E \right) \left[ u_1(r_1) u_2(r_2) \pm
u_2(r_1) u_1(r_2) \right] = 0\nonumber
\end{equation}

Define

\begin{equation}
I_{12} = I_{21} = \int d{\bf r}\, u_1(r)u_2(r), \nonumber
\end{equation}

\begin{equation}
H_{ij} = \int d{\bf r}\, u_i(r)(-\frac{1}{2}\nabla - \frac{1}{r})u_j(r), \nonumber
\end{equation}

\begin{equation}
G_{ij}(r) = \int d{\bf r}^\prime u_i(r^\prime)\frac{1}{|{\bf r} - {\bf r}\prime|}u_j(r^\prime)\nonumber
\end{equation}

Then the above equations become the pair of integro-differential equations

\begin{equation}
[ H_0 - E + H_{22}+G_{22}(r)]u_1(r) = \mp [ I_{12}(H_0-E) + H_{12}+G_{12}(r)]u_2(r)\nonumber
\end{equation}

\begin{equation}
[H_0-E+H_{11}+G_{11}(r)]u_2(r) = \mp [I_{12}(H_0-E) + H_{12}+G_{12}(r)]u_1(r)\nonumber
\end{equation}

These must be solved self-consistently for the "constants" <math>I_{12}</math> and
<math>H_{ij}</math> and the function <math>G_{ij}(r)</math>.

The Hartree Fock energy is <math>E \simeq -2.87\ldots </math>a.u. while the exact energy is <math>E = -2.903724\ldots </math>a.u.

The difference is called the "correlation energy" because it arises from the
way in which the motion of one electron is correlated to the other. The
Hartree Fock equations only describe how one electron moves in the average field
provided by the other (mean-field theory).

==Configuration Interaction==
Expand
\begin{equation}
\psi({\bf r}_1,{\bf r}_2)= C_0u^{(s)}_1(r_1)u^{(s)}_1(r_2) + C_1u^{(P)}_1({\bf r}_1)u^{(P)}_1({\bf r}_2){\cal Y}^0_{1,1,0}(\hat{\bf r}_1, \hat{\bf r}_2)+C_2u^{(d)}_1({\bf r}_1)u^{(d)}_2({\bf r}_2){\cal Y}^0_{2,2,0}(\hat{\bf r}_1, \hat{\bf r}_2)+... \pm {\rm\ exchange}
\end{equation}
where
\begin{equation}
{\cal Y}^M_{l_1,l_2,L}(\hat{\bf r}_1, \hat{\bf r}_2)=\sum_{m_1,m_2}Y^{m_1}_{l_1}({\bf r}_1)Y^{m_2}_{l_2}({\bf r}_2)\times <l_1l_2m_1m_2\mid LM>
\end{equation}

This works, but is slowly convergent, and very laborious. The best CI calculations are accurate to <math> ~10^{-7}</math> a.u.

==Hylleraas Coordinates==
[E.A. Hylleraas, Z. Phys. <math>{\bf 48}, 469(1928)</math> and <math>{\bf 54}, 347(1929)</math>]
suggested using the co-ordinates <math>r_1</math>, <math>r_2</math> and <math>r_{12}</math> or equivalently

\begin{eqnarray}
s &=& r_1 + r_2, \nonumber\\
t &=& r_1-r_2, \nonumber\\
u &=& r_{12}\nonumber
\end{eqnarray}

and writing the trial functions in the form

\begin{equation}
\Psi({\bf r}_1,{\bf r}_2) = \sum^{i+j+k\leq
N}_{i,j,k}c_{i,j,k}r_1^{i+l_1}r_2^{j+l_2}r_{12}^ke^{-\alpha r_1 - \beta r_2}
\mathcal{Y}^M_{l_1,l_2,L}(\hat{r}_1,\hat{r}_2)\pm \text{exchange}\nonumber
\end{equation}

Diagonalizing H in this non-orthogonal basis set is equivalent to solving
<math> \begin{eqnarray}
\frac{\partial E}{\partial c_{i,j,k}} = 0\nonumber
\end{eqnarray} </math>
for fixed <math>\alpha</math> and <math>\beta</math>.

The diagonalization must be repeated for different values of <math>\alpha</math> and
<math>\beta</math> in order to optimize the non-linear parameters.

==Completeness==
The completeness of the above basis set can be shown by first writing
\begin{equation}
r_{12}^2 = r_1^2 + r_2^2 - 2r_1r_2\cos\theta_{12}
\end{equation}
and, from the spherical harmonic addition theorem,
\begin{equation}
\cos(\theta_{12})=\frac{4\pi}{3}\sum^1_{m=-1}Y^{m*}_1(\theta_1,\varphi_1)Y^m_1(\theta_2,\varphi_2)
\end{equation}
Consider first the S-states. The <math>r_{12}^0</math> terms are like the ss terms in a CI calculation. The
<math>r_{12}^2</math> terms bring in p-p type contributions, and the higher powers bring
in d-d, f-f etc. type terms. In general

\begin{equation}
P_l(\cos\theta_{12}) = \frac{4\pi}{2l+1}\sum^l_{m=-l}{Y^{m}_l}^*(\theta_1,\varphi_1)Y^m_l(\theta_2,\varphi_2)\nonumber
\end{equation}

For P-states, one would have similarly

\begin{array}{lr}
r_{12}^0 & (sp)P\nonumber\\
r_{12}^2 & (pd)P\nonumber\\
r_{12}^4 & (df)P\nonumber\\
\vdots & \vdots\nonumber
\end{array}

For D-states

\begin{array}{lr}
r_{12}^0 & (sd)D & (pp^\prime)D\nonumber\\
r_{12}^2 & (pf)D & (dd^\prime)D\nonumber\\
r_{12}^4 & (dg)D & (ff^\prime)D\nonumber\\
\vdots & \vdots & \vdots\nonumber
\end{array}

In this case, since there are two "lowest-order" couplings to form a
D-state, both must be present in the basis set, i.e.

\begin{equation}
\Psi(r_2,r_2) = \sum c_{ijk}r_1^ir_2^{j+2}r_{12}^ke^{-\alpha r_1-\beta
r_2}\mathcal{Y}^M_{022}(\hat{r}_1,\hat{r}_2)
+\sum d_{ijk}r_1^{i+1}r_2^{j+1}r_{12^k}e^{-\alpha^\prime r_1 - \beta^\prime
r_2}\mathcal{Y}^M_{112}(\hat{r}_1,\hat{r}_2)\nonumber
\end{equation}

For F-states, one would need <math>(sf)F</math> and <math>(pd)F</math> terms.

For G-states, one would need <math>(sg)G</math>, <math>(pf)G</math> and <math>(dd^\prime)G</math> terms.

Completeness of the radial functions can be proven by considering the
Sturm-Liouville problem

\begin{equation}
\left(-\frac{1}{2}\nabla^2-\frac{\lambda}{r_s}-E\right)\psi({\bf r}) = 0\nonumber
\end{equation}

or

\begin{equation}
\left(-\frac{1}{2}\frac{1}{r^2}\frac{\partial}{\partial r}\left({r^2}\frac{\partial}{\partial r}\right) -
\frac{l\left(l+1\right)}{2r^2} - \frac{\lambda}{r} - E\right)u(r) = 0.\nonumber
\end{equation}

For fixed E and variable <math>\lambda</math> (nuclear charge).

The eigenvalues are <math>\lambda_n = (E/E_n)^{1/2}</math>, where <math>E_n =- \frac{1}{2n^2}</math>

[[File:Pic2.jpg|500px]]

and the eigenfunctions are

\begin{equation}
u_{nl}(r) =
\frac{1}{(2l+1)!}\left(\frac{(n+l)!}{(n-l-1)2!}\right)^{1/2}(2\alpha)^{3/2}e^{-\alpha
r}\nonumber
\times \ \left( 2 \alpha r \right)^l\,_{1}F_{1}\left(-n+l+1,2l+2; 2 \alpha r \right)
\end{equation}

with <math>\alpha = (-2E)^{1/2}</math> and <math>n\geq l+1</math>. The confluent hypergeometric function <math> _1F_1(a,b,;z)</math> then denotes a finite polynomial since <math>a=-n+l+1</math> is a negative integer or zero.

Unlike the hydrogen spectrum, which has both a discrete part for <math>E<0</math> and a
continuous part for <math>E>0</math>, this forms an entirely discrete set of finite
polynomials, called Sturmian functions. They are orthogonal with respect to
the potential, i.e.

\begin{equation}
\int^\infty_0 r^2dr \left( u_{n^\prime l}(r)\frac{1}{r}u_{nl}(r) \right) = \delta_{n,n^\prime}\nonumber
\end{equation}

Since they become complete in the limit <math>n\rightarrow\infty</math>, this assures the
completeness of the variational basis set.

[See also B Klahn and W.A. Bingel Theo. Chim. Acta (Berlin) 44, 9 and 27
(1977)].

==Solutions of the Eigenvalue Problem==

For convenience, write

\[
\Psi({\bf r}_1,{\bf r}_2) = \sum^N_{m=1}c_m\varphi_m\nonumber
\]

where <math>m</math> represents the <math>m</math>'th combination of <math>i,j,k</math> and

\begin{equation}
\varphi_{ijk}=r_1^ir_2^jr_{12}^ke^{-\alpha r_1 - \beta
r_2}\mathcal{Y}^M_{l_1,l_2,L}(\hat{r}_1,\hat{r}_2) \pm \mbox{exchange}\nonumber
\end{equation}

\begin{align}

&
\left( \begin{array}{lr}
\cos(\theta) & \sin(\theta)\\
-\sin(\theta) & \cos(\theta)
\end{array} \right)
\left( \begin{array}{lr}
H_{11} & H_{12} \\
H_{12} & H_{22}
\end{array}\right)
\left( \begin{array}{lr}
\cos(\theta) & -\sin(\theta)\\
\sin(\theta) & \cos(\theta)
\end{array} \right)
\\[5pt]
= & \left(\begin{array}{lr}
cH_{11}+sH_{12} & cH_{12} + sH_{22}\\
-sH_{11} + cH_{12} & -sH_{12} + cH_{22}
\end{array}\right)
\left( \begin{array}{cc}
c & -s \\
s & c
\end{array}\right)
\\[5pt]
= & \left(\begin{array}{lr}
c^2H_{11}+s^2H_{22} + 2csH_{12} & (c^2-s^2)H_{12}+cs(H_{22}-H_{11})\\
(c^2-s^2)H_{12}+cs(H_{22}-H_{11}) & s^2H_{11}+c^2H_{22}-2csH_{12}
\end{array}\right)
\end{align}

Therefore

\begin{equation}
\left[\cos^2(\theta)-\sin^2(\theta)\right]H_{12} =
\cos(\theta)\sin(\theta)(H_{11}-H_{22})\nonumber
\end{equation}

and

\begin{equation}
\tan(2\theta) = \frac{2H_{12}}{H_{11}-H_{22}}\nonumber
\end{equation}

i.e.

\begin{equation}
\cos(\theta)=\left(\frac{r+\omega}{2r}\right)^{1/2}\nonumber
\end{equation}

\begin{equation}
\sin(\theta)=-\mbox{sgn}(H_{12})\left(\frac{r-\omega}{2r}\right)^{1/2}\nonumber
\end{equation}

where

\begin{equation}
\omega = H_{22}-H_{11}\nonumber
\end{equation}

\begin{equation}
r=\left(\omega^2+4H_{12}^2\right)^{1/2}\nonumber
\end{equation}

\begin{equation}
E_1=\frac{1}{2}\left(H_{11}+H_{22}-r\right)\nonumber
\end{equation}

\begin{equation}
E_2=\frac{1}{2}\left(H_{11}+H_{22}+r\right)\nonumber
\end{equation}

===Brute Force Method===

- Gives all the eigenvalues and eigenvectors, but it is slow

- First orthonormalize the <math>N</math>-dimensional basis set; i.e. form linear combinations

\begin{equation}
\Phi_m = \sum_{n=1}^N\varphi_nR_{nm}\nonumber
\end{equation}

such that
\begin{equation}
\langle \Phi_m | \Phi_n \rangle = \delta_{m,n}
\end{equation}

This can be done by finding an orthogonal tranformation T such that

\begin{equation}
T^TOT = I = \left(
\begin{array}{ccccc}
I_1 & 0 & 0 & \ldots & 0 \\
0 & I_2 & 0 & \ldots & 0 \\
0 & 0 & I_3 & \ldots & 0 \\
\vdots & \vdots & \vdots & \ddots & \vdots \\
0 & 0 & 0 & \ldots & I_N
\end{array}\right);
\end{equation}

\begin{equation}
O_{mn} = \langle \varphi_m | \varphi_n \rangle \nonumber
\end{equation}

and then applying a scale change matrix

\begin{equation}
S = \left(\begin{array}{ccccc}
\frac{1}{I_1^{1/2}} & 0 & 0 & \ldots & 0\\
0 & \frac{1}{I_2^{1/2}} & 0 & \ldots & 0 \\
0 & 0 & \frac{1}{I_3^{1/2}} & \ldots & 0 \\
\vdots & \vdots & \vdots & \ddots & \vdots \\
0 & 0 & 0 & \ldots & I_N^{1/2}
\end{array}\right)= S^T\nonumber
\end{equation}

Then

\begin{equation}
S^TT^TOTS = 1
\end{equation}

i.e.

\begin{equation}
R^TOR = 1
\end{equation}

with <math>R=TS</math>.

If H is the matrix with elements <math>H_{mn} = \langle \varphi_m | \varphi_n \rangle </math>, then H
expressed in the <math>\Phi_m</math> basis set is

\begin{equation}
H^\prime = R^THR.\nonumber
\end{equation}

We next diagonalize <math>H^\prime</math> by finding an orthogonal transformation W such
that

\begin{equation}
W^T H^\prime W = \lambda = \left(
\begin{array}{cccc}
\lambda_1 & 0 & \ldots & 0 \\
0 & \lambda_2 & \ldots & 0 \\
\vdots & \vdots & \ddots & \vdots \\
0 & 0 & \ldots & \lambda_N
\end{array}\right)\nonumber
\end{equation}

The q'th eigenvector is

\begin{equation}
\Psi^{(q)} = \sum_{n=1}^N\Phi_n W_{n,q} = \sum_{n,n^\prime} \varphi_{n^\prime} R_{n^\prime ,n} W_{n,q} \nonumber
\end{equation}

i.e.

\begin{equation}
c_{n^\prime}^{(q)} = \sum_{n=1}^N R_{n^\prime n} W_{n,q}
\end{equation}

===The Power Method===

- Based on the observation that if H has one eigenvalue, <math>\lambda_M</math>, much bigger than all the rest, and <math>\chi = \left(
\begin{array}{c}a_1\\a_2\\\vdots\end{array}\right)</math>
is an arbitrary starting vector, then <math>\chi = \sum_{q=1}^N x_q\Psi^{(q)}</math>.

\begin{equation}
(H)^n\chi = \sum_{q=1}^N x_q \lambda^n_q\Psi^{(q)}\nonumber\rightarrow x_M\lambda_M^n\Psi^{(M)}\nonumber
\end{equation}

provided <math>x_M\neq 0</math>.

To pick out the eigenvector corresponding to any eigenvalue, write the original problem in the form

\begin{align}
H\Psi & = \lambda O \Psi \nonumber \\
\left( H - \lambda_q O \right) \Psi & = \left( \lambda - \lambda_q \right) O \Psi \nonumber
\end{align}

Therefore,

\begin{equation}
G \Psi = \frac{1}{\lambda - \lambda_q } \Psi \nonumber \\
\end{equation}

where <math>G=(H-\lambda_qO)^{-1}O</math> with eigenvalues <math>\frac{1}{\lambda_n - \lambda_q}</math>.

By picking <math>\lambda_q</math> close to any one of the <math>\lambda_n</math>, say
<math>\lambda_{n^\prime}</math>, then <math>\frac{1}{\lambda_n-\lambda_q}</math> is much larger for
<math>n=n^\prime</math> than for any other value. The sequence is then

\begin{array}{c}
\chi_1=G\chi\nonumber\\
\chi_2=G\chi_1\nonumber\\
\chi_3=G\chi_2\nonumber\\
\vdots\nonumber
\end{array}

until the ratios of components in <math>\chi_n</math> stop changing.

- To avoid matrix inversion and multiplication, note that the sequence is equivalent to
\begin{equation}
F \chi_n = \left( \lambda - \lambda_q \right) O \chi_{n-1} \nonumber \\
\end{equation}
where <math>F = H-\lambda_qO</math>. The factor of <math>\left( \lambda - \lambda_q \right)</math> can be
dropped because this only affects the normalization of <math>\chi_n</math>. To find <math>\chi_n</math>, solve
\begin{equation}
F \chi_n = O \chi_{n-1} \nonumber
\end{equation}
(N equations in N unknowns). Then
\begin{equation}
\lambda = \frac{ \langle \chi_n | H | \chi_n \rangle}{\langle \chi_n | \chi_n \rangle} \nonumber
\end{equation}

==Matrix Elements of H==

\begin{equation}
H=-\frac{1}{2}\nabla^2_1 -\frac{1}{2}\nabla^2_2 - \frac{1}{r_1} - \frac{1}{r_2} +\frac{Z^{-1}}{r_{12}}\nonumber
\end{equation}

Taking <math>r_1, r_2</math> and <math>r_{12}</math> as independent variables,

\begin{equation}
\nabla_1^2 = \frac{1}{r_1^2}\frac{\partial}{\partial r_1}\left(r_1^2\frac{\partial}{\partial r_1}\right)+ \frac{1}{r_{12}^2}\frac{\partial}{\partial r_{12}}
\left(r_{12}^2\frac{\partial}{\partial r_{12}}\right)\nonumber -\frac{l_1(l_1+1)}{r_1^2}+2(r_1 - r_2\cos(\theta_{12}))\frac{1}{r_{12}}\frac{\partial^2}{\partial r_1 \partial r_{12}} \nonumber - 2(\nabla_1^Y \cdot {\bf r}_2)\frac{1}{r_{12}}\frac{\partial}{\partial r_{12}}\nonumber
\end{equation}

where <math>\nabla_1^Y</math> acts only on the spherical harmonic part of the wave function and the diagram

[[File:Pic3.jpg|400px]]

defines the complete set of 6 independent variables is <math>r_1</math>, <math>r_2</math>, <math>r_{12}</math>, <math>\theta_1</math>, <math>\varphi_1</math>, <math>\chi</math>.

If <math>r_{12}</math> were not an independent variable, then one could take the volume element to be

\begin{equation}
d\tau = r_1^2dr_1\sin(\theta_1)d\theta_1d\varphi_1r_2^2dr_2\sin(\theta_2)d\theta_2d\varphi_2.\nonumber
\end{equation}

However, <math>\theta_2</math> and <math>\varphi_2</math> are no longer independent variables. To eliminate them, take the point <math>{\bf r}_1</math> as the
origin of a new polar co-ordinate system, and write

\begin{equation}
d\tau=-r_1^2dr_1\sin(\theta_1)d\theta_1d\varphi_1r_{12}^2dr_{12}\sin(\psi)d\psi d\chi\nonumber
\end{equation}

and use

\begin{equation}
r_2^2=r_1^2+r_{12}^2 +2r_1r_{12}\cos(\psi)
\end{equation}

Then for fixed <math>r_1</math> and <math>r_{12}</math>,

\begin{equation}
2r_2dr_2 = -2r_1r_{12}\sin(\psi)d\psi
\end{equation}

Thus

\begin{equation}
d\tau= r_1dr_1r_2dr_2r_{12}dr_{12}\sin(\theta_1)d\theta_1d\varphi_1d\chi
\end{equation}

The basic type of integral to be calculated is

\begin{align}
I \left( l_1, m_1, l_2, m_2; R \right) & = \int \sin \left( \theta_1 \right) d\theta_1 d\varphi_1 d\chi Y^{m_1}_{l_1} \left( \theta_1, \varphi_1 \right)^{*} Y^{m_2}_{l_2} \left( \theta_2, \varphi_2 \right) \\
& \times \int r_1dr_1r_2dr_2r_{12}dr_{12}R \left( r_1, r_2, r_{12} \right)
\end{align}

Consider first the angular integral. <math>Y^{m_2}_{l_2} \left( \theta_2, \varphi_2 \right)</math> can be expressed in terms of the independent variables <math>\theta_1, \varphi_1, \chi</math> by use of the rotation matrix relation
\begin{equation}
Y^{m_2}_{l_2}( \theta_2, \varphi_2) = \sum_m \mathcal{D}^{(l_2)}_{m_2,m}(\varphi_1,\theta_1,\chi)^*Y^m_{l_2}( \theta_{12}, \varphi)
\end{equation}
where <math>\theta, \varphi</math> are the polar angles of <math>{\bf r}_2</math> relative to <math>{\bf r}_1</math>. The angular integral is then

\begin{align}
I_{ang} & = \int^{2 \pi}_{0} d\chi \int^{2\pi}_{0} d\varphi_1 \int^{\pi}_{0} \sin \left( \theta_1 \right) d\theta_{1} Y^{m_1}_{l_1} \left( \theta_1, \varphi_1 \right)^{*} \\
& \times \sum_m \mathcal{D}^{(l_2)}_{m_2,m} \left( \varphi_1,\theta_1,\chi \right)^{*} Y^{m}_{l_2} \left( \theta_{12},\varphi \right) \\
\end{align}

Use

\begin{equation}
Y^{m_1}_{l_1} \left( \theta_1, \varphi_1 \right)^{*} = \sqrt{\frac{2l_1+1}{4\pi}}\mathcal{D}^{(l_1)}_{m_1,0} \left( \varphi_1, \theta_1, \chi \right)
\end{equation}

together with the orthogonality property of the rotation matrices (Brink and Satchler, p 147)

\begin{align}
& \int \mathcal{D}^{(j)*}_{m,m^\prime}\mathcal{D}^{(J)}_{M,M^\prime}\sin(\theta_1)d\theta_1d\varphi_1d\chi \\
& = \frac{8\pi^2}{2j+1}\delta_{jJ}\delta_{mM}\delta_{m^\prime M^\prime}\nonumber
\end{align}

to obtain

\begin{align}
I_{ang} & = \sqrt{ \frac{2l_1+1}{4\pi}} \frac{8\pi^2}{2l_1+1} \delta_{l_1,l_2} \delta_{m_1,m_2} Y^{0}_{l_2} \left( \theta_{12}, \varphi \right) \\
& = 2 \pi \delta_{l_1,l_2} \delta_{m_1,m_2} P_{l_2} \left( \cos \theta_{12} \right)
\end{align}

since

\begin{equation}
Y^0_{l_2} \left( \theta_{12}, \varphi \right) = \sqrt{\frac{2l_1+1}{4\pi}} P_{l_2} \left( \cos \left( \theta_{12} \right) \right)
\end{equation}

Note that <math>P_{l_2} \left( \cos \theta \right)</math> is just a short hand expression for a radial function because

\begin{equation}
\cos \theta_{12} = \frac{r_1^2 + r_2^2 - r_{12}^2}{2r_1r_2}
\end{equation}

The original integral is thus

\begin{equation}
I \left( l_1, m_1, l_2, m_2; R \right) = 2 \pi \delta_{l_1,l_2} \delta_{m_1,m_2} \int^{\infty}_{0} r_{1} dr_{1} \int^{\infty}_{0} r_{2} dr_{2} \int^{r_{1} + r_{2}}_{|r_{1} - r_{2}|} r_{12} dr_{12} R \left( r_{1}, r_{2}, r_{12} \right) P_{l_2} \left( \cos \theta_{12} \right)
\end{equation}

where again

\begin{equation}
\cos \theta_{12} = \frac{r_{1}^{2} +r^{2}_{2} - r_{12}^{2}}{2 r_{1} r_{2}}
\end{equation}

is a purely radial function.

The above would become quite complicated for large <math>l_2</math> because <math>P_{l_2}(\cos \theta_{12})</math> contains terms up to <math>\left( \cos \theta_{12} \right)^{l_2}</math>. However, recursion relations exist which allow any integral containing <math>P_l \left( \cos \theta_{12} \right)</math> in terms of those containing just <math>P_0 \left( \cos \theta_{12} \right) = 1</math> and <math>P_1 \left( \cos \theta_{12} \right) = \cos \theta_{12}</math>.

==Radial Integrals and Recursion Relations==

The basic radial integral is [see G.W.F. Drake, Phys. Rev. A 18, 820 (1978)]

\begin{align}
I_0 \left( a, b, c \right) & =
\int^{\infty}_{0} r_{1} dr_{1} \int^{\infty}_{r_{1}} r_{2} dr_{2} \int^{r_{1}+r_{2}}_{r_{2} - r_{1}} r_{12} dr_{12} r_{1}^{a} r_{2}^{b} r_{12}^{c} e^{-\alpha r_{1} - \beta r_{2}} + \int^{\infty}_{0} r_{2} dr_{2} \int^{\infty}_{r_{2}} r_{1} dr_{1} \int^{r_{1} + r_{2}}_{r_{1} - r_{2}} r_{12} dr_{12} r_{1}^{a} r_{2}^{b} r_{12}^{c} e^{- \alpha r_{1} - \beta r_{2}} \\
& = \frac{2}{c+2} \sum^{[(c+1)/2]}_{i=0}
\left(\begin{array}{c}
c+2 \\
2i+1
\end{array} \right)
\left\{ \frac{q!}{\beta^{q+1} \left( \alpha + \beta \right)^{p+1}} \sum_{j=0}^{q} \frac{\left( p + j \right)!}{j!} \left( \frac{\beta}{\alpha + \beta} \right)^j + \frac{q^{\prime}!}{\alpha^{q^{\prime}+1} \left( \alpha + \beta \right)^{p^{\prime}+1}} \sum^{q^{\prime}}_{j=0} \frac{\left( p^{\prime} + j \right)!}{j!} \left( \frac{\alpha}{\alpha + \beta} \right)^j \right\}
\end{align}

where

\begin{array}{ll}
p = a + 2i + 2 & p^{\prime} = b + 2i + 2 \\
q = b + c - 2i + 2 & q^{\prime} = a + c - 2i + 2
\end{array}

The above applies for <math> a, b \geq -2, c\geq -1</math>.
<math>[x]</math> means "greatest integer in" <math>x</math>.

Then

\begin{align}
I_1 \left( a, b, c \right) & = \int d\tau_{r} r_{1}^{a} r_{2}^{b} r_{12}^{c} e^{ -\alpha r_{1} - \beta r_{2}} P_{1}\left( \cos \theta \right) \\
& = \frac{1}{2} \left( I_{0} \left( a + 1, b - 1, c \right) + I_{0} \left( a - 1, b + 1, c \right) - I_{0} \left( a - 1, b - 1, c + 2 \right) \right)
\end{align}

===The Radial Recursion Relation===

Recall that the full integral is

\begin{equation}
I(l_1m_1,l_2m_2;R)=2\pi\delta_{l_1,l_2}\delta_{m_1,m_2}I_{l_2}(R)
\end{equation}

where

\begin{equation}
I_{l_2}(R) = \int d\tau_r R(r_1,r_2,r_{12})P_{l_2}(\cos \theta_{12})
\end{equation}

To obtain the recursion relation, use

\begin{equation}
P_l(x) = \frac{[P^{\prime}_{l+1}(x)-P^{\prime}_{l-1}(x)]}{2l+1}
\end{equation}

with

\begin{equation}
P_{l+1}^\prime (x) = \frac{d}{dx}P_{l+1}(x)
\end{equation}

Here <math>x=\cos\theta_{12}</math> and

\begin{align}
\frac{d}{d \cos \theta_{12}} & = \frac{dr_{12}}{d \cos \theta_{12}} \frac{d}{dr_{12}} \\
& = -\frac{r_{1} r_{2}}{r_{12}} \frac{d}{dr_{12}}
\end{align}

Then,

\begin{equation}
I_{l}\left( R \right) = - \int d\tau_r R \frac{r_{1} r_{2}}{r_{12}} \frac{d}{dr_{12}} \frac{ \left[ P_{l+1} \left( \cos \theta_{12} \right) - P_{l-1} \left( \cos \theta_{12} \right) \right]}{2l+1}
\end{equation}

The <math>r_{12}</math> part of the integral is

\begin{align}
& \int^{r_{1} + r_{2}}_{|r_{1} - r_{2}|} r_{12} dr_{12} R \frac{r_{1} r_{2}}{r_{12}} \frac{d}{dr_{12}} \left[P_{l+1}-P_{l-1} \right] \\
& = R r_{1} r_{2} \left[ P_{l+1} - P_{l-1} \right]^{r_{1} + r_{2}}_{|r_{1} - r_{2}|} - \int^{r_{1} + r_{2}}_{|r_{1} - r_{2}|} r_{12} dr_{12} \left( \frac{d}{dr_{12}} R \right) \frac{r_{1} r_{2}}{r_{12}} \frac{ \left[ P_{l+1} - P_{l-1} \right]}{2l+1}
\end{align}

The integrated term vanishes because

\begin{equation}
\cos \theta_{12} = \frac{r_{1}^{2} + r_{2}^{2} - r_{12}^{2}}{2 r_{1} r_{2}}
\begin{array}{lll}
& = -1 & {\rm when} & r_{12}^{2} = \left( r_{1} + r_{2} \right)^{2} \\
& = 1 & {\rm when} & r_{12}^{2} = \left( r_{1} - r_{2} \right)^{2}
\end{array}
\end{equation}

and <math>P_l(1)=1</math>, <math>P_{l}(-1)=(-1)^{l}</math>.

Thus,

\begin{equation}
I_{l+1}\left(\frac{r_1r_2}{r_{12}}R^\prime\right) = \left( 2l+1 \right) I_{l}(R) - I_{l-1} \left( \frac{r_{1} r_{2}}{r_{12}} R^{\prime} \right)
\end{equation}

If

\begin{equation}
R = r_{1}^{a-1} r_{2}^{b-1} r_{12}^{c+2} e^{-\alpha r_{1} - \beta r_{2}}
\end{equation}

then [G.W.F. Drake, Phys. Rev. A 18, 820 (1978)]

\begin{equation}
I_{l+1}(r_1^ar_2^br_{12}^c) =
\frac{2l+1}{c+2} I_{l} \left( r_{1}^{a-1} r_{2}^{b-2} r_{12}^{c+2} \right)+I_{l-1} \left( r_{1}^{a} r_{2}^{b} r_{12}^{c} \right)
\end{equation}

For the case <math>c=-2</math>, take

\begin{equation}
R = r_{1}^{a-1} r_{2}^{b-1} \ln r_{12} e^{ -\alpha r_{1} -\beta r_{2}}
\end{equation}

Then

\begin{equation}
I_{l+1} \left( r_{1}^{a} r_{2}^{b} r_{12}^{c} \right) = I_{l} \left( r_{1}^{a-1} r_{2}^{b-1} \ln r_{12} \right) \left( 2l + 1 \right) - I_{l-1} \left( r_{1}^{a} r_{2}^{b} r_{12}^{-2} \right)
\end{equation}

This allows all <math>I_{l}</math> integrals to be calculated from tables of <math>I_0</math> and
<math>I_1</math> integrals.

===The General Integral===

The above results for the angular and radial integrals can now be combined into a general formula for integrals of the type

\begin{equation}
I = \int \int d{\bf r}_{1} d{\bf r}_{2} R_{1} \mathcal{Y}^{M^{\prime}}_{l^{\prime}_{1} l^{\prime}_{2} L^{\prime}} \left( \hat{r}_{1}, \hat{r}_{2} \right) T_{k_{1} k_{2} K}^{Q} \left( {\bf r}_{1},{\bf r}_{2} \right) R_{2} \mathcal{Y}_{l_{1} l_{2} L}^{M} \left( \hat{r}_{1}, \hat{r}_{2} \right)
\end{equation}

where

\begin{equation}
\mathcal{Y}^{M}_{l_{1} l_{2} L} \left( \hat{r}_{1}, \hat{r}_{2} \right) = \sum_{m_1,m_2} \langle l_{1} l_{2} m_{1} m_{2} | LM \rangle Y^{m_{1}}_{l_{1}} \left( \hat{r}_{1} \right) Y^{m_{2}}_{l_{2}} \left( \hat{r}_{2} \right)
\end{equation}

and

\begin{equation}
T^Q_{k_{1} k_{2} K} \left( {\bf r}_{1}, {\bf r}_{2} \right) = \sum_{q_{1}, q_{2}} \langle k_{1} k_{2} q_{1} q_{2} | KQ \rangle Y_{k_{1}}^{q_{1}} \left( \hat{r}_{1} \right) Y^{q_{2}}_{k_{2}} \left( \hat{r}_{2} \right)
\end{equation}

The basic idea is to make repeated use of the formula

\begin{equation}
Y^{m_{1}}_{l_{1}} \left( \hat{r}_{1} \right) Y^{m_{2}}_{l_{2}} \left( \hat{r}_{1} \right) = \sum_{l m} \left( \frac{\left( 2l_{1}+1 \right) \left(2l_{2}+1 \right) \left(2l+1 \right)}{4 \pi} \right)^{1/2}
\times \left( \begin{array}{ccc}
l_1 & l_2 & l \\
m_1 & m_2 & m
\end{array} \right) \left(\begin{array}{ccc}
l_1 & l_2 & l \\
0 & 0 & 0
\end{array} \right)
Y^{m*}_{l} \left( \hat{r}_{1} \right)
\end{equation}

where

\begin{equation}
Y^{m*}(\hat{r}) = (-1)^mY_l^{-m}(\hat{r})
\end{equation}

and

\begin{equation}
\left(\begin{array}{ccc}
l_1 & l_2 & l \\
m_1 & m_2 & m
\end{array} \right) = \frac{\left( -1 \right)^{l_{1} - l_{2} - m}}{\left( 2l + 1 \right)^{1/2}} \left( l_{1} l_{2} m_{1} m_{2} | l, -m \right)
\end{equation}

is a 3-<math>j</math> symbol. In particular, write

\begin{array}{cc}
Y^{m_1^\prime}_{l_1^\prime}(\hat{r}_1)^{*} \underbrace{Y_{k_{1}}^{q_{1}} \left( \hat{r}_{1} \right) Y^{m_{1}}_{l_{1}} \left( \hat{r}_{1} \right)} & = & \sum_{\Lambda M}(\ldots)Y^{M}_\Lambda(\hat{r}_1)\\
\ \ \ \ \ \ \ \sum_{\lambda_1\mu_1}Y_{\lambda_1}^{\mu_1}(\hat{r}_1) & & \\
& & \\
Y^{m_{2}^{\prime}}_{l_{2}^{\prime}} \left( \hat{r}_{2} \right) \underbrace{Y^{q_{2}}_{k_{2}} \left( \hat{r}_{2} \right) Y_{l_{2}}^{m_{2}} \left( \hat{r}_{2} \right)} & = & \sum_{\Lambda^{\prime} M^{\prime}} \left( \ldots \right)Y_{\Lambda^{\prime}}^{M^{\prime}*} \left( \hat{r}_{2} \right) \\
\ \ \ \ \ \ \ \ \ \sum_{\lambda_{2} \mu_{2}}Y_{\lambda_{2}}^{\mu_{2}} \left( \hat{r}_{2} \right) & &
\end{array}

The angular integral then gives a factor of <math>2 \pi \delta_{\Lambda, \Lambda^{\prime}} \delta_{M, M^{\prime}} P_{\Lambda} \left( \cos \theta_{12} \right)</math>. The total integral therefore reduces to the form

\begin{equation}
I = \sum_{\Lambda} C_{\Lambda} I_{\Lambda} \left( R_{1} R_{2} \right)
\end{equation}

where <math>C_{\Lambda} = \sum_{\lambda_{1}, \lambda_{2}} C_{ \lambda_{1}, \lambda_{2}, \Lambda}</math>. For further details and derivations, including graphical representations, see G.W.F. Drake, Phys. Rev. A <math>{\bf 18}</math>, 820 (1978).

==Graphical Representation==

[[File:GraphRep1.jpg|500px]]

==Matrix Elements of H==
Recall that

\begin{equation}
H = -\frac{1}{2} \nabla^{2}_{1} - \frac{1}{2} \nabla^{2}_{2} - \frac{1}{r_1} - \frac{1}{r_2} + \frac{Z^{-1}}{r_{12}}
\end{equation}

Consider matrix elements of

\begin{equation}
\nabla^2_1 = \frac{1}{r_1^2} \frac{\partial}{\partial r_1}\left(r_1^2\frac{\partial}{\partial r_1} \right) + \frac{1}{r^2}\frac{\partial}{\partial r}\left(r^2\frac{\partial}{\partial r}\right) - \frac{(\vec{l}_1^Y)^2}{r_1^2} + \frac{2 \left( r_1 - r_2 \cos \theta \right)}{r} \frac{\partial^2}{\partial r_1 \partial r} - 2 \left( \nabla^Y_1 \cdot {\bf r}_2 \right) \frac{1}{r} \frac{\partial}{\partial r}
\end{equation}

where <math>r \equiv r_{12}</math> and <math>\cos\theta \equiv \cos\theta_{12}</math>. Also in what follows, define <math>\hat{\nabla}^Y_1 \equiv r_1\nabla^Y_1</math> where <math>\nabla^Y</math> operates only on the spherical harmonic part of the wave function, and similarly for <math>\vec{l}_1^Y</math>.

A general matrix element is

\begin{equation}
\langle r_1^{a^\prime} r_2^{b^\prime} r_{12}^{c^\prime} e^{-\alpha^\prime r_1 - \beta^\prime r_2} \mathcal{Y}^{M^\prime}_{l^\prime_1 l^\prime_2 L^\prime} \left( \hat{r}_1, \hat{r}_2 \right) \left| \nabla^2_1 \right| r_1^a r_2^b r_{12}^c e^{-\alpha r_1 - \beta r_2} \mathcal{Y}^M_{l_1 l_2 L} \left( \hat{r}_1, \hat{r}_2 \right) \rangle
\end{equation}

Since <math>\nabla^2_1</math> is rotationally invariant, this vanishes unless <math>L=L^\prime</math>, <math>M=M^\prime</math>. Also <math>\nabla_1^2</math> is Hermitian, so that the result must be the same whether it operates to the right or left, even though the results look very different. In fact, requiring the results to be the same yields some interesting and useful integral identities as follows:

Put

\begin{equation}
\mathcal{F} = F \mathcal{Y}^M_{l_1 l_2 L} \left( \hat{r}_1, \hat{r_2} \right)
\end{equation}

and

\begin{equation}
F = r_1^a r_2^b r_{12}^c e^{-\alpha r_1 - \beta r_2}
\end{equation}

Then

\begin{equation}
\nabla^2_1 \mathcal{F} = \left\{ \frac{1}{r_1^2} \left[ a \left( a + 1 \right) - l_1 \left( l_1 + 1 \right) \right] + \frac{c \left( c + 1 \right)}{r^2} + \alpha^2\frac{2\alpha \left( a + 1 \right)}{r_1} + \frac{2 \left(r_1 - r_2 \cos \theta \right) }{r_1r^2}c \left[ a - \alpha r_1 \right] - \frac{2c}{r^2} \left( \hat{\nabla}^Y_1 \cdot \hat{r}_2 \right) \frac{r_2}{r_1} \right\} \mathcal{F}
\end{equation}

and

\begin{align}
\langle \mathcal{F}^\prime \left| \nabla^2_1 \right| \mathcal{F} \rangle & = \sum_\Lambda \int d\tau_r F^\prime C_\Lambda \left( 1 \right) P_\Lambda \left( \cos \theta \right) \times \left\{ \frac{1}{r_1^2} \left[ a \left( a + 1 \right) - l_1 \left( l_1 + 1 \right) \right] -\frac{2 \alpha \left( a + 1 \right)}{r_1} + \frac{c \left( c + 1 \right)}{r^2} +\alpha^2 + \frac{2 \left( r_2 - r_2 \cos \theta \right)}{r_1r^2}c \left( a - \alpha r_1 \right) \right\} F \\
& + \sum_\Lambda \int d\tau_r F^\prime C_\Lambda \left( \hat{\nabla}^Y_1 \cdot \hat{r}_2 \right) P_\Lambda \left( \cos \theta \right) \left( \frac{-2c r_2}{r_1 r^2} \right) F
\end{align}

where

\begin{equation}
\int d\tau_r = \int^\infty_0 r_1 dr_1 \int^\infty_0 r_2 dr_2 \int^{r_1+r_2}_{\left| r_1 - r_2 \right|} r dr
\end{equation}

For brevity, let the sum over <math>\Lambda</math> and the radial integrations be understood, and let <math>\left( \nabla_1^2 \right)</math> stand for the terms which appear in the integrand. Then operating to the right gives

\begin{equation}
(\nabla_1^2)_R = \frac{1}{r_1^2} \left[ a \left( a + 1 \right) - l_1 \left( l_1 + 1 \right) \right] + \frac{c \left( c + 1 \right)}{r^2} + \alpha^2 - \frac{2 \alpha \left( a + 1 \right)}{r_1}
+ \frac{2 \left( r_1 - r_2 \cos \theta \right)}{r_1 r^2}c \left( a - \alpha r_1 \right) - \frac{2c r_2}{r_1 r^2} \left( \hat{\nabla}^Y_1 \cdot \hat{r}_2 \right)
\end{equation}

Operating to the left gives

\begin{equation}
(\nabla_1^2)_L = \frac{1}{r_1^2} \left[a^\prime \left( a^\prime + 1 \right) - l_1^\prime \left( l_1^\prime + 1 \right) \right] + \frac{c^\prime \left( c^\prime + 1 \right)}{r^2} + \alpha^{\prime 2} -\frac{2 \alpha^\prime \left( a^\prime + 1 \right)}{r_1}
+ \frac{2 \left( r_1 - r_2 \cos \theta \right)}{r_1 r^2} c^\prime \left( a^\prime - \alpha^\prime r_1 \right) - \frac{2 c^\prime r_2}{r_1 r^2} \left( \hat{\nabla}^{Y^\prime}_1 \cdot \hat{r}_2 \right)
\end{equation}

Now put

\begin{array}{cc}
a_+ = a+a^\prime, & \hat{\nabla}_1^+ = \hat{\nabla}_1^Y + \hat{\nabla}_1^{Y\prime} \\
a_{-} = a-a^\prime, & \hat{\nabla}_1^{-} = \hat{\nabla}_1^Y - \hat{\nabla}_1^{Y\prime}
\end{array}

etc., and substitute <math>a^\prime = a_+ - a</math>, <math>c^\prime = c_+ - c</math>, and <math>\alpha^\prime = \alpha_+ - \alpha</math> in <math>(\nabla^2_1)_L</math>. If <math>a_+</math>, <math>c_+</math> and <math>\alpha_+</math> are held fixed, then the equation

\begin{equation}
(\nabla^2_1)_R = (\nabla^2_1)_L
\end{equation}

must be true for arbitrary <math>a</math>, <math>c</math>, and <math>\alpha</math>. Their coefficients must thus vanish.

This yields the integral relations

\begin{equation}
\label{I}
\frac{\left( r_1 - r_2 \cos \theta \right)}{r_1 r^2} = \frac{1}{c_+} \left( \frac{- \left( a_+ + 1 \right)}{r_1^2} + \frac{\alpha_+}{r_1} \right) \tag{I}
\end{equation}

from the coefficient of <math>a</math> and

\begin{equation}
\label{II}
\frac{ \left( r_1 - r_2 \cos \theta \right) \left( a_+ - \alpha_+ r_1 \right)}{r_1 r^2} = \frac{r_2}{r_1 r^2}\left( \hat{r}_2 \cdot \hat{\nabla}_1^+ \right) - \frac{ \left( c_+ + 1 \right)}{r^2} \tag{II}
\end{equation}

from the coefficient of <math>c</math>. The coefficient of <math>\alpha</math> gives an equation equivalent to (I).

Furthermore, if can be show that (see problem)

\begin{align}
& \sum_\Lambda \int d\tau_r \frac{r^c}{r^2} C_\Lambda \left( \hat{r}_2 \cdot \hat{\nabla}_1^Y \right) P_\Lambda \left( \cos \theta \right) \\
= & \sum_\Lambda \int d\tau_r \frac{r^c}{c r_1 r_2}C_\Lambda \left( 1 \right)P_\Lambda \left( \cos \theta \right) \times \left( \frac{l_1^\prime \left( l_1^\prime + 1 \right) - l_1 \left( l_1 + 1 \right) -\Lambda \left( \Lambda + 1 \right)}{2} \right)
\end{align}

and similarly for <math>\left( \hat{r}_2 \cdot \hat{\nabla}^{Y\prime}_1 \right)</math> with <math>l_1</math> and <math>l_1^\prime</math> interchangeable, then it follows that

\begin{equation}
\frac{r^{c_+}}{r^2} \left( \hat{r}_2 \cdot \hat{\nabla}_1^+ \right) = - \frac{r^{c_+}}{c_+ r_1 r_2} \Lambda \left( \Lambda + 1 \right)
\end{equation}

and

\begin{equation}
\frac{r^{c_+}}{r^2} \left( \hat{r}_2 \cdot \hat{\nabla}_1^- \right) = \frac{r^{c_+}}{c_+ r_1 r_2} \left[ l_1^\prime \left( l_1^\prime + 1 \right) - l_1 \left( l_1 + 1 \right) \right]
\end{equation}

where equality applies after integration and summation over <math>\Lambda</math>.

Thus (II) becomes

\begin{equation}\label{IIa}
\frac{ \left( r_1 - r_2 \cos \theta \right) \left( a_+ - \alpha_+ r_1 \right)}{r_1 r^2} = - \frac{\Lambda \left( \Lambda + 1 \right)}{c_+ r^2_1} - \frac{ \left( c_+ + 1 \right)}{r^2} \tag{IIa}
\end{equation}

===Problem===

Prove the integral relation

\begin{align}
& \sum_\Lambda \int d\tau_r f \left( r_1, r_2 \right) \frac{1}{r} \left( \frac{d}{dr} q(r) \right) C_\Lambda \left( \hat{r}_2 \cdot \hat{\nabla}^Y_1 \right) P_\Lambda \left( \cos \theta \right) \\
= & \sum_\Lambda \int d\tau_r \frac{ f \left( r_1, r_2 \right)}{r_1, r_2} q(r) C_\Lambda \left( 1 \right) P_\Lambda \left( \cos \theta \right) \left(\frac{l_1^\prime \left( l_1^\prime + 1 \right) - l_1 \left( l_1 + 1 \right) - \Lambda \left( \Lambda + 1 \right)}{2}\right)
\end{align}

where the coefficients <math>C_\Lambda \left( 1 \right)</math> are the angular coefficients from the overlap integral

\begin{align}
\int & d\Omega\, \mathcal{Y}^{M*}_{l_1^\prime l_2^\prime L}(\hat{r}_1,\hat{r}_2)\, \mathcal{Y}^{M}_{l_1 l_2 L}(\hat{r}_1,\hat{r}_2)\\
= & \sum_\Lambda C_\Lambda(1)P_\Lambda(\cos\theta_{12}).
\end{align}

Hint: Use the fact that <math>l_1^2</math> is Hermitian so that
<math>
\int d\tau (l_1^2\mathcal{Y}^\prime)^*q(r)\mathcal{Y} = \int d\tau
\mathcal{Y}^{\prime *}l_1^2(q(r)\mathcal{Y})
</math>
with <math>\ \vec{l}_1= \frac{1}{i}\vec{r}_1\times\nabla_1</math>.

It is also useful to use
<math>\begin{eqnarray}\ \
\cos\theta\, P_L(\cos\theta) &=& \frac{1}{2L+1}[L\,P_{L-1}(\cos\theta) + (L+1)P_{L+1}(\cos\theta)]
\\
(\cos^2\theta-1)P_L(\cos\theta) &=&
\frac{L(L-1)}{(2L-1)(2L+1)}P_{L-2}(\cos\theta)\nonumber
-\frac{2(L^2+L-1)}{(2L-1)(2L+3)}P_L(\cos\theta) + \frac{(L+1)(L+2)}{(2L+1)(2L+3)}P_{L+2}(\cos\theta)\nonumber\\
&=& \frac{L(L-1)}{(2L-1)(2L+1)}[P_{L-2}(\cos\theta) - P_L(\cos\theta)]
+ \frac{(L+1)(L+2)}{(2L+1)(2L+3)}[P_{L+2}(\cos\theta) - P_L(\cos\theta)]
\end{eqnarray}</math>

together with a double application of the integral recursion relation
<math>
I_{L+1}\left(\frac{1}{r}\frac{d}{dr}q(r)\right) =
(2L+1)I_L\left(\frac{1}{r_1r_2}q(r)\right)+I_{L-1}\left(\frac{1}{r}\frac{d}{dr}q(r)\right)
</math>

Of course <math>l_1^2\mathcal{Y}^M_{l_1l_2L}(\hat{r}_1,\hat{r}_2)=l_1(l_1+1)\mathcal{Y}^M_{l_1l_2L}(\hat{r}_1,\hat{r}_2)</math>. Begin by expanding
<math>
l_1^2(g\mathcal{Y}) = \mathcal{Y}l_1^2g + g\,l_1^2\mathcal{Y} + 2(\vec{l}_1g)\cdot(\vec{l}_1\mathcal{Y})
</math>, and show that

<math>
\begin{eqnarray}\ \
(\vec{l}_1g)\cdot(\vec{l}_1\mathcal{Y}) & = & \frac{r_1^2}{r}\frac{dg}{dr}\vec{r}_2\cdot \nabla_1\mathcal{Y}\,\,\,\,\mbox{ and}\\
l_1^2g(r) &=& 2r_1r_2\cos\theta\frac{1}{r}\frac{dg}{dr} + r_1^2r_2^2(\cos^2\theta - 1)
\frac{1}{r}\frac{d}{dr}\left(\frac{1}{r}\frac{dg}{dr}\right).
\end{eqnarray}
</math>

The proof amounts to showing that the term <math>l_1^2g(r)</math> can be replaced by
<math>\Lambda(\Lambda+1)g(r)</math> after multiplying by <math>P_\Lambda(\cos\theta)</math>
and integrating by parts with respect to the radial integrations over <math>r_1</math>, <math>r_2</math> and
<math>r \equiv |\vec{r}_1 - \vec{r}_2|</math>. Remember that <math>\cos\theta</math> is just a short-hand notation for the radial function <math>(r_1^2 + r_2^2 - r^2)/(2r_1r_2)</math>.

==General Hermitian Property==

Each combination of terms of the form
\begin{equation}
<f>_R = a^2f_1+af_2+abf_3+b^2f_4+bf_5+af_6\nabla_1^Y+bf_7\nabla^Y_2+f_8(Y)
\end{equation}
can be rewritten
\begin{equation}
<f>_L= (a_+-a)^2f_1+(a_+-a)f_2 + (a_+-a)(b_+-b)f_3+(b_+-b)^2f_4
+(b_+-b)f_5+(a_+-a)f_6\nabla^{Y\prime}_1+(b_+-b)f_7\nabla^{Y\prime}_2+f_8)Y^\prime
\end{equation}
where as usual <math>\nabla_1^Y</math> acts only on the spherical harmonic part of the wave function denoted for short by <math>Y</math>.
Since these must be equal for arbitrary <math>a</math> and <math>b</math>,
\begin{equation}
a_+^2f_1+a_+f_2+_+b_+f_3+b_+f_4+b_+f_5+a_+f_6\nabla^{Y\prime}_1 +
b_+f_7\nabla^{Y\prime}_2+f_8(Y^\prime)-f_8(Y)=0
\end{equation}

Adding the corresponding expression with <math>Y</math> and <math>Y^\prime</math> interchanged
yields
<math>\label{III}
a_+^2f_1+a_+f_2+a_+b_+f_3+b_+^2f_4+b_+f_5+\frac{1}{2}a_+f_6\nabla^+_1+\frac{1}{2}b_+f_7\nabla^+_2=0\tag{III}
</math>

Subtracting gives
\begin{eqnarray*}
f_8(Y)-f_8(Y^\prime)&=&-\frac{1}{2}[a_+f_6\nabla^-_1+b_+f_7\nabla_2^-]\tag{IV}\\
a[-2a_+f_1-2f_2-b_+f_3-f_6\nabla^+_1]&=&0\tag{V}\\
b[-2b_+f_4-2f_5-a_+f_3-f_7\nabla^+_2]&=&0\tag{VI}\\
\end{eqnarray*}

Adding the two forms gives

\begin{eqnarray}
<f>_R+<f>_L=\frac{1}{2}(a_+^2+a_-^2)f_1+a_+f_2+\frac{1}{2}(a_+b_++a_-b_-)f_3\nonumber\\
\ +\frac{1}{2}(b_+^2+b_-^2)f_4+b_+f_5+\frac{1}{2}f_6(a_+\nabla_1^++a_-\nabla^-_1)\nonumber\\
\ +\frac{1}{2}(b_+\nabla_2^++b_-\nabla^-_2)+f_8(Y)+f_8(Y^\prime)\nonumber
\end{eqnarray}

Subtracting <math>\frac{x}{2}\times({\rm I})</math> gives

\begin{eqnarray}\ \
<f>_R+<f>_L = \frac{1}{2}[(1-x)a_+^2 + a_-^2]f_1 +
(1-\frac{x}{2})a_+f_2+\frac{1}{2}[(1-x)a_+b_++a_-b_-]f_3\nonumber\\
+\frac{1}{2}[(1-x)b_+^2+b_-^2]f_4+(1-\frac{x}{2})f_5b_++\frac{1}{2}f_6[(1-\frac{x}{2})a_+\nabla_1^++a_-\nabla^-_1]\nonumber\\
+\frac{1}{2}[(1-\frac{x}{2})b_+\nabla^+_2+b_-\nabla^-_2]+f_8(Y)+f_8(Y^\prime)\nonumber
\end{eqnarray}

If <math>x=1</math>,
\begin{eqnarray}
<f>_R+<f>_L &=&
\frac{1}{2}[a_-^2f_1+a_+f_2+a_-b_-f_3+b_-^2f_4+b_+f_5\nonumber\\
&+&\left(\frac{1}{2}a_+\nabla^+_1+a_-\nabla^-_1\right)f_6+\left(\frac{1}{2}b_+\nabla^+_2+b_-\nabla^-_2\right)f_7]+f_8(Y)+f_8(Y^\prime)\nonumber
\end{eqnarray}

The General Hermitian Property for arbitrary <math>x</math> gives
\begin{eqnarray}\ \
\nabla_1^2=
\frac{1}{4}\{\frac{1}{r_1^2}[(1-x)a_+^2+a_-^2+2(1-\frac{x}{2})a_+-2[l_1(l_1+1)+l_1^\prime(l_1^\prime+1)]\nonumber\\
-\frac{2}{r_1}[(1-x)\alpha_+a_++\alpha_-a_-+2(1-\frac{x}{2})\alpha_+]+(1-x)\alpha_+^2
+ \alpha^2_-]\nonumber\\
+\frac{2(r_1-r_2\cos\theta)}{r_1r^2}[(1-x)(a_+-\alpha_+r_1)c_++(a_--\alpha_-r_2)c_-]\nonumber\\
-\frac{2r_2}{r_1r^2}[(1-\frac{x}{2})c_+\hat{r}_2\cdot\hat{\nabla}_1^++c_-\hat{r}_2\hat{\nabla}_1^-]+\frac{(1-x)c_+^2+c_-^2+2(1-\frac{x}{2})c_+}{r^2}\}\nonumber
\end{eqnarray}.

Use
<math>\begin{eqnarray}\ \
\frac{-2r_2}{r_1r^2}[(1-\frac{x}{2})c_+\hat{r}_2\hat{\nabla}_1^++c_-\hat{r}_2\hat{\nabla}_1^-]=\frac{2}{r_1^2}\left[(1-\frac{x}{2})\Lambda(\Lambda+1)-\frac{c_-}{c_+}[l_1^\prime(l_1^\prime+1)+l_1(l_1+1)]\right]\end{eqnarray}</math>,

<math>\begin{eqnarray}
\frac{2(r_1-r_2\cos\theta)}{r_1r^2}(a_--\alpha_-r_1)c_- = 2\frac{c_-}{c_+}\left[\frac{-1}{r_1^2}[a_-(a_++1)]+\frac{1}{r_1}[a_-\alpha_++\alpha_-(a_++2)]-\alpha_-\alpha_+\right]\nonumber
\end{eqnarray}</math>,

and
<math>\begin{eqnarray}\ \
\frac{2(r_1-r_2\cos\theta)}{r_1r^2}(a_+-\alpha_+r_1)c_+ = 2\left[-\frac{1}{r_1^2}[a_+(a_++1)]+\frac{1}{r_1}[a_+\alpha_++\alpha_+(a_++2)]-\alpha_+^2\right]
\end{eqnarray}</math>.

Substituting into <math>\nabla_1^2</math> gives

\begin{eqnarray}
\nabla^2_1 = \frac{1}{4}\{\frac{1}{r_1^2}[-(1-x)a_+^2 + a_-^2 +xa_+ +
2(1-\frac{x}{2})\Lambda(\Lambda+1)\nonumber\\
-2l_1(l_1+1)(1-\frac{c_-}{c_+}) -
2l_1^\prime(l_1^\prime+1)(1+\frac{c_-}{c_+})-\frac{2c_-a_-}{c_+}(a_++1)]\nonumber\\
- \frac{2}{r_1}\left[-(1-x)\alpha_+(a_++2)+\alpha_-a_-+2(1-\frac{x}{2})\alpha_+
-\frac{c_-}{c_+}[a_-\alpha_++\alpha_-(a_++2)]\right]\nonumber\\
-(1-x)\alpha_+^2+\alpha_-^2-2\frac{c_-}{c_+}\alpha_-\alpha_+ +\frac{1}{r^2}\left[(1-x)c_+^2+c_-^2+2(1-\frac{x}{2})c_+\right]\}\nonumber
\end{eqnarray}

This has the form
<math>
\nabla_1^2 =
\frac{1}{4}\left[\frac{A_1}{r_1^2}+\frac{B_1}{r_1}+\frac{C_1}{r^2}+D_1\right]
</math>
with

<math>\begin{array}{ccc}
A_1=-(1-x)a_+^2+a_-^2
+xa_++2(1-\frac{x}{2})\Lambda(\Lambda+1)-2l_1(l_1+1)(1-\frac{c_-}{c_+})\
-2l_1^\prime(l_1^\prime+1)(1+\frac{c_-}{c_+})-\frac{2c_-a_-}{c_+}(a_++1) \ \
\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \
\end{array}
</math>

<math>
B_1 =
2\left[(1-x)\alpha_+(a_++2)-\alpha_-a_--2(1-\frac{x}{2})\alpha_++\frac{c_-}{c_+}[a_-\alpha_++\alpha_-(a_++2)]\right]
</math>

<math>
C_1 = (1-x)c_+^2+ c_-^2 +2(1-\frac{x}{2})c_+
</math>

<math>
D_1 = -(1-x)\alpha_+^2 +\alpha_-^2 - 2\frac{c_-}{c_+}\alpha_-\alpha_+
</math>

The complete Hamiltonian is then (in Z-scaled a.u.)

<math>\begin{eqnarray}
H =-\frac{1}{2}\nabla^2_1 -\frac{1}{2}\nabla^2_2 -\frac{1}{r_1} -\frac{1}{r_2} +\frac{Z^{-1}}{r}\nonumber\\
= -\frac{1}{2}\nabla^2_1 -\frac{1}{2}\nabla^2_2 -\frac{1}{r_1} -\frac{Z-1}{Zr_2}+Z^{-1}\left(\frac{1}{r}-\frac{1}{r_2}\right)\nonumber\\
\ =-\frac{1}{8}\left[\frac{A_1}{r_1^2}+\frac{B_1+8}{r_1}+\frac{C_1}{r^2}+D_1+D_2 +\frac{A_2}{r_2^2} + \frac{B_2 + 8(Z-1)/Z}{r_2}+\frac{C_2}{r^2}\right]\nonumber
+ Z^{-1}\left(\frac{1}{r}-\frac{1}{r_2}\right)\nonumber
\end{eqnarray}</math>

The screened hydrogenic energy is
<math>\begin{eqnarray}
E_{SH}=-\frac{1}{2}\left[\frac{1}{n_1^2}+ \left(\frac{Z-1}{Z}\right)^2\frac{1}{n_2^2}\right]
\end{eqnarray}</math>
so that

<math>\begin{eqnarray}
H-E_{SH} &=& -\frac{1}{8}[
\frac{A_1}{r_1^2}+\frac{B_1+8}{r_1}+\frac{C_1}{r^2} + D_1 + D_2
-\frac{4}{n_1^2} -\left(\frac{Z-1}{Z}\right)^2\frac{4}{n_2^2}\nonumber
+\frac{A_2}{r_2}+\frac{B_2+8(Z-1)/Z}{r_2}+\frac{C_2}{r^2}] +
Z^{-1}\left(\frac{1}{r} -\frac{1}{r_2}\right)\nonumber
\end{eqnarray}</math>

==Optimization of Nonlinear Parameters==
- The traditional method of performing Hylleraas calculations is to write the
basis set in the form

<math>
\Psi = \sum_{i,j,k}c_{i,j,k}\varphi_{i,j,k}(\alpha,\beta)\pm exchange
</math>

with

<math>
\varphi_{i,j,k}(\alpha,\beta) = r_1^ir_2^jr_{12}^ke^{-\alpha r_1 - \beta
r_2}\mathcal{Y}^M_{l_1l_2L}.
</math>

The usual procedure is to set <math>\alpha = Z</math> so that it represents the inner
<math>1s</math> electron, and then to vary <math>\beta</math> so as to minimize the energy.

[[File:OptNon.jpg|500px]]

Since <math>\beta</math> appears in <math>\Psi</math> as a non-linear parameter, the entire
calculation must be repeated for each value of <math>\beta</math>. However, the minimum
becomes progressively smaller as the basis set is enlarged.

[[File:OptNon02.jpg|500px]]

*Difficulties

1. If the basis set is constructed so that <math> i +j +j \leq N</math>, the the
number of terms is
<math>
(N+1)(N+2)(N+3)/6
</math>
N=14 already gives 680 terms and an accuracy of about 1 part in <math>10^{10}</math> for
low-lying states. A substantial improvement in accuracy would require much
larger basis sets, together with multiple precision arithmetic to avoid loss
of significant figures when high powers are included.

2. The accuracy rapidly deteriorates as one goes to more highly excited
states - about 1 significant figure is lost each time the principle quantum
number is increased.

*Cure

We have found that writing basis sets in the form
\begin{eqnarray}
\Psi &=& \sum_{i,j,k}\left(c^{(1)}_{i,j,k}\varphi_{ijk}(\alpha_1\beta_1) +
c_{ijk}^{(2)}\varphi_{ijk}(\alpha_2\beta_2)\right) \pm exchange \nonumber\\
&=& \psi({\bf r}_1,{\bf r}_2) \pm \psi({\bf r}_2,{\bf r}_1)\nonumber
\end{eqnarray}
so that each combination of powers is included twice with different nonlinear
parameters gives a dramatic improvement in accuracy for basis sets of about
the same total size.

However, the optimization of the nonlinear parameters is now much more
difficult, and an automated procedure is needed.

If
<math>
E = \frac{<\Psi|H|\Psi>}{<\Psi|\Psi>}</math> and we assume <math><\Psi|\Psi> = 1</math>, then

<math>
\frac{\partial E}{\partial\alpha_t} = -2 <\Psi|H-E|r_1\psi({\bf r}_1,{\bf
r}_2;\alpha_t)\pm r_21\psi({\bf r}_2,{\bf
r}_1;\alpha_t)>
</math>
where
<math>
\psi({\bf r}_1,{\bf r}_2;\alpha_t) =
\sum_{i,j,k}c_{ijk}^{(t)}\varphi_{ijk}(\alpha_t\beta_t)
</math>

*Problem

1. Prove the above.

2. Prove that there is no contribution from the implicit dependence of
<math>c_{ijk}^{(t)}</math> on <math>\alpha_t</math> if the linear parameters have been optimized.

==The Screened Hydrogenic Term==

If the Hamiltonian is written in the form (in Z-scaled a.u.)
<math>\begin{eqnarray}\
H = H_0({\bf r}_1,Z) + H_0({\bf r}_2,Z-1) +
Z^{-1}\left(\frac{1}{r}-\frac{1}{r_2}\right)
\end{eqnarray}</math>
with
<math>\
H_0({\bf r}_1,Z) = -\frac{1}{2}\nabla_1^2 - \frac{1}{r_1}\nonumber
</math> and
<math>
H_0({\bf r}_2,Z-1) = -\frac{1}{2}\nabla_2^2 - \frac{Z-1}Z{r_2}\nonumber
</math>,
then the eigenvectors of <math>H_0({\bf r},Z) +H_0({\bf r}_2,Z-1)</math> are products of
hydrogenic orbitals
<math>
\Psi_0=\psi_0(1s,Z)\psi_0(nl,Z-1)
</math>
and the eigenvalue is
<math>\begin{eqnarray}
E_{SH} = \left[-\frac{1}{2} - \left(\frac{Z-1}{Z}\right)^2\frac{1}{2n^2}\right]Z^2 a.u.
\end{eqnarray}</math>
called the screened hydrogenic eigenvalue. From highly excited states,
<math>E_{SH}</math> and <math>\Psi_0</math>are already excellent approximations.

For example, for the 1s8d states, the energies are

<math>\begin{eqnarray}\\
E(1s8d^1D)&=& -2.007\,816\,512\,563\,81 {\rm\ a.u.}\nonumber\\
E(1s8d^3D)&=& - 2.007\,817\,934\,711\,71 {\rm\ a.u.}\nonumber\\
E_{SH} &=& -2.007\,812\,5 {\rm\ a.u.} \nonumber
\end{eqnarray}</math>

It is therefore advantageous to include the screened hydrogenic terms in the
basis set so that the complete trial function becomes
<math>
\Psi = c_0\Psi_0 +
\sum_{ijk}\left[c_{ijk}^{(1)}\varphi_{ijk}(\alpha_1,\beta_1)+c_{ijk}^{(2)}\varphi(\alpha_2,\beta_2)\right]\pm
{\rm exchange}
</math>

Also, the variational principal can be re-expressed in the form
<math>\begin{eqnarray}
E= E_{SH}+\frac{<\Psi|H-E_{SH}|\Psi>}{<\Psi|\Psi>}
\end{eqnarray}</math>
so that the <math>E_{SH}</math> term can be cancelled analytically from the matrix
elements.

For example
<math>
<\Psi_0|H-E_{SH}|\Psi_0> = 0
</math>

and
<math>
<\varphi_{ijk}|H-E_{SH}|\Psi_0> =
Z^{-1}<\varphi_{ijk}|\frac{1}{r}-\frac{1}{r_2}|\Psi_0>
</math>

Recall that
<math>\begin{eqnarray}\ \
I_0(a,b,c) = \frac{2}{c+2}\sum^{[c+1]/2}_{i=0}\left(\begin{array}{c}
c+2\\
2i+1
\end{array}\right) \left[f(p,q;\beta) + f(p^\prime,q^\prime;\alpha)\right]
\end{eqnarray}</math>
where

<math>\begin{eqnarray}
f(p,q;x) =
\frac{q!}{x^{q+1}(\alpha+\beta)^{p+1}}\sum^p_{j=0}\frac{(p+j)!}{j!}\left(\frac{x}{\alpha+\beta}\right)^j
\end{eqnarray}</math>

<math>
\begin{array}{cc}
p=a+2i+2 & p^\prime = b+2i+2\\
q= b+c-2i+2 & q^\prime =a+c-2i+2
\end{array}
</math>

If <math>\beta\ll\alpha</math>, the <math>f(p,q;\beta)\gg f(p^\prime,q^\prime;\alpha)</math> and the
<math>i=0</math> term is the dominant contribution to <math>f(p,q;\beta)</math>. However, since
this term depends only on the sum of powers <math>b+c</math> for <math>r_2</math>, it cancels exactly
from the matrix element of
<math>
\frac{1}{r_{12}} - \frac{1}{r_2}
</math>
and can therefore
be omitted in the calculations of integrands, thereby saving many significant
figures. This is especially valuable when <math>b+c</math> is large and <math>a</math> is small.

==Small Corrections==
*Mass Polarization
If the nuclear mass is not taken to be infinite, then the Hamiltonian is
<math>\begin{eqnarray}\ \
H = \frac{P_N^2}{2M} + \sum_{i=1}\frac{p_i^2}{2m} + V
\end{eqnarray}</math>

[[File:SmallCor.jpg|500px]]

where <math>M =M_A-nm</math> is the nuclear mass and <math>m</math> is the electron mass.

Change variables to the C of M
<math>\begin{eqnarray}
{\bf R} = \frac{1}{M+nm}\left[M{\bf r}_N + m({\bf r}_1 + {\bf r}_2 +
\ldots)\right]
\end{eqnarray}</math>

and relative variables

<math>
{\bf s}_i = {\bf r}_i - {\bf r}_N.
</math>

Then

<math>\begin{eqnarray}\ \ \ \ \
H = \frac{1}{2(M+nm)}{\bf p}_R^2+\frac{1}{2m}\sum_{i=1}^n{\bf p}_{s_i}^2
+\frac{1}{2M}\sum_{i,k}{\bf p}_{s_i}\cdot {\bf p}_{s_k} + V({\bf s}_1, {\bf
s}_2, \ldots)\nonumber
= \frac{1}{M+nm}{\bf p}_R^2+\frac{1}{2\mu}\sum_{i=1}^n{\bf
p}_{s_i}^2+\frac{1}{2M}\sum_{i\neq k}{\bf p}_{s_i}\cdot {\bf p}_{s_k}+V({\bf s}_1, {\bf
s}_2, \ldots)\nonumber
\end{eqnarray}</math>

where
<math>\begin{eqnarray}
\mu = \left(\frac{1}{M} + \frac{1}{m}\right)^{-1}
\end{eqnarray}</math>

If <math>n=2</math>, the Schrödinger equation is
<math>\begin{eqnarray}\ \ \
\left[ -\frac{\hbar^2}{2\mu}(\nabla^2_{s_1}+ \nabla^2_{s_2}) -
\frac{\hbar^2}{M}\nabla_{s_1}\nabla_{s_2}-\frac{Ze^2}{s_1}-\frac{Ze^2}{s_2}+\frac{e^2}{s_{12}}\right]\psi
= E\psi
\end{eqnarray}</math>.

Define the reduced mass Bohr radius
<math>\begin{eqnarray}
a_\mu = \frac{\hbar^2}{\mu e^2} = \frac{m}{\mu}a_0
\end{eqnarray}</math>
and

<math>\begin{eqnarray}
\rho_i = \frac{Zs_i}{a_\mu}\nonumber\\
\varepsilon = \frac{E}{Z^2(e^2/a_\mu)}\nonumber
\end{eqnarray}</math>

The Schrödinger equation then becomes
<math>\begin{eqnarray}\ \ \
\left[
-\frac{1}{2}(\nabla^2_{\rho_1}+\nabla^2_{\rho_2})-\frac{\mu}{M}\vec{\nabla}_{\rho_1}\cdot\vec{\nabla}_{\rho_2} - \frac{1}{\rho_1}-\frac{1}{\rho_2}+\frac{Z^{-1}}{\rho_{12}}\right]\psi = \varepsilon \psi
\end{eqnarray}</math>

The effects of finite mass are thus

1. A reduced mass correction to all energy levels of <math>(e^2/a_\mu)/(e^2/a_0)
= \mu/m</math>, i.e.
<math>\begin{eqnarray}
E_M=\frac{\mu}{m}E_\infty
\end{eqnarray}</math>

2. A specific mass shift given in first order perturbation theory by
<math>\begin{eqnarray}
\Delta\varepsilon =
-\frac{\mu}{M}<\psi|\nabla_{\rho_1}\cdot\nabla_{\rho_2}|\psi>
\end{eqnarray}</math>
or

<math>\begin{eqnarray}
\Delta\varepsilon =
-\frac{\mu}{M}<\psi|\nabla_{\rho_1}\cdot\nabla_{\rho_2}|\psi>\frac{e^2}{a_\mu}\nonumber\\
= -\frac{\mu^2}{mM}<\psi|\nabla_{\rho_1}\cdot\nabla_{\rho_2}|\psi>\frac{e^2}{a_0}\nonumber
\end{eqnarray}</math>

Since <math>\nabla_{\rho_1}\cdot\nabla_{\rho_2}</math> has the same angular properties as
<math>{\bf\rho}_1 \cdot {\bf\rho}_2</math>, the operator is like the product of two
dipole operators. For product type wave functions of the form
$
\psi = \psi(1s)\psi(nl) \pm \mbox{exchange},
$
the matrix element vanishes for all but P-states.

=Radiative Transitions=

==Transition Integrals and the Line Strength==

For an electric dipole transition <math>\gamma'\,L'\,S'\,J'\,M' \rightarrow \gamma\,L\,S\,J\,M</math>, the line strength is defined by
\begin{equation}
S = \sum_{M',\mu,M}|\langle\gamma'\,L'\,S'\,J'\,M'|r_\mu|\gamma\,L\,S\,J\,M\rangle|^2
\end{equation}
Using the Wigner-Eckart theorem, this is related to the reduced matrix element by
\begin{eqnarray}
S(J',J) = |\langle\gamma'\,L'\,S'\,J'||r||\gamma\,L\,S\,J\,M\rangle|^2
\sum_{M',\mu,M}\left|\left(\matrix{J' & 1 & J \cr
-M'& \mu &M \cr}\right)\right|^2
\end{eqnarray}
Using the sum rule for the 3-<math>j</math> symbols
\begin{equation}
\sum_{M',\mu,M}\left|\left(\matrix{J' & 1 & J \cr
-M'& \mu &M \cr}\right)\right|^2 = 1
\end{equation}
we see that the line strength is identical to the reduced matrix element
\begin{equation}
S(J',J) = |\langle\gamma'\,L'\,S'\,J'||r||\gamma\,L\,S\,J\,M\rangle|^2
\end{equation}
For a nonrelativistic calculation in <math>LS</math> coupling, <math>S = S'</math> and we can next strip out the dependence on <math>J</math> and <math>J'</math> by use of Eq.\ (7.1.7) of Edmonds { Angular Momentum in Quantum Mechanics} with <math>k = 1</math> to obtain
\begin{equation}
S(J',J) = (-1)^{L'+S+J+1}[(2J+1)(2J'+1)]^{1/2}\left\{
\matrix{L' & J'& S\cr
J & L & 1\cr}\right\}\langle\gamma' L'|| r || \gamma L\rangle
\end{equation}
where <math>\langle\gamma' L'|| r || \gamma L\rangle</math> is the reduced matrix element if spin is neglected. Note that if spin is neglected, then the line-strength is simply
\begin{equation}
S(L',L) = |\langle\gamma' L'|| r || \gamma L\rangle|^2
\end{equation}
To calculate the reduced matrix element, use a particular example along with the Wigner-Eckart theorem
such as
\begin{equation}
\langle\gamma'\, L'|| r || \gamma\, L\rangle = (-1)^{L'}\langle\gamma'\, L'\, 0| z | \gamma\, L\, 0\rangle{\Large /}\left(\matrix{L' & 1 & L \cr
0 & 0 & 0 \cr}\right)
\end{equation}

===Correlated Two-electron Integrals===
For the case of two-electron integrals, <math>z = z_1 + z_2</math> and <math>r = r_1 + r_2</math>. As discussed in G.W.F. Drake, Phys. Rev. A 18, 820 (1978), the general integral in Hylleraas coordinates for the integral over the <math>z_1</math> term can be written in the form
\begin{equation}
\label{int1}
\langle\gamma'\, L'\, 0| z_1 | \gamma\, L\, 0\rangle = \sum_\Lambda C_\Lambda I_\Lambda(\gamma',\gamma)
\end{equation}
where <math>I_\Lambda(\gamma',\gamma)</math> stands for the purely radial integral
\begin{equation}
\label{radial}
I_\Lambda(\gamma',\gamma) = \int_0^\infty r_1\,dr_1\int_0^\infty r_2\,dr_2\int_{|r_1-r_2|}^{r_1+r_2}r_{12}\, dr_{12}
R_{\gamma'}^*\,r_1\, R_\gamma P_\Lambda(\cos\theta_{12})
\end{equation}
and <math>\cos\theta_{12} = (r_1^2 + r_2^2 - r_{12}^2)/(2r_1r_2)</math> is a short-hand notation for a radial function of <math>r_1</math>, <math>r_2</math> and <math>r_{12}</math>. Also, <math>P_\Lambda(\cos\theta_{12})</math> is a Legendre polynomial,
and, <math>R_\gamma = R_\gamma(r_1,r_2,r_{12})</math> is the radial part of the Hylleraas wave function for state <math>\gamma</math> such
that the total wave function is <math>\psi({\bf r}_1,{\bf r}_2) = R_\gamma(r_1,r_2,r_{12}){\cal Y}_{l_1,l_2,L}^M(\hat{\bf r}_1,\hat{\bf r}_2)\pm {\rm exchange}</math>, normalized to unity, and <math>{\cal Y}_{l_1,l_2,L}^M(\hat{\bf r}_1,\hat{\bf r}_2)</math> is a vector coupled product of spherical harmonics. The coefficents <math>C_\Lambda</math> in Eq.\ (\ref{int1}) can be calculated from Eq.\ (32) of the above 1978 paper. They need to be calculated separately for he direct-direct and the direct-exchange contributions. For the reduced matrix element of <math>z_1</math>, the leading 3-<math>j</math> symbol can be dropped, leaving (for the case <math>k = 1</math>)
\begin{eqnarray}
C_\Lambda^{\rm reduced} &=& \frac{(-1)^{l_2'+L'+L+1}}{2} (l_1,l_1',l_2,l_2',L,L')^{1/2}
\sum_{\lambda={\rm max}(l_1-1,0)}^{l_1+1} (\lambda,\Lambda,1)
\left(\matrix{l_1 & 1 & \lambda \cr
0 & 0 & 0 \cr}\right)
\left(\matrix{l_1'& \lambda & \Lambda \cr
0 & 0 & 0 \cr}\right)\nonumber\\
&\times&\left(\matrix{l_2' & 1_2 & \Lambda \cr
0 & 0 & 0 \cr}\right)
\left\{\matrix{L' & \lambda & l_2 \cr
\Lambda& l_2' & l_1' \cr}\right\}
\left\{\matrix{L' & \lambda & l_2 \cr
l_1 & L & 1 \cr}\right\}
\end{eqnarray}
where the notation <math>(a,b,c,\cdots)</math> means <math>(2a+1)(2b+1)(2c+1)\cdots</math>.
A similar expression can be obtained for the angular coefficients for the <math>z_2</math> term by interchanging the subscripts 1 and 2 throughout and multiplying by <math>(-1)^{L+L'+1}</math>.

===Example===
As an example, for the <math>1s^2\;^1S - 1s2p\;^1P</math> transition of helium, the only nonvanishing <math>C_\Lambda</math> angular coefficients are <math>C_1 = 1/2</math> for the direct-direct term (i.e. <math>P_1(\cos\theta_{12}) = \cos\theta_{12}</math> is included in the radial integral (\ref{radial})) and <math>C_0 = 1/2</math> for the direct-exchange term (i.e. <math>P_0(\cos\theta_{12}) = 1</math> is included in the radial integral (\ref{radial})). The results for infinite nuclear mass are
\begin{equation}
S(1\;^1S-2\;^1P) = (0.781\,127\,158\,667)^2
\end{equation}
\begin{equation}f(1\;^1S-2\;^1P) = -0.276\,164\,703\,736
\end{equation}
where
\begin{equation}
f(1\;^1S-2\;^1P) = \frac{2}{3}[E(1\;^1S) - E(2\;^1P)]S(1\;^1S-2\;^1P)
\end{equation}
is the absorption oscillator strength.

Theory Notes

2020-12-14T16:09:11Z

Drake:

<center>
<H1>Notes on solving the Schrödinger equation in Hylleraas coordinates for heliumlike atoms
Gordon W.F. Drake
Department of Physics, University of Windsor
Windsor, Ontario, Canada N9B 3P4
(Transcribed from hand-written notes and edited by Lauren Moffatt. Last revised November 22, 2017 to correct a missing subscript in the expression for $B_1$ in the Hamiltonian in Sect. 11 General Hermitian Property.)
</H1>
</center>

==Introduction==
These notes describe some of the technical details involved in solving the Schrödinger equation for a heliumlike atom of nuclear charge Z in correlated Hylleraas coordinates. The standard for computational purposes is to express all quantities in a form that is valid for electronic states of arbitrary angular momentum, and to express matrix elements of the Hamiltonian in a manifestly Hermitian symmetrized form. Further information and derivations of the mathematical techniques for angular integrals and radial recursion relations can be found in the paper
G.W.F. Drake, Phys. Rev. A <math>{\bf 18}</math>, 820 (1978).

The starting point is the two-electron Schrödinger equation for infinite nuclear mass
\begin{equation} \left[-\frac{\hbar^2}{2m}(\nabla^2_1 +\nabla^2_2) - \frac{Ze^2}{r_1} -
\frac{Ze^2}{r_2}+\frac{e^2}{r_{12}} \right]\psi = E\psi\nonumber
\end{equation}
where <math>m</math> is the electron mass, and <math>r_{12} = |{\bf r}_1 - {\bf r}_2|</math> (see diagram below).

[[File:Pic1.jpg|250px]]

Begin by rescaling distances and energies so that the Schrödinger equation can be expressed in a dimensionless form. The dimensionless Z-scaled distance is defined by <math>\begin{eqnarray} \rho = \frac{Zr}{a_0} \end{eqnarray} </math> where <math> \begin{eqnarray} a_0 = \frac{\hbar^2}{me^2} \end{eqnarray} </math> is the atomic unit (a.u.) of distance equal to the Bohr radius <math>0.529\, 177\, 210\, 92(17) \times 10^{-10}</math> m. Then

\begin{equation}[-\frac{\hbar^2}{2m}Z^2(\frac{me^2}{\hbar^2})^2(\nabla^2_{\rho_1}+\nabla^2_{\rho_2})
- Z^2\frac{e^2}{a_0}\rho^{-1}_1 - Z^2\frac{e^2}{a_0}\rho^{-1}_2 +
\frac{e^2}{a_0}Z\rho^{-1}_{12}]\psi= E\psi\nonumber\end{equation}

But <math>\begin{eqnarray} \frac{\hbar^2}{m}(\frac{me^2}{\hbar^2})^2 = \frac{e^2}{a_0} \end{eqnarray} </math> is the hartree atomic unit of of energy <math>(E_h = 27.211\,385\,05(60)</math> eV, or equivalently <math>E_h/(hc) = 2194.746\,313\,708(11)</math> m<math>^{-1})</math>. Therefore, after multiplying through by <math>a_0/(Ze)^2</math>, the problem to be solved in Z-scaled dimensionless units becomes

\begin{equation} \left[-\frac{1}{2}(\nabla^2_{\rho_1}+\nabla^2_{\rho_2}) -
\frac{1}{\rho_1} - \frac{1}{\rho_2} + \frac{Z^{-1}}{\rho_{12}}\right]\psi =
\varepsilon\psi\nonumber \end{equation}

where <math>\begin{eqnarray}\varepsilon = \frac{Ea_0}{(Ze)^2} \end{eqnarray}</math> is the energy in Z-scaled atomic units. For convenience, rewrite this in the conventional form
\begin{equation}
H\psi = \varepsilon\psi\nonumber \end{equation}
where (using <math>r</math> in place of <math>\rho</math> for the Z-scaled distance)
\begin{equation}
H = -\frac{1}{2}(\nabla^2_1+\nabla^2_2) - \frac{1}{r_1}-\frac{1}{r_2} + \frac{Z^{-1}}{r_{12}}
\nonumber\end{equation}
is the atomic Hamiltonian for infinite nuclear mass.

==The Hartree Fock Method==

For purposes of comparison, and to define the correlation energy, assume that <math>\psi({\bf r}_1,{\bf r}_2)</math> can be written in the separable product form
\begin{equation}
\psi({\bf r}_1,{\bf r}_2) = \frac{1}{\sqrt{2}} \left[ u_1(r_1) u_2(r_2) \pm u_2(r_1) u_1(r_2) \right] \nonumber
\end{equation}
for the <math>1s^2\ ^1\!S</math> ground state. Because of the <math>\frac{1}{r_{12}}</math> term in the Schrödinger equation, the exact solution connot be expressed in this form as a separable product. However, the Hartree-Fock approximation corresponds to finding the best possible solution to the Schrödinger equation
\begin{equation}
H\psi({\bf r}_1,{\bf r}_2) = E\psi({\bf r}_1,{\bf r}_2)\nonumber
\end{equation}
that can nevertheless be expressed in this separable product form, where as before
\begin{equation}
H = -\frac{1}{2}(\nabla^2_1+\nabla^2_2) - \frac{1}{r_1}- \frac{1}{r_2} + \frac{Z^{-1}}{r_{12}}\nonumber
\end{equation}
is the full two-electron Hamiltonian. To find the best solution, substitute into <math>\left\langle\psi \lvert H-E \rvert \psi\right\rangle</math> and require this expression to be stationary
with respect to arbitrary infinitesimal variations <math>\delta u_1</math> and <math>\delta
u_2</math> in <math>u_1</math> and <math>u_2</math> respectively; i.e.

\begin{equation}
\frac{1}{2} \left\langle \delta u_1(r_1)u_2(r_2) \pm u_2(r_1)\delta
u_1(r_2) \lvert H-E \rvert u_1(r_1) u_2(r_2)\pm u_2(r_1) u_1(r_2)\right\rangle\nonumber
\end{equation}

\begin{equation}
=\int\delta u_1(r_1) d{\bf r}_1 \left\{ \int d{\bf r}_2 u_2(r_2) \left( H - E \right) \left[ u_1(r_1)u_2(r_2) \pm u_2(r_1) u_1(r_2) \right] \right\}\nonumber
\end{equation}

\begin{equation}
= 0 \nonumber
\end{equation}

for arbitrary <math> \delta u_1(r_1).\nonumber </math> Therefore <math>\{\int d{\bf r}_2 \ldots \} = 0</math>.

Similarly, the coefficient of <math>\delta u_2</math> would give

\begin{equation}
\int d{\bf r}_1 u_1(r_1) \left( H-E \right) \left[ u_1(r_1) u_2(r_2) \pm
u_2(r_1) u_1(r_2) \right] = 0\nonumber
\end{equation}

Define

\begin{equation}
I_{12} = I_{21} = \int d{\bf r}\, u_1(r)u_2(r), \nonumber
\end{equation}

\begin{equation}
H_{ij} = \int d{\bf r}\, u_i(r)(-\frac{1}{2}\nabla - \frac{1}{r})u_j(r), \nonumber
\end{equation}

\begin{equation}
G_{ij}(r) = \int d{\bf r}^\prime u_i(r^\prime)\frac{1}{|{\bf r} - {\bf r}\prime|}u_j(r^\prime)\nonumber
\end{equation}

Then the above equations become the pair of integro-differential equations

\begin{equation}
[ H_0 - E + H_{22}+G_{22}(r)]u_1(r) = \mp [ I_{12}(H_0-E) + H_{12}+G_{12}(r)]u_2(r)\nonumber
\end{equation}

\begin{equation}
[H_0-E+H_{11}+G_{11}(r)]u_2(r) = \mp [I_{12}(H_0-E) + H_{12}+G_{12}(r)]u_1(r)\nonumber
\end{equation}

These must be solved self-consistently for the "constants" <math>I_{12}</math> and
<math>H_{ij}</math> and the function <math>G_{ij}(r)</math>.

The Hartree Fock energy is <math>E \simeq -2.87\ldots </math>a.u. while the exact energy is <math>E = -2.903724\ldots </math>a.u.

The difference is called the "correlation energy" because it arises from the
way in which the motion of one electron is correlated to the other. The
Hartree Fock equations only describe how one electron moves in the average field
provided by the other (mean-field theory).

==Configuration Interaction==
Expand
\begin{equation}
\psi({\bf r}_1,{\bf r}_2)= C_0u^{(s)}_1(r_1)u^{(s)}_1(r_2) + C_1u^{(P)}_1({\bf r}_1)u^{(P)}_1({\bf r}_2){\cal Y}^0_{1,1,0}(\hat{\bf r}_1, \hat{\bf r}_2)+C_2u^{(d)}_1({\bf r}_1)u^{(d)}_2({\bf r}_2){\cal Y}^0_{2,2,0}(\hat{\bf r}_1, \hat{\bf r}_2)+... \pm {\rm\ exchange}
\end{equation}
where
\begin{equation}
{\cal Y}^M_{l_1,l_2,L}(\hat{\bf r}_1, \hat{\bf r}_2)=\sum_{m_1,m_2}Y^{m_1}_{l_1}({\bf r}_1)Y^{m_2}_{l_2}({\bf r}_2)\times <l_1l_2m_1m_2\mid LM>
\end{equation}

This works, but is slowly convergent, and very laborious. The best CI calculations are accurate to <math> ~10^{-7}</math> a.u.

==Hylleraas Coordinates==
[E.A. Hylleraas, Z. Phys. <math>{\bf 48}, 469(1928)</math> and <math>{\bf 54}, 347(1929)</math>]
suggested using the co-ordinates <math>r_1</math>, <math>r_2</math> and <math>r_{12}</math> or equivalently

\begin{eqnarray}
s &=& r_1 + r_2, \nonumber\\
t &=& r_1-r_2, \nonumber\\
u &=& r_{12}\nonumber
\end{eqnarray}

and writing the trial functions in the form

\begin{equation}
\Psi({\bf r}_1,{\bf r}_2) = \sum^{i+j+k\leq
N}_{i,j,k}c_{i,j,k}r_1^{i+l_1}r_2^{j+l_2}r_{12}^ke^{-\alpha r_1 - \beta r_2}
\mathcal{Y}^M_{l_1,l_2,L}(\hat{r}_1,\hat{r}_2)\pm \text{exchange}\nonumber
\end{equation}

Diagonalizing H in this non-orthogonal basis set is equivalent to solving
<math> \begin{eqnarray}
\frac{\partial E}{\partial c_{i,j,k}} = 0\nonumber
\end{eqnarray} </math>
for fixed <math>\alpha</math> and <math>\beta</math>.

The diagonalization must be repeated for different values of <math>\alpha</math> and
<math>\beta</math> in order to optimize the non-linear parameters.

==Completeness==
The completeness of the above basis set can be shown by first writing
\begin{equation}
r_{12}^2 = r_1^2 + r_2^2 - 2r_1r_2\cos\theta_{12}
\end{equation}
and, from the spherical harmonic addition theorem,
\begin{equation}
\cos(\theta_{12})=\frac{4\pi}{3}\sum^1_{m=-1}Y^{m*}_1(\theta_1,\varphi_1)Y^m_1(\theta_2,\varphi_2)
\end{equation}
Consider first the S-states. The <math>r_{12}^0</math> terms are like the ss terms in a CI calculation. The
<math>r_{12}^2</math> terms bring in p-p type contributions, and the higher powers bring
in d-d, f-f etc. type terms. In general

\begin{equation}
P_l(\cos\theta_{12}) = \frac{4\pi}{2l+1}\sum^l_{m=-l}{Y^{m}_l}^*(\theta_1,\varphi_1)Y^m_l(\theta_2,\varphi_2)\nonumber
\end{equation}

For P-states, one would have similarly

\begin{array}{lr}
r_{12}^0 & (sp)P\nonumber\\
r_{12}^2 & (pd)P\nonumber\\
r_{12}^4 & (df)P\nonumber\\
\vdots & \vdots\nonumber
\end{array}

For D-states

\begin{array}{lr}
r_{12}^0 & (sd)D & (pp^\prime)D\nonumber\\
r_{12}^2 & (pf)D & (dd^\prime)D\nonumber\\
r_{12}^4 & (dg)D & (ff^\prime)D\nonumber\\
\vdots & \vdots & \vdots\nonumber
\end{array}

In this case, since there are two "lowest-order" couplings to form a
D-state, both must be present in the basis set, i.e.

\begin{equation}
\Psi(r_2,r_2) = \sum c_{ijk}r_1^ir_2^{j+2}r_{12}^ke^{-\alpha r_1-\beta
r_2}\mathcal{Y}^M_{022}(\hat{r}_1,\hat{r}_2)
+\sum d_{ijk}r_1^{i+1}r_2^{j+1}r_{12^k}e^{-\alpha^\prime r_1 - \beta^\prime
r_2}\mathcal{Y}^M_{112}(\hat{r}_1,\hat{r}_2)\nonumber
\end{equation}

For F-states, one would need <math>(sf)F</math> and <math>(pd)F</math> terms.

For G-states, one would need <math>(sg)G</math>, <math>(pf)G</math> and <math>(dd^\prime)G</math> terms.

Completeness of the radial functions can be proven by considering the
Sturm-Liouville problem

\begin{equation}
\left(-\frac{1}{2}\nabla^2-\frac{\lambda}{r_s}-E\right)\psi({\bf r}) = 0\nonumber
\end{equation}

or

\begin{equation}
\left(-\frac{1}{2}\frac{1}{r^2}\frac{\partial}{\partial r}\left({r^2}\frac{\partial}{\partial r}\right) -
\frac{l\left(l+1\right)}{2r^2} - \frac{\lambda}{r} - E\right)u(r) = 0.\nonumber
\end{equation}

For fixed E and variable <math>\lambda</math> (nuclear charge).

The eigenvalues are <math>\lambda_n = (E/E_n)^{1/2}</math>, where <math>E_n =- \frac{1}{2n^2}</math>

[[File:Pic2.jpg|500px]]

and the eigenfunctions are

\begin{equation}
u_{nl}(r) =
\frac{1}{(2l+1)!}\left(\frac{(n+l)!}{(n-l-1)2!}\right)^{1/2}(2\alpha)^{3/2}e^{-\alpha
r}\nonumber
\times \ \left( 2 \alpha r \right)^l\,_{1}F_{1}\left(-n+l+1,2l+2; 2 \alpha r \right)
\end{equation}

with <math>\alpha = (-2E)^{1/2}</math> and <math>n\geq l+1</math>. The confluent hypergeometric function <math> _1F_1(a,b,;z)</math> then denotes a finite polynomial since <math>a=-n+l+1</math> is a negative integer or zero.

Unlike the hydrogen spectrum, which has both a discrete part for <math>E<0</math> and a
continuous part for <math>E>0</math>, this forms an entirely discrete set of finite
polynomials, called Sturmian functions. They are orthogonal with respect to
the potential, i.e.

\begin{equation}
\int^\infty_0 r^2dr \left( u_{n^\prime l}(r)\frac{1}{r}u_{nl}(r) \right) = \delta_{n,n^\prime}\nonumber
\end{equation}

Since they become complete in the limit <math>n\rightarrow\infty</math>, this assures the
completeness of the variational basis set.

[See also B Klahn and W.A. Bingel Theo. Chim. Acta (Berlin) 44, 9 and 27
(1977)].

==Solutions of the Eigenvalue Problem==

For convenience, write

\[
\Psi({\bf r}_1,{\bf r}_2) = \sum^N_{m=1}c_m\varphi_m\nonumber
\]

where <math>m</math> represents the <math>m</math>'th combination of <math>i,j,k</math> and

\begin{equation}
\varphi_{ijk}=r_1^ir_2^jr_{12}^ke^{-\alpha r_1 - \beta
r_2}\mathcal{Y}^M_{l_1,l_2,L}(\hat{r}_1,\hat{r}_2) \pm \mbox{exchange}\nonumber
\end{equation}

\begin{align}

&
\left( \begin{array}{lr}
\cos(\theta) & \sin(\theta)\\
-\sin(\theta) & \cos(\theta)
\end{array} \right)
\left( \begin{array}{lr}
H_{11} & H_{12} \\
H_{12} & H_{22}
\end{array}\right)
\left( \begin{array}{lr}
\cos(\theta) & -\sin(\theta)\\
\sin(\theta) & \cos(\theta)
\end{array} \right)
\\[5pt]
= & \left(\begin{array}{lr}
cH_{11}+sH_{12} & cH_{12} + sH_{22}\\
-sH_{11} + cH_{12} & -sH_{12} + cH_{22}
\end{array}\right)
\left( \begin{array}{cc}
c & -s \\
s & c
\end{array}\right)
\\[5pt]
= & \left(\begin{array}{lr}
c^2H_{11}+s^2H_{22} + 2csH_{12} & (c^2-s^2)H_{12}+cs(H_{22}-H_{11})\\
(c^2-s^2)H_{12}+cs(H_{22}-H_{11}) & s^2H_{11}+c^2H_{22}-2csH_{12}
\end{array}\right)
\end{align}

Therefore

\begin{equation}
\left[\cos^2(\theta)-\sin^2(\theta)\right]H_{12} =
\cos(\theta)\sin(\theta)(H_{11}-H_{22})\nonumber
\end{equation}

and

\begin{equation}
\tan(2\theta) = \frac{2H_{12}}{H_{11}-H_{22}}\nonumber
\end{equation}

i.e.

\begin{equation}
\cos(\theta)=\left(\frac{r+\omega}{2r}\right)^{1/2}\nonumber
\end{equation}

\begin{equation}
\sin(\theta)=-\mbox{sgn}(H_{12})\left(\frac{r-\omega}{2r}\right)^{1/2}\nonumber
\end{equation}

where

\begin{equation}
\omega = H_{22}-H_{11}\nonumber
\end{equation}

\begin{equation}
r=\left(\omega^2+4H_{12}^2\right)^{1/2}\nonumber
\end{equation}

\begin{equation}
E_1=\frac{1}{2}\left(H_{11}+H_{22}-r\right)\nonumber
\end{equation}

\begin{equation}
E_2=\frac{1}{2}\left(H_{11}+H_{22}+r\right)\nonumber
\end{equation}

===Brute Force Method===

- Gives all the eigenvalues and eigenvectors, but it is slow

- First orthonormalize the <math>N</math>-dimensional basis set; i.e. form linear combinations

\begin{equation}
\Phi_m = \sum_{n=1}^N\varphi_nR_{nm}\nonumber
\end{equation}

such that
\begin{equation}
\langle \Phi_m | \Phi_n \rangle = \delta_{m,n}
\end{equation}

This can be done by finding an orthogonal tranformation T such that

\begin{equation}
T^TOT = I = \left(
\begin{array}{ccccc}
I_1 & 0 & 0 & \ldots & 0 \\
0 & I_2 & 0 & \ldots & 0 \\
0 & 0 & I_3 & \ldots & 0 \\
\vdots & \vdots & \vdots & \ddots & \vdots \\
0 & 0 & 0 & \ldots & I_N
\end{array}\right);
\end{equation}

\begin{equation}
O_{mn} = \langle \varphi_m | \varphi_n \rangle \nonumber
\end{equation}

and then applying a scale change matrix

\begin{equation}
S = \left(\begin{array}{ccccc}
\frac{1}{I_1^{1/2}} & 0 & 0 & \ldots & 0\\
0 & \frac{1}{I_2^{1/2}} & 0 & \ldots & 0 \\
0 & 0 & \frac{1}{I_3^{1/2}} & \ldots & 0 \\
\vdots & \vdots & \vdots & \ddots & \vdots \\
0 & 0 & 0 & \ldots & I_N^{1/2}
\end{array}\right)= S^T\nonumber
\end{equation}

Then

\begin{equation}
S^TT^TOTS = 1
\end{equation}

i.e.

\begin{equation}
R^TOR = 1
\end{equation}

with <math>R=TS</math>.

If H is the matrix with elements <math>H_{mn} = \langle \varphi_m | \varphi_n \rangle </math>, then H
expressed in the <math>\Phi_m</math> basis set is

\begin{equation}
H^\prime = R^THR.\nonumber
\end{equation}

We next diagonalize <math>H^\prime</math> by finding an orthogonal transformation W such
that

\begin{equation}
W^T H^\prime W = \lambda = \left(
\begin{array}{cccc}
\lambda_1 & 0 & \ldots & 0 \\
0 & \lambda_2 & \ldots & 0 \\
\vdots & \vdots & \ddots & \vdots \\
0 & 0 & \ldots & \lambda_N
\end{array}\right)\nonumber
\end{equation}

The q'th eigenvector is

\begin{equation}
\Psi^{(q)} = \sum_{n=1}^N\Phi_n W_{n,q} = \sum_{n,n^\prime} \varphi_{n^\prime} R_{n^\prime ,n} W_{n,q} \nonumber
\end{equation}

i.e.

\begin{equation}
c_{n^\prime}^{(q)} = \sum_{n=1}^N R_{n^\prime n} W_{n,q}
\end{equation}

===The Power Method===

- Based on the observation that if H has one eigenvalue, <math>\lambda_M</math>, much bigger than all the rest, and <math>\chi = \left(
\begin{array}{c}a_1\\a_2\\\vdots\end{array}\right)</math>
is an arbitrary starting vector, then <math>\chi = \sum_{q=1}^N x_q\Psi^{(q)}</math>.

\begin{equation}
(H)^n\chi = \sum_{q=1}^N x_q \lambda^n_q\Psi^{(q)}\nonumber\rightarrow x_M\lambda_M^n\Psi^{(M)}\nonumber
\end{equation}

provided <math>x_M\neq 0</math>.

To pick out the eigenvector corresponding to any eigenvalue, write the original problem in the form

\begin{align}
H\Psi & = \lambda O \Psi \nonumber \\
\left( H - \lambda_q O \right) \Psi & = \left( \lambda - \lambda_q \right) O \Psi \nonumber
\end{align}

Therefore,

\begin{equation}
G \Psi = \frac{1}{\lambda - \lambda_q } \Psi \nonumber \\
\end{equation}

where <math>G=(H-\lambda_qO)^{-1}O</math> with eigenvalues <math>\frac{1}{\lambda_n - \lambda_q}</math>.

By picking <math>\lambda_q</math> close to any one of the <math>\lambda_n</math>, say
<math>\lambda_{n^\prime}</math>, then <math>\frac{1}{\lambda_n-\lambda_q}</math> is much larger for
<math>n=n^\prime</math> than for any other value. The sequence is then

\begin{array}{c}
\chi_1=G\chi\nonumber\\
\chi_2=G\chi_1\nonumber\\
\chi_3=G\chi_2\nonumber\\
\vdots\nonumber
\end{array}

until the ratios of components in <math>\chi_n</math> stop changing.

- To avoid matrix inversion and multiplication, note that the sequence is equivalent to
\begin{equation}
F \chi_n = \left( \lambda - \lambda_q \right) O \chi_{n-1} \nonumber \\
\end{equation}
where <math>F = H-\lambda_qO</math>. The factor of <math>\left( \lambda - \lambda_q \right)</math> can be
dropped because this only affects the normalization of <math>\chi_n</math>. To find <math>\chi_n</math>, solve
\begin{equation}
F \chi_n = O \chi_{n-1} \nonumber
\end{equation}
(N equations in N unknowns). Then
\begin{equation}
\lambda = \frac{ \langle \chi_n | H | \chi_n \rangle}{\langle \chi_n | \chi_n \rangle} \nonumber
\end{equation}

==Matrix Elements of H==

\begin{equation}
H=-\frac{1}{2}\nabla^2_1 -\frac{1}{2}\nabla^2_2 - \frac{1}{r_1} - \frac{1}{r_2} +\frac{Z^{-1}}{r_{12}}\nonumber
\end{equation}

Taking <math>r_1, r_2</math> and <math>r_{12}</math> as independent variables,

\begin{equation}
\nabla_1^2 = \frac{1}{r_1^2}\frac{\partial}{\partial r_1}\left(r_1^2\frac{\partial}{\partial r_1}\right)+ \frac{1}{r_{12}^2}\frac{\partial}{\partial r_{12}}
\left(r_{12}^2\frac{\partial}{\partial r_{12}}\right)\nonumber -\frac{l_1(l_1+1)}{r_1^2}+2(r_1 - r_2\cos(\theta_{12}))\frac{1}{r_{12}}\frac{\partial^2}{\partial r_1 \partial r_{12}} \nonumber - 2(\nabla_1^Y \cdot {\bf r}_2)\frac{1}{r_{12}}\frac{\partial}{\partial r_{12}}\nonumber
\end{equation}

where <math>\nabla_1^Y</math> acts only on the spherical harmonic part of the wave function and the diagram

[[File:Pic3.jpg|400px]]

defines the complete set of 6 independent variables is <math>r_1</math>, <math>r_2</math>, <math>r_{12}</math>, <math>\theta_1</math>, <math>\varphi_1</math>, <math>\chi</math>.

If <math>r_{12}</math> were not an independent variable, then one could take the volume element to be

\begin{equation}
d\tau = r_1^2dr_1\sin(\theta_1)d\theta_1d\varphi_1r_2^2dr_2\sin(\theta_2)d\theta_2d\varphi_2.\nonumber
\end{equation}

However, <math>\theta_2</math> and <math>\varphi_2</math> are no longer independent variables. To eliminate them, take the point <math>{\bf r}_1</math> as the
origin of a new polar co-ordinate system, and write

\begin{equation}
d\tau=-r_1^2dr_1\sin(\theta_1)d\theta_1d\varphi_1r_{12}^2dr_{12}\sin(\psi)d\psi d\chi\nonumber
\end{equation}

and use

\begin{equation}
r_2^2=r_1^2+r_{12}^2 +2r_1r_{12}\cos(\psi)
\end{equation}

Then for fixed <math>r_1</math> and <math>r_{12}</math>,

\begin{equation}
2r_2dr_2 = -2r_1r_{12}\sin(\psi)d\psi
\end{equation}

Thus

\begin{equation}
d\tau= r_1dr_1r_2dr_2r_{12}dr_{12}\sin(\theta_1)d\theta_1d\varphi_1d\chi
\end{equation}

The basic type of integral to be calculated is

\begin{align}
I \left( l_1, m_1, l_2, m_2; R \right) & = \int \sin \left( \theta_1 \right) d\theta_1 d\varphi_1 d\chi Y^{m_1}_{l_1} \left( \theta_1, \varphi_1 \right)^{*} Y^{m_2}_{l_2} \left( \theta_2, \varphi_2 \right) \\
& \times \int r_1dr_1r_2dr_2r_{12}dr_{12}R \left( r_1, r_2, r_{12} \right)
\end{align}

Consider first the angular integral. <math>Y^{m_2}_{l_2} \left( \theta_2, \varphi_2 \right)</math> can be expressed in terms of the independent variables <math>\theta_1, \varphi_1, \chi</math> by use of the rotation matrix relation
\begin{equation}
Y^{m_2}_{l_2}( \theta_2, \varphi_2) = \sum_m \mathcal{D}^{(l_2)}_{m_2,m}(\varphi_1,\theta_1,\chi)^*Y^m_{l_2}( \theta_{12}, \varphi)
\end{equation}
where <math>\theta, \varphi</math> are the polar angles of <math>{\bf r}_2</math> relative to <math>{\bf r}_1</math>. The angular integral is then

\begin{align}
I_{ang} & = \int^{2 \pi}_{0} d\chi \int^{2\pi}_{0} d\varphi_1 \int^{\pi}_{0} \sin \left( \theta_1 \right) d\theta_{1} Y^{m_1}_{l_1} \left( \theta_1, \varphi_1 \right)^{*} \\
& \times \sum_m \mathcal{D}^{(l_2)}_{m_2,m} \left( \varphi_1,\theta_1,\chi \right)^{*} Y^{m}_{l_2} \left( \theta_{12},\varphi \right) \\
\end{align}

Use

\begin{equation}
Y^{m_1}_{l_1} \left( \theta_1, \varphi_1 \right)^{*} = \sqrt{\frac{2l_1+1}{4\pi}}\mathcal{D}^{(l_1)}_{m_1,0} \left( \varphi_1, \theta_1, \chi \right)
\end{equation}

together with the orthogonality property of the rotation matrices (Brink and Satchler, p 147)

\begin{align}
& \int \mathcal{D}^{(j)*}_{m,m^\prime}\mathcal{D}^{(J)}_{M,M^\prime}\sin(\theta_1)d\theta_1d\varphi_1d\chi \\
& = \frac{8\pi^2}{2j+1}\delta_{jJ}\delta_{mM}\delta_{m^\prime M^\prime}\nonumber
\end{align}

to obtain

\begin{align}
I_{ang} & = \sqrt{ \frac{2l_1+1}{4\pi}} \frac{8\pi^2}{2l_1+1} \delta_{l_1,l_2} \delta_{m_1,m_2} Y^{0}_{l_2} \left( \theta_{12}, \varphi \right) \\
& = 2 \pi \delta_{l_1,l_2} \delta_{m_1,m_2} P_{l_2} \left( \cos \theta_{12} \right)
\end{align}

since

\begin{equation}
Y^0_{l_2} \left( \theta_{12}, \varphi \right) = \sqrt{\frac{2l_1+1}{4\pi}} P_{l_2} \left( \cos \left( \theta_{12} \right) \right)
\end{equation}

Note that <math>P_{l_2} \left( \cos \theta \right)</math> is just a short hand expression for a radial function because

\begin{equation}
\cos \theta_{12} = \frac{r_1^2 + r_2^2 - r_{12}^2}{2r_1r_2}
\end{equation}

The original integral is thus

\begin{equation}
I \left( l_1, m_1, l_2, m_2; R \right) = 2 \pi \delta_{l_1,l_2} \delta_{m_1,m_2} \int^{\infty}_{0} r_{1} dr_{1} \int^{\infty}_{0} r_{2} dr_{2} \int^{r_{1} + r_{2}}_{|r_{1} - r_{2}|} r_{12} dr_{12} R \left( r_{1}, r_{2}, r_{12} \right) P_{l_2} \left( \cos \theta_{12} \right)
\end{equation}

where again

\begin{equation}
\cos \theta_{12} = \frac{r_{1}^{2} +r^{2}_{2} - r_{12}^{2}}{2 r_{1} r_{2}}
\end{equation}

is a purely radial function.

The above would become quite complicated for large <math>l_2</math> because <math>P_{l_2}(\cos \theta_{12})</math> contains terms up to <math>\left( \cos \theta_{12} \right)^{l_2}</math>. However, recursion relations exist which allow any integral containing <math>P_l \left( \cos \theta_{12} \right)</math> in terms of those containing just <math>P_0 \left( \cos \theta_{12} \right) = 1</math> and <math>P_1 \left( \cos \theta_{12} \right) = \cos \theta_{12}</math>.

==Radial Integrals and Recursion Relations==

The basic radial integral is [see G.W.F. Drake, Phys. Rev. A 18, 820 (1978)]

\begin{align}
I_0 \left( a, b, c \right) & =
\int^{\infty}_{0} r_{1} dr_{1} \int^{\infty}_{r_{1}} r_{2} dr_{2} \int^{r_{1}+r_{2}}_{r_{2} - r_{1}} r_{12} dr_{12} r_{1}^{a} r_{2}^{b} r_{12}^{c} e^{-\alpha r_{1} - \beta r_{2}} + \int^{\infty}_{0} r_{2} dr_{2} \int^{\infty}_{r_{2}} r_{1} dr_{1} \int^{r_{1} + r_{2}}_{r_{1} - r_{2}} r_{12} dr_{12} r_{1}^{a} r_{2}^{b} r_{12}^{c} e^{- \alpha r_{1} - \beta r_{2}} \\
& = \frac{2}{c+2} \sum^{[(c+1)/2]}_{i=0}
\left(\begin{array}{c}
c+2 \\
2i+1
\end{array} \right)
\left\{ \frac{q!}{\beta^{q+1} \left( \alpha + \beta \right)^{p+1}} \sum_{j=0}^{q} \frac{\left( p + j \right)!}{j!} \left( \frac{\beta}{\alpha + \beta} \right)^j + \frac{q^{\prime}!}{\alpha^{q^{\prime}+1} \left( \alpha + \beta \right)^{p^{\prime}+1}} \sum^{q^{\prime}}_{j=0} \frac{\left( p^{\prime} + j \right)!}{j!} \left( \frac{\alpha}{\alpha + \beta} \right)^j \right\}
\end{align}

where

\begin{array}{ll}
p = a + 2i + 2 & p^{\prime} = b + 2i + 2 \\
q = b + c - 2i + 2 & q^{\prime} = a + c - 2i + 2
\end{array}

The above applies for <math> a, b \geq -2, c\geq -1</math>.
<math>[x]</math> means "greatest integer in" <math>x</math>.

Then

\begin{align}
I_1 \left( a, b, c \right) & = \int d\tau_{r} r_{1}^{a} r_{2}^{b} r_{12}^{c} e^{ -\alpha r_{1} - \beta r_{2}} P_{1}\left( \cos \theta \right) \\
& = \frac{1}{2} \left( I_{0} \left( a + 1, b - 1, c \right) + I_{0} \left( a - 1, b + 1, c \right) - I_{0} \left( a - 1, b - 1, c + 2 \right) \right)
\end{align}

===The Radial Recursion Relation===

Recall that the full integral is

\begin{equation}
I(l_1m_1,l_2m_2;R)=2\pi\delta_{l_1,l_2}\delta_{m_1,m_2}I_{l_2}(R)
\end{equation}

where

\begin{equation}
I_{l_2}(R) = \int d\tau_r R(r_1,r_2,r_{12})P_{l_2}(\cos \theta_{12})
\end{equation}

To obtain the recursion relation, use

\begin{equation}
P_l(x) = \frac{[P^{\prime}_{l+1}(x)-P^{\prime}_{l-1}(x)]}{2l+1}
\end{equation}

with

\begin{equation}
P_{l+1}^\prime (x) = \frac{d}{dx}P_{l+1}(x)
\end{equation}

Here <math>x=\cos\theta_{12}</math> and

\begin{align}
\frac{d}{d \cos \theta_{12}} & = \frac{dr_{12}}{d \cos \theta_{12}} \frac{d}{dr_{12}} \\
& = -\frac{r_{1} r_{2}}{r_{12}} \frac{d}{dr_{12}}
\end{align}

Then,

\begin{equation}
I_{l}\left( R \right) = - \int d\tau_r R \frac{r_{1} r_{2}}{r_{12}} \frac{d}{dr_{12}} \frac{ \left[ P_{l+1} \left( \cos \theta_{12} \right) - P_{l-1} \left( \cos \theta_{12} \right) \right]}{2l+1}
\end{equation}

The <math>r_{12}</math> part of the integral is

\begin{align}
& \int^{r_{1} + r_{2}}_{|r_{1} - r_{2}|} r_{12} dr_{12} R \frac{r_{1} r_{2}}{r_{12}} \frac{d}{dr_{12}} \left[P_{l+1}-P_{l-1} \right] \\
& = R r_{1} r_{2} \left[ P_{l+1} - P_{l-1} \right]^{r_{1} + r_{2}}_{|r_{1} - r_{2}|} - \int^{r_{1} + r_{2}}_{|r_{1} - r_{2}|} r_{12} dr_{12} \left( \frac{d}{dr_{12}} R \right) \frac{r_{1} r_{2}}{r_{12}} \frac{ \left[ P_{l+1} - P_{l-1} \right]}{2l+1}
\end{align}

The integrated term vanishes because

\begin{equation}
\cos \theta_{12} = \frac{r_{1}^{2} + r_{2}^{2} - r_{12}^{2}}{2 r_{1} r_{2}}
\begin{array}{lll}
& = -1 & {\rm when} & r_{12}^{2} = \left( r_{1} + r_{2} \right)^{2} \\
& = 1 & {\rm when} & r_{12}^{2} = \left( r_{1} - r_{2} \right)^{2}
\end{array}
\end{equation}

and <math>P_l(1)=1</math>, <math>P_{l}(-1)=(-1)^{l}</math>.

Thus,

\begin{equation}
I_{l+1}\left(\frac{r_1r_2}{r_{12}}R^\prime\right) = \left( 2l+1 \right) I_{l}(R) - I_{l-1} \left( \frac{r_{1} r_{2}}{r_{12}} R^{\prime} \right)
\end{equation}

If

\begin{equation}
R = r_{1}^{a-1} r_{2}^{b-1} r_{12}^{c+2} e^{-\alpha r_{1} - \beta r_{2}}
\end{equation}

then [G.W.F. Drake, Phys. Rev. A 18, 820 (1978)]

\begin{equation}
I_{l+1}(r_1^ar_2^br_{12}^c) =
\frac{2l+1}{c+2} I_{l} \left( r_{1}^{a-1} r_{2}^{b-2} r_{12}^{c+2} \right)+I_{l-1} \left( r_{1}^{a} r_{2}^{b} r_{12}^{c} \right)
\end{equation}

For the case <math>c=-2</math>, take

\begin{equation}
R = r_{1}^{a-1} r_{2}^{b-1} \ln r_{12} e^{ -\alpha r_{1} -\beta r_{2}}
\end{equation}

Then

\begin{equation}
I_{l+1} \left( r_{1}^{a} r_{2}^{b} r_{12}^{c} \right) = I_{l} \left( r_{1}^{a-1} r_{2}^{b-1} \ln r_{12} \right) \left( 2l + 1 \right) - I_{l-1} \left( r_{1}^{a} r_{2}^{b} r_{12}^{-2} \right)
\end{equation}

This allows all <math>I_{l}</math> integrals to be calculated from tables of <math>I_0</math> and
<math>I_1</math> integrals.

===The General Integral===

The above results for the angular and radial integrals can now be combined into a general formula for integrals of the type

\begin{equation}
I = \int \int d{\bf r}_{1} d{\bf r}_{2} R_{1} \mathcal{Y}^{M^{\prime}}_{l^{\prime}_{1} l^{\prime}_{2} L^{\prime}} \left( \hat{r}_{1}, \hat{r}_{2} \right) T_{k_{1} k_{2} K}^{Q} \left( {\bf r}_{1},{\bf r}_{2} \right) R_{2} \mathcal{Y}_{l_{1} l_{2} L}^{M} \left( \hat{r}_{1}, \hat{r}_{2} \right)
\end{equation}

where

\begin{equation}
\mathcal{Y}^{M}_{l_{1} l_{2} L} \left( \hat{r}_{1}, \hat{r}_{2} \right) = \sum_{m_1,m_2} \langle l_{1} l_{2} m_{1} m_{2} | LM \rangle Y^{m_{1}}_{l_{1}} \left( \hat{r}_{1} \right) Y^{m_{2}}_{l_{2}} \left( \hat{r}_{2} \right)
\end{equation}

and

\begin{equation}
T^Q_{k_{1} k_{2} K} \left( {\bf r}_{1}, {\bf r}_{2} \right) = \sum_{q_{1}, q_{2}} \langle k_{1} k_{2} q_{1} q_{2} | KQ \rangle Y_{k_{1}}^{q_{1}} \left( \hat{r}_{1} \right) Y^{q_{2}}_{k_{2}} \left( \hat{r}_{2} \right)
\end{equation}

The basic idea is to make repeated use of the formula

\begin{equation}
Y^{m_{1}}_{l_{1}} \left( \hat{r}_{1} \right) Y^{m_{2}}_{l_{2}} \left( \hat{r}_{1} \right) = \sum_{l m} \left( \frac{\left( 2l_{1}+1 \right) \left(2l_{2}+1 \right) \left(2l+1 \right)}{4 \pi} \right)^{1/2}
\times \left( \begin{array}{ccc}
l_1 & l_2 & l \\
m_1 & m_2 & m
\end{array} \right) \left(\begin{array}{ccc}
l_1 & l_2 & l \\
0 & 0 & 0
\end{array} \right)
Y^{m*}_{l} \left( \hat{r}_{1} \right)
\end{equation}

where

\begin{equation}
Y^{m*}(\hat{r}) = (-1)^mY_l^{-m}(\hat{r})
\end{equation}

and

\begin{equation}
\left(\begin{array}{ccc}
l_1 & l_2 & l \\
m_1 & m_2 & m
\end{array} \right) = \frac{\left( -1 \right)^{l_{1} - l_{2} - m}}{\left( 2l + 1 \right)^{1/2}} \left( l_{1} l_{2} m_{1} m_{2} | l, -m \right)
\end{equation}

is a 3-<math>j</math> symbol. In particular, write

\begin{array}{cc}
Y^{m_1^\prime}_{l_1^\prime}(\hat{r}_1)^{*} \underbrace{Y_{k_{1}}^{q_{1}} \left( \hat{r}_{1} \right) Y^{m_{1}}_{l_{1}} \left( \hat{r}_{1} \right)} & = & \sum_{\Lambda M}(\ldots)Y^{M}_\Lambda(\hat{r}_1)\\
\ \ \ \ \ \ \ \sum_{\lambda_1\mu_1}Y_{\lambda_1}^{\mu_1}(\hat{r}_1) & & \\
& & \\
Y^{m_{2}^{\prime}}_{l_{2}^{\prime}} \left( \hat{r}_{2} \right) \underbrace{Y^{q_{2}}_{k_{2}} \left( \hat{r}_{2} \right) Y_{l_{2}}^{m_{2}} \left( \hat{r}_{2} \right)} & = & \sum_{\Lambda^{\prime} M^{\prime}} \left( \ldots \right)Y_{\Lambda^{\prime}}^{M^{\prime}*} \left( \hat{r}_{2} \right) \\
\ \ \ \ \ \ \ \ \ \sum_{\lambda_{2} \mu_{2}}Y_{\lambda_{2}}^{\mu_{2}} \left( \hat{r}_{2} \right) & &
\end{array}

The angular integral then gives a factor of <math>2 \pi \delta_{\Lambda, \Lambda^{\prime}} \delta_{M, M^{\prime}} P_{\Lambda} \left( \cos \theta_{12} \right)</math>. The total integral therefore reduces to the form

\begin{equation}
I = \sum_{\Lambda} C_{\Lambda} I_{\Lambda} \left( R_{1} R_{2} \right)
\end{equation}

where <math>C_{\Lambda} = \sum_{\lambda_{1}, \lambda_{2}} C_{ \lambda_{1}, \lambda_{2}, \Lambda}</math>. For further details and derivations, including graphical representations, see G.W.F. Drake, Phys. Rev. A <math>{\bf 18}</math>, 820 (1978).

==Graphical Representation==

[[File:GraphRep1.jpg|500px]]

==Matrix Elements of H==
Recall that

\begin{equation}
H = -\frac{1}{2} \nabla^{2}_{1} - \frac{1}{2} \nabla^{2}_{2} - \frac{1}{r_1} - \frac{1}{r_2} + \frac{Z^{-1}}{r_{12}}
\end{equation}

Consider matrix elements of

\begin{equation}
\nabla^2_1 = \frac{1}{r_1^2} \frac{\partial}{\partial r_1}\left(r_1^2\frac{\partial}{\partial r_1} \right) + \frac{1}{r^2}\frac{\partial}{\partial r}\left(r^2\frac{\partial}{\partial r}\right) - \frac{(\vec{l}_1^Y)^2}{r_1^2} + \frac{2 \left( r_1 - r_2 \cos \theta \right)}{r} \frac{\partial^2}{\partial r_1 \partial r} - 2 \left( \nabla^Y_1 \cdot {\bf r}_2 \right) \frac{1}{r} \frac{\partial}{\partial r}
\end{equation}

where <math>r \equiv r_{12}</math> and <math>\cos\theta \equiv \cos\theta_{12}</math>. Also in what follows, define <math>\hat{\nabla}^Y_1 \equiv r_1\nabla^Y_1</math> where <math>\nabla^Y</math> operates only on the spherical harmonic part of the wave function, and similarly for <math>\vec{l}_1^Y</math>.

A general matrix element is

\begin{equation}
\langle r_1^{a^\prime} r_2^{b^\prime} r_{12}^{c^\prime} e^{-\alpha^\prime r_1 - \beta^\prime r_2} \mathcal{Y}^{M^\prime}_{l^\prime_1 l^\prime_2 L^\prime} \left( \hat{r}_1, \hat{r}_2 \right) \left| \nabla^2_1 \right| r_1^a r_2^b r_{12}^c e^{-\alpha r_1 - \beta r_2} \mathcal{Y}^M_{l_1 l_2 L} \left( \hat{r}_1, \hat{r}_2 \right) \rangle
\end{equation}

Since <math>\nabla^2_1</math> is rotationally invariant, this vanishes unless <math>L=L^\prime</math>, <math>M=M^\prime</math>. Also <math>\nabla_1^2</math> is Hermitian, so that the result must be the same whether it operates to the right or left, even though the results look very different. In fact, requiring the results to be the same yields some interesting and useful integral identities as follows:

Put

\begin{equation}
\mathcal{F} = F \mathcal{Y}^M_{l_1 l_2 L} \left( \hat{r}_1, \hat{r_2} \right)
\end{equation}

and

\begin{equation}
F = r_1^a r_2^b r_{12}^c e^{-\alpha r_1 - \beta r_2}
\end{equation}

Then

\begin{equation}
\nabla^2_1 \mathcal{F} = \left\{ \frac{1}{r_1^2} \left[ a \left( a + 1 \right) - l_1 \left( l_1 + 1 \right) \right] + \frac{c \left( c + 1 \right)}{r^2} + \alpha^2\frac{2\alpha \left( a + 1 \right)}{r_1} + \frac{2 \left(r_1 - r_2 \cos \theta \right) }{r_1r^2}c \left[ a - \alpha r_1 \right] - \frac{2c}{r^2} \left( \hat{\nabla}^Y_1 \cdot \hat{r}_2 \right) \frac{r_2}{r_1} \right\} \mathcal{F}
\end{equation}

and

\begin{align}
\langle \mathcal{F}^\prime \left| \nabla^2_1 \right| \mathcal{F} \rangle & = \sum_\Lambda \int d\tau_r F^\prime C_\Lambda \left( 1 \right) P_\Lambda \left( \cos \theta \right) \times \left\{ \frac{1}{r_1^2} \left[ a \left( a + 1 \right) - l_1 \left( l_1 + 1 \right) \right] -\frac{2 \alpha \left( a + 1 \right)}{r_1} + \frac{c \left( c + 1 \right)}{r^2} +\alpha^2 + \frac{2 \left( r_2 - r_2 \cos \theta \right)}{r_1r^2}c \left( a - \alpha r_1 \right) \right\} F \\
& + \sum_\Lambda \int d\tau_r F^\prime C_\Lambda \left( \hat{\nabla}^Y_1 \cdot \hat{r}_2 \right) P_\Lambda \left( \cos \theta \right) \left( \frac{-2c r_2}{r_1 r^2} \right) F
\end{align}

where

\begin{equation}
\int d\tau_r = \int^\infty_0 r_1 dr_1 \int^\infty_0 r_2 dr_2 \int^{r_1+r_2}_{\left| r_1 - r_2 \right|} r dr
\end{equation}

For brevity, let the sum over <math>\Lambda</math> and the radial integrations be understood, and let <math>\left( \nabla_1^2 \right)</math> stand for the terms which appear in the integrand. Then operating to the right gives

\begin{equation}
(\nabla_1^2)_R = \frac{1}{r_1^2} \left[ a \left( a + 1 \right) - l_1 \left( l_1 + 1 \right) \right] + \frac{c \left( c + 1 \right)}{r^2} + \alpha^2 - \frac{2 \alpha \left( a + 1 \right)}{r_1}
+ \frac{2 \left( r_1 - r_2 \cos \theta \right)}{r_1 r^2}c \left( a - \alpha r_1 \right) - \frac{2c r_2}{r_1 r^2} \left( \hat{\nabla}^Y_1 \cdot \hat{r}_2 \right)
\end{equation}

Operating to the left gives

\begin{equation}
(\nabla_1^2)_L = \frac{1}{r_1^2} \left[a^\prime \left( a^\prime + 1 \right) - l_1^\prime \left( l_1^\prime + 1 \right) \right] + \frac{c^\prime \left( c^\prime + 1 \right)}{r^2} + \alpha^{\prime 2} -\frac{2 \alpha^\prime \left( a^\prime + 1 \right)}{r_1}
+ \frac{2 \left( r_1 - r_2 \cos \theta \right)}{r_1 r^2} c^\prime \left( a^\prime - \alpha^\prime r_1 \right) - \frac{2 c^\prime r_2}{r_1 r^2} \left( \hat{\nabla}^{Y^\prime}_1 \cdot \hat{r}_2 \right)
\end{equation}

Now put

\begin{array}{cc}
a_+ = a+a^\prime, & \hat{\nabla}_1^+ = \hat{\nabla}_1^Y + \hat{\nabla}_1^{Y\prime} \\
a_{-} = a-a^\prime, & \hat{\nabla}_1^{-} = \hat{\nabla}_1^Y - \hat{\nabla}_1^{Y\prime}
\end{array}

etc., and substitute <math>a^\prime = a_+ - a</math>, <math>c^\prime = c_+ - c</math>, and <math>\alpha^\prime = \alpha_+ - \alpha</math> in <math>(\nabla^2_1)_L</math>. If <math>a_+</math>, <math>c_+</math> and <math>\alpha_+</math> are held fixed, then the equation

\begin{equation}
(\nabla^2_1)_R = (\nabla^2_1)_L
\end{equation}

must be true for arbitrary <math>a</math>, <math>c</math>, and <math>\alpha</math>. Their coefficients must thus vanish.

This yields the integral relations

\begin{equation}
\label{I}
\frac{\left( r_1 - r_2 \cos \theta \right)}{r_1 r^2} = \frac{1}{c_+} \left( \frac{- \left( a_+ + 1 \right)}{r_1^2} + \frac{\alpha_+}{r_1} \right) \tag{I}
\end{equation}

from the coefficient of <math>a</math> and

\begin{equation}
\label{II}
\frac{ \left( r_1 - r_2 \cos \theta \right) \left( a_+ - \alpha_+ r_1 \right)}{r_1 r^2} = \frac{r_2}{r_1 r^2}\left( \hat{r}_2 \cdot \hat{\nabla}_1^+ \right) - \frac{ \left( c_+ + 1 \right)}{r^2} \tag{II}
\end{equation}

from the coefficient of <math>c</math>. The coefficient of <math>\alpha</math> gives an equation equivalent to (I).

Furthermore, if can be show that (see problem)

\begin{align}
& \sum_\Lambda \int d\tau_r \frac{r^c}{r^2} C_\Lambda \left( \hat{r}_2 \cdot \hat{\nabla}_1^Y \right) P_\Lambda \left( \cos \theta \right) \\
= & \sum_\Lambda \int d\tau_r \frac{r^c}{c r_1 r_2}C_\Lambda \left( 1 \right)P_\Lambda \left( \cos \theta \right) \times \left( \frac{l_1^\prime \left( l_1^\prime + 1 \right) - l_1 \left( l_1 + 1 \right) -\Lambda \left( \Lambda + 1 \right)}{2} \right)
\end{align}

and similarly for <math>\left( \hat{r}_2 \cdot \hat{\nabla}^{Y\prime}_1 \right)</math> with <math>l_1</math> and <math>l_1^\prime</math> interchangeable, then it follows that

\begin{equation}
\frac{r^{c_+}}{r^2} \left( \hat{r}_2 \cdot \hat{\nabla}_1^+ \right) = - \frac{r^{c_+}}{c_+ r_1 r_2} \Lambda \left( \Lambda + 1 \right)
\end{equation}

and

\begin{equation}
\frac{r^{c_+}}{r^2} \left( \hat{r}_2 \cdot \hat{\nabla}_1^- \right) = \frac{r^{c_+}}{c_+ r_1 r_2} \left[ l_1^\prime \left( l_1^\prime + 1 \right) - l_1 \left( l_1 + 1 \right) \right]
\end{equation}

where equality applies after integration and summation over <math>\Lambda</math>.

Thus (II) becomes

\begin{equation}\label{IIa}
\frac{ \left( r_1 - r_2 \cos \theta \right) \left( a_+ - \alpha_+ r_1 \right)}{r_1 r^2} = - \frac{\Lambda \left( \Lambda + 1 \right)}{c_+ r^2_1} - \frac{ \left( c_+ + 1 \right)}{r^2} \tag{IIa}
\end{equation}

===Problem===

Prove the integral relation

\begin{align}
& \sum_\Lambda \int d\tau_r f \left( r_1, r_2 \right) \frac{1}{r} \left( \frac{d}{dr} q(r) \right) C_\Lambda \left( \hat{r}_2 \cdot \hat{\nabla}^Y_1 \right) P_\Lambda \left( \cos \theta \right) \\
= & \sum_\Lambda \int d\tau_r \frac{ f \left( r_1, r_2 \right)}{r_1, r_2} q(r) C_\Lambda \left( 1 \right) P_\Lambda \left( \cos \theta \right) \left(\frac{l_1^\prime \left( l_1^\prime + 1 \right) - l_1 \left( l_1 + 1 \right) - \Lambda \left( \Lambda + 1 \right)}{2}\right)
\end{align}

where the coefficients <math>C_\Lambda \left( 1 \right)</math> are the angular coefficients from the overlap integral

\begin{align}
\int & d\Omega\, \mathcal{Y}^{M*}_{l_1^\prime l_2^\prime L}(\hat{r}_1,\hat{r}_2)\, \mathcal{Y}^{M}_{l_1 l_2 L}(\hat{r}_1,\hat{r}_2)\\
= & \sum_\Lambda C_\Lambda(1)P_\Lambda(\cos\theta_{12}).
\end{align}

Hint: Use the fact that <math>l_1^2</math> is Hermitian so that
<math>
\int d\tau (l_1^2\mathcal{Y}^\prime)^*q(r)\mathcal{Y} = \int d\tau
\mathcal{Y}^{\prime *}l_1^2(q(r)\mathcal{Y})
</math>
with <math>\ \vec{l}_1= \frac{1}{i}\vec{r}_1\times\nabla_1</math>.

It is also useful to use
<math>\begin{eqnarray}\ \
\cos\theta\, P_L(\cos\theta) &=& \frac{1}{2L+1}[L\,P_{L-1}(\cos\theta) + (L+1)P_{L+1}(\cos\theta)]
\\
(\cos^2\theta-1)P_L(\cos\theta) &=&
\frac{L(L-1)}{(2L-1)(2L+1)}P_{L-2}(\cos\theta)\nonumber
-\frac{2(L^2+L-1)}{(2L-1)(2L+3)}P_L(\cos\theta) + \frac{(L+1)(L+2)}{(2L+1)(2L+3)}P_{L+2}(\cos\theta)\nonumber\\
&=& \frac{L(L-1)}{(2L-1)(2L+1)}[P_{L-2}(\cos\theta) - P_L(\cos\theta)]
+ \frac{(L+1)(L+2)}{(2L+1)(2L+3)}[P_{L+2}(\cos\theta) - P_L(\cos\theta)]
\end{eqnarray}</math>

together with a double application of the integral recursion relation
<math>
I_{L+1}\left(\frac{1}{r}\frac{d}{dr}q(r)\right) =
(2L+1)I_L\left(\frac{1}{r_1r_2}q(r)\right)+I_{L-1}\left(\frac{1}{r}\frac{d}{dr}q(r)\right)
</math>

Of course <math>l_1^2\mathcal{Y}^M_{l_1l_2L}(\hat{r}_1,\hat{r}_2)=l_1(l_1+1)\mathcal{Y}^M_{l_1l_2L}(\hat{r}_1,\hat{r}_2)</math>. Begin by expanding
<math>
l_1^2(g\mathcal{Y}) = \mathcal{Y}l_1^2g + g\,l_1^2\mathcal{Y} + 2(\vec{l}_1g)\cdot(\vec{l}_1\mathcal{Y})
</math>, and show that

<math>
\begin{eqnarray}\ \
(\vec{l}_1g)\cdot(\vec{l}_1\mathcal{Y}) & = & \frac{r_1^2}{r}\frac{dg}{dr}\vec{r}_2\cdot \nabla_1\mathcal{Y}\,\,\,\,\mbox{ and}\\
l_1^2g(r) &=& 2r_1r_2\cos\theta\frac{1}{r}\frac{dg}{dr} + r_1^2r_2^2(\cos^2\theta - 1)
\frac{1}{r}\frac{d}{dr}\left(\frac{1}{r}\frac{dg}{dr}\right).
\end{eqnarray}
</math>

The proof amounts to showing that the term <math>l_1^2g(r)</math> can be replaced by
<math>\Lambda(\Lambda+1)g(r)</math> after multiplying by <math>P_\Lambda(\cos\theta)</math>
and integrating by parts with respect to the radial integrations over <math>r_1</math>, <math>r_2</math> and
<math>r \equiv |\vec{r}_1 - \vec{r}_2|</math>. Remember that <math>\cos\theta</math> is just a short-hand notation for the radial function <math>(r_1^2 + r_2^2 - r^2)/(2r_1r_2)</math>.

==General Hermitian Property==

Each combination of terms of the form
\begin{equation}
<f>_R = a^2f_1+af_2+abf_3+b^2f_4+bf_5+af_6\nabla_1^Y+bf_7\nabla^Y_2+f_8(Y)
\end{equation}
can be rewritten
\begin{equation}
<f>_L= (a_+-a)^2f_1+(a_+-a)f_2 + (a_+-a)(b_+-b)f_3+(b_+-b)^2f_4
+(b_+-b)f_5+(a_+-a)f_6\nabla^{Y\prime}_1+(b_+-b)f_7\nabla^{Y\prime}_2+f_8)Y^\prime
\end{equation}
where as usual <math>\nabla_1^Y</math> acts only on the spherical harmonic part of the wave function denoted for short by <math>Y</math>.
Since these must be equal for arbitrary <math>a</math> and <math>b</math>,
\begin{equation}
a_+^2f_1+a_+f_2+_+b_+f_3+b_+f_4+b_+f_5+a_+f_6\nabla^{Y\prime}_1 +
b_+f_7\nabla^{Y\prime}_2+f_8(Y^\prime)-f_8(Y)=0
\end{equation}

Adding the corresponding expression with <math>Y</math> and <math>Y^\prime</math> interchanged
yields
<math>\label{III}
a_+^2f_1+a_+f_2+a_+b_+f_3+b_+^2f_4+b_+f_5+\frac{1}{2}a_+f_6\nabla^+_1+\frac{1}{2}b_+f_7\nabla^+_2=0\tag{III}
</math>

Subtracting gives
\begin{eqnarray*}
f_8(Y)-f_8(Y^\prime)&=&-\frac{1}{2}[a_+f_6\nabla^-_1+b_+f_7\nabla_2^-]\tag{IV}\\
a[-2a_+f_1-2f_2-b_+f_3-f_6\nabla^+_1]&=&0\tag{V}\\
b[-2b_+f_4-2f_5-a_+f_3-f_7\nabla^+_2]&=&0\tag{VI}\\
\end{eqnarray*}

Adding the two forms gives

\begin{eqnarray}
<f>_R+<f>_L=\frac{1}{2}(a_+^2+a_-^2)f_1+a_+f_2+\frac{1}{2}(a_+b_++a_-b_-)f_3\nonumber\\
\ +\frac{1}{2}(b_+^2+b_-^2)f_4+b_+f_5+\frac{1}{2}f_6(a_+\nabla_1^++a_-\nabla^-_1)\nonumber\\
\ +\frac{1}{2}(b_+\nabla_2^++b_-\nabla^-_2)+f_8(Y)+f_8(Y^\prime)\nonumber
\end{eqnarray}

Subtracting <math>\frac{x}{2}\times({\rm I})</math> gives

\begin{eqnarray}\ \
<f>_R+<f>_L = \frac{1}{2}[(1-x)a_+^2 + a_-^2]f_1 +
(1-\frac{x}{2})a_+f_2+\frac{1}{2}[(1-x)a_+b_++a_-b_-]f_3\nonumber\\
+\frac{1}{2}[(1-x)b_+^2+b_-^2]f_4+(1-\frac{x}{2})f_5b_++\frac{1}{2}f_6[(1-\frac{x}{2})a_+\nabla_1^++a_-\nabla^-_1]\nonumber\\
+\frac{1}{2}[(1-\frac{x}{2})b_+\nabla^+_2+b_-\nabla^-_2]+f_8(Y)+f_8(Y^\prime)\nonumber
\end{eqnarray}

If <math>x=1</math>,
\begin{eqnarray}
<f>_R+<f>_L &=&
\frac{1}{2}[a_-^2f_1+a_+f_2+a_-b_-f_3+b_-^2f_4+b_+f_5\nonumber\\
&+&\left(\frac{1}{2}a_+\nabla^+_1+a_-\nabla^-_1\right)f_6+\left(\frac{1}{2}b_+\nabla^+_2+b_-\nabla^-_2\right)f_7]+f_8(Y)+f_8(Y^\prime)\nonumber
\end{eqnarray}

The General Hermitian Property for arbitrary <math>x</math> gives
\begin{eqnarray}\ \
\nabla_1^2=
\frac{1}{4}\{\frac{1}{r_1^2}[(1-x)a_+^2+a_-^2+2(1-\frac{x}{2})a_+-2[l_1(l_1+1)+l_1^\prime(l_1^\prime+1)]\nonumber\\
-\frac{2}{r_1}[(1-x)\alpha_+a_++\alpha_-a_-+2(1-\frac{x}{2})\alpha_+]+(1-x)\alpha_+^2
+ \alpha^2_-]\nonumber\\
+\frac{2(r_1-r_2\cos\theta)}{r_1r^2}[(1-x)(a_+-\alpha_+r_1)c_++(a_--\alpha_-r_2)c_-]\nonumber\\
-\frac{2r_2}{r_1r^2}[(1-\frac{x}{2})c_+\hat{r}_2\cdot\hat{\nabla}_1^++c_-\hat{r}_2\hat{\nabla}_1^-]+\frac{(1-x)c_+^2+c_-^2+2(1-\frac{x}{2})c_+}{r^2}\}\nonumber
\end{eqnarray}.

Use
<math>\begin{eqnarray}\ \
\frac{-2r_2}{r_1r^2}[(1-\frac{x}{2})c_+\hat{r}_2\hat{\nabla}_1^++c_-\hat{r}_2\hat{\nabla}_1^-]=\frac{2}{r_1^2}\left[(1-\frac{x}{2})\Lambda(\Lambda+1)-\frac{c_-}{c_+}[l_1^\prime(l_1^\prime+1)+l_1(l_1+1)]\right]\end{eqnarray}</math>,

<math>\begin{eqnarray}
\frac{2(r_1-r_2\cos\theta)}{r_1r^2}(a_--\alpha_-r_1)c_- = 2\frac{c_-}{c_+}\left[\frac{-1}{r_1^2}[a_-(a_++1)]+\frac{1}{r_1}[a_-\alpha_++\alpha_-(a_++2)]-\alpha_-\alpha_+\right]\nonumber
\end{eqnarray}</math>,

and
<math>\begin{eqnarray}\ \
\frac{2(r_1-r_2\cos\theta)}{r_1r^2}(a_+-\alpha_+r_1)c_+ = 2\left[-\frac{1}{r_1^2}[a_+(a_++1)]+\frac{1}{r_1}[a_+\alpha_++\alpha_+(a_++2)]-\alpha_+^2\right]
\end{eqnarray}</math>.

Substituting into <math>\nabla_1^2</math> gives

\begin{eqnarray}
\nabla^2_1 = \frac{1}{4}\{\frac{1}{r_1^2}[-(1-x)a_+^2 + a_-^2 +xa_+ +
2(1-\frac{x}{2})\Lambda(\Lambda+1)\nonumber\\
-2l_1(l_1+1)(1-\frac{c_-}{c_+}) -
2l_1^\prime(l_1^\prime+1)(1+\frac{c_-}{c_+})-\frac{2c_-a_-}{c_+}(a_++1)]\nonumber\\
- \frac{2}{r_1}\left[-(1-x)\alpha_+(a_++2)+\alpha_-a_-+2(1-\frac{x}{2})\alpha_+
-\frac{c_-}{c_+}[a_-\alpha_++\alpha_-(a_++2)]\right]\nonumber\\
-(1-x)\alpha_+^2+\alpha_-^2-2\frac{c_-}{c_+}\alpha_-\alpha_+ +\frac{1}{r^2}\left[(1-x)c_+^2+c_-^2+2(1-\frac{x}{2})c_+\right]\}\nonumber
\end{eqnarray}

This has the form
<math>
\nabla_1^2 =
\frac{1}{4}\left[\frac{A_1}{r_1^2}+\frac{B_1}{r_1}+\frac{C_1}{r^2}+D_1\right]
</math>
with

<math>\begin{array}{ccc}
A_1=-(1-x)a_+^2+a_-^2
+xa_++2(1-\frac{x}{2})\Lambda(\Lambda+1)-2l_1(l_1+1)(1-\frac{c_-}{c_+})\
-2l_1^\prime(l_1^\prime+1)(1+\frac{c_-}{c_+})-\frac{2c_-a_-}{c_+}(a_++1) \ \
\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \
\end{array}
</math>

<math>
B_1 =
2\left[(1-x)\alpha_+(a_++2)-\alpha_-a_--2(1-\frac{x}{2})\alpha_++\frac{c_-}{c_+}[a_-\alpha_++\alpha_-(a_++2)]\right]
</math>

<math>
C_1 = (1-x)c_+^2+ c_-^2 +2(1-\frac{x}{2})c_+
</math>

<math>
D_1 = -1(1-x)\alpha_+^2 +\alpha_-^2 - 2\frac{c_-}{c_+}\alpha_-\alpha_+
</math>

The complete Hamiltonian is then (in Z-scaled a.u.)

<math>\begin{eqnarray}
H =-\frac{1}{2}\nabla^2_1 -\frac{1}{2}\nabla^2_2 -\frac{1}{r_1} -\frac{1}{r_2} +\frac{Z^{-1}}{r}\nonumber\\
= -\frac{1}{2}\nabla^2_1 -\frac{1}{2}\nabla^2_2 -\frac{1}{r_1} -\frac{Z-1}{Zr_2}+Z^{-1}\left(\frac{1}{r}-\frac{1}{r_2}\right)\nonumber\\
\ =-\frac{1}{8}\left[\frac{A_1}{r_1^2}+\frac{B_1+8}{r_1}+\frac{C_1}{r^2}+D_1+D_2 +\frac{A_2}{r_2^2} + \frac{B_2 + 8(Z-1)/Z}{r_2}+\frac{C_2}{r^2}\right]\nonumber
+ Z^{-1}\left(\frac{1}{r}-\frac{1}{r_2}\right)\nonumber
\end{eqnarray}</math>

The screened hydrogenic energy is
<math>\begin{eqnarray}
E_{SH}=-\frac{1}{2}\left[\frac{1}{n_1^2}+ \left(\frac{Z-1}{Z}\right)^2\frac{1}{n_2^2}\right]
\end{eqnarray}</math>
so that

<math>\begin{eqnarray}
H-E_{SH} &=& -\frac{1}{8}[
\frac{A_1}{r_1^2}+\frac{B_1+8}{r_1}+\frac{C_1}{r^2} + D_1 + D_2
-\frac{4}{n_1^2} -\left(\frac{Z-1}{Z}\right)^2\frac{4}{n_2^2}\nonumber
+\frac{A_2}{r_2}+\frac{B_2+8(Z-1)/Z}{r_2}+\frac{C_2}{r^2}] +
Z^{-1}\left(\frac{1}{r} -\frac{1}{r_2}\right)\nonumber
\end{eqnarray}</math>

==Optimization of Nonlinear Parameters==
- The traditional method of performing Hylleraas calculations is to write the
basis set in the form

<math>
\Psi = \sum_{i,j,k}c_{i,j,k}\varphi_{i,j,k}(\alpha,\beta)\pm exchange
</math>

with

<math>
\varphi_{i,j,k}(\alpha,\beta) = r_1^ir_2^jr_{12}^ke^{-\alpha r_1 - \beta
r_2}\mathcal{Y}^M_{l_1l_2L}.
</math>

The usual procedure is to set <math>\alpha = Z</math> so that it represents the inner
<math>1s</math> electron, and then to vary <math>\beta</math> so as to minimize the energy.

[[File:OptNon.jpg|500px]]

Since <math>\beta</math> appears in <math>\Psi</math> as a non-linear parameter, the entire
calculation must be repeated for each value of <math>\beta</math>. However, the minimum
becomes progressively smaller as the basis set is enlarged.

[[File:OptNon02.jpg|500px]]

*Difficulties

1. If the basis set is constructed so that <math> i +j +j \leq N</math>, the the
number of terms is
<math>
(N+1)(N+2)(N+3)/6
</math>
N=14 already gives 680 terms and an accuracy of about 1 part in <math>10^{10}</math> for
low-lying states. A substantial improvement in accuracy would require much
larger basis sets, together with multiple precision arithmetic to avoid loss
of significant figures when high powers are included.

2. The accuracy rapidly deteriorates as one goes to more highly excited
states - about 1 significant figure is lost each time the principle quantum
number is increased.

*Cure

We have found that writing basis sets in the form
\begin{eqnarray}
\Psi &=& \sum_{i,j,k}\left(c^{(1)}_{i,j,k}\varphi_{ijk}(\alpha_1\beta_1) +
c_{ijk}^{(2)}\varphi_{ijk}(\alpha_2\beta_2)\right) \pm exchange \nonumber\\
&=& \psi({\bf r}_1,{\bf r}_2) \pm \psi({\bf r}_2,{\bf r}_1)\nonumber
\end{eqnarray}
so that each combination of powers is included twice with different nonlinear
parameters gives a dramatic improvement in accuracy for basis sets of about
the same total size.

However, the optimization of the nonlinear parameters is now much more
difficult, and an automated procedure is needed.

If
<math>
E = \frac{<\Psi|H|\Psi>}{<\Psi|\Psi>}</math> and we assume <math><\Psi|\Psi> = 1</math>, then

<math>
\frac{\partial E}{\partial\alpha_t} = -2 <\Psi|H-E|r_1\psi({\bf r}_1,{\bf
r}_2;\alpha_t)\pm r_21\psi({\bf r}_2,{\bf
r}_1;\alpha_t)>
</math>
where
<math>
\psi({\bf r}_1,{\bf r}_2;\alpha_t) =
\sum_{i,j,k}c_{ijk}^{(t)}\varphi_{ijk}(\alpha_t\beta_t)
</math>

*Problem

1. Prove the above.

2. Prove that there is no contribution from the implicit dependence of
<math>c_{ijk}^{(t)}</math> on <math>\alpha_t</math> if the linear parameters have been optimized.

==The Screened Hydrogenic Term==

If the Hamiltonian is written in the form (in Z-scaled a.u.)
<math>\begin{eqnarray}\
H = H_0({\bf r}_1,Z) + H_0({\bf r}_2,Z-1) +
Z^{-1}\left(\frac{1}{r}-\frac{1}{r_2}\right)
\end{eqnarray}</math>
with
<math>\
H_0({\bf r}_1,Z) = -\frac{1}{2}\nabla_1^2 - \frac{1}{r_1}\nonumber
</math> and
<math>
H_0({\bf r}_2,Z-1) = -\frac{1}{2}\nabla_2^2 - \frac{Z-1}Z{r_2}\nonumber
</math>,
then the eigenvectors of <math>H_0({\bf r},Z) +H_0({\bf r}_2,Z-1)</math> are products of
hydrogenic orbitals
<math>
\Psi_0=\psi_0(1s,Z)\psi_0(nl,Z-1)
</math>
and the eigenvalue is
<math>\begin{eqnarray}
E_{SH} = \left[-\frac{1}{2} - \left(\frac{Z-1}{Z}\right)^2\frac{1}{2n^2}\right]Z^2 a.u.
\end{eqnarray}</math>
called the screened hydrogenic eigenvalue. From highly excited states,
<math>E_{SH}</math> and <math>\Psi_0</math>are already excellent approximations.

For example, for the 1s8d states, the energies are

<math>\begin{eqnarray}\\
E(1s8d^1D)&=& -2.007\,816\,512\,563\,81 {\rm\ a.u.}\nonumber\\
E(1s8d^3D)&=& - 2.007\,817\,934\,711\,71 {\rm\ a.u.}\nonumber\\
E_{SH} &=& -2.007\,812\,5 {\rm\ a.u.} \nonumber
\end{eqnarray}</math>

It is therefore advantageous to include the screened hydrogenic terms in the
basis set so that the complete trial function becomes
<math>
\Psi = c_0\Psi_0 +
\sum_{ijk}\left[c_{ijk}^{(1)}\varphi_{ijk}(\alpha_1,\beta_1)+c_{ijk}^{(2)}\varphi(\alpha_2,\beta_2)\right]\pm
{\rm exchange}
</math>

Also, the variational principal can be re-expressed in the form
<math>\begin{eqnarray}
E= E_{SH}+\frac{<\Psi|H-E_{SH}|\Psi>}{<\Psi|\Psi>}
\end{eqnarray}</math>
so that the <math>E_{SH}</math> term can be cancelled analytically from the matrix
elements.

For example
<math>
<\Psi_0|H-E_{SH}|\Psi_0> = 0
</math>

and
<math>
<\varphi_{ijk}|H-E_{SH}|\Psi_0> =
Z^{-1}<\varphi_{ijk}|\frac{1}{r}-\frac{1}{r_2}|\Psi_0>
</math>

Recall that
<math>\begin{eqnarray}\ \
I_0(a,b,c) = \frac{2}{c+2}\sum^{[c+1]/2}_{i=0}\left(\begin{array}{c}
c+2\\
2i+1
\end{array}\right) \left[f(p,q;\beta) + f(p^\prime,q^\prime;\alpha)\right]
\end{eqnarray}</math>
where

<math>\begin{eqnarray}
f(p,q;x) =
\frac{q!}{x^{q+1}(\alpha+\beta)^{p+1}}\sum^p_{j=0}\frac{(p+j)!}{j!}\left(\frac{x}{\alpha+\beta}\right)^j
\end{eqnarray}</math>

<math>
\begin{array}{cc}
p=a+2i+2 & p^\prime = b+2i+2\\
q= b+c-2i+2 & q^\prime =a+c-2i+2
\end{array}
</math>

If <math>\beta\ll\alpha</math>, the <math>f(p,q;\beta)\gg f(p^\prime,q^\prime;\alpha)</math> and the
<math>i=0</math> term is the dominant contribution to <math>f(p,q;\beta)</math>. However, since
this term depends only on the sum of powers <math>b+c</math> for <math>r_2</math>, it cancels exactly
from the matrix element of
<math>
\frac{1}{r_{12}} - \frac{1}{r_2}
</math>
and can therefore
be omitted in the calculations of integrands, thereby saving many significant
figures. This is especially valuable when <math>b+c</math> is large and <math>a</math> is small.

==Small Corrections==
*Mass Polarization
If the nuclear mass is not taken to be infinite, then the Hamiltonian is
<math>\begin{eqnarray}\ \
H = \frac{P_N^2}{2M} + \sum_{i=1}\frac{p_i^2}{2m} + V
\end{eqnarray}</math>

[[File:SmallCor.jpg|500px]]

where <math>M =M_A-nm</math> is the nuclear mass and <math>m</math> is the electron mass.

Change variables to the C of M
<math>\begin{eqnarray}
{\bf R} = \frac{1}{M+nm}\left[M{\bf r}_N + m({\bf r}_1 + {\bf r}_2 +
\ldots)\right]
\end{eqnarray}</math>

and relative variables

<math>
{\bf s}_i = {\bf r}_i - {\bf r}_N.
</math>

Then

<math>\begin{eqnarray}\ \ \ \ \
H = \frac{1}{2(M+nm)}{\bf p}_R^2+\frac{1}{2m}\sum_{i=1}^n{\bf p}_{s_i}^2
+\frac{1}{2M}\sum_{i,k}{\bf p}_{s_i}\cdot {\bf p}_{s_k} + V({\bf s}_1, {\bf
s}_2, \ldots)\nonumber
= \frac{1}{M+nm}{\bf p}_R^2+\frac{1}{2\mu}\sum_{i=1}^n{\bf
p}_{s_i}^2+\frac{1}{2M}\sum_{i\neq k}{\bf p}_{s_i}\cdot {\bf p}_{s_k}+V({\bf s}_1, {\bf
s}_2, \ldots)\nonumber
\end{eqnarray}</math>

where
<math>\begin{eqnarray}
\mu = \left(\frac{1}{M} + \frac{1}{m}\right)^{-1}
\end{eqnarray}</math>

If <math>n=2</math>, the Schrödinger equation is
<math>\begin{eqnarray}\ \ \
\left[ -\frac{\hbar^2}{2\mu}(\nabla^2_{s_1}+ \nabla^2_{s_2}) -
\frac{\hbar^2}{M}\nabla_{s_1}\nabla_{s_2}-\frac{Ze^2}{s_1}-\frac{Ze^2}{s_2}+\frac{e^2}{s_{12}}\right]\psi
= E\psi
\end{eqnarray}</math>.

Define the reduced mass Bohr radius
<math>\begin{eqnarray}
a_\mu = \frac{\hbar^2}{\mu e^2} = \frac{m}{\mu}a_0
\end{eqnarray}</math>
and

<math>\begin{eqnarray}
\rho_i = \frac{Zs_i}{a_\mu}\nonumber\\
\varepsilon = \frac{E}{Z^2(e^2/a_\mu)}\nonumber
\end{eqnarray}</math>

The Schrödinger equation then becomes
<math>\begin{eqnarray}\ \ \
\left[
-\frac{1}{2}(\nabla^2_{\rho_1}+\nabla^2_{\rho_2})-\frac{\mu}{M}\vec{\nabla}_{\rho_1}\cdot\vec{\nabla}_{\rho_2} - \frac{1}{\rho_1}-\frac{1}{\rho_2}+\frac{Z^{-1}}{\rho_{12}}\right]\psi = \varepsilon \psi
\end{eqnarray}</math>

The effects of finite mass are thus

1. A reduced mass correction to all energy levels of <math>(e^2/a_\mu)/(e^2/a_0)
= \mu/m</math>, i.e.
<math>\begin{eqnarray}
E_M=\frac{\mu}{m}E_\infty
\end{eqnarray}</math>

2. A specific mass shift given in first order perturbation theory by
<math>\begin{eqnarray}
\Delta\varepsilon =
-\frac{\mu}{M}<\psi|\nabla_{\rho_1}\cdot\nabla_{\rho_2}|\psi>
\end{eqnarray}</math>
or

<math>\begin{eqnarray}
\Delta\varepsilon =
-\frac{\mu}{M}<\psi|\nabla_{\rho_1}\cdot\nabla_{\rho_2}|\psi>\frac{e^2}{a_\mu}\nonumber\\
= -\frac{\mu^2}{mM}<\psi|\nabla_{\rho_1}\cdot\nabla_{\rho_2}|\psi>\frac{e^2}{a_0}\nonumber
\end{eqnarray}</math>

Since <math>\nabla_{\rho_1}\cdot\nabla_{\rho_2}</math> has the same angular properties as
<math>{\bf\rho}_1 \cdot {\bf\rho}_2</math>, the operator is like the product of two
dipole operators. For product type wave functions of the form
$
\psi = \psi(1s)\psi(nl) \pm \mbox{exchange},
$
the matrix element vanishes for all but P-states.

=Radiative Transitions=

==Transition Integrals and the Line Strength==

For an electric dipole transition <math>\gamma'\,L'\,S'\,J'\,M' \rightarrow \gamma\,L\,S\,J\,M</math>, the line strength is defined by
\begin{equation}
S = \sum_{M',\mu,M}|\langle\gamma'\,L'\,S'\,J'\,M'|r_\mu|\gamma\,L\,S\,J\,M\rangle|^2
\end{equation}
Using the Wigner-Eckart theorem, this is related to the reduced matrix element by
\begin{eqnarray}
S(J',J) = |\langle\gamma'\,L'\,S'\,J'||r||\gamma\,L\,S\,J\,M\rangle|^2
\sum_{M',\mu,M}\left|\left(\matrix{J' & 1 & J \cr
-M'& \mu &M \cr}\right)\right|^2
\end{eqnarray}
Using the sum rule for the 3-<math>j</math> symbols
\begin{equation}
\sum_{M',\mu,M}\left|\left(\matrix{J' & 1 & J \cr
-M'& \mu &M \cr}\right)\right|^2 = 1
\end{equation}
we see that the line strength is identical to the reduced matrix element
\begin{equation}
S(J',J) = |\langle\gamma'\,L'\,S'\,J'||r||\gamma\,L\,S\,J\,M\rangle|^2
\end{equation}
For a nonrelativistic calculation in <math>LS</math> coupling, <math>S = S'</math> and we can next strip out the dependence on <math>J</math> and <math>J'</math> by use of Eq.\ (7.1.7) of Edmonds { Angular Momentum in Quantum Mechanics} with <math>k = 1</math> to obtain
\begin{equation}
S(J',J) = (-1)^{L'+S+J+1}[(2J+1)(2J'+1)]^{1/2}\left\{
\matrix{L' & J'& S\cr
J & L & 1\cr}\right\}\langle\gamma' L'|| r || \gamma L\rangle
\end{equation}
where <math>\langle\gamma' L'|| r || \gamma L\rangle</math> is the reduced matrix element if spin is neglected. Note that if spin is neglected, then the line-strength is simply
\begin{equation}
S(L',L) = |\langle\gamma' L'|| r || \gamma L\rangle|^2
\end{equation}
To calculate the reduced matrix element, use a particular example along with the Wigner-Eckart theorem
such as
\begin{equation}
\langle\gamma'\, L'|| r || \gamma\, L\rangle = (-1)^{L'}\langle\gamma'\, L'\, 0| z | \gamma\, L\, 0\rangle{\Large /}\left(\matrix{L' & 1 & L \cr
0 & 0 & 0 \cr}\right)
\end{equation}

===Correlated Two-electron Integrals===
For the case of two-electron integrals, <math>z = z_1 + z_2</math> and <math>r = r_1 + r_2</math>. As discussed in G.W.F. Drake, Phys. Rev. A 18, 820 (1978), the general integral in Hylleraas coordinates for the integral over the <math>z_1</math> term can be written in the form
\begin{equation}
\label{int1}
\langle\gamma'\, L'\, 0| z_1 | \gamma\, L\, 0\rangle = \sum_\Lambda C_\Lambda I_\Lambda(\gamma',\gamma)
\end{equation}
where <math>I_\Lambda(\gamma',\gamma)</math> stands for the purely radial integral
\begin{equation}
\label{radial}
I_\Lambda(\gamma',\gamma) = \int_0^\infty r_1\,dr_1\int_0^\infty r_2\,dr_2\int_{|r_1-r_2|}^{r_1+r_2}r_{12}\, dr_{12}
R_{\gamma'}^*\,r_1\, R_\gamma P_\Lambda(\cos\theta_{12})
\end{equation}
and <math>\cos\theta_{12} = (r_1^2 + r_2^2 - r_{12}^2)/(2r_1r_2)</math> is a short-hand notation for a radial function of <math>r_1</math>, <math>r_2</math> and <math>r_{12}</math>. Also, <math>P_\Lambda(\cos\theta_{12})</math> is a Legendre polynomial,
and, <math>R_\gamma = R_\gamma(r_1,r_2,r_{12})</math> is the radial part of the Hylleraas wave function for state <math>\gamma</math> such
that the total wave function is <math>\psi({\bf r}_1,{\bf r}_2) = R_\gamma(r_1,r_2,r_{12}){\cal Y}_{l_1,l_2,L}^M(\hat{\bf r}_1,\hat{\bf r}_2)\pm {\rm exchange}</math>, normalized to unity, and <math>{\cal Y}_{l_1,l_2,L}^M(\hat{\bf r}_1,\hat{\bf r}_2)</math> is a vector coupled product of spherical harmonics. The coefficents <math>C_\Lambda</math> in Eq.\ (\ref{int1}) can be calculated from Eq.\ (32) of the above 1978 paper. They need to be calculated separately for he direct-direct and the direct-exchange contributions. For the reduced matrix element of <math>z_1</math>, the leading 3-<math>j</math> symbol can be dropped, leaving (for the case <math>k = 1</math>)
\begin{eqnarray}
C_\Lambda^{\rm reduced} &=& \frac{(-1)^{l_2'+L'+L+1}}{2} (l_1,l_1',l_2,l_2',L,L')^{1/2}
\sum_{\lambda={\rm max}(l_1-1,0)}^{l_1+1} (\lambda,\Lambda,1)
\left(\matrix{l_1 & 1 & \lambda \cr
0 & 0 & 0 \cr}\right)
\left(\matrix{l_1'& \lambda & \Lambda \cr
0 & 0 & 0 \cr}\right)\nonumber\\
&\times&\left(\matrix{l_2' & 1_2 & \Lambda \cr
0 & 0 & 0 \cr}\right)
\left\{\matrix{L' & \lambda & l_2 \cr
\Lambda& l_2' & l_1' \cr}\right\}
\left\{\matrix{L' & \lambda & l_2 \cr
l_1 & L & 1 \cr}\right\}
\end{eqnarray}
where the notation <math>(a,b,c,\cdots)</math> means <math>(2a+1)(2b+1)(2c+1)\cdots</math>.
A similar expression can be obtained for the angular coefficients for the <math>z_2</math> term by interchanging the subscripts 1 and 2 throughout and multiplying by <math>(-1)^{L+L'+1}</math>.

===Example===
As an example, for the <math>1s^2\;^1S - 1s2p\;^1P</math> transition of helium, the only nonvanishing <math>C_\Lambda</math> angular coefficients are <math>C_1 = 1/2</math> for the direct-direct term (i.e. <math>P_1(\cos\theta_{12}) = \cos\theta_{12}</math> is included in the radial integral (\ref{radial})) and <math>C_0 = 1/2</math> for the direct-exchange term (i.e. <math>P_0(\cos\theta_{12}) = 1</math> is included in the radial integral (\ref{radial})). The results for infinite nuclear mass are
\begin{equation}
S(1\;^1S-2\;^1P) = (0.781\,127\,158\,667)^2
\end{equation}
\begin{equation}f(1\;^1S-2\;^1P) = -0.276\,164\,703\,736
\end{equation}
where
\begin{equation}
f(1\;^1S-2\;^1P) = \frac{2}{3}[E(1\;^1S) - E(2\;^1P)]S(1\;^1S-2\;^1P)
\end{equation}
is the absorption oscillator strength.

Theory Notes

2020-12-14T05:36:59Z

Drake: /* General Hermitian Property */

<center>
<H1>Notes on solving the Schrödinger equation in Hylleraas coordinates for heliumlike atoms
Gordon W.F. Drake
Department of Physics, University of Windsor
Windsor, Ontario, Canada N9B 3P4
(Transcribed from hand-written notes and edited by Lauren Moffatt. Last revised November 22, 2017 to correct a missing subscript in the expression for $B_1$ in the Hamiltonian in Sect. 11 General Hermitian Property.)
</H1>
</center>

==Introduction==
These notes describe some of the technical details involved in solving the Schrödinger equation for a heliumlike atom of nuclear charge Z in correlated Hylleraas coordinates. The standard for computational purposes is to express all quantities in a form that is valid for electronic states of arbitrary angular momentum, and to express matrix elements of the Hamiltonian in a manifestly Hermitian symmetrized form. Further information and derivations of the mathematical techniques for angular integrals and radial recursion relations can be found in the paper
G.W.F. Drake, Phys. Rev. A <math>{\bf 18}</math>, 820 (1978).

The starting point is the two-electron Schrödinger equation for infinite nuclear mass
\begin{equation} \left[-\frac{\hbar^2}{2m}(\nabla^2_1 +\nabla^2_2) - \frac{Ze^2}{r_1} -
\frac{Ze^2}{r_2}+\frac{e^2}{r_{12}} \right]\psi = E\psi\nonumber
\end{equation}
where <math>m</math> is the electron mass, and <math>r_{12} = |{\bf r}_1 - {\bf r}_2|</math> (see diagram below).

[[File:Pic1.jpg|250px]]

Begin by rescaling distances and energies so that the Schrödinger equation can be expressed in a dimensionless form. The dimensionless Z-scaled distance is defined by <math>\begin{eqnarray} \rho = \frac{Zr}{a_0} \end{eqnarray} </math> where <math> \begin{eqnarray} a_0 = \frac{\hbar^2}{me^2} \end{eqnarray} </math> is the atomic unit (a.u.) of distance equal to the Bohr radius <math>0.529\, 177\, 210\, 92(17) \times 10^{-10}</math> m. Then

\begin{equation}[-\frac{\hbar^2}{2m}Z^2(\frac{me^2}{\hbar^2})^2(\nabla^2_{\rho_1}+\nabla^2_{\rho_2})
- Z^2\frac{e^2}{a_0}\rho^{-1}_1 - Z^2\frac{e^2}{a_0}\rho^{-1}_2 +
\frac{e^2}{a_0}Z\rho^{-1}_{12}]\psi= E\psi\nonumber\end{equation}

But <math>\begin{eqnarray} \frac{\hbar^2}{m}(\frac{me^2}{\hbar^2})^2 = \frac{e^2}{a_0} \end{eqnarray} </math> is the hartree atomic unit of of energy <math>(E_h = 27.211\,385\,05(60)</math> eV, or equivalently <math>E_h/(hc) = 2194.746\,313\,708(11)</math> m<math>^{-1})</math>. Therefore, after multiplying through by <math>a_0/(Ze)^2</math>, the problem to be solved in Z-scaled dimensionless units becomes

\begin{equation} \left[-\frac{1}{2}(\nabla^2_{\rho_1}+\nabla^2_{\rho_2}) -
\frac{1}{\rho_1} - \frac{1}{\rho_2} + \frac{Z^{-1}}{\rho_{12}}\right]\psi =
\varepsilon\psi\nonumber \end{equation}

where <math>\begin{eqnarray}\varepsilon = \frac{Ea_0}{(Ze)^2} \end{eqnarray}</math> is the energy in Z-scaled atomic units. For convenience, rewrite this in the conventional form
\begin{equation}
H\psi = \varepsilon\psi\nonumber \end{equation}
where (using <math>r</math> in place of <math>\rho</math> for the Z-scaled distance)
\begin{equation}
H = -\frac{1}{2}(\nabla^2_1+\nabla^2_2) - \frac{1}{r_1}-\frac{1}{r_2} + \frac{Z^{-1}}{r_{12}}
\nonumber\end{equation}
is the atomic Hamiltonian for infinite nuclear mass.

==The Hartree Fock Method==

For purposes of comparison, and to define the correlation energy, assume that <math>\psi({\bf r}_1,{\bf r}_2)</math> can be written in the separable product form
\begin{equation}
\psi({\bf r}_1,{\bf r}_2) = \frac{1}{\sqrt{2}} \left[ u_1(r_1) u_2(r_2) \pm u_2(r_1) u_1(r_2) \right] \nonumber
\end{equation}
for the <math>1s^2\ ^1\!S</math> ground state. Because of the <math>\frac{1}{r_{12}}</math> term in the Schrödinger equation, the exact solution connot be expressed in this form as a separable product. However, the Hartree-Fock approximation corresponds to finding the best possible solution to the Schrödinger equation
\begin{equation}
H\psi({\bf r}_1,{\bf r}_2) = E\psi({\bf r}_1,{\bf r}_2)\nonumber
\end{equation}
that can nevertheless be expressed in this separable product form, where as before
\begin{equation}
H = -\frac{1}{2}(\nabla^2_1+\nabla^2_2) - \frac{1}{r_1}- \frac{1}{r_2} + \frac{Z^{-1}}{r_{12}}\nonumber
\end{equation}
is the full two-electron Hamiltonian. To find the best solution, substitute into <math>\left\langle\psi \lvert H-E \rvert \psi\right\rangle</math> and require this expression to be stationary
with respect to arbitrary infinitesimal variations <math>\delta u_1</math> and <math>\delta
u_2</math> in <math>u_1</math> and <math>u_2</math> respectively; i.e.

\begin{equation}
\frac{1}{2} \left\langle \delta u_1(r_1)u_2(r_2) \pm u_2(r_1)\delta
u_1(r_2) \lvert H-E \rvert u_1(r_1) u_2(r_2)\pm u_2(r_1) u_1(r_2)\right\rangle\nonumber
\end{equation}

\begin{equation}
=\int\delta u_1(r_1) d{\bf r}_1 \left\{ \int d{\bf r}_2 u_2(r_2) \left( H - E \right) \left[ u_1(r_1)u_2(r_2) \pm u_2(r_1) u_1(r_2) \right] \right\}\nonumber
\end{equation}

\begin{equation}
= 0 \nonumber
\end{equation}

for arbitrary <math> \delta u_1(r_1).\nonumber </math> Therefore <math>\{\int d{\bf r}_2 \ldots \} = 0</math>.

Similarly, the coefficient of <math>\delta u_2</math> would give

\begin{equation}
\int d{\bf r}_1 u_1(r_1) \left( H-E \right) \left[ u_1(r_1) u_2(r_2) \pm
u_2(r_1) u_1(r_2) \right] = 0\nonumber
\end{equation}

Define

\begin{equation}
I_{12} = I_{21} = \int d{\bf r}\, u_1(r)u_2(r), \nonumber
\end{equation}

\begin{equation}
H_{ij} = \int d{\bf r}\, u_i(r)(-\frac{1}{2}\nabla - \frac{1}{r})u_j(r), \nonumber
\end{equation}

\begin{equation}
G_{ij}(r) = \int d{\bf r}^\prime u_i(r^\prime)\frac{1}{|{\bf r} - {\bf r}\prime|}u_j(r^\prime)\nonumber
\end{equation}

Then the above equations become the pair of integro-differential equations

\begin{equation}
[ H_0 - E + H_{22}+G_{22}(r)]u_1(r) = \mp [ I_{12}(H_0-E) + H_{12}+G_{12}(r)]u_2(r)\nonumber
\end{equation}

\begin{equation}
[H_0-E+H_{11}+G_{11}(r)]u_2(r) = \mp [I_{12}(H_0-E) + H_{12}+G_{12}(r)]u_1(r)\nonumber
\end{equation}

These must be solved self-consistently for the "constants" <math>I_{12}</math> and
<math>H_{ij}</math> and the function <math>G_{ij}(r)</math>.

The Hartree Fock energy is <math>E \simeq -2.87\ldots </math>a.u. while the exact energy is <math>E = -2.903724\ldots </math>a.u.

The difference is called the "correlation energy" because it arises from the
way in which the motion of one electron is correlated to the other. The
Hartree Fock equations only describe how one electron moves in the average field
provided by the other (mean-field theory).

==Configuration Interaction==
Expand
\begin{equation}
\psi({\bf r}_1,{\bf r}_2)= C_0u^{(s)}_1(r_1)u^{(s)}_1(r_2) + C_1u^{(P)}_1({\bf r}_1)u^{(P)}_1({\bf r}_2){\cal Y}^0_{1,1,0}(\hat{\bf r}_1, \hat{\bf r}_2)+C_2u^{(d)}_1({\bf r}_1)u^{(d)}_2({\bf r}_2){\cal Y}^0_{2,2,0}(\hat{\bf r}_1, \hat{\bf r}_2)+... \pm {\rm\ exchange}
\end{equation}
where
\begin{equation}
{\cal Y}^M_{l_1,l_2,L}(\hat{\bf r}_1, \hat{\bf r}_2)=\sum_{m_1,m_2}Y^{m_1}_{l_1}({\bf r}_1)Y^{m_2}_{l_2}({\bf r}_2)\times <l_1l_2m_1m_2\mid LM>
\end{equation}

This works, but is slowly convergent, and very laborious. The best CI calculations are accurate to <math> ~10^{-7}</math> a.u.

==Hylleraas Coordinates==
[E.A. Hylleraas, Z. Phys. <math>{\bf 48}, 469(1928)</math> and <math>{\bf 54}, 347(1929)</math>]
suggested using the co-ordinates <math>r_1</math>, <math>r_2</math> and <math>r_{12}</math> or equivalently

\begin{eqnarray}
s &=& r_1 + r_2, \nonumber\\
t &=& r_1-r_2, \nonumber\\
u &=& r_{12}\nonumber
\end{eqnarray}

and writing the trial functions in the form

\begin{equation}
\Psi({\bf r}_1,{\bf r}_2) = \sum^{i+j+k\leq
N}_{i,j,k}c_{i,j,k}r_1^{i+l_1}r_2^{j+l_2}r_{12}^ke^{-\alpha r_1 - \beta r_2}
\mathcal{Y}^M_{l_1,l_2,L}(\hat{r}_1,\hat{r}_2)\pm \text{exchange}\nonumber
\end{equation}

Diagonalizing H in this non-orthogonal basis set is equivalent to solving
<math> \begin{eqnarray}
\frac{\partial E}{\partial c_{i,j,k}} = 0\nonumber
\end{eqnarray} </math>
for fixed <math>\alpha</math> and <math>\beta</math>.

The diagonalization must be repeated for different values of <math>\alpha</math> and
<math>\beta</math> in order to optimize the non-linear parameters.

==Completeness==
The completeness of the above basis set can be shown by first writing
\begin{equation}
r_{12}^2 = r_1^2 + r_2^2 - 2r_1r_2\cos\theta_{12}
\end{equation}
and, from the spherical harmonic addition theorem,
\begin{equation}
\cos(\theta_{12})=\frac{4\pi}{3}\sum^1_{m=-1}Y^{m*}_1(\theta_1,\varphi_1)Y^m_1(\theta_2,\varphi_2)
\end{equation}
Consider first the S-states. The <math>r_{12}^0</math> terms are like the ss terms in a CI calculation. The
<math>r_{12}^2</math> terms bring in p-p type contributions, and the higher powers bring
in d-d, f-f etc. type terms. In general

\begin{equation}
P_l(\cos\theta_{12}) = \frac{4\pi}{2l+1}\sum^l_{m=-l}{Y^{m}_l}^*(\theta_1,\varphi_1)Y^m_l(\theta_2,\varphi_2)\nonumber
\end{equation}

For P-states, one would have similarly

\begin{array}{lr}
r_{12}^0 & (sp)P\nonumber\\
r_{12}^2 & (pd)P\nonumber\\
r_{12}^4 & (df)P\nonumber\\
\vdots & \vdots\nonumber
\end{array}

For D-states

\begin{array}{lr}
r_{12}^0 & (sd)D & (pp^\prime)D\nonumber\\
r_{12}^2 & (pf)D & (dd^\prime)D\nonumber\\
r_{12}^4 & (dg)D & (ff^\prime)D\nonumber\\
\vdots & \vdots & \vdots\nonumber
\end{array}

In this case, since there are two "lowest-order" couplings to form a
D-state, both must be present in the basis set, i.e.

\begin{equation}
\Psi(r_2,r_2) = \sum c_{ijk}r_1^ir_2^{j+2}r_{12}^ke^{-\alpha r_1-\beta
r_2}\mathcal{Y}^M_{022}(\hat{r}_1,\hat{r}_2)
+\sum d_{ijk}r_1^{i+1}r_2^{j+1}r_{12^k}e^{-\alpha^\prime r_1 - \beta^\prime
r_2}\mathcal{Y}^M_{112}(\hat{r}_1,\hat{r}_2)\nonumber
\end{equation}

For F-states, one would need <math>(sf)F</math> and <math>(pd)F</math> terms.

For G-states, one would need <math>(sg)G</math>, <math>(pf)G</math> and <math>(dd^\prime)G</math> terms.

Completeness of the radial functions can be proven by considering the
Sturm-Liouville problem

\begin{equation}
\left(-\frac{1}{2}\nabla^2-\frac{\lambda}{r_s}-E\right)\psi({\bf r}) = 0\nonumber
\end{equation}

or

\begin{equation}
\left(-\frac{1}{2}\frac{1}{r^2}\frac{\partial}{\partial r}\left({r^2}\frac{\partial}{\partial r}\right) -
\frac{l\left(l+1\right)}{2r^2} - \frac{\lambda}{r} - E\right)u(r) = 0.\nonumber
\end{equation}

For fixed E and variable <math>\lambda</math> (nuclear charge).

The eigenvalues are <math>\lambda_n = (E/E_n)^{1/2}</math>, where <math>E_n =- \frac{1}{2n^2}</math>

[[File:Pic2.jpg|500px]]

and the eigenfunctions are

\begin{equation}
u_{nl}(r) =
\frac{1}{(2l+1)!}\left(\frac{(n+l)!}{(n-l-1)2!}\right)^{1/2}(2\alpha)^{3/2}e^{-\alpha
r}\nonumber
\times \ \left( 2 \alpha r \right)^l\,_{1}F_{1}\left(-n+l+1,2l+2; 2 \alpha r \right)
\end{equation}

with <math>\alpha = (-2E)^{1/2}</math> and <math>n\geq l+1</math>. The confluent hypergeometric function <math> _1F_1(a,b,;z)</math> then denotes a finite polynomial since <math>a=-n+l+1</math> is a negative integer or zero.

Unlike the hydrogen spectrum, which has both a discrete part for <math>E<0</math> and a
continuous part for <math>E>0</math>, this forms an entirely discrete set of finite
polynomials, called Sturmian functions. They are orthogonal with respect to
the potential, i.e.

\begin{equation}
\int^\infty_0 r^2dr \left( u_{n^\prime l}(r)\frac{1}{r}u_{nl}(r) \right) = \delta_{n,n^\prime}\nonumber
\end{equation}

Since they become complete in the limit <math>n\rightarrow\infty</math>, this assures the
completeness of the variational basis set.

[See also B Klahn and W.A. Bingel Theo. Chim. Acta (Berlin) 44, 9 and 27
(1977)].

==Solutions of the Eigenvalue Problem==

For convenience, write

\[
\Psi({\bf r}_1,{\bf r}_2) = \sum^N_{m=1}c_m\varphi_m\nonumber
\]

where <math>m</math> represents the <math>m</math>'th combination of <math>i,j,k</math> and

\begin{equation}
\varphi_{ijk}=r_1^ir_2^jr_{12}^ke^{-\alpha r_1 - \beta
r_2}\mathcal{Y}^M_{l_1,l_2,L}(\hat{r}_1,\hat{r}_2) \pm \mbox{exchange}\nonumber
\end{equation}

\begin{align}

&
\left( \begin{array}{lr}
\cos(\theta) & \sin(\theta)\\
-\sin(\theta) & \cos(\theta)
\end{array} \right)
\left( \begin{array}{lr}
H_{11} & H_{12} \\
H_{12} & H_{22}
\end{array}\right)
\left( \begin{array}{lr}
\cos(\theta) & -\sin(\theta)\\
\sin(\theta) & \cos(\theta)
\end{array} \right)
\\[5pt]
= & \left(\begin{array}{lr}
cH_{11}+sH_{12} & cH_{12} + sH_{22}\\
-sH_{11} + cH_{12} & -sH_{12} + cH_{22}
\end{array}\right)
\left( \begin{array}{cc}
c & -s \\
s & c
\end{array}\right)
\\[5pt]
= & \left(\begin{array}{lr}
c^2H_{11}+s^2H_{22} + 2csH_{12} & (c^2-s^2)H_{12}+cs(H_{22}-H_{11})\\
(c^2-s^2)H_{12}+cs(H_{22}-H_{11}) & s^2H_{11}+c^2H_{22}-2csH_{12}
\end{array}\right)
\end{align}

Therefore

\begin{equation}
\left[\cos^2(\theta)-\sin^2(\theta)\right]H_{12} =
\cos(\theta)\sin(\theta)(H_{11}-H_{22})\nonumber
\end{equation}

and

\begin{equation}
\tan(2\theta) = \frac{2H_{12}}{H_{11}-H_{22}}\nonumber
\end{equation}

i.e.

\begin{equation}
\cos(\theta)=\left(\frac{r+\omega}{2r}\right)^{1/2}\nonumber
\end{equation}

\begin{equation}
\sin(\theta)=-\mbox{sgn}(H_{12})\left(\frac{r-\omega}{2r}\right)^{1/2}\nonumber
\end{equation}

where

\begin{equation}
\omega = H_{22}-H_{11}\nonumber
\end{equation}

\begin{equation}
r=\left(\omega^2+4H_{12}^2\right)^{1/2}\nonumber
\end{equation}

\begin{equation}
E_1=\frac{1}{2}\left(H_{11}+H_{22}-r\right)\nonumber
\end{equation}

\begin{equation}
E_2=\frac{1}{2}\left(H_{11}+H_{22}+r\right)\nonumber
\end{equation}

===Brute Force Method===

- Gives all the eigenvalues and eigenvectors, but it is slow

- First orthonormalize the <math>N</math>-dimensional basis set; i.e. form linear combinations

\begin{equation}
\Phi_m = \sum_{n=1}^N\varphi_nR_{nm}\nonumber
\end{equation}

such that
\begin{equation}
\langle \Phi_m | \Phi_n \rangle = \delta_{m,n}
\end{equation}

This can be done by finding an orthogonal tranformation T such that

\begin{equation}
T^TOT = I = \left(
\begin{array}{ccccc}
I_1 & 0 & 0 & \ldots & 0 \\
0 & I_2 & 0 & \ldots & 0 \\
0 & 0 & I_3 & \ldots & 0 \\
\vdots & \vdots & \vdots & \ddots & \vdots \\
0 & 0 & 0 & \ldots & I_N
\end{array}\right);
\end{equation}

\begin{equation}
O_{mn} = \langle \varphi_m | \varphi_n \rangle \nonumber
\end{equation}

and then applying a scale change matrix

\begin{equation}
S = \left(\begin{array}{ccccc}
\frac{1}{I_1^{1/2}} & 0 & 0 & \ldots & 0\\
0 & \frac{1}{I_2^{1/2}} & 0 & \ldots & 0 \\
0 & 0 & \frac{1}{I_3^{1/2}} & \ldots & 0 \\
\vdots & \vdots & \vdots & \ddots & \vdots \\
0 & 0 & 0 & \ldots & I_N^{1/2}
\end{array}\right)= S^T\nonumber
\end{equation}

Then

\begin{equation}
S^TT^TOTS = 1
\end{equation}

i.e.

\begin{equation}
R^TOR = 1
\end{equation}

with <math>R=TS</math>.

If H is the matrix with elements <math>H_{mn} = \langle \varphi_m | \varphi_n \rangle </math>, then H
expressed in the <math>\Phi_m</math> basis set is

\begin{equation}
H^\prime = R^THR.\nonumber
\end{equation}

We next diagonalize <math>H^\prime</math> by finding an orthogonal transformation W such
that

\begin{equation}
W^T H^\prime W = \lambda = \left(
\begin{array}{cccc}
\lambda_1 & 0 & \ldots & 0 \\
0 & \lambda_2 & \ldots & 0 \\
\vdots & \vdots & \ddots & \vdots \\
0 & 0 & \ldots & \lambda_N
\end{array}\right)\nonumber
\end{equation}

The q'th eigenvector is

\begin{equation}
\Psi^{(q)} = \sum_{n=1}^N\Phi_n W_{n,q} = \sum_{n,n^\prime} \varphi_{n^\prime} R_{n^\prime ,n} W_{n,q} \nonumber
\end{equation}

i.e.

\begin{equation}
c_{n^\prime}^{(q)} = \sum_{n=1}^N R_{n^\prime n} W_{n,q}
\end{equation}

===The Power Method===

- Based on the observation that if H has one eigenvalue, <math>\lambda_M</math>, much bigger than all the rest, and <math>\chi = \left(
\begin{array}{c}a_1\\a_2\\\vdots\end{array}\right)</math>
is an arbitrary starting vector, then <math>\chi = \sum_{q=1}^N x_q\Psi^{(q)}</math>.

\begin{equation}
(H)^n\chi = \sum_{q=1}^N x_q \lambda^n_q\Psi^{(q)}\nonumber\rightarrow x_M\lambda_M^n\Psi^{(M)}\nonumber
\end{equation}

provided <math>x_M\neq 0</math>.

To pick out the eigenvector corresponding to any eigenvalue, write the original problem in the form

\begin{align}
H\Psi & = \lambda O \Psi \nonumber \\
\left( H - \lambda_q O \right) \Psi & = \left( \lambda - \lambda_q \right) O \Psi \nonumber
\end{align}

Therefore,

\begin{equation}
G \Psi = \frac{1}{\lambda - \lambda_q } \Psi \nonumber \\
\end{equation}

where <math>G=(H-\lambda_qO)^{-1}O</math> with eigenvalues <math>\frac{1}{\lambda_n - \lambda_q}</math>.

By picking <math>\lambda_q</math> close to any one of the <math>\lambda_n</math>, say
<math>\lambda_{n^\prime}</math>, then <math>\frac{1}{\lambda_n-\lambda_q}</math> is much larger for
<math>n=n^\prime</math> than for any other value. The sequence is then

\begin{array}{c}
\chi_1=G\chi\nonumber\\
\chi_2=G\chi_1\nonumber\\
\chi_3=G\chi_2\nonumber\\
\vdots\nonumber
\end{array}

until the ratios of components in <math>\chi_n</math> stop changing.

- To avoid matrix inversion and multiplication, note that the sequence is equivalent to
\begin{equation}
F \chi_n = \left( \lambda - \lambda_q \right) O \chi_{n-1} \nonumber \\
\end{equation}
where <math>F = H-\lambda_qO</math>. The factor of <math>\left( \lambda - \lambda_q \right)</math> can be
dropped because this only affects the normalization of <math>\chi_n</math>. To find <math>\chi_n</math>, solve
\begin{equation}
F \chi_n = O \chi_{n-1} \nonumber
\end{equation}
(N equations in N unknowns). Then
\begin{equation}
\lambda = \frac{ \langle \chi_n | H | \chi_n \rangle}{\langle \chi_n | \chi_n \rangle} \nonumber
\end{equation}

==Matrix Elements of H==

\begin{equation}
H=-\frac{1}{2}\nabla^2_1 -\frac{1}{2}\nabla^2_2 - \frac{1}{r_1} - \frac{1}{r_2} +\frac{Z^{-1}}{r_{12}}\nonumber
\end{equation}

Taking <math>r_1, r_2</math> and <math>r_{12}</math> as independent variables,

\begin{equation}
\nabla_1^2 = \frac{1}{r_1^2}\frac{\partial}{\partial r_1}\left(r_1^2\frac{\partial}{\partial r_1}\right)+ \frac{1}{r_{12}^2}\frac{\partial}{\partial r_{12}}
\left(r_{12}^2\frac{\partial}{\partial r_{12}}\right)\nonumber -\frac{l_1(l_1+1)}{r_1^2}+2(r_1 - r_2\cos(\theta_{12}))\frac{1}{r_{12}}\frac{\partial^2}{\partial r_1 \partial r_{12}} \nonumber - 2(\nabla_1^Y \cdot {\bf r}_2)\frac{1}{r_{12}}\frac{\partial}{\partial r_{12}}\nonumber
\end{equation}

where <math>\nabla_1^Y</math> acts only on the spherical harmonic part of the wave function and the diagram

[[File:Pic3.jpg|400px]]

defines the complete set of 6 independent variables is <math>r_1</math>, <math>r_2</math>, <math>r_{12}</math>, <math>\theta_1</math>, <math>\varphi_1</math>, <math>\chi</math>.

If <math>r_{12}</math> were not an independent variable, then one could take the volume element to be

\begin{equation}
d\tau = r_1^2dr_1\sin(\theta_1)d\theta_1d\varphi_1r_2^2dr_2\sin(\theta_2)d\theta_2d\varphi_2.\nonumber
\end{equation}

However, <math>\theta_2</math> and <math>\varphi_2</math> are no longer independent variables. To eliminate them, take the point <math>{\bf r}_1</math> as the
origin of a new polar co-ordinate system, and write

\begin{equation}
d\tau=-r_1^2dr_1\sin(\theta_1)d\theta_1d\varphi_1r_{12}^2dr_{12}\sin(\psi)d\psi d\chi\nonumber
\end{equation}

and use

\begin{equation}
r_2^2=r_1^2+r_{12}^2 +2r_1r_{12}\cos(\psi)
\end{equation}

Then for fixed <math>r_1</math> and <math>r_{12}</math>,

\begin{equation}
2r_2dr_2 = -2r_1r_{12}\sin(\psi)d\psi
\end{equation}

Thus

\begin{equation}
d\tau= r_1dr_1r_2dr_2r_{12}dr_{12}\sin(\theta_1)d\theta_1d\varphi_1d\chi
\end{equation}

The basic type of integral to be calculated is

\begin{align}
I \left( l_1, m_1, l_2, m_2; R \right) & = \int \sin \left( \theta_1 \right) d\theta_1 d\varphi_1 d\chi Y^{m_1}_{l_1} \left( \theta_1, \varphi_1 \right)^{*} Y^{m_2}_{l_2} \left( \theta_2, \varphi_2 \right) \\
& \times \int r_1dr_1r_2dr_2r_{12}dr_{12}R \left( r_1, r_2, r_{12} \right)
\end{align}

Consider first the angular integral. <math>Y^{m_2}_{l_2} \left( \theta_2, \varphi_2 \right)</math> can be expressed in terms of the independent variables <math>\theta_1, \varphi_1, \chi</math> by use of the rotation matrix relation
\begin{equation}
Y^{m_2}_{l_2}( \theta_2, \varphi_2) = \sum_m \mathcal{D}^{(l_2)}_{m_2,m}(\varphi_1,\theta_1,\chi)^*Y^m_{l_2}( \theta_{12}, \varphi)
\end{equation}
where <math>\theta, \varphi</math> are the polar angles of <math>{\bf r}_2</math> relative to <math>{\bf r}_1</math>. The angular integral is then

\begin{align}
I_{ang} & = \int^{2 \pi}_{0} d\chi \int^{2\pi}_{0} d\varphi_1 \int^{\pi}_{0} \sin \left( \theta_1 \right) d\theta_{1} Y^{m_1}_{l_1} \left( \theta_1, \varphi_1 \right)^{*} \\
& \times \sum_m \mathcal{D}^{(l_2)}_{m_2,m} \left( \varphi_1,\theta_1,\chi \right)^{*} Y^{m}_{l_2} \left( \theta_{12},\varphi \right) \\
\end{align}

Use

\begin{equation}
Y^{m_1}_{l_1} \left( \theta_1, \varphi_1 \right)^{*} = \sqrt{\frac{2l_1+1}{4\pi}}\mathcal{D}^{(l_1)}_{m_1,0} \left( \varphi_1, \theta_1, \chi \right)
\end{equation}

together with the orthogonality property of the rotation matrices (Brink and Satchler, p 147)

\begin{align}
& \int \mathcal{D}^{(j)*}_{m,m^\prime}\mathcal{D}^{(J)}_{M,M^\prime}\sin(\theta_1)d\theta_1d\varphi_1d\chi \\
& = \frac{8\pi^2}{2j+1}\delta_{jJ}\delta_{mM}\delta_{m^\prime M^\prime}\nonumber
\end{align}

to obtain

\begin{align}
I_{ang} & = \sqrt{ \frac{2l_1+1}{4\pi}} \frac{8\pi^2}{2l_1+1} \delta_{l_1,l_2} \delta_{m_1,m_2} Y^{0}_{l_2} \left( \theta_{12}, \varphi \right) \\
& = 2 \pi \delta_{l_1,l_2} \delta_{m_1,m_2} P_{l_2} \left( \cos \theta_{12} \right)
\end{align}

since

\begin{equation}
Y^0_{l_2} \left( \theta_{12}, \varphi \right) = \sqrt{\frac{2l_1+1}{4\pi}} P_{l_2} \left( \cos \left( \theta_{12} \right) \right)
\end{equation}

Note that <math>P_{l_2} \left( \cos \theta \right)</math> is just a short hand expression for a radial function because

\begin{equation}
\cos \theta_{12} = \frac{r_1^2 + r_2^2 - r_{12}^2}{2r_1r_2}
\end{equation}

The original integral is thus

\begin{equation}
I \left( l_1, m_1, l_2, m_2; R \right) = 2 \pi \delta_{l_1,l_2} \delta_{m_1,m_2} \int^{\infty}_{0} r_{1} dr_{1} \int^{\infty}_{0} r_{2} dr_{2} \int^{r_{1} + r_{2}}_{|r_{1} - r_{2}|} r_{12} dr_{12} R \left( r_{1}, r_{2}, r_{12} \right) P_{l_2} \left( \cos \theta_{12} \right)
\end{equation}

where again

\begin{equation}
\cos \theta_{12} = \frac{r_{1}^{2} +r^{2}_{2} - r_{12}^{2}}{2 r_{1} r_{2}}
\end{equation}

is a purely radial function.

The above would become quite complicated for large <math>l_2</math> because <math>P_{l_2}(\cos \theta_{12})</math> contains terms up to <math>\left( \cos \theta_{12} \right)^{l_2}</math>. However, recursion relations exist which allow any integral containing <math>P_l \left( \cos \theta_{12} \right)</math> in terms of those containing just <math>P_0 \left( \cos \theta_{12} \right) = 1</math> and <math>P_1 \left( \cos \theta_{12} \right) = \cos \theta_{12}</math>.

==Radial Integrals and Recursion Relations==

The basic radial integral is [see G.W.F. Drake, Phys. Rev. A 18, 820 (1978)]

\begin{align}
I_0 \left( a, b, c \right) & =
\int^{\infty}_{0} r_{1} dr_{1} \int^{\infty}_{r_{1}} r_{2} dr_{2} \int^{r_{1}+r_{2}}_{r_{2} - r_{1}} r_{12} dr_{12} r_{1}^{a} r_{2}^{b} r_{12}^{c} e^{-\alpha r_{1} - \beta r_{2}} + \int^{\infty}_{0} r_{2} dr_{2} \int^{\infty}_{r_{2}} r_{1} dr_{1} \int^{r_{1} + r_{2}}_{r_{1} - r_{2}} r_{12} dr_{12} r_{1}^{a} r_{2}^{b} r_{12}^{c} e^{- \alpha r_{1} - \beta r_{2}} \\
& = \frac{2}{c+2} \sum^{[(c+1)/2]}_{i=0}
\left(\begin{array}{c}
c+2 \\
2i+1
\end{array} \right)
\left\{ \frac{q!}{\beta^{q+1} \left( \alpha + \beta \right)^{p+1}} \sum_{j=0}^{q} \frac{\left( p + j \right)!}{j!} \left( \frac{\beta}{\alpha + \beta} \right)^j + \frac{q^{\prime}!}{\alpha^{q^{\prime}+1} \left( \alpha + \beta \right)^{p^{\prime}+1}} \sum^{q^{\prime}}_{j=0} \frac{\left( p^{\prime} + j \right)!}{j!} \left( \frac{\alpha}{\alpha + \beta} \right)^j \right\}
\end{align}

where

\begin{array}{ll}
p = a + 2i + 2 & p^{\prime} = b + 2i + 2 \\
q = b + c - 2i + 2 & q^{\prime} = a + c - 2i + 2
\end{array}

The above applies for <math> a, b \geq -2, c\geq -1</math>.
<math>[x]</math> means "greatest integer in" <math>x</math>.

Then

\begin{align}
I_1 \left( a, b, c \right) & = \int d\tau_{r} r_{1}^{a} r_{2}^{b} r_{12}^{c} e^{ -\alpha r_{1} - \beta r_{2}} P_{1}\left( \cos \theta \right) \\
& = \frac{1}{2} \left( I_{0} \left( a + 1, b - 1, c \right) + I_{0} \left( a - 1, b + 1, c \right) - I_{0} \left( a - 1, b - 1, c + 2 \right) \right)
\end{align}

===The Radial Recursion Relation===

Recall that the full integral is

\begin{equation}
I(l_1m_1,l_2m_2;R)=2\pi\delta_{l_1,l_2}\delta_{m_1,m_2}I_{l_2}(R)
\end{equation}

where

\begin{equation}
I_{l_2}(R) = \int d\tau_r R(r_1,r_2,r_{12})P_{l_2}(\cos \theta_{12})
\end{equation}

To obtain the recursion relation, use

\begin{equation}
P_l(x) = \frac{[P^{\prime}_{l+1}(x)-P^{\prime}_{l-1}(x)]}{2l+1}
\end{equation}

with

\begin{equation}
P_{l+1}^\prime (x) = \frac{d}{dx}P_{l+1}(x)
\end{equation}

Here <math>x=\cos\theta_{12}</math> and

\begin{align}
\frac{d}{d \cos \theta_{12}} & = \frac{dr_{12}}{d \cos \theta_{12}} \frac{d}{dr_{12}} \\
& = -\frac{r_{1} r_{2}}{r_{12}} \frac{d}{dr_{12}}
\end{align}

Then,

\begin{equation}
I_{l}\left( R \right) = - \int d\tau_r R \frac{r_{1} r_{2}}{r_{12}} \frac{d}{dr_{12}} \frac{ \left[ P_{l+1} \left( \cos \theta_{12} \right) - P_{l-1} \left( \cos \theta_{12} \right) \right]}{2l+1}
\end{equation}

The <math>r_{12}</math> part of the integral is

\begin{align}
& \int^{r_{1} + r_{2}}_{|r_{1} - r_{2}|} r_{12} dr_{12} R \frac{r_{1} r_{2}}{r_{12}} \frac{d}{dr_{12}} \left[P_{l+1}-P_{l-1} \right] \\
& = R r_{1} r_{2} \left[ P_{l+1} - P_{l-1} \right]^{r_{1} + r_{2}}_{|r_{1} - r_{2}|} - \int^{r_{1} + r_{2}}_{|r_{1} - r_{2}|} r_{12} dr_{12} \left( \frac{d}{dr_{12}} R \right) \frac{r_{1} r_{2}}{r_{12}} \frac{ \left[ P_{l+1} - P_{l-1} \right]}{2l+1}
\end{align}

The integrated term vanishes because

\begin{equation}
\cos \theta_{12} = \frac{r_{1}^{2} + r_{2}^{2} - r_{12}^{2}}{2 r_{1} r_{2}}
\begin{array}{lll}
& = -1 & {\rm when} & r_{12}^{2} = \left( r_{1} + r_{2} \right)^{2} \\
& = 1 & {\rm when} & r_{12}^{2} = \left( r_{1} - r_{2} \right)^{2}
\end{array}
\end{equation}

and <math>P_l(1)=1</math>, <math>P_{l}(-1)=(-1)^{l}</math>.

Thus,

\begin{equation}
I_{l+1}\left(\frac{r_1r_2}{r_{12}}R^\prime\right) = \left( 2l+1 \right) I_{l}(R) - I_{l-1} \left( \frac{r_{1} r_{2}}{r_{12}} R^{\prime} \right)
\end{equation}

If

\begin{equation}
R = r_{1}^{a-1} r_{2}^{b-1} r_{12}^{c+2} e^{-\alpha r_{1} - \beta r_{2}}
\end{equation}

then [G.W.F. Drake, Phys. Rev. A 18, 820 (1978)]

\begin{equation}
I_{l+1}(r_1^ar_2^br_{12}^c) =
\frac{2l+1}{c+2} I_{l} \left( r_{1}^{a-1} r_{2}^{b-2} r_{12}^{c+2} \right)+I_{l-1} \left( r_{1}^{a} r_{2}^{b} r_{12}^{c} \right)
\end{equation}

For the case <math>c=-2</math>, take

\begin{equation}
R = r_{1}^{a-1} r_{2}^{b-1} \ln r_{12} e^{ -\alpha r_{1} -\beta r_{2}}
\end{equation}

Then

\begin{equation}
I_{l+1} \left( r_{1}^{a} r_{2}^{b} r_{12}^{c} \right) = I_{l} \left( r_{1}^{a-1} r_{2}^{b-1} \ln r_{12} \right) \left( 2l + 1 \right) - I_{l-1} \left( r_{1}^{a} r_{2}^{b} r_{12}^{-2} \right)
\end{equation}

This allows all <math>I_{l}</math> integrals to be calculated from tables of <math>I_0</math> and
<math>I_1</math> integrals.

===The General Integral===

The above results for the angular and radial integrals can now be combined into a general formula for integrals of the type

\begin{equation}
I = \int \int d{\bf r}_{1} d{\bf r}_{2} R_{1} \mathcal{Y}^{M^{\prime}}_{l^{\prime}_{1} l^{\prime}_{2} L^{\prime}} \left( \hat{r}_{1}, \hat{r}_{2} \right) T_{k_{1} k_{2} K}^{Q} \left( {\bf r}_{1},{\bf r}_{2} \right) R_{2} \mathcal{Y}_{l_{1} l_{2} L}^{M} \left( \hat{r}_{1}, \hat{r}_{2} \right)
\end{equation}

where

\begin{equation}
\mathcal{Y}^{M}_{l_{1} l_{2} L} \left( \hat{r}_{1}, \hat{r}_{2} \right) = \sum_{m_1,m_2} \langle l_{1} l_{2} m_{1} m_{2} | LM \rangle Y^{m_{1}}_{l_{1}} \left( \hat{r}_{1} \right) Y^{m_{2}}_{l_{2}} \left( \hat{r}_{2} \right)
\end{equation}

and

\begin{equation}
T^Q_{k_{1} k_{2} K} \left( {\bf r}_{1}, {\bf r}_{2} \right) = \sum_{q_{1}, q_{2}} \langle k_{1} k_{2} q_{1} q_{2} | KQ \rangle Y_{k_{1}}^{q_{1}} \left( \hat{r}_{1} \right) Y^{q_{2}}_{k_{2}} \left( \hat{r}_{2} \right)
\end{equation}

The basic idea is to make repeated use of the formula

\begin{equation}
Y^{m_{1}}_{l_{1}} \left( \hat{r}_{1} \right) Y^{m_{2}}_{l_{2}} \left( \hat{r}_{1} \right) = \sum_{l m} \left( \frac{\left( 2l_{1}+1 \right) \left(2l_{2}+1 \right) \left(2l+1 \right)}{4 \pi} \right)^{1/2}
\times \left( \begin{array}{ccc}
l_1 & l_2 & l \\
m_1 & m_2 & m
\end{array} \right) \left(\begin{array}{ccc}
l_1 & l_2 & l \\
0 & 0 & 0
\end{array} \right)
Y^{m*}_{l} \left( \hat{r}_{1} \right)
\end{equation}

where

\begin{equation}
Y^{m*}(\hat{r}) = (-1)^mY_l^{-m}(\hat{r})
\end{equation}

and

\begin{equation}
\left(\begin{array}{ccc}
l_1 & l_2 & l \\
m_1 & m_2 & m
\end{array} \right) = \frac{\left( -1 \right)^{l_{1} - l_{2} - m}}{\left( 2l + 1 \right)^{1/2}} \left( l_{1} l_{2} m_{1} m_{2} | l, -m \right)
\end{equation}

is a 3-<math>j</math> symbol. In particular, write

\begin{array}{cc}
Y^{m_1^\prime}_{l_1^\prime}(\hat{r}_1)^{*} \underbrace{Y_{k_{1}}^{q_{1}} \left( \hat{r}_{1} \right) Y^{m_{1}}_{l_{1}} \left( \hat{r}_{1} \right)} & = & \sum_{\Lambda M}(\ldots)Y^{M}_\Lambda(\hat{r}_1)\\
\ \ \ \ \ \ \ \sum_{\lambda_1\mu_1}Y_{\lambda_1}^{\mu_1}(\hat{r}_1) & & \\
& & \\
Y^{m_{2}^{\prime}}_{l_{2}^{\prime}} \left( \hat{r}_{2} \right) \underbrace{Y^{q_{2}}_{k_{2}} \left( \hat{r}_{2} \right) Y_{l_{2}}^{m_{2}} \left( \hat{r}_{2} \right)} & = & \sum_{\Lambda^{\prime} M^{\prime}} \left( \ldots \right)Y_{\Lambda^{\prime}}^{M^{\prime}*} \left( \hat{r}_{2} \right) \\
\ \ \ \ \ \ \ \ \ \sum_{\lambda_{2} \mu_{2}}Y_{\lambda_{2}}^{\mu_{2}} \left( \hat{r}_{2} \right) & &
\end{array}

The angular integral then gives a factor of <math>2 \pi \delta_{\Lambda, \Lambda^{\prime}} \delta_{M, M^{\prime}} P_{\Lambda} \left( \cos \theta_{12} \right)</math>. The total integral therefore reduces to the form

\begin{equation}
I = \sum_{\Lambda} C_{\Lambda} I_{\Lambda} \left( R_{1} R_{2} \right)
\end{equation}

where <math>C_{\Lambda} = \sum_{\lambda_{1}, \lambda_{2}} C_{ \lambda_{1}, \lambda_{2}, \Lambda}</math>. For further details and derivations, including graphical representations, see G.W.F. Drake, Phys. Rev. A <math>{\bf 18}</math>, 820 (1978).

==Graphical Representation==

[[File:GraphRep1.jpg|500px]]

==Matrix Elements of H==
Recall that

\begin{equation}
H = -\frac{1}{2} \nabla^{2}_{1} - \frac{1}{2} \nabla^{2}_{2} - \frac{1}{r_1} - \frac{1}{r_2} + \frac{Z^{-1}}{r_{12}}
\end{equation}

Consider matrix elements of

\begin{equation}
\nabla^2_1 = \frac{1}{r_1^2} \frac{\partial}{\partial r_1}\left(r_1^2\frac{\partial}{\partial r_1} \right) + \frac{1}{r^2}\frac{\partial}{\partial r}\left(r^2\frac{\partial}{\partial r}\right) - \frac{(\vec{l}_1^Y)^2}{r_1^2} + \frac{2 \left( r_1 - r_2 \cos \theta \right)}{r} \frac{\partial^2}{\partial r_1 \partial r} - 2 \left( \nabla^Y_1 \cdot {\bf r}_2 \right) \frac{1}{r} \frac{\partial}{\partial r}
\end{equation}

where <math>r \equiv r_{12}</math> and <math>\cos\theta \equiv \cos\theta_{12}</math>. Also in what follows, define <math>\hat{\nabla}^Y_1 \equiv r_1\nabla^Y_1</math> where <math>\nabla^Y</math> operates only on the spherical harmonic part of the wave function, and similarly for <math>\vec{l}_1^Y</math>.

A general matrix element is

\begin{equation}
\langle r_1^{a^\prime} r_2^{b^\prime} r_{12}^{c^\prime} e^{-\alpha^\prime r_1 - \beta^\prime r_2} \mathcal{Y}^{M^\prime}_{l^\prime_1 l^\prime_2 L^\prime} \left( \hat{r}_1, \hat{r}_2 \right) \left| \nabla^2_1 \right| r_1^a r_2^b r_{12}^c e^{-\alpha r_1 - \beta r_2} \mathcal{Y}^M_{l_1 l_2 L} \left( \hat{r}_1, \hat{r}_2 \right) \rangle
\end{equation}

Since <math>\nabla^2_1</math> is rotationally invariant, this vanishes unless <math>L=L^\prime</math>, <math>M=M^\prime</math>. Also <math>\nabla_1^2</math> is Hermitian, so that the result must be the same whether it operates to the right or left, even though the results look very different. In fact, requiring the results to be the same yields some interesting and useful integral identities as follows:

Put

\begin{equation}
\mathcal{F} = F \mathcal{Y}^M_{l_1 l_2 L} \left( \hat{r}_1, \hat{r_2} \right)
\end{equation}

and

\begin{equation}
F = r_1^a r_2^b r_{12}^c e^{-\alpha r_1 - \beta r_2}
\end{equation}

Then

\begin{equation}
\nabla^2_1 \mathcal{F} = \left\{ \frac{1}{r_1^2} \left[ a \left( a + 1 \right) - l_1 \left( l_1 + 1 \right) \right] + \frac{c \left( c + 1 \right)}{r^2} + \alpha^2\frac{2\alpha \left( a + 1 \right)}{r_1} + \frac{2 \left(r_1 - r_2 \cos \theta \right) }{r_1r^2}c \left[ a - \alpha r_1 \right] - \frac{2c}{r^2} \left( \hat{\nabla}^Y_1 \cdot \hat{r}_2 \right) \frac{r_2}{r_1} \right\} \mathcal{F}
\end{equation}

and

\begin{align}
\langle \mathcal{F}^\prime \left| \nabla^2_1 \right| \mathcal{F} \rangle & = \sum_\Lambda \int d\tau_r F^\prime C_\Lambda \left( 1 \right) P_\Lambda \left( \cos \theta \right) \times \left\{ \frac{1}{r_1^2} \left[ a \left( a + 1 \right) - l_1 \left( l_1 + 1 \right) \right] -\frac{2 \alpha \left( a + 1 \right)}{r_1} + \frac{c \left( c + 1 \right)}{r^2} +\alpha^2 + \frac{2 \left( r_2 - r_2 \cos \theta \right)}{r_1r^2}c \left( a - \alpha r_1 \right) \right\} F \\
& + \sum_\Lambda \int d\tau_r F^\prime C_\Lambda \left( \hat{\nabla}^Y_1 \cdot \hat{r}_2 \right) P_\Lambda \left( \cos \theta \right) \left( \frac{-2c r_2}{r_1 r^2} \right) F
\end{align}

where

\begin{equation}
\int d\tau_r = \int^\infty_0 r_1 dr_1 \int^\infty_0 r_2 dr_2 \int^{r_1+r_2}_{\left| r_1 - r_2 \right|} r dr
\end{equation}

For brevity, let the sum over <math>\Lambda</math> and the radial integrations be understood, and let <math>\left( \nabla_1^2 \right)</math> stand for the terms which appear in the integrand. Then operating to the right gives

\begin{equation}
(\nabla_1^2)_R = \frac{1}{r_1^2} \left[ a \left( a + 1 \right) - l_1 \left( l_1 + 1 \right) \right] + \frac{c \left( c + 1 \right)}{r^2} + \alpha^2 - \frac{2 \alpha \left( a + 1 \right)}{r_1}
+ \frac{2 \left( r_1 - r_2 \cos \theta \right)}{r_1 r^2}c \left( a - \alpha r_1 \right) - \frac{2c r_2}{r_1 r^2} \left( \hat{\nabla}^Y_1 \cdot \hat{r}_2 \right)
\end{equation}

Operating to the left gives

\begin{equation}
(\nabla_1^2)_L = \frac{1}{r_1^2} \left[a^\prime \left( a^\prime + 1 \right) - l_1^\prime \left( l_1^\prime + 1 \right) \right] + \frac{c^\prime \left( c^\prime + 1 \right)}{r^2} + \alpha^{\prime 2} -\frac{2 \alpha^\prime \left( a^\prime + 1 \right)}{r_1}
+ \frac{2 \left( r_1 - r_2 \cos \theta \right)}{r_1 r^2} c^\prime \left( a^\prime - \alpha^\prime r_1 \right) - \frac{2 c^\prime r_2}{r_1 r^2} \left( \hat{\nabla}^{Y^\prime}_1 \cdot \hat{r}_2 \right)
\end{equation}

Now put

\begin{array}{cc}
a_+ = a+a^\prime, & \hat{\nabla}_1^+ = \hat{\nabla}_1^Y + \hat{\nabla}_1^{Y\prime} \\
a_{-} = a-a^\prime, & \hat{\nabla}_1^{-} = \hat{\nabla}_1^Y - \hat{\nabla}_1^{Y\prime}
\end{array}

etc., and substitute <math>a^\prime = a_+ - a</math>, <math>c^\prime = c_+ - c</math>, and <math>\alpha^\prime = \alpha_+ - \alpha</math> in <math>(\nabla^2_1)_L</math>. If <math>a_+</math>, <math>c_+</math> and <math>\alpha_+</math> are held fixed, then the equation

\begin{equation}
(\nabla^2_1)_R = (\nabla^2_1)_L
\end{equation}

must be true for arbitrary <math>a</math>, <math>c</math>, and <math>\alpha</math>. Their coefficients must thus vanish.

This yields the integral relations

\begin{equation}
\label{I}
\frac{\left( r_1 - r_2 \cos \theta \right)}{r_1 r^2} = \frac{1}{c_+} \left( \frac{- \left( a_+ + 1 \right)}{r_1^2} + \frac{\alpha_+}{r_1} \right) \tag{I}
\end{equation}

from the coefficient of <math>a</math> and

\begin{equation}
\label{II}
\frac{ \left( r_1 - r_2 \cos \theta \right) \left( a_+ - \alpha_+ r_1 \right)}{r_1 r^2} = \frac{r_2}{r_1 r^2}\left( \hat{r}_2 \cdot \hat{\nabla}_1^+ \right) - \frac{ \left( c_+ + 1 \right)}{r^2} \tag{II}
\end{equation}

from the coefficient of <math>c</math>. The coefficient of <math>\alpha</math> gives an equation equivalent to (I).

Furthermore, if can be show that (see problem)

\begin{align}
& \sum_\Lambda \int d\tau_r \frac{r^c}{r^2} C_\Lambda \left( \hat{r}_2 \cdot \hat{\nabla}_1^Y \right) P_\Lambda \left( \cos \theta \right) \\
= & \sum_\Lambda \int d\tau_r \frac{r^c}{c r_1 r_2}C_\Lambda \left( 1 \right)P_\Lambda \left( \cos \theta \right) \times \left( \frac{l_1^\prime \left( l_1^\prime + 1 \right) - l_1 \left( l_1 + 1 \right) -\Lambda \left( \Lambda + 1 \right)}{2} \right)
\end{align}

and similarly for <math>\left( \hat{r}_2 \cdot \hat{\nabla}^{Y\prime}_1 \right)</math> with <math>l_1</math> and <math>l_1^\prime</math> interchangeable, then it follows that

\begin{equation}
\frac{r^{c_+}}{r^2} \left( \hat{r}_2 \cdot \hat{\nabla}_1^+ \right) = - \frac{r^{c_+}}{c_+ r_1 r_2} \Lambda \left( \Lambda + 1 \right)
\end{equation}

and

\begin{equation}
\frac{r^{c_+}}{r^2} \left( \hat{r}_2 \cdot \hat{\nabla}_1^- \right) = \frac{r^{c_+}}{c_+ r_1 r_2} \left[ l_1^\prime \left( l_1^\prime + 1 \right) - l_1 \left( l_1 + 1 \right) \right]
\end{equation}

where equality applies after integration and summation over <math>\Lambda</math>.

Thus (II) becomes

\begin{equation}\label{IIa}
\frac{ \left( r_1 - r_2 \cos \theta \right) \left( a_+ - \alpha_+ r_1 \right)}{r_1 r^2} = - \frac{\Lambda \left( \Lambda + 1 \right)}{c_+ r^2_1} - \frac{ \left( c_+ + 1 \right)}{r^2} \tag{IIa}
\end{equation}

===Problem===

Prove the integral relation

\begin{align}
& \sum_\Lambda \int d\tau_r f \left( r_1, r_2 \right) \frac{1}{r} \left( \frac{d}{dr} q(r) \right) C_\Lambda \left( \hat{r}_2 \cdot \hat{\nabla}^Y_1 \right) P_\Lambda \left( \cos \theta \right) \\
= & \sum_\Lambda \int d\tau_r \frac{ f \left( r_1, r_2 \right)}{r_1, r_2} q(r) C_\Lambda \left( 1 \right) P_\Lambda \left( \cos \theta \right) \left(\frac{l_1^\prime \left( l_1^\prime + 1 \right) - l_1 \left( l_1 + 1 \right) - \Lambda \left( \Lambda + 1 \right)}{2}\right)
\end{align}

where the coefficients <math>C_\Lambda \left( 1 \right)</math> are the angular coefficients from the overlap integral

\begin{align}
\int & d\Omega\, \mathcal{Y}^{M*}_{l_1^\prime l_2^\prime L}(\hat{r}_1,\hat{r}_2)\, \mathcal{Y}^{M}_{l_1 l_2 L}(\hat{r}_1,\hat{r}_2)\\
= & \sum_\Lambda C_\Lambda(1)P_\Lambda(\cos\theta_{12}).
\end{align}

Hint: Use the fact that <math>l_1^2</math> is Hermitian so that
<math>
\int d\tau (l_1^2\mathcal{Y}^\prime)^*q(r)\mathcal{Y} = \int d\tau
\mathcal{Y}^{\prime *}l_1^2(q(r)\mathcal{Y})
</math>
with <math>\ \vec{l}_1= \frac{1}{i}\vec{r}_1\times\nabla_1</math>.

It is also useful to use
<math>\begin{eqnarray}\ \
\cos\theta\, P_L(\cos\theta) &=& \frac{1}{2L+1}[L\,P_{L-1}(\cos\theta) + (L+1)P_{L+1}(\cos\theta)]
\\
(\cos^2\theta-1)P_L(\cos\theta) &=&
\frac{L(L-1)}{(2L-1)(2L+1)}P_{L-2}(\cos\theta)\nonumber
-\frac{2(L^2+L-1)}{(2L-1)(2L+3)}P_L(\cos\theta) + \frac{(L+1)(L+2)}{(2L+1)(2L+3)}P_{L+2}(\cos\theta)\nonumber\\
&=& \frac{L(L-1)}{(2L-1)(2L+1)}[P_{L-2}(\cos\theta) - P_L(\cos\theta)]
+ \frac{(L+1)(L+2)}{(2L+1)(2L+3)}[P_{L+2}(\cos\theta) - P_L(\cos\theta)]
\end{eqnarray}</math>

together with a double application of the integral recursion relation
<math>
I_{L+1}\left(\frac{1}{r}\frac{d}{dr}q(r)\right) =
(2L+1)I_L\left(\frac{1}{r_1r_2}q(r)\right)+I_{L-1}\left(\frac{1}{r}\frac{d}{dr}q(r)\right)
</math>

Of course <math>l_1^2\mathcal{Y}^M_{l_1l_2L}(\hat{r}_1,\hat{r}_2)=l_1(l_1+1)\mathcal{Y}^M_{l_1l_2L}(\hat{r}_1,\hat{r}_2)</math>. Begin by expanding
<math>
l_1^2(g\mathcal{Y}) = \mathcal{Y}l_1^2g + g\,l_1^2\mathcal{Y} + 2(\vec{l}_1g)\cdot(\vec{l}_1\mathcal{Y})
</math>, and show that

<math>
\begin{eqnarray}\ \
(\vec{l}_1g)\cdot(\vec{l}_1\mathcal{Y}) & = & \frac{r_1^2}{r}\frac{dg}{dr}\vec{r}_2\cdot \nabla_1\mathcal{Y}\,\,\,\,\mbox{ and}\\
l_1^2g(r) &=& 2r_1r_2\cos\theta\frac{1}{r}\frac{dg}{dr} + r_1^2r_2^2(\cos^2\theta - 1)
\frac{1}{r}\frac{d}{dr}\left(\frac{1}{r}\frac{dg}{dr}\right).
\end{eqnarray}
</math>

The proof amounts to showing that the term <math>l_1^2g(r)</math> can be replaced by
<math>\Lambda(\Lambda+1)g(r)</math> after multiplying by <math>P_\Lambda(\cos\theta)</math>
and integrating by parts with respect to the radial integrations over <math>r_1</math>, <math>r_2</math> and
<math>r \equiv |\vec{r}_1 - \vec{r}_2|</math>. Remember that <math>\cos\theta</math> is just a short-hand notation for the radial function <math>(r_1^2 + r_2^2 - r^2)/(2r_1r_2)</math>.

==General Hermitian Property==

Each combination of terms of the form
\begin{equation}
<f>_R = a^2f_1+af_2+abf_3+b^2f_4+bf_5+af_6\nabla_1^Y+bf_7\nabla^Y_2+f_8(Y)
\end{equation}
can be rewritten
\begin{equation}
<f>_L= (a_+-a)^2f_1+(a_+-a)f_2 + (a_+-a)(b_+-b)f_3+(b_+-b)^2f_4
+(b_+-b)f_5+(a_+-a)f_6\nabla^{Y\prime}_1+(b_+-b)f_7\nabla^{Y\prime}_2+f_8)Y^\prime
\end{equation}
where as usual <math>\nabla_1^Y</math> acts only on the spherical harmonic part of the wave function denoted for short by <math>Y</math>.
Since these must be equal for arbitrary <math>a</math> and <math>b</math>,
\begin{equation}
a_+^2f_1+a_+f_2+_+b_+f_3+b_+f_4+b_+f_5+a_+f_6\nabla^{Y\prime}_1 +
b_+f_7\nabla^{Y\prime}_2+f_8(Y^\prime)-f_8(Y)=0
\end{equation}

Adding the corresponding expression with <math>Y</math> and <math>Y^\prime</math> interchanged
yields
<math>\label{III}
a_+^2f_1+a_+f_2+a_+b_+f_3+b_+^2f_4+b_+f_5+\frac{1}{2}a_+f_6\nabla^+_1+\frac{1}{2}b_+f_7\nabla^+_2=0\tag{III}
</math>

Subtracting gives
\begin{eqnarray*}
f_8(Y)-f_8(Y^\prime)&=&-\frac{1}{2}[a_+f_6\nabla^-_1+b_+f_7\nabla_2^-]\tag{IV}\\
a[-2a_+f_1-2f_2-b_+f_3-f_6\nabla^+_1]&=&0\tag{V}\\
b[-2b_+f_4-2f_5-a_+f_3-f_7\nabla^+_2]&=&0\tag{VI}\\
\end{eqnarray*}

Adding the two forms gives

\begin{eqnarray}
<f>_R+<f>_L=\frac{1}{2}(a_+^2+a_-^2)f_1+a_+f_2+\frac{1}{2}(a_+b_++a_-b_-)f_3\nonumber\\
\ +\frac{1}{2}(b_+^2+b_-^2)f_4+b_+f_5+\frac{1}{2}f_6(a_+\nabla_1^++a_-\nabla^-_1)\nonumber\\
\ +\frac{1}{2}(b_+\nabla_2^++b_-\nabla^-_2)+f_8(Y)+f_8(Y^\prime)\nonumber
\end{eqnarray}

Subtracting <math>\frac{x}{2}\times({\rm I})</math> gives

\begin{eqnarray}\ \
<f>_R+<f>_L = \frac{1}{2}[(1-x)a_+^2 + a_-^2]f_1 +
(1-\frac{x}{2})a_+f_2+\frac{1}{2}[(1-x)a_+b_++a_-b_-]f_3\nonumber\\
+\frac{1}{2}[(1-x)b_+^2+b_-^2]f_4+(1-\frac{x}{2})f_5b_++\frac{1}{2}f_6[(1-\frac{x}{2})a_+\nabla_1^++a_-\nabla^-_1]\nonumber\\
+\frac{1}{2}[(1-\frac{x}{2})b_+\nabla^+_2+b_-\nabla^-_2]+f_8(Y)+f_8(Y^\prime)\nonumber
\end{eqnarray}

If <math>x=1</math>,
\begin{eqnarray}
<f>_R+<f>_L &=&
\frac{1}{2}[a_-^2f_1+a_+f_2+a_-b_-f_3+b_-^2f_4+b_+f_5\nonumber\\
&+&\left(\frac{1}{2}a_+\nabla^+_1+a_-\nabla^-_1\right)f_6+\left(\frac{1}{2}b_+\nabla^+_2+b_-\nabla^-_2\right)f_7]+f_8(Y)+f_8(Y^\prime)\nonumber
\end{eqnarray}

The General Hermitian Property for arbitrary <math>x</math> gives
\begin{eqnarray}\ \
\nabla_1^2=
\frac{1}{4}\{\frac{1}{r_1^2}[(1-x)a_+^2+a_-^2+2(1-\frac{x}{2})a_+-2[l_1(l_1+1)+l_1^\prime(l_1^\prime+1)]\nonumber\\
-\frac{2}{r_1}[(1-x)\alpha_+a_++\alpha_-a_-+2(1-\frac{x}{2})\alpha_+]+(1-x)\alpha_+^2
+ \alpha^2_-]\nonumber\\
+\frac{2(r_1-r_2\cos\theta)}{r_1r^2}[(1-x)(a_+-\alpha_+r_1)c_++(a_--\alpha_-r_2)c_-]\nonumber\\
-\frac{2r_2}{r_1r^2}[(1-\frac{x}{2})c_+\hat{r}_2\cdot\hat{\nabla}_1^++c_-\hat{r}_2\hat{\nabla}_1^-]+\frac{(1-x)c_+^2+c_-^2+2(1-\frac{x}{2})c_+}{r^2}\}\nonumber
\end{eqnarray}.

Use
<math>\begin{eqnarray}\ \
\frac{-2r_2}{r_1r^2}[(1-\frac{x}{2})c_+\hat{r}_2\hat{\nabla}_1^++c_-\hat{r}_2\hat{\nabla}_1^-]=\frac{2}{r_1^2}\left[(1-\frac{x}{2})\Lambda(\Lambda+1)-\frac{c_-}{c_+}[l_1^\prime(l_1^\prime+1)+l_1(l_1+1)]\right]\end{eqnarray}</math>,

<math>\begin{eqnarray}
\frac{2(r_1-r_2\cos\theta)}{r_1r^2}(a_--\alpha_-r_1)c_- = 2\frac{c_-}{c_+}\left[\frac{-1}{r_1^2}[a_-(a_++1)]+\frac{1}{r_1}[a_-\alpha_++\alpha_-(a_++2)]-\alpha_-\alpha_+\right]\nonumber
\end{eqnarray}</math>,

and
<math>\begin{eqnarray}\ \
\frac{2(r_1-r_2\cos\theta)}{r_1r^2}(a_+-\alpha_+r_1)c_+ = 2\left[-\frac{1}{r_1^2}[a_+(a_++1)]+\frac{1}{r_1}[a_+\alpha_++\alpha_+(a_++2)]-\alpha_+^2\right]
\end{eqnarray}</math>.

Substituting into <math>\nabla_1^2</math> gives

\begin{eqnarray}
\nabla^2_1 = \frac{1}{4}\{\frac{1}{r_1^2}[-(1-x)a_+^2 + a_-^2 +xa_+ +
2(1-\frac{x}{2})\Lambda(\Lambda+1)\nonumber\\
-2l_1(l_1+1)(1-\frac{c_-}{c_+}) -
2l_1^\prime(l_1^\prime+1)(1+\frac{c_-}{c_+})-\frac{2c_-a_-}{c_+}(a_++1)]\nonumber\\
- \frac{2}{r_1}\left[-(1-x)\alpha_+(a_++2)+\alpha_-a_-+2(1-\frac{x}{2})\alpha_+
-\frac{c_-}{c_+}[a_-\alpha_++\alpha_-(a_++2)]\right]\nonumber\\
-(1-x)\alpha_+^2+\alpha_-^2-2\frac{c_-}{c_+}\alpha_-\alpha_+ +\frac{1}{r^2}\left[(1-x)c_+^2+c_-^2+2(1-\frac{x}{2})c_+\right]\}\nonumber
\end{eqnarray}

This has the form
<math>
\nabla_1^2 =
\frac{1}{4}\left[\frac{A_1}{r_1^2}+\frac{B_1}{r_1}+\frac{C_1}{r^2}+D_1\right]
</math>
with

<math>\begin{array}{ccc}
A_1=-(1-x)a_+^2+a_-^2
+xa_++2(1-\frac{x}{2})\Lambda(\Lambda+1)-2l_1(l_1+1)(1-\frac{c_-}{c_+})\
-2l_1^\prime(l_1^\prime+1)(1+\frac{c_-}{c_+})-\frac{2c_-a_-}{c_+}(a_++1) \ \
\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \
\end{array}
</math>

<math>
B_1 =
2\left[(1-x)\alpha_+(a_++2)-\alpha_-a_--2(1-\frac{x}{2})\alpha_++\frac{c_-}{c_+}[a_-\alpha_++\alpha_-(a_++2)]\right]
</math>

<math>
C_1 = (1-x)c_+^2+ c_-^2 +2(1-\frac{x}{2})c_+
</math>

<math>
D_1 = -1(1-x)\alpha_+^2 +\alpha_-^2 - 2\frac{c_-}{c_+}\alpha_-\alpha_+
</math>

The complete Hamiltonian is then (in Z-scaled a.u.)

<math>\begin{eqnarray}
H =-\frac{1}{2}\nabla^2_1 -\frac{1}{2}\nabla^2_2 -\frac{1}{r_1} -\frac{1}{r_2} +\frac{Z^{-1}}{r}\nonumber\\
= -\frac{1}{2}\nabla^2_1 -\frac{1}{2}\nabla^2_2 -\frac{1}{r_1} -\frac{Z-1}{Zr_2}+Z^{-1}\left(\frac{1}{r}-\frac{1}{r_2}\right)\nonumber\\
\ =-\frac{1}{8}\left[\frac{A_1}{r_1^2}+\frac{B_1+8}{r_1}+\frac{C_1}{r^2}+D_1+D_2 +\frac{A_2}{r_2^2} + \frac{B_2 + 8(Z-1)/Z}{r_2}+\frac{C_2}{r^2}\right]\nonumber
+ Z^{-1}\left(\frac{1}{r}-\frac{1}{r_2}\right)\nonumber
\end{eqnarray}</math>

The screened hydrogenic energy is
<math>\begin{eqnarray}
E_{SH}=-\frac{1}{2}\left[\frac{1}{n_1^2}+ \left(\frac{Z-1}{Z}\right)^2\frac{1}{n_2^2}\right]
\end{eqnarray}</math>
so that

<math>\begin{eqnarray}
H-E_{SH} &=& -\frac{1}{8}[
\frac{A_1}{r_1^2}+\frac{B_1+8}{r_1}+\frac{C_1}{r^2} + D_1 + D_2
-\frac{4}{n_1^2} -\left(\frac{Z-1}{Z}\right)^2\frac{4}{n_2^2}\nonumber
+\frac{A_2}{r_2}+\frac{B_2+8(Z-1)/Z}{r_2}+\frac{C_2}{r^2}] +
Z^{-1}\left(\frac{1}{r} -\frac{1}{r_2}\right)\nonumber
\end{eqnarray}</math>

==Optimization of Nonlinear Parameters==
- The traditional method of performing Hylleraas calculations is to write the
basis set in the form

<math>
\Psi = \sum_{i,j,k}c_{i,j,k}\varphi_{i,j,k}(\alpha,\beta)\pm exchange
</math>

with

<math>
\varphi_{i,j,k}(\alpha,\beta) = r_1^ir_2^jr_{12}^ke^{-\alpha r_1 - \beta
r_2}\mathcal{Y}^M_{l_1l_2L}.
</math>

The usual procedure is to set <math>\alpha = Z</math> so that it represents the inner
<math>1s</math> electron, and then to vary <math>\beta</math> so as to minimize the energy.

[[File:OptNon.jpg|500px]]

Since <math>\beta</math> appears in <math>\Psi</math> as a non-linear parameter, the entire
calculation must be repeated for each value of <math>\beta</math>. However, the minimum
becomes progressively smaller as the basis set is enlarged.

[[File:OptNon02.jpg|500px]]

*Difficulties

1. If the basis set is constructed so that <math> i +j +j \leq N</math>, the the
number of terms is
<math>
(N+1)(N+2)(N+3)/6
</math>
N=14 already gives 680 terms and an accuracy of about 1 part in <math>10^{10}</math> for
low-lying states. A substantial improvement in accuracy would require much
larger basis sets, together with multiple precision arithmetic to avoid loss
of significant figures when high powers are included.

2. The accuracy rapidly deteriorates as one goes to more highly excited
states - about 1 significant figure is lost each time the principle quantum
number is increased.

*Cure

We have found that writing basis sets in the form
\begin{eqnarray}
\Psi &=& \sum_{i,j,k}\left(c^{(1)}_{i,j,k}\varphi_{ijk}(\alpha_1\beta_1) +
c_{ijk}^{(2)}\varphi_{ijk}(\alpha_2\beta_2)\right) \pm exchange \nonumber\\
&=& \psi({\bf r}_1,{\bf r}_2) \pm \psi({\bf r}_2,{\bf r}_1)\nonumber
\end{eqnarray}
so that each combination of powers is included twice with different nonlinear
parameters gives a dramatic improvement in accuracy for basis sets of about
the same total size.

However, the optimization of the nonlinear parameters is now much more
difficult, and an automated procedure is needed.

If
<math>
E = \frac{<\Psi|H|\Psi>}{<\Psi|\Psi>}</math> and we assume <math><\Psi|\Psi> = 1</math>, then

<math>
\frac{\partial E}{\partial\alpha_t} = -2 <\Psi|H-E|r_1\psi({\bf r}_1,{\bf
r}_2;\alpha_t)\pm r_21\psi({\bf r}_2,{\bf
r}_1;\alpha_t)>
</math>
where
<math>
\psi({\bf r}_1,{\bf r}_2;\alpha_t) =
\sum_{i,j,k}c_{ijk}^{(t)}\varphi_{ijk}(\alpha_t\beta_t)
</math>

*Problem

1. Prove the above.

2. Prove that there is no contribution from the implicit dependence of
<math>c_{ijk}^{(t)}</math> on <math>\alpha_t</math> if the linear parameters have been optimized.

==The Screened Hydrogenic Term==

If the Hamiltonian is written in the form (in Z-scaled a.u.)
<math>\begin{eqnarray}\
H = H_0({\bf r}_1,Z) + H_0({\bf r}_2,Z-1) +
Z^{-1}\left(\frac{1}{r}-\frac{1}{r_2}\right)
\end{eqnarray}</math>
with
<math>\
H_0({\bf r}_1,Z) = -\frac{1}{2}\nabla_1^2 - \frac{1}{r_1}\nonumber
</math> and
<math>
H_0({\bf r}_2,Z-1) = -\frac{1}{2}\nabla_2^2 - \frac{Z-1}Z{r_2}\nonumber
</math>,
then the eigenvectors of <math>H_0({\bf r},Z) +H_0({\bf r}_2,Z-1)</math> are products of
hydrogenic orbitals
<math>
\Psi_0=\psi_0(1s,Z)\psi_0(nl,Z-1)
</math>
and the eigenvalue is
<math>\begin{eqnarray}
E_{SH} = \left[-\frac{1}{2} - \left(\frac{Z-1}{Z}\right)^2\frac{1}{2n^2}\right]Z^2 a.u.
\end{eqnarray}</math>
called the screened hydrogenic eigenvalue. From highly excited states,
<math>E_{SH}</math> and <math>\Psi_0</math>are already excellent approximations.

For example, for the 1s8d states, the energies are

<math>\begin{eqnarray}\\
E(1s8d^1D)&=& -2.007\,816\,512\,563\,81 {\rm\ a.u.}\nonumber\\
E(1s8d^3D)&=& - 2.007\,817\,934\,711\,71 {\rm\ a.u.}\nonumber\\
E_{SH} &=& -2.007\,812\,5 {\rm\ a.u.} \nonumber
\end{eqnarray}</math>

It is therefore advantageous to include the screened hydrogenic terms in the
basis set so that the complete trial function becomes
<math>
\Psi = c_0\Psi_0 +
\sum_{ijk}\left[c_{ijk}^{(1)}\varphi_{ijk}(\alpha_1,\beta_1)+c_{ijk}^{(2)}\varphi(\alpha_2,\beta_2)\right]\pm
{\rm exchange}
</math>

Also, the variational principal can be re-expressed in the form
<math>\begin{eqnarray}
E= E_{SH}+\frac{<\Psi|H-E_{SH}|\Psi>}{<\Psi|\Psi>}
\end{eqnarray}</math>
so that the <math>E_{SH}</math> term can be cancelled analytically from the matrix
elements.

For example
<math>
<\Psi_0|H-E_{SH}|\Psi_0> = 0
</math>

and
<math>
<\varphi_{ijk}|H-E_{SH}|\Psi_0> =
Z^{-1}<\varphi_{ijk}|\frac{1}{r}-\frac{1}{r_2}|\Psi_0>
</math>

Recall that
<math>\begin{eqnarray}\ \
I_0(a,b,c) = \frac{2}{c+2}\sum^{[c+1]/2}_{i=0}\left(\begin{array}{c}
c+2\\
2i+1
\end{array}\right) \left[f(p,q;\beta) + f(p^\prime,q^\prime;\alpha)\right]
\end{eqnarray}</math>
where

<math>\begin{eqnarray}
f(p,q;x) =
\frac{q!}{x^{q+1}(\alpha+\beta)^{p+1}}\sum^p_{j=0}\frac{(p+j)!}{j!}\left(\frac{x}{\alpha+\beta}\right)^j
\end{eqnarray}</math>

<math>
\begin{array}{cc}
p=a+2i+2 & p^\prime = b+2i+2\\
q= b+c-2i+2 & q^\prime =a+c-2i+2
\end{array}
</math>

If <math>\beta\ll\alpha</math>, the <math>f(p,q;\beta)\gg f(p^\prime,q^\prime;\alpha)</math> and the
<math>i=0</math> term is the dominant contribution to <math>f(p,q;\beta)</math>. However, since
this term depends only on the sum of powers <math>b+c</math> for <math>r_2</math>, it cancels exactly
from the matrix element of
<math>
\frac{1}{r_{12}} - \frac{1}{r_2}
</math>
and can therefore
be omitted in the calculations of integrands, thereby saving many significant
figures. This is especially valuable when <math>b+c</math> is large and <math>a</math> is small.

==Small Corrections==
*Mass Polarization
If the nuclear mass is not taken to be infinite, then the Hamiltonian is
<math>\begin{eqnarray}\ \
H = \frac{P_N^2}{2M} + \sum_{i=1}\frac{p_i^2}{2m} + V
\end{eqnarray}</math>

[[File:SmallCor.jpg|500px]]

where <math>M =M_A-nm</math> is the nuclear mass and <math>m</math> is the electron mass.

Change variables to the C of M
<math>\begin{eqnarray}
{\bf R} = \frac{1}{M+nm}\left[M{\bf r}_N + m({\bf r}_1 + {\bf r}_2 +
\ldots)\right]
\end{eqnarray}</math>

and relative variables

<math>
{\bf s}_i = {\bf r}_i - {\bf r}_N.
</math>

Then

<math>\begin{eqnarray}\ \ \ \ \
H = \frac{1}{2(M+nm)}{\bf p}_R^2+\frac{1}{2m}\sum_{i=1}^n{\bf p}_{s_i}^2
+\frac{1}{2M}\sum_{i,k}{\bf p}_{s_i}\cdot {\bf p}_{s_k} + V({\bf s}_1, {\bf
s}_2, \ldots)\nonumber
= \frac{1}{M+nm}{\bf p}_R^2+\frac{1}{2\mu}\sum_{i=1}^n{\bf
p}_{s_i}^2+\frac{1}{2M}\sum_{i\neq k}{\bf p}_{s_i}\cdot {\bf p}_{s_k}+V({\bf s}_1, {\bf
s}_2, \ldots)\nonumber
\end{eqnarray}</math>

where
<math>\begin{eqnarray}
\mu = \left(\frac{1}{M} + \frac{1}{m}\right)^{-1}
\end{eqnarray}</math>

If <math>n=2</math>, the Schrödinger equation is
<math>\begin{eqnarray}\ \ \
\left[ -\frac{\hbar^2}{2\mu}(\nabla^2_{s_1}+ \nabla^2_{s_2}) -
\frac{\hbar^2}{M}\nabla_{s_1}\nabla_{s_2}-\frac{Ze^2}{s_1}-\frac{Ze^2}{s_2}+\frac{e^2}{s_{12}}\right]\psi
= E\psi
\end{eqnarray}</math>.

Define the reduced mass Bohr radius
<math>\begin{eqnarray}
a_\mu = \frac{\hbar^2}{\mu e^2} = \frac{m}{\mu}a_0
\end{eqnarray}</math>
and

<math>\begin{eqnarray}
\rho_i = \frac{Zs_i}{a_\mu}\nonumber\\
\varepsilon = \frac{E}{Z^2(e^2/a_\mu)}\nonumber
\end{eqnarray}</math>

The Schrödinger equation then becomes
<math>\begin{eqnarray}\ \ \
\left[
-\frac{1}{2}(\nabla^2_{\rho_1}+\nabla^2_{\rho_2})-\frac{\mu}{M}\vec{\nabla}_{\rho_1}\cdot\vec{\nabla}_{\rho_2} - \frac{1}{\rho_1}-\frac{1}{\rho_2}+\frac{Z^{-1}}{\rho_{12}}\right]\psi = \varepsilon \psi
\end{eqnarray}</math>

The effects of finite mass are thus

1. A reduced mass correction to all energy levels of <math>(e^2/a_\mu)/(e^2/a_0)
= \mu/m</math>, i.e.
<math>\begin{eqnarray}
E_M=\frac{\mu}{m}E_\infty
\end{eqnarray}</math>

2. A specific mass shift given in first order perturbation theory by
<math>\begin{eqnarray}
\Delta\varepsilon =
-\frac{\mu}{M}<\psi|\nabla_{\rho_1}\cdot\nabla_{\rho_2}|\psi>
\end{eqnarray}</math>
or

<math>\begin{eqnarray}
\Delta\varepsilon =
-\frac{\mu}{M}<\psi|\nabla_{\rho_1}\cdot\nabla_{\rho_2}|\psi>\frac{e^2}{a_\mu}\nonumber\\
= -\frac{\mu^2}{mM}<\psi|\nabla_{\rho_1}\cdot\nabla_{\rho_2}|\psi>\frac{e^2}{a_0}\nonumber
\end{eqnarray}</math>

Since <math>\nabla_{\rho_1}\cdot\nabla_{\rho_2}</math> has the same angular properties as
<math>{\bf\rho}_1 \cdot {\bf\rho}_2</math>, the operator is like the product of two
dipole operators. For product type wave functions of the form
$
\psi = \psi(1s)\psi(nl) \pm \mbox{exchange},
$
the matrix element vanishes for all but P-states.

=Radiative Transitions=

==Transition Integrals and the Line Strength==

For an electric dipole transition <math>\gamma'\,L'\,S'\,J'\,M' \rightarrow \gamma\,L\,S\,J\,M</math>, the line strength is defined by
\begin{equation}
S = \sum_{M',\mu,M}|\langle\gamma'\,L'\,S'\,J'\,M'|r_\mu|\gamma\,L\,S\,J\,M\rangle|^2
\end{equation}
Using the Wigner-Eckart theorem, this is related to the reduced matrix element by
\begin{eqnarray}
S(J',J) = |\langle\gamma'\,L'\,S'\,J'||r||\gamma\,L\,S\,J\,M\rangle|^2
\sum_{M',\mu,M}\left|\left(\matrix{J' & 1 & J \cr
-M'& \mu &M \cr}\right)\right|^2
\end{eqnarray}
Using the sum rule for the 3-<math>j</math> symbols
\begin{equation}
\sum_{M',\mu,M}\left|\left(\matrix{J' & 1 & J \cr
-M'& \mu &M \cr}\right)\right|^2 = 1
\end{equation}
we see that the line strength is identical to the reduced matrix element
\begin{equation}
S(J',J) = |\langle\gamma'\,L'\,S'\,J'||r||\gamma\,L\,S\,J\,M\rangle|^2
\end{equation}
For a nonrelativistic calculation in <math>LS</math> coupling, <math>S = S'</math> and we can next strip out the dependence on <math>J</math> and <math>J'</math> by use of Eq.\ (7.1.7) of Edmonds { Angular Momentum in Quantum Mechanics} with <math>k = 1</math> to obtain
\begin{equation}
S(J',J) = (-1)^{L'+S+J+1}[(2J+1)(2J'+1)]^{1/2}\left\{
\matrix{L' & J'& S\cr
J & L & 1\cr}\right\}\langle\gamma' L'|| r || \gamma L\rangle
\end{equation}
where <math>\langle\gamma' L'|| r || \gamma L\rangle</math> is the reduced matrix element if spin is neglected. Note that if spin is neglected, then the line-strength is simply
\begin{equation}
S(L',L) = |\langle\gamma' L'|| r || \gamma L\rangle|^2
\end{equation}
To calculate the reduced matrix element, use a particular example along with the Wigner-Eckart theorem
such as
\begin{equation}
\langle\gamma'\, L'|| r || \gamma\, L\rangle = (-1)^{L'}\langle\gamma'\, L'\, 0| z | \gamma\, L\, 0\rangle{\Large /}\left(\matrix{L' & 1 & L \cr
0 & 0 & 0 \cr}\right)
\end{equation}

===Correlated Two-electron Integrals===
For the case of two-electron integrals, <math>z = z_1 + z_2</math> and <math>r = r_1 + r_2</math>. As discussed in G.W.F. Drake, Phys. Rev. A 18, 820 (1978), the general integral in Hylleraas coordinates for the integral over the <math>z_1</math> term can be written in the form
\begin{equation}
\label{int1}
\langle\gamma'\, L'\, 0| z_1 | \gamma\, L\, 0\rangle = \sum_\Lambda C_\Lambda I_\Lambda(\gamma',\gamma)
\end{equation}
where <math>I_\Lambda(\gamma',\gamma)</math> stands for the purely radial integral
\begin{equation}
\label{radial}
I_\Lambda(\gamma',\gamma) = \int_0^\infty r_1\,dr_1\int_0^\infty r_2\,dr_2\int_{|r_1-r_2|}^{r_1+r_2}r_{12}\, dr_{12}
R_{\gamma'}^*\,r_1\, R_\gamma P_\Lambda(\cos\theta_{12})
\end{equation}
and <math>\cos\theta_{12} = (r_1^2 + r_2^2 - r_{12}^2)/(2r_1r_2)</math> is a short-hand notation for a radial function of <math>r_1</math>, <math>r_2</math> and <math>r_{12}</math>. Also, <math>P_\Lambda(\cos\theta_{12})</math> is a Legendre polynomial,
and, <math>R_\gamma = R_\gamma(r_1,r_2,r_{12})</math> is the radial part of the Hylleraas wave function for state <math>\gamma</math> such
that the total wave function is <math>\psi({\bf r}_1,{\bf r}_2) = R_\gamma(r_1,r_2,r_{12}){\cal Y}_{l_1,l_2,L}^M(\hat{\bf r}_1,\hat{\bf r}_2)\pm {\rm exchange}</math>, normalized to unity, and <math>{\cal Y}_{l_1,l_2,L}^M(\hat{\bf r}_1,\hat{\bf r}_2)</math> is a vector coupled product of spherical harmonics. The coefficents <math>C_\Lambda</math> in Eq.\ (\ref{int1}) can be calculated from Eq.\ (32) of the above 1978 paper. They need to be calculated separately for he direct-direct and the direct-exchange contributions. For the reduced matrix element of <math>z_1</math>, the leading 3-<math>j</math> symbol can be dropped, leaving (for the case <math>k = 1</math>)
\begin{eqnarray}
C_\Lambda^{\rm reduced} &=& \frac{(-1)^{l_2'+L'+L+1}}{2} (l_1,l_1',l_2,l_2',L,L')^{1/2}
\sum_{\lambda={\rm max}(l_1-1,0)}^{l_1+1} (\lambda,\Lambda,1)
\left(\matrix{l_1 & 1 & \lambda \cr
0 & 0 & 0 \cr}\right)
\left(\matrix{l_1'& \lambda & \Lambda \cr
0 & 0 & 0 \cr}\right)\nonumber\\
&\times&\left(\matrix{l_2' & 1_2 & \Lambda \cr
0 & 0 & 0 \cr}\right)
\left\{\matrix{L' & \lambda & l_2 \cr
\Lambda& l_2' & l_1' \cr}\right\}
\left\{\matrix{L' & \lambda & l_2 \cr
l_1 & L & 1 \cr}\right\}
\end{eqnarray}
where the notation <math>(a,b,c,\cdots)</math> means <math>(2a+1)(2b+1)(2c+1)\cdots</math>.
A similar expression can be obtained for the angular coefficients for the <math>z_2</math> term by interchanging the subscripts 1 and 2 throughout and multiplying by <math>(-1)^{L+L'+1}</math>.

===Example===
As an example, for the <math>1s^2\;^1S - 1s2p\;^1P</math> transition of helium, the only nonvanishing <math>C_\Lambda</math> angular coefficients are <math>C_1 = 1/2</math> for the direct-direct term (i.e. <math>P_1(\cos\theta_{12}) = \cos\theta_{12}</math> is included in the radial integral (\ref{radial})) and <math>C_0 = 1/2</math> for the direct-exchange term (i.e. <math>P_0(\cos\theta_{12}) = 1</math> is included in the radial integral (\ref{radial})). The results for infinite nuclear mass are
\begin{equation}
S(1\;^1S-2\;^1P) = (0.781\,127\,158\,667)^2
\end{equation}
\begin{equation}f(1\;^1S-2\;^1P) = -0.276\,164\,703\,736
\end{equation}
where
\begin{equation}
f(1\;^1S-2\;^1P) = \frac{2}{3}[E(1\;^1S) - E(2\;^1P)]S(1\;^1S-2\;^1P)
\end{equation}
is the absorption oscillator strength.

===Test===
This is a test edit

Theory Notes

2020-12-14T05:34:16Z

Drake: /* General Hermitian Property */

<center>
<H1>Notes on solving the Schrödinger equation in Hylleraas coordinates for heliumlike atoms
Gordon W.F. Drake
Department of Physics, University of Windsor
Windsor, Ontario, Canada N9B 3P4
(Transcribed from hand-written notes and edited by Lauren Moffatt. Last revised November 22, 2017 to correct a missing subscript in the expression for $B_1$ in the Hamiltonian in Sect. 11 General Hermitian Property.)
</H1>
</center>

==Introduction==
These notes describe some of the technical details involved in solving the Schrödinger equation for a heliumlike atom of nuclear charge Z in correlated Hylleraas coordinates. The standard for computational purposes is to express all quantities in a form that is valid for electronic states of arbitrary angular momentum, and to express matrix elements of the Hamiltonian in a manifestly Hermitian symmetrized form. Further information and derivations of the mathematical techniques for angular integrals and radial recursion relations can be found in the paper
G.W.F. Drake, Phys. Rev. A <math>{\bf 18}</math>, 820 (1978).

The starting point is the two-electron Schrödinger equation for infinite nuclear mass
\begin{equation} \left[-\frac{\hbar^2}{2m}(\nabla^2_1 +\nabla^2_2) - \frac{Ze^2}{r_1} -
\frac{Ze^2}{r_2}+\frac{e^2}{r_{12}} \right]\psi = E\psi\nonumber
\end{equation}
where <math>m</math> is the electron mass, and <math>r_{12} = |{\bf r}_1 - {\bf r}_2|</math> (see diagram below).

[[File:Pic1.jpg|250px]]

Begin by rescaling distances and energies so that the Schrödinger equation can be expressed in a dimensionless form. The dimensionless Z-scaled distance is defined by <math>\begin{eqnarray} \rho = \frac{Zr}{a_0} \end{eqnarray} </math> where <math> \begin{eqnarray} a_0 = \frac{\hbar^2}{me^2} \end{eqnarray} </math> is the atomic unit (a.u.) of distance equal to the Bohr radius <math>0.529\, 177\, 210\, 92(17) \times 10^{-10}</math> m. Then

\begin{equation}[-\frac{\hbar^2}{2m}Z^2(\frac{me^2}{\hbar^2})^2(\nabla^2_{\rho_1}+\nabla^2_{\rho_2})
- Z^2\frac{e^2}{a_0}\rho^{-1}_1 - Z^2\frac{e^2}{a_0}\rho^{-1}_2 +
\frac{e^2}{a_0}Z\rho^{-1}_{12}]\psi= E\psi\nonumber\end{equation}

But <math>\begin{eqnarray} \frac{\hbar^2}{m}(\frac{me^2}{\hbar^2})^2 = \frac{e^2}{a_0} \end{eqnarray} </math> is the hartree atomic unit of of energy <math>(E_h = 27.211\,385\,05(60)</math> eV, or equivalently <math>E_h/(hc) = 2194.746\,313\,708(11)</math> m<math>^{-1})</math>. Therefore, after multiplying through by <math>a_0/(Ze)^2</math>, the problem to be solved in Z-scaled dimensionless units becomes

\begin{equation} \left[-\frac{1}{2}(\nabla^2_{\rho_1}+\nabla^2_{\rho_2}) -
\frac{1}{\rho_1} - \frac{1}{\rho_2} + \frac{Z^{-1}}{\rho_{12}}\right]\psi =
\varepsilon\psi\nonumber \end{equation}

where <math>\begin{eqnarray}\varepsilon = \frac{Ea_0}{(Ze)^2} \end{eqnarray}</math> is the energy in Z-scaled atomic units. For convenience, rewrite this in the conventional form
\begin{equation}
H\psi = \varepsilon\psi\nonumber \end{equation}
where (using <math>r</math> in place of <math>\rho</math> for the Z-scaled distance)
\begin{equation}
H = -\frac{1}{2}(\nabla^2_1+\nabla^2_2) - \frac{1}{r_1}-\frac{1}{r_2} + \frac{Z^{-1}}{r_{12}}
\nonumber\end{equation}
is the atomic Hamiltonian for infinite nuclear mass.

==The Hartree Fock Method==

For purposes of comparison, and to define the correlation energy, assume that <math>\psi({\bf r}_1,{\bf r}_2)</math> can be written in the separable product form
\begin{equation}
\psi({\bf r}_1,{\bf r}_2) = \frac{1}{\sqrt{2}} \left[ u_1(r_1) u_2(r_2) \pm u_2(r_1) u_1(r_2) \right] \nonumber
\end{equation}
for the <math>1s^2\ ^1\!S</math> ground state. Because of the <math>\frac{1}{r_{12}}</math> term in the Schrödinger equation, the exact solution connot be expressed in this form as a separable product. However, the Hartree-Fock approximation corresponds to finding the best possible solution to the Schrödinger equation
\begin{equation}
H\psi({\bf r}_1,{\bf r}_2) = E\psi({\bf r}_1,{\bf r}_2)\nonumber
\end{equation}
that can nevertheless be expressed in this separable product form, where as before
\begin{equation}
H = -\frac{1}{2}(\nabla^2_1+\nabla^2_2) - \frac{1}{r_1}- \frac{1}{r_2} + \frac{Z^{-1}}{r_{12}}\nonumber
\end{equation}
is the full two-electron Hamiltonian. To find the best solution, substitute into <math>\left\langle\psi \lvert H-E \rvert \psi\right\rangle</math> and require this expression to be stationary
with respect to arbitrary infinitesimal variations <math>\delta u_1</math> and <math>\delta
u_2</math> in <math>u_1</math> and <math>u_2</math> respectively; i.e.

\begin{equation}
\frac{1}{2} \left\langle \delta u_1(r_1)u_2(r_2) \pm u_2(r_1)\delta
u_1(r_2) \lvert H-E \rvert u_1(r_1) u_2(r_2)\pm u_2(r_1) u_1(r_2)\right\rangle\nonumber
\end{equation}

\begin{equation}
=\int\delta u_1(r_1) d{\bf r}_1 \left\{ \int d{\bf r}_2 u_2(r_2) \left( H - E \right) \left[ u_1(r_1)u_2(r_2) \pm u_2(r_1) u_1(r_2) \right] \right\}\nonumber
\end{equation}

\begin{equation}
= 0 \nonumber
\end{equation}

for arbitrary <math> \delta u_1(r_1).\nonumber </math> Therefore <math>\{\int d{\bf r}_2 \ldots \} = 0</math>.

Similarly, the coefficient of <math>\delta u_2</math> would give

\begin{equation}
\int d{\bf r}_1 u_1(r_1) \left( H-E \right) \left[ u_1(r_1) u_2(r_2) \pm
u_2(r_1) u_1(r_2) \right] = 0\nonumber
\end{equation}

Define

\begin{equation}
I_{12} = I_{21} = \int d{\bf r}\, u_1(r)u_2(r), \nonumber
\end{equation}

\begin{equation}
H_{ij} = \int d{\bf r}\, u_i(r)(-\frac{1}{2}\nabla - \frac{1}{r})u_j(r), \nonumber
\end{equation}

\begin{equation}
G_{ij}(r) = \int d{\bf r}^\prime u_i(r^\prime)\frac{1}{|{\bf r} - {\bf r}\prime|}u_j(r^\prime)\nonumber
\end{equation}

Then the above equations become the pair of integro-differential equations

\begin{equation}
[ H_0 - E + H_{22}+G_{22}(r)]u_1(r) = \mp [ I_{12}(H_0-E) + H_{12}+G_{12}(r)]u_2(r)\nonumber
\end{equation}

\begin{equation}
[H_0-E+H_{11}+G_{11}(r)]u_2(r) = \mp [I_{12}(H_0-E) + H_{12}+G_{12}(r)]u_1(r)\nonumber
\end{equation}

These must be solved self-consistently for the "constants" <math>I_{12}</math> and
<math>H_{ij}</math> and the function <math>G_{ij}(r)</math>.

The Hartree Fock energy is <math>E \simeq -2.87\ldots </math>a.u. while the exact energy is <math>E = -2.903724\ldots </math>a.u.

The difference is called the "correlation energy" because it arises from the
way in which the motion of one electron is correlated to the other. The
Hartree Fock equations only describe how one electron moves in the average field
provided by the other (mean-field theory).

==Configuration Interaction==
Expand
\begin{equation}
\psi({\bf r}_1,{\bf r}_2)= C_0u^{(s)}_1(r_1)u^{(s)}_1(r_2) + C_1u^{(P)}_1({\bf r}_1)u^{(P)}_1({\bf r}_2){\cal Y}^0_{1,1,0}(\hat{\bf r}_1, \hat{\bf r}_2)+C_2u^{(d)}_1({\bf r}_1)u^{(d)}_2({\bf r}_2){\cal Y}^0_{2,2,0}(\hat{\bf r}_1, \hat{\bf r}_2)+... \pm {\rm\ exchange}
\end{equation}
where
\begin{equation}
{\cal Y}^M_{l_1,l_2,L}(\hat{\bf r}_1, \hat{\bf r}_2)=\sum_{m_1,m_2}Y^{m_1}_{l_1}({\bf r}_1)Y^{m_2}_{l_2}({\bf r}_2)\times <l_1l_2m_1m_2\mid LM>
\end{equation}

This works, but is slowly convergent, and very laborious. The best CI calculations are accurate to <math> ~10^{-7}</math> a.u.

==Hylleraas Coordinates==
[E.A. Hylleraas, Z. Phys. <math>{\bf 48}, 469(1928)</math> and <math>{\bf 54}, 347(1929)</math>]
suggested using the co-ordinates <math>r_1</math>, <math>r_2</math> and <math>r_{12}</math> or equivalently

\begin{eqnarray}
s &=& r_1 + r_2, \nonumber\\
t &=& r_1-r_2, \nonumber\\
u &=& r_{12}\nonumber
\end{eqnarray}

and writing the trial functions in the form

\begin{equation}
\Psi({\bf r}_1,{\bf r}_2) = \sum^{i+j+k\leq
N}_{i,j,k}c_{i,j,k}r_1^{i+l_1}r_2^{j+l_2}r_{12}^ke^{-\alpha r_1 - \beta r_2}
\mathcal{Y}^M_{l_1,l_2,L}(\hat{r}_1,\hat{r}_2)\pm \text{exchange}\nonumber
\end{equation}

Diagonalizing H in this non-orthogonal basis set is equivalent to solving
<math> \begin{eqnarray}
\frac{\partial E}{\partial c_{i,j,k}} = 0\nonumber
\end{eqnarray} </math>
for fixed <math>\alpha</math> and <math>\beta</math>.

The diagonalization must be repeated for different values of <math>\alpha</math> and
<math>\beta</math> in order to optimize the non-linear parameters.

==Completeness==
The completeness of the above basis set can be shown by first writing
\begin{equation}
r_{12}^2 = r_1^2 + r_2^2 - 2r_1r_2\cos\theta_{12}
\end{equation}
and, from the spherical harmonic addition theorem,
\begin{equation}
\cos(\theta_{12})=\frac{4\pi}{3}\sum^1_{m=-1}Y^{m*}_1(\theta_1,\varphi_1)Y^m_1(\theta_2,\varphi_2)
\end{equation}
Consider first the S-states. The <math>r_{12}^0</math> terms are like the ss terms in a CI calculation. The
<math>r_{12}^2</math> terms bring in p-p type contributions, and the higher powers bring
in d-d, f-f etc. type terms. In general

\begin{equation}
P_l(\cos\theta_{12}) = \frac{4\pi}{2l+1}\sum^l_{m=-l}{Y^{m}_l}^*(\theta_1,\varphi_1)Y^m_l(\theta_2,\varphi_2)\nonumber
\end{equation}

For P-states, one would have similarly

\begin{array}{lr}
r_{12}^0 & (sp)P\nonumber\\
r_{12}^2 & (pd)P\nonumber\\
r_{12}^4 & (df)P\nonumber\\
\vdots & \vdots\nonumber
\end{array}

For D-states

\begin{array}{lr}
r_{12}^0 & (sd)D & (pp^\prime)D\nonumber\\
r_{12}^2 & (pf)D & (dd^\prime)D\nonumber\\
r_{12}^4 & (dg)D & (ff^\prime)D\nonumber\\
\vdots & \vdots & \vdots\nonumber
\end{array}

In this case, since there are two "lowest-order" couplings to form a
D-state, both must be present in the basis set, i.e.

\begin{equation}
\Psi(r_2,r_2) = \sum c_{ijk}r_1^ir_2^{j+2}r_{12}^ke^{-\alpha r_1-\beta
r_2}\mathcal{Y}^M_{022}(\hat{r}_1,\hat{r}_2)
+\sum d_{ijk}r_1^{i+1}r_2^{j+1}r_{12^k}e^{-\alpha^\prime r_1 - \beta^\prime
r_2}\mathcal{Y}^M_{112}(\hat{r}_1,\hat{r}_2)\nonumber
\end{equation}

For F-states, one would need <math>(sf)F</math> and <math>(pd)F</math> terms.

For G-states, one would need <math>(sg)G</math>, <math>(pf)G</math> and <math>(dd^\prime)G</math> terms.

Completeness of the radial functions can be proven by considering the
Sturm-Liouville problem

\begin{equation}
\left(-\frac{1}{2}\nabla^2-\frac{\lambda}{r_s}-E\right)\psi({\bf r}) = 0\nonumber
\end{equation}

or

\begin{equation}
\left(-\frac{1}{2}\frac{1}{r^2}\frac{\partial}{\partial r}\left({r^2}\frac{\partial}{\partial r}\right) -
\frac{l\left(l+1\right)}{2r^2} - \frac{\lambda}{r} - E\right)u(r) = 0.\nonumber
\end{equation}

For fixed E and variable <math>\lambda</math> (nuclear charge).

The eigenvalues are <math>\lambda_n = (E/E_n)^{1/2}</math>, where <math>E_n =- \frac{1}{2n^2}</math>

[[File:Pic2.jpg|500px]]

and the eigenfunctions are

\begin{equation}
u_{nl}(r) =
\frac{1}{(2l+1)!}\left(\frac{(n+l)!}{(n-l-1)2!}\right)^{1/2}(2\alpha)^{3/2}e^{-\alpha
r}\nonumber
\times \ \left( 2 \alpha r \right)^l\,_{1}F_{1}\left(-n+l+1,2l+2; 2 \alpha r \right)
\end{equation}

with <math>\alpha = (-2E)^{1/2}</math> and <math>n\geq l+1</math>. The confluent hypergeometric function <math> _1F_1(a,b,;z)</math> then denotes a finite polynomial since <math>a=-n+l+1</math> is a negative integer or zero.

Unlike the hydrogen spectrum, which has both a discrete part for <math>E<0</math> and a
continuous part for <math>E>0</math>, this forms an entirely discrete set of finite
polynomials, called Sturmian functions. They are orthogonal with respect to
the potential, i.e.

\begin{equation}
\int^\infty_0 r^2dr \left( u_{n^\prime l}(r)\frac{1}{r}u_{nl}(r) \right) = \delta_{n,n^\prime}\nonumber
\end{equation}

Since they become complete in the limit <math>n\rightarrow\infty</math>, this assures the
completeness of the variational basis set.

[See also B Klahn and W.A. Bingel Theo. Chim. Acta (Berlin) 44, 9 and 27
(1977)].

==Solutions of the Eigenvalue Problem==

For convenience, write

\[
\Psi({\bf r}_1,{\bf r}_2) = \sum^N_{m=1}c_m\varphi_m\nonumber
\]

where <math>m</math> represents the <math>m</math>'th combination of <math>i,j,k</math> and

\begin{equation}
\varphi_{ijk}=r_1^ir_2^jr_{12}^ke^{-\alpha r_1 - \beta
r_2}\mathcal{Y}^M_{l_1,l_2,L}(\hat{r}_1,\hat{r}_2) \pm \mbox{exchange}\nonumber
\end{equation}

\begin{align}

&
\left( \begin{array}{lr}
\cos(\theta) & \sin(\theta)\\
-\sin(\theta) & \cos(\theta)
\end{array} \right)
\left( \begin{array}{lr}
H_{11} & H_{12} \\
H_{12} & H_{22}
\end{array}\right)
\left( \begin{array}{lr}
\cos(\theta) & -\sin(\theta)\\
\sin(\theta) & \cos(\theta)
\end{array} \right)
\\[5pt]
= & \left(\begin{array}{lr}
cH_{11}+sH_{12} & cH_{12} + sH_{22}\\
-sH_{11} + cH_{12} & -sH_{12} + cH_{22}
\end{array}\right)
\left( \begin{array}{cc}
c & -s \\
s & c
\end{array}\right)
\\[5pt]
= & \left(\begin{array}{lr}
c^2H_{11}+s^2H_{22} + 2csH_{12} & (c^2-s^2)H_{12}+cs(H_{22}-H_{11})\\
(c^2-s^2)H_{12}+cs(H_{22}-H_{11}) & s^2H_{11}+c^2H_{22}-2csH_{12}
\end{array}\right)
\end{align}

Therefore

\begin{equation}
\left[\cos^2(\theta)-\sin^2(\theta)\right]H_{12} =
\cos(\theta)\sin(\theta)(H_{11}-H_{22})\nonumber
\end{equation}

and

\begin{equation}
\tan(2\theta) = \frac{2H_{12}}{H_{11}-H_{22}}\nonumber
\end{equation}

i.e.

\begin{equation}
\cos(\theta)=\left(\frac{r+\omega}{2r}\right)^{1/2}\nonumber
\end{equation}

\begin{equation}
\sin(\theta)=-\mbox{sgn}(H_{12})\left(\frac{r-\omega}{2r}\right)^{1/2}\nonumber
\end{equation}

where

\begin{equation}
\omega = H_{22}-H_{11}\nonumber
\end{equation}

\begin{equation}
r=\left(\omega^2+4H_{12}^2\right)^{1/2}\nonumber
\end{equation}

\begin{equation}
E_1=\frac{1}{2}\left(H_{11}+H_{22}-r\right)\nonumber
\end{equation}

\begin{equation}
E_2=\frac{1}{2}\left(H_{11}+H_{22}+r\right)\nonumber
\end{equation}

===Brute Force Method===

- Gives all the eigenvalues and eigenvectors, but it is slow

- First orthonormalize the <math>N</math>-dimensional basis set; i.e. form linear combinations

\begin{equation}
\Phi_m = \sum_{n=1}^N\varphi_nR_{nm}\nonumber
\end{equation}

such that
\begin{equation}
\langle \Phi_m | \Phi_n \rangle = \delta_{m,n}
\end{equation}

This can be done by finding an orthogonal tranformation T such that

\begin{equation}
T^TOT = I = \left(
\begin{array}{ccccc}
I_1 & 0 & 0 & \ldots & 0 \\
0 & I_2 & 0 & \ldots & 0 \\
0 & 0 & I_3 & \ldots & 0 \\
\vdots & \vdots & \vdots & \ddots & \vdots \\
0 & 0 & 0 & \ldots & I_N
\end{array}\right);
\end{equation}

\begin{equation}
O_{mn} = \langle \varphi_m | \varphi_n \rangle \nonumber
\end{equation}

and then applying a scale change matrix

\begin{equation}
S = \left(\begin{array}{ccccc}
\frac{1}{I_1^{1/2}} & 0 & 0 & \ldots & 0\\
0 & \frac{1}{I_2^{1/2}} & 0 & \ldots & 0 \\
0 & 0 & \frac{1}{I_3^{1/2}} & \ldots & 0 \\
\vdots & \vdots & \vdots & \ddots & \vdots \\
0 & 0 & 0 & \ldots & I_N^{1/2}
\end{array}\right)= S^T\nonumber
\end{equation}

Then

\begin{equation}
S^TT^TOTS = 1
\end{equation}

i.e.

\begin{equation}
R^TOR = 1
\end{equation}

with <math>R=TS</math>.

If H is the matrix with elements <math>H_{mn} = \langle \varphi_m | \varphi_n \rangle </math>, then H
expressed in the <math>\Phi_m</math> basis set is

\begin{equation}
H^\prime = R^THR.\nonumber
\end{equation}

We next diagonalize <math>H^\prime</math> by finding an orthogonal transformation W such
that

\begin{equation}
W^T H^\prime W = \lambda = \left(
\begin{array}{cccc}
\lambda_1 & 0 & \ldots & 0 \\
0 & \lambda_2 & \ldots & 0 \\
\vdots & \vdots & \ddots & \vdots \\
0 & 0 & \ldots & \lambda_N
\end{array}\right)\nonumber
\end{equation}

The q'th eigenvector is

\begin{equation}
\Psi^{(q)} = \sum_{n=1}^N\Phi_n W_{n,q} = \sum_{n,n^\prime} \varphi_{n^\prime} R_{n^\prime ,n} W_{n,q} \nonumber
\end{equation}

i.e.

\begin{equation}
c_{n^\prime}^{(q)} = \sum_{n=1}^N R_{n^\prime n} W_{n,q}
\end{equation}

===The Power Method===

- Based on the observation that if H has one eigenvalue, <math>\lambda_M</math>, much bigger than all the rest, and <math>\chi = \left(
\begin{array}{c}a_1\\a_2\\\vdots\end{array}\right)</math>
is an arbitrary starting vector, then <math>\chi = \sum_{q=1}^N x_q\Psi^{(q)}</math>.

\begin{equation}
(H)^n\chi = \sum_{q=1}^N x_q \lambda^n_q\Psi^{(q)}\nonumber\rightarrow x_M\lambda_M^n\Psi^{(M)}\nonumber
\end{equation}

provided <math>x_M\neq 0</math>.

To pick out the eigenvector corresponding to any eigenvalue, write the original problem in the form

\begin{align}
H\Psi & = \lambda O \Psi \nonumber \\
\left( H - \lambda_q O \right) \Psi & = \left( \lambda - \lambda_q \right) O \Psi \nonumber
\end{align}

Therefore,

\begin{equation}
G \Psi = \frac{1}{\lambda - \lambda_q } \Psi \nonumber \\
\end{equation}

where <math>G=(H-\lambda_qO)^{-1}O</math> with eigenvalues <math>\frac{1}{\lambda_n - \lambda_q}</math>.

By picking <math>\lambda_q</math> close to any one of the <math>\lambda_n</math>, say
<math>\lambda_{n^\prime}</math>, then <math>\frac{1}{\lambda_n-\lambda_q}</math> is much larger for
<math>n=n^\prime</math> than for any other value. The sequence is then

\begin{array}{c}
\chi_1=G\chi\nonumber\\
\chi_2=G\chi_1\nonumber\\
\chi_3=G\chi_2\nonumber\\
\vdots\nonumber
\end{array}

until the ratios of components in <math>\chi_n</math> stop changing.

- To avoid matrix inversion and multiplication, note that the sequence is equivalent to
\begin{equation}
F \chi_n = \left( \lambda - \lambda_q \right) O \chi_{n-1} \nonumber \\
\end{equation}
where <math>F = H-\lambda_qO</math>. The factor of <math>\left( \lambda - \lambda_q \right)</math> can be
dropped because this only affects the normalization of <math>\chi_n</math>. To find <math>\chi_n</math>, solve
\begin{equation}
F \chi_n = O \chi_{n-1} \nonumber
\end{equation}
(N equations in N unknowns). Then
\begin{equation}
\lambda = \frac{ \langle \chi_n | H | \chi_n \rangle}{\langle \chi_n | \chi_n \rangle} \nonumber
\end{equation}

==Matrix Elements of H==

\begin{equation}
H=-\frac{1}{2}\nabla^2_1 -\frac{1}{2}\nabla^2_2 - \frac{1}{r_1} - \frac{1}{r_2} +\frac{Z^{-1}}{r_{12}}\nonumber
\end{equation}

Taking <math>r_1, r_2</math> and <math>r_{12}</math> as independent variables,

\begin{equation}
\nabla_1^2 = \frac{1}{r_1^2}\frac{\partial}{\partial r_1}\left(r_1^2\frac{\partial}{\partial r_1}\right)+ \frac{1}{r_{12}^2}\frac{\partial}{\partial r_{12}}
\left(r_{12}^2\frac{\partial}{\partial r_{12}}\right)\nonumber -\frac{l_1(l_1+1)}{r_1^2}+2(r_1 - r_2\cos(\theta_{12}))\frac{1}{r_{12}}\frac{\partial^2}{\partial r_1 \partial r_{12}} \nonumber - 2(\nabla_1^Y \cdot {\bf r}_2)\frac{1}{r_{12}}\frac{\partial}{\partial r_{12}}\nonumber
\end{equation}

where <math>\nabla_1^Y</math> acts only on the spherical harmonic part of the wave function and the diagram

[[File:Pic3.jpg|400px]]

defines the complete set of 6 independent variables is <math>r_1</math>, <math>r_2</math>, <math>r_{12}</math>, <math>\theta_1</math>, <math>\varphi_1</math>, <math>\chi</math>.

If <math>r_{12}</math> were not an independent variable, then one could take the volume element to be

\begin{equation}
d\tau = r_1^2dr_1\sin(\theta_1)d\theta_1d\varphi_1r_2^2dr_2\sin(\theta_2)d\theta_2d\varphi_2.\nonumber
\end{equation}

However, <math>\theta_2</math> and <math>\varphi_2</math> are no longer independent variables. To eliminate them, take the point <math>{\bf r}_1</math> as the
origin of a new polar co-ordinate system, and write

\begin{equation}
d\tau=-r_1^2dr_1\sin(\theta_1)d\theta_1d\varphi_1r_{12}^2dr_{12}\sin(\psi)d\psi d\chi\nonumber
\end{equation}

and use

\begin{equation}
r_2^2=r_1^2+r_{12}^2 +2r_1r_{12}\cos(\psi)
\end{equation}

Then for fixed <math>r_1</math> and <math>r_{12}</math>,

\begin{equation}
2r_2dr_2 = -2r_1r_{12}\sin(\psi)d\psi
\end{equation}

Thus

\begin{equation}
d\tau= r_1dr_1r_2dr_2r_{12}dr_{12}\sin(\theta_1)d\theta_1d\varphi_1d\chi
\end{equation}

The basic type of integral to be calculated is

\begin{align}
I \left( l_1, m_1, l_2, m_2; R \right) & = \int \sin \left( \theta_1 \right) d\theta_1 d\varphi_1 d\chi Y^{m_1}_{l_1} \left( \theta_1, \varphi_1 \right)^{*} Y^{m_2}_{l_2} \left( \theta_2, \varphi_2 \right) \\
& \times \int r_1dr_1r_2dr_2r_{12}dr_{12}R \left( r_1, r_2, r_{12} \right)
\end{align}

Consider first the angular integral. <math>Y^{m_2}_{l_2} \left( \theta_2, \varphi_2 \right)</math> can be expressed in terms of the independent variables <math>\theta_1, \varphi_1, \chi</math> by use of the rotation matrix relation
\begin{equation}
Y^{m_2}_{l_2}( \theta_2, \varphi_2) = \sum_m \mathcal{D}^{(l_2)}_{m_2,m}(\varphi_1,\theta_1,\chi)^*Y^m_{l_2}( \theta_{12}, \varphi)
\end{equation}
where <math>\theta, \varphi</math> are the polar angles of <math>{\bf r}_2</math> relative to <math>{\bf r}_1</math>. The angular integral is then

\begin{align}
I_{ang} & = \int^{2 \pi}_{0} d\chi \int^{2\pi}_{0} d\varphi_1 \int^{\pi}_{0} \sin \left( \theta_1 \right) d\theta_{1} Y^{m_1}_{l_1} \left( \theta_1, \varphi_1 \right)^{*} \\
& \times \sum_m \mathcal{D}^{(l_2)}_{m_2,m} \left( \varphi_1,\theta_1,\chi \right)^{*} Y^{m}_{l_2} \left( \theta_{12},\varphi \right) \\
\end{align}

Use

\begin{equation}
Y^{m_1}_{l_1} \left( \theta_1, \varphi_1 \right)^{*} = \sqrt{\frac{2l_1+1}{4\pi}}\mathcal{D}^{(l_1)}_{m_1,0} \left( \varphi_1, \theta_1, \chi \right)
\end{equation}

together with the orthogonality property of the rotation matrices (Brink and Satchler, p 147)

\begin{align}
& \int \mathcal{D}^{(j)*}_{m,m^\prime}\mathcal{D}^{(J)}_{M,M^\prime}\sin(\theta_1)d\theta_1d\varphi_1d\chi \\
& = \frac{8\pi^2}{2j+1}\delta_{jJ}\delta_{mM}\delta_{m^\prime M^\prime}\nonumber
\end{align}

to obtain

\begin{align}
I_{ang} & = \sqrt{ \frac{2l_1+1}{4\pi}} \frac{8\pi^2}{2l_1+1} \delta_{l_1,l_2} \delta_{m_1,m_2} Y^{0}_{l_2} \left( \theta_{12}, \varphi \right) \\
& = 2 \pi \delta_{l_1,l_2} \delta_{m_1,m_2} P_{l_2} \left( \cos \theta_{12} \right)
\end{align}

since

\begin{equation}
Y^0_{l_2} \left( \theta_{12}, \varphi \right) = \sqrt{\frac{2l_1+1}{4\pi}} P_{l_2} \left( \cos \left( \theta_{12} \right) \right)
\end{equation}

Note that <math>P_{l_2} \left( \cos \theta \right)</math> is just a short hand expression for a radial function because

\begin{equation}
\cos \theta_{12} = \frac{r_1^2 + r_2^2 - r_{12}^2}{2r_1r_2}
\end{equation}

The original integral is thus

\begin{equation}
I \left( l_1, m_1, l_2, m_2; R \right) = 2 \pi \delta_{l_1,l_2} \delta_{m_1,m_2} \int^{\infty}_{0} r_{1} dr_{1} \int^{\infty}_{0} r_{2} dr_{2} \int^{r_{1} + r_{2}}_{|r_{1} - r_{2}|} r_{12} dr_{12} R \left( r_{1}, r_{2}, r_{12} \right) P_{l_2} \left( \cos \theta_{12} \right)
\end{equation}

where again

\begin{equation}
\cos \theta_{12} = \frac{r_{1}^{2} +r^{2}_{2} - r_{12}^{2}}{2 r_{1} r_{2}}
\end{equation}

is a purely radial function.

The above would become quite complicated for large <math>l_2</math> because <math>P_{l_2}(\cos \theta_{12})</math> contains terms up to <math>\left( \cos \theta_{12} \right)^{l_2}</math>. However, recursion relations exist which allow any integral containing <math>P_l \left( \cos \theta_{12} \right)</math> in terms of those containing just <math>P_0 \left( \cos \theta_{12} \right) = 1</math> and <math>P_1 \left( \cos \theta_{12} \right) = \cos \theta_{12}</math>.

==Radial Integrals and Recursion Relations==

The basic radial integral is [see G.W.F. Drake, Phys. Rev. A 18, 820 (1978)]

\begin{align}
I_0 \left( a, b, c \right) & =
\int^{\infty}_{0} r_{1} dr_{1} \int^{\infty}_{r_{1}} r_{2} dr_{2} \int^{r_{1}+r_{2}}_{r_{2} - r_{1}} r_{12} dr_{12} r_{1}^{a} r_{2}^{b} r_{12}^{c} e^{-\alpha r_{1} - \beta r_{2}} + \int^{\infty}_{0} r_{2} dr_{2} \int^{\infty}_{r_{2}} r_{1} dr_{1} \int^{r_{1} + r_{2}}_{r_{1} - r_{2}} r_{12} dr_{12} r_{1}^{a} r_{2}^{b} r_{12}^{c} e^{- \alpha r_{1} - \beta r_{2}} \\
& = \frac{2}{c+2} \sum^{[(c+1)/2]}_{i=0}
\left(\begin{array}{c}
c+2 \\
2i+1
\end{array} \right)
\left\{ \frac{q!}{\beta^{q+1} \left( \alpha + \beta \right)^{p+1}} \sum_{j=0}^{q} \frac{\left( p + j \right)!}{j!} \left( \frac{\beta}{\alpha + \beta} \right)^j + \frac{q^{\prime}!}{\alpha^{q^{\prime}+1} \left( \alpha + \beta \right)^{p^{\prime}+1}} \sum^{q^{\prime}}_{j=0} \frac{\left( p^{\prime} + j \right)!}{j!} \left( \frac{\alpha}{\alpha + \beta} \right)^j \right\}
\end{align}

where

\begin{array}{ll}
p = a + 2i + 2 & p^{\prime} = b + 2i + 2 \\
q = b + c - 2i + 2 & q^{\prime} = a + c - 2i + 2
\end{array}

The above applies for <math> a, b \geq -2, c\geq -1</math>.
<math>[x]</math> means "greatest integer in" <math>x</math>.

Then

\begin{align}
I_1 \left( a, b, c \right) & = \int d\tau_{r} r_{1}^{a} r_{2}^{b} r_{12}^{c} e^{ -\alpha r_{1} - \beta r_{2}} P_{1}\left( \cos \theta \right) \\
& = \frac{1}{2} \left( I_{0} \left( a + 1, b - 1, c \right) + I_{0} \left( a - 1, b + 1, c \right) - I_{0} \left( a - 1, b - 1, c + 2 \right) \right)
\end{align}

===The Radial Recursion Relation===

Recall that the full integral is

\begin{equation}
I(l_1m_1,l_2m_2;R)=2\pi\delta_{l_1,l_2}\delta_{m_1,m_2}I_{l_2}(R)
\end{equation}

where

\begin{equation}
I_{l_2}(R) = \int d\tau_r R(r_1,r_2,r_{12})P_{l_2}(\cos \theta_{12})
\end{equation}

To obtain the recursion relation, use

\begin{equation}
P_l(x) = \frac{[P^{\prime}_{l+1}(x)-P^{\prime}_{l-1}(x)]}{2l+1}
\end{equation}

with

\begin{equation}
P_{l+1}^\prime (x) = \frac{d}{dx}P_{l+1}(x)
\end{equation}

Here <math>x=\cos\theta_{12}</math> and

\begin{align}
\frac{d}{d \cos \theta_{12}} & = \frac{dr_{12}}{d \cos \theta_{12}} \frac{d}{dr_{12}} \\
& = -\frac{r_{1} r_{2}}{r_{12}} \frac{d}{dr_{12}}
\end{align}

Then,

\begin{equation}
I_{l}\left( R \right) = - \int d\tau_r R \frac{r_{1} r_{2}}{r_{12}} \frac{d}{dr_{12}} \frac{ \left[ P_{l+1} \left( \cos \theta_{12} \right) - P_{l-1} \left( \cos \theta_{12} \right) \right]}{2l+1}
\end{equation}

The <math>r_{12}</math> part of the integral is

\begin{align}
& \int^{r_{1} + r_{2}}_{|r_{1} - r_{2}|} r_{12} dr_{12} R \frac{r_{1} r_{2}}{r_{12}} \frac{d}{dr_{12}} \left[P_{l+1}-P_{l-1} \right] \\
& = R r_{1} r_{2} \left[ P_{l+1} - P_{l-1} \right]^{r_{1} + r_{2}}_{|r_{1} - r_{2}|} - \int^{r_{1} + r_{2}}_{|r_{1} - r_{2}|} r_{12} dr_{12} \left( \frac{d}{dr_{12}} R \right) \frac{r_{1} r_{2}}{r_{12}} \frac{ \left[ P_{l+1} - P_{l-1} \right]}{2l+1}
\end{align}

The integrated term vanishes because

\begin{equation}
\cos \theta_{12} = \frac{r_{1}^{2} + r_{2}^{2} - r_{12}^{2}}{2 r_{1} r_{2}}
\begin{array}{lll}
& = -1 & {\rm when} & r_{12}^{2} = \left( r_{1} + r_{2} \right)^{2} \\
& = 1 & {\rm when} & r_{12}^{2} = \left( r_{1} - r_{2} \right)^{2}
\end{array}
\end{equation}

and <math>P_l(1)=1</math>, <math>P_{l}(-1)=(-1)^{l}</math>.

Thus,

\begin{equation}
I_{l+1}\left(\frac{r_1r_2}{r_{12}}R^\prime\right) = \left( 2l+1 \right) I_{l}(R) - I_{l-1} \left( \frac{r_{1} r_{2}}{r_{12}} R^{\prime} \right)
\end{equation}

If

\begin{equation}
R = r_{1}^{a-1} r_{2}^{b-1} r_{12}^{c+2} e^{-\alpha r_{1} - \beta r_{2}}
\end{equation}

then [G.W.F. Drake, Phys. Rev. A 18, 820 (1978)]

\begin{equation}
I_{l+1}(r_1^ar_2^br_{12}^c) =
\frac{2l+1}{c+2} I_{l} \left( r_{1}^{a-1} r_{2}^{b-2} r_{12}^{c+2} \right)+I_{l-1} \left( r_{1}^{a} r_{2}^{b} r_{12}^{c} \right)
\end{equation}

For the case <math>c=-2</math>, take

\begin{equation}
R = r_{1}^{a-1} r_{2}^{b-1} \ln r_{12} e^{ -\alpha r_{1} -\beta r_{2}}
\end{equation}

Then

\begin{equation}
I_{l+1} \left( r_{1}^{a} r_{2}^{b} r_{12}^{c} \right) = I_{l} \left( r_{1}^{a-1} r_{2}^{b-1} \ln r_{12} \right) \left( 2l + 1 \right) - I_{l-1} \left( r_{1}^{a} r_{2}^{b} r_{12}^{-2} \right)
\end{equation}

This allows all <math>I_{l}</math> integrals to be calculated from tables of <math>I_0</math> and
<math>I_1</math> integrals.

===The General Integral===

The above results for the angular and radial integrals can now be combined into a general formula for integrals of the type

\begin{equation}
I = \int \int d{\bf r}_{1} d{\bf r}_{2} R_{1} \mathcal{Y}^{M^{\prime}}_{l^{\prime}_{1} l^{\prime}_{2} L^{\prime}} \left( \hat{r}_{1}, \hat{r}_{2} \right) T_{k_{1} k_{2} K}^{Q} \left( {\bf r}_{1},{\bf r}_{2} \right) R_{2} \mathcal{Y}_{l_{1} l_{2} L}^{M} \left( \hat{r}_{1}, \hat{r}_{2} \right)
\end{equation}

where

\begin{equation}
\mathcal{Y}^{M}_{l_{1} l_{2} L} \left( \hat{r}_{1}, \hat{r}_{2} \right) = \sum_{m_1,m_2} \langle l_{1} l_{2} m_{1} m_{2} | LM \rangle Y^{m_{1}}_{l_{1}} \left( \hat{r}_{1} \right) Y^{m_{2}}_{l_{2}} \left( \hat{r}_{2} \right)
\end{equation}

and

\begin{equation}
T^Q_{k_{1} k_{2} K} \left( {\bf r}_{1}, {\bf r}_{2} \right) = \sum_{q_{1}, q_{2}} \langle k_{1} k_{2} q_{1} q_{2} | KQ \rangle Y_{k_{1}}^{q_{1}} \left( \hat{r}_{1} \right) Y^{q_{2}}_{k_{2}} \left( \hat{r}_{2} \right)
\end{equation}

The basic idea is to make repeated use of the formula

\begin{equation}
Y^{m_{1}}_{l_{1}} \left( \hat{r}_{1} \right) Y^{m_{2}}_{l_{2}} \left( \hat{r}_{1} \right) = \sum_{l m} \left( \frac{\left( 2l_{1}+1 \right) \left(2l_{2}+1 \right) \left(2l+1 \right)}{4 \pi} \right)^{1/2}
\times \left( \begin{array}{ccc}
l_1 & l_2 & l \\
m_1 & m_2 & m
\end{array} \right) \left(\begin{array}{ccc}
l_1 & l_2 & l \\
0 & 0 & 0
\end{array} \right)
Y^{m*}_{l} \left( \hat{r}_{1} \right)
\end{equation}

where

\begin{equation}
Y^{m*}(\hat{r}) = (-1)^mY_l^{-m}(\hat{r})
\end{equation}

and

\begin{equation}
\left(\begin{array}{ccc}
l_1 & l_2 & l \\
m_1 & m_2 & m
\end{array} \right) = \frac{\left( -1 \right)^{l_{1} - l_{2} - m}}{\left( 2l + 1 \right)^{1/2}} \left( l_{1} l_{2} m_{1} m_{2} | l, -m \right)
\end{equation}

is a 3-<math>j</math> symbol. In particular, write

\begin{array}{cc}
Y^{m_1^\prime}_{l_1^\prime}(\hat{r}_1)^{*} \underbrace{Y_{k_{1}}^{q_{1}} \left( \hat{r}_{1} \right) Y^{m_{1}}_{l_{1}} \left( \hat{r}_{1} \right)} & = & \sum_{\Lambda M}(\ldots)Y^{M}_\Lambda(\hat{r}_1)\\
\ \ \ \ \ \ \ \sum_{\lambda_1\mu_1}Y_{\lambda_1}^{\mu_1}(\hat{r}_1) & & \\
& & \\
Y^{m_{2}^{\prime}}_{l_{2}^{\prime}} \left( \hat{r}_{2} \right) \underbrace{Y^{q_{2}}_{k_{2}} \left( \hat{r}_{2} \right) Y_{l_{2}}^{m_{2}} \left( \hat{r}_{2} \right)} & = & \sum_{\Lambda^{\prime} M^{\prime}} \left( \ldots \right)Y_{\Lambda^{\prime}}^{M^{\prime}*} \left( \hat{r}_{2} \right) \\
\ \ \ \ \ \ \ \ \ \sum_{\lambda_{2} \mu_{2}}Y_{\lambda_{2}}^{\mu_{2}} \left( \hat{r}_{2} \right) & &
\end{array}

The angular integral then gives a factor of <math>2 \pi \delta_{\Lambda, \Lambda^{\prime}} \delta_{M, M^{\prime}} P_{\Lambda} \left( \cos \theta_{12} \right)</math>. The total integral therefore reduces to the form

\begin{equation}
I = \sum_{\Lambda} C_{\Lambda} I_{\Lambda} \left( R_{1} R_{2} \right)
\end{equation}

where <math>C_{\Lambda} = \sum_{\lambda_{1}, \lambda_{2}} C_{ \lambda_{1}, \lambda_{2}, \Lambda}</math>. For further details and derivations, including graphical representations, see G.W.F. Drake, Phys. Rev. A <math>{\bf 18}</math>, 820 (1978).

==Graphical Representation==

[[File:GraphRep1.jpg|500px]]

==Matrix Elements of H==
Recall that

\begin{equation}
H = -\frac{1}{2} \nabla^{2}_{1} - \frac{1}{2} \nabla^{2}_{2} - \frac{1}{r_1} - \frac{1}{r_2} + \frac{Z^{-1}}{r_{12}}
\end{equation}

Consider matrix elements of

\begin{equation}
\nabla^2_1 = \frac{1}{r_1^2} \frac{\partial}{\partial r_1}\left(r_1^2\frac{\partial}{\partial r_1} \right) + \frac{1}{r^2}\frac{\partial}{\partial r}\left(r^2\frac{\partial}{\partial r}\right) - \frac{(\vec{l}_1^Y)^2}{r_1^2} + \frac{2 \left( r_1 - r_2 \cos \theta \right)}{r} \frac{\partial^2}{\partial r_1 \partial r} - 2 \left( \nabla^Y_1 \cdot {\bf r}_2 \right) \frac{1}{r} \frac{\partial}{\partial r}
\end{equation}

where <math>r \equiv r_{12}</math> and <math>\cos\theta \equiv \cos\theta_{12}</math>. Also in what follows, define <math>\hat{\nabla}^Y_1 \equiv r_1\nabla^Y_1</math> where <math>\nabla^Y</math> operates only on the spherical harmonic part of the wave function, and similarly for <math>\vec{l}_1^Y</math>.

A general matrix element is

\begin{equation}
\langle r_1^{a^\prime} r_2^{b^\prime} r_{12}^{c^\prime} e^{-\alpha^\prime r_1 - \beta^\prime r_2} \mathcal{Y}^{M^\prime}_{l^\prime_1 l^\prime_2 L^\prime} \left( \hat{r}_1, \hat{r}_2 \right) \left| \nabla^2_1 \right| r_1^a r_2^b r_{12}^c e^{-\alpha r_1 - \beta r_2} \mathcal{Y}^M_{l_1 l_2 L} \left( \hat{r}_1, \hat{r}_2 \right) \rangle
\end{equation}

Since <math>\nabla^2_1</math> is rotationally invariant, this vanishes unless <math>L=L^\prime</math>, <math>M=M^\prime</math>. Also <math>\nabla_1^2</math> is Hermitian, so that the result must be the same whether it operates to the right or left, even though the results look very different. In fact, requiring the results to be the same yields some interesting and useful integral identities as follows:

Put

\begin{equation}
\mathcal{F} = F \mathcal{Y}^M_{l_1 l_2 L} \left( \hat{r}_1, \hat{r_2} \right)
\end{equation}

and

\begin{equation}
F = r_1^a r_2^b r_{12}^c e^{-\alpha r_1 - \beta r_2}
\end{equation}

Then

\begin{equation}
\nabla^2_1 \mathcal{F} = \left\{ \frac{1}{r_1^2} \left[ a \left( a + 1 \right) - l_1 \left( l_1 + 1 \right) \right] + \frac{c \left( c + 1 \right)}{r^2} + \alpha^2\frac{2\alpha \left( a + 1 \right)}{r_1} + \frac{2 \left(r_1 - r_2 \cos \theta \right) }{r_1r^2}c \left[ a - \alpha r_1 \right] - \frac{2c}{r^2} \left( \hat{\nabla}^Y_1 \cdot \hat{r}_2 \right) \frac{r_2}{r_1} \right\} \mathcal{F}
\end{equation}

and

\begin{align}
\langle \mathcal{F}^\prime \left| \nabla^2_1 \right| \mathcal{F} \rangle & = \sum_\Lambda \int d\tau_r F^\prime C_\Lambda \left( 1 \right) P_\Lambda \left( \cos \theta \right) \times \left\{ \frac{1}{r_1^2} \left[ a \left( a + 1 \right) - l_1 \left( l_1 + 1 \right) \right] -\frac{2 \alpha \left( a + 1 \right)}{r_1} + \frac{c \left( c + 1 \right)}{r^2} +\alpha^2 + \frac{2 \left( r_2 - r_2 \cos \theta \right)}{r_1r^2}c \left( a - \alpha r_1 \right) \right\} F \\
& + \sum_\Lambda \int d\tau_r F^\prime C_\Lambda \left( \hat{\nabla}^Y_1 \cdot \hat{r}_2 \right) P_\Lambda \left( \cos \theta \right) \left( \frac{-2c r_2}{r_1 r^2} \right) F
\end{align}

where

\begin{equation}
\int d\tau_r = \int^\infty_0 r_1 dr_1 \int^\infty_0 r_2 dr_2 \int^{r_1+r_2}_{\left| r_1 - r_2 \right|} r dr
\end{equation}

For brevity, let the sum over <math>\Lambda</math> and the radial integrations be understood, and let <math>\left( \nabla_1^2 \right)</math> stand for the terms which appear in the integrand. Then operating to the right gives

\begin{equation}
(\nabla_1^2)_R = \frac{1}{r_1^2} \left[ a \left( a + 1 \right) - l_1 \left( l_1 + 1 \right) \right] + \frac{c \left( c + 1 \right)}{r^2} + \alpha^2 - \frac{2 \alpha \left( a + 1 \right)}{r_1}
+ \frac{2 \left( r_1 - r_2 \cos \theta \right)}{r_1 r^2}c \left( a - \alpha r_1 \right) - \frac{2c r_2}{r_1 r^2} \left( \hat{\nabla}^Y_1 \cdot \hat{r}_2 \right)
\end{equation}

Operating to the left gives

\begin{equation}
(\nabla_1^2)_L = \frac{1}{r_1^2} \left[a^\prime \left( a^\prime + 1 \right) - l_1^\prime \left( l_1^\prime + 1 \right) \right] + \frac{c^\prime \left( c^\prime + 1 \right)}{r^2} + \alpha^{\prime 2} -\frac{2 \alpha^\prime \left( a^\prime + 1 \right)}{r_1}
+ \frac{2 \left( r_1 - r_2 \cos \theta \right)}{r_1 r^2} c^\prime \left( a^\prime - \alpha^\prime r_1 \right) - \frac{2 c^\prime r_2}{r_1 r^2} \left( \hat{\nabla}^{Y^\prime}_1 \cdot \hat{r}_2 \right)
\end{equation}

Now put

\begin{array}{cc}
a_+ = a+a^\prime, & \hat{\nabla}_1^+ = \hat{\nabla}_1^Y + \hat{\nabla}_1^{Y\prime} \\
a_{-} = a-a^\prime, & \hat{\nabla}_1^{-} = \hat{\nabla}_1^Y - \hat{\nabla}_1^{Y\prime}
\end{array}

etc., and substitute <math>a^\prime = a_+ - a</math>, <math>c^\prime = c_+ - c</math>, and <math>\alpha^\prime = \alpha_+ - \alpha</math> in <math>(\nabla^2_1)_L</math>. If <math>a_+</math>, <math>c_+</math> and <math>\alpha_+</math> are held fixed, then the equation

\begin{equation}
(\nabla^2_1)_R = (\nabla^2_1)_L
\end{equation}

must be true for arbitrary <math>a</math>, <math>c</math>, and <math>\alpha</math>. Their coefficients must thus vanish.

This yields the integral relations

\begin{equation}
\label{I}
\frac{\left( r_1 - r_2 \cos \theta \right)}{r_1 r^2} = \frac{1}{c_+} \left( \frac{- \left( a_+ + 1 \right)}{r_1^2} + \frac{\alpha_+}{r_1} \right) \tag{I}
\end{equation}

from the coefficient of <math>a</math> and

\begin{equation}
\label{II}
\frac{ \left( r_1 - r_2 \cos \theta \right) \left( a_+ - \alpha_+ r_1 \right)}{r_1 r^2} = \frac{r_2}{r_1 r^2}\left( \hat{r}_2 \cdot \hat{\nabla}_1^+ \right) - \frac{ \left( c_+ + 1 \right)}{r^2} \tag{II}
\end{equation}

from the coefficient of <math>c</math>. The coefficient of <math>\alpha</math> gives an equation equivalent to (I).

Furthermore, if can be show that (see problem)

\begin{align}
& \sum_\Lambda \int d\tau_r \frac{r^c}{r^2} C_\Lambda \left( \hat{r}_2 \cdot \hat{\nabla}_1^Y \right) P_\Lambda \left( \cos \theta \right) \\
= & \sum_\Lambda \int d\tau_r \frac{r^c}{c r_1 r_2}C_\Lambda \left( 1 \right)P_\Lambda \left( \cos \theta \right) \times \left( \frac{l_1^\prime \left( l_1^\prime + 1 \right) - l_1 \left( l_1 + 1 \right) -\Lambda \left( \Lambda + 1 \right)}{2} \right)
\end{align}

and similarly for <math>\left( \hat{r}_2 \cdot \hat{\nabla}^{Y\prime}_1 \right)</math> with <math>l_1</math> and <math>l_1^\prime</math> interchangeable, then it follows that

\begin{equation}
\frac{r^{c_+}}{r^2} \left( \hat{r}_2 \cdot \hat{\nabla}_1^+ \right) = - \frac{r^{c_+}}{c_+ r_1 r_2} \Lambda \left( \Lambda + 1 \right)
\end{equation}

and

\begin{equation}
\frac{r^{c_+}}{r^2} \left( \hat{r}_2 \cdot \hat{\nabla}_1^- \right) = \frac{r^{c_+}}{c_+ r_1 r_2} \left[ l_1^\prime \left( l_1^\prime + 1 \right) - l_1 \left( l_1 + 1 \right) \right]
\end{equation}

where equality applies after integration and summation over <math>\Lambda</math>.

Thus (II) becomes

\begin{equation}\label{IIa}
\frac{ \left( r_1 - r_2 \cos \theta \right) \left( a_+ - \alpha_+ r_1 \right)}{r_1 r^2} = - \frac{\Lambda \left( \Lambda + 1 \right)}{c_+ r^2_1} - \frac{ \left( c_+ + 1 \right)}{r^2} \tag{IIa}
\end{equation}

===Problem===

Prove the integral relation

\begin{align}
& \sum_\Lambda \int d\tau_r f \left( r_1, r_2 \right) \frac{1}{r} \left( \frac{d}{dr} q(r) \right) C_\Lambda \left( \hat{r}_2 \cdot \hat{\nabla}^Y_1 \right) P_\Lambda \left( \cos \theta \right) \\
= & \sum_\Lambda \int d\tau_r \frac{ f \left( r_1, r_2 \right)}{r_1, r_2} q(r) C_\Lambda \left( 1 \right) P_\Lambda \left( \cos \theta \right) \left(\frac{l_1^\prime \left( l_1^\prime + 1 \right) - l_1 \left( l_1 + 1 \right) - \Lambda \left( \Lambda + 1 \right)}{2}\right)
\end{align}

where the coefficients <math>C_\Lambda \left( 1 \right)</math> are the angular coefficients from the overlap integral

\begin{align}
\int & d\Omega\, \mathcal{Y}^{M*}_{l_1^\prime l_2^\prime L}(\hat{r}_1,\hat{r}_2)\, \mathcal{Y}^{M}_{l_1 l_2 L}(\hat{r}_1,\hat{r}_2)\\
= & \sum_\Lambda C_\Lambda(1)P_\Lambda(\cos\theta_{12}).
\end{align}

Hint: Use the fact that <math>l_1^2</math> is Hermitian so that
<math>
\int d\tau (l_1^2\mathcal{Y}^\prime)^*q(r)\mathcal{Y} = \int d\tau
\mathcal{Y}^{\prime *}l_1^2(q(r)\mathcal{Y})
</math>
with <math>\ \vec{l}_1= \frac{1}{i}\vec{r}_1\times\nabla_1</math>.

It is also useful to use
<math>\begin{eqnarray}\ \
\cos\theta\, P_L(\cos\theta) &=& \frac{1}{2L+1}[L\,P_{L-1}(\cos\theta) + (L+1)P_{L+1}(\cos\theta)]
\\
(\cos^2\theta-1)P_L(\cos\theta) &=&
\frac{L(L-1)}{(2L-1)(2L+1)}P_{L-2}(\cos\theta)\nonumber
-\frac{2(L^2+L-1)}{(2L-1)(2L+3)}P_L(\cos\theta) + \frac{(L+1)(L+2)}{(2L+1)(2L+3)}P_{L+2}(\cos\theta)\nonumber\\
&=& \frac{L(L-1)}{(2L-1)(2L+1)}[P_{L-2}(\cos\theta) - P_L(\cos\theta)]
+ \frac{(L+1)(L+2)}{(2L+1)(2L+3)}[P_{L+2}(\cos\theta) - P_L(\cos\theta)]
\end{eqnarray}</math>

together with a double application of the integral recursion relation
<math>
I_{L+1}\left(\frac{1}{r}\frac{d}{dr}q(r)\right) =
(2L+1)I_L\left(\frac{1}{r_1r_2}q(r)\right)+I_{L-1}\left(\frac{1}{r}\frac{d}{dr}q(r)\right)
</math>

Of course <math>l_1^2\mathcal{Y}^M_{l_1l_2L}(\hat{r}_1,\hat{r}_2)=l_1(l_1+1)\mathcal{Y}^M_{l_1l_2L}(\hat{r}_1,\hat{r}_2)</math>. Begin by expanding
<math>
l_1^2(g\mathcal{Y}) = \mathcal{Y}l_1^2g + g\,l_1^2\mathcal{Y} + 2(\vec{l}_1g)\cdot(\vec{l}_1\mathcal{Y})
</math>, and show that

<math>
\begin{eqnarray}\ \
(\vec{l}_1g)\cdot(\vec{l}_1\mathcal{Y}) & = & \frac{r_1^2}{r}\frac{dg}{dr}\vec{r}_2\cdot \nabla_1\mathcal{Y}\,\,\,\,\mbox{ and}\\
l_1^2g(r) &=& 2r_1r_2\cos\theta\frac{1}{r}\frac{dg}{dr} + r_1^2r_2^2(\cos^2\theta - 1)
\frac{1}{r}\frac{d}{dr}\left(\frac{1}{r}\frac{dg}{dr}\right).
\end{eqnarray}
</math>

The proof amounts to showing that the term <math>l_1^2g(r)</math> can be replaced by
<math>\Lambda(\Lambda+1)g(r)</math> after multiplying by <math>P_\Lambda(\cos\theta)</math>
and integrating by parts with respect to the radial integrations over <math>r_1</math>, <math>r_2</math> and
<math>r \equiv |\vec{r}_1 - \vec{r}_2|</math>. Remember that <math>\cos\theta</math> is just a short-hand notation for the radial function <math>(r_1^2 + r_2^2 - r^2)/(2r_1r_2)</math>.

==General Hermitian Property==

Each combination of terms of the form
\begin{equation}
<f>_R = a^2f_1+af_2+abf_3+b^2f_4+bf_5+af_6\nabla_1^Y+bf_7\nabla^Y_2+f_8(Y)
\end{equation}
can be rewritten
\begin{equation}
<f>_L= (a_+-a)^2f_1+(a_+-a)f_2 + (a_+-a)(b_+-b)f_3+(b_+-b)^2f_4
+(b_+-b)f_5+(a_+-a)f_6\nabla^{Y\prime}_1+(b_+-b)f_7\nabla^{Y\prime}_2+f_8)Y^\prime
\end{equation}
where as usual <math>\nabla_1^Y</math> acts only on the spherical harmonic part of the wave function denoted for short by <math>Y</math>.
Since these must be equal for arbitrary <math>a</math> and <math>b</math>,
\begin{equation}
a_+^2f_1+a_+f_2+_+b_+f_3+b_+f_4+b_+f_5+a_+f_6\nabla^{Y\prime}_1 +
b_+f_7\nabla^{Y\prime}_2+f_8(Y^\prime)-f_8(Y)=0
\end{equation}

Adding the corresponding expression with <math>Y</math> and <math>Y^\prime</math> interchanged
yields
<math>\label{III}
a_+^2f_1+a_+f_2+a_+b_+f_3+b_+^2f_4+b_+f_5+\frac{1}{2}a_+f_6\nabla^+_1+\frac{1}{2}b_+f_7\nabla^+_2=0\tag{III}
</math>

Subtracting gives
\begin{eqnarray*}
f_8(Y)-f_8(Y^\prime)=-\frac{1}{2}[a_+f_6\nabla^-_1+b_+f_7\nabla_2^-]\tag{IV}\\
a[-2a_+f_1-2f_2-b_+f_3-f_6\nabla^+_1]=0\tag{V}\\
b[-2b_+f_4-2f_5-a_+f_3-f_7\nabla^+_2]=0\tag{VI}\\
\end{eqnarray*}

Adding the two forms gives

\begin{eqnarray}
<f>_R+<f>_L=\frac{1}{2}(a_+^2+a_-^2)f_1+a_+f_2+\frac{1}{2}(a_+b_++a_-b_-)f_3\nonumber\\
\ +\frac{1}{2}(b_+^2+b_-^2)f_4+b_+f_5+\frac{1}{2}f_6(a_+\nabla_1^++a_-\nabla^-_1)\nonumber\\
\ +\frac{1}{2}(b_+\nabla_2^++b_-\nabla^-_2)+f_8(Y)+f_8(Y^\prime)\nonumber
\end{eqnarray}

Subtracting <math>\frac{x}{2}\times({\rm I})</math> gives

\begin{eqnarray}\ \
<f>_R+<f>_L = \frac{1}{2}[(1-x)a_+^2 + a_-^2]f_1 +
(1-\frac{x}{2})a_+f_2+\frac{1}{2}[(1-x)a_+b_++a_-b_-]f_3\nonumber\\
+\frac{1}{2}[(1-x)b_+^2+b_-^2]f_4+(1-\frac{x}{2})f_5b_++\frac{1}{2}f_6[(1-\frac{x}{2})a_+\nabla_1^++a_-\nabla^-_1]\nonumber\\
+\frac{1}{2}[(1-\frac{x}{2})b_+\nabla^+_2+b_-\nabla^-_2]+f_8(Y)+f_8(Y^\prime)\nonumber
\end{eqnarray}

If <math>x=1</math>,
\begin{eqnarray}
<f>_R+<f>_L &=&
\frac{1}{2}[a_-^2f_1+a_+f_2+a_-b_-f_3+b_-^2f_4+b_+f_5\nonumber\\
&+&\left(\frac{1}{2}a_+\nabla^+_1+a_-\nabla^-_1\right)f_6+\left(\frac{1}{2}b_+\nabla^+_2+b_-\nabla^-_2\right)f_7]+f_8(Y)+f_8(Y^\prime)\nonumber
\end{eqnarray}

The General Hermitian Property for arbitrary <math>x</math> gives
\begin{eqnarray}\ \
\nabla_1^2=
\frac{1}{4}\{\frac{1}{r_1^2}[(1-x)a_+^2+a_-^2+2(1-\frac{x}{2})a_+-2[l_1(l_1+1)+l_1^\prime(l_1^\prime+1)]\nonumber\\
-\frac{2}{r_1}[(1-x)\alpha_+a_++\alpha_-a_-+2(1-\frac{x}{2})\alpha_+]+(1-x)\alpha_+^2
+ \alpha^2_-]\nonumber\\
+\frac{2(r_1-r_2\cos\theta)}{r_1r^2}[(1-x)(a_+-\alpha_+r_1)c_++(a_--\alpha_-r_2)c_-]\nonumber\\
-\frac{2r_2}{r_1r^2}[(1-\frac{x}{2})c_+\hat{r}_2\cdot\hat{\nabla}_1^++c_-\hat{r}_2\hat{\nabla}_1^-]+\frac{(1-x)c_+^2+c_-^2+2(1-\frac{x}{2})c_+}{r^2}\}\nonumber
\end{eqnarray}.

Use
<math>\begin{eqnarray}\ \
\frac{-2r_2}{r_1r^2}[(1-\frac{x}{2})c_+\hat{r}_2\hat{\nabla}_1^++c_-\hat{r}_2\hat{\nabla}_1^-]=\frac{2}{r_1^2}\left[(1-\frac{x}{2})\Lambda(\Lambda+1)-\frac{c_-}{c_+}[l_1^\prime(l_1^\prime+1)+l_1(l_1+1)]\right]\end{eqnarray}</math>,

<math>\begin{eqnarray}
\frac{2(r_1-r_2\cos\theta)}{r_1r^2}(a_--\alpha_-r_1)c_- = 2\frac{c_-}{c_+}\left[\frac{-1}{r_1^2}[a_-(a_++1)]+\frac{1}{r_1}[a_-\alpha_++\alpha_-(a_++2)]-\alpha_-\alpha_+\right]\nonumber
\end{eqnarray}</math>,

and
<math>\begin{eqnarray}\ \
\frac{2(r_1-r_2\cos\theta)}{r_1r^2}(a_+-\alpha_+r_1)c_+ = 2\left[-\frac{1}{r_1^2}[a_+(a_++1)]+\frac{1}{r_1}[a_+\alpha_++\alpha_+(a_++2)]-\alpha_+^2\right]
\end{eqnarray}</math>.

Substituting into <math>\nabla_1^2</math> gives

\begin{eqnarray}
\nabla^2_1 = \frac{1}{4}\{\frac{1}{r_1^2}[-(1-x)a_+^2 + a_-^2 +xa_+ +
2(1-\frac{x}{2})\Lambda(\Lambda+1)\nonumber\\
-2l_1(l_1+1)(1-\frac{c_-}{c_+}) -
2l_1^\prime(l_1^\prime+1)(1+\frac{c_-}{c_+})-\frac{2c_-a_-}{c_+}(a_++1)]\nonumber\\
- \frac{2}{r_1}\left[-(1-x)\alpha_+(a_++2)+\alpha_-a_-+2(1-\frac{x}{2})\alpha_+
-\frac{c_-}{c_+}[a_-\alpha_++\alpha_-(a_++2)]\right]\nonumber\\
-(1-x)\alpha_+^2+\alpha_-^2-2\frac{c_-}{c_+}\alpha_-\alpha_+ +\frac{1}{r^2}\left[(1-x)c_+^2+c_-^2+2(1-\frac{x}{2})c_+\right]\}\nonumber
\end{eqnarray}

This has the form
<math>
\nabla_1^2 =
\frac{1}{4}\left[\frac{A_1}{r_1^2}+\frac{B_1}{r_1}+\frac{C_1}{r^2}+D_1\right]
</math>
with

<math>\begin{array}{ccc}
A_1=-(1-x)a_+^2+a_-^2
+xa_++2(1-\frac{x}{2})\Lambda(\Lambda+1)-2l_1(l_1+1)(1-\frac{c_-}{c_+})\
-2l_1^\prime(l_1^\prime+1)(1+\frac{c_-}{c_+})-\frac{2c_-a_-}{c_+}(a_++1) \ \
\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \
\end{array}
</math>

<math>
B_1 =
2\left[(1-x)\alpha_+(a_++2)-\alpha_-a_--2(1-\frac{x}{2})\alpha_++\frac{c_-}{c_+}[a_-\alpha_++\alpha_-(a_++2)]\right]
</math>

<math>
C_1 = (1-x)c_+^2+ c_-^2 +2(1-\frac{x}{2})c_+
</math>

<math>
D_1 = -1(1-x)\alpha_+^2 +\alpha_-^2 - 2\frac{c_-}{c_+}\alpha_-\alpha_+
</math>

The complete Hamiltonian is then (in Z-scaled a.u.)

<math>\begin{eqnarray}
H =-\frac{1}{2}\nabla^2_1 -\frac{1}{2}\nabla^2_2 -\frac{1}{r_1} -\frac{1}{r_2} +\frac{Z^{-1}}{r}\nonumber\\
= -\frac{1}{2}\nabla^2_1 -\frac{1}{2}\nabla^2_2 -\frac{1}{r_1} -\frac{Z-1}{Zr_2}+Z^{-1}\left(\frac{1}{r}-\frac{1}{r_2}\right)\nonumber\\
\ =-\frac{1}{8}\left[\frac{A_1}{r_1^2}+\frac{B_1+8}{r_1}+\frac{C_1}{r^2}+D_1+D_2 +\frac{A_2}{r_2^2} + \frac{B_2 + 8(Z-1)/Z}{r_2}+\frac{C_2}{r^2}\right]\nonumber
+ Z^{-1}\left(\frac{1}{r}-\frac{1}{r_2}\right)\nonumber
\end{eqnarray}</math>

The screened hydrogenic energy is
<math>\begin{eqnarray}
E_{SH}=-\frac{1}{2}\left[\frac{1}{n_1^2}+ \left(\frac{Z-1}{Z}\right)^2\frac{1}{n_2^2}\right]
\end{eqnarray}</math>
so that

<math>\begin{eqnarray}
H-E_{SH} &=& -\frac{1}{8}[
\frac{A_1}{r_1^2}+\frac{B_1+8}{r_1}+\frac{C_1}{r^2} + D_1 + D_2
-\frac{4}{n_1^2} -\left(\frac{Z-1}{Z}\right)^2\frac{4}{n_2^2}\nonumber
+\frac{A_2}{r_2}+\frac{B_2+8(Z-1)/Z}{r_2}+\frac{C_2}{r^2}] +
Z^{-1}\left(\frac{1}{r} -\frac{1}{r_2}\right)\nonumber
\end{eqnarray}</math>

==Optimization of Nonlinear Parameters==
- The traditional method of performing Hylleraas calculations is to write the
basis set in the form

<math>
\Psi = \sum_{i,j,k}c_{i,j,k}\varphi_{i,j,k}(\alpha,\beta)\pm exchange
</math>

with

<math>
\varphi_{i,j,k}(\alpha,\beta) = r_1^ir_2^jr_{12}^ke^{-\alpha r_1 - \beta
r_2}\mathcal{Y}^M_{l_1l_2L}.
</math>

The usual procedure is to set <math>\alpha = Z</math> so that it represents the inner
<math>1s</math> electron, and then to vary <math>\beta</math> so as to minimize the energy.

[[File:OptNon.jpg|500px]]

Since <math>\beta</math> appears in <math>\Psi</math> as a non-linear parameter, the entire
calculation must be repeated for each value of <math>\beta</math>. However, the minimum
becomes progressively smaller as the basis set is enlarged.

[[File:OptNon02.jpg|500px]]

*Difficulties

1. If the basis set is constructed so that <math> i +j +j \leq N</math>, the the
number of terms is
<math>
(N+1)(N+2)(N+3)/6
</math>
N=14 already gives 680 terms and an accuracy of about 1 part in <math>10^{10}</math> for
low-lying states. A substantial improvement in accuracy would require much
larger basis sets, together with multiple precision arithmetic to avoid loss
of significant figures when high powers are included.

2. The accuracy rapidly deteriorates as one goes to more highly excited
states - about 1 significant figure is lost each time the principle quantum
number is increased.

*Cure

We have found that writing basis sets in the form
\begin{eqnarray}
\Psi &=& \sum_{i,j,k}\left(c^{(1)}_{i,j,k}\varphi_{ijk}(\alpha_1\beta_1) +
c_{ijk}^{(2)}\varphi_{ijk}(\alpha_2\beta_2)\right) \pm exchange \nonumber\\
&=& \psi({\bf r}_1,{\bf r}_2) \pm \psi({\bf r}_2,{\bf r}_1)\nonumber
\end{eqnarray}
so that each combination of powers is included twice with different nonlinear
parameters gives a dramatic improvement in accuracy for basis sets of about
the same total size.

However, the optimization of the nonlinear parameters is now much more
difficult, and an automated procedure is needed.

If
<math>
E = \frac{<\Psi|H|\Psi>}{<\Psi|\Psi>}</math> and we assume <math><\Psi|\Psi> = 1</math>, then

<math>
\frac{\partial E}{\partial\alpha_t} = -2 <\Psi|H-E|r_1\psi({\bf r}_1,{\bf
r}_2;\alpha_t)\pm r_21\psi({\bf r}_2,{\bf
r}_1;\alpha_t)>
</math>
where
<math>
\psi({\bf r}_1,{\bf r}_2;\alpha_t) =
\sum_{i,j,k}c_{ijk}^{(t)}\varphi_{ijk}(\alpha_t\beta_t)
</math>

*Problem

1. Prove the above.

2. Prove that there is no contribution from the implicit dependence of
<math>c_{ijk}^{(t)}</math> on <math>\alpha_t</math> if the linear parameters have been optimized.

==The Screened Hydrogenic Term==

If the Hamiltonian is written in the form (in Z-scaled a.u.)
<math>\begin{eqnarray}\
H = H_0({\bf r}_1,Z) + H_0({\bf r}_2,Z-1) +
Z^{-1}\left(\frac{1}{r}-\frac{1}{r_2}\right)
\end{eqnarray}</math>
with
<math>\
H_0({\bf r}_1,Z) = -\frac{1}{2}\nabla_1^2 - \frac{1}{r_1}\nonumber
</math> and
<math>
H_0({\bf r}_2,Z-1) = -\frac{1}{2}\nabla_2^2 - \frac{Z-1}Z{r_2}\nonumber
</math>,
then the eigenvectors of <math>H_0({\bf r},Z) +H_0({\bf r}_2,Z-1)</math> are products of
hydrogenic orbitals
<math>
\Psi_0=\psi_0(1s,Z)\psi_0(nl,Z-1)
</math>
and the eigenvalue is
<math>\begin{eqnarray}
E_{SH} = \left[-\frac{1}{2} - \left(\frac{Z-1}{Z}\right)^2\frac{1}{2n^2}\right]Z^2 a.u.
\end{eqnarray}</math>
called the screened hydrogenic eigenvalue. From highly excited states,
<math>E_{SH}</math> and <math>\Psi_0</math>are already excellent approximations.

For example, for the 1s8d states, the energies are

<math>\begin{eqnarray}\\
E(1s8d^1D)&=& -2.007\,816\,512\,563\,81 {\rm\ a.u.}\nonumber\\
E(1s8d^3D)&=& - 2.007\,817\,934\,711\,71 {\rm\ a.u.}\nonumber\\
E_{SH} &=& -2.007\,812\,5 {\rm\ a.u.} \nonumber
\end{eqnarray}</math>

It is therefore advantageous to include the screened hydrogenic terms in the
basis set so that the complete trial function becomes
<math>
\Psi = c_0\Psi_0 +
\sum_{ijk}\left[c_{ijk}^{(1)}\varphi_{ijk}(\alpha_1,\beta_1)+c_{ijk}^{(2)}\varphi(\alpha_2,\beta_2)\right]\pm
{\rm exchange}
</math>

Also, the variational principal can be re-expressed in the form
<math>\begin{eqnarray}
E= E_{SH}+\frac{<\Psi|H-E_{SH}|\Psi>}{<\Psi|\Psi>}
\end{eqnarray}</math>
so that the <math>E_{SH}</math> term can be cancelled analytically from the matrix
elements.

For example
<math>
<\Psi_0|H-E_{SH}|\Psi_0> = 0
</math>

and
<math>
<\varphi_{ijk}|H-E_{SH}|\Psi_0> =
Z^{-1}<\varphi_{ijk}|\frac{1}{r}-\frac{1}{r_2}|\Psi_0>
</math>

Recall that
<math>\begin{eqnarray}\ \
I_0(a,b,c) = \frac{2}{c+2}\sum^{[c+1]/2}_{i=0}\left(\begin{array}{c}
c+2\\
2i+1
\end{array}\right) \left[f(p,q;\beta) + f(p^\prime,q^\prime;\alpha)\right]
\end{eqnarray}</math>
where

<math>\begin{eqnarray}
f(p,q;x) =
\frac{q!}{x^{q+1}(\alpha+\beta)^{p+1}}\sum^p_{j=0}\frac{(p+j)!}{j!}\left(\frac{x}{\alpha+\beta}\right)^j
\end{eqnarray}</math>

<math>
\begin{array}{cc}
p=a+2i+2 & p^\prime = b+2i+2\\
q= b+c-2i+2 & q^\prime =a+c-2i+2
\end{array}
</math>

If <math>\beta\ll\alpha</math>, the <math>f(p,q;\beta)\gg f(p^\prime,q^\prime;\alpha)</math> and the
<math>i=0</math> term is the dominant contribution to <math>f(p,q;\beta)</math>. However, since
this term depends only on the sum of powers <math>b+c</math> for <math>r_2</math>, it cancels exactly
from the matrix element of
<math>
\frac{1}{r_{12}} - \frac{1}{r_2}
</math>
and can therefore
be omitted in the calculations of integrands, thereby saving many significant
figures. This is especially valuable when <math>b+c</math> is large and <math>a</math> is small.

==Small Corrections==
*Mass Polarization
If the nuclear mass is not taken to be infinite, then the Hamiltonian is
<math>\begin{eqnarray}\ \
H = \frac{P_N^2}{2M} + \sum_{i=1}\frac{p_i^2}{2m} + V
\end{eqnarray}</math>

[[File:SmallCor.jpg|500px]]

where <math>M =M_A-nm</math> is the nuclear mass and <math>m</math> is the electron mass.

Change variables to the C of M
<math>\begin{eqnarray}
{\bf R} = \frac{1}{M+nm}\left[M{\bf r}_N + m({\bf r}_1 + {\bf r}_2 +
\ldots)\right]
\end{eqnarray}</math>

and relative variables

<math>
{\bf s}_i = {\bf r}_i - {\bf r}_N.
</math>

Then

<math>\begin{eqnarray}\ \ \ \ \
H = \frac{1}{2(M+nm)}{\bf p}_R^2+\frac{1}{2m}\sum_{i=1}^n{\bf p}_{s_i}^2
+\frac{1}{2M}\sum_{i,k}{\bf p}_{s_i}\cdot {\bf p}_{s_k} + V({\bf s}_1, {\bf
s}_2, \ldots)\nonumber
= \frac{1}{M+nm}{\bf p}_R^2+\frac{1}{2\mu}\sum_{i=1}^n{\bf
p}_{s_i}^2+\frac{1}{2M}\sum_{i\neq k}{\bf p}_{s_i}\cdot {\bf p}_{s_k}+V({\bf s}_1, {\bf
s}_2, \ldots)\nonumber
\end{eqnarray}</math>

where
<math>\begin{eqnarray}
\mu = \left(\frac{1}{M} + \frac{1}{m}\right)^{-1}
\end{eqnarray}</math>

If <math>n=2</math>, the Schrödinger equation is
<math>\begin{eqnarray}\ \ \
\left[ -\frac{\hbar^2}{2\mu}(\nabla^2_{s_1}+ \nabla^2_{s_2}) -
\frac{\hbar^2}{M}\nabla_{s_1}\nabla_{s_2}-\frac{Ze^2}{s_1}-\frac{Ze^2}{s_2}+\frac{e^2}{s_{12}}\right]\psi
= E\psi
\end{eqnarray}</math>.

Define the reduced mass Bohr radius
<math>\begin{eqnarray}
a_\mu = \frac{\hbar^2}{\mu e^2} = \frac{m}{\mu}a_0
\end{eqnarray}</math>
and

<math>\begin{eqnarray}
\rho_i = \frac{Zs_i}{a_\mu}\nonumber\\
\varepsilon = \frac{E}{Z^2(e^2/a_\mu)}\nonumber
\end{eqnarray}</math>

The Schrödinger equation then becomes
<math>\begin{eqnarray}\ \ \
\left[
-\frac{1}{2}(\nabla^2_{\rho_1}+\nabla^2_{\rho_2})-\frac{\mu}{M}\vec{\nabla}_{\rho_1}\cdot\vec{\nabla}_{\rho_2} - \frac{1}{\rho_1}-\frac{1}{\rho_2}+\frac{Z^{-1}}{\rho_{12}}\right]\psi = \varepsilon \psi
\end{eqnarray}</math>

The effects of finite mass are thus

1. A reduced mass correction to all energy levels of <math>(e^2/a_\mu)/(e^2/a_0)
= \mu/m</math>, i.e.
<math>\begin{eqnarray}
E_M=\frac{\mu}{m}E_\infty
\end{eqnarray}</math>

2. A specific mass shift given in first order perturbation theory by
<math>\begin{eqnarray}
\Delta\varepsilon =
-\frac{\mu}{M}<\psi|\nabla_{\rho_1}\cdot\nabla_{\rho_2}|\psi>
\end{eqnarray}</math>
or

<math>\begin{eqnarray}
\Delta\varepsilon =
-\frac{\mu}{M}<\psi|\nabla_{\rho_1}\cdot\nabla_{\rho_2}|\psi>\frac{e^2}{a_\mu}\nonumber\\
= -\frac{\mu^2}{mM}<\psi|\nabla_{\rho_1}\cdot\nabla_{\rho_2}|\psi>\frac{e^2}{a_0}\nonumber
\end{eqnarray}</math>

Since <math>\nabla_{\rho_1}\cdot\nabla_{\rho_2}</math> has the same angular properties as
<math>{\bf\rho}_1 \cdot {\bf\rho}_2</math>, the operator is like the product of two
dipole operators. For product type wave functions of the form
$
\psi = \psi(1s)\psi(nl) \pm \mbox{exchange},
$
the matrix element vanishes for all but P-states.

=Radiative Transitions=

==Transition Integrals and the Line Strength==

For an electric dipole transition <math>\gamma'\,L'\,S'\,J'\,M' \rightarrow \gamma\,L\,S\,J\,M</math>, the line strength is defined by
\begin{equation}
S = \sum_{M',\mu,M}|\langle\gamma'\,L'\,S'\,J'\,M'|r_\mu|\gamma\,L\,S\,J\,M\rangle|^2
\end{equation}
Using the Wigner-Eckart theorem, this is related to the reduced matrix element by
\begin{eqnarray}
S(J',J) = |\langle\gamma'\,L'\,S'\,J'||r||\gamma\,L\,S\,J\,M\rangle|^2
\sum_{M',\mu,M}\left|\left(\matrix{J' & 1 & J \cr
-M'& \mu &M \cr}\right)\right|^2
\end{eqnarray}
Using the sum rule for the 3-<math>j</math> symbols
\begin{equation}
\sum_{M',\mu,M}\left|\left(\matrix{J' & 1 & J \cr
-M'& \mu &M \cr}\right)\right|^2 = 1
\end{equation}
we see that the line strength is identical to the reduced matrix element
\begin{equation}
S(J',J) = |\langle\gamma'\,L'\,S'\,J'||r||\gamma\,L\,S\,J\,M\rangle|^2
\end{equation}
For a nonrelativistic calculation in <math>LS</math> coupling, <math>S = S'</math> and we can next strip out the dependence on <math>J</math> and <math>J'</math> by use of Eq.\ (7.1.7) of Edmonds { Angular Momentum in Quantum Mechanics} with <math>k = 1</math> to obtain
\begin{equation}
S(J',J) = (-1)^{L'+S+J+1}[(2J+1)(2J'+1)]^{1/2}\left\{
\matrix{L' & J'& S\cr
J & L & 1\cr}\right\}\langle\gamma' L'|| r || \gamma L\rangle
\end{equation}
where <math>\langle\gamma' L'|| r || \gamma L\rangle</math> is the reduced matrix element if spin is neglected. Note that if spin is neglected, then the line-strength is simply
\begin{equation}
S(L',L) = |\langle\gamma' L'|| r || \gamma L\rangle|^2
\end{equation}
To calculate the reduced matrix element, use a particular example along with the Wigner-Eckart theorem
such as
\begin{equation}
\langle\gamma'\, L'|| r || \gamma\, L\rangle = (-1)^{L'}\langle\gamma'\, L'\, 0| z | \gamma\, L\, 0\rangle{\Large /}\left(\matrix{L' & 1 & L \cr
0 & 0 & 0 \cr}\right)
\end{equation}

===Correlated Two-electron Integrals===
For the case of two-electron integrals, <math>z = z_1 + z_2</math> and <math>r = r_1 + r_2</math>. As discussed in G.W.F. Drake, Phys. Rev. A 18, 820 (1978), the general integral in Hylleraas coordinates for the integral over the <math>z_1</math> term can be written in the form
\begin{equation}
\label{int1}
\langle\gamma'\, L'\, 0| z_1 | \gamma\, L\, 0\rangle = \sum_\Lambda C_\Lambda I_\Lambda(\gamma',\gamma)
\end{equation}
where <math>I_\Lambda(\gamma',\gamma)</math> stands for the purely radial integral
\begin{equation}
\label{radial}
I_\Lambda(\gamma',\gamma) = \int_0^\infty r_1\,dr_1\int_0^\infty r_2\,dr_2\int_{|r_1-r_2|}^{r_1+r_2}r_{12}\, dr_{12}
R_{\gamma'}^*\,r_1\, R_\gamma P_\Lambda(\cos\theta_{12})
\end{equation}
and <math>\cos\theta_{12} = (r_1^2 + r_2^2 - r_{12}^2)/(2r_1r_2)</math> is a short-hand notation for a radial function of <math>r_1</math>, <math>r_2</math> and <math>r_{12}</math>. Also, <math>P_\Lambda(\cos\theta_{12})</math> is a Legendre polynomial,
and, <math>R_\gamma = R_\gamma(r_1,r_2,r_{12})</math> is the radial part of the Hylleraas wave function for state <math>\gamma</math> such
that the total wave function is <math>\psi({\bf r}_1,{\bf r}_2) = R_\gamma(r_1,r_2,r_{12}){\cal Y}_{l_1,l_2,L}^M(\hat{\bf r}_1,\hat{\bf r}_2)\pm {\rm exchange}</math>, normalized to unity, and <math>{\cal Y}_{l_1,l_2,L}^M(\hat{\bf r}_1,\hat{\bf r}_2)</math> is a vector coupled product of spherical harmonics. The coefficents <math>C_\Lambda</math> in Eq.\ (\ref{int1}) can be calculated from Eq.\ (32) of the above 1978 paper. They need to be calculated separately for he direct-direct and the direct-exchange contributions. For the reduced matrix element of <math>z_1</math>, the leading 3-<math>j</math> symbol can be dropped, leaving (for the case <math>k = 1</math>)
\begin{eqnarray}
C_\Lambda^{\rm reduced} &=& \frac{(-1)^{l_2'+L'+L+1}}{2} (l_1,l_1',l_2,l_2',L,L')^{1/2}
\sum_{\lambda={\rm max}(l_1-1,0)}^{l_1+1} (\lambda,\Lambda,1)
\left(\matrix{l_1 & 1 & \lambda \cr
0 & 0 & 0 \cr}\right)
\left(\matrix{l_1'& \lambda & \Lambda \cr
0 & 0 & 0 \cr}\right)\nonumber\\
&\times&\left(\matrix{l_2' & 1_2 & \Lambda \cr
0 & 0 & 0 \cr}\right)
\left\{\matrix{L' & \lambda & l_2 \cr
\Lambda& l_2' & l_1' \cr}\right\}
\left\{\matrix{L' & \lambda & l_2 \cr
l_1 & L & 1 \cr}\right\}
\end{eqnarray}
where the notation <math>(a,b,c,\cdots)</math> means <math>(2a+1)(2b+1)(2c+1)\cdots</math>.
A similar expression can be obtained for the angular coefficients for the <math>z_2</math> term by interchanging the subscripts 1 and 2 throughout and multiplying by <math>(-1)^{L+L'+1}</math>.

===Example===
As an example, for the <math>1s^2\;^1S - 1s2p\;^1P</math> transition of helium, the only nonvanishing <math>C_\Lambda</math> angular coefficients are <math>C_1 = 1/2</math> for the direct-direct term (i.e. <math>P_1(\cos\theta_{12}) = \cos\theta_{12}</math> is included in the radial integral (\ref{radial})) and <math>C_0 = 1/2</math> for the direct-exchange term (i.e. <math>P_0(\cos\theta_{12}) = 1</math> is included in the radial integral (\ref{radial})). The results for infinite nuclear mass are
\begin{equation}
S(1\;^1S-2\;^1P) = (0.781\,127\,158\,667)^2
\end{equation}
\begin{equation}f(1\;^1S-2\;^1P) = -0.276\,164\,703\,736
\end{equation}
where
\begin{equation}
f(1\;^1S-2\;^1P) = \frac{2}{3}[E(1\;^1S) - E(2\;^1P)]S(1\;^1S-2\;^1P)
\end{equation}
is the absorption oscillator strength.

===Test===
This is a test edit

Theory Notes

2020-12-14T05:30:51Z

Drake: /* General Hermitian Property */

<center>
<H1>Notes on solving the Schrödinger equation in Hylleraas coordinates for heliumlike atoms
Gordon W.F. Drake
Department of Physics, University of Windsor
Windsor, Ontario, Canada N9B 3P4
(Transcribed from hand-written notes and edited by Lauren Moffatt. Last revised November 22, 2017 to correct a missing subscript in the expression for $B_1$ in the Hamiltonian in Sect. 11 General Hermitian Property.)
</H1>
</center>

==Introduction==
These notes describe some of the technical details involved in solving the Schrödinger equation for a heliumlike atom of nuclear charge Z in correlated Hylleraas coordinates. The standard for computational purposes is to express all quantities in a form that is valid for electronic states of arbitrary angular momentum, and to express matrix elements of the Hamiltonian in a manifestly Hermitian symmetrized form. Further information and derivations of the mathematical techniques for angular integrals and radial recursion relations can be found in the paper
G.W.F. Drake, Phys. Rev. A <math>{\bf 18}</math>, 820 (1978).

The starting point is the two-electron Schrödinger equation for infinite nuclear mass
\begin{equation} \left[-\frac{\hbar^2}{2m}(\nabla^2_1 +\nabla^2_2) - \frac{Ze^2}{r_1} -
\frac{Ze^2}{r_2}+\frac{e^2}{r_{12}} \right]\psi = E\psi\nonumber
\end{equation}
where <math>m</math> is the electron mass, and <math>r_{12} = |{\bf r}_1 - {\bf r}_2|</math> (see diagram below).

[[File:Pic1.jpg|250px]]

Begin by rescaling distances and energies so that the Schrödinger equation can be expressed in a dimensionless form. The dimensionless Z-scaled distance is defined by <math>\begin{eqnarray} \rho = \frac{Zr}{a_0} \end{eqnarray} </math> where <math> \begin{eqnarray} a_0 = \frac{\hbar^2}{me^2} \end{eqnarray} </math> is the atomic unit (a.u.) of distance equal to the Bohr radius <math>0.529\, 177\, 210\, 92(17) \times 10^{-10}</math> m. Then

\begin{equation}[-\frac{\hbar^2}{2m}Z^2(\frac{me^2}{\hbar^2})^2(\nabla^2_{\rho_1}+\nabla^2_{\rho_2})
- Z^2\frac{e^2}{a_0}\rho^{-1}_1 - Z^2\frac{e^2}{a_0}\rho^{-1}_2 +
\frac{e^2}{a_0}Z\rho^{-1}_{12}]\psi= E\psi\nonumber\end{equation}

But <math>\begin{eqnarray} \frac{\hbar^2}{m}(\frac{me^2}{\hbar^2})^2 = \frac{e^2}{a_0} \end{eqnarray} </math> is the hartree atomic unit of of energy <math>(E_h = 27.211\,385\,05(60)</math> eV, or equivalently <math>E_h/(hc) = 2194.746\,313\,708(11)</math> m<math>^{-1})</math>. Therefore, after multiplying through by <math>a_0/(Ze)^2</math>, the problem to be solved in Z-scaled dimensionless units becomes

\begin{equation} \left[-\frac{1}{2}(\nabla^2_{\rho_1}+\nabla^2_{\rho_2}) -
\frac{1}{\rho_1} - \frac{1}{\rho_2} + \frac{Z^{-1}}{\rho_{12}}\right]\psi =
\varepsilon\psi\nonumber \end{equation}

where <math>\begin{eqnarray}\varepsilon = \frac{Ea_0}{(Ze)^2} \end{eqnarray}</math> is the energy in Z-scaled atomic units. For convenience, rewrite this in the conventional form
\begin{equation}
H\psi = \varepsilon\psi\nonumber \end{equation}
where (using <math>r</math> in place of <math>\rho</math> for the Z-scaled distance)
\begin{equation}
H = -\frac{1}{2}(\nabla^2_1+\nabla^2_2) - \frac{1}{r_1}-\frac{1}{r_2} + \frac{Z^{-1}}{r_{12}}
\nonumber\end{equation}
is the atomic Hamiltonian for infinite nuclear mass.

==The Hartree Fock Method==

For purposes of comparison, and to define the correlation energy, assume that <math>\psi({\bf r}_1,{\bf r}_2)</math> can be written in the separable product form
\begin{equation}
\psi({\bf r}_1,{\bf r}_2) = \frac{1}{\sqrt{2}} \left[ u_1(r_1) u_2(r_2) \pm u_2(r_1) u_1(r_2) \right] \nonumber
\end{equation}
for the <math>1s^2\ ^1\!S</math> ground state. Because of the <math>\frac{1}{r_{12}}</math> term in the Schrödinger equation, the exact solution connot be expressed in this form as a separable product. However, the Hartree-Fock approximation corresponds to finding the best possible solution to the Schrödinger equation
\begin{equation}
H\psi({\bf r}_1,{\bf r}_2) = E\psi({\bf r}_1,{\bf r}_2)\nonumber
\end{equation}
that can nevertheless be expressed in this separable product form, where as before
\begin{equation}
H = -\frac{1}{2}(\nabla^2_1+\nabla^2_2) - \frac{1}{r_1}- \frac{1}{r_2} + \frac{Z^{-1}}{r_{12}}\nonumber
\end{equation}
is the full two-electron Hamiltonian. To find the best solution, substitute into <math>\left\langle\psi \lvert H-E \rvert \psi\right\rangle</math> and require this expression to be stationary
with respect to arbitrary infinitesimal variations <math>\delta u_1</math> and <math>\delta
u_2</math> in <math>u_1</math> and <math>u_2</math> respectively; i.e.

\begin{equation}
\frac{1}{2} \left\langle \delta u_1(r_1)u_2(r_2) \pm u_2(r_1)\delta
u_1(r_2) \lvert H-E \rvert u_1(r_1) u_2(r_2)\pm u_2(r_1) u_1(r_2)\right\rangle\nonumber
\end{equation}

\begin{equation}
=\int\delta u_1(r_1) d{\bf r}_1 \left\{ \int d{\bf r}_2 u_2(r_2) \left( H - E \right) \left[ u_1(r_1)u_2(r_2) \pm u_2(r_1) u_1(r_2) \right] \right\}\nonumber
\end{equation}

\begin{equation}
= 0 \nonumber
\end{equation}

for arbitrary <math> \delta u_1(r_1).\nonumber </math> Therefore <math>\{\int d{\bf r}_2 \ldots \} = 0</math>.

Similarly, the coefficient of <math>\delta u_2</math> would give

\begin{equation}
\int d{\bf r}_1 u_1(r_1) \left( H-E \right) \left[ u_1(r_1) u_2(r_2) \pm
u_2(r_1) u_1(r_2) \right] = 0\nonumber
\end{equation}

Define

\begin{equation}
I_{12} = I_{21} = \int d{\bf r}\, u_1(r)u_2(r), \nonumber
\end{equation}

\begin{equation}
H_{ij} = \int d{\bf r}\, u_i(r)(-\frac{1}{2}\nabla - \frac{1}{r})u_j(r), \nonumber
\end{equation}

\begin{equation}
G_{ij}(r) = \int d{\bf r}^\prime u_i(r^\prime)\frac{1}{|{\bf r} - {\bf r}\prime|}u_j(r^\prime)\nonumber
\end{equation}

Then the above equations become the pair of integro-differential equations

\begin{equation}
[ H_0 - E + H_{22}+G_{22}(r)]u_1(r) = \mp [ I_{12}(H_0-E) + H_{12}+G_{12}(r)]u_2(r)\nonumber
\end{equation}

\begin{equation}
[H_0-E+H_{11}+G_{11}(r)]u_2(r) = \mp [I_{12}(H_0-E) + H_{12}+G_{12}(r)]u_1(r)\nonumber
\end{equation}

These must be solved self-consistently for the "constants" <math>I_{12}</math> and
<math>H_{ij}</math> and the function <math>G_{ij}(r)</math>.

The Hartree Fock energy is <math>E \simeq -2.87\ldots </math>a.u. while the exact energy is <math>E = -2.903724\ldots </math>a.u.

The difference is called the "correlation energy" because it arises from the
way in which the motion of one electron is correlated to the other. The
Hartree Fock equations only describe how one electron moves in the average field
provided by the other (mean-field theory).

==Configuration Interaction==
Expand
\begin{equation}
\psi({\bf r}_1,{\bf r}_2)= C_0u^{(s)}_1(r_1)u^{(s)}_1(r_2) + C_1u^{(P)}_1({\bf r}_1)u^{(P)}_1({\bf r}_2){\cal Y}^0_{1,1,0}(\hat{\bf r}_1, \hat{\bf r}_2)+C_2u^{(d)}_1({\bf r}_1)u^{(d)}_2({\bf r}_2){\cal Y}^0_{2,2,0}(\hat{\bf r}_1, \hat{\bf r}_2)+... \pm {\rm\ exchange}
\end{equation}
where
\begin{equation}
{\cal Y}^M_{l_1,l_2,L}(\hat{\bf r}_1, \hat{\bf r}_2)=\sum_{m_1,m_2}Y^{m_1}_{l_1}({\bf r}_1)Y^{m_2}_{l_2}({\bf r}_2)\times <l_1l_2m_1m_2\mid LM>
\end{equation}

This works, but is slowly convergent, and very laborious. The best CI calculations are accurate to <math> ~10^{-7}</math> a.u.

==Hylleraas Coordinates==
[E.A. Hylleraas, Z. Phys. <math>{\bf 48}, 469(1928)</math> and <math>{\bf 54}, 347(1929)</math>]
suggested using the co-ordinates <math>r_1</math>, <math>r_2</math> and <math>r_{12}</math> or equivalently

\begin{eqnarray}
s &=& r_1 + r_2, \nonumber\\
t &=& r_1-r_2, \nonumber\\
u &=& r_{12}\nonumber
\end{eqnarray}

and writing the trial functions in the form

\begin{equation}
\Psi({\bf r}_1,{\bf r}_2) = \sum^{i+j+k\leq
N}_{i,j,k}c_{i,j,k}r_1^{i+l_1}r_2^{j+l_2}r_{12}^ke^{-\alpha r_1 - \beta r_2}
\mathcal{Y}^M_{l_1,l_2,L}(\hat{r}_1,\hat{r}_2)\pm \text{exchange}\nonumber
\end{equation}

Diagonalizing H in this non-orthogonal basis set is equivalent to solving
<math> \begin{eqnarray}
\frac{\partial E}{\partial c_{i,j,k}} = 0\nonumber
\end{eqnarray} </math>
for fixed <math>\alpha</math> and <math>\beta</math>.

The diagonalization must be repeated for different values of <math>\alpha</math> and
<math>\beta</math> in order to optimize the non-linear parameters.

==Completeness==
The completeness of the above basis set can be shown by first writing
\begin{equation}
r_{12}^2 = r_1^2 + r_2^2 - 2r_1r_2\cos\theta_{12}
\end{equation}
and, from the spherical harmonic addition theorem,
\begin{equation}
\cos(\theta_{12})=\frac{4\pi}{3}\sum^1_{m=-1}Y^{m*}_1(\theta_1,\varphi_1)Y^m_1(\theta_2,\varphi_2)
\end{equation}
Consider first the S-states. The <math>r_{12}^0</math> terms are like the ss terms in a CI calculation. The
<math>r_{12}^2</math> terms bring in p-p type contributions, and the higher powers bring
in d-d, f-f etc. type terms. In general

\begin{equation}
P_l(\cos\theta_{12}) = \frac{4\pi}{2l+1}\sum^l_{m=-l}{Y^{m}_l}^*(\theta_1,\varphi_1)Y^m_l(\theta_2,\varphi_2)\nonumber
\end{equation}

For P-states, one would have similarly

\begin{array}{lr}
r_{12}^0 & (sp)P\nonumber\\
r_{12}^2 & (pd)P\nonumber\\
r_{12}^4 & (df)P\nonumber\\
\vdots & \vdots\nonumber
\end{array}

For D-states

\begin{array}{lr}
r_{12}^0 & (sd)D & (pp^\prime)D\nonumber\\
r_{12}^2 & (pf)D & (dd^\prime)D\nonumber\\
r_{12}^4 & (dg)D & (ff^\prime)D\nonumber\\
\vdots & \vdots & \vdots\nonumber
\end{array}

In this case, since there are two "lowest-order" couplings to form a
D-state, both must be present in the basis set, i.e.

\begin{equation}
\Psi(r_2,r_2) = \sum c_{ijk}r_1^ir_2^{j+2}r_{12}^ke^{-\alpha r_1-\beta
r_2}\mathcal{Y}^M_{022}(\hat{r}_1,\hat{r}_2)
+\sum d_{ijk}r_1^{i+1}r_2^{j+1}r_{12^k}e^{-\alpha^\prime r_1 - \beta^\prime
r_2}\mathcal{Y}^M_{112}(\hat{r}_1,\hat{r}_2)\nonumber
\end{equation}

For F-states, one would need <math>(sf)F</math> and <math>(pd)F</math> terms.

For G-states, one would need <math>(sg)G</math>, <math>(pf)G</math> and <math>(dd^\prime)G</math> terms.

Completeness of the radial functions can be proven by considering the
Sturm-Liouville problem

\begin{equation}
\left(-\frac{1}{2}\nabla^2-\frac{\lambda}{r_s}-E\right)\psi({\bf r}) = 0\nonumber
\end{equation}

or

\begin{equation}
\left(-\frac{1}{2}\frac{1}{r^2}\frac{\partial}{\partial r}\left({r^2}\frac{\partial}{\partial r}\right) -
\frac{l\left(l+1\right)}{2r^2} - \frac{\lambda}{r} - E\right)u(r) = 0.\nonumber
\end{equation}

For fixed E and variable <math>\lambda</math> (nuclear charge).

The eigenvalues are <math>\lambda_n = (E/E_n)^{1/2}</math>, where <math>E_n =- \frac{1}{2n^2}</math>

[[File:Pic2.jpg|500px]]

and the eigenfunctions are

\begin{equation}
u_{nl}(r) =
\frac{1}{(2l+1)!}\left(\frac{(n+l)!}{(n-l-1)2!}\right)^{1/2}(2\alpha)^{3/2}e^{-\alpha
r}\nonumber
\times \ \left( 2 \alpha r \right)^l\,_{1}F_{1}\left(-n+l+1,2l+2; 2 \alpha r \right)
\end{equation}

with <math>\alpha = (-2E)^{1/2}</math> and <math>n\geq l+1</math>. The confluent hypergeometric function <math> _1F_1(a,b,;z)</math> then denotes a finite polynomial since <math>a=-n+l+1</math> is a negative integer or zero.

Unlike the hydrogen spectrum, which has both a discrete part for <math>E<0</math> and a
continuous part for <math>E>0</math>, this forms an entirely discrete set of finite
polynomials, called Sturmian functions. They are orthogonal with respect to
the potential, i.e.

\begin{equation}
\int^\infty_0 r^2dr \left( u_{n^\prime l}(r)\frac{1}{r}u_{nl}(r) \right) = \delta_{n,n^\prime}\nonumber
\end{equation}

Since they become complete in the limit <math>n\rightarrow\infty</math>, this assures the
completeness of the variational basis set.

[See also B Klahn and W.A. Bingel Theo. Chim. Acta (Berlin) 44, 9 and 27
(1977)].

==Solutions of the Eigenvalue Problem==

For convenience, write

\[
\Psi({\bf r}_1,{\bf r}_2) = \sum^N_{m=1}c_m\varphi_m\nonumber
\]

where <math>m</math> represents the <math>m</math>'th combination of <math>i,j,k</math> and

\begin{equation}
\varphi_{ijk}=r_1^ir_2^jr_{12}^ke^{-\alpha r_1 - \beta
r_2}\mathcal{Y}^M_{l_1,l_2,L}(\hat{r}_1,\hat{r}_2) \pm \mbox{exchange}\nonumber
\end{equation}

\begin{align}

&
\left( \begin{array}{lr}
\cos(\theta) & \sin(\theta)\\
-\sin(\theta) & \cos(\theta)
\end{array} \right)
\left( \begin{array}{lr}
H_{11} & H_{12} \\
H_{12} & H_{22}
\end{array}\right)
\left( \begin{array}{lr}
\cos(\theta) & -\sin(\theta)\\
\sin(\theta) & \cos(\theta)
\end{array} \right)
\\[5pt]
= & \left(\begin{array}{lr}
cH_{11}+sH_{12} & cH_{12} + sH_{22}\\
-sH_{11} + cH_{12} & -sH_{12} + cH_{22}
\end{array}\right)
\left( \begin{array}{cc}
c & -s \\
s & c
\end{array}\right)
\\[5pt]
= & \left(\begin{array}{lr}
c^2H_{11}+s^2H_{22} + 2csH_{12} & (c^2-s^2)H_{12}+cs(H_{22}-H_{11})\\
(c^2-s^2)H_{12}+cs(H_{22}-H_{11}) & s^2H_{11}+c^2H_{22}-2csH_{12}
\end{array}\right)
\end{align}

Therefore

\begin{equation}
\left[\cos^2(\theta)-\sin^2(\theta)\right]H_{12} =
\cos(\theta)\sin(\theta)(H_{11}-H_{22})\nonumber
\end{equation}

and

\begin{equation}
\tan(2\theta) = \frac{2H_{12}}{H_{11}-H_{22}}\nonumber
\end{equation}

i.e.

\begin{equation}
\cos(\theta)=\left(\frac{r+\omega}{2r}\right)^{1/2}\nonumber
\end{equation}

\begin{equation}
\sin(\theta)=-\mbox{sgn}(H_{12})\left(\frac{r-\omega}{2r}\right)^{1/2}\nonumber
\end{equation}

where

\begin{equation}
\omega = H_{22}-H_{11}\nonumber
\end{equation}

\begin{equation}
r=\left(\omega^2+4H_{12}^2\right)^{1/2}\nonumber
\end{equation}

\begin{equation}
E_1=\frac{1}{2}\left(H_{11}+H_{22}-r\right)\nonumber
\end{equation}

\begin{equation}
E_2=\frac{1}{2}\left(H_{11}+H_{22}+r\right)\nonumber
\end{equation}

===Brute Force Method===

- Gives all the eigenvalues and eigenvectors, but it is slow

- First orthonormalize the <math>N</math>-dimensional basis set; i.e. form linear combinations

\begin{equation}
\Phi_m = \sum_{n=1}^N\varphi_nR_{nm}\nonumber
\end{equation}

such that
\begin{equation}
\langle \Phi_m | \Phi_n \rangle = \delta_{m,n}
\end{equation}

This can be done by finding an orthogonal tranformation T such that

\begin{equation}
T^TOT = I = \left(
\begin{array}{ccccc}
I_1 & 0 & 0 & \ldots & 0 \\
0 & I_2 & 0 & \ldots & 0 \\
0 & 0 & I_3 & \ldots & 0 \\
\vdots & \vdots & \vdots & \ddots & \vdots \\
0 & 0 & 0 & \ldots & I_N
\end{array}\right);
\end{equation}

\begin{equation}
O_{mn} = \langle \varphi_m | \varphi_n \rangle \nonumber
\end{equation}

and then applying a scale change matrix

\begin{equation}
S = \left(\begin{array}{ccccc}
\frac{1}{I_1^{1/2}} & 0 & 0 & \ldots & 0\\
0 & \frac{1}{I_2^{1/2}} & 0 & \ldots & 0 \\
0 & 0 & \frac{1}{I_3^{1/2}} & \ldots & 0 \\
\vdots & \vdots & \vdots & \ddots & \vdots \\
0 & 0 & 0 & \ldots & I_N^{1/2}
\end{array}\right)= S^T\nonumber
\end{equation}

Then

\begin{equation}
S^TT^TOTS = 1
\end{equation}

i.e.

\begin{equation}
R^TOR = 1
\end{equation}

with <math>R=TS</math>.

If H is the matrix with elements <math>H_{mn} = \langle \varphi_m | \varphi_n \rangle </math>, then H
expressed in the <math>\Phi_m</math> basis set is

\begin{equation}
H^\prime = R^THR.\nonumber
\end{equation}

We next diagonalize <math>H^\prime</math> by finding an orthogonal transformation W such
that

\begin{equation}
W^T H^\prime W = \lambda = \left(
\begin{array}{cccc}
\lambda_1 & 0 & \ldots & 0 \\
0 & \lambda_2 & \ldots & 0 \\
\vdots & \vdots & \ddots & \vdots \\
0 & 0 & \ldots & \lambda_N
\end{array}\right)\nonumber
\end{equation}

The q'th eigenvector is

\begin{equation}
\Psi^{(q)} = \sum_{n=1}^N\Phi_n W_{n,q} = \sum_{n,n^\prime} \varphi_{n^\prime} R_{n^\prime ,n} W_{n,q} \nonumber
\end{equation}

i.e.

\begin{equation}
c_{n^\prime}^{(q)} = \sum_{n=1}^N R_{n^\prime n} W_{n,q}
\end{equation}

===The Power Method===

- Based on the observation that if H has one eigenvalue, <math>\lambda_M</math>, much bigger than all the rest, and <math>\chi = \left(
\begin{array}{c}a_1\\a_2\\\vdots\end{array}\right)</math>
is an arbitrary starting vector, then <math>\chi = \sum_{q=1}^N x_q\Psi^{(q)}</math>.

\begin{equation}
(H)^n\chi = \sum_{q=1}^N x_q \lambda^n_q\Psi^{(q)}\nonumber\rightarrow x_M\lambda_M^n\Psi^{(M)}\nonumber
\end{equation}

provided <math>x_M\neq 0</math>.

To pick out the eigenvector corresponding to any eigenvalue, write the original problem in the form

\begin{align}
H\Psi & = \lambda O \Psi \nonumber \\
\left( H - \lambda_q O \right) \Psi & = \left( \lambda - \lambda_q \right) O \Psi \nonumber
\end{align}

Therefore,

\begin{equation}
G \Psi = \frac{1}{\lambda - \lambda_q } \Psi \nonumber \\
\end{equation}

where <math>G=(H-\lambda_qO)^{-1}O</math> with eigenvalues <math>\frac{1}{\lambda_n - \lambda_q}</math>.

By picking <math>\lambda_q</math> close to any one of the <math>\lambda_n</math>, say
<math>\lambda_{n^\prime}</math>, then <math>\frac{1}{\lambda_n-\lambda_q}</math> is much larger for
<math>n=n^\prime</math> than for any other value. The sequence is then

\begin{array}{c}
\chi_1=G\chi\nonumber\\
\chi_2=G\chi_1\nonumber\\
\chi_3=G\chi_2\nonumber\\
\vdots\nonumber
\end{array}

until the ratios of components in <math>\chi_n</math> stop changing.

- To avoid matrix inversion and multiplication, note that the sequence is equivalent to
\begin{equation}
F \chi_n = \left( \lambda - \lambda_q \right) O \chi_{n-1} \nonumber \\
\end{equation}
where <math>F = H-\lambda_qO</math>. The factor of <math>\left( \lambda - \lambda_q \right)</math> can be
dropped because this only affects the normalization of <math>\chi_n</math>. To find <math>\chi_n</math>, solve
\begin{equation}
F \chi_n = O \chi_{n-1} \nonumber
\end{equation}
(N equations in N unknowns). Then
\begin{equation}
\lambda = \frac{ \langle \chi_n | H | \chi_n \rangle}{\langle \chi_n | \chi_n \rangle} \nonumber
\end{equation}

==Matrix Elements of H==

\begin{equation}
H=-\frac{1}{2}\nabla^2_1 -\frac{1}{2}\nabla^2_2 - \frac{1}{r_1} - \frac{1}{r_2} +\frac{Z^{-1}}{r_{12}}\nonumber
\end{equation}

Taking <math>r_1, r_2</math> and <math>r_{12}</math> as independent variables,

\begin{equation}
\nabla_1^2 = \frac{1}{r_1^2}\frac{\partial}{\partial r_1}\left(r_1^2\frac{\partial}{\partial r_1}\right)+ \frac{1}{r_{12}^2}\frac{\partial}{\partial r_{12}}
\left(r_{12}^2\frac{\partial}{\partial r_{12}}\right)\nonumber -\frac{l_1(l_1+1)}{r_1^2}+2(r_1 - r_2\cos(\theta_{12}))\frac{1}{r_{12}}\frac{\partial^2}{\partial r_1 \partial r_{12}} \nonumber - 2(\nabla_1^Y \cdot {\bf r}_2)\frac{1}{r_{12}}\frac{\partial}{\partial r_{12}}\nonumber
\end{equation}

where <math>\nabla_1^Y</math> acts only on the spherical harmonic part of the wave function and the diagram

[[File:Pic3.jpg|400px]]

defines the complete set of 6 independent variables is <math>r_1</math>, <math>r_2</math>, <math>r_{12}</math>, <math>\theta_1</math>, <math>\varphi_1</math>, <math>\chi</math>.

If <math>r_{12}</math> were not an independent variable, then one could take the volume element to be

\begin{equation}
d\tau = r_1^2dr_1\sin(\theta_1)d\theta_1d\varphi_1r_2^2dr_2\sin(\theta_2)d\theta_2d\varphi_2.\nonumber
\end{equation}

However, <math>\theta_2</math> and <math>\varphi_2</math> are no longer independent variables. To eliminate them, take the point <math>{\bf r}_1</math> as the
origin of a new polar co-ordinate system, and write

\begin{equation}
d\tau=-r_1^2dr_1\sin(\theta_1)d\theta_1d\varphi_1r_{12}^2dr_{12}\sin(\psi)d\psi d\chi\nonumber
\end{equation}

and use

\begin{equation}
r_2^2=r_1^2+r_{12}^2 +2r_1r_{12}\cos(\psi)
\end{equation}

Then for fixed <math>r_1</math> and <math>r_{12}</math>,

\begin{equation}
2r_2dr_2 = -2r_1r_{12}\sin(\psi)d\psi
\end{equation}

Thus

\begin{equation}
d\tau= r_1dr_1r_2dr_2r_{12}dr_{12}\sin(\theta_1)d\theta_1d\varphi_1d\chi
\end{equation}

The basic type of integral to be calculated is

\begin{align}
I \left( l_1, m_1, l_2, m_2; R \right) & = \int \sin \left( \theta_1 \right) d\theta_1 d\varphi_1 d\chi Y^{m_1}_{l_1} \left( \theta_1, \varphi_1 \right)^{*} Y^{m_2}_{l_2} \left( \theta_2, \varphi_2 \right) \\
& \times \int r_1dr_1r_2dr_2r_{12}dr_{12}R \left( r_1, r_2, r_{12} \right)
\end{align}

Consider first the angular integral. <math>Y^{m_2}_{l_2} \left( \theta_2, \varphi_2 \right)</math> can be expressed in terms of the independent variables <math>\theta_1, \varphi_1, \chi</math> by use of the rotation matrix relation
\begin{equation}
Y^{m_2}_{l_2}( \theta_2, \varphi_2) = \sum_m \mathcal{D}^{(l_2)}_{m_2,m}(\varphi_1,\theta_1,\chi)^*Y^m_{l_2}( \theta_{12}, \varphi)
\end{equation}
where <math>\theta, \varphi</math> are the polar angles of <math>{\bf r}_2</math> relative to <math>{\bf r}_1</math>. The angular integral is then

\begin{align}
I_{ang} & = \int^{2 \pi}_{0} d\chi \int^{2\pi}_{0} d\varphi_1 \int^{\pi}_{0} \sin \left( \theta_1 \right) d\theta_{1} Y^{m_1}_{l_1} \left( \theta_1, \varphi_1 \right)^{*} \\
& \times \sum_m \mathcal{D}^{(l_2)}_{m_2,m} \left( \varphi_1,\theta_1,\chi \right)^{*} Y^{m}_{l_2} \left( \theta_{12},\varphi \right) \\
\end{align}

Use

\begin{equation}
Y^{m_1}_{l_1} \left( \theta_1, \varphi_1 \right)^{*} = \sqrt{\frac{2l_1+1}{4\pi}}\mathcal{D}^{(l_1)}_{m_1,0} \left( \varphi_1, \theta_1, \chi \right)
\end{equation}

together with the orthogonality property of the rotation matrices (Brink and Satchler, p 147)

\begin{align}
& \int \mathcal{D}^{(j)*}_{m,m^\prime}\mathcal{D}^{(J)}_{M,M^\prime}\sin(\theta_1)d\theta_1d\varphi_1d\chi \\
& = \frac{8\pi^2}{2j+1}\delta_{jJ}\delta_{mM}\delta_{m^\prime M^\prime}\nonumber
\end{align}

to obtain

\begin{align}
I_{ang} & = \sqrt{ \frac{2l_1+1}{4\pi}} \frac{8\pi^2}{2l_1+1} \delta_{l_1,l_2} \delta_{m_1,m_2} Y^{0}_{l_2} \left( \theta_{12}, \varphi \right) \\
& = 2 \pi \delta_{l_1,l_2} \delta_{m_1,m_2} P_{l_2} \left( \cos \theta_{12} \right)
\end{align}

since

\begin{equation}
Y^0_{l_2} \left( \theta_{12}, \varphi \right) = \sqrt{\frac{2l_1+1}{4\pi}} P_{l_2} \left( \cos \left( \theta_{12} \right) \right)
\end{equation}

Note that <math>P_{l_2} \left( \cos \theta \right)</math> is just a short hand expression for a radial function because

\begin{equation}
\cos \theta_{12} = \frac{r_1^2 + r_2^2 - r_{12}^2}{2r_1r_2}
\end{equation}

The original integral is thus

\begin{equation}
I \left( l_1, m_1, l_2, m_2; R \right) = 2 \pi \delta_{l_1,l_2} \delta_{m_1,m_2} \int^{\infty}_{0} r_{1} dr_{1} \int^{\infty}_{0} r_{2} dr_{2} \int^{r_{1} + r_{2}}_{|r_{1} - r_{2}|} r_{12} dr_{12} R \left( r_{1}, r_{2}, r_{12} \right) P_{l_2} \left( \cos \theta_{12} \right)
\end{equation}

where again

\begin{equation}
\cos \theta_{12} = \frac{r_{1}^{2} +r^{2}_{2} - r_{12}^{2}}{2 r_{1} r_{2}}
\end{equation}

is a purely radial function.

The above would become quite complicated for large <math>l_2</math> because <math>P_{l_2}(\cos \theta_{12})</math> contains terms up to <math>\left( \cos \theta_{12} \right)^{l_2}</math>. However, recursion relations exist which allow any integral containing <math>P_l \left( \cos \theta_{12} \right)</math> in terms of those containing just <math>P_0 \left( \cos \theta_{12} \right) = 1</math> and <math>P_1 \left( \cos \theta_{12} \right) = \cos \theta_{12}</math>.

==Radial Integrals and Recursion Relations==

The basic radial integral is [see G.W.F. Drake, Phys. Rev. A 18, 820 (1978)]

\begin{align}
I_0 \left( a, b, c \right) & =
\int^{\infty}_{0} r_{1} dr_{1} \int^{\infty}_{r_{1}} r_{2} dr_{2} \int^{r_{1}+r_{2}}_{r_{2} - r_{1}} r_{12} dr_{12} r_{1}^{a} r_{2}^{b} r_{12}^{c} e^{-\alpha r_{1} - \beta r_{2}} + \int^{\infty}_{0} r_{2} dr_{2} \int^{\infty}_{r_{2}} r_{1} dr_{1} \int^{r_{1} + r_{2}}_{r_{1} - r_{2}} r_{12} dr_{12} r_{1}^{a} r_{2}^{b} r_{12}^{c} e^{- \alpha r_{1} - \beta r_{2}} \\
& = \frac{2}{c+2} \sum^{[(c+1)/2]}_{i=0}
\left(\begin{array}{c}
c+2 \\
2i+1
\end{array} \right)
\left\{ \frac{q!}{\beta^{q+1} \left( \alpha + \beta \right)^{p+1}} \sum_{j=0}^{q} \frac{\left( p + j \right)!}{j!} \left( \frac{\beta}{\alpha + \beta} \right)^j + \frac{q^{\prime}!}{\alpha^{q^{\prime}+1} \left( \alpha + \beta \right)^{p^{\prime}+1}} \sum^{q^{\prime}}_{j=0} \frac{\left( p^{\prime} + j \right)!}{j!} \left( \frac{\alpha}{\alpha + \beta} \right)^j \right\}
\end{align}

where

\begin{array}{ll}
p = a + 2i + 2 & p^{\prime} = b + 2i + 2 \\
q = b + c - 2i + 2 & q^{\prime} = a + c - 2i + 2
\end{array}

The above applies for <math> a, b \geq -2, c\geq -1</math>.
<math>[x]</math> means "greatest integer in" <math>x</math>.

Then

\begin{align}
I_1 \left( a, b, c \right) & = \int d\tau_{r} r_{1}^{a} r_{2}^{b} r_{12}^{c} e^{ -\alpha r_{1} - \beta r_{2}} P_{1}\left( \cos \theta \right) \\
& = \frac{1}{2} \left( I_{0} \left( a + 1, b - 1, c \right) + I_{0} \left( a - 1, b + 1, c \right) - I_{0} \left( a - 1, b - 1, c + 2 \right) \right)
\end{align}

===The Radial Recursion Relation===

Recall that the full integral is

\begin{equation}
I(l_1m_1,l_2m_2;R)=2\pi\delta_{l_1,l_2}\delta_{m_1,m_2}I_{l_2}(R)
\end{equation}

where

\begin{equation}
I_{l_2}(R) = \int d\tau_r R(r_1,r_2,r_{12})P_{l_2}(\cos \theta_{12})
\end{equation}

To obtain the recursion relation, use

\begin{equation}
P_l(x) = \frac{[P^{\prime}_{l+1}(x)-P^{\prime}_{l-1}(x)]}{2l+1}
\end{equation}

with

\begin{equation}
P_{l+1}^\prime (x) = \frac{d}{dx}P_{l+1}(x)
\end{equation}

Here <math>x=\cos\theta_{12}</math> and

\begin{align}
\frac{d}{d \cos \theta_{12}} & = \frac{dr_{12}}{d \cos \theta_{12}} \frac{d}{dr_{12}} \\
& = -\frac{r_{1} r_{2}}{r_{12}} \frac{d}{dr_{12}}
\end{align}

Then,

\begin{equation}
I_{l}\left( R \right) = - \int d\tau_r R \frac{r_{1} r_{2}}{r_{12}} \frac{d}{dr_{12}} \frac{ \left[ P_{l+1} \left( \cos \theta_{12} \right) - P_{l-1} \left( \cos \theta_{12} \right) \right]}{2l+1}
\end{equation}

The <math>r_{12}</math> part of the integral is

\begin{align}
& \int^{r_{1} + r_{2}}_{|r_{1} - r_{2}|} r_{12} dr_{12} R \frac{r_{1} r_{2}}{r_{12}} \frac{d}{dr_{12}} \left[P_{l+1}-P_{l-1} \right] \\
& = R r_{1} r_{2} \left[ P_{l+1} - P_{l-1} \right]^{r_{1} + r_{2}}_{|r_{1} - r_{2}|} - \int^{r_{1} + r_{2}}_{|r_{1} - r_{2}|} r_{12} dr_{12} \left( \frac{d}{dr_{12}} R \right) \frac{r_{1} r_{2}}{r_{12}} \frac{ \left[ P_{l+1} - P_{l-1} \right]}{2l+1}
\end{align}

The integrated term vanishes because

\begin{equation}
\cos \theta_{12} = \frac{r_{1}^{2} + r_{2}^{2} - r_{12}^{2}}{2 r_{1} r_{2}}
\begin{array}{lll}
& = -1 & {\rm when} & r_{12}^{2} = \left( r_{1} + r_{2} \right)^{2} \\
& = 1 & {\rm when} & r_{12}^{2} = \left( r_{1} - r_{2} \right)^{2}
\end{array}
\end{equation}

and <math>P_l(1)=1</math>, <math>P_{l}(-1)=(-1)^{l}</math>.

Thus,

\begin{equation}
I_{l+1}\left(\frac{r_1r_2}{r_{12}}R^\prime\right) = \left( 2l+1 \right) I_{l}(R) - I_{l-1} \left( \frac{r_{1} r_{2}}{r_{12}} R^{\prime} \right)
\end{equation}

If

\begin{equation}
R = r_{1}^{a-1} r_{2}^{b-1} r_{12}^{c+2} e^{-\alpha r_{1} - \beta r_{2}}
\end{equation}

then [G.W.F. Drake, Phys. Rev. A 18, 820 (1978)]

\begin{equation}
I_{l+1}(r_1^ar_2^br_{12}^c) =
\frac{2l+1}{c+2} I_{l} \left( r_{1}^{a-1} r_{2}^{b-2} r_{12}^{c+2} \right)+I_{l-1} \left( r_{1}^{a} r_{2}^{b} r_{12}^{c} \right)
\end{equation}

For the case <math>c=-2</math>, take

\begin{equation}
R = r_{1}^{a-1} r_{2}^{b-1} \ln r_{12} e^{ -\alpha r_{1} -\beta r_{2}}
\end{equation}

Then

\begin{equation}
I_{l+1} \left( r_{1}^{a} r_{2}^{b} r_{12}^{c} \right) = I_{l} \left( r_{1}^{a-1} r_{2}^{b-1} \ln r_{12} \right) \left( 2l + 1 \right) - I_{l-1} \left( r_{1}^{a} r_{2}^{b} r_{12}^{-2} \right)
\end{equation}

This allows all <math>I_{l}</math> integrals to be calculated from tables of <math>I_0</math> and
<math>I_1</math> integrals.

===The General Integral===

The above results for the angular and radial integrals can now be combined into a general formula for integrals of the type

\begin{equation}
I = \int \int d{\bf r}_{1} d{\bf r}_{2} R_{1} \mathcal{Y}^{M^{\prime}}_{l^{\prime}_{1} l^{\prime}_{2} L^{\prime}} \left( \hat{r}_{1}, \hat{r}_{2} \right) T_{k_{1} k_{2} K}^{Q} \left( {\bf r}_{1},{\bf r}_{2} \right) R_{2} \mathcal{Y}_{l_{1} l_{2} L}^{M} \left( \hat{r}_{1}, \hat{r}_{2} \right)
\end{equation}

where

\begin{equation}
\mathcal{Y}^{M}_{l_{1} l_{2} L} \left( \hat{r}_{1}, \hat{r}_{2} \right) = \sum_{m_1,m_2} \langle l_{1} l_{2} m_{1} m_{2} | LM \rangle Y^{m_{1}}_{l_{1}} \left( \hat{r}_{1} \right) Y^{m_{2}}_{l_{2}} \left( \hat{r}_{2} \right)
\end{equation}

and

\begin{equation}
T^Q_{k_{1} k_{2} K} \left( {\bf r}_{1}, {\bf r}_{2} \right) = \sum_{q_{1}, q_{2}} \langle k_{1} k_{2} q_{1} q_{2} | KQ \rangle Y_{k_{1}}^{q_{1}} \left( \hat{r}_{1} \right) Y^{q_{2}}_{k_{2}} \left( \hat{r}_{2} \right)
\end{equation}

The basic idea is to make repeated use of the formula

\begin{equation}
Y^{m_{1}}_{l_{1}} \left( \hat{r}_{1} \right) Y^{m_{2}}_{l_{2}} \left( \hat{r}_{1} \right) = \sum_{l m} \left( \frac{\left( 2l_{1}+1 \right) \left(2l_{2}+1 \right) \left(2l+1 \right)}{4 \pi} \right)^{1/2}
\times \left( \begin{array}{ccc}
l_1 & l_2 & l \\
m_1 & m_2 & m
\end{array} \right) \left(\begin{array}{ccc}
l_1 & l_2 & l \\
0 & 0 & 0
\end{array} \right)
Y^{m*}_{l} \left( \hat{r}_{1} \right)
\end{equation}

where

\begin{equation}
Y^{m*}(\hat{r}) = (-1)^mY_l^{-m}(\hat{r})
\end{equation}

and

\begin{equation}
\left(\begin{array}{ccc}
l_1 & l_2 & l \\
m_1 & m_2 & m
\end{array} \right) = \frac{\left( -1 \right)^{l_{1} - l_{2} - m}}{\left( 2l + 1 \right)^{1/2}} \left( l_{1} l_{2} m_{1} m_{2} | l, -m \right)
\end{equation}

is a 3-<math>j</math> symbol. In particular, write

\begin{array}{cc}
Y^{m_1^\prime}_{l_1^\prime}(\hat{r}_1)^{*} \underbrace{Y_{k_{1}}^{q_{1}} \left( \hat{r}_{1} \right) Y^{m_{1}}_{l_{1}} \left( \hat{r}_{1} \right)} & = & \sum_{\Lambda M}(\ldots)Y^{M}_\Lambda(\hat{r}_1)\\
\ \ \ \ \ \ \ \sum_{\lambda_1\mu_1}Y_{\lambda_1}^{\mu_1}(\hat{r}_1) & & \\
& & \\
Y^{m_{2}^{\prime}}_{l_{2}^{\prime}} \left( \hat{r}_{2} \right) \underbrace{Y^{q_{2}}_{k_{2}} \left( \hat{r}_{2} \right) Y_{l_{2}}^{m_{2}} \left( \hat{r}_{2} \right)} & = & \sum_{\Lambda^{\prime} M^{\prime}} \left( \ldots \right)Y_{\Lambda^{\prime}}^{M^{\prime}*} \left( \hat{r}_{2} \right) \\
\ \ \ \ \ \ \ \ \ \sum_{\lambda_{2} \mu_{2}}Y_{\lambda_{2}}^{\mu_{2}} \left( \hat{r}_{2} \right) & &
\end{array}

The angular integral then gives a factor of <math>2 \pi \delta_{\Lambda, \Lambda^{\prime}} \delta_{M, M^{\prime}} P_{\Lambda} \left( \cos \theta_{12} \right)</math>. The total integral therefore reduces to the form

\begin{equation}
I = \sum_{\Lambda} C_{\Lambda} I_{\Lambda} \left( R_{1} R_{2} \right)
\end{equation}

where <math>C_{\Lambda} = \sum_{\lambda_{1}, \lambda_{2}} C_{ \lambda_{1}, \lambda_{2}, \Lambda}</math>. For further details and derivations, including graphical representations, see G.W.F. Drake, Phys. Rev. A <math>{\bf 18}</math>, 820 (1978).

==Graphical Representation==

[[File:GraphRep1.jpg|500px]]

==Matrix Elements of H==
Recall that

\begin{equation}
H = -\frac{1}{2} \nabla^{2}_{1} - \frac{1}{2} \nabla^{2}_{2} - \frac{1}{r_1} - \frac{1}{r_2} + \frac{Z^{-1}}{r_{12}}
\end{equation}

Consider matrix elements of

\begin{equation}
\nabla^2_1 = \frac{1}{r_1^2} \frac{\partial}{\partial r_1}\left(r_1^2\frac{\partial}{\partial r_1} \right) + \frac{1}{r^2}\frac{\partial}{\partial r}\left(r^2\frac{\partial}{\partial r}\right) - \frac{(\vec{l}_1^Y)^2}{r_1^2} + \frac{2 \left( r_1 - r_2 \cos \theta \right)}{r} \frac{\partial^2}{\partial r_1 \partial r} - 2 \left( \nabla^Y_1 \cdot {\bf r}_2 \right) \frac{1}{r} \frac{\partial}{\partial r}
\end{equation}

where <math>r \equiv r_{12}</math> and <math>\cos\theta \equiv \cos\theta_{12}</math>. Also in what follows, define <math>\hat{\nabla}^Y_1 \equiv r_1\nabla^Y_1</math> where <math>\nabla^Y</math> operates only on the spherical harmonic part of the wave function, and similarly for <math>\vec{l}_1^Y</math>.

A general matrix element is

\begin{equation}
\langle r_1^{a^\prime} r_2^{b^\prime} r_{12}^{c^\prime} e^{-\alpha^\prime r_1 - \beta^\prime r_2} \mathcal{Y}^{M^\prime}_{l^\prime_1 l^\prime_2 L^\prime} \left( \hat{r}_1, \hat{r}_2 \right) \left| \nabla^2_1 \right| r_1^a r_2^b r_{12}^c e^{-\alpha r_1 - \beta r_2} \mathcal{Y}^M_{l_1 l_2 L} \left( \hat{r}_1, \hat{r}_2 \right) \rangle
\end{equation}

Since <math>\nabla^2_1</math> is rotationally invariant, this vanishes unless <math>L=L^\prime</math>, <math>M=M^\prime</math>. Also <math>\nabla_1^2</math> is Hermitian, so that the result must be the same whether it operates to the right or left, even though the results look very different. In fact, requiring the results to be the same yields some interesting and useful integral identities as follows:

Put

\begin{equation}
\mathcal{F} = F \mathcal{Y}^M_{l_1 l_2 L} \left( \hat{r}_1, \hat{r_2} \right)
\end{equation}

and

\begin{equation}
F = r_1^a r_2^b r_{12}^c e^{-\alpha r_1 - \beta r_2}
\end{equation}

Then

\begin{equation}
\nabla^2_1 \mathcal{F} = \left\{ \frac{1}{r_1^2} \left[ a \left( a + 1 \right) - l_1 \left( l_1 + 1 \right) \right] + \frac{c \left( c + 1 \right)}{r^2} + \alpha^2\frac{2\alpha \left( a + 1 \right)}{r_1} + \frac{2 \left(r_1 - r_2 \cos \theta \right) }{r_1r^2}c \left[ a - \alpha r_1 \right] - \frac{2c}{r^2} \left( \hat{\nabla}^Y_1 \cdot \hat{r}_2 \right) \frac{r_2}{r_1} \right\} \mathcal{F}
\end{equation}

and

\begin{align}
\langle \mathcal{F}^\prime \left| \nabla^2_1 \right| \mathcal{F} \rangle & = \sum_\Lambda \int d\tau_r F^\prime C_\Lambda \left( 1 \right) P_\Lambda \left( \cos \theta \right) \times \left\{ \frac{1}{r_1^2} \left[ a \left( a + 1 \right) - l_1 \left( l_1 + 1 \right) \right] -\frac{2 \alpha \left( a + 1 \right)}{r_1} + \frac{c \left( c + 1 \right)}{r^2} +\alpha^2 + \frac{2 \left( r_2 - r_2 \cos \theta \right)}{r_1r^2}c \left( a - \alpha r_1 \right) \right\} F \\
& + \sum_\Lambda \int d\tau_r F^\prime C_\Lambda \left( \hat{\nabla}^Y_1 \cdot \hat{r}_2 \right) P_\Lambda \left( \cos \theta \right) \left( \frac{-2c r_2}{r_1 r^2} \right) F
\end{align}

where

\begin{equation}
\int d\tau_r = \int^\infty_0 r_1 dr_1 \int^\infty_0 r_2 dr_2 \int^{r_1+r_2}_{\left| r_1 - r_2 \right|} r dr
\end{equation}

For brevity, let the sum over <math>\Lambda</math> and the radial integrations be understood, and let <math>\left( \nabla_1^2 \right)</math> stand for the terms which appear in the integrand. Then operating to the right gives

\begin{equation}
(\nabla_1^2)_R = \frac{1}{r_1^2} \left[ a \left( a + 1 \right) - l_1 \left( l_1 + 1 \right) \right] + \frac{c \left( c + 1 \right)}{r^2} + \alpha^2 - \frac{2 \alpha \left( a + 1 \right)}{r_1}
+ \frac{2 \left( r_1 - r_2 \cos \theta \right)}{r_1 r^2}c \left( a - \alpha r_1 \right) - \frac{2c r_2}{r_1 r^2} \left( \hat{\nabla}^Y_1 \cdot \hat{r}_2 \right)
\end{equation}

Operating to the left gives

\begin{equation}
(\nabla_1^2)_L = \frac{1}{r_1^2} \left[a^\prime \left( a^\prime + 1 \right) - l_1^\prime \left( l_1^\prime + 1 \right) \right] + \frac{c^\prime \left( c^\prime + 1 \right)}{r^2} + \alpha^{\prime 2} -\frac{2 \alpha^\prime \left( a^\prime + 1 \right)}{r_1}
+ \frac{2 \left( r_1 - r_2 \cos \theta \right)}{r_1 r^2} c^\prime \left( a^\prime - \alpha^\prime r_1 \right) - \frac{2 c^\prime r_2}{r_1 r^2} \left( \hat{\nabla}^{Y^\prime}_1 \cdot \hat{r}_2 \right)
\end{equation}

Now put

\begin{array}{cc}
a_+ = a+a^\prime, & \hat{\nabla}_1^+ = \hat{\nabla}_1^Y + \hat{\nabla}_1^{Y\prime} \\
a_{-} = a-a^\prime, & \hat{\nabla}_1^{-} = \hat{\nabla}_1^Y - \hat{\nabla}_1^{Y\prime}
\end{array}

etc., and substitute <math>a^\prime = a_+ - a</math>, <math>c^\prime = c_+ - c</math>, and <math>\alpha^\prime = \alpha_+ - \alpha</math> in <math>(\nabla^2_1)_L</math>. If <math>a_+</math>, <math>c_+</math> and <math>\alpha_+</math> are held fixed, then the equation

\begin{equation}
(\nabla^2_1)_R = (\nabla^2_1)_L
\end{equation}

must be true for arbitrary <math>a</math>, <math>c</math>, and <math>\alpha</math>. Their coefficients must thus vanish.

This yields the integral relations

\begin{equation}
\label{I}
\frac{\left( r_1 - r_2 \cos \theta \right)}{r_1 r^2} = \frac{1}{c_+} \left( \frac{- \left( a_+ + 1 \right)}{r_1^2} + \frac{\alpha_+}{r_1} \right) \tag{I}
\end{equation}

from the coefficient of <math>a</math> and

\begin{equation}
\label{II}
\frac{ \left( r_1 - r_2 \cos \theta \right) \left( a_+ - \alpha_+ r_1 \right)}{r_1 r^2} = \frac{r_2}{r_1 r^2}\left( \hat{r}_2 \cdot \hat{\nabla}_1^+ \right) - \frac{ \left( c_+ + 1 \right)}{r^2} \tag{II}
\end{equation}

from the coefficient of <math>c</math>. The coefficient of <math>\alpha</math> gives an equation equivalent to (I).

Furthermore, if can be show that (see problem)

\begin{align}
& \sum_\Lambda \int d\tau_r \frac{r^c}{r^2} C_\Lambda \left( \hat{r}_2 \cdot \hat{\nabla}_1^Y \right) P_\Lambda \left( \cos \theta \right) \\
= & \sum_\Lambda \int d\tau_r \frac{r^c}{c r_1 r_2}C_\Lambda \left( 1 \right)P_\Lambda \left( \cos \theta \right) \times \left( \frac{l_1^\prime \left( l_1^\prime + 1 \right) - l_1 \left( l_1 + 1 \right) -\Lambda \left( \Lambda + 1 \right)}{2} \right)
\end{align}

and similarly for <math>\left( \hat{r}_2 \cdot \hat{\nabla}^{Y\prime}_1 \right)</math> with <math>l_1</math> and <math>l_1^\prime</math> interchangeable, then it follows that

\begin{equation}
\frac{r^{c_+}}{r^2} \left( \hat{r}_2 \cdot \hat{\nabla}_1^+ \right) = - \frac{r^{c_+}}{c_+ r_1 r_2} \Lambda \left( \Lambda + 1 \right)
\end{equation}

and

\begin{equation}
\frac{r^{c_+}}{r^2} \left( \hat{r}_2 \cdot \hat{\nabla}_1^- \right) = \frac{r^{c_+}}{c_+ r_1 r_2} \left[ l_1^\prime \left( l_1^\prime + 1 \right) - l_1 \left( l_1 + 1 \right) \right]
\end{equation}

where equality applies after integration and summation over <math>\Lambda</math>.

Thus (II) becomes

\begin{equation}\label{IIa}
\frac{ \left( r_1 - r_2 \cos \theta \right) \left( a_+ - \alpha_+ r_1 \right)}{r_1 r^2} = - \frac{\Lambda \left( \Lambda + 1 \right)}{c_+ r^2_1} - \frac{ \left( c_+ + 1 \right)}{r^2} \tag{IIa}
\end{equation}

===Problem===

Prove the integral relation

\begin{align}
& \sum_\Lambda \int d\tau_r f \left( r_1, r_2 \right) \frac{1}{r} \left( \frac{d}{dr} q(r) \right) C_\Lambda \left( \hat{r}_2 \cdot \hat{\nabla}^Y_1 \right) P_\Lambda \left( \cos \theta \right) \\
= & \sum_\Lambda \int d\tau_r \frac{ f \left( r_1, r_2 \right)}{r_1, r_2} q(r) C_\Lambda \left( 1 \right) P_\Lambda \left( \cos \theta \right) \left(\frac{l_1^\prime \left( l_1^\prime + 1 \right) - l_1 \left( l_1 + 1 \right) - \Lambda \left( \Lambda + 1 \right)}{2}\right)
\end{align}

where the coefficients <math>C_\Lambda \left( 1 \right)</math> are the angular coefficients from the overlap integral

\begin{align}
\int & d\Omega\, \mathcal{Y}^{M*}_{l_1^\prime l_2^\prime L}(\hat{r}_1,\hat{r}_2)\, \mathcal{Y}^{M}_{l_1 l_2 L}(\hat{r}_1,\hat{r}_2)\\
= & \sum_\Lambda C_\Lambda(1)P_\Lambda(\cos\theta_{12}).
\end{align}

Hint: Use the fact that <math>l_1^2</math> is Hermitian so that
<math>
\int d\tau (l_1^2\mathcal{Y}^\prime)^*q(r)\mathcal{Y} = \int d\tau
\mathcal{Y}^{\prime *}l_1^2(q(r)\mathcal{Y})
</math>
with <math>\ \vec{l}_1= \frac{1}{i}\vec{r}_1\times\nabla_1</math>.

It is also useful to use
<math>\begin{eqnarray}\ \
\cos\theta\, P_L(\cos\theta) &=& \frac{1}{2L+1}[L\,P_{L-1}(\cos\theta) + (L+1)P_{L+1}(\cos\theta)]
\\
(\cos^2\theta-1)P_L(\cos\theta) &=&
\frac{L(L-1)}{(2L-1)(2L+1)}P_{L-2}(\cos\theta)\nonumber
-\frac{2(L^2+L-1)}{(2L-1)(2L+3)}P_L(\cos\theta) + \frac{(L+1)(L+2)}{(2L+1)(2L+3)}P_{L+2}(\cos\theta)\nonumber\\
&=& \frac{L(L-1)}{(2L-1)(2L+1)}[P_{L-2}(\cos\theta) - P_L(\cos\theta)]
+ \frac{(L+1)(L+2)}{(2L+1)(2L+3)}[P_{L+2}(\cos\theta) - P_L(\cos\theta)]
\end{eqnarray}</math>

together with a double application of the integral recursion relation
<math>
I_{L+1}\left(\frac{1}{r}\frac{d}{dr}q(r)\right) =
(2L+1)I_L\left(\frac{1}{r_1r_2}q(r)\right)+I_{L-1}\left(\frac{1}{r}\frac{d}{dr}q(r)\right)
</math>

Of course <math>l_1^2\mathcal{Y}^M_{l_1l_2L}(\hat{r}_1,\hat{r}_2)=l_1(l_1+1)\mathcal{Y}^M_{l_1l_2L}(\hat{r}_1,\hat{r}_2)</math>. Begin by expanding
<math>
l_1^2(g\mathcal{Y}) = \mathcal{Y}l_1^2g + g\,l_1^2\mathcal{Y} + 2(\vec{l}_1g)\cdot(\vec{l}_1\mathcal{Y})
</math>, and show that

<math>
\begin{eqnarray}\ \
(\vec{l}_1g)\cdot(\vec{l}_1\mathcal{Y}) & = & \frac{r_1^2}{r}\frac{dg}{dr}\vec{r}_2\cdot \nabla_1\mathcal{Y}\,\,\,\,\mbox{ and}\\
l_1^2g(r) &=& 2r_1r_2\cos\theta\frac{1}{r}\frac{dg}{dr} + r_1^2r_2^2(\cos^2\theta - 1)
\frac{1}{r}\frac{d}{dr}\left(\frac{1}{r}\frac{dg}{dr}\right).
\end{eqnarray}
</math>

The proof amounts to showing that the term <math>l_1^2g(r)</math> can be replaced by
<math>\Lambda(\Lambda+1)g(r)</math> after multiplying by <math>P_\Lambda(\cos\theta)</math>
and integrating by parts with respect to the radial integrations over <math>r_1</math>, <math>r_2</math> and
<math>r \equiv |\vec{r}_1 - \vec{r}_2|</math>. Remember that <math>\cos\theta</math> is just a short-hand notation for the radial function <math>(r_1^2 + r_2^2 - r^2)/(2r_1r_2)</math>.

==General Hermitian Property==

Each combination of terms of the form
\begin{equation}
<f>_R = a^2f_1+af_2+abf_3+b^2f_4+bf_5+af_6\nabla_1^Y+bf_7\nabla^Y_2+f_8(Y)
\end{equation}
can be rewritten
\begin{equation}
<f>_L= (a_+-a)^2f_1+(a_+-a)f_2 + (a_+-a)(b_+-b)f_3+(b_+-b)^2f_4
+(b_+-b)f_5+(a_+-a)f_6\nabla^{Y\prime}_1+(b_+-b)f_7\nabla^{Y\prime}_2+f_8)Y^\prime
\end{equation}
where as usual <math>\nabla_1^Y</math> acts only on the spherical harmonic part of the wave function denoted for short by <math>Y</math>.
Since these must be equal for arbitrary <math>a</math> and <math>b</math>,
\begin{equation}
a_+^2f_1+a_+f_2+_+b_+f_3+b_+f_4+b_+f_5+a_+f_6\nabla^{Y\prime}_1 +
b_+f_7\nabla^{Y\prime}_2+f_8(Y^\prime)-f_8(Y)=0
\end{equation}

Adding the corresponding expression with <math>Y</math> and <math>Y^\prime</math> interchanged
yields
<math>\label{III}
a_+^2f_1+a_+f_2+a_+b_+f_3+b_+^2f_4+b_+f_5+\frac{1}{2}a_+f_6\nabla^+_1+\frac{1}{2}b_+f_7\nabla^+_2=0\tag{III}
</math>

Subtracting gives
<math>\label{IV}
f_8(Y)-f_8(Y^\prime)=-\frac{1}{2}[a_+f_6\nabla^-_1+b_+f_7\nabla_2^-]\tag{IV}
</math>\\
<math>\label{V}
a[-2a_+f_1-2f_2-b_+f_3-f_6\nabla^+_1]=0\tag{V}
</math>\\
<math>\label{VI}
b[-2b_+f_4-2f_5-a_+f_3-f_7\nabla^+_2]=0\tag{VI}
</math>

Adding the two forms gives

\begin{eqnarray}
<f>_R+<f>_L=\frac{1}{2}(a_+^2+a_-^2)f_1+a_+f_2+\frac{1}{2}(a_+b_++a_-b_-)f_3\nonumber\\
\ +\frac{1}{2}(b_+^2+b_-^2)f_4+b_+f_5+\frac{1}{2}f_6(a_+\nabla_1^++a_-\nabla^-_1)\nonumber\\
\ +\frac{1}{2}(b_+\nabla_2^++b_-\nabla^-_2)+f_8(Y)+f_8(Y^\prime)\nonumber
\end{eqnarray}

Subtracting <math>\frac{x}{2}\times({\rm I})</math> gives

\begin{eqnarray}\ \
<f>_R+<f>_L = \frac{1}{2}[(1-x)a_+^2 + a_-^2]f_1 +
(1-\frac{x}{2})a_+f_2+\frac{1}{2}[(1-x)a_+b_++a_-b_-]f_3\nonumber\\
+\frac{1}{2}[(1-x)b_+^2+b_-^2]f_4+(1-\frac{x}{2})f_5b_++\frac{1}{2}f_6[(1-\frac{x}{2})a_+\nabla_1^++a_-\nabla^-_1]\nonumber\\
+\frac{1}{2}[(1-\frac{x}{2})b_+\nabla^+_2+b_-\nabla^-_2]+f_8(Y)+f_8(Y^\prime)\nonumber
\end{eqnarray}

If <math>x=1</math>,
\begin{eqnarray}
<f>_R+<f>_L &=&
\frac{1}{2}[a_-^2f_1+a_+f_2+a_-b_-f_3+b_-^2f_4+b_+f_5\nonumber\\
&+&\left(\frac{1}{2}a_+\nabla^+_1+a_-\nabla^-_1\right)f_6+\left(\frac{1}{2}b_+\nabla^+_2+b_-\nabla^-_2\right)f_7]+f_8(Y)+f_8(Y^\prime)\nonumber
\end{eqnarray}

The General Hermitian Property for arbitrary <math>x</math> gives
\begin{eqnarray}\ \
\nabla_1^2=
\frac{1}{4}\{\frac{1}{r_1^2}[(1-x)a_+^2+a_-^2+2(1-\frac{x}{2})a_+-2[l_1(l_1+1)+l_1^\prime(l_1^\prime+1)]\nonumber\\
-\frac{2}{r_1}[(1-x)\alpha_+a_++\alpha_-a_-+2(1-\frac{x}{2})\alpha_+]+(1-x)\alpha_+^2
+ \alpha^2_-]\nonumber\\
+\frac{2(r_1-r_2\cos\theta)}{r_1r^2}[(1-x)(a_+-\alpha_+r_1)c_++(a_--\alpha_-r_2)c_-]\nonumber\\
-\frac{2r_2}{r_1r^2}[(1-\frac{x}{2})c_+\hat{r}_2\cdot\hat{\nabla}_1^++c_-\hat{r}_2\hat{\nabla}_1^-]+\frac{(1-x)c_+^2+c_-^2+2(1-\frac{x}{2})c_+}{r^2}\}\nonumber
\end{eqnarray}.

Use
<math>\begin{eqnarray}\ \
\frac{-2r_2}{r_1r^2}[(1-\frac{x}{2})c_+\hat{r}_2\hat{\nabla}_1^++c_-\hat{r}_2\hat{\nabla}_1^-]=\frac{2}{r_1^2}\left[(1-\frac{x}{2})\Lambda(\Lambda+1)-\frac{c_-}{c_+}[l_1^\prime(l_1^\prime+1)+l_1(l_1+1)]\right]\end{eqnarray}</math>,

<math>\begin{eqnarray}
\frac{2(r_1-r_2\cos\theta)}{r_1r^2}(a_--\alpha_-r_1)c_- = 2\frac{c_-}{c_+}\left[\frac{-1}{r_1^2}[a_-(a_++1)]+\frac{1}{r_1}[a_-\alpha_++\alpha_-(a_++2)]-\alpha_-\alpha_+\right]\nonumber
\end{eqnarray}</math>,

and
<math>\begin{eqnarray}\ \
\frac{2(r_1-r_2\cos\theta)}{r_1r^2}(a_+-\alpha_+r_1)c_+ = 2\left[-\frac{1}{r_1^2}[a_+(a_++1)]+\frac{1}{r_1}[a_+\alpha_++\alpha_+(a_++2)]-\alpha_+^2\right]
\end{eqnarray}</math>.

Substituting into <math>\nabla_1^2</math> gives

\begin{eqnarray}
\nabla^2_1 = \frac{1}{4}\{\frac{1}{r_1^2}[-(1-x)a_+^2 + a_-^2 +xa_+ +
2(1-\frac{x}{2})\Lambda(\Lambda+1)\nonumber\\
-2l_1(l_1+1)(1-\frac{c_-}{c_+}) -
2l_1^\prime(l_1^\prime+1)(1+\frac{c_-}{c_+})-\frac{2c_-a_-}{c_+}(a_++1)]\nonumber\\
- \frac{2}{r_1}\left[-(1-x)\alpha_+(a_++2)+\alpha_-a_-+2(1-\frac{x}{2})\alpha_+
-\frac{c_-}{c_+}[a_-\alpha_++\alpha_-(a_++2)]\right]\nonumber\\
-(1-x)\alpha_+^2+\alpha_-^2-2\frac{c_-}{c_+}\alpha_-\alpha_+ +\frac{1}{r^2}\left[(1-x)c_+^2+c_-^2+2(1-\frac{x}{2})c_+\right]\}\nonumber
\end{eqnarray}

This has the form
<math>
\nabla_1^2 =
\frac{1}{4}\left[\frac{A_1}{r_1^2}+\frac{B_1}{r_1}+\frac{C_1}{r^2}+D_1\right]
</math>
with

<math>\begin{array}{ccc}
A_1=-(1-x)a_+^2+a_-^2
+xa_++2(1-\frac{x}{2})\Lambda(\Lambda+1)-2l_1(l_1+1)(1-\frac{c_-}{c_+})\
-2l_1^\prime(l_1^\prime+1)(1+\frac{c_-}{c_+})-\frac{2c_-a_-}{c_+}(a_++1) \ \
\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \
\end{array}
</math>

<math>
B_1 =
2\left[(1-x)\alpha_+(a_++2)-\alpha_-a_--2(1-\frac{x}{2})\alpha_++\frac{c_-}{c_+}[a_-\alpha_++\alpha_-(a_++2)]\right]
</math>

<math>
C_1 = (1-x)c_+^2+ c_-^2 +2(1-\frac{x}{2})c_+
</math>

<math>
D_1 = -1(1-x)\alpha_+^2 +\alpha_-^2 - 2\frac{c_-}{c_+}\alpha_-\alpha_+
</math>

The complete Hamiltonian is then (in Z-scaled a.u.)

<math>\begin{eqnarray}
H =-\frac{1}{2}\nabla^2_1 -\frac{1}{2}\nabla^2_2 -\frac{1}{r_1} -\frac{1}{r_2} +\frac{Z^{-1}}{r}\nonumber\\
= -\frac{1}{2}\nabla^2_1 -\frac{1}{2}\nabla^2_2 -\frac{1}{r_1} -\frac{Z-1}{Zr_2}+Z^{-1}\left(\frac{1}{r}-\frac{1}{r_2}\right)\nonumber\\
\ =-\frac{1}{8}\left[\frac{A_1}{r_1^2}+\frac{B_1+8}{r_1}+\frac{C_1}{r^2}+D_1+D_2 +\frac{A_2}{r_2^2} + \frac{B_2 + 8(Z-1)/Z}{r_2}+\frac{C_2}{r^2}\right]\nonumber
+ Z^{-1}\left(\frac{1}{r}-\frac{1}{r_2}\right)\nonumber
\end{eqnarray}</math>

The screened hydrogenic energy is
<math>\begin{eqnarray}
E_{SH}=-\frac{1}{2}\left[\frac{1}{n_1^2}+ \left(\frac{Z-1}{Z}\right)^2\frac{1}{n_2^2}\right]
\end{eqnarray}</math>
so that

<math>\begin{eqnarray}
H-E_{SH} &=& -\frac{1}{8}[
\frac{A_1}{r_1^2}+\frac{B_1+8}{r_1}+\frac{C_1}{r^2} + D_1 + D_2
-\frac{4}{n_1^2} -\left(\frac{Z-1}{Z}\right)^2\frac{4}{n_2^2}\nonumber
+\frac{A_2}{r_2}+\frac{B_2+8(Z-1)/Z}{r_2}+\frac{C_2}{r^2}] +
Z^{-1}\left(\frac{1}{r} -\frac{1}{r_2}\right)\nonumber
\end{eqnarray}</math>

==Optimization of Nonlinear Parameters==
- The traditional method of performing Hylleraas calculations is to write the
basis set in the form

<math>
\Psi = \sum_{i,j,k}c_{i,j,k}\varphi_{i,j,k}(\alpha,\beta)\pm exchange
</math>

with

<math>
\varphi_{i,j,k}(\alpha,\beta) = r_1^ir_2^jr_{12}^ke^{-\alpha r_1 - \beta
r_2}\mathcal{Y}^M_{l_1l_2L}.
</math>

The usual procedure is to set <math>\alpha = Z</math> so that it represents the inner
<math>1s</math> electron, and then to vary <math>\beta</math> so as to minimize the energy.

[[File:OptNon.jpg|500px]]

Since <math>\beta</math> appears in <math>\Psi</math> as a non-linear parameter, the entire
calculation must be repeated for each value of <math>\beta</math>. However, the minimum
becomes progressively smaller as the basis set is enlarged.

[[File:OptNon02.jpg|500px]]

*Difficulties

1. If the basis set is constructed so that <math> i +j +j \leq N</math>, the the
number of terms is
<math>
(N+1)(N+2)(N+3)/6
</math>
N=14 already gives 680 terms and an accuracy of about 1 part in <math>10^{10}</math> for
low-lying states. A substantial improvement in accuracy would require much
larger basis sets, together with multiple precision arithmetic to avoid loss
of significant figures when high powers are included.

2. The accuracy rapidly deteriorates as one goes to more highly excited
states - about 1 significant figure is lost each time the principle quantum
number is increased.

*Cure

We have found that writing basis sets in the form
\begin{eqnarray}
\Psi &=& \sum_{i,j,k}\left(c^{(1)}_{i,j,k}\varphi_{ijk}(\alpha_1\beta_1) +
c_{ijk}^{(2)}\varphi_{ijk}(\alpha_2\beta_2)\right) \pm exchange \nonumber\\
&=& \psi({\bf r}_1,{\bf r}_2) \pm \psi({\bf r}_2,{\bf r}_1)\nonumber
\end{eqnarray}
so that each combination of powers is included twice with different nonlinear
parameters gives a dramatic improvement in accuracy for basis sets of about
the same total size.

However, the optimization of the nonlinear parameters is now much more
difficult, and an automated procedure is needed.

If
<math>
E = \frac{<\Psi|H|\Psi>}{<\Psi|\Psi>}</math> and we assume <math><\Psi|\Psi> = 1</math>, then

<math>
\frac{\partial E}{\partial\alpha_t} = -2 <\Psi|H-E|r_1\psi({\bf r}_1,{\bf
r}_2;\alpha_t)\pm r_21\psi({\bf r}_2,{\bf
r}_1;\alpha_t)>
</math>
where
<math>
\psi({\bf r}_1,{\bf r}_2;\alpha_t) =
\sum_{i,j,k}c_{ijk}^{(t)}\varphi_{ijk}(\alpha_t\beta_t)
</math>

*Problem

1. Prove the above.

2. Prove that there is no contribution from the implicit dependence of
<math>c_{ijk}^{(t)}</math> on <math>\alpha_t</math> if the linear parameters have been optimized.

==The Screened Hydrogenic Term==

If the Hamiltonian is written in the form (in Z-scaled a.u.)
<math>\begin{eqnarray}\
H = H_0({\bf r}_1,Z) + H_0({\bf r}_2,Z-1) +
Z^{-1}\left(\frac{1}{r}-\frac{1}{r_2}\right)
\end{eqnarray}</math>
with
<math>\
H_0({\bf r}_1,Z) = -\frac{1}{2}\nabla_1^2 - \frac{1}{r_1}\nonumber
</math> and
<math>
H_0({\bf r}_2,Z-1) = -\frac{1}{2}\nabla_2^2 - \frac{Z-1}Z{r_2}\nonumber
</math>,
then the eigenvectors of <math>H_0({\bf r},Z) +H_0({\bf r}_2,Z-1)</math> are products of
hydrogenic orbitals
<math>
\Psi_0=\psi_0(1s,Z)\psi_0(nl,Z-1)
</math>
and the eigenvalue is
<math>\begin{eqnarray}
E_{SH} = \left[-\frac{1}{2} - \left(\frac{Z-1}{Z}\right)^2\frac{1}{2n^2}\right]Z^2 a.u.
\end{eqnarray}</math>
called the screened hydrogenic eigenvalue. From highly excited states,
<math>E_{SH}</math> and <math>\Psi_0</math>are already excellent approximations.

For example, for the 1s8d states, the energies are

<math>\begin{eqnarray}\\
E(1s8d^1D)&=& -2.007\,816\,512\,563\,81 {\rm\ a.u.}\nonumber\\
E(1s8d^3D)&=& - 2.007\,817\,934\,711\,71 {\rm\ a.u.}\nonumber\\
E_{SH} &=& -2.007\,812\,5 {\rm\ a.u.} \nonumber
\end{eqnarray}</math>

It is therefore advantageous to include the screened hydrogenic terms in the
basis set so that the complete trial function becomes
<math>
\Psi = c_0\Psi_0 +
\sum_{ijk}\left[c_{ijk}^{(1)}\varphi_{ijk}(\alpha_1,\beta_1)+c_{ijk}^{(2)}\varphi(\alpha_2,\beta_2)\right]\pm
{\rm exchange}
</math>

Also, the variational principal can be re-expressed in the form
<math>\begin{eqnarray}
E= E_{SH}+\frac{<\Psi|H-E_{SH}|\Psi>}{<\Psi|\Psi>}
\end{eqnarray}</math>
so that the <math>E_{SH}</math> term can be cancelled analytically from the matrix
elements.

For example
<math>
<\Psi_0|H-E_{SH}|\Psi_0> = 0
</math>

and
<math>
<\varphi_{ijk}|H-E_{SH}|\Psi_0> =
Z^{-1}<\varphi_{ijk}|\frac{1}{r}-\frac{1}{r_2}|\Psi_0>
</math>

Recall that
<math>\begin{eqnarray}\ \
I_0(a,b,c) = \frac{2}{c+2}\sum^{[c+1]/2}_{i=0}\left(\begin{array}{c}
c+2\\
2i+1
\end{array}\right) \left[f(p,q;\beta) + f(p^\prime,q^\prime;\alpha)\right]
\end{eqnarray}</math>
where

<math>\begin{eqnarray}
f(p,q;x) =
\frac{q!}{x^{q+1}(\alpha+\beta)^{p+1}}\sum^p_{j=0}\frac{(p+j)!}{j!}\left(\frac{x}{\alpha+\beta}\right)^j
\end{eqnarray}</math>

<math>
\begin{array}{cc}
p=a+2i+2 & p^\prime = b+2i+2\\
q= b+c-2i+2 & q^\prime =a+c-2i+2
\end{array}
</math>

If <math>\beta\ll\alpha</math>, the <math>f(p,q;\beta)\gg f(p^\prime,q^\prime;\alpha)</math> and the
<math>i=0</math> term is the dominant contribution to <math>f(p,q;\beta)</math>. However, since
this term depends only on the sum of powers <math>b+c</math> for <math>r_2</math>, it cancels exactly
from the matrix element of
<math>
\frac{1}{r_{12}} - \frac{1}{r_2}
</math>
and can therefore
be omitted in the calculations of integrands, thereby saving many significant
figures. This is especially valuable when <math>b+c</math> is large and <math>a</math> is small.

==Small Corrections==
*Mass Polarization
If the nuclear mass is not taken to be infinite, then the Hamiltonian is
<math>\begin{eqnarray}\ \
H = \frac{P_N^2}{2M} + \sum_{i=1}\frac{p_i^2}{2m} + V
\end{eqnarray}</math>

[[File:SmallCor.jpg|500px]]

where <math>M =M_A-nm</math> is the nuclear mass and <math>m</math> is the electron mass.

Change variables to the C of M
<math>\begin{eqnarray}
{\bf R} = \frac{1}{M+nm}\left[M{\bf r}_N + m({\bf r}_1 + {\bf r}_2 +
\ldots)\right]
\end{eqnarray}</math>

and relative variables

<math>
{\bf s}_i = {\bf r}_i - {\bf r}_N.
</math>

Then

<math>\begin{eqnarray}\ \ \ \ \
H = \frac{1}{2(M+nm)}{\bf p}_R^2+\frac{1}{2m}\sum_{i=1}^n{\bf p}_{s_i}^2
+\frac{1}{2M}\sum_{i,k}{\bf p}_{s_i}\cdot {\bf p}_{s_k} + V({\bf s}_1, {\bf
s}_2, \ldots)\nonumber
= \frac{1}{M+nm}{\bf p}_R^2+\frac{1}{2\mu}\sum_{i=1}^n{\bf
p}_{s_i}^2+\frac{1}{2M}\sum_{i\neq k}{\bf p}_{s_i}\cdot {\bf p}_{s_k}+V({\bf s}_1, {\bf
s}_2, \ldots)\nonumber
\end{eqnarray}</math>

where
<math>\begin{eqnarray}
\mu = \left(\frac{1}{M} + \frac{1}{m}\right)^{-1}
\end{eqnarray}</math>

If <math>n=2</math>, the Schrödinger equation is
<math>\begin{eqnarray}\ \ \
\left[ -\frac{\hbar^2}{2\mu}(\nabla^2_{s_1}+ \nabla^2_{s_2}) -
\frac{\hbar^2}{M}\nabla_{s_1}\nabla_{s_2}-\frac{Ze^2}{s_1}-\frac{Ze^2}{s_2}+\frac{e^2}{s_{12}}\right]\psi
= E\psi
\end{eqnarray}</math>.

Define the reduced mass Bohr radius
<math>\begin{eqnarray}
a_\mu = \frac{\hbar^2}{\mu e^2} = \frac{m}{\mu}a_0
\end{eqnarray}</math>
and

<math>\begin{eqnarray}
\rho_i = \frac{Zs_i}{a_\mu}\nonumber\\
\varepsilon = \frac{E}{Z^2(e^2/a_\mu)}\nonumber
\end{eqnarray}</math>

The Schrödinger equation then becomes
<math>\begin{eqnarray}\ \ \
\left[
-\frac{1}{2}(\nabla^2_{\rho_1}+\nabla^2_{\rho_2})-\frac{\mu}{M}\vec{\nabla}_{\rho_1}\cdot\vec{\nabla}_{\rho_2} - \frac{1}{\rho_1}-\frac{1}{\rho_2}+\frac{Z^{-1}}{\rho_{12}}\right]\psi = \varepsilon \psi
\end{eqnarray}</math>

The effects of finite mass are thus

1. A reduced mass correction to all energy levels of <math>(e^2/a_\mu)/(e^2/a_0)
= \mu/m</math>, i.e.
<math>\begin{eqnarray}
E_M=\frac{\mu}{m}E_\infty
\end{eqnarray}</math>

2. A specific mass shift given in first order perturbation theory by
<math>\begin{eqnarray}
\Delta\varepsilon =
-\frac{\mu}{M}<\psi|\nabla_{\rho_1}\cdot\nabla_{\rho_2}|\psi>
\end{eqnarray}</math>
or

<math>\begin{eqnarray}
\Delta\varepsilon =
-\frac{\mu}{M}<\psi|\nabla_{\rho_1}\cdot\nabla_{\rho_2}|\psi>\frac{e^2}{a_\mu}\nonumber\\
= -\frac{\mu^2}{mM}<\psi|\nabla_{\rho_1}\cdot\nabla_{\rho_2}|\psi>\frac{e^2}{a_0}\nonumber
\end{eqnarray}</math>

Since <math>\nabla_{\rho_1}\cdot\nabla_{\rho_2}</math> has the same angular properties as
<math>{\bf\rho}_1 \cdot {\bf\rho}_2</math>, the operator is like the product of two
dipole operators. For product type wave functions of the form
$
\psi = \psi(1s)\psi(nl) \pm \mbox{exchange},
$
the matrix element vanishes for all but P-states.

=Radiative Transitions=

==Transition Integrals and the Line Strength==

For an electric dipole transition <math>\gamma'\,L'\,S'\,J'\,M' \rightarrow \gamma\,L\,S\,J\,M</math>, the line strength is defined by
\begin{equation}
S = \sum_{M',\mu,M}|\langle\gamma'\,L'\,S'\,J'\,M'|r_\mu|\gamma\,L\,S\,J\,M\rangle|^2
\end{equation}
Using the Wigner-Eckart theorem, this is related to the reduced matrix element by
\begin{eqnarray}
S(J',J) = |\langle\gamma'\,L'\,S'\,J'||r||\gamma\,L\,S\,J\,M\rangle|^2
\sum_{M',\mu,M}\left|\left(\matrix{J' & 1 & J \cr
-M'& \mu &M \cr}\right)\right|^2
\end{eqnarray}
Using the sum rule for the 3-<math>j</math> symbols
\begin{equation}
\sum_{M',\mu,M}\left|\left(\matrix{J' & 1 & J \cr
-M'& \mu &M \cr}\right)\right|^2 = 1
\end{equation}
we see that the line strength is identical to the reduced matrix element
\begin{equation}
S(J',J) = |\langle\gamma'\,L'\,S'\,J'||r||\gamma\,L\,S\,J\,M\rangle|^2
\end{equation}
For a nonrelativistic calculation in <math>LS</math> coupling, <math>S = S'</math> and we can next strip out the dependence on <math>J</math> and <math>J'</math> by use of Eq.\ (7.1.7) of Edmonds { Angular Momentum in Quantum Mechanics} with <math>k = 1</math> to obtain
\begin{equation}
S(J',J) = (-1)^{L'+S+J+1}[(2J+1)(2J'+1)]^{1/2}\left\{
\matrix{L' & J'& S\cr
J & L & 1\cr}\right\}\langle\gamma' L'|| r || \gamma L\rangle
\end{equation}
where <math>\langle\gamma' L'|| r || \gamma L\rangle</math> is the reduced matrix element if spin is neglected. Note that if spin is neglected, then the line-strength is simply
\begin{equation}
S(L',L) = |\langle\gamma' L'|| r || \gamma L\rangle|^2
\end{equation}
To calculate the reduced matrix element, use a particular example along with the Wigner-Eckart theorem
such as
\begin{equation}
\langle\gamma'\, L'|| r || \gamma\, L\rangle = (-1)^{L'}\langle\gamma'\, L'\, 0| z | \gamma\, L\, 0\rangle{\Large /}\left(\matrix{L' & 1 & L \cr
0 & 0 & 0 \cr}\right)
\end{equation}

===Correlated Two-electron Integrals===
For the case of two-electron integrals, <math>z = z_1 + z_2</math> and <math>r = r_1 + r_2</math>. As discussed in G.W.F. Drake, Phys. Rev. A 18, 820 (1978), the general integral in Hylleraas coordinates for the integral over the <math>z_1</math> term can be written in the form
\begin{equation}
\label{int1}
\langle\gamma'\, L'\, 0| z_1 | \gamma\, L\, 0\rangle = \sum_\Lambda C_\Lambda I_\Lambda(\gamma',\gamma)
\end{equation}
where <math>I_\Lambda(\gamma',\gamma)</math> stands for the purely radial integral
\begin{equation}
\label{radial}
I_\Lambda(\gamma',\gamma) = \int_0^\infty r_1\,dr_1\int_0^\infty r_2\,dr_2\int_{|r_1-r_2|}^{r_1+r_2}r_{12}\, dr_{12}
R_{\gamma'}^*\,r_1\, R_\gamma P_\Lambda(\cos\theta_{12})
\end{equation}
and <math>\cos\theta_{12} = (r_1^2 + r_2^2 - r_{12}^2)/(2r_1r_2)</math> is a short-hand notation for a radial function of <math>r_1</math>, <math>r_2</math> and <math>r_{12}</math>. Also, <math>P_\Lambda(\cos\theta_{12})</math> is a Legendre polynomial,
and, <math>R_\gamma = R_\gamma(r_1,r_2,r_{12})</math> is the radial part of the Hylleraas wave function for state <math>\gamma</math> such
that the total wave function is <math>\psi({\bf r}_1,{\bf r}_2) = R_\gamma(r_1,r_2,r_{12}){\cal Y}_{l_1,l_2,L}^M(\hat{\bf r}_1,\hat{\bf r}_2)\pm {\rm exchange}</math>, normalized to unity, and <math>{\cal Y}_{l_1,l_2,L}^M(\hat{\bf r}_1,\hat{\bf r}_2)</math> is a vector coupled product of spherical harmonics. The coefficents <math>C_\Lambda</math> in Eq.\ (\ref{int1}) can be calculated from Eq.\ (32) of the above 1978 paper. They need to be calculated separately for he direct-direct and the direct-exchange contributions. For the reduced matrix element of <math>z_1</math>, the leading 3-<math>j</math> symbol can be dropped, leaving (for the case <math>k = 1</math>)
\begin{eqnarray}
C_\Lambda^{\rm reduced} &=& \frac{(-1)^{l_2'+L'+L+1}}{2} (l_1,l_1',l_2,l_2',L,L')^{1/2}
\sum_{\lambda={\rm max}(l_1-1,0)}^{l_1+1} (\lambda,\Lambda,1)
\left(\matrix{l_1 & 1 & \lambda \cr
0 & 0 & 0 \cr}\right)
\left(\matrix{l_1'& \lambda & \Lambda \cr
0 & 0 & 0 \cr}\right)\nonumber\\
&\times&\left(\matrix{l_2' & 1_2 & \Lambda \cr
0 & 0 & 0 \cr}\right)
\left\{\matrix{L' & \lambda & l_2 \cr
\Lambda& l_2' & l_1' \cr}\right\}
\left\{\matrix{L' & \lambda & l_2 \cr
l_1 & L & 1 \cr}\right\}
\end{eqnarray}
where the notation <math>(a,b,c,\cdots)</math> means <math>(2a+1)(2b+1)(2c+1)\cdots</math>.
A similar expression can be obtained for the angular coefficients for the <math>z_2</math> term by interchanging the subscripts 1 and 2 throughout and multiplying by <math>(-1)^{L+L'+1}</math>.

===Example===
As an example, for the <math>1s^2\;^1S - 1s2p\;^1P</math> transition of helium, the only nonvanishing <math>C_\Lambda</math> angular coefficients are <math>C_1 = 1/2</math> for the direct-direct term (i.e. <math>P_1(\cos\theta_{12}) = \cos\theta_{12}</math> is included in the radial integral (\ref{radial})) and <math>C_0 = 1/2</math> for the direct-exchange term (i.e. <math>P_0(\cos\theta_{12}) = 1</math> is included in the radial integral (\ref{radial})). The results for infinite nuclear mass are
\begin{equation}
S(1\;^1S-2\;^1P) = (0.781\,127\,158\,667)^2
\end{equation}
\begin{equation}f(1\;^1S-2\;^1P) = -0.276\,164\,703\,736
\end{equation}
where
\begin{equation}
f(1\;^1S-2\;^1P) = \frac{2}{3}[E(1\;^1S) - E(2\;^1P)]S(1\;^1S-2\;^1P)
\end{equation}
is the absorption oscillator strength.

===Test===
This is a test edit

Theory Notes

2020-12-14T05:25:57Z

Drake: Undo revision 800 by Drake (talk)

<center>
<H1>Notes on solving the Schrödinger equation in Hylleraas coordinates for heliumlike atoms
Gordon W.F. Drake
Department of Physics, University of Windsor
Windsor, Ontario, Canada N9B 3P4
(Transcribed from hand-written notes and edited by Lauren Moffatt. Last revised November 22, 2017 to correct a missing subscript in the expression for $B_1$ in the Hamiltonian in Sect. 11 General Hermitian Property.)
</H1>
</center>

==Introduction==
These notes describe some of the technical details involved in solving the Schrödinger equation for a heliumlike atom of nuclear charge Z in correlated Hylleraas coordinates. The standard for computational purposes is to express all quantities in a form that is valid for electronic states of arbitrary angular momentum, and to express matrix elements of the Hamiltonian in a manifestly Hermitian symmetrized form. Further information and derivations of the mathematical techniques for angular integrals and radial recursion relations can be found in the paper
G.W.F. Drake, Phys. Rev. A <math>{\bf 18}</math>, 820 (1978).

The starting point is the two-electron Schrödinger equation for infinite nuclear mass
\begin{equation} \left[-\frac{\hbar^2}{2m}(\nabla^2_1 +\nabla^2_2) - \frac{Ze^2}{r_1} -
\frac{Ze^2}{r_2}+\frac{e^2}{r_{12}} \right]\psi = E\psi\nonumber
\end{equation}
where <math>m</math> is the electron mass, and <math>r_{12} = |{\bf r}_1 - {\bf r}_2|</math> (see diagram below).

[[File:Pic1.jpg|250px]]

Begin by rescaling distances and energies so that the Schrödinger equation can be expressed in a dimensionless form. The dimensionless Z-scaled distance is defined by <math>\begin{eqnarray} \rho = \frac{Zr}{a_0} \end{eqnarray} </math> where <math> \begin{eqnarray} a_0 = \frac{\hbar^2}{me^2} \end{eqnarray} </math> is the atomic unit (a.u.) of distance equal to the Bohr radius <math>0.529\, 177\, 210\, 92(17) \times 10^{-10}</math> m. Then

\begin{equation}[-\frac{\hbar^2}{2m}Z^2(\frac{me^2}{\hbar^2})^2(\nabla^2_{\rho_1}+\nabla^2_{\rho_2})
- Z^2\frac{e^2}{a_0}\rho^{-1}_1 - Z^2\frac{e^2}{a_0}\rho^{-1}_2 +
\frac{e^2}{a_0}Z\rho^{-1}_{12}]\psi= E\psi\nonumber\end{equation}

But <math>\begin{eqnarray} \frac{\hbar^2}{m}(\frac{me^2}{\hbar^2})^2 = \frac{e^2}{a_0} \end{eqnarray} </math> is the hartree atomic unit of of energy <math>(E_h = 27.211\,385\,05(60)</math> eV, or equivalently <math>E_h/(hc) = 2194.746\,313\,708(11)</math> m<math>^{-1})</math>. Therefore, after multiplying through by <math>a_0/(Ze)^2</math>, the problem to be solved in Z-scaled dimensionless units becomes

\begin{equation} \left[-\frac{1}{2}(\nabla^2_{\rho_1}+\nabla^2_{\rho_2}) -
\frac{1}{\rho_1} - \frac{1}{\rho_2} + \frac{Z^{-1}}{\rho_{12}}\right]\psi =
\varepsilon\psi\nonumber \end{equation}

where <math>\begin{eqnarray}\varepsilon = \frac{Ea_0}{(Ze)^2} \end{eqnarray}</math> is the energy in Z-scaled atomic units. For convenience, rewrite this in the conventional form
\begin{equation}
H\psi = \varepsilon\psi\nonumber \end{equation}
where (using <math>r</math> in place of <math>\rho</math> for the Z-scaled distance)
\begin{equation}
H = -\frac{1}{2}(\nabla^2_1+\nabla^2_2) - \frac{1}{r_1}-\frac{1}{r_2} + \frac{Z^{-1}}{r_{12}}
\nonumber\end{equation}
is the atomic Hamiltonian for infinite nuclear mass.

==The Hartree Fock Method==

For purposes of comparison, and to define the correlation energy, assume that <math>\psi({\bf r}_1,{\bf r}_2)</math> can be written in the separable product form
\begin{equation}
\psi({\bf r}_1,{\bf r}_2) = \frac{1}{\sqrt{2}} \left[ u_1(r_1) u_2(r_2) \pm u_2(r_1) u_1(r_2) \right] \nonumber
\end{equation}
for the <math>1s^2\ ^1\!S</math> ground state. Because of the <math>\frac{1}{r_{12}}</math> term in the Schrödinger equation, the exact solution connot be expressed in this form as a separable product. However, the Hartree-Fock approximation corresponds to finding the best possible solution to the Schrödinger equation
\begin{equation}
H\psi({\bf r}_1,{\bf r}_2) = E\psi({\bf r}_1,{\bf r}_2)\nonumber
\end{equation}
that can nevertheless be expressed in this separable product form, where as before
\begin{equation}
H = -\frac{1}{2}(\nabla^2_1+\nabla^2_2) - \frac{1}{r_1}- \frac{1}{r_2} + \frac{Z^{-1}}{r_{12}}\nonumber
\end{equation}
is the full two-electron Hamiltonian. To find the best solution, substitute into <math>\left\langle\psi \lvert H-E \rvert \psi\right\rangle</math> and require this expression to be stationary
with respect to arbitrary infinitesimal variations <math>\delta u_1</math> and <math>\delta
u_2</math> in <math>u_1</math> and <math>u_2</math> respectively; i.e.

\begin{equation}
\frac{1}{2} \left\langle \delta u_1(r_1)u_2(r_2) \pm u_2(r_1)\delta
u_1(r_2) \lvert H-E \rvert u_1(r_1) u_2(r_2)\pm u_2(r_1) u_1(r_2)\right\rangle\nonumber
\end{equation}

\begin{equation}
=\int\delta u_1(r_1) d{\bf r}_1 \left\{ \int d{\bf r}_2 u_2(r_2) \left( H - E \right) \left[ u_1(r_1)u_2(r_2) \pm u_2(r_1) u_1(r_2) \right] \right\}\nonumber
\end{equation}

\begin{equation}
= 0 \nonumber
\end{equation}

for arbitrary <math> \delta u_1(r_1).\nonumber </math> Therefore <math>\{\int d{\bf r}_2 \ldots \} = 0</math>.

Similarly, the coefficient of <math>\delta u_2</math> would give

\begin{equation}
\int d{\bf r}_1 u_1(r_1) \left( H-E \right) \left[ u_1(r_1) u_2(r_2) \pm
u_2(r_1) u_1(r_2) \right] = 0\nonumber
\end{equation}

Define

\begin{equation}
I_{12} = I_{21} = \int d{\bf r}\, u_1(r)u_2(r), \nonumber
\end{equation}

\begin{equation}
H_{ij} = \int d{\bf r}\, u_i(r)(-\frac{1}{2}\nabla - \frac{1}{r})u_j(r), \nonumber
\end{equation}

\begin{equation}
G_{ij}(r) = \int d{\bf r}^\prime u_i(r^\prime)\frac{1}{|{\bf r} - {\bf r}\prime|}u_j(r^\prime)\nonumber
\end{equation}

Then the above equations become the pair of integro-differential equations

\begin{equation}
[ H_0 - E + H_{22}+G_{22}(r)]u_1(r) = \mp [ I_{12}(H_0-E) + H_{12}+G_{12}(r)]u_2(r)\nonumber
\end{equation}

\begin{equation}
[H_0-E+H_{11}+G_{11}(r)]u_2(r) = \mp [I_{12}(H_0-E) + H_{12}+G_{12}(r)]u_1(r)\nonumber
\end{equation}

These must be solved self-consistently for the "constants" <math>I_{12}</math> and
<math>H_{ij}</math> and the function <math>G_{ij}(r)</math>.

The Hartree Fock energy is <math>E \simeq -2.87\ldots </math>a.u. while the exact energy is <math>E = -2.903724\ldots </math>a.u.

The difference is called the "correlation energy" because it arises from the
way in which the motion of one electron is correlated to the other. The
Hartree Fock equations only describe how one electron moves in the average field
provided by the other (mean-field theory).

==Configuration Interaction==
Expand
\begin{equation}
\psi({\bf r}_1,{\bf r}_2)= C_0u^{(s)}_1(r_1)u^{(s)}_1(r_2) + C_1u^{(P)}_1({\bf r}_1)u^{(P)}_1({\bf r}_2){\cal Y}^0_{1,1,0}(\hat{\bf r}_1, \hat{\bf r}_2)+C_2u^{(d)}_1({\bf r}_1)u^{(d)}_2({\bf r}_2){\cal Y}^0_{2,2,0}(\hat{\bf r}_1, \hat{\bf r}_2)+... \pm {\rm\ exchange}
\end{equation}
where
\begin{equation}
{\cal Y}^M_{l_1,l_2,L}(\hat{\bf r}_1, \hat{\bf r}_2)=\sum_{m_1,m_2}Y^{m_1}_{l_1}({\bf r}_1)Y^{m_2}_{l_2}({\bf r}_2)\times <l_1l_2m_1m_2\mid LM>
\end{equation}

This works, but is slowly convergent, and very laborious. The best CI calculations are accurate to <math> ~10^{-7}</math> a.u.

==Hylleraas Coordinates==
[E.A. Hylleraas, Z. Phys. <math>{\bf 48}, 469(1928)</math> and <math>{\bf 54}, 347(1929)</math>]
suggested using the co-ordinates <math>r_1</math>, <math>r_2</math> and <math>r_{12}</math> or equivalently

\begin{eqnarray}
s &=& r_1 + r_2, \nonumber\\
t &=& r_1-r_2, \nonumber\\
u &=& r_{12}\nonumber
\end{eqnarray}

and writing the trial functions in the form

\begin{equation}
\Psi({\bf r}_1,{\bf r}_2) = \sum^{i+j+k\leq
N}_{i,j,k}c_{i,j,k}r_1^{i+l_1}r_2^{j+l_2}r_{12}^ke^{-\alpha r_1 - \beta r_2}
\mathcal{Y}^M_{l_1,l_2,L}(\hat{r}_1,\hat{r}_2)\pm \text{exchange}\nonumber
\end{equation}

Diagonalizing H in this non-orthogonal basis set is equivalent to solving
<math> \begin{eqnarray}
\frac{\partial E}{\partial c_{i,j,k}} = 0\nonumber
\end{eqnarray} </math>
for fixed <math>\alpha</math> and <math>\beta</math>.

The diagonalization must be repeated for different values of <math>\alpha</math> and
<math>\beta</math> in order to optimize the non-linear parameters.

==Completeness==
The completeness of the above basis set can be shown by first writing
\begin{equation}
r_{12}^2 = r_1^2 + r_2^2 - 2r_1r_2\cos\theta_{12}
\end{equation}
and, from the spherical harmonic addition theorem,
\begin{equation}
\cos(\theta_{12})=\frac{4\pi}{3}\sum^1_{m=-1}Y^{m*}_1(\theta_1,\varphi_1)Y^m_1(\theta_2,\varphi_2)
\end{equation}
Consider first the S-states. The <math>r_{12}^0</math> terms are like the ss terms in a CI calculation. The
<math>r_{12}^2</math> terms bring in p-p type contributions, and the higher powers bring
in d-d, f-f etc. type terms. In general

\begin{equation}
P_l(\cos\theta_{12}) = \frac{4\pi}{2l+1}\sum^l_{m=-l}{Y^{m}_l}^*(\theta_1,\varphi_1)Y^m_l(\theta_2,\varphi_2)\nonumber
\end{equation}

For P-states, one would have similarly

\begin{array}{lr}
r_{12}^0 & (sp)P\nonumber\\
r_{12}^2 & (pd)P\nonumber\\
r_{12}^4 & (df)P\nonumber\\
\vdots & \vdots\nonumber
\end{array}

For D-states

\begin{array}{lr}
r_{12}^0 & (sd)D & (pp^\prime)D\nonumber\\
r_{12}^2 & (pf)D & (dd^\prime)D\nonumber\\
r_{12}^4 & (dg)D & (ff^\prime)D\nonumber\\
\vdots & \vdots & \vdots\nonumber
\end{array}

In this case, since there are two "lowest-order" couplings to form a
D-state, both must be present in the basis set, i.e.

\begin{equation}
\Psi(r_2,r_2) = \sum c_{ijk}r_1^ir_2^{j+2}r_{12}^ke^{-\alpha r_1-\beta
r_2}\mathcal{Y}^M_{022}(\hat{r}_1,\hat{r}_2)
+\sum d_{ijk}r_1^{i+1}r_2^{j+1}r_{12^k}e^{-\alpha^\prime r_1 - \beta^\prime
r_2}\mathcal{Y}^M_{112}(\hat{r}_1,\hat{r}_2)\nonumber
\end{equation}

For F-states, one would need <math>(sf)F</math> and <math>(pd)F</math> terms.

For G-states, one would need <math>(sg)G</math>, <math>(pf)G</math> and <math>(dd^\prime)G</math> terms.

Completeness of the radial functions can be proven by considering the
Sturm-Liouville problem

\begin{equation}
\left(-\frac{1}{2}\nabla^2-\frac{\lambda}{r_s}-E\right)\psi({\bf r}) = 0\nonumber
\end{equation}

or

\begin{equation}
\left(-\frac{1}{2}\frac{1}{r^2}\frac{\partial}{\partial r}\left({r^2}\frac{\partial}{\partial r}\right) -
\frac{l\left(l+1\right)}{2r^2} - \frac{\lambda}{r} - E\right)u(r) = 0.\nonumber
\end{equation}

For fixed E and variable <math>\lambda</math> (nuclear charge).

The eigenvalues are <math>\lambda_n = (E/E_n)^{1/2}</math>, where <math>E_n =- \frac{1}{2n^2}</math>

[[File:Pic2.jpg|500px]]

and the eigenfunctions are

\begin{equation}
u_{nl}(r) =
\frac{1}{(2l+1)!}\left(\frac{(n+l)!}{(n-l-1)2!}\right)^{1/2}(2\alpha)^{3/2}e^{-\alpha
r}\nonumber
\times \ \left( 2 \alpha r \right)^l\,_{1}F_{1}\left(-n+l+1,2l+2; 2 \alpha r \right)
\end{equation}

with <math>\alpha = (-2E)^{1/2}</math> and <math>n\geq l+1</math>. The confluent hypergeometric function <math> _1F_1(a,b,;z)</math> then denotes a finite polynomial since <math>a=-n+l+1</math> is a negative integer or zero.

Unlike the hydrogen spectrum, which has both a discrete part for <math>E<0</math> and a
continuous part for <math>E>0</math>, this forms an entirely discrete set of finite
polynomials, called Sturmian functions. They are orthogonal with respect to
the potential, i.e.

\begin{equation}
\int^\infty_0 r^2dr \left( u_{n^\prime l}(r)\frac{1}{r}u_{nl}(r) \right) = \delta_{n,n^\prime}\nonumber
\end{equation}

Since they become complete in the limit <math>n\rightarrow\infty</math>, this assures the
completeness of the variational basis set.

[See also B Klahn and W.A. Bingel Theo. Chim. Acta (Berlin) 44, 9 and 27
(1977)].

==Solutions of the Eigenvalue Problem==

For convenience, write

\[
\Psi({\bf r}_1,{\bf r}_2) = \sum^N_{m=1}c_m\varphi_m\nonumber
\]

where <math>m</math> represents the <math>m</math>'th combination of <math>i,j,k</math> and

\begin{equation}
\varphi_{ijk}=r_1^ir_2^jr_{12}^ke^{-\alpha r_1 - \beta
r_2}\mathcal{Y}^M_{l_1,l_2,L}(\hat{r}_1,\hat{r}_2) \pm \mbox{exchange}\nonumber
\end{equation}

\begin{align}

&
\left( \begin{array}{lr}
\cos(\theta) & \sin(\theta)\\
-\sin(\theta) & \cos(\theta)
\end{array} \right)
\left( \begin{array}{lr}
H_{11} & H_{12} \\
H_{12} & H_{22}
\end{array}\right)
\left( \begin{array}{lr}
\cos(\theta) & -\sin(\theta)\\
\sin(\theta) & \cos(\theta)
\end{array} \right)
\\[5pt]
= & \left(\begin{array}{lr}
cH_{11}+sH_{12} & cH_{12} + sH_{22}\\
-sH_{11} + cH_{12} & -sH_{12} + cH_{22}
\end{array}\right)
\left( \begin{array}{cc}
c & -s \\
s & c
\end{array}\right)
\\[5pt]
= & \left(\begin{array}{lr}
c^2H_{11}+s^2H_{22} + 2csH_{12} & (c^2-s^2)H_{12}+cs(H_{22}-H_{11})\\
(c^2-s^2)H_{12}+cs(H_{22}-H_{11}) & s^2H_{11}+c^2H_{22}-2csH_{12}
\end{array}\right)
\end{align}

Therefore

\begin{equation}
\left[\cos^2(\theta)-\sin^2(\theta)\right]H_{12} =
\cos(\theta)\sin(\theta)(H_{11}-H_{22})\nonumber
\end{equation}

and

\begin{equation}
\tan(2\theta) = \frac{2H_{12}}{H_{11}-H_{22}}\nonumber
\end{equation}

i.e.

\begin{equation}
\cos(\theta)=\left(\frac{r+\omega}{2r}\right)^{1/2}\nonumber
\end{equation}

\begin{equation}
\sin(\theta)=-\mbox{sgn}(H_{12})\left(\frac{r-\omega}{2r}\right)^{1/2}\nonumber
\end{equation}

where

\begin{equation}
\omega = H_{22}-H_{11}\nonumber
\end{equation}

\begin{equation}
r=\left(\omega^2+4H_{12}^2\right)^{1/2}\nonumber
\end{equation}

\begin{equation}
E_1=\frac{1}{2}\left(H_{11}+H_{22}-r\right)\nonumber
\end{equation}

\begin{equation}
E_2=\frac{1}{2}\left(H_{11}+H_{22}+r\right)\nonumber
\end{equation}

===Brute Force Method===

- Gives all the eigenvalues and eigenvectors, but it is slow

- First orthonormalize the <math>N</math>-dimensional basis set; i.e. form linear combinations

\begin{equation}
\Phi_m = \sum_{n=1}^N\varphi_nR_{nm}\nonumber
\end{equation}

such that
\begin{equation}
\langle \Phi_m | \Phi_n \rangle = \delta_{m,n}
\end{equation}

This can be done by finding an orthogonal tranformation T such that

\begin{equation}
T^TOT = I = \left(
\begin{array}{ccccc}
I_1 & 0 & 0 & \ldots & 0 \\
0 & I_2 & 0 & \ldots & 0 \\
0 & 0 & I_3 & \ldots & 0 \\
\vdots & \vdots & \vdots & \ddots & \vdots \\
0 & 0 & 0 & \ldots & I_N
\end{array}\right);
\end{equation}

\begin{equation}
O_{mn} = \langle \varphi_m | \varphi_n \rangle \nonumber
\end{equation}

and then applying a scale change matrix

\begin{equation}
S = \left(\begin{array}{ccccc}
\frac{1}{I_1^{1/2}} & 0 & 0 & \ldots & 0\\
0 & \frac{1}{I_2^{1/2}} & 0 & \ldots & 0 \\
0 & 0 & \frac{1}{I_3^{1/2}} & \ldots & 0 \\
\vdots & \vdots & \vdots & \ddots & \vdots \\
0 & 0 & 0 & \ldots & I_N^{1/2}
\end{array}\right)= S^T\nonumber
\end{equation}

Then

\begin{equation}
S^TT^TOTS = 1
\end{equation}

i.e.

\begin{equation}
R^TOR = 1
\end{equation}

with <math>R=TS</math>.

If H is the matrix with elements <math>H_{mn} = \langle \varphi_m | \varphi_n \rangle </math>, then H
expressed in the <math>\Phi_m</math> basis set is

\begin{equation}
H^\prime = R^THR.\nonumber
\end{equation}

We next diagonalize <math>H^\prime</math> by finding an orthogonal transformation W such
that

\begin{equation}
W^T H^\prime W = \lambda = \left(
\begin{array}{cccc}
\lambda_1 & 0 & \ldots & 0 \\
0 & \lambda_2 & \ldots & 0 \\
\vdots & \vdots & \ddots & \vdots \\
0 & 0 & \ldots & \lambda_N
\end{array}\right)\nonumber
\end{equation}

The q'th eigenvector is

\begin{equation}
\Psi^{(q)} = \sum_{n=1}^N\Phi_n W_{n,q} = \sum_{n,n^\prime} \varphi_{n^\prime} R_{n^\prime ,n} W_{n,q} \nonumber
\end{equation}

i.e.

\begin{equation}
c_{n^\prime}^{(q)} = \sum_{n=1}^N R_{n^\prime n} W_{n,q}
\end{equation}

===The Power Method===

- Based on the observation that if H has one eigenvalue, <math>\lambda_M</math>, much bigger than all the rest, and <math>\chi = \left(
\begin{array}{c}a_1\\a_2\\\vdots\end{array}\right)</math>
is an arbitrary starting vector, then <math>\chi = \sum_{q=1}^N x_q\Psi^{(q)}</math>.

\begin{equation}
(H)^n\chi = \sum_{q=1}^N x_q \lambda^n_q\Psi^{(q)}\nonumber\rightarrow x_M\lambda_M^n\Psi^{(M)}\nonumber
\end{equation}

provided <math>x_M\neq 0</math>.

To pick out the eigenvector corresponding to any eigenvalue, write the original problem in the form

\begin{align}
H\Psi & = \lambda O \Psi \nonumber \\
\left( H - \lambda_q O \right) \Psi & = \left( \lambda - \lambda_q \right) O \Psi \nonumber
\end{align}

Therefore,

\begin{equation}
G \Psi = \frac{1}{\lambda - \lambda_q } \Psi \nonumber \\
\end{equation}

where <math>G=(H-\lambda_qO)^{-1}O</math> with eigenvalues <math>\frac{1}{\lambda_n - \lambda_q}</math>.

By picking <math>\lambda_q</math> close to any one of the <math>\lambda_n</math>, say
<math>\lambda_{n^\prime}</math>, then <math>\frac{1}{\lambda_n-\lambda_q}</math> is much larger for
<math>n=n^\prime</math> than for any other value. The sequence is then

\begin{array}{c}
\chi_1=G\chi\nonumber\\
\chi_2=G\chi_1\nonumber\\
\chi_3=G\chi_2\nonumber\\
\vdots\nonumber
\end{array}

until the ratios of components in <math>\chi_n</math> stop changing.

- To avoid matrix inversion and multiplication, note that the sequence is equivalent to
\begin{equation}
F \chi_n = \left( \lambda - \lambda_q \right) O \chi_{n-1} \nonumber \\
\end{equation}
where <math>F = H-\lambda_qO</math>. The factor of <math>\left( \lambda - \lambda_q \right)</math> can be
dropped because this only affects the normalization of <math>\chi_n</math>. To find <math>\chi_n</math>, solve
\begin{equation}
F \chi_n = O \chi_{n-1} \nonumber
\end{equation}
(N equations in N unknowns). Then
\begin{equation}
\lambda = \frac{ \langle \chi_n | H | \chi_n \rangle}{\langle \chi_n | \chi_n \rangle} \nonumber
\end{equation}

==Matrix Elements of H==

\begin{equation}
H=-\frac{1}{2}\nabla^2_1 -\frac{1}{2}\nabla^2_2 - \frac{1}{r_1} - \frac{1}{r_2} +\frac{Z^{-1}}{r_{12}}\nonumber
\end{equation}

Taking <math>r_1, r_2</math> and <math>r_{12}</math> as independent variables,

\begin{equation}
\nabla_1^2 = \frac{1}{r_1^2}\frac{\partial}{\partial r_1}\left(r_1^2\frac{\partial}{\partial r_1}\right)+ \frac{1}{r_{12}^2}\frac{\partial}{\partial r_{12}}
\left(r_{12}^2\frac{\partial}{\partial r_{12}}\right)\nonumber -\frac{l_1(l_1+1)}{r_1^2}+2(r_1 - r_2\cos(\theta_{12}))\frac{1}{r_{12}}\frac{\partial^2}{\partial r_1 \partial r_{12}} \nonumber - 2(\nabla_1^Y \cdot {\bf r}_2)\frac{1}{r_{12}}\frac{\partial}{\partial r_{12}}\nonumber
\end{equation}

where <math>\nabla_1^Y</math> acts only on the spherical harmonic part of the wave function and the diagram

[[File:Pic3.jpg|400px]]

defines the complete set of 6 independent variables is <math>r_1</math>, <math>r_2</math>, <math>r_{12}</math>, <math>\theta_1</math>, <math>\varphi_1</math>, <math>\chi</math>.

If <math>r_{12}</math> were not an independent variable, then one could take the volume element to be

\begin{equation}
d\tau = r_1^2dr_1\sin(\theta_1)d\theta_1d\varphi_1r_2^2dr_2\sin(\theta_2)d\theta_2d\varphi_2.\nonumber
\end{equation}

However, <math>\theta_2</math> and <math>\varphi_2</math> are no longer independent variables. To eliminate them, take the point <math>{\bf r}_1</math> as the
origin of a new polar co-ordinate system, and write

\begin{equation}
d\tau=-r_1^2dr_1\sin(\theta_1)d\theta_1d\varphi_1r_{12}^2dr_{12}\sin(\psi)d\psi d\chi\nonumber
\end{equation}

and use

\begin{equation}
r_2^2=r_1^2+r_{12}^2 +2r_1r_{12}\cos(\psi)
\end{equation}

Then for fixed <math>r_1</math> and <math>r_{12}</math>,

\begin{equation}
2r_2dr_2 = -2r_1r_{12}\sin(\psi)d\psi
\end{equation}

Thus

\begin{equation}
d\tau= r_1dr_1r_2dr_2r_{12}dr_{12}\sin(\theta_1)d\theta_1d\varphi_1d\chi
\end{equation}

The basic type of integral to be calculated is

\begin{align}
I \left( l_1, m_1, l_2, m_2; R \right) & = \int \sin \left( \theta_1 \right) d\theta_1 d\varphi_1 d\chi Y^{m_1}_{l_1} \left( \theta_1, \varphi_1 \right)^{*} Y^{m_2}_{l_2} \left( \theta_2, \varphi_2 \right) \\
& \times \int r_1dr_1r_2dr_2r_{12}dr_{12}R \left( r_1, r_2, r_{12} \right)
\end{align}

Consider first the angular integral. <math>Y^{m_2}_{l_2} \left( \theta_2, \varphi_2 \right)</math> can be expressed in terms of the independent variables <math>\theta_1, \varphi_1, \chi</math> by use of the rotation matrix relation
\begin{equation}
Y^{m_2}_{l_2}( \theta_2, \varphi_2) = \sum_m \mathcal{D}^{(l_2)}_{m_2,m}(\varphi_1,\theta_1,\chi)^*Y^m_{l_2}( \theta_{12}, \varphi)
\end{equation}
where <math>\theta, \varphi</math> are the polar angles of <math>{\bf r}_2</math> relative to <math>{\bf r}_1</math>. The angular integral is then

\begin{align}
I_{ang} & = \int^{2 \pi}_{0} d\chi \int^{2\pi}_{0} d\varphi_1 \int^{\pi}_{0} \sin \left( \theta_1 \right) d\theta_{1} Y^{m_1}_{l_1} \left( \theta_1, \varphi_1 \right)^{*} \\
& \times \sum_m \mathcal{D}^{(l_2)}_{m_2,m} \left( \varphi_1,\theta_1,\chi \right)^{*} Y^{m}_{l_2} \left( \theta_{12},\varphi \right) \\
\end{align}

Use

\begin{equation}
Y^{m_1}_{l_1} \left( \theta_1, \varphi_1 \right)^{*} = \sqrt{\frac{2l_1+1}{4\pi}}\mathcal{D}^{(l_1)}_{m_1,0} \left( \varphi_1, \theta_1, \chi \right)
\end{equation}

together with the orthogonality property of the rotation matrices (Brink and Satchler, p 147)

\begin{align}
& \int \mathcal{D}^{(j)*}_{m,m^\prime}\mathcal{D}^{(J)}_{M,M^\prime}\sin(\theta_1)d\theta_1d\varphi_1d\chi \\
& = \frac{8\pi^2}{2j+1}\delta_{jJ}\delta_{mM}\delta_{m^\prime M^\prime}\nonumber
\end{align}

to obtain

\begin{align}
I_{ang} & = \sqrt{ \frac{2l_1+1}{4\pi}} \frac{8\pi^2}{2l_1+1} \delta_{l_1,l_2} \delta_{m_1,m_2} Y^{0}_{l_2} \left( \theta_{12}, \varphi \right) \\
& = 2 \pi \delta_{l_1,l_2} \delta_{m_1,m_2} P_{l_2} \left( \cos \theta_{12} \right)
\end{align}

since

\begin{equation}
Y^0_{l_2} \left( \theta_{12}, \varphi \right) = \sqrt{\frac{2l_1+1}{4\pi}} P_{l_2} \left( \cos \left( \theta_{12} \right) \right)
\end{equation}

Note that <math>P_{l_2} \left( \cos \theta \right)</math> is just a short hand expression for a radial function because

\begin{equation}
\cos \theta_{12} = \frac{r_1^2 + r_2^2 - r_{12}^2}{2r_1r_2}
\end{equation}

The original integral is thus

\begin{equation}
I \left( l_1, m_1, l_2, m_2; R \right) = 2 \pi \delta_{l_1,l_2} \delta_{m_1,m_2} \int^{\infty}_{0} r_{1} dr_{1} \int^{\infty}_{0} r_{2} dr_{2} \int^{r_{1} + r_{2}}_{|r_{1} - r_{2}|} r_{12} dr_{12} R \left( r_{1}, r_{2}, r_{12} \right) P_{l_2} \left( \cos \theta_{12} \right)
\end{equation}

where again

\begin{equation}
\cos \theta_{12} = \frac{r_{1}^{2} +r^{2}_{2} - r_{12}^{2}}{2 r_{1} r_{2}}
\end{equation}

is a purely radial function.

The above would become quite complicated for large <math>l_2</math> because <math>P_{l_2}(\cos \theta_{12})</math> contains terms up to <math>\left( \cos \theta_{12} \right)^{l_2}</math>. However, recursion relations exist which allow any integral containing <math>P_l \left( \cos \theta_{12} \right)</math> in terms of those containing just <math>P_0 \left( \cos \theta_{12} \right) = 1</math> and <math>P_1 \left( \cos \theta_{12} \right) = \cos \theta_{12}</math>.

==Radial Integrals and Recursion Relations==

The basic radial integral is [see G.W.F. Drake, Phys. Rev. A 18, 820 (1978)]

\begin{align}
I_0 \left( a, b, c \right) & =
\int^{\infty}_{0} r_{1} dr_{1} \int^{\infty}_{r_{1}} r_{2} dr_{2} \int^{r_{1}+r_{2}}_{r_{2} - r_{1}} r_{12} dr_{12} r_{1}^{a} r_{2}^{b} r_{12}^{c} e^{-\alpha r_{1} - \beta r_{2}} + \int^{\infty}_{0} r_{2} dr_{2} \int^{\infty}_{r_{2}} r_{1} dr_{1} \int^{r_{1} + r_{2}}_{r_{1} - r_{2}} r_{12} dr_{12} r_{1}^{a} r_{2}^{b} r_{12}^{c} e^{- \alpha r_{1} - \beta r_{2}} \\
& = \frac{2}{c+2} \sum^{[(c+1)/2]}_{i=0}
\left(\begin{array}{c}
c+2 \\
2i+1
\end{array} \right)
\left\{ \frac{q!}{\beta^{q+1} \left( \alpha + \beta \right)^{p+1}} \sum_{j=0}^{q} \frac{\left( p + j \right)!}{j!} \left( \frac{\beta}{\alpha + \beta} \right)^j + \frac{q^{\prime}!}{\alpha^{q^{\prime}+1} \left( \alpha + \beta \right)^{p^{\prime}+1}} \sum^{q^{\prime}}_{j=0} \frac{\left( p^{\prime} + j \right)!}{j!} \left( \frac{\alpha}{\alpha + \beta} \right)^j \right\}
\end{align}

where

\begin{array}{ll}
p = a + 2i + 2 & p^{\prime} = b + 2i + 2 \\
q = b + c - 2i + 2 & q^{\prime} = a + c - 2i + 2
\end{array}

The above applies for <math> a, b \geq -2, c\geq -1</math>.
<math>[x]</math> means "greatest integer in" <math>x</math>.

Then

\begin{align}
I_1 \left( a, b, c \right) & = \int d\tau_{r} r_{1}^{a} r_{2}^{b} r_{12}^{c} e^{ -\alpha r_{1} - \beta r_{2}} P_{1}\left( \cos \theta \right) \\
& = \frac{1}{2} \left( I_{0} \left( a + 1, b - 1, c \right) + I_{0} \left( a - 1, b + 1, c \right) - I_{0} \left( a - 1, b - 1, c + 2 \right) \right)
\end{align}

===The Radial Recursion Relation===

Recall that the full integral is

\begin{equation}
I(l_1m_1,l_2m_2;R)=2\pi\delta_{l_1,l_2}\delta_{m_1,m_2}I_{l_2}(R)
\end{equation}

where

\begin{equation}
I_{l_2}(R) = \int d\tau_r R(r_1,r_2,r_{12})P_{l_2}(\cos \theta_{12})
\end{equation}

To obtain the recursion relation, use

\begin{equation}
P_l(x) = \frac{[P^{\prime}_{l+1}(x)-P^{\prime}_{l-1}(x)]}{2l+1}
\end{equation}

with

\begin{equation}
P_{l+1}^\prime (x) = \frac{d}{dx}P_{l+1}(x)
\end{equation}

Here <math>x=\cos\theta_{12}</math> and

\begin{align}
\frac{d}{d \cos \theta_{12}} & = \frac{dr_{12}}{d \cos \theta_{12}} \frac{d}{dr_{12}} \\
& = -\frac{r_{1} r_{2}}{r_{12}} \frac{d}{dr_{12}}
\end{align}

Then,

\begin{equation}
I_{l}\left( R \right) = - \int d\tau_r R \frac{r_{1} r_{2}}{r_{12}} \frac{d}{dr_{12}} \frac{ \left[ P_{l+1} \left( \cos \theta_{12} \right) - P_{l-1} \left( \cos \theta_{12} \right) \right]}{2l+1}
\end{equation}

The <math>r_{12}</math> part of the integral is

\begin{align}
& \int^{r_{1} + r_{2}}_{|r_{1} - r_{2}|} r_{12} dr_{12} R \frac{r_{1} r_{2}}{r_{12}} \frac{d}{dr_{12}} \left[P_{l+1}-P_{l-1} \right] \\
& = R r_{1} r_{2} \left[ P_{l+1} - P_{l-1} \right]^{r_{1} + r_{2}}_{|r_{1} - r_{2}|} - \int^{r_{1} + r_{2}}_{|r_{1} - r_{2}|} r_{12} dr_{12} \left( \frac{d}{dr_{12}} R \right) \frac{r_{1} r_{2}}{r_{12}} \frac{ \left[ P_{l+1} - P_{l-1} \right]}{2l+1}
\end{align}

The integrated term vanishes because

\begin{equation}
\cos \theta_{12} = \frac{r_{1}^{2} + r_{2}^{2} - r_{12}^{2}}{2 r_{1} r_{2}}
\begin{array}{lll}
& = -1 & {\rm when} & r_{12}^{2} = \left( r_{1} + r_{2} \right)^{2} \\
& = 1 & {\rm when} & r_{12}^{2} = \left( r_{1} - r_{2} \right)^{2}
\end{array}
\end{equation}

and <math>P_l(1)=1</math>, <math>P_{l}(-1)=(-1)^{l}</math>.

Thus,

\begin{equation}
I_{l+1}\left(\frac{r_1r_2}{r_{12}}R^\prime\right) = \left( 2l+1 \right) I_{l}(R) - I_{l-1} \left( \frac{r_{1} r_{2}}{r_{12}} R^{\prime} \right)
\end{equation}

If

\begin{equation}
R = r_{1}^{a-1} r_{2}^{b-1} r_{12}^{c+2} e^{-\alpha r_{1} - \beta r_{2}}
\end{equation}

then [G.W.F. Drake, Phys. Rev. A 18, 820 (1978)]

\begin{equation}
I_{l+1}(r_1^ar_2^br_{12}^c) =
\frac{2l+1}{c+2} I_{l} \left( r_{1}^{a-1} r_{2}^{b-2} r_{12}^{c+2} \right)+I_{l-1} \left( r_{1}^{a} r_{2}^{b} r_{12}^{c} \right)
\end{equation}

For the case <math>c=-2</math>, take

\begin{equation}
R = r_{1}^{a-1} r_{2}^{b-1} \ln r_{12} e^{ -\alpha r_{1} -\beta r_{2}}
\end{equation}

Then

\begin{equation}
I_{l+1} \left( r_{1}^{a} r_{2}^{b} r_{12}^{c} \right) = I_{l} \left( r_{1}^{a-1} r_{2}^{b-1} \ln r_{12} \right) \left( 2l + 1 \right) - I_{l-1} \left( r_{1}^{a} r_{2}^{b} r_{12}^{-2} \right)
\end{equation}

This allows all <math>I_{l}</math> integrals to be calculated from tables of <math>I_0</math> and
<math>I_1</math> integrals.

===The General Integral===

The above results for the angular and radial integrals can now be combined into a general formula for integrals of the type

\begin{equation}
I = \int \int d{\bf r}_{1} d{\bf r}_{2} R_{1} \mathcal{Y}^{M^{\prime}}_{l^{\prime}_{1} l^{\prime}_{2} L^{\prime}} \left( \hat{r}_{1}, \hat{r}_{2} \right) T_{k_{1} k_{2} K}^{Q} \left( {\bf r}_{1},{\bf r}_{2} \right) R_{2} \mathcal{Y}_{l_{1} l_{2} L}^{M} \left( \hat{r}_{1}, \hat{r}_{2} \right)
\end{equation}

where

\begin{equation}
\mathcal{Y}^{M}_{l_{1} l_{2} L} \left( \hat{r}_{1}, \hat{r}_{2} \right) = \sum_{m_1,m_2} \langle l_{1} l_{2} m_{1} m_{2} | LM \rangle Y^{m_{1}}_{l_{1}} \left( \hat{r}_{1} \right) Y^{m_{2}}_{l_{2}} \left( \hat{r}_{2} \right)
\end{equation}

and

\begin{equation}
T^Q_{k_{1} k_{2} K} \left( {\bf r}_{1}, {\bf r}_{2} \right) = \sum_{q_{1}, q_{2}} \langle k_{1} k_{2} q_{1} q_{2} | KQ \rangle Y_{k_{1}}^{q_{1}} \left( \hat{r}_{1} \right) Y^{q_{2}}_{k_{2}} \left( \hat{r}_{2} \right)
\end{equation}

The basic idea is to make repeated use of the formula

\begin{equation}
Y^{m_{1}}_{l_{1}} \left( \hat{r}_{1} \right) Y^{m_{2}}_{l_{2}} \left( \hat{r}_{1} \right) = \sum_{l m} \left( \frac{\left( 2l_{1}+1 \right) \left(2l_{2}+1 \right) \left(2l+1 \right)}{4 \pi} \right)^{1/2}
\times \left( \begin{array}{ccc}
l_1 & l_2 & l \\
m_1 & m_2 & m
\end{array} \right) \left(\begin{array}{ccc}
l_1 & l_2 & l \\
0 & 0 & 0
\end{array} \right)
Y^{m*}_{l} \left( \hat{r}_{1} \right)
\end{equation}

where

\begin{equation}
Y^{m*}(\hat{r}) = (-1)^mY_l^{-m}(\hat{r})
\end{equation}

and

\begin{equation}
\left(\begin{array}{ccc}
l_1 & l_2 & l \\
m_1 & m_2 & m
\end{array} \right) = \frac{\left( -1 \right)^{l_{1} - l_{2} - m}}{\left( 2l + 1 \right)^{1/2}} \left( l_{1} l_{2} m_{1} m_{2} | l, -m \right)
\end{equation}

is a 3-<math>j</math> symbol. In particular, write

\begin{array}{cc}
Y^{m_1^\prime}_{l_1^\prime}(\hat{r}_1)^{*} \underbrace{Y_{k_{1}}^{q_{1}} \left( \hat{r}_{1} \right) Y^{m_{1}}_{l_{1}} \left( \hat{r}_{1} \right)} & = & \sum_{\Lambda M}(\ldots)Y^{M}_\Lambda(\hat{r}_1)\\
\ \ \ \ \ \ \ \sum_{\lambda_1\mu_1}Y_{\lambda_1}^{\mu_1}(\hat{r}_1) & & \\
& & \\
Y^{m_{2}^{\prime}}_{l_{2}^{\prime}} \left( \hat{r}_{2} \right) \underbrace{Y^{q_{2}}_{k_{2}} \left( \hat{r}_{2} \right) Y_{l_{2}}^{m_{2}} \left( \hat{r}_{2} \right)} & = & \sum_{\Lambda^{\prime} M^{\prime}} \left( \ldots \right)Y_{\Lambda^{\prime}}^{M^{\prime}*} \left( \hat{r}_{2} \right) \\
\ \ \ \ \ \ \ \ \ \sum_{\lambda_{2} \mu_{2}}Y_{\lambda_{2}}^{\mu_{2}} \left( \hat{r}_{2} \right) & &
\end{array}

The angular integral then gives a factor of <math>2 \pi \delta_{\Lambda, \Lambda^{\prime}} \delta_{M, M^{\prime}} P_{\Lambda} \left( \cos \theta_{12} \right)</math>. The total integral therefore reduces to the form

\begin{equation}
I = \sum_{\Lambda} C_{\Lambda} I_{\Lambda} \left( R_{1} R_{2} \right)
\end{equation}

where <math>C_{\Lambda} = \sum_{\lambda_{1}, \lambda_{2}} C_{ \lambda_{1}, \lambda_{2}, \Lambda}</math>. For further details and derivations, including graphical representations, see G.W.F. Drake, Phys. Rev. A <math>{\bf 18}</math>, 820 (1978).

==Graphical Representation==

[[File:GraphRep1.jpg|500px]]

==Matrix Elements of H==
Recall that

\begin{equation}
H = -\frac{1}{2} \nabla^{2}_{1} - \frac{1}{2} \nabla^{2}_{2} - \frac{1}{r_1} - \frac{1}{r_2} + \frac{Z^{-1}}{r_{12}}
\end{equation}

Consider matrix elements of

\begin{equation}
\nabla^2_1 = \frac{1}{r_1^2} \frac{\partial}{\partial r_1}\left(r_1^2\frac{\partial}{\partial r_1} \right) + \frac{1}{r^2}\frac{\partial}{\partial r}\left(r^2\frac{\partial}{\partial r}\right) - \frac{(\vec{l}_1^Y)^2}{r_1^2} + \frac{2 \left( r_1 - r_2 \cos \theta \right)}{r} \frac{\partial^2}{\partial r_1 \partial r} - 2 \left( \nabla^Y_1 \cdot {\bf r}_2 \right) \frac{1}{r} \frac{\partial}{\partial r}
\end{equation}

where <math>r \equiv r_{12}</math> and <math>\cos\theta \equiv \cos\theta_{12}</math>. Also in what follows, define <math>\hat{\nabla}^Y_1 \equiv r_1\nabla^Y_1</math> where <math>\nabla^Y</math> operates only on the spherical harmonic part of the wave function, and similarly for <math>\vec{l}_1^Y</math>.

A general matrix element is

\begin{equation}
\langle r_1^{a^\prime} r_2^{b^\prime} r_{12}^{c^\prime} e^{-\alpha^\prime r_1 - \beta^\prime r_2} \mathcal{Y}^{M^\prime}_{l^\prime_1 l^\prime_2 L^\prime} \left( \hat{r}_1, \hat{r}_2 \right) \left| \nabla^2_1 \right| r_1^a r_2^b r_{12}^c e^{-\alpha r_1 - \beta r_2} \mathcal{Y}^M_{l_1 l_2 L} \left( \hat{r}_1, \hat{r}_2 \right) \rangle
\end{equation}

Since <math>\nabla^2_1</math> is rotationally invariant, this vanishes unless <math>L=L^\prime</math>, <math>M=M^\prime</math>. Also <math>\nabla_1^2</math> is Hermitian, so that the result must be the same whether it operates to the right or left, even though the results look very different. In fact, requiring the results to be the same yields some interesting and useful integral identities as follows:

Put

\begin{equation}
\mathcal{F} = F \mathcal{Y}^M_{l_1 l_2 L} \left( \hat{r}_1, \hat{r_2} \right)
\end{equation}

and

\begin{equation}
F = r_1^a r_2^b r_{12}^c e^{-\alpha r_1 - \beta r_2}
\end{equation}

Then

\begin{equation}
\nabla^2_1 \mathcal{F} = \left\{ \frac{1}{r_1^2} \left[ a \left( a + 1 \right) - l_1 \left( l_1 + 1 \right) \right] + \frac{c \left( c + 1 \right)}{r^2} + \alpha^2\frac{2\alpha \left( a + 1 \right)}{r_1} + \frac{2 \left(r_1 - r_2 \cos \theta \right) }{r_1r^2}c \left[ a - \alpha r_1 \right] - \frac{2c}{r^2} \left( \hat{\nabla}^Y_1 \cdot \hat{r}_2 \right) \frac{r_2}{r_1} \right\} \mathcal{F}
\end{equation}

and

\begin{align}
\langle \mathcal{F}^\prime \left| \nabla^2_1 \right| \mathcal{F} \rangle & = \sum_\Lambda \int d\tau_r F^\prime C_\Lambda \left( 1 \right) P_\Lambda \left( \cos \theta \right) \times \left\{ \frac{1}{r_1^2} \left[ a \left( a + 1 \right) - l_1 \left( l_1 + 1 \right) \right] -\frac{2 \alpha \left( a + 1 \right)}{r_1} + \frac{c \left( c + 1 \right)}{r^2} +\alpha^2 + \frac{2 \left( r_2 - r_2 \cos \theta \right)}{r_1r^2}c \left( a - \alpha r_1 \right) \right\} F \\
& + \sum_\Lambda \int d\tau_r F^\prime C_\Lambda \left( \hat{\nabla}^Y_1 \cdot \hat{r}_2 \right) P_\Lambda \left( \cos \theta \right) \left( \frac{-2c r_2}{r_1 r^2} \right) F
\end{align}

where

\begin{equation}
\int d\tau_r = \int^\infty_0 r_1 dr_1 \int^\infty_0 r_2 dr_2 \int^{r_1+r_2}_{\left| r_1 - r_2 \right|} r dr
\end{equation}

For brevity, let the sum over <math>\Lambda</math> and the radial integrations be understood, and let <math>\left( \nabla_1^2 \right)</math> stand for the terms which appear in the integrand. Then operating to the right gives

\begin{equation}
(\nabla_1^2)_R = \frac{1}{r_1^2} \left[ a \left( a + 1 \right) - l_1 \left( l_1 + 1 \right) \right] + \frac{c \left( c + 1 \right)}{r^2} + \alpha^2 - \frac{2 \alpha \left( a + 1 \right)}{r_1}
+ \frac{2 \left( r_1 - r_2 \cos \theta \right)}{r_1 r^2}c \left( a - \alpha r_1 \right) - \frac{2c r_2}{r_1 r^2} \left( \hat{\nabla}^Y_1 \cdot \hat{r}_2 \right)
\end{equation}

Operating to the left gives

\begin{equation}
(\nabla_1^2)_L = \frac{1}{r_1^2} \left[a^\prime \left( a^\prime + 1 \right) - l_1^\prime \left( l_1^\prime + 1 \right) \right] + \frac{c^\prime \left( c^\prime + 1 \right)}{r^2} + \alpha^{\prime 2} -\frac{2 \alpha^\prime \left( a^\prime + 1 \right)}{r_1}
+ \frac{2 \left( r_1 - r_2 \cos \theta \right)}{r_1 r^2} c^\prime \left( a^\prime - \alpha^\prime r_1 \right) - \frac{2 c^\prime r_2}{r_1 r^2} \left( \hat{\nabla}^{Y^\prime}_1 \cdot \hat{r}_2 \right)
\end{equation}

Now put

\begin{array}{cc}
a_+ = a+a^\prime, & \hat{\nabla}_1^+ = \hat{\nabla}_1^Y + \hat{\nabla}_1^{Y\prime} \\
a_{-} = a-a^\prime, & \hat{\nabla}_1^{-} = \hat{\nabla}_1^Y - \hat{\nabla}_1^{Y\prime}
\end{array}

etc., and substitute <math>a^\prime = a_+ - a</math>, <math>c^\prime = c_+ - c</math>, and <math>\alpha^\prime = \alpha_+ - \alpha</math> in <math>(\nabla^2_1)_L</math>. If <math>a_+</math>, <math>c_+</math> and <math>\alpha_+</math> are held fixed, then the equation

\begin{equation}
(\nabla^2_1)_R = (\nabla^2_1)_L
\end{equation}

must be true for arbitrary <math>a</math>, <math>c</math>, and <math>\alpha</math>. Their coefficients must thus vanish.

This yields the integral relations

\begin{equation}
\label{I}
\frac{\left( r_1 - r_2 \cos \theta \right)}{r_1 r^2} = \frac{1}{c_+} \left( \frac{- \left( a_+ + 1 \right)}{r_1^2} + \frac{\alpha_+}{r_1} \right) \tag{I}
\end{equation}

from the coefficient of <math>a</math> and

\begin{equation}
\label{II}
\frac{ \left( r_1 - r_2 \cos \theta \right) \left( a_+ - \alpha_+ r_1 \right)}{r_1 r^2} = \frac{r_2}{r_1 r^2}\left( \hat{r}_2 \cdot \hat{\nabla}_1^+ \right) - \frac{ \left( c_+ + 1 \right)}{r^2} \tag{II}
\end{equation}

from the coefficient of <math>c</math>. The coefficient of <math>\alpha</math> gives an equation equivalent to (I).

Furthermore, if can be show that (see problem)

\begin{align}
& \sum_\Lambda \int d\tau_r \frac{r^c}{r^2} C_\Lambda \left( \hat{r}_2 \cdot \hat{\nabla}_1^Y \right) P_\Lambda \left( \cos \theta \right) \\
= & \sum_\Lambda \int d\tau_r \frac{r^c}{c r_1 r_2}C_\Lambda \left( 1 \right)P_\Lambda \left( \cos \theta \right) \times \left( \frac{l_1^\prime \left( l_1^\prime + 1 \right) - l_1 \left( l_1 + 1 \right) -\Lambda \left( \Lambda + 1 \right)}{2} \right)
\end{align}

and similarly for <math>\left( \hat{r}_2 \cdot \hat{\nabla}^{Y\prime}_1 \right)</math> with <math>l_1</math> and <math>l_1^\prime</math> interchangeable, then it follows that

\begin{equation}
\frac{r^{c_+}}{r^2} \left( \hat{r}_2 \cdot \hat{\nabla}_1^+ \right) = - \frac{r^{c_+}}{c_+ r_1 r_2} \Lambda \left( \Lambda + 1 \right)
\end{equation}

and

\begin{equation}
\frac{r^{c_+}}{r^2} \left( \hat{r}_2 \cdot \hat{\nabla}_1^- \right) = \frac{r^{c_+}}{c_+ r_1 r_2} \left[ l_1^\prime \left( l_1^\prime + 1 \right) - l_1 \left( l_1 + 1 \right) \right]
\end{equation}

where equality applies after integration and summation over <math>\Lambda</math>.

Thus (II) becomes

\begin{equation}\label{IIa}
\frac{ \left( r_1 - r_2 \cos \theta \right) \left( a_+ - \alpha_+ r_1 \right)}{r_1 r^2} = - \frac{\Lambda \left( \Lambda + 1 \right)}{c_+ r^2_1} - \frac{ \left( c_+ + 1 \right)}{r^2} \tag{IIa}
\end{equation}

===Problem===

Prove the integral relation

\begin{align}
& \sum_\Lambda \int d\tau_r f \left( r_1, r_2 \right) \frac{1}{r} \left( \frac{d}{dr} q(r) \right) C_\Lambda \left( \hat{r}_2 \cdot \hat{\nabla}^Y_1 \right) P_\Lambda \left( \cos \theta \right) \\
= & \sum_\Lambda \int d\tau_r \frac{ f \left( r_1, r_2 \right)}{r_1, r_2} q(r) C_\Lambda \left( 1 \right) P_\Lambda \left( \cos \theta \right) \left(\frac{l_1^\prime \left( l_1^\prime + 1 \right) - l_1 \left( l_1 + 1 \right) - \Lambda \left( \Lambda + 1 \right)}{2}\right)
\end{align}

where the coefficients <math>C_\Lambda \left( 1 \right)</math> are the angular coefficients from the overlap integral

\begin{align}
\int & d\Omega\, \mathcal{Y}^{M*}_{l_1^\prime l_2^\prime L}(\hat{r}_1,\hat{r}_2)\, \mathcal{Y}^{M}_{l_1 l_2 L}(\hat{r}_1,\hat{r}_2)\\
= & \sum_\Lambda C_\Lambda(1)P_\Lambda(\cos\theta_{12}).
\end{align}

Hint: Use the fact that <math>l_1^2</math> is Hermitian so that
<math>
\int d\tau (l_1^2\mathcal{Y}^\prime)^*q(r)\mathcal{Y} = \int d\tau
\mathcal{Y}^{\prime *}l_1^2(q(r)\mathcal{Y})
</math>
with <math>\ \vec{l}_1= \frac{1}{i}\vec{r}_1\times\nabla_1</math>.

It is also useful to use
<math>\begin{eqnarray}\ \
\cos\theta\, P_L(\cos\theta) &=& \frac{1}{2L+1}[L\,P_{L-1}(\cos\theta) + (L+1)P_{L+1}(\cos\theta)]
\\
(\cos^2\theta-1)P_L(\cos\theta) &=&
\frac{L(L-1)}{(2L-1)(2L+1)}P_{L-2}(\cos\theta)\nonumber
-\frac{2(L^2+L-1)}{(2L-1)(2L+3)}P_L(\cos\theta) + \frac{(L+1)(L+2)}{(2L+1)(2L+3)}P_{L+2}(\cos\theta)\nonumber\\
&=& \frac{L(L-1)}{(2L-1)(2L+1)}[P_{L-2}(\cos\theta) - P_L(\cos\theta)]
+ \frac{(L+1)(L+2)}{(2L+1)(2L+3)}[P_{L+2}(\cos\theta) - P_L(\cos\theta)]
\end{eqnarray}</math>

together with a double application of the integral recursion relation
<math>
I_{L+1}\left(\frac{1}{r}\frac{d}{dr}q(r)\right) =
(2L+1)I_L\left(\frac{1}{r_1r_2}q(r)\right)+I_{L-1}\left(\frac{1}{r}\frac{d}{dr}q(r)\right)
</math>

Of course <math>l_1^2\mathcal{Y}^M_{l_1l_2L}(\hat{r}_1,\hat{r}_2)=l_1(l_1+1)\mathcal{Y}^M_{l_1l_2L}(\hat{r}_1,\hat{r}_2)</math>. Begin by expanding
<math>
l_1^2(g\mathcal{Y}) = \mathcal{Y}l_1^2g + g\,l_1^2\mathcal{Y} + 2(\vec{l}_1g)\cdot(\vec{l}_1\mathcal{Y})
</math>, and show that

<math>
\begin{eqnarray}\ \
(\vec{l}_1g)\cdot(\vec{l}_1\mathcal{Y}) & = & \frac{r_1^2}{r}\frac{dg}{dr}\vec{r}_2\cdot \nabla_1\mathcal{Y}\,\,\,\,\mbox{ and}\\
l_1^2g(r) &=& 2r_1r_2\cos\theta\frac{1}{r}\frac{dg}{dr} + r_1^2r_2^2(\cos^2\theta - 1)
\frac{1}{r}\frac{d}{dr}\left(\frac{1}{r}\frac{dg}{dr}\right).
\end{eqnarray}
</math>

The proof amounts to showing that the term <math>l_1^2g(r)</math> can be replaced by
<math>\Lambda(\Lambda+1)g(r)</math> after multiplying by <math>P_\Lambda(\cos\theta)</math>
and integrating by parts with respect to the radial integrations over <math>r_1</math>, <math>r_2</math> and
<math>r \equiv |\vec{r}_1 - \vec{r}_2|</math>. Remember that <math>\cos\theta</math> is just a short-hand notation for the radial function <math>(r_1^2 + r_2^2 - r^2)/(2r_1r_2)</math>.

==General Hermitian Property==

Each combination of terms of the form
\begin{equation}
<f>_R = a^2f_1+af_2+abf_3+b^2f_4+bf_5+af_6\nabla_1^Y+bf_7\nabla^Y_2+f_8(Y)
\end{equation}
can be rewritten
\begin{equation}
<f>_L= (a_+-a)^2f_1+(a_+-a)f_2 + (a_+-a)(b_+-b)f_3+(b_+-b)^2f_4
+(b_+-b)f_5+(a_+-a)f_6\nabla^{Y\prime}_1+(b_+-b)f_7\nabla^{Y\prime}_2+f_8)Y^\prime
\end{equation}
where as usual <math>\nabla_1^Y</math> acts only on the spherical harmonic part of the wave function denoted for short by <math>Y</math>.
Since these must be equal for arbitrary <math>a</math> and <math>b</math>,
\begin{equation}
a_+^2f_1+a_+f_2+_+b_+f_3+b_+f_4+b_+f_5+a_+f_6\nabla^{Y\prime}_1 +
b_+f_7\nabla^{Y\prime}_2+f_8(Y^\prime)-f_8(Y)=0
\end{equation}

Adding the corresponding expression with <math>Y</math> and <math>Y^\prime</math> interchanged
yields
<math>\label{III}
a_+^2f_1+a_+f_2+a_+b_+f_3+b_+^2f_4+b_+f_5+\frac{1}{2}a_+f_6\nabla^+_1+\frac{1}{2}b_+f_7\nabla^+_2=0\tag{III}
</math>

Subtracting gives
<math>\label{IV}
f_8(Y)-f_8(Y^\prime)=-\frac{1}{2}[a_+f_6\nabla^-_1+b_+f_7\nabla_2^-]\tag{IV}
</math>
<math>\label{V}
a[-2a_+f_1-2f_2-b_+f_3-f_6\nabla^+_1]=0\tag{V}
</math>
<math>\label{VI}
b[-2b_+f_4-2f_5-a_+f_3-f_7\nabla^+_2]=0\tag{VI}
</math>

Adding the two forms gives

\begin{eqnarray}
<f>_R+<f>_L=\frac{1}{2}(a_+^2+a_-^2)f_1+a_+f_2+\frac{1}{2}(a_+b_++a_-b_-)f_3\nonumber\\
\ +\frac{1}{2}(b_+^2+b_-^2)f_4+b_+f_5+\frac{1}{2}f_6(a_+\nabla_1^++a_-\nabla^-_1)\nonumber\\
\ +\frac{1}{2}(b_+\nabla_2^++b_-\nabla^-_2)+f_8(Y)+f_8(Y^\prime)\nonumber
\end{eqnarray}

Subtracting <math>\frac{x}{2}\times({\rm I})</math> gives

\begin{eqnarray}\ \
<f>_R+<f>_L = \frac{1}{2}[(1-x)a_+^2 + a_-^2]f_1 +
(1-\frac{x}{2})a_+f_2+\frac{1}{2}[(1-x)a_+b_++a_-b_-]f_3\nonumber\\
+\frac{1}{2}[(1-x)b_+^2+b_-^2]f_4+(1-\frac{x}{2})f_5b_++\frac{1}{2}f_6[(1-\frac{x}{2})a_+\nabla_1^++a_-\nabla^-_1]\nonumber\\
+\frac{1}{2}[(1-\frac{x}{2})b_+\nabla^+_2+b_-\nabla^-_2]+f_8(Y)+f_8(Y^\prime)\nonumber
\end{eqnarray}

If <math>x=1</math>,
\begin{eqnarray}
<f>_R+<f>_L &=&
\frac{1}{2}[a_-^2f_1+a_+f_2+a_-b_-f_3+b_-^2f_4+b_+f_5\nonumber\\
&+&\left(\frac{1}{2}a_+\nabla^+_1+a_-\nabla^-_1\right)f_6+\left(\frac{1}{2}b_+\nabla^+_2+b_-\nabla^-_2\right)f_7]+f_8(Y)+f_8(Y^\prime)\nonumber
\end{eqnarray}

The General Hermitian Property for arbitrary <math>x</math> gives
\begin{eqnarray}\ \
\nabla_1^2=
\frac{1}{4}\{\frac{1}{r_1^2}[(1-x)a_+^2+a_-^2+2(1-\frac{x}{2})a_+-2[l_1(l_1+1)+l_1^\prime(l_1^\prime+1)]\nonumber\\
-\frac{2}{r_1}[(1-x)\alpha_+a_++\alpha_-a_-+2(1-\frac{x}{2})\alpha_+]+(1-x)\alpha_+^2
+ \alpha^2_-]\nonumber\\
+\frac{2(r_1-r_2\cos\theta)}{r_1r^2}[(1-x)(a_+-\alpha_+r_1)c_++(a_--\alpha_-r_2)c_-]\nonumber\\
-\frac{2r_2}{r_1r^2}[(1-\frac{x}{2})c_+\hat{r}_2\cdot\hat{\nabla}_1^++c_-\hat{r}_2\hat{\nabla}_1^-]+\frac{(1-x)c_+^2+c_-^2+2(1-\frac{x}{2})c_+}{r^2}\}\nonumber
\end{eqnarray}.

Use
<math>\begin{eqnarray}\ \
\frac{-2r_2}{r_1r^2}[(1-\frac{x}{2})c_+\hat{r}_2\hat{\nabla}_1^++c_-\hat{r}_2\hat{\nabla}_1^-]=\frac{2}{r_1^2}\left[(1-\frac{x}{2})\Lambda(\Lambda+1)-\frac{c_-}{c_+}[l_1^\prime(l_1^\prime+1)+l_1(l_1+1)]\right]\end{eqnarray}</math>,

<math>\begin{eqnarray}
\frac{2(r_1-r_2\cos\theta)}{r_1r^2}(a_--\alpha_-r_1)c_- = 2\frac{c_-}{c_+}\left[\frac{-1}{r_1^2}[a_-(a_++1)]+\frac{1}{r_1}[a_-\alpha_++\alpha_-(a_++2)]-\alpha_-\alpha_+\right]\nonumber
\end{eqnarray}</math>,

and
<math>\begin{eqnarray}\ \
\frac{2(r_1-r_2\cos\theta)}{r_1r^2}(a_+-\alpha_+r_1)c_+ = 2\left[-\frac{1}{r_1^2}[a_+(a_++1)]+\frac{1}{r_1}[a_+\alpha_++\alpha_+(a_++2)]-\alpha_+^2\right]
\end{eqnarray}</math>.

Substituting into <math>\nabla_1^2</math> gives

\begin{eqnarray}
\nabla^2_1 = \frac{1}{4}\{\frac{1}{r_1^2}[-(1-x)a_+^2 + a_-^2 +xa_+ +
2(1-\frac{x}{2})\Lambda(\Lambda+1)\nonumber\\
-2l_1(l_1+1)(1-\frac{c_-}{c_+}) -
2l_1^\prime(l_1^\prime+1)(1+\frac{c_-}{c_+})-\frac{2c_-a_-}{c_+}(a_++1)]\nonumber\\
- \frac{2}{r_1}\left[-(1-x)\alpha_+(a_++2)+\alpha_-a_-+2(1-\frac{x}{2})\alpha_+
-\frac{c_-}{c_+}[a_-\alpha_++\alpha_-(a_++2)]\right]\nonumber\\
-(1-x)\alpha_+^2+\alpha_-^2-2\frac{c_-}{c_+}\alpha_-\alpha_+ +\frac{1}{r^2}\left[(1-x)c_+^2+c_-^2+2(1-\frac{x}{2})c_+\right]\}\nonumber
\end{eqnarray}

This has the form
<math>
\nabla_1^2 =
\frac{1}{4}\left[\frac{A_1}{r_1^2}+\frac{B_1}{r_1}+\frac{C_1}{r^2}+D_1\right]
</math>
with

<math>\begin{array}{ccc}
A_1=-(1-x)a_+^2+a_-^2
+xa_++2(1-\frac{x}{2})\Lambda(\Lambda+1)-2l_1(l_1+1)(1-\frac{c_-}{c_+})\
-2l_1^\prime(l_1^\prime+1)(1+\frac{c_-}{c_+})-\frac{2c_-a_-}{c_+}(a_++1) \ \
\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \
\end{array}
</math>

<math>
B_1 =
2\left[(1-x)\alpha_+(a_++2)-\alpha_-a_--2(1-\frac{x}{2})\alpha_++\frac{c_-}{c_+}[a_-\alpha_++\alpha_-(a_++2)]\right]
</math>

<math>
C_1 = (1-x)c_+^2+ c_-^2 +2(1-\frac{x}{2})c_+
</math>

<math>
D_1 = -1(1-x)\alpha_+^2 +\alpha_-^2 - 2\frac{c_-}{c_+}\alpha_-\alpha_+
</math>

The complete Hamiltonian is then (in Z-scaled a.u.)

<math>\begin{eqnarray}
H =-\frac{1}{2}\nabla^2_1 -\frac{1}{2}\nabla^2_2 -\frac{1}{r_1} -\frac{1}{r_2} +\frac{Z^{-1}}{r}\nonumber\\
= -\frac{1}{2}\nabla^2_1 -\frac{1}{2}\nabla^2_2 -\frac{1}{r_1} -\frac{Z-1}{Zr_2}+Z^{-1}\left(\frac{1}{r}-\frac{1}{r_2}\right)\nonumber\\
\ =-\frac{1}{8}\left[\frac{A_1}{r_1^2}+\frac{B_1+8}{r_1}+\frac{C_1}{r^2}+D_1+D_2 +\frac{A_2}{r_2^2} + \frac{B_2 + 8(Z-1)/Z}{r_2}+\frac{C_2}{r^2}\right]\nonumber
+ Z^{-1}\left(\frac{1}{r}-\frac{1}{r_2}\right)\nonumber
\end{eqnarray}</math>

The screened hydrogenic energy is
<math>\begin{eqnarray}
E_{SH}=-\frac{1}{2}\left[\frac{1}{n_1^2}+ \left(\frac{Z-1}{Z}\right)^2\frac{1}{n_2^2}\right]
\end{eqnarray}</math>
so that

<math>\begin{eqnarray}
H-E_{SH} &=& -\frac{1}{8}[
\frac{A_1}{r_1^2}+\frac{B_1+8}{r_1}+\frac{C_1}{r^2} + D_1 + D_2
-\frac{4}{n_1^2} -\left(\frac{Z-1}{Z}\right)^2\frac{4}{n_2^2}\nonumber
+\frac{A_2}{r_2}+\frac{B_2+8(Z-1)/Z}{r_2}+\frac{C_2}{r^2}] +
Z^{-1}\left(\frac{1}{r} -\frac{1}{r_2}\right)\nonumber
\end{eqnarray}</math>

==Optimization of Nonlinear Parameters==
- The traditional method of performing Hylleraas calculations is to write the
basis set in the form

<math>
\Psi = \sum_{i,j,k}c_{i,j,k}\varphi_{i,j,k}(\alpha,\beta)\pm exchange
</math>

with

<math>
\varphi_{i,j,k}(\alpha,\beta) = r_1^ir_2^jr_{12}^ke^{-\alpha r_1 - \beta
r_2}\mathcal{Y}^M_{l_1l_2L}.
</math>

The usual procedure is to set <math>\alpha = Z</math> so that it represents the inner
<math>1s</math> electron, and then to vary <math>\beta</math> so as to minimize the energy.

[[File:OptNon.jpg|500px]]

Since <math>\beta</math> appears in <math>\Psi</math> as a non-linear parameter, the entire
calculation must be repeated for each value of <math>\beta</math>. However, the minimum
becomes progressively smaller as the basis set is enlarged.

[[File:OptNon02.jpg|500px]]

*Difficulties

1. If the basis set is constructed so that <math> i +j +j \leq N</math>, the the
number of terms is
<math>
(N+1)(N+2)(N+3)/6
</math>
N=14 already gives 680 terms and an accuracy of about 1 part in <math>10^{10}</math> for
low-lying states. A substantial improvement in accuracy would require much
larger basis sets, together with multiple precision arithmetic to avoid loss
of significant figures when high powers are included.

2. The accuracy rapidly deteriorates as one goes to more highly excited
states - about 1 significant figure is lost each time the principle quantum
number is increased.

*Cure

We have found that writing basis sets in the form
\begin{eqnarray}
\Psi &=& \sum_{i,j,k}\left(c^{(1)}_{i,j,k}\varphi_{ijk}(\alpha_1\beta_1) +
c_{ijk}^{(2)}\varphi_{ijk}(\alpha_2\beta_2)\right) \pm exchange \nonumber\\
&=& \psi({\bf r}_1,{\bf r}_2) \pm \psi({\bf r}_2,{\bf r}_1)\nonumber
\end{eqnarray}
so that each combination of powers is included twice with different nonlinear
parameters gives a dramatic improvement in accuracy for basis sets of about
the same total size.

However, the optimization of the nonlinear parameters is now much more
difficult, and an automated procedure is needed.

If
<math>
E = \frac{<\Psi|H|\Psi>}{<\Psi|\Psi>}</math> and we assume <math><\Psi|\Psi> = 1</math>, then

<math>
\frac{\partial E}{\partial\alpha_t} = -2 <\Psi|H-E|r_1\psi({\bf r}_1,{\bf
r}_2;\alpha_t)\pm r_21\psi({\bf r}_2,{\bf
r}_1;\alpha_t)>
</math>
where
<math>
\psi({\bf r}_1,{\bf r}_2;\alpha_t) =
\sum_{i,j,k}c_{ijk}^{(t)}\varphi_{ijk}(\alpha_t\beta_t)
</math>

*Problem

1. Prove the above.

2. Prove that there is no contribution from the implicit dependence of
<math>c_{ijk}^{(t)}</math> on <math>\alpha_t</math> if the linear parameters have been optimized.

==The Screened Hydrogenic Term==

If the Hamiltonian is written in the form (in Z-scaled a.u.)
<math>\begin{eqnarray}\
H = H_0({\bf r}_1,Z) + H_0({\bf r}_2,Z-1) +
Z^{-1}\left(\frac{1}{r}-\frac{1}{r_2}\right)
\end{eqnarray}</math>
with
<math>\
H_0({\bf r}_1,Z) = -\frac{1}{2}\nabla_1^2 - \frac{1}{r_1}\nonumber
</math> and
<math>
H_0({\bf r}_2,Z-1) = -\frac{1}{2}\nabla_2^2 - \frac{Z-1}Z{r_2}\nonumber
</math>,
then the eigenvectors of <math>H_0({\bf r},Z) +H_0({\bf r}_2,Z-1)</math> are products of
hydrogenic orbitals
<math>
\Psi_0=\psi_0(1s,Z)\psi_0(nl,Z-1)
</math>
and the eigenvalue is
<math>\begin{eqnarray}
E_{SH} = \left[-\frac{1}{2} - \left(\frac{Z-1}{Z}\right)^2\frac{1}{2n^2}\right]Z^2 a.u.
\end{eqnarray}</math>
called the screened hydrogenic eigenvalue. From highly excited states,
<math>E_{SH}</math> and <math>\Psi_0</math>are already excellent approximations.

For example, for the 1s8d states, the energies are

<math>\begin{eqnarray}\\
E(1s8d^1D)&=& -2.007\,816\,512\,563\,81 {\rm\ a.u.}\nonumber\\
E(1s8d^3D)&=& - 2.007\,817\,934\,711\,71 {\rm\ a.u.}\nonumber\\
E_{SH} &=& -2.007\,812\,5 {\rm\ a.u.} \nonumber
\end{eqnarray}</math>

It is therefore advantageous to include the screened hydrogenic terms in the
basis set so that the complete trial function becomes
<math>
\Psi = c_0\Psi_0 +
\sum_{ijk}\left[c_{ijk}^{(1)}\varphi_{ijk}(\alpha_1,\beta_1)+c_{ijk}^{(2)}\varphi(\alpha_2,\beta_2)\right]\pm
{\rm exchange}
</math>

Also, the variational principal can be re-expressed in the form
<math>\begin{eqnarray}
E= E_{SH}+\frac{<\Psi|H-E_{SH}|\Psi>}{<\Psi|\Psi>}
\end{eqnarray}</math>
so that the <math>E_{SH}</math> term can be cancelled analytically from the matrix
elements.

For example
<math>
<\Psi_0|H-E_{SH}|\Psi_0> = 0
</math>

and
<math>
<\varphi_{ijk}|H-E_{SH}|\Psi_0> =
Z^{-1}<\varphi_{ijk}|\frac{1}{r}-\frac{1}{r_2}|\Psi_0>
</math>

Recall that
<math>\begin{eqnarray}\ \
I_0(a,b,c) = \frac{2}{c+2}\sum^{[c+1]/2}_{i=0}\left(\begin{array}{c}
c+2\\
2i+1
\end{array}\right) \left[f(p,q;\beta) + f(p^\prime,q^\prime;\alpha)\right]
\end{eqnarray}</math>
where

<math>\begin{eqnarray}
f(p,q;x) =
\frac{q!}{x^{q+1}(\alpha+\beta)^{p+1}}\sum^p_{j=0}\frac{(p+j)!}{j!}\left(\frac{x}{\alpha+\beta}\right)^j
\end{eqnarray}</math>

<math>
\begin{array}{cc}
p=a+2i+2 & p^\prime = b+2i+2\\
q= b+c-2i+2 & q^\prime =a+c-2i+2
\end{array}
</math>

If <math>\beta\ll\alpha</math>, the <math>f(p,q;\beta)\gg f(p^\prime,q^\prime;\alpha)</math> and the
<math>i=0</math> term is the dominant contribution to <math>f(p,q;\beta)</math>. However, since
this term depends only on the sum of powers <math>b+c</math> for <math>r_2</math>, it cancels exactly
from the matrix element of
<math>
\frac{1}{r_{12}} - \frac{1}{r_2}
</math>
and can therefore
be omitted in the calculations of integrands, thereby saving many significant
figures. This is especially valuable when <math>b+c</math> is large and <math>a</math> is small.

==Small Corrections==
*Mass Polarization
If the nuclear mass is not taken to be infinite, then the Hamiltonian is
<math>\begin{eqnarray}\ \
H = \frac{P_N^2}{2M} + \sum_{i=1}\frac{p_i^2}{2m} + V
\end{eqnarray}</math>

[[File:SmallCor.jpg|500px]]

where <math>M =M_A-nm</math> is the nuclear mass and <math>m</math> is the electron mass.

Change variables to the C of M
<math>\begin{eqnarray}
{\bf R} = \frac{1}{M+nm}\left[M{\bf r}_N + m({\bf r}_1 + {\bf r}_2 +
\ldots)\right]
\end{eqnarray}</math>

and relative variables

<math>
{\bf s}_i = {\bf r}_i - {\bf r}_N.
</math>

Then

<math>\begin{eqnarray}\ \ \ \ \
H = \frac{1}{2(M+nm)}{\bf p}_R^2+\frac{1}{2m}\sum_{i=1}^n{\bf p}_{s_i}^2
+\frac{1}{2M}\sum_{i,k}{\bf p}_{s_i}\cdot {\bf p}_{s_k} + V({\bf s}_1, {\bf
s}_2, \ldots)\nonumber
= \frac{1}{M+nm}{\bf p}_R^2+\frac{1}{2\mu}\sum_{i=1}^n{\bf
p}_{s_i}^2+\frac{1}{2M}\sum_{i\neq k}{\bf p}_{s_i}\cdot {\bf p}_{s_k}+V({\bf s}_1, {\bf
s}_2, \ldots)\nonumber
\end{eqnarray}</math>

where
<math>\begin{eqnarray}
\mu = \left(\frac{1}{M} + \frac{1}{m}\right)^{-1}
\end{eqnarray}</math>

If <math>n=2</math>, the Schrödinger equation is
<math>\begin{eqnarray}\ \ \
\left[ -\frac{\hbar^2}{2\mu}(\nabla^2_{s_1}+ \nabla^2_{s_2}) -
\frac{\hbar^2}{M}\nabla_{s_1}\nabla_{s_2}-\frac{Ze^2}{s_1}-\frac{Ze^2}{s_2}+\frac{e^2}{s_{12}}\right]\psi
= E\psi
\end{eqnarray}</math>.

Define the reduced mass Bohr radius
<math>\begin{eqnarray}
a_\mu = \frac{\hbar^2}{\mu e^2} = \frac{m}{\mu}a_0
\end{eqnarray}</math>
and

<math>\begin{eqnarray}
\rho_i = \frac{Zs_i}{a_\mu}\nonumber\\
\varepsilon = \frac{E}{Z^2(e^2/a_\mu)}\nonumber
\end{eqnarray}</math>

The Schrödinger equation then becomes
<math>\begin{eqnarray}\ \ \
\left[
-\frac{1}{2}(\nabla^2_{\rho_1}+\nabla^2_{\rho_2})-\frac{\mu}{M}\vec{\nabla}_{\rho_1}\cdot\vec{\nabla}_{\rho_2} - \frac{1}{\rho_1}-\frac{1}{\rho_2}+\frac{Z^{-1}}{\rho_{12}}\right]\psi = \varepsilon \psi
\end{eqnarray}</math>

The effects of finite mass are thus

1. A reduced mass correction to all energy levels of <math>(e^2/a_\mu)/(e^2/a_0)
= \mu/m</math>, i.e.
<math>\begin{eqnarray}
E_M=\frac{\mu}{m}E_\infty
\end{eqnarray}</math>

2. A specific mass shift given in first order perturbation theory by
<math>\begin{eqnarray}
\Delta\varepsilon =
-\frac{\mu}{M}<\psi|\nabla_{\rho_1}\cdot\nabla_{\rho_2}|\psi>
\end{eqnarray}</math>
or

<math>\begin{eqnarray}
\Delta\varepsilon =
-\frac{\mu}{M}<\psi|\nabla_{\rho_1}\cdot\nabla_{\rho_2}|\psi>\frac{e^2}{a_\mu}\nonumber\\
= -\frac{\mu^2}{mM}<\psi|\nabla_{\rho_1}\cdot\nabla_{\rho_2}|\psi>\frac{e^2}{a_0}\nonumber
\end{eqnarray}</math>

Since <math>\nabla_{\rho_1}\cdot\nabla_{\rho_2}</math> has the same angular properties as
<math>{\bf\rho}_1 \cdot {\bf\rho}_2</math>, the operator is like the product of two
dipole operators. For product type wave functions of the form
$
\psi = \psi(1s)\psi(nl) \pm \mbox{exchange},
$
the matrix element vanishes for all but P-states.

=Radiative Transitions=

==Transition Integrals and the Line Strength==

For an electric dipole transition <math>\gamma'\,L'\,S'\,J'\,M' \rightarrow \gamma\,L\,S\,J\,M</math>, the line strength is defined by
\begin{equation}
S = \sum_{M',\mu,M}|\langle\gamma'\,L'\,S'\,J'\,M'|r_\mu|\gamma\,L\,S\,J\,M\rangle|^2
\end{equation}
Using the Wigner-Eckart theorem, this is related to the reduced matrix element by
\begin{eqnarray}
S(J',J) = |\langle\gamma'\,L'\,S'\,J'||r||\gamma\,L\,S\,J\,M\rangle|^2
\sum_{M',\mu,M}\left|\left(\matrix{J' & 1 & J \cr
-M'& \mu &M \cr}\right)\right|^2
\end{eqnarray}
Using the sum rule for the 3-<math>j</math> symbols
\begin{equation}
\sum_{M',\mu,M}\left|\left(\matrix{J' & 1 & J \cr
-M'& \mu &M \cr}\right)\right|^2 = 1
\end{equation}
we see that the line strength is identical to the reduced matrix element
\begin{equation}
S(J',J) = |\langle\gamma'\,L'\,S'\,J'||r||\gamma\,L\,S\,J\,M\rangle|^2
\end{equation}
For a nonrelativistic calculation in <math>LS</math> coupling, <math>S = S'</math> and we can next strip out the dependence on <math>J</math> and <math>J'</math> by use of Eq.\ (7.1.7) of Edmonds { Angular Momentum in Quantum Mechanics} with <math>k = 1</math> to obtain
\begin{equation}
S(J',J) = (-1)^{L'+S+J+1}[(2J+1)(2J'+1)]^{1/2}\left\{
\matrix{L' & J'& S\cr
J & L & 1\cr}\right\}\langle\gamma' L'|| r || \gamma L\rangle
\end{equation}
where <math>\langle\gamma' L'|| r || \gamma L\rangle</math> is the reduced matrix element if spin is neglected. Note that if spin is neglected, then the line-strength is simply
\begin{equation}
S(L',L) = |\langle\gamma' L'|| r || \gamma L\rangle|^2
\end{equation}
To calculate the reduced matrix element, use a particular example along with the Wigner-Eckart theorem
such as
\begin{equation}
\langle\gamma'\, L'|| r || \gamma\, L\rangle = (-1)^{L'}\langle\gamma'\, L'\, 0| z | \gamma\, L\, 0\rangle{\Large /}\left(\matrix{L' & 1 & L \cr
0 & 0 & 0 \cr}\right)
\end{equation}

===Correlated Two-electron Integrals===
For the case of two-electron integrals, <math>z = z_1 + z_2</math> and <math>r = r_1 + r_2</math>. As discussed in G.W.F. Drake, Phys. Rev. A 18, 820 (1978), the general integral in Hylleraas coordinates for the integral over the <math>z_1</math> term can be written in the form
\begin{equation}
\label{int1}
\langle\gamma'\, L'\, 0| z_1 | \gamma\, L\, 0\rangle = \sum_\Lambda C_\Lambda I_\Lambda(\gamma',\gamma)
\end{equation}
where <math>I_\Lambda(\gamma',\gamma)</math> stands for the purely radial integral
\begin{equation}
\label{radial}
I_\Lambda(\gamma',\gamma) = \int_0^\infty r_1\,dr_1\int_0^\infty r_2\,dr_2\int_{|r_1-r_2|}^{r_1+r_2}r_{12}\, dr_{12}
R_{\gamma'}^*\,r_1\, R_\gamma P_\Lambda(\cos\theta_{12})
\end{equation}
and <math>\cos\theta_{12} = (r_1^2 + r_2^2 - r_{12}^2)/(2r_1r_2)</math> is a short-hand notation for a radial function of <math>r_1</math>, <math>r_2</math> and <math>r_{12}</math>. Also, <math>P_\Lambda(\cos\theta_{12})</math> is a Legendre polynomial,
and, <math>R_\gamma = R_\gamma(r_1,r_2,r_{12})</math> is the radial part of the Hylleraas wave function for state <math>\gamma</math> such
that the total wave function is <math>\psi({\bf r}_1,{\bf r}_2) = R_\gamma(r_1,r_2,r_{12}){\cal Y}_{l_1,l_2,L}^M(\hat{\bf r}_1,\hat{\bf r}_2)\pm {\rm exchange}</math>, normalized to unity, and <math>{\cal Y}_{l_1,l_2,L}^M(\hat{\bf r}_1,\hat{\bf r}_2)</math> is a vector coupled product of spherical harmonics. The coefficents <math>C_\Lambda</math> in Eq.\ (\ref{int1}) can be calculated from Eq.\ (32) of the above 1978 paper. They need to be calculated separately for he direct-direct and the direct-exchange contributions. For the reduced matrix element of <math>z_1</math>, the leading 3-<math>j</math> symbol can be dropped, leaving (for the case <math>k = 1</math>)
\begin{eqnarray}
C_\Lambda^{\rm reduced} &=& \frac{(-1)^{l_2'+L'+L+1}}{2} (l_1,l_1',l_2,l_2',L,L')^{1/2}
\sum_{\lambda={\rm max}(l_1-1,0)}^{l_1+1} (\lambda,\Lambda,1)
\left(\matrix{l_1 & 1 & \lambda \cr
0 & 0 & 0 \cr}\right)
\left(\matrix{l_1'& \lambda & \Lambda \cr
0 & 0 & 0 \cr}\right)\nonumber\\
&\times&\left(\matrix{l_2' & 1_2 & \Lambda \cr
0 & 0 & 0 \cr}\right)
\left\{\matrix{L' & \lambda & l_2 \cr
\Lambda& l_2' & l_1' \cr}\right\}
\left\{\matrix{L' & \lambda & l_2 \cr
l_1 & L & 1 \cr}\right\}
\end{eqnarray}
where the notation <math>(a,b,c,\cdots)</math> means <math>(2a+1)(2b+1)(2c+1)\cdots</math>.
A similar expression can be obtained for the angular coefficients for the <math>z_2</math> term by interchanging the subscripts 1 and 2 throughout and multiplying by <math>(-1)^{L+L'+1}</math>.

===Example===
As an example, for the <math>1s^2\;^1S - 1s2p\;^1P</math> transition of helium, the only nonvanishing <math>C_\Lambda</math> angular coefficients are <math>C_1 = 1/2</math> for the direct-direct term (i.e. <math>P_1(\cos\theta_{12}) = \cos\theta_{12}</math> is included in the radial integral (\ref{radial})) and <math>C_0 = 1/2</math> for the direct-exchange term (i.e. <math>P_0(\cos\theta_{12}) = 1</math> is included in the radial integral (\ref{radial})). The results for infinite nuclear mass are
\begin{equation}
S(1\;^1S-2\;^1P) = (0.781\,127\,158\,667)^2
\end{equation}
\begin{equation}f(1\;^1S-2\;^1P) = -0.276\,164\,703\,736
\end{equation}
where
\begin{equation}
f(1\;^1S-2\;^1P) = \frac{2}{3}[E(1\;^1S) - E(2\;^1P)]S(1\;^1S-2\;^1P)
\end{equation}
is the absorption oscillator strength.

===Test===
This is a test edit

Theory Notes

2020-12-14T05:25:03Z

Drake: /* General Hermitian Property */

<center>
<H1>Notes on solving the Schrödinger equation in Hylleraas coordinates for heliumlike atoms
Gordon W.F. Drake
Department of Physics, University of Windsor
Windsor, Ontario, Canada N9B 3P4
(Transcribed from hand-written notes and edited by Lauren Moffatt. Last revised November 22, 2017 to correct a missing subscript in the expression for $B_1$ in the Hamiltonian in Sect. 11 General Hermitian Property.)
</H1>
</center>

==Introduction==
These notes describe some of the technical details involved in solving the Schrödinger equation for a heliumlike atom of nuclear charge Z in correlated Hylleraas coordinates. The standard for computational purposes is to express all quantities in a form that is valid for electronic states of arbitrary angular momentum, and to express matrix elements of the Hamiltonian in a manifestly Hermitian symmetrized form. Further information and derivations of the mathematical techniques for angular integrals and radial recursion relations can be found in the paper
G.W.F. Drake, Phys. Rev. A <math>{\bf 18}</math>, 820 (1978).

The starting point is the two-electron Schrödinger equation for infinite nuclear mass
\begin{equation} \left[-\frac{\hbar^2}{2m}(\nabla^2_1 +\nabla^2_2) - \frac{Ze^2}{r_1} -
\frac{Ze^2}{r_2}+\frac{e^2}{r_{12}} \right]\psi = E\psi\nonumber
\end{equation}
where <math>m</math> is the electron mass, and <math>r_{12} = |{\bf r}_1 - {\bf r}_2|</math> (see diagram below).

[[File:Pic1.jpg|250px]]

Begin by rescaling distances and energies so that the Schrödinger equation can be expressed in a dimensionless form. The dimensionless Z-scaled distance is defined by <math>\begin{eqnarray} \rho = \frac{Zr}{a_0} \end{eqnarray} </math> where <math> \begin{eqnarray} a_0 = \frac{\hbar^2}{me^2} \end{eqnarray} </math> is the atomic unit (a.u.) of distance equal to the Bohr radius <math>0.529\, 177\, 210\, 92(17) \times 10^{-10}</math> m. Then

\begin{equation}[-\frac{\hbar^2}{2m}Z^2(\frac{me^2}{\hbar^2})^2(\nabla^2_{\rho_1}+\nabla^2_{\rho_2})
- Z^2\frac{e^2}{a_0}\rho^{-1}_1 - Z^2\frac{e^2}{a_0}\rho^{-1}_2 +
\frac{e^2}{a_0}Z\rho^{-1}_{12}]\psi= E\psi\nonumber\end{equation}

But <math>\begin{eqnarray} \frac{\hbar^2}{m}(\frac{me^2}{\hbar^2})^2 = \frac{e^2}{a_0} \end{eqnarray} </math> is the hartree atomic unit of of energy <math>(E_h = 27.211\,385\,05(60)</math> eV, or equivalently <math>E_h/(hc) = 2194.746\,313\,708(11)</math> m<math>^{-1})</math>. Therefore, after multiplying through by <math>a_0/(Ze)^2</math>, the problem to be solved in Z-scaled dimensionless units becomes

\begin{equation} \left[-\frac{1}{2}(\nabla^2_{\rho_1}+\nabla^2_{\rho_2}) -
\frac{1}{\rho_1} - \frac{1}{\rho_2} + \frac{Z^{-1}}{\rho_{12}}\right]\psi =
\varepsilon\psi\nonumber \end{equation}

where <math>\begin{eqnarray}\varepsilon = \frac{Ea_0}{(Ze)^2} \end{eqnarray}</math> is the energy in Z-scaled atomic units. For convenience, rewrite this in the conventional form
\begin{equation}
H\psi = \varepsilon\psi\nonumber \end{equation}
where (using <math>r</math> in place of <math>\rho</math> for the Z-scaled distance)
\begin{equation}
H = -\frac{1}{2}(\nabla^2_1+\nabla^2_2) - \frac{1}{r_1}-\frac{1}{r_2} + \frac{Z^{-1}}{r_{12}}
\nonumber\end{equation}
is the atomic Hamiltonian for infinite nuclear mass.

==The Hartree Fock Method==

For purposes of comparison, and to define the correlation energy, assume that <math>\psi({\bf r}_1,{\bf r}_2)</math> can be written in the separable product form
\begin{equation}
\psi({\bf r}_1,{\bf r}_2) = \frac{1}{\sqrt{2}} \left[ u_1(r_1) u_2(r_2) \pm u_2(r_1) u_1(r_2) \right] \nonumber
\end{equation}
for the <math>1s^2\ ^1\!S</math> ground state. Because of the <math>\frac{1}{r_{12}}</math> term in the Schrödinger equation, the exact solution connot be expressed in this form as a separable product. However, the Hartree-Fock approximation corresponds to finding the best possible solution to the Schrödinger equation
\begin{equation}
H\psi({\bf r}_1,{\bf r}_2) = E\psi({\bf r}_1,{\bf r}_2)\nonumber
\end{equation}
that can nevertheless be expressed in this separable product form, where as before
\begin{equation}
H = -\frac{1}{2}(\nabla^2_1+\nabla^2_2) - \frac{1}{r_1}- \frac{1}{r_2} + \frac{Z^{-1}}{r_{12}}\nonumber
\end{equation}
is the full two-electron Hamiltonian. To find the best solution, substitute into <math>\left\langle\psi \lvert H-E \rvert \psi\right\rangle</math> and require this expression to be stationary
with respect to arbitrary infinitesimal variations <math>\delta u_1</math> and <math>\delta
u_2</math> in <math>u_1</math> and <math>u_2</math> respectively; i.e.

\begin{equation}
\frac{1}{2} \left\langle \delta u_1(r_1)u_2(r_2) \pm u_2(r_1)\delta
u_1(r_2) \lvert H-E \rvert u_1(r_1) u_2(r_2)\pm u_2(r_1) u_1(r_2)\right\rangle\nonumber
\end{equation}

\begin{equation}
=\int\delta u_1(r_1) d{\bf r}_1 \left\{ \int d{\bf r}_2 u_2(r_2) \left( H - E \right) \left[ u_1(r_1)u_2(r_2) \pm u_2(r_1) u_1(r_2) \right] \right\}\nonumber
\end{equation}

\begin{equation}
= 0 \nonumber
\end{equation}

for arbitrary <math> \delta u_1(r_1).\nonumber </math> Therefore <math>\{\int d{\bf r}_2 \ldots \} = 0</math>.

Similarly, the coefficient of <math>\delta u_2</math> would give

\begin{equation}
\int d{\bf r}_1 u_1(r_1) \left( H-E \right) \left[ u_1(r_1) u_2(r_2) \pm
u_2(r_1) u_1(r_2) \right] = 0\nonumber
\end{equation}

Define

\begin{equation}
I_{12} = I_{21} = \int d{\bf r}\, u_1(r)u_2(r), \nonumber
\end{equation}

\begin{equation}
H_{ij} = \int d{\bf r}\, u_i(r)(-\frac{1}{2}\nabla - \frac{1}{r})u_j(r), \nonumber
\end{equation}

\begin{equation}
G_{ij}(r) = \int d{\bf r}^\prime u_i(r^\prime)\frac{1}{|{\bf r} - {\bf r}\prime|}u_j(r^\prime)\nonumber
\end{equation}

Then the above equations become the pair of integro-differential equations

\begin{equation}
[ H_0 - E + H_{22}+G_{22}(r)]u_1(r) = \mp [ I_{12}(H_0-E) + H_{12}+G_{12}(r)]u_2(r)\nonumber
\end{equation}

\begin{equation}
[H_0-E+H_{11}+G_{11}(r)]u_2(r) = \mp [I_{12}(H_0-E) + H_{12}+G_{12}(r)]u_1(r)\nonumber
\end{equation}

These must be solved self-consistently for the "constants" <math>I_{12}</math> and
<math>H_{ij}</math> and the function <math>G_{ij}(r)</math>.

The Hartree Fock energy is <math>E \simeq -2.87\ldots </math>a.u. while the exact energy is <math>E = -2.903724\ldots </math>a.u.

The difference is called the "correlation energy" because it arises from the
way in which the motion of one electron is correlated to the other. The
Hartree Fock equations only describe how one electron moves in the average field
provided by the other (mean-field theory).

==Configuration Interaction==
Expand
\begin{equation}
\psi({\bf r}_1,{\bf r}_2)= C_0u^{(s)}_1(r_1)u^{(s)}_1(r_2) + C_1u^{(P)}_1({\bf r}_1)u^{(P)}_1({\bf r}_2){\cal Y}^0_{1,1,0}(\hat{\bf r}_1, \hat{\bf r}_2)+C_2u^{(d)}_1({\bf r}_1)u^{(d)}_2({\bf r}_2){\cal Y}^0_{2,2,0}(\hat{\bf r}_1, \hat{\bf r}_2)+... \pm {\rm\ exchange}
\end{equation}
where
\begin{equation}
{\cal Y}^M_{l_1,l_2,L}(\hat{\bf r}_1, \hat{\bf r}_2)=\sum_{m_1,m_2}Y^{m_1}_{l_1}({\bf r}_1)Y^{m_2}_{l_2}({\bf r}_2)\times <l_1l_2m_1m_2\mid LM>
\end{equation}

This works, but is slowly convergent, and very laborious. The best CI calculations are accurate to <math> ~10^{-7}</math> a.u.

==Hylleraas Coordinates==
[E.A. Hylleraas, Z. Phys. <math>{\bf 48}, 469(1928)</math> and <math>{\bf 54}, 347(1929)</math>]
suggested using the co-ordinates <math>r_1</math>, <math>r_2</math> and <math>r_{12}</math> or equivalently

\begin{eqnarray}
s &=& r_1 + r_2, \nonumber\\
t &=& r_1-r_2, \nonumber\\
u &=& r_{12}\nonumber
\end{eqnarray}

and writing the trial functions in the form

\begin{equation}
\Psi({\bf r}_1,{\bf r}_2) = \sum^{i+j+k\leq
N}_{i,j,k}c_{i,j,k}r_1^{i+l_1}r_2^{j+l_2}r_{12}^ke^{-\alpha r_1 - \beta r_2}
\mathcal{Y}^M_{l_1,l_2,L}(\hat{r}_1,\hat{r}_2)\pm \text{exchange}\nonumber
\end{equation}

Diagonalizing H in this non-orthogonal basis set is equivalent to solving
<math> \begin{eqnarray}
\frac{\partial E}{\partial c_{i,j,k}} = 0\nonumber
\end{eqnarray} </math>
for fixed <math>\alpha</math> and <math>\beta</math>.

The diagonalization must be repeated for different values of <math>\alpha</math> and
<math>\beta</math> in order to optimize the non-linear parameters.

==Completeness==
The completeness of the above basis set can be shown by first writing
\begin{equation}
r_{12}^2 = r_1^2 + r_2^2 - 2r_1r_2\cos\theta_{12}
\end{equation}
and, from the spherical harmonic addition theorem,
\begin{equation}
\cos(\theta_{12})=\frac{4\pi}{3}\sum^1_{m=-1}Y^{m*}_1(\theta_1,\varphi_1)Y^m_1(\theta_2,\varphi_2)
\end{equation}
Consider first the S-states. The <math>r_{12}^0</math> terms are like the ss terms in a CI calculation. The
<math>r_{12}^2</math> terms bring in p-p type contributions, and the higher powers bring
in d-d, f-f etc. type terms. In general

\begin{equation}
P_l(\cos\theta_{12}) = \frac{4\pi}{2l+1}\sum^l_{m=-l}{Y^{m}_l}^*(\theta_1,\varphi_1)Y^m_l(\theta_2,\varphi_2)\nonumber
\end{equation}

For P-states, one would have similarly

\begin{array}{lr}
r_{12}^0 & (sp)P\nonumber\\
r_{12}^2 & (pd)P\nonumber\\
r_{12}^4 & (df)P\nonumber\\
\vdots & \vdots\nonumber
\end{array}

For D-states

\begin{array}{lr}
r_{12}^0 & (sd)D & (pp^\prime)D\nonumber\\
r_{12}^2 & (pf)D & (dd^\prime)D\nonumber\\
r_{12}^4 & (dg)D & (ff^\prime)D\nonumber\\
\vdots & \vdots & \vdots\nonumber
\end{array}

In this case, since there are two "lowest-order" couplings to form a
D-state, both must be present in the basis set, i.e.

\begin{equation}
\Psi(r_2,r_2) = \sum c_{ijk}r_1^ir_2^{j+2}r_{12}^ke^{-\alpha r_1-\beta
r_2}\mathcal{Y}^M_{022}(\hat{r}_1,\hat{r}_2)
+\sum d_{ijk}r_1^{i+1}r_2^{j+1}r_{12^k}e^{-\alpha^\prime r_1 - \beta^\prime
r_2}\mathcal{Y}^M_{112}(\hat{r}_1,\hat{r}_2)\nonumber
\end{equation}

For F-states, one would need <math>(sf)F</math> and <math>(pd)F</math> terms.

For G-states, one would need <math>(sg)G</math>, <math>(pf)G</math> and <math>(dd^\prime)G</math> terms.

Completeness of the radial functions can be proven by considering the
Sturm-Liouville problem

\begin{equation}
\left(-\frac{1}{2}\nabla^2-\frac{\lambda}{r_s}-E\right)\psi({\bf r}) = 0\nonumber
\end{equation}

or

\begin{equation}
\left(-\frac{1}{2}\frac{1}{r^2}\frac{\partial}{\partial r}\left({r^2}\frac{\partial}{\partial r}\right) -
\frac{l\left(l+1\right)}{2r^2} - \frac{\lambda}{r} - E\right)u(r) = 0.\nonumber
\end{equation}

For fixed E and variable <math>\lambda</math> (nuclear charge).

The eigenvalues are <math>\lambda_n = (E/E_n)^{1/2}</math>, where <math>E_n =- \frac{1}{2n^2}</math>

[[File:Pic2.jpg|500px]]

and the eigenfunctions are

\begin{equation}
u_{nl}(r) =
\frac{1}{(2l+1)!}\left(\frac{(n+l)!}{(n-l-1)2!}\right)^{1/2}(2\alpha)^{3/2}e^{-\alpha
r}\nonumber
\times \ \left( 2 \alpha r \right)^l\,_{1}F_{1}\left(-n+l+1,2l+2; 2 \alpha r \right)
\end{equation}

with <math>\alpha = (-2E)^{1/2}</math> and <math>n\geq l+1</math>. The confluent hypergeometric function <math> _1F_1(a,b,;z)</math> then denotes a finite polynomial since <math>a=-n+l+1</math> is a negative integer or zero.

Unlike the hydrogen spectrum, which has both a discrete part for <math>E<0</math> and a
continuous part for <math>E>0</math>, this forms an entirely discrete set of finite
polynomials, called Sturmian functions. They are orthogonal with respect to
the potential, i.e.

\begin{equation}
\int^\infty_0 r^2dr \left( u_{n^\prime l}(r)\frac{1}{r}u_{nl}(r) \right) = \delta_{n,n^\prime}\nonumber
\end{equation}

Since they become complete in the limit <math>n\rightarrow\infty</math>, this assures the
completeness of the variational basis set.

[See also B Klahn and W.A. Bingel Theo. Chim. Acta (Berlin) 44, 9 and 27
(1977)].

==Solutions of the Eigenvalue Problem==

For convenience, write

\[
\Psi({\bf r}_1,{\bf r}_2) = \sum^N_{m=1}c_m\varphi_m\nonumber
\]

where <math>m</math> represents the <math>m</math>'th combination of <math>i,j,k</math> and

\begin{equation}
\varphi_{ijk}=r_1^ir_2^jr_{12}^ke^{-\alpha r_1 - \beta
r_2}\mathcal{Y}^M_{l_1,l_2,L}(\hat{r}_1,\hat{r}_2) \pm \mbox{exchange}\nonumber
\end{equation}

\begin{align}

&
\left( \begin{array}{lr}
\cos(\theta) & \sin(\theta)\\
-\sin(\theta) & \cos(\theta)
\end{array} \right)
\left( \begin{array}{lr}
H_{11} & H_{12} \\
H_{12} & H_{22}
\end{array}\right)
\left( \begin{array}{lr}
\cos(\theta) & -\sin(\theta)\\
\sin(\theta) & \cos(\theta)
\end{array} \right)
\\[5pt]
= & \left(\begin{array}{lr}
cH_{11}+sH_{12} & cH_{12} + sH_{22}\\
-sH_{11} + cH_{12} & -sH_{12} + cH_{22}
\end{array}\right)
\left( \begin{array}{cc}
c & -s \\
s & c
\end{array}\right)
\\[5pt]
= & \left(\begin{array}{lr}
c^2H_{11}+s^2H_{22} + 2csH_{12} & (c^2-s^2)H_{12}+cs(H_{22}-H_{11})\\
(c^2-s^2)H_{12}+cs(H_{22}-H_{11}) & s^2H_{11}+c^2H_{22}-2csH_{12}
\end{array}\right)
\end{align}

Therefore

\begin{equation}
\left[\cos^2(\theta)-\sin^2(\theta)\right]H_{12} =
\cos(\theta)\sin(\theta)(H_{11}-H_{22})\nonumber
\end{equation}

and

\begin{equation}
\tan(2\theta) = \frac{2H_{12}}{H_{11}-H_{22}}\nonumber
\end{equation}

i.e.

\begin{equation}
\cos(\theta)=\left(\frac{r+\omega}{2r}\right)^{1/2}\nonumber
\end{equation}

\begin{equation}
\sin(\theta)=-\mbox{sgn}(H_{12})\left(\frac{r-\omega}{2r}\right)^{1/2}\nonumber
\end{equation}

where

\begin{equation}
\omega = H_{22}-H_{11}\nonumber
\end{equation}

\begin{equation}
r=\left(\omega^2+4H_{12}^2\right)^{1/2}\nonumber
\end{equation}

\begin{equation}
E_1=\frac{1}{2}\left(H_{11}+H_{22}-r\right)\nonumber
\end{equation}

\begin{equation}
E_2=\frac{1}{2}\left(H_{11}+H_{22}+r\right)\nonumber
\end{equation}

===Brute Force Method===

- Gives all the eigenvalues and eigenvectors, but it is slow

- First orthonormalize the <math>N</math>-dimensional basis set; i.e. form linear combinations

\begin{equation}
\Phi_m = \sum_{n=1}^N\varphi_nR_{nm}\nonumber
\end{equation}

such that
\begin{equation}
\langle \Phi_m | \Phi_n \rangle = \delta_{m,n}
\end{equation}

This can be done by finding an orthogonal tranformation T such that

\begin{equation}
T^TOT = I = \left(
\begin{array}{ccccc}
I_1 & 0 & 0 & \ldots & 0 \\
0 & I_2 & 0 & \ldots & 0 \\
0 & 0 & I_3 & \ldots & 0 \\
\vdots & \vdots & \vdots & \ddots & \vdots \\
0 & 0 & 0 & \ldots & I_N
\end{array}\right);
\end{equation}

\begin{equation}
O_{mn} = \langle \varphi_m | \varphi_n \rangle \nonumber
\end{equation}

and then applying a scale change matrix

\begin{equation}
S = \left(\begin{array}{ccccc}
\frac{1}{I_1^{1/2}} & 0 & 0 & \ldots & 0\\
0 & \frac{1}{I_2^{1/2}} & 0 & \ldots & 0 \\
0 & 0 & \frac{1}{I_3^{1/2}} & \ldots & 0 \\
\vdots & \vdots & \vdots & \ddots & \vdots \\
0 & 0 & 0 & \ldots & I_N^{1/2}
\end{array}\right)= S^T\nonumber
\end{equation}

Then

\begin{equation}
S^TT^TOTS = 1
\end{equation}

i.e.

\begin{equation}
R^TOR = 1
\end{equation}

with <math>R=TS</math>.

If H is the matrix with elements <math>H_{mn} = \langle \varphi_m | \varphi_n \rangle </math>, then H
expressed in the <math>\Phi_m</math> basis set is

\begin{equation}
H^\prime = R^THR.\nonumber
\end{equation}

We next diagonalize <math>H^\prime</math> by finding an orthogonal transformation W such
that

\begin{equation}
W^T H^\prime W = \lambda = \left(
\begin{array}{cccc}
\lambda_1 & 0 & \ldots & 0 \\
0 & \lambda_2 & \ldots & 0 \\
\vdots & \vdots & \ddots & \vdots \\
0 & 0 & \ldots & \lambda_N
\end{array}\right)\nonumber
\end{equation}

The q'th eigenvector is

\begin{equation}
\Psi^{(q)} = \sum_{n=1}^N\Phi_n W_{n,q} = \sum_{n,n^\prime} \varphi_{n^\prime} R_{n^\prime ,n} W_{n,q} \nonumber
\end{equation}

i.e.

\begin{equation}
c_{n^\prime}^{(q)} = \sum_{n=1}^N R_{n^\prime n} W_{n,q}
\end{equation}

===The Power Method===

- Based on the observation that if H has one eigenvalue, <math>\lambda_M</math>, much bigger than all the rest, and <math>\chi = \left(
\begin{array}{c}a_1\\a_2\\\vdots\end{array}\right)</math>
is an arbitrary starting vector, then <math>\chi = \sum_{q=1}^N x_q\Psi^{(q)}</math>.

\begin{equation}
(H)^n\chi = \sum_{q=1}^N x_q \lambda^n_q\Psi^{(q)}\nonumber\rightarrow x_M\lambda_M^n\Psi^{(M)}\nonumber
\end{equation}

provided <math>x_M\neq 0</math>.

To pick out the eigenvector corresponding to any eigenvalue, write the original problem in the form

\begin{align}
H\Psi & = \lambda O \Psi \nonumber \\
\left( H - \lambda_q O \right) \Psi & = \left( \lambda - \lambda_q \right) O \Psi \nonumber
\end{align}

Therefore,

\begin{equation}
G \Psi = \frac{1}{\lambda - \lambda_q } \Psi \nonumber \\
\end{equation}

where <math>G=(H-\lambda_qO)^{-1}O</math> with eigenvalues <math>\frac{1}{\lambda_n - \lambda_q}</math>.

By picking <math>\lambda_q</math> close to any one of the <math>\lambda_n</math>, say
<math>\lambda_{n^\prime}</math>, then <math>\frac{1}{\lambda_n-\lambda_q}</math> is much larger for
<math>n=n^\prime</math> than for any other value. The sequence is then

\begin{array}{c}
\chi_1=G\chi\nonumber\\
\chi_2=G\chi_1\nonumber\\
\chi_3=G\chi_2\nonumber\\
\vdots\nonumber
\end{array}

until the ratios of components in <math>\chi_n</math> stop changing.

- To avoid matrix inversion and multiplication, note that the sequence is equivalent to
\begin{equation}
F \chi_n = \left( \lambda - \lambda_q \right) O \chi_{n-1} \nonumber \\
\end{equation}
where <math>F = H-\lambda_qO</math>. The factor of <math>\left( \lambda - \lambda_q \right)</math> can be
dropped because this only affects the normalization of <math>\chi_n</math>. To find <math>\chi_n</math>, solve
\begin{equation}
F \chi_n = O \chi_{n-1} \nonumber
\end{equation}
(N equations in N unknowns). Then
\begin{equation}
\lambda = \frac{ \langle \chi_n | H | \chi_n \rangle}{\langle \chi_n | \chi_n \rangle} \nonumber
\end{equation}

==Matrix Elements of H==

\begin{equation}
H=-\frac{1}{2}\nabla^2_1 -\frac{1}{2}\nabla^2_2 - \frac{1}{r_1} - \frac{1}{r_2} +\frac{Z^{-1}}{r_{12}}\nonumber
\end{equation}

Taking <math>r_1, r_2</math> and <math>r_{12}</math> as independent variables,

\begin{equation}
\nabla_1^2 = \frac{1}{r_1^2}\frac{\partial}{\partial r_1}\left(r_1^2\frac{\partial}{\partial r_1}\right)+ \frac{1}{r_{12}^2}\frac{\partial}{\partial r_{12}}
\left(r_{12}^2\frac{\partial}{\partial r_{12}}\right)\nonumber -\frac{l_1(l_1+1)}{r_1^2}+2(r_1 - r_2\cos(\theta_{12}))\frac{1}{r_{12}}\frac{\partial^2}{\partial r_1 \partial r_{12}} \nonumber - 2(\nabla_1^Y \cdot {\bf r}_2)\frac{1}{r_{12}}\frac{\partial}{\partial r_{12}}\nonumber
\end{equation}

where <math>\nabla_1^Y</math> acts only on the spherical harmonic part of the wave function and the diagram

[[File:Pic3.jpg|400px]]

defines the complete set of 6 independent variables is <math>r_1</math>, <math>r_2</math>, <math>r_{12}</math>, <math>\theta_1</math>, <math>\varphi_1</math>, <math>\chi</math>.

If <math>r_{12}</math> were not an independent variable, then one could take the volume element to be

\begin{equation}
d\tau = r_1^2dr_1\sin(\theta_1)d\theta_1d\varphi_1r_2^2dr_2\sin(\theta_2)d\theta_2d\varphi_2.\nonumber
\end{equation}

However, <math>\theta_2</math> and <math>\varphi_2</math> are no longer independent variables. To eliminate them, take the point <math>{\bf r}_1</math> as the
origin of a new polar co-ordinate system, and write

\begin{equation}
d\tau=-r_1^2dr_1\sin(\theta_1)d\theta_1d\varphi_1r_{12}^2dr_{12}\sin(\psi)d\psi d\chi\nonumber
\end{equation}

and use

\begin{equation}
r_2^2=r_1^2+r_{12}^2 +2r_1r_{12}\cos(\psi)
\end{equation}

Then for fixed <math>r_1</math> and <math>r_{12}</math>,

\begin{equation}
2r_2dr_2 = -2r_1r_{12}\sin(\psi)d\psi
\end{equation}

Thus

\begin{equation}
d\tau= r_1dr_1r_2dr_2r_{12}dr_{12}\sin(\theta_1)d\theta_1d\varphi_1d\chi
\end{equation}

The basic type of integral to be calculated is

\begin{align}
I \left( l_1, m_1, l_2, m_2; R \right) & = \int \sin \left( \theta_1 \right) d\theta_1 d\varphi_1 d\chi Y^{m_1}_{l_1} \left( \theta_1, \varphi_1 \right)^{*} Y^{m_2}_{l_2} \left( \theta_2, \varphi_2 \right) \\
& \times \int r_1dr_1r_2dr_2r_{12}dr_{12}R \left( r_1, r_2, r_{12} \right)
\end{align}

Consider first the angular integral. <math>Y^{m_2}_{l_2} \left( \theta_2, \varphi_2 \right)</math> can be expressed in terms of the independent variables <math>\theta_1, \varphi_1, \chi</math> by use of the rotation matrix relation
\begin{equation}
Y^{m_2}_{l_2}( \theta_2, \varphi_2) = \sum_m \mathcal{D}^{(l_2)}_{m_2,m}(\varphi_1,\theta_1,\chi)^*Y^m_{l_2}( \theta_{12}, \varphi)
\end{equation}
where <math>\theta, \varphi</math> are the polar angles of <math>{\bf r}_2</math> relative to <math>{\bf r}_1</math>. The angular integral is then

\begin{align}
I_{ang} & = \int^{2 \pi}_{0} d\chi \int^{2\pi}_{0} d\varphi_1 \int^{\pi}_{0} \sin \left( \theta_1 \right) d\theta_{1} Y^{m_1}_{l_1} \left( \theta_1, \varphi_1 \right)^{*} \\
& \times \sum_m \mathcal{D}^{(l_2)}_{m_2,m} \left( \varphi_1,\theta_1,\chi \right)^{*} Y^{m}_{l_2} \left( \theta_{12},\varphi \right) \\
\end{align}

Use

\begin{equation}
Y^{m_1}_{l_1} \left( \theta_1, \varphi_1 \right)^{*} = \sqrt{\frac{2l_1+1}{4\pi}}\mathcal{D}^{(l_1)}_{m_1,0} \left( \varphi_1, \theta_1, \chi \right)
\end{equation}

together with the orthogonality property of the rotation matrices (Brink and Satchler, p 147)

\begin{align}
& \int \mathcal{D}^{(j)*}_{m,m^\prime}\mathcal{D}^{(J)}_{M,M^\prime}\sin(\theta_1)d\theta_1d\varphi_1d\chi \\
& = \frac{8\pi^2}{2j+1}\delta_{jJ}\delta_{mM}\delta_{m^\prime M^\prime}\nonumber
\end{align}

to obtain

\begin{align}
I_{ang} & = \sqrt{ \frac{2l_1+1}{4\pi}} \frac{8\pi^2}{2l_1+1} \delta_{l_1,l_2} \delta_{m_1,m_2} Y^{0}_{l_2} \left( \theta_{12}, \varphi \right) \\
& = 2 \pi \delta_{l_1,l_2} \delta_{m_1,m_2} P_{l_2} \left( \cos \theta_{12} \right)
\end{align}

since

\begin{equation}
Y^0_{l_2} \left( \theta_{12}, \varphi \right) = \sqrt{\frac{2l_1+1}{4\pi}} P_{l_2} \left( \cos \left( \theta_{12} \right) \right)
\end{equation}

Note that <math>P_{l_2} \left( \cos \theta \right)</math> is just a short hand expression for a radial function because

\begin{equation}
\cos \theta_{12} = \frac{r_1^2 + r_2^2 - r_{12}^2}{2r_1r_2}
\end{equation}

The original integral is thus

\begin{equation}
I \left( l_1, m_1, l_2, m_2; R \right) = 2 \pi \delta_{l_1,l_2} \delta_{m_1,m_2} \int^{\infty}_{0} r_{1} dr_{1} \int^{\infty}_{0} r_{2} dr_{2} \int^{r_{1} + r_{2}}_{|r_{1} - r_{2}|} r_{12} dr_{12} R \left( r_{1}, r_{2}, r_{12} \right) P_{l_2} \left( \cos \theta_{12} \right)
\end{equation}

where again

\begin{equation}
\cos \theta_{12} = \frac{r_{1}^{2} +r^{2}_{2} - r_{12}^{2}}{2 r_{1} r_{2}}
\end{equation}

is a purely radial function.

The above would become quite complicated for large <math>l_2</math> because <math>P_{l_2}(\cos \theta_{12})</math> contains terms up to <math>\left( \cos \theta_{12} \right)^{l_2}</math>. However, recursion relations exist which allow any integral containing <math>P_l \left( \cos \theta_{12} \right)</math> in terms of those containing just <math>P_0 \left( \cos \theta_{12} \right) = 1</math> and <math>P_1 \left( \cos \theta_{12} \right) = \cos \theta_{12}</math>.

==Radial Integrals and Recursion Relations==

The basic radial integral is [see G.W.F. Drake, Phys. Rev. A 18, 820 (1978)]

\begin{align}
I_0 \left( a, b, c \right) & =
\int^{\infty}_{0} r_{1} dr_{1} \int^{\infty}_{r_{1}} r_{2} dr_{2} \int^{r_{1}+r_{2}}_{r_{2} - r_{1}} r_{12} dr_{12} r_{1}^{a} r_{2}^{b} r_{12}^{c} e^{-\alpha r_{1} - \beta r_{2}} + \int^{\infty}_{0} r_{2} dr_{2} \int^{\infty}_{r_{2}} r_{1} dr_{1} \int^{r_{1} + r_{2}}_{r_{1} - r_{2}} r_{12} dr_{12} r_{1}^{a} r_{2}^{b} r_{12}^{c} e^{- \alpha r_{1} - \beta r_{2}} \\
& = \frac{2}{c+2} \sum^{[(c+1)/2]}_{i=0}
\left(\begin{array}{c}
c+2 \\
2i+1
\end{array} \right)
\left\{ \frac{q!}{\beta^{q+1} \left( \alpha + \beta \right)^{p+1}} \sum_{j=0}^{q} \frac{\left( p + j \right)!}{j!} \left( \frac{\beta}{\alpha + \beta} \right)^j + \frac{q^{\prime}!}{\alpha^{q^{\prime}+1} \left( \alpha + \beta \right)^{p^{\prime}+1}} \sum^{q^{\prime}}_{j=0} \frac{\left( p^{\prime} + j \right)!}{j!} \left( \frac{\alpha}{\alpha + \beta} \right)^j \right\}
\end{align}

where

\begin{array}{ll}
p = a + 2i + 2 & p^{\prime} = b + 2i + 2 \\
q = b + c - 2i + 2 & q^{\prime} = a + c - 2i + 2
\end{array}

The above applies for <math> a, b \geq -2, c\geq -1</math>.
<math>[x]</math> means "greatest integer in" <math>x</math>.

Then

\begin{align}
I_1 \left( a, b, c \right) & = \int d\tau_{r} r_{1}^{a} r_{2}^{b} r_{12}^{c} e^{ -\alpha r_{1} - \beta r_{2}} P_{1}\left( \cos \theta \right) \\
& = \frac{1}{2} \left( I_{0} \left( a + 1, b - 1, c \right) + I_{0} \left( a - 1, b + 1, c \right) - I_{0} \left( a - 1, b - 1, c + 2 \right) \right)
\end{align}

===The Radial Recursion Relation===

Recall that the full integral is

\begin{equation}
I(l_1m_1,l_2m_2;R)=2\pi\delta_{l_1,l_2}\delta_{m_1,m_2}I_{l_2}(R)
\end{equation}

where

\begin{equation}
I_{l_2}(R) = \int d\tau_r R(r_1,r_2,r_{12})P_{l_2}(\cos \theta_{12})
\end{equation}

To obtain the recursion relation, use

\begin{equation}
P_l(x) = \frac{[P^{\prime}_{l+1}(x)-P^{\prime}_{l-1}(x)]}{2l+1}
\end{equation}

with

\begin{equation}
P_{l+1}^\prime (x) = \frac{d}{dx}P_{l+1}(x)
\end{equation}

Here <math>x=\cos\theta_{12}</math> and

\begin{align}
\frac{d}{d \cos \theta_{12}} & = \frac{dr_{12}}{d \cos \theta_{12}} \frac{d}{dr_{12}} \\
& = -\frac{r_{1} r_{2}}{r_{12}} \frac{d}{dr_{12}}
\end{align}

Then,

\begin{equation}
I_{l}\left( R \right) = - \int d\tau_r R \frac{r_{1} r_{2}}{r_{12}} \frac{d}{dr_{12}} \frac{ \left[ P_{l+1} \left( \cos \theta_{12} \right) - P_{l-1} \left( \cos \theta_{12} \right) \right]}{2l+1}
\end{equation}

The <math>r_{12}</math> part of the integral is

\begin{align}
& \int^{r_{1} + r_{2}}_{|r_{1} - r_{2}|} r_{12} dr_{12} R \frac{r_{1} r_{2}}{r_{12}} \frac{d}{dr_{12}} \left[P_{l+1}-P_{l-1} \right] \\
& = R r_{1} r_{2} \left[ P_{l+1} - P_{l-1} \right]^{r_{1} + r_{2}}_{|r_{1} - r_{2}|} - \int^{r_{1} + r_{2}}_{|r_{1} - r_{2}|} r_{12} dr_{12} \left( \frac{d}{dr_{12}} R \right) \frac{r_{1} r_{2}}{r_{12}} \frac{ \left[ P_{l+1} - P_{l-1} \right]}{2l+1}
\end{align}

The integrated term vanishes because

\begin{equation}
\cos \theta_{12} = \frac{r_{1}^{2} + r_{2}^{2} - r_{12}^{2}}{2 r_{1} r_{2}}
\begin{array}{lll}
& = -1 & {\rm when} & r_{12}^{2} = \left( r_{1} + r_{2} \right)^{2} \\
& = 1 & {\rm when} & r_{12}^{2} = \left( r_{1} - r_{2} \right)^{2}
\end{array}
\end{equation}

and <math>P_l(1)=1</math>, <math>P_{l}(-1)=(-1)^{l}</math>.

Thus,

\begin{equation}
I_{l+1}\left(\frac{r_1r_2}{r_{12}}R^\prime\right) = \left( 2l+1 \right) I_{l}(R) - I_{l-1} \left( \frac{r_{1} r_{2}}{r_{12}} R^{\prime} \right)
\end{equation}

If

\begin{equation}
R = r_{1}^{a-1} r_{2}^{b-1} r_{12}^{c+2} e^{-\alpha r_{1} - \beta r_{2}}
\end{equation}

then [G.W.F. Drake, Phys. Rev. A 18, 820 (1978)]

\begin{equation}
I_{l+1}(r_1^ar_2^br_{12}^c) =
\frac{2l+1}{c+2} I_{l} \left( r_{1}^{a-1} r_{2}^{b-2} r_{12}^{c+2} \right)+I_{l-1} \left( r_{1}^{a} r_{2}^{b} r_{12}^{c} \right)
\end{equation}

For the case <math>c=-2</math>, take

\begin{equation}
R = r_{1}^{a-1} r_{2}^{b-1} \ln r_{12} e^{ -\alpha r_{1} -\beta r_{2}}
\end{equation}

Then

\begin{equation}
I_{l+1} \left( r_{1}^{a} r_{2}^{b} r_{12}^{c} \right) = I_{l} \left( r_{1}^{a-1} r_{2}^{b-1} \ln r_{12} \right) \left( 2l + 1 \right) - I_{l-1} \left( r_{1}^{a} r_{2}^{b} r_{12}^{-2} \right)
\end{equation}

This allows all <math>I_{l}</math> integrals to be calculated from tables of <math>I_0</math> and
<math>I_1</math> integrals.

===The General Integral===

The above results for the angular and radial integrals can now be combined into a general formula for integrals of the type

\begin{equation}
I = \int \int d{\bf r}_{1} d{\bf r}_{2} R_{1} \mathcal{Y}^{M^{\prime}}_{l^{\prime}_{1} l^{\prime}_{2} L^{\prime}} \left( \hat{r}_{1}, \hat{r}_{2} \right) T_{k_{1} k_{2} K}^{Q} \left( {\bf r}_{1},{\bf r}_{2} \right) R_{2} \mathcal{Y}_{l_{1} l_{2} L}^{M} \left( \hat{r}_{1}, \hat{r}_{2} \right)
\end{equation}

where

\begin{equation}
\mathcal{Y}^{M}_{l_{1} l_{2} L} \left( \hat{r}_{1}, \hat{r}_{2} \right) = \sum_{m_1,m_2} \langle l_{1} l_{2} m_{1} m_{2} | LM \rangle Y^{m_{1}}_{l_{1}} \left( \hat{r}_{1} \right) Y^{m_{2}}_{l_{2}} \left( \hat{r}_{2} \right)
\end{equation}

and

\begin{equation}
T^Q_{k_{1} k_{2} K} \left( {\bf r}_{1}, {\bf r}_{2} \right) = \sum_{q_{1}, q_{2}} \langle k_{1} k_{2} q_{1} q_{2} | KQ \rangle Y_{k_{1}}^{q_{1}} \left( \hat{r}_{1} \right) Y^{q_{2}}_{k_{2}} \left( \hat{r}_{2} \right)
\end{equation}

The basic idea is to make repeated use of the formula

\begin{equation}
Y^{m_{1}}_{l_{1}} \left( \hat{r}_{1} \right) Y^{m_{2}}_{l_{2}} \left( \hat{r}_{1} \right) = \sum_{l m} \left( \frac{\left( 2l_{1}+1 \right) \left(2l_{2}+1 \right) \left(2l+1 \right)}{4 \pi} \right)^{1/2}
\times \left( \begin{array}{ccc}
l_1 & l_2 & l \\
m_1 & m_2 & m
\end{array} \right) \left(\begin{array}{ccc}
l_1 & l_2 & l \\
0 & 0 & 0
\end{array} \right)
Y^{m*}_{l} \left( \hat{r}_{1} \right)
\end{equation}

where

\begin{equation}
Y^{m*}(\hat{r}) = (-1)^mY_l^{-m}(\hat{r})
\end{equation}

and

\begin{equation}
\left(\begin{array}{ccc}
l_1 & l_2 & l \\
m_1 & m_2 & m
\end{array} \right) = \frac{\left( -1 \right)^{l_{1} - l_{2} - m}}{\left( 2l + 1 \right)^{1/2}} \left( l_{1} l_{2} m_{1} m_{2} | l, -m \right)
\end{equation}

is a 3-<math>j</math> symbol. In particular, write

\begin{array}{cc}
Y^{m_1^\prime}_{l_1^\prime}(\hat{r}_1)^{*} \underbrace{Y_{k_{1}}^{q_{1}} \left( \hat{r}_{1} \right) Y^{m_{1}}_{l_{1}} \left( \hat{r}_{1} \right)} & = & \sum_{\Lambda M}(\ldots)Y^{M}_\Lambda(\hat{r}_1)\\
\ \ \ \ \ \ \ \sum_{\lambda_1\mu_1}Y_{\lambda_1}^{\mu_1}(\hat{r}_1) & & \\
& & \\
Y^{m_{2}^{\prime}}_{l_{2}^{\prime}} \left( \hat{r}_{2} \right) \underbrace{Y^{q_{2}}_{k_{2}} \left( \hat{r}_{2} \right) Y_{l_{2}}^{m_{2}} \left( \hat{r}_{2} \right)} & = & \sum_{\Lambda^{\prime} M^{\prime}} \left( \ldots \right)Y_{\Lambda^{\prime}}^{M^{\prime}*} \left( \hat{r}_{2} \right) \\
\ \ \ \ \ \ \ \ \ \sum_{\lambda_{2} \mu_{2}}Y_{\lambda_{2}}^{\mu_{2}} \left( \hat{r}_{2} \right) & &
\end{array}

The angular integral then gives a factor of <math>2 \pi \delta_{\Lambda, \Lambda^{\prime}} \delta_{M, M^{\prime}} P_{\Lambda} \left( \cos \theta_{12} \right)</math>. The total integral therefore reduces to the form

\begin{equation}
I = \sum_{\Lambda} C_{\Lambda} I_{\Lambda} \left( R_{1} R_{2} \right)
\end{equation}

where <math>C_{\Lambda} = \sum_{\lambda_{1}, \lambda_{2}} C_{ \lambda_{1}, \lambda_{2}, \Lambda}</math>. For further details and derivations, including graphical representations, see G.W.F. Drake, Phys. Rev. A <math>{\bf 18}</math>, 820 (1978).

==Graphical Representation==

[[File:GraphRep1.jpg|500px]]

==Matrix Elements of H==
Recall that

\begin{equation}
H = -\frac{1}{2} \nabla^{2}_{1} - \frac{1}{2} \nabla^{2}_{2} - \frac{1}{r_1} - \frac{1}{r_2} + \frac{Z^{-1}}{r_{12}}
\end{equation}

Consider matrix elements of

\begin{equation}
\nabla^2_1 = \frac{1}{r_1^2} \frac{\partial}{\partial r_1}\left(r_1^2\frac{\partial}{\partial r_1} \right) + \frac{1}{r^2}\frac{\partial}{\partial r}\left(r^2\frac{\partial}{\partial r}\right) - \frac{(\vec{l}_1^Y)^2}{r_1^2} + \frac{2 \left( r_1 - r_2 \cos \theta \right)}{r} \frac{\partial^2}{\partial r_1 \partial r} - 2 \left( \nabla^Y_1 \cdot {\bf r}_2 \right) \frac{1}{r} \frac{\partial}{\partial r}
\end{equation}

where <math>r \equiv r_{12}</math> and <math>\cos\theta \equiv \cos\theta_{12}</math>. Also in what follows, define <math>\hat{\nabla}^Y_1 \equiv r_1\nabla^Y_1</math> where <math>\nabla^Y</math> operates only on the spherical harmonic part of the wave function, and similarly for <math>\vec{l}_1^Y</math>.

A general matrix element is

\begin{equation}
\langle r_1^{a^\prime} r_2^{b^\prime} r_{12}^{c^\prime} e^{-\alpha^\prime r_1 - \beta^\prime r_2} \mathcal{Y}^{M^\prime}_{l^\prime_1 l^\prime_2 L^\prime} \left( \hat{r}_1, \hat{r}_2 \right) \left| \nabla^2_1 \right| r_1^a r_2^b r_{12}^c e^{-\alpha r_1 - \beta r_2} \mathcal{Y}^M_{l_1 l_2 L} \left( \hat{r}_1, \hat{r}_2 \right) \rangle
\end{equation}

Since <math>\nabla^2_1</math> is rotationally invariant, this vanishes unless <math>L=L^\prime</math>, <math>M=M^\prime</math>. Also <math>\nabla_1^2</math> is Hermitian, so that the result must be the same whether it operates to the right or left, even though the results look very different. In fact, requiring the results to be the same yields some interesting and useful integral identities as follows:

Put

\begin{equation}
\mathcal{F} = F \mathcal{Y}^M_{l_1 l_2 L} \left( \hat{r}_1, \hat{r_2} \right)
\end{equation}

and

\begin{equation}
F = r_1^a r_2^b r_{12}^c e^{-\alpha r_1 - \beta r_2}
\end{equation}

Then

\begin{equation}
\nabla^2_1 \mathcal{F} = \left\{ \frac{1}{r_1^2} \left[ a \left( a + 1 \right) - l_1 \left( l_1 + 1 \right) \right] + \frac{c \left( c + 1 \right)}{r^2} + \alpha^2\frac{2\alpha \left( a + 1 \right)}{r_1} + \frac{2 \left(r_1 - r_2 \cos \theta \right) }{r_1r^2}c \left[ a - \alpha r_1 \right] - \frac{2c}{r^2} \left( \hat{\nabla}^Y_1 \cdot \hat{r}_2 \right) \frac{r_2}{r_1} \right\} \mathcal{F}
\end{equation}

and

\begin{align}
\langle \mathcal{F}^\prime \left| \nabla^2_1 \right| \mathcal{F} \rangle & = \sum_\Lambda \int d\tau_r F^\prime C_\Lambda \left( 1 \right) P_\Lambda \left( \cos \theta \right) \times \left\{ \frac{1}{r_1^2} \left[ a \left( a + 1 \right) - l_1 \left( l_1 + 1 \right) \right] -\frac{2 \alpha \left( a + 1 \right)}{r_1} + \frac{c \left( c + 1 \right)}{r^2} +\alpha^2 + \frac{2 \left( r_2 - r_2 \cos \theta \right)}{r_1r^2}c \left( a - \alpha r_1 \right) \right\} F \\
& + \sum_\Lambda \int d\tau_r F^\prime C_\Lambda \left( \hat{\nabla}^Y_1 \cdot \hat{r}_2 \right) P_\Lambda \left( \cos \theta \right) \left( \frac{-2c r_2}{r_1 r^2} \right) F
\end{align}

where

\begin{equation}
\int d\tau_r = \int^\infty_0 r_1 dr_1 \int^\infty_0 r_2 dr_2 \int^{r_1+r_2}_{\left| r_1 - r_2 \right|} r dr
\end{equation}

For brevity, let the sum over <math>\Lambda</math> and the radial integrations be understood, and let <math>\left( \nabla_1^2 \right)</math> stand for the terms which appear in the integrand. Then operating to the right gives

\begin{equation}
(\nabla_1^2)_R = \frac{1}{r_1^2} \left[ a \left( a + 1 \right) - l_1 \left( l_1 + 1 \right) \right] + \frac{c \left( c + 1 \right)}{r^2} + \alpha^2 - \frac{2 \alpha \left( a + 1 \right)}{r_1}
+ \frac{2 \left( r_1 - r_2 \cos \theta \right)}{r_1 r^2}c \left( a - \alpha r_1 \right) - \frac{2c r_2}{r_1 r^2} \left( \hat{\nabla}^Y_1 \cdot \hat{r}_2 \right)
\end{equation}

Operating to the left gives

\begin{equation}
(\nabla_1^2)_L = \frac{1}{r_1^2} \left[a^\prime \left( a^\prime + 1 \right) - l_1^\prime \left( l_1^\prime + 1 \right) \right] + \frac{c^\prime \left( c^\prime + 1 \right)}{r^2} + \alpha^{\prime 2} -\frac{2 \alpha^\prime \left( a^\prime + 1 \right)}{r_1}
+ \frac{2 \left( r_1 - r_2 \cos \theta \right)}{r_1 r^2} c^\prime \left( a^\prime - \alpha^\prime r_1 \right) - \frac{2 c^\prime r_2}{r_1 r^2} \left( \hat{\nabla}^{Y^\prime}_1 \cdot \hat{r}_2 \right)
\end{equation}

Now put

\begin{array}{cc}
a_+ = a+a^\prime, & \hat{\nabla}_1^+ = \hat{\nabla}_1^Y + \hat{\nabla}_1^{Y\prime} \\
a_{-} = a-a^\prime, & \hat{\nabla}_1^{-} = \hat{\nabla}_1^Y - \hat{\nabla}_1^{Y\prime}
\end{array}

etc., and substitute <math>a^\prime = a_+ - a</math>, <math>c^\prime = c_+ - c</math>, and <math>\alpha^\prime = \alpha_+ - \alpha</math> in <math>(\nabla^2_1)_L</math>. If <math>a_+</math>, <math>c_+</math> and <math>\alpha_+</math> are held fixed, then the equation

\begin{equation}
(\nabla^2_1)_R = (\nabla^2_1)_L
\end{equation}

must be true for arbitrary <math>a</math>, <math>c</math>, and <math>\alpha</math>. Their coefficients must thus vanish.

This yields the integral relations

\begin{equation}
\label{I}
\frac{\left( r_1 - r_2 \cos \theta \right)}{r_1 r^2} = \frac{1}{c_+} \left( \frac{- \left( a_+ + 1 \right)}{r_1^2} + \frac{\alpha_+}{r_1} \right) \tag{I}
\end{equation}

from the coefficient of <math>a</math> and

\begin{equation}
\label{II}
\frac{ \left( r_1 - r_2 \cos \theta \right) \left( a_+ - \alpha_+ r_1 \right)}{r_1 r^2} = \frac{r_2}{r_1 r^2}\left( \hat{r}_2 \cdot \hat{\nabla}_1^+ \right) - \frac{ \left( c_+ + 1 \right)}{r^2} \tag{II}
\end{equation}

from the coefficient of <math>c</math>. The coefficient of <math>\alpha</math> gives an equation equivalent to (I).

Furthermore, if can be show that (see problem)

\begin{align}
& \sum_\Lambda \int d\tau_r \frac{r^c}{r^2} C_\Lambda \left( \hat{r}_2 \cdot \hat{\nabla}_1^Y \right) P_\Lambda \left( \cos \theta \right) \\
= & \sum_\Lambda \int d\tau_r \frac{r^c}{c r_1 r_2}C_\Lambda \left( 1 \right)P_\Lambda \left( \cos \theta \right) \times \left( \frac{l_1^\prime \left( l_1^\prime + 1 \right) - l_1 \left( l_1 + 1 \right) -\Lambda \left( \Lambda + 1 \right)}{2} \right)
\end{align}

and similarly for <math>\left( \hat{r}_2 \cdot \hat{\nabla}^{Y\prime}_1 \right)</math> with <math>l_1</math> and <math>l_1^\prime</math> interchangeable, then it follows that

\begin{equation}
\frac{r^{c_+}}{r^2} \left( \hat{r}_2 \cdot \hat{\nabla}_1^+ \right) = - \frac{r^{c_+}}{c_+ r_1 r_2} \Lambda \left( \Lambda + 1 \right)
\end{equation}

and

\begin{equation}
\frac{r^{c_+}}{r^2} \left( \hat{r}_2 \cdot \hat{\nabla}_1^- \right) = \frac{r^{c_+}}{c_+ r_1 r_2} \left[ l_1^\prime \left( l_1^\prime + 1 \right) - l_1 \left( l_1 + 1 \right) \right]
\end{equation}

where equality applies after integration and summation over <math>\Lambda</math>.

Thus (II) becomes

\begin{equation}\label{IIa}
\frac{ \left( r_1 - r_2 \cos \theta \right) \left( a_+ - \alpha_+ r_1 \right)}{r_1 r^2} = - \frac{\Lambda \left( \Lambda + 1 \right)}{c_+ r^2_1} - \frac{ \left( c_+ + 1 \right)}{r^2} \tag{IIa}
\end{equation}

===Problem===

Prove the integral relation

\begin{align}
& \sum_\Lambda \int d\tau_r f \left( r_1, r_2 \right) \frac{1}{r} \left( \frac{d}{dr} q(r) \right) C_\Lambda \left( \hat{r}_2 \cdot \hat{\nabla}^Y_1 \right) P_\Lambda \left( \cos \theta \right) \\
= & \sum_\Lambda \int d\tau_r \frac{ f \left( r_1, r_2 \right)}{r_1, r_2} q(r) C_\Lambda \left( 1 \right) P_\Lambda \left( \cos \theta \right) \left(\frac{l_1^\prime \left( l_1^\prime + 1 \right) - l_1 \left( l_1 + 1 \right) - \Lambda \left( \Lambda + 1 \right)}{2}\right)
\end{align}

where the coefficients <math>C_\Lambda \left( 1 \right)</math> are the angular coefficients from the overlap integral

\begin{align}
\int & d\Omega\, \mathcal{Y}^{M*}_{l_1^\prime l_2^\prime L}(\hat{r}_1,\hat{r}_2)\, \mathcal{Y}^{M}_{l_1 l_2 L}(\hat{r}_1,\hat{r}_2)\\
= & \sum_\Lambda C_\Lambda(1)P_\Lambda(\cos\theta_{12}).
\end{align}

Hint: Use the fact that <math>l_1^2</math> is Hermitian so that
<math>
\int d\tau (l_1^2\mathcal{Y}^\prime)^*q(r)\mathcal{Y} = \int d\tau
\mathcal{Y}^{\prime *}l_1^2(q(r)\mathcal{Y})
</math>
with <math>\ \vec{l}_1= \frac{1}{i}\vec{r}_1\times\nabla_1</math>.

It is also useful to use
<math>\begin{eqnarray}\ \
\cos\theta\, P_L(\cos\theta) &=& \frac{1}{2L+1}[L\,P_{L-1}(\cos\theta) + (L+1)P_{L+1}(\cos\theta)]
\\
(\cos^2\theta-1)P_L(\cos\theta) &=&
\frac{L(L-1)}{(2L-1)(2L+1)}P_{L-2}(\cos\theta)\nonumber
-\frac{2(L^2+L-1)}{(2L-1)(2L+3)}P_L(\cos\theta) + \frac{(L+1)(L+2)}{(2L+1)(2L+3)}P_{L+2}(\cos\theta)\nonumber\\
&=& \frac{L(L-1)}{(2L-1)(2L+1)}[P_{L-2}(\cos\theta) - P_L(\cos\theta)]
+ \frac{(L+1)(L+2)}{(2L+1)(2L+3)}[P_{L+2}(\cos\theta) - P_L(\cos\theta)]
\end{eqnarray}</math>

together with a double application of the integral recursion relation
<math>
I_{L+1}\left(\frac{1}{r}\frac{d}{dr}q(r)\right) =
(2L+1)I_L\left(\frac{1}{r_1r_2}q(r)\right)+I_{L-1}\left(\frac{1}{r}\frac{d}{dr}q(r)\right)
</math>

Of course <math>l_1^2\mathcal{Y}^M_{l_1l_2L}(\hat{r}_1,\hat{r}_2)=l_1(l_1+1)\mathcal{Y}^M_{l_1l_2L}(\hat{r}_1,\hat{r}_2)</math>. Begin by expanding
<math>
l_1^2(g\mathcal{Y}) = \mathcal{Y}l_1^2g + g\,l_1^2\mathcal{Y} + 2(\vec{l}_1g)\cdot(\vec{l}_1\mathcal{Y})
</math>, and show that

<math>
\begin{eqnarray}\ \
(\vec{l}_1g)\cdot(\vec{l}_1\mathcal{Y}) & = & \frac{r_1^2}{r}\frac{dg}{dr}\vec{r}_2\cdot \nabla_1\mathcal{Y}\,\,\,\,\mbox{ and}\\
l_1^2g(r) &=& 2r_1r_2\cos\theta\frac{1}{r}\frac{dg}{dr} + r_1^2r_2^2(\cos^2\theta - 1)
\frac{1}{r}\frac{d}{dr}\left(\frac{1}{r}\frac{dg}{dr}\right).
\end{eqnarray}
</math>

The proof amounts to showing that the term <math>l_1^2g(r)</math> can be replaced by
<math>\Lambda(\Lambda+1)g(r)</math> after multiplying by <math>P_\Lambda(\cos\theta)</math>
and integrating by parts with respect to the radial integrations over <math>r_1</math>, <math>r_2</math> and
<math>r \equiv |\vec{r}_1 - \vec{r}_2|</math>. Remember that <math>\cos\theta</math> is just a short-hand notation for the radial function <math>(r_1^2 + r_2^2 - r^2)/(2r_1r_2)</math>.

==General Hermitian Property==

Each combination of terms of the form
\begin{equation}
<f>_R = a^2f_1+af_2+abf_3+b^2f_4+bf_5+af_6\nabla_1^Y+bf_7\nabla^Y_2+f_8(Y)
\end{equation}
can be rewritten
\begin{equation}
<f>_L= (a_+-a)^2f_1+(a_+-a)f_2 + (a_+-a)(b_+-b)f_3+(b_+-b)^2f_4
+(b_+-b)f_5+(a_+-a)f_6\nabla^{Y\prime}_1+(b_+-b)f_7\nabla^{Y\prime}_2+f_8)Y^\prime
\end{equation}
where as usual <math>\nabla_1^Y</math> acts only on the spherical harmonic part of the wave function denoted for short by <math>Y</math>.
Since these must be equal for arbitrary <math>a</math> and <math>b</math>,
\begin{equation}
a_+^2f_1+a_+f_2+_+b_+f_3+b_+f_4+b_+f_5+a_+f_6\nabla^{Y\prime}_1 +
b_+f_7\nabla^{Y\prime}_2+f_8(Y^\prime)-f_8(Y)=0
\end{equation}

Adding the corresponding expression with <math>Y</math> and <math>Y^\prime</math> interchanged
yields
<math>\label{III}
a_+^2f_1+a_+f_2+a_+b_+f_3+b_+^2f_4+b_+f_5+\frac{1}{2}a_+f_6\nabla^+_1+\frac{1}{2}b_+f_7\nabla^+_2=0\tag{III}
</math>

Subtracting gives
\[
f_8(Y)-f_8(Y^\prime)=-\frac{1}{2}[a_+f_6\nabla^-_1+b_+f_7\nabla_2^-]\tag{IV}
\]
<math>\label{V}
a[-2a_+f_1-2f_2-b_+f_3-f_6\nabla^+_1]=0\tag{V}
</math>
<math>\label{VI}
b[-2b_+f_4-2f_5-a_+f_3-f_7\nabla^+_2]=0\tag{VI}
</math>

Adding the two forms gives

\begin{eqnarray}
<f>_R+<f>_L=\frac{1}{2}(a_+^2+a_-^2)f_1+a_+f_2+\frac{1}{2}(a_+b_++a_-b_-)f_3\nonumber\\
\ +\frac{1}{2}(b_+^2+b_-^2)f_4+b_+f_5+\frac{1}{2}f_6(a_+\nabla_1^++a_-\nabla^-_1)\nonumber\\
\ +\frac{1}{2}(b_+\nabla_2^++b_-\nabla^-_2)+f_8(Y)+f_8(Y^\prime)\nonumber
\end{eqnarray}

Subtracting <math>\frac{x}{2}\times({\rm I})</math> gives

\begin{eqnarray}\ \
<f>_R+<f>_L = \frac{1}{2}[(1-x)a_+^2 + a_-^2]f_1 +
(1-\frac{x}{2})a_+f_2+\frac{1}{2}[(1-x)a_+b_++a_-b_-]f_3\nonumber\\
+\frac{1}{2}[(1-x)b_+^2+b_-^2]f_4+(1-\frac{x}{2})f_5b_++\frac{1}{2}f_6[(1-\frac{x}{2})a_+\nabla_1^++a_-\nabla^-_1]\nonumber\\
+\frac{1}{2}[(1-\frac{x}{2})b_+\nabla^+_2+b_-\nabla^-_2]+f_8(Y)+f_8(Y^\prime)\nonumber
\end{eqnarray}

If <math>x=1</math>,
\begin{eqnarray}
<f>_R+<f>_L &=&
\frac{1}{2}[a_-^2f_1+a_+f_2+a_-b_-f_3+b_-^2f_4+b_+f_5\nonumber\\
&+&\left(\frac{1}{2}a_+\nabla^+_1+a_-\nabla^-_1\right)f_6+\left(\frac{1}{2}b_+\nabla^+_2+b_-\nabla^-_2\right)f_7]+f_8(Y)+f_8(Y^\prime)\nonumber
\end{eqnarray}

The General Hermitian Property for arbitrary <math>x</math> gives
\begin{eqnarray}\ \
\nabla_1^2=
\frac{1}{4}\{\frac{1}{r_1^2}[(1-x)a_+^2+a_-^2+2(1-\frac{x}{2})a_+-2[l_1(l_1+1)+l_1^\prime(l_1^\prime+1)]\nonumber\\
-\frac{2}{r_1}[(1-x)\alpha_+a_++\alpha_-a_-+2(1-\frac{x}{2})\alpha_+]+(1-x)\alpha_+^2
+ \alpha^2_-]\nonumber\\
+\frac{2(r_1-r_2\cos\theta)}{r_1r^2}[(1-x)(a_+-\alpha_+r_1)c_++(a_--\alpha_-r_2)c_-]\nonumber\\
-\frac{2r_2}{r_1r^2}[(1-\frac{x}{2})c_+\hat{r}_2\cdot\hat{\nabla}_1^++c_-\hat{r}_2\hat{\nabla}_1^-]+\frac{(1-x)c_+^2+c_-^2+2(1-\frac{x}{2})c_+}{r^2}\}\nonumber
\end{eqnarray}.

Use
<math>\begin{eqnarray}\ \
\frac{-2r_2}{r_1r^2}[(1-\frac{x}{2})c_+\hat{r}_2\hat{\nabla}_1^++c_-\hat{r}_2\hat{\nabla}_1^-]=\frac{2}{r_1^2}\left[(1-\frac{x}{2})\Lambda(\Lambda+1)-\frac{c_-}{c_+}[l_1^\prime(l_1^\prime+1)+l_1(l_1+1)]\right]\end{eqnarray}</math>,

<math>\begin{eqnarray}
\frac{2(r_1-r_2\cos\theta)}{r_1r^2}(a_--\alpha_-r_1)c_- = 2\frac{c_-}{c_+}\left[\frac{-1}{r_1^2}[a_-(a_++1)]+\frac{1}{r_1}[a_-\alpha_++\alpha_-(a_++2)]-\alpha_-\alpha_+\right]\nonumber
\end{eqnarray}</math>,

and
<math>\begin{eqnarray}\ \
\frac{2(r_1-r_2\cos\theta)}{r_1r^2}(a_+-\alpha_+r_1)c_+ = 2\left[-\frac{1}{r_1^2}[a_+(a_++1)]+\frac{1}{r_1}[a_+\alpha_++\alpha_+(a_++2)]-\alpha_+^2\right]
\end{eqnarray}</math>.

Substituting into <math>\nabla_1^2</math> gives

\begin{eqnarray}
\nabla^2_1 = \frac{1}{4}\{\frac{1}{r_1^2}[-(1-x)a_+^2 + a_-^2 +xa_+ +
2(1-\frac{x}{2})\Lambda(\Lambda+1)\nonumber\\
-2l_1(l_1+1)(1-\frac{c_-}{c_+}) -
2l_1^\prime(l_1^\prime+1)(1+\frac{c_-}{c_+})-\frac{2c_-a_-}{c_+}(a_++1)]\nonumber\\
- \frac{2}{r_1}\left[-(1-x)\alpha_+(a_++2)+\alpha_-a_-+2(1-\frac{x}{2})\alpha_+
-\frac{c_-}{c_+}[a_-\alpha_++\alpha_-(a_++2)]\right]\nonumber\\
-(1-x)\alpha_+^2+\alpha_-^2-2\frac{c_-}{c_+}\alpha_-\alpha_+ +\frac{1}{r^2}\left[(1-x)c_+^2+c_-^2+2(1-\frac{x}{2})c_+\right]\}\nonumber
\end{eqnarray}

This has the form
<math>
\nabla_1^2 =
\frac{1}{4}\left[\frac{A_1}{r_1^2}+\frac{B_1}{r_1}+\frac{C_1}{r^2}+D_1\right]
</math>
with

<math>\begin{array}{ccc}
A_1=-(1-x)a_+^2+a_-^2
+xa_++2(1-\frac{x}{2})\Lambda(\Lambda+1)-2l_1(l_1+1)(1-\frac{c_-}{c_+})\
-2l_1^\prime(l_1^\prime+1)(1+\frac{c_-}{c_+})-\frac{2c_-a_-}{c_+}(a_++1) \ \
\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \
\end{array}
</math>

<math>
B_1 =
2\left[(1-x)\alpha_+(a_++2)-\alpha_-a_--2(1-\frac{x}{2})\alpha_++\frac{c_-}{c_+}[a_-\alpha_++\alpha_-(a_++2)]\right]
</math>

<math>
C_1 = (1-x)c_+^2+ c_-^2 +2(1-\frac{x}{2})c_+
</math>

<math>
D_1 = -1(1-x)\alpha_+^2 +\alpha_-^2 - 2\frac{c_-}{c_+}\alpha_-\alpha_+
</math>

The complete Hamiltonian is then (in Z-scaled a.u.)

<math>\begin{eqnarray}
H =-\frac{1}{2}\nabla^2_1 -\frac{1}{2}\nabla^2_2 -\frac{1}{r_1} -\frac{1}{r_2} +\frac{Z^{-1}}{r}\nonumber\\
= -\frac{1}{2}\nabla^2_1 -\frac{1}{2}\nabla^2_2 -\frac{1}{r_1} -\frac{Z-1}{Zr_2}+Z^{-1}\left(\frac{1}{r}-\frac{1}{r_2}\right)\nonumber\\
\ =-\frac{1}{8}\left[\frac{A_1}{r_1^2}+\frac{B_1+8}{r_1}+\frac{C_1}{r^2}+D_1+D_2 +\frac{A_2}{r_2^2} + \frac{B_2 + 8(Z-1)/Z}{r_2}+\frac{C_2}{r^2}\right]\nonumber
+ Z^{-1}\left(\frac{1}{r}-\frac{1}{r_2}\right)\nonumber
\end{eqnarray}</math>

The screened hydrogenic energy is
<math>\begin{eqnarray}
E_{SH}=-\frac{1}{2}\left[\frac{1}{n_1^2}+ \left(\frac{Z-1}{Z}\right)^2\frac{1}{n_2^2}\right]
\end{eqnarray}</math>
so that

<math>\begin{eqnarray}
H-E_{SH} &=& -\frac{1}{8}[
\frac{A_1}{r_1^2}+\frac{B_1+8}{r_1}+\frac{C_1}{r^2} + D_1 + D_2
-\frac{4}{n_1^2} -\left(\frac{Z-1}{Z}\right)^2\frac{4}{n_2^2}\nonumber
+\frac{A_2}{r_2}+\frac{B_2+8(Z-1)/Z}{r_2}+\frac{C_2}{r^2}] +
Z^{-1}\left(\frac{1}{r} -\frac{1}{r_2}\right)\nonumber
\end{eqnarray}</math>

==Optimization of Nonlinear Parameters==
- The traditional method of performing Hylleraas calculations is to write the
basis set in the form

<math>
\Psi = \sum_{i,j,k}c_{i,j,k}\varphi_{i,j,k}(\alpha,\beta)\pm exchange
</math>

with

<math>
\varphi_{i,j,k}(\alpha,\beta) = r_1^ir_2^jr_{12}^ke^{-\alpha r_1 - \beta
r_2}\mathcal{Y}^M_{l_1l_2L}.
</math>

The usual procedure is to set <math>\alpha = Z</math> so that it represents the inner
<math>1s</math> electron, and then to vary <math>\beta</math> so as to minimize the energy.

[[File:OptNon.jpg|500px]]

Since <math>\beta</math> appears in <math>\Psi</math> as a non-linear parameter, the entire
calculation must be repeated for each value of <math>\beta</math>. However, the minimum
becomes progressively smaller as the basis set is enlarged.

[[File:OptNon02.jpg|500px]]

*Difficulties

1. If the basis set is constructed so that <math> i +j +j \leq N</math>, the the
number of terms is
<math>
(N+1)(N+2)(N+3)/6
</math>
N=14 already gives 680 terms and an accuracy of about 1 part in <math>10^{10}</math> for
low-lying states. A substantial improvement in accuracy would require much
larger basis sets, together with multiple precision arithmetic to avoid loss
of significant figures when high powers are included.

2. The accuracy rapidly deteriorates as one goes to more highly excited
states - about 1 significant figure is lost each time the principle quantum
number is increased.

*Cure

We have found that writing basis sets in the form
\begin{eqnarray}
\Psi &=& \sum_{i,j,k}\left(c^{(1)}_{i,j,k}\varphi_{ijk}(\alpha_1\beta_1) +
c_{ijk}^{(2)}\varphi_{ijk}(\alpha_2\beta_2)\right) \pm exchange \nonumber\\
&=& \psi({\bf r}_1,{\bf r}_2) \pm \psi({\bf r}_2,{\bf r}_1)\nonumber
\end{eqnarray}
so that each combination of powers is included twice with different nonlinear
parameters gives a dramatic improvement in accuracy for basis sets of about
the same total size.

However, the optimization of the nonlinear parameters is now much more
difficult, and an automated procedure is needed.

If
<math>
E = \frac{<\Psi|H|\Psi>}{<\Psi|\Psi>}</math> and we assume <math><\Psi|\Psi> = 1</math>, then

<math>
\frac{\partial E}{\partial\alpha_t} = -2 <\Psi|H-E|r_1\psi({\bf r}_1,{\bf
r}_2;\alpha_t)\pm r_21\psi({\bf r}_2,{\bf
r}_1;\alpha_t)>
</math>
where
<math>
\psi({\bf r}_1,{\bf r}_2;\alpha_t) =
\sum_{i,j,k}c_{ijk}^{(t)}\varphi_{ijk}(\alpha_t\beta_t)
</math>

*Problem

1. Prove the above.

2. Prove that there is no contribution from the implicit dependence of
<math>c_{ijk}^{(t)}</math> on <math>\alpha_t</math> if the linear parameters have been optimized.

==The Screened Hydrogenic Term==

If the Hamiltonian is written in the form (in Z-scaled a.u.)
<math>\begin{eqnarray}\
H = H_0({\bf r}_1,Z) + H_0({\bf r}_2,Z-1) +
Z^{-1}\left(\frac{1}{r}-\frac{1}{r_2}\right)
\end{eqnarray}</math>
with
<math>\
H_0({\bf r}_1,Z) = -\frac{1}{2}\nabla_1^2 - \frac{1}{r_1}\nonumber
</math> and
<math>
H_0({\bf r}_2,Z-1) = -\frac{1}{2}\nabla_2^2 - \frac{Z-1}Z{r_2}\nonumber
</math>,
then the eigenvectors of <math>H_0({\bf r},Z) +H_0({\bf r}_2,Z-1)</math> are products of
hydrogenic orbitals
<math>
\Psi_0=\psi_0(1s,Z)\psi_0(nl,Z-1)
</math>
and the eigenvalue is
<math>\begin{eqnarray}
E_{SH} = \left[-\frac{1}{2} - \left(\frac{Z-1}{Z}\right)^2\frac{1}{2n^2}\right]Z^2 a.u.
\end{eqnarray}</math>
called the screened hydrogenic eigenvalue. From highly excited states,
<math>E_{SH}</math> and <math>\Psi_0</math>are already excellent approximations.

For example, for the 1s8d states, the energies are

<math>\begin{eqnarray}\\
E(1s8d^1D)&=& -2.007\,816\,512\,563\,81 {\rm\ a.u.}\nonumber\\
E(1s8d^3D)&=& - 2.007\,817\,934\,711\,71 {\rm\ a.u.}\nonumber\\
E_{SH} &=& -2.007\,812\,5 {\rm\ a.u.} \nonumber
\end{eqnarray}</math>

It is therefore advantageous to include the screened hydrogenic terms in the
basis set so that the complete trial function becomes
<math>
\Psi = c_0\Psi_0 +
\sum_{ijk}\left[c_{ijk}^{(1)}\varphi_{ijk}(\alpha_1,\beta_1)+c_{ijk}^{(2)}\varphi(\alpha_2,\beta_2)\right]\pm
{\rm exchange}
</math>

Also, the variational principal can be re-expressed in the form
<math>\begin{eqnarray}
E= E_{SH}+\frac{<\Psi|H-E_{SH}|\Psi>}{<\Psi|\Psi>}
\end{eqnarray}</math>
so that the <math>E_{SH}</math> term can be cancelled analytically from the matrix
elements.

For example
<math>
<\Psi_0|H-E_{SH}|\Psi_0> = 0
</math>

and
<math>
<\varphi_{ijk}|H-E_{SH}|\Psi_0> =
Z^{-1}<\varphi_{ijk}|\frac{1}{r}-\frac{1}{r_2}|\Psi_0>
</math>

Recall that
<math>\begin{eqnarray}\ \
I_0(a,b,c) = \frac{2}{c+2}\sum^{[c+1]/2}_{i=0}\left(\begin{array}{c}
c+2\\
2i+1
\end{array}\right) \left[f(p,q;\beta) + f(p^\prime,q^\prime;\alpha)\right]
\end{eqnarray}</math>
where

<math>\begin{eqnarray}
f(p,q;x) =
\frac{q!}{x^{q+1}(\alpha+\beta)^{p+1}}\sum^p_{j=0}\frac{(p+j)!}{j!}\left(\frac{x}{\alpha+\beta}\right)^j
\end{eqnarray}</math>

<math>
\begin{array}{cc}
p=a+2i+2 & p^\prime = b+2i+2\\
q= b+c-2i+2 & q^\prime =a+c-2i+2
\end{array}
</math>

If <math>\beta\ll\alpha</math>, the <math>f(p,q;\beta)\gg f(p^\prime,q^\prime;\alpha)</math> and the
<math>i=0</math> term is the dominant contribution to <math>f(p,q;\beta)</math>. However, since
this term depends only on the sum of powers <math>b+c</math> for <math>r_2</math>, it cancels exactly
from the matrix element of
<math>
\frac{1}{r_{12}} - \frac{1}{r_2}
</math>
and can therefore
be omitted in the calculations of integrands, thereby saving many significant
figures. This is especially valuable when <math>b+c</math> is large and <math>a</math> is small.

==Small Corrections==
*Mass Polarization
If the nuclear mass is not taken to be infinite, then the Hamiltonian is
<math>\begin{eqnarray}\ \
H = \frac{P_N^2}{2M} + \sum_{i=1}\frac{p_i^2}{2m} + V
\end{eqnarray}</math>

[[File:SmallCor.jpg|500px]]

where <math>M =M_A-nm</math> is the nuclear mass and <math>m</math> is the electron mass.

Change variables to the C of M
<math>\begin{eqnarray}
{\bf R} = \frac{1}{M+nm}\left[M{\bf r}_N + m({\bf r}_1 + {\bf r}_2 +
\ldots)\right]
\end{eqnarray}</math>

and relative variables

<math>
{\bf s}_i = {\bf r}_i - {\bf r}_N.
</math>

Then

<math>\begin{eqnarray}\ \ \ \ \
H = \frac{1}{2(M+nm)}{\bf p}_R^2+\frac{1}{2m}\sum_{i=1}^n{\bf p}_{s_i}^2
+\frac{1}{2M}\sum_{i,k}{\bf p}_{s_i}\cdot {\bf p}_{s_k} + V({\bf s}_1, {\bf
s}_2, \ldots)\nonumber
= \frac{1}{M+nm}{\bf p}_R^2+\frac{1}{2\mu}\sum_{i=1}^n{\bf
p}_{s_i}^2+\frac{1}{2M}\sum_{i\neq k}{\bf p}_{s_i}\cdot {\bf p}_{s_k}+V({\bf s}_1, {\bf
s}_2, \ldots)\nonumber
\end{eqnarray}</math>

where
<math>\begin{eqnarray}
\mu = \left(\frac{1}{M} + \frac{1}{m}\right)^{-1}
\end{eqnarray}</math>

If <math>n=2</math>, the Schrödinger equation is
<math>\begin{eqnarray}\ \ \
\left[ -\frac{\hbar^2}{2\mu}(\nabla^2_{s_1}+ \nabla^2_{s_2}) -
\frac{\hbar^2}{M}\nabla_{s_1}\nabla_{s_2}-\frac{Ze^2}{s_1}-\frac{Ze^2}{s_2}+\frac{e^2}{s_{12}}\right]\psi
= E\psi
\end{eqnarray}</math>.

Define the reduced mass Bohr radius
<math>\begin{eqnarray}
a_\mu = \frac{\hbar^2}{\mu e^2} = \frac{m}{\mu}a_0
\end{eqnarray}</math>
and

<math>\begin{eqnarray}
\rho_i = \frac{Zs_i}{a_\mu}\nonumber\\
\varepsilon = \frac{E}{Z^2(e^2/a_\mu)}\nonumber
\end{eqnarray}</math>

The Schrödinger equation then becomes
<math>\begin{eqnarray}\ \ \
\left[
-\frac{1}{2}(\nabla^2_{\rho_1}+\nabla^2_{\rho_2})-\frac{\mu}{M}\vec{\nabla}_{\rho_1}\cdot\vec{\nabla}_{\rho_2} - \frac{1}{\rho_1}-\frac{1}{\rho_2}+\frac{Z^{-1}}{\rho_{12}}\right]\psi = \varepsilon \psi
\end{eqnarray}</math>

The effects of finite mass are thus

1. A reduced mass correction to all energy levels of <math>(e^2/a_\mu)/(e^2/a_0)
= \mu/m</math>, i.e.
<math>\begin{eqnarray}
E_M=\frac{\mu}{m}E_\infty
\end{eqnarray}</math>

2. A specific mass shift given in first order perturbation theory by
<math>\begin{eqnarray}
\Delta\varepsilon =
-\frac{\mu}{M}<\psi|\nabla_{\rho_1}\cdot\nabla_{\rho_2}|\psi>
\end{eqnarray}</math>
or

<math>\begin{eqnarray}
\Delta\varepsilon =
-\frac{\mu}{M}<\psi|\nabla_{\rho_1}\cdot\nabla_{\rho_2}|\psi>\frac{e^2}{a_\mu}\nonumber\\
= -\frac{\mu^2}{mM}<\psi|\nabla_{\rho_1}\cdot\nabla_{\rho_2}|\psi>\frac{e^2}{a_0}\nonumber
\end{eqnarray}</math>

Since <math>\nabla_{\rho_1}\cdot\nabla_{\rho_2}</math> has the same angular properties as
<math>{\bf\rho}_1 \cdot {\bf\rho}_2</math>, the operator is like the product of two
dipole operators. For product type wave functions of the form
$
\psi = \psi(1s)\psi(nl) \pm \mbox{exchange},
$
the matrix element vanishes for all but P-states.

=Radiative Transitions=

==Transition Integrals and the Line Strength==

For an electric dipole transition <math>\gamma'\,L'\,S'\,J'\,M' \rightarrow \gamma\,L\,S\,J\,M</math>, the line strength is defined by
\begin{equation}
S = \sum_{M',\mu,M}|\langle\gamma'\,L'\,S'\,J'\,M'|r_\mu|\gamma\,L\,S\,J\,M\rangle|^2
\end{equation}
Using the Wigner-Eckart theorem, this is related to the reduced matrix element by
\begin{eqnarray}
S(J',J) = |\langle\gamma'\,L'\,S'\,J'||r||\gamma\,L\,S\,J\,M\rangle|^2
\sum_{M',\mu,M}\left|\left(\matrix{J' & 1 & J \cr
-M'& \mu &M \cr}\right)\right|^2
\end{eqnarray}
Using the sum rule for the 3-<math>j</math> symbols
\begin{equation}
\sum_{M',\mu,M}\left|\left(\matrix{J' & 1 & J \cr
-M'& \mu &M \cr}\right)\right|^2 = 1
\end{equation}
we see that the line strength is identical to the reduced matrix element
\begin{equation}
S(J',J) = |\langle\gamma'\,L'\,S'\,J'||r||\gamma\,L\,S\,J\,M\rangle|^2
\end{equation}
For a nonrelativistic calculation in <math>LS</math> coupling, <math>S = S'</math> and we can next strip out the dependence on <math>J</math> and <math>J'</math> by use of Eq.\ (7.1.7) of Edmonds { Angular Momentum in Quantum Mechanics} with <math>k = 1</math> to obtain
\begin{equation}
S(J',J) = (-1)^{L'+S+J+1}[(2J+1)(2J'+1)]^{1/2}\left\{
\matrix{L' & J'& S\cr
J & L & 1\cr}\right\}\langle\gamma' L'|| r || \gamma L\rangle
\end{equation}
where <math>\langle\gamma' L'|| r || \gamma L\rangle</math> is the reduced matrix element if spin is neglected. Note that if spin is neglected, then the line-strength is simply
\begin{equation}
S(L',L) = |\langle\gamma' L'|| r || \gamma L\rangle|^2
\end{equation}
To calculate the reduced matrix element, use a particular example along with the Wigner-Eckart theorem
such as
\begin{equation}
\langle\gamma'\, L'|| r || \gamma\, L\rangle = (-1)^{L'}\langle\gamma'\, L'\, 0| z | \gamma\, L\, 0\rangle{\Large /}\left(\matrix{L' & 1 & L \cr
0 & 0 & 0 \cr}\right)
\end{equation}

===Correlated Two-electron Integrals===
For the case of two-electron integrals, <math>z = z_1 + z_2</math> and <math>r = r_1 + r_2</math>. As discussed in G.W.F. Drake, Phys. Rev. A 18, 820 (1978), the general integral in Hylleraas coordinates for the integral over the <math>z_1</math> term can be written in the form
\begin{equation}
\label{int1}
\langle\gamma'\, L'\, 0| z_1 | \gamma\, L\, 0\rangle = \sum_\Lambda C_\Lambda I_\Lambda(\gamma',\gamma)
\end{equation}
where <math>I_\Lambda(\gamma',\gamma)</math> stands for the purely radial integral
\begin{equation}
\label{radial}
I_\Lambda(\gamma',\gamma) = \int_0^\infty r_1\,dr_1\int_0^\infty r_2\,dr_2\int_{|r_1-r_2|}^{r_1+r_2}r_{12}\, dr_{12}
R_{\gamma'}^*\,r_1\, R_\gamma P_\Lambda(\cos\theta_{12})
\end{equation}
and <math>\cos\theta_{12} = (r_1^2 + r_2^2 - r_{12}^2)/(2r_1r_2)</math> is a short-hand notation for a radial function of <math>r_1</math>, <math>r_2</math> and <math>r_{12}</math>. Also, <math>P_\Lambda(\cos\theta_{12})</math> is a Legendre polynomial,
and, <math>R_\gamma = R_\gamma(r_1,r_2,r_{12})</math> is the radial part of the Hylleraas wave function for state <math>\gamma</math> such
that the total wave function is <math>\psi({\bf r}_1,{\bf r}_2) = R_\gamma(r_1,r_2,r_{12}){\cal Y}_{l_1,l_2,L}^M(\hat{\bf r}_1,\hat{\bf r}_2)\pm {\rm exchange}</math>, normalized to unity, and <math>{\cal Y}_{l_1,l_2,L}^M(\hat{\bf r}_1,\hat{\bf r}_2)</math> is a vector coupled product of spherical harmonics. The coefficents <math>C_\Lambda</math> in Eq.\ (\ref{int1}) can be calculated from Eq.\ (32) of the above 1978 paper. They need to be calculated separately for he direct-direct and the direct-exchange contributions. For the reduced matrix element of <math>z_1</math>, the leading 3-<math>j</math> symbol can be dropped, leaving (for the case <math>k = 1</math>)
\begin{eqnarray}
C_\Lambda^{\rm reduced} &=& \frac{(-1)^{l_2'+L'+L+1}}{2} (l_1,l_1',l_2,l_2',L,L')^{1/2}
\sum_{\lambda={\rm max}(l_1-1,0)}^{l_1+1} (\lambda,\Lambda,1)
\left(\matrix{l_1 & 1 & \lambda \cr
0 & 0 & 0 \cr}\right)
\left(\matrix{l_1'& \lambda & \Lambda \cr
0 & 0 & 0 \cr}\right)\nonumber\\
&\times&\left(\matrix{l_2' & 1_2 & \Lambda \cr
0 & 0 & 0 \cr}\right)
\left\{\matrix{L' & \lambda & l_2 \cr
\Lambda& l_2' & l_1' \cr}\right\}
\left\{\matrix{L' & \lambda & l_2 \cr
l_1 & L & 1 \cr}\right\}
\end{eqnarray}
where the notation <math>(a,b,c,\cdots)</math> means <math>(2a+1)(2b+1)(2c+1)\cdots</math>.
A similar expression can be obtained for the angular coefficients for the <math>z_2</math> term by interchanging the subscripts 1 and 2 throughout and multiplying by <math>(-1)^{L+L'+1}</math>.

===Example===
As an example, for the <math>1s^2\;^1S - 1s2p\;^1P</math> transition of helium, the only nonvanishing <math>C_\Lambda</math> angular coefficients are <math>C_1 = 1/2</math> for the direct-direct term (i.e. <math>P_1(\cos\theta_{12}) = \cos\theta_{12}</math> is included in the radial integral (\ref{radial})) and <math>C_0 = 1/2</math> for the direct-exchange term (i.e. <math>P_0(\cos\theta_{12}) = 1</math> is included in the radial integral (\ref{radial})). The results for infinite nuclear mass are
\begin{equation}
S(1\;^1S-2\;^1P) = (0.781\,127\,158\,667)^2
\end{equation}
\begin{equation}f(1\;^1S-2\;^1P) = -0.276\,164\,703\,736
\end{equation}
where
\begin{equation}
f(1\;^1S-2\;^1P) = \frac{2}{3}[E(1\;^1S) - E(2\;^1P)]S(1\;^1S-2\;^1P)
\end{equation}
is the absorption oscillator strength.

===Test===
This is a test edit

Theory Notes

2020-12-14T05:21:56Z

Drake: Undo revision 798 by Drake (talk)

<center>
<H1>Notes on solving the Schrödinger equation in Hylleraas coordinates for heliumlike atoms
Gordon W.F. Drake
Department of Physics, University of Windsor
Windsor, Ontario, Canada N9B 3P4
(Transcribed from hand-written notes and edited by Lauren Moffatt. Last revised November 22, 2017 to correct a missing subscript in the expression for $B_1$ in the Hamiltonian in Sect. 11 General Hermitian Property.)
</H1>
</center>

==Introduction==
These notes describe some of the technical details involved in solving the Schrödinger equation for a heliumlike atom of nuclear charge Z in correlated Hylleraas coordinates. The standard for computational purposes is to express all quantities in a form that is valid for electronic states of arbitrary angular momentum, and to express matrix elements of the Hamiltonian in a manifestly Hermitian symmetrized form. Further information and derivations of the mathematical techniques for angular integrals and radial recursion relations can be found in the paper
G.W.F. Drake, Phys. Rev. A <math>{\bf 18}</math>, 820 (1978).

The starting point is the two-electron Schrödinger equation for infinite nuclear mass
\begin{equation} \left[-\frac{\hbar^2}{2m}(\nabla^2_1 +\nabla^2_2) - \frac{Ze^2}{r_1} -
\frac{Ze^2}{r_2}+\frac{e^2}{r_{12}} \right]\psi = E\psi\nonumber
\end{equation}
where <math>m</math> is the electron mass, and <math>r_{12} = |{\bf r}_1 - {\bf r}_2|</math> (see diagram below).

[[File:Pic1.jpg|250px]]

Begin by rescaling distances and energies so that the Schrödinger equation can be expressed in a dimensionless form. The dimensionless Z-scaled distance is defined by <math>\begin{eqnarray} \rho = \frac{Zr}{a_0} \end{eqnarray} </math> where <math> \begin{eqnarray} a_0 = \frac{\hbar^2}{me^2} \end{eqnarray} </math> is the atomic unit (a.u.) of distance equal to the Bohr radius <math>0.529\, 177\, 210\, 92(17) \times 10^{-10}</math> m. Then

\begin{equation}[-\frac{\hbar^2}{2m}Z^2(\frac{me^2}{\hbar^2})^2(\nabla^2_{\rho_1}+\nabla^2_{\rho_2})
- Z^2\frac{e^2}{a_0}\rho^{-1}_1 - Z^2\frac{e^2}{a_0}\rho^{-1}_2 +
\frac{e^2}{a_0}Z\rho^{-1}_{12}]\psi= E\psi\nonumber\end{equation}

But <math>\begin{eqnarray} \frac{\hbar^2}{m}(\frac{me^2}{\hbar^2})^2 = \frac{e^2}{a_0} \end{eqnarray} </math> is the hartree atomic unit of of energy <math>(E_h = 27.211\,385\,05(60)</math> eV, or equivalently <math>E_h/(hc) = 2194.746\,313\,708(11)</math> m<math>^{-1})</math>. Therefore, after multiplying through by <math>a_0/(Ze)^2</math>, the problem to be solved in Z-scaled dimensionless units becomes

\begin{equation} \left[-\frac{1}{2}(\nabla^2_{\rho_1}+\nabla^2_{\rho_2}) -
\frac{1}{\rho_1} - \frac{1}{\rho_2} + \frac{Z^{-1}}{\rho_{12}}\right]\psi =
\varepsilon\psi\nonumber \end{equation}

where <math>\begin{eqnarray}\varepsilon = \frac{Ea_0}{(Ze)^2} \end{eqnarray}</math> is the energy in Z-scaled atomic units. For convenience, rewrite this in the conventional form
\begin{equation}
H\psi = \varepsilon\psi\nonumber \end{equation}
where (using <math>r</math> in place of <math>\rho</math> for the Z-scaled distance)
\begin{equation}
H = -\frac{1}{2}(\nabla^2_1+\nabla^2_2) - \frac{1}{r_1}-\frac{1}{r_2} + \frac{Z^{-1}}{r_{12}}
\nonumber\end{equation}
is the atomic Hamiltonian for infinite nuclear mass.

==The Hartree Fock Method==

For purposes of comparison, and to define the correlation energy, assume that <math>\psi({\bf r}_1,{\bf r}_2)</math> can be written in the separable product form
\begin{equation}
\psi({\bf r}_1,{\bf r}_2) = \frac{1}{\sqrt{2}} \left[ u_1(r_1) u_2(r_2) \pm u_2(r_1) u_1(r_2) \right] \nonumber
\end{equation}
for the <math>1s^2\ ^1\!S</math> ground state. Because of the <math>\frac{1}{r_{12}}</math> term in the Schrödinger equation, the exact solution connot be expressed in this form as a separable product. However, the Hartree-Fock approximation corresponds to finding the best possible solution to the Schrödinger equation
\begin{equation}
H\psi({\bf r}_1,{\bf r}_2) = E\psi({\bf r}_1,{\bf r}_2)\nonumber
\end{equation}
that can nevertheless be expressed in this separable product form, where as before
\begin{equation}
H = -\frac{1}{2}(\nabla^2_1+\nabla^2_2) - \frac{1}{r_1}- \frac{1}{r_2} + \frac{Z^{-1}}{r_{12}}\nonumber
\end{equation}
is the full two-electron Hamiltonian. To find the best solution, substitute into <math>\left\langle\psi \lvert H-E \rvert \psi\right\rangle</math> and require this expression to be stationary
with respect to arbitrary infinitesimal variations <math>\delta u_1</math> and <math>\delta
u_2</math> in <math>u_1</math> and <math>u_2</math> respectively; i.e.

\begin{equation}
\frac{1}{2} \left\langle \delta u_1(r_1)u_2(r_2) \pm u_2(r_1)\delta
u_1(r_2) \lvert H-E \rvert u_1(r_1) u_2(r_2)\pm u_2(r_1) u_1(r_2)\right\rangle\nonumber
\end{equation}

\begin{equation}
=\int\delta u_1(r_1) d{\bf r}_1 \left\{ \int d{\bf r}_2 u_2(r_2) \left( H - E \right) \left[ u_1(r_1)u_2(r_2) \pm u_2(r_1) u_1(r_2) \right] \right\}\nonumber
\end{equation}

\begin{equation}
= 0 \nonumber
\end{equation}

for arbitrary <math> \delta u_1(r_1).\nonumber </math> Therefore <math>\{\int d{\bf r}_2 \ldots \} = 0</math>.

Similarly, the coefficient of <math>\delta u_2</math> would give

\begin{equation}
\int d{\bf r}_1 u_1(r_1) \left( H-E \right) \left[ u_1(r_1) u_2(r_2) \pm
u_2(r_1) u_1(r_2) \right] = 0\nonumber
\end{equation}

Define

\begin{equation}
I_{12} = I_{21} = \int d{\bf r}\, u_1(r)u_2(r), \nonumber
\end{equation}

\begin{equation}
H_{ij} = \int d{\bf r}\, u_i(r)(-\frac{1}{2}\nabla - \frac{1}{r})u_j(r), \nonumber
\end{equation}

\begin{equation}
G_{ij}(r) = \int d{\bf r}^\prime u_i(r^\prime)\frac{1}{|{\bf r} - {\bf r}\prime|}u_j(r^\prime)\nonumber
\end{equation}

Then the above equations become the pair of integro-differential equations

\begin{equation}
[ H_0 - E + H_{22}+G_{22}(r)]u_1(r) = \mp [ I_{12}(H_0-E) + H_{12}+G_{12}(r)]u_2(r)\nonumber
\end{equation}

\begin{equation}
[H_0-E+H_{11}+G_{11}(r)]u_2(r) = \mp [I_{12}(H_0-E) + H_{12}+G_{12}(r)]u_1(r)\nonumber
\end{equation}

These must be solved self-consistently for the "constants" <math>I_{12}</math> and
<math>H_{ij}</math> and the function <math>G_{ij}(r)</math>.

The Hartree Fock energy is <math>E \simeq -2.87\ldots </math>a.u. while the exact energy is <math>E = -2.903724\ldots </math>a.u.

The difference is called the "correlation energy" because it arises from the
way in which the motion of one electron is correlated to the other. The
Hartree Fock equations only describe how one electron moves in the average field
provided by the other (mean-field theory).

==Configuration Interaction==
Expand
\begin{equation}
\psi({\bf r}_1,{\bf r}_2)= C_0u^{(s)}_1(r_1)u^{(s)}_1(r_2) + C_1u^{(P)}_1({\bf r}_1)u^{(P)}_1({\bf r}_2){\cal Y}^0_{1,1,0}(\hat{\bf r}_1, \hat{\bf r}_2)+C_2u^{(d)}_1({\bf r}_1)u^{(d)}_2({\bf r}_2){\cal Y}^0_{2,2,0}(\hat{\bf r}_1, \hat{\bf r}_2)+... \pm {\rm\ exchange}
\end{equation}
where
\begin{equation}
{\cal Y}^M_{l_1,l_2,L}(\hat{\bf r}_1, \hat{\bf r}_2)=\sum_{m_1,m_2}Y^{m_1}_{l_1}({\bf r}_1)Y^{m_2}_{l_2}({\bf r}_2)\times <l_1l_2m_1m_2\mid LM>
\end{equation}

This works, but is slowly convergent, and very laborious. The best CI calculations are accurate to <math> ~10^{-7}</math> a.u.

==Hylleraas Coordinates==
[E.A. Hylleraas, Z. Phys. <math>{\bf 48}, 469(1928)</math> and <math>{\bf 54}, 347(1929)</math>]
suggested using the co-ordinates <math>r_1</math>, <math>r_2</math> and <math>r_{12}</math> or equivalently

\begin{eqnarray}
s &=& r_1 + r_2, \nonumber\\
t &=& r_1-r_2, \nonumber\\
u &=& r_{12}\nonumber
\end{eqnarray}

and writing the trial functions in the form

\begin{equation}
\Psi({\bf r}_1,{\bf r}_2) = \sum^{i+j+k\leq
N}_{i,j,k}c_{i,j,k}r_1^{i+l_1}r_2^{j+l_2}r_{12}^ke^{-\alpha r_1 - \beta r_2}
\mathcal{Y}^M_{l_1,l_2,L}(\hat{r}_1,\hat{r}_2)\pm \text{exchange}\nonumber
\end{equation}

Diagonalizing H in this non-orthogonal basis set is equivalent to solving
<math> \begin{eqnarray}
\frac{\partial E}{\partial c_{i,j,k}} = 0\nonumber
\end{eqnarray} </math>
for fixed <math>\alpha</math> and <math>\beta</math>.

The diagonalization must be repeated for different values of <math>\alpha</math> and
<math>\beta</math> in order to optimize the non-linear parameters.

==Completeness==
The completeness of the above basis set can be shown by first writing
\begin{equation}
r_{12}^2 = r_1^2 + r_2^2 - 2r_1r_2\cos\theta_{12}
\end{equation}
and, from the spherical harmonic addition theorem,
\begin{equation}
\cos(\theta_{12})=\frac{4\pi}{3}\sum^1_{m=-1}Y^{m*}_1(\theta_1,\varphi_1)Y^m_1(\theta_2,\varphi_2)
\end{equation}
Consider first the S-states. The <math>r_{12}^0</math> terms are like the ss terms in a CI calculation. The
<math>r_{12}^2</math> terms bring in p-p type contributions, and the higher powers bring
in d-d, f-f etc. type terms. In general

\begin{equation}
P_l(\cos\theta_{12}) = \frac{4\pi}{2l+1}\sum^l_{m=-l}{Y^{m}_l}^*(\theta_1,\varphi_1)Y^m_l(\theta_2,\varphi_2)\nonumber
\end{equation}

For P-states, one would have similarly

\begin{array}{lr}
r_{12}^0 & (sp)P\nonumber\\
r_{12}^2 & (pd)P\nonumber\\
r_{12}^4 & (df)P\nonumber\\
\vdots & \vdots\nonumber
\end{array}

For D-states

\begin{array}{lr}
r_{12}^0 & (sd)D & (pp^\prime)D\nonumber\\
r_{12}^2 & (pf)D & (dd^\prime)D\nonumber\\
r_{12}^4 & (dg)D & (ff^\prime)D\nonumber\\
\vdots & \vdots & \vdots\nonumber
\end{array}

In this case, since there are two "lowest-order" couplings to form a
D-state, both must be present in the basis set, i.e.

\begin{equation}
\Psi(r_2,r_2) = \sum c_{ijk}r_1^ir_2^{j+2}r_{12}^ke^{-\alpha r_1-\beta
r_2}\mathcal{Y}^M_{022}(\hat{r}_1,\hat{r}_2)
+\sum d_{ijk}r_1^{i+1}r_2^{j+1}r_{12^k}e^{-\alpha^\prime r_1 - \beta^\prime
r_2}\mathcal{Y}^M_{112}(\hat{r}_1,\hat{r}_2)\nonumber
\end{equation}

For F-states, one would need <math>(sf)F</math> and <math>(pd)F</math> terms.

For G-states, one would need <math>(sg)G</math>, <math>(pf)G</math> and <math>(dd^\prime)G</math> terms.

Completeness of the radial functions can be proven by considering the
Sturm-Liouville problem

\begin{equation}
\left(-\frac{1}{2}\nabla^2-\frac{\lambda}{r_s}-E\right)\psi({\bf r}) = 0\nonumber
\end{equation}

or

\begin{equation}
\left(-\frac{1}{2}\frac{1}{r^2}\frac{\partial}{\partial r}\left({r^2}\frac{\partial}{\partial r}\right) -
\frac{l\left(l+1\right)}{2r^2} - \frac{\lambda}{r} - E\right)u(r) = 0.\nonumber
\end{equation}

For fixed E and variable <math>\lambda</math> (nuclear charge).

The eigenvalues are <math>\lambda_n = (E/E_n)^{1/2}</math>, where <math>E_n =- \frac{1}{2n^2}</math>

[[File:Pic2.jpg|500px]]

and the eigenfunctions are

\begin{equation}
u_{nl}(r) =
\frac{1}{(2l+1)!}\left(\frac{(n+l)!}{(n-l-1)2!}\right)^{1/2}(2\alpha)^{3/2}e^{-\alpha
r}\nonumber
\times \ \left( 2 \alpha r \right)^l\,_{1}F_{1}\left(-n+l+1,2l+2; 2 \alpha r \right)
\end{equation}

with <math>\alpha = (-2E)^{1/2}</math> and <math>n\geq l+1</math>. The confluent hypergeometric function <math> _1F_1(a,b,;z)</math> then denotes a finite polynomial since <math>a=-n+l+1</math> is a negative integer or zero.

Unlike the hydrogen spectrum, which has both a discrete part for <math>E<0</math> and a
continuous part for <math>E>0</math>, this forms an entirely discrete set of finite
polynomials, called Sturmian functions. They are orthogonal with respect to
the potential, i.e.

\begin{equation}
\int^\infty_0 r^2dr \left( u_{n^\prime l}(r)\frac{1}{r}u_{nl}(r) \right) = \delta_{n,n^\prime}\nonumber
\end{equation}

Since they become complete in the limit <math>n\rightarrow\infty</math>, this assures the
completeness of the variational basis set.

[See also B Klahn and W.A. Bingel Theo. Chim. Acta (Berlin) 44, 9 and 27
(1977)].

==Solutions of the Eigenvalue Problem==

For convenience, write

\[
\Psi({\bf r}_1,{\bf r}_2) = \sum^N_{m=1}c_m\varphi_m\nonumber
\]

where <math>m</math> represents the <math>m</math>'th combination of <math>i,j,k</math> and

\begin{equation}
\varphi_{ijk}=r_1^ir_2^jr_{12}^ke^{-\alpha r_1 - \beta
r_2}\mathcal{Y}^M_{l_1,l_2,L}(\hat{r}_1,\hat{r}_2) \pm \mbox{exchange}\nonumber
\end{equation}

\begin{align}

&
\left( \begin{array}{lr}
\cos(\theta) & \sin(\theta)\\
-\sin(\theta) & \cos(\theta)
\end{array} \right)
\left( \begin{array}{lr}
H_{11} & H_{12} \\
H_{12} & H_{22}
\end{array}\right)
\left( \begin{array}{lr}
\cos(\theta) & -\sin(\theta)\\
\sin(\theta) & \cos(\theta)
\end{array} \right)
\\[5pt]
= & \left(\begin{array}{lr}
cH_{11}+sH_{12} & cH_{12} + sH_{22}\\
-sH_{11} + cH_{12} & -sH_{12} + cH_{22}
\end{array}\right)
\left( \begin{array}{cc}
c & -s \\
s & c
\end{array}\right)
\\[5pt]
= & \left(\begin{array}{lr}
c^2H_{11}+s^2H_{22} + 2csH_{12} & (c^2-s^2)H_{12}+cs(H_{22}-H_{11})\\
(c^2-s^2)H_{12}+cs(H_{22}-H_{11}) & s^2H_{11}+c^2H_{22}-2csH_{12}
\end{array}\right)
\end{align}

Therefore

\begin{equation}
\left[\cos^2(\theta)-\sin^2(\theta)\right]H_{12} =
\cos(\theta)\sin(\theta)(H_{11}-H_{22})\nonumber
\end{equation}

and

\begin{equation}
\tan(2\theta) = \frac{2H_{12}}{H_{11}-H_{22}}\nonumber
\end{equation}

i.e.

\begin{equation}
\cos(\theta)=\left(\frac{r+\omega}{2r}\right)^{1/2}\nonumber
\end{equation}

\begin{equation}
\sin(\theta)=-\mbox{sgn}(H_{12})\left(\frac{r-\omega}{2r}\right)^{1/2}\nonumber
\end{equation}

where

\begin{equation}
\omega = H_{22}-H_{11}\nonumber
\end{equation}

\begin{equation}
r=\left(\omega^2+4H_{12}^2\right)^{1/2}\nonumber
\end{equation}

\begin{equation}
E_1=\frac{1}{2}\left(H_{11}+H_{22}-r\right)\nonumber
\end{equation}

\begin{equation}
E_2=\frac{1}{2}\left(H_{11}+H_{22}+r\right)\nonumber
\end{equation}

===Brute Force Method===

- Gives all the eigenvalues and eigenvectors, but it is slow

- First orthonormalize the <math>N</math>-dimensional basis set; i.e. form linear combinations

\begin{equation}
\Phi_m = \sum_{n=1}^N\varphi_nR_{nm}\nonumber
\end{equation}

such that
\begin{equation}
\langle \Phi_m | \Phi_n \rangle = \delta_{m,n}
\end{equation}

This can be done by finding an orthogonal tranformation T such that

\begin{equation}
T^TOT = I = \left(
\begin{array}{ccccc}
I_1 & 0 & 0 & \ldots & 0 \\
0 & I_2 & 0 & \ldots & 0 \\
0 & 0 & I_3 & \ldots & 0 \\
\vdots & \vdots & \vdots & \ddots & \vdots \\
0 & 0 & 0 & \ldots & I_N
\end{array}\right);
\end{equation}

\begin{equation}
O_{mn} = \langle \varphi_m | \varphi_n \rangle \nonumber
\end{equation}

and then applying a scale change matrix

\begin{equation}
S = \left(\begin{array}{ccccc}
\frac{1}{I_1^{1/2}} & 0 & 0 & \ldots & 0\\
0 & \frac{1}{I_2^{1/2}} & 0 & \ldots & 0 \\
0 & 0 & \frac{1}{I_3^{1/2}} & \ldots & 0 \\
\vdots & \vdots & \vdots & \ddots & \vdots \\
0 & 0 & 0 & \ldots & I_N^{1/2}
\end{array}\right)= S^T\nonumber
\end{equation}

Then

\begin{equation}
S^TT^TOTS = 1
\end{equation}

i.e.

\begin{equation}
R^TOR = 1
\end{equation}

with <math>R=TS</math>.

If H is the matrix with elements <math>H_{mn} = \langle \varphi_m | \varphi_n \rangle </math>, then H
expressed in the <math>\Phi_m</math> basis set is

\begin{equation}
H^\prime = R^THR.\nonumber
\end{equation}

We next diagonalize <math>H^\prime</math> by finding an orthogonal transformation W such
that

\begin{equation}
W^T H^\prime W = \lambda = \left(
\begin{array}{cccc}
\lambda_1 & 0 & \ldots & 0 \\
0 & \lambda_2 & \ldots & 0 \\
\vdots & \vdots & \ddots & \vdots \\
0 & 0 & \ldots & \lambda_N
\end{array}\right)\nonumber
\end{equation}

The q'th eigenvector is

\begin{equation}
\Psi^{(q)} = \sum_{n=1}^N\Phi_n W_{n,q} = \sum_{n,n^\prime} \varphi_{n^\prime} R_{n^\prime ,n} W_{n,q} \nonumber
\end{equation}

i.e.

\begin{equation}
c_{n^\prime}^{(q)} = \sum_{n=1}^N R_{n^\prime n} W_{n,q}
\end{equation}

===The Power Method===

- Based on the observation that if H has one eigenvalue, <math>\lambda_M</math>, much bigger than all the rest, and <math>\chi = \left(
\begin{array}{c}a_1\\a_2\\\vdots\end{array}\right)</math>
is an arbitrary starting vector, then <math>\chi = \sum_{q=1}^N x_q\Psi^{(q)}</math>.

\begin{equation}
(H)^n\chi = \sum_{q=1}^N x_q \lambda^n_q\Psi^{(q)}\nonumber\rightarrow x_M\lambda_M^n\Psi^{(M)}\nonumber
\end{equation}

provided <math>x_M\neq 0</math>.

To pick out the eigenvector corresponding to any eigenvalue, write the original problem in the form

\begin{align}
H\Psi & = \lambda O \Psi \nonumber \\
\left( H - \lambda_q O \right) \Psi & = \left( \lambda - \lambda_q \right) O \Psi \nonumber
\end{align}

Therefore,

\begin{equation}
G \Psi = \frac{1}{\lambda - \lambda_q } \Psi \nonumber \\
\end{equation}

where <math>G=(H-\lambda_qO)^{-1}O</math> with eigenvalues <math>\frac{1}{\lambda_n - \lambda_q}</math>.

By picking <math>\lambda_q</math> close to any one of the <math>\lambda_n</math>, say
<math>\lambda_{n^\prime}</math>, then <math>\frac{1}{\lambda_n-\lambda_q}</math> is much larger for
<math>n=n^\prime</math> than for any other value. The sequence is then

\begin{array}{c}
\chi_1=G\chi\nonumber\\
\chi_2=G\chi_1\nonumber\\
\chi_3=G\chi_2\nonumber\\
\vdots\nonumber
\end{array}

until the ratios of components in <math>\chi_n</math> stop changing.

- To avoid matrix inversion and multiplication, note that the sequence is equivalent to
\begin{equation}
F \chi_n = \left( \lambda - \lambda_q \right) O \chi_{n-1} \nonumber \\
\end{equation}
where <math>F = H-\lambda_qO</math>. The factor of <math>\left( \lambda - \lambda_q \right)</math> can be
dropped because this only affects the normalization of <math>\chi_n</math>. To find <math>\chi_n</math>, solve
\begin{equation}
F \chi_n = O \chi_{n-1} \nonumber
\end{equation}
(N equations in N unknowns). Then
\begin{equation}
\lambda = \frac{ \langle \chi_n | H | \chi_n \rangle}{\langle \chi_n | \chi_n \rangle} \nonumber
\end{equation}

==Matrix Elements of H==

\begin{equation}
H=-\frac{1}{2}\nabla^2_1 -\frac{1}{2}\nabla^2_2 - \frac{1}{r_1} - \frac{1}{r_2} +\frac{Z^{-1}}{r_{12}}\nonumber
\end{equation}

Taking <math>r_1, r_2</math> and <math>r_{12}</math> as independent variables,

\begin{equation}
\nabla_1^2 = \frac{1}{r_1^2}\frac{\partial}{\partial r_1}\left(r_1^2\frac{\partial}{\partial r_1}\right)+ \frac{1}{r_{12}^2}\frac{\partial}{\partial r_{12}}
\left(r_{12}^2\frac{\partial}{\partial r_{12}}\right)\nonumber -\frac{l_1(l_1+1)}{r_1^2}+2(r_1 - r_2\cos(\theta_{12}))\frac{1}{r_{12}}\frac{\partial^2}{\partial r_1 \partial r_{12}} \nonumber - 2(\nabla_1^Y \cdot {\bf r}_2)\frac{1}{r_{12}}\frac{\partial}{\partial r_{12}}\nonumber
\end{equation}

where <math>\nabla_1^Y</math> acts only on the spherical harmonic part of the wave function and the diagram

[[File:Pic3.jpg|400px]]

defines the complete set of 6 independent variables is <math>r_1</math>, <math>r_2</math>, <math>r_{12}</math>, <math>\theta_1</math>, <math>\varphi_1</math>, <math>\chi</math>.

If <math>r_{12}</math> were not an independent variable, then one could take the volume element to be

\begin{equation}
d\tau = r_1^2dr_1\sin(\theta_1)d\theta_1d\varphi_1r_2^2dr_2\sin(\theta_2)d\theta_2d\varphi_2.\nonumber
\end{equation}

However, <math>\theta_2</math> and <math>\varphi_2</math> are no longer independent variables. To eliminate them, take the point <math>{\bf r}_1</math> as the
origin of a new polar co-ordinate system, and write

\begin{equation}
d\tau=-r_1^2dr_1\sin(\theta_1)d\theta_1d\varphi_1r_{12}^2dr_{12}\sin(\psi)d\psi d\chi\nonumber
\end{equation}

and use

\begin{equation}
r_2^2=r_1^2+r_{12}^2 +2r_1r_{12}\cos(\psi)
\end{equation}

Then for fixed <math>r_1</math> and <math>r_{12}</math>,

\begin{equation}
2r_2dr_2 = -2r_1r_{12}\sin(\psi)d\psi
\end{equation}

Thus

\begin{equation}
d\tau= r_1dr_1r_2dr_2r_{12}dr_{12}\sin(\theta_1)d\theta_1d\varphi_1d\chi
\end{equation}

The basic type of integral to be calculated is

\begin{align}
I \left( l_1, m_1, l_2, m_2; R \right) & = \int \sin \left( \theta_1 \right) d\theta_1 d\varphi_1 d\chi Y^{m_1}_{l_1} \left( \theta_1, \varphi_1 \right)^{*} Y^{m_2}_{l_2} \left( \theta_2, \varphi_2 \right) \\
& \times \int r_1dr_1r_2dr_2r_{12}dr_{12}R \left( r_1, r_2, r_{12} \right)
\end{align}

Consider first the angular integral. <math>Y^{m_2}_{l_2} \left( \theta_2, \varphi_2 \right)</math> can be expressed in terms of the independent variables <math>\theta_1, \varphi_1, \chi</math> by use of the rotation matrix relation
\begin{equation}
Y^{m_2}_{l_2}( \theta_2, \varphi_2) = \sum_m \mathcal{D}^{(l_2)}_{m_2,m}(\varphi_1,\theta_1,\chi)^*Y^m_{l_2}( \theta_{12}, \varphi)
\end{equation}
where <math>\theta, \varphi</math> are the polar angles of <math>{\bf r}_2</math> relative to <math>{\bf r}_1</math>. The angular integral is then

\begin{align}
I_{ang} & = \int^{2 \pi}_{0} d\chi \int^{2\pi}_{0} d\varphi_1 \int^{\pi}_{0} \sin \left( \theta_1 \right) d\theta_{1} Y^{m_1}_{l_1} \left( \theta_1, \varphi_1 \right)^{*} \\
& \times \sum_m \mathcal{D}^{(l_2)}_{m_2,m} \left( \varphi_1,\theta_1,\chi \right)^{*} Y^{m}_{l_2} \left( \theta_{12},\varphi \right) \\
\end{align}

Use

\begin{equation}
Y^{m_1}_{l_1} \left( \theta_1, \varphi_1 \right)^{*} = \sqrt{\frac{2l_1+1}{4\pi}}\mathcal{D}^{(l_1)}_{m_1,0} \left( \varphi_1, \theta_1, \chi \right)
\end{equation}

together with the orthogonality property of the rotation matrices (Brink and Satchler, p 147)

\begin{align}
& \int \mathcal{D}^{(j)*}_{m,m^\prime}\mathcal{D}^{(J)}_{M,M^\prime}\sin(\theta_1)d\theta_1d\varphi_1d\chi \\
& = \frac{8\pi^2}{2j+1}\delta_{jJ}\delta_{mM}\delta_{m^\prime M^\prime}\nonumber
\end{align}

to obtain

\begin{align}
I_{ang} & = \sqrt{ \frac{2l_1+1}{4\pi}} \frac{8\pi^2}{2l_1+1} \delta_{l_1,l_2} \delta_{m_1,m_2} Y^{0}_{l_2} \left( \theta_{12}, \varphi \right) \\
& = 2 \pi \delta_{l_1,l_2} \delta_{m_1,m_2} P_{l_2} \left( \cos \theta_{12} \right)
\end{align}

since

\begin{equation}
Y^0_{l_2} \left( \theta_{12}, \varphi \right) = \sqrt{\frac{2l_1+1}{4\pi}} P_{l_2} \left( \cos \left( \theta_{12} \right) \right)
\end{equation}

Note that <math>P_{l_2} \left( \cos \theta \right)</math> is just a short hand expression for a radial function because

\begin{equation}
\cos \theta_{12} = \frac{r_1^2 + r_2^2 - r_{12}^2}{2r_1r_2}
\end{equation}

The original integral is thus

\begin{equation}
I \left( l_1, m_1, l_2, m_2; R \right) = 2 \pi \delta_{l_1,l_2} \delta_{m_1,m_2} \int^{\infty}_{0} r_{1} dr_{1} \int^{\infty}_{0} r_{2} dr_{2} \int^{r_{1} + r_{2}}_{|r_{1} - r_{2}|} r_{12} dr_{12} R \left( r_{1}, r_{2}, r_{12} \right) P_{l_2} \left( \cos \theta_{12} \right)
\end{equation}

where again

\begin{equation}
\cos \theta_{12} = \frac{r_{1}^{2} +r^{2}_{2} - r_{12}^{2}}{2 r_{1} r_{2}}
\end{equation}

is a purely radial function.

The above would become quite complicated for large <math>l_2</math> because <math>P_{l_2}(\cos \theta_{12})</math> contains terms up to <math>\left( \cos \theta_{12} \right)^{l_2}</math>. However, recursion relations exist which allow any integral containing <math>P_l \left( \cos \theta_{12} \right)</math> in terms of those containing just <math>P_0 \left( \cos \theta_{12} \right) = 1</math> and <math>P_1 \left( \cos \theta_{12} \right) = \cos \theta_{12}</math>.

==Radial Integrals and Recursion Relations==

The basic radial integral is [see G.W.F. Drake, Phys. Rev. A 18, 820 (1978)]

\begin{align}
I_0 \left( a, b, c \right) & =
\int^{\infty}_{0} r_{1} dr_{1} \int^{\infty}_{r_{1}} r_{2} dr_{2} \int^{r_{1}+r_{2}}_{r_{2} - r_{1}} r_{12} dr_{12} r_{1}^{a} r_{2}^{b} r_{12}^{c} e^{-\alpha r_{1} - \beta r_{2}} + \int^{\infty}_{0} r_{2} dr_{2} \int^{\infty}_{r_{2}} r_{1} dr_{1} \int^{r_{1} + r_{2}}_{r_{1} - r_{2}} r_{12} dr_{12} r_{1}^{a} r_{2}^{b} r_{12}^{c} e^{- \alpha r_{1} - \beta r_{2}} \\
& = \frac{2}{c+2} \sum^{[(c+1)/2]}_{i=0}
\left(\begin{array}{c}
c+2 \\
2i+1
\end{array} \right)
\left\{ \frac{q!}{\beta^{q+1} \left( \alpha + \beta \right)^{p+1}} \sum_{j=0}^{q} \frac{\left( p + j \right)!}{j!} \left( \frac{\beta}{\alpha + \beta} \right)^j + \frac{q^{\prime}!}{\alpha^{q^{\prime}+1} \left( \alpha + \beta \right)^{p^{\prime}+1}} \sum^{q^{\prime}}_{j=0} \frac{\left( p^{\prime} + j \right)!}{j!} \left( \frac{\alpha}{\alpha + \beta} \right)^j \right\}
\end{align}

where

\begin{array}{ll}
p = a + 2i + 2 & p^{\prime} = b + 2i + 2 \\
q = b + c - 2i + 2 & q^{\prime} = a + c - 2i + 2
\end{array}

The above applies for <math> a, b \geq -2, c\geq -1</math>.
<math>[x]</math> means "greatest integer in" <math>x</math>.

Then

\begin{align}
I_1 \left( a, b, c \right) & = \int d\tau_{r} r_{1}^{a} r_{2}^{b} r_{12}^{c} e^{ -\alpha r_{1} - \beta r_{2}} P_{1}\left( \cos \theta \right) \\
& = \frac{1}{2} \left( I_{0} \left( a + 1, b - 1, c \right) + I_{0} \left( a - 1, b + 1, c \right) - I_{0} \left( a - 1, b - 1, c + 2 \right) \right)
\end{align}

===The Radial Recursion Relation===

Recall that the full integral is

\begin{equation}
I(l_1m_1,l_2m_2;R)=2\pi\delta_{l_1,l_2}\delta_{m_1,m_2}I_{l_2}(R)
\end{equation}

where

\begin{equation}
I_{l_2}(R) = \int d\tau_r R(r_1,r_2,r_{12})P_{l_2}(\cos \theta_{12})
\end{equation}

To obtain the recursion relation, use

\begin{equation}
P_l(x) = \frac{[P^{\prime}_{l+1}(x)-P^{\prime}_{l-1}(x)]}{2l+1}
\end{equation}

with

\begin{equation}
P_{l+1}^\prime (x) = \frac{d}{dx}P_{l+1}(x)
\end{equation}

Here <math>x=\cos\theta_{12}</math> and

\begin{align}
\frac{d}{d \cos \theta_{12}} & = \frac{dr_{12}}{d \cos \theta_{12}} \frac{d}{dr_{12}} \\
& = -\frac{r_{1} r_{2}}{r_{12}} \frac{d}{dr_{12}}
\end{align}

Then,

\begin{equation}
I_{l}\left( R \right) = - \int d\tau_r R \frac{r_{1} r_{2}}{r_{12}} \frac{d}{dr_{12}} \frac{ \left[ P_{l+1} \left( \cos \theta_{12} \right) - P_{l-1} \left( \cos \theta_{12} \right) \right]}{2l+1}
\end{equation}

The <math>r_{12}</math> part of the integral is

\begin{align}
& \int^{r_{1} + r_{2}}_{|r_{1} - r_{2}|} r_{12} dr_{12} R \frac{r_{1} r_{2}}{r_{12}} \frac{d}{dr_{12}} \left[P_{l+1}-P_{l-1} \right] \\
& = R r_{1} r_{2} \left[ P_{l+1} - P_{l-1} \right]^{r_{1} + r_{2}}_{|r_{1} - r_{2}|} - \int^{r_{1} + r_{2}}_{|r_{1} - r_{2}|} r_{12} dr_{12} \left( \frac{d}{dr_{12}} R \right) \frac{r_{1} r_{2}}{r_{12}} \frac{ \left[ P_{l+1} - P_{l-1} \right]}{2l+1}
\end{align}

The integrated term vanishes because

\begin{equation}
\cos \theta_{12} = \frac{r_{1}^{2} + r_{2}^{2} - r_{12}^{2}}{2 r_{1} r_{2}}
\begin{array}{lll}
& = -1 & {\rm when} & r_{12}^{2} = \left( r_{1} + r_{2} \right)^{2} \\
& = 1 & {\rm when} & r_{12}^{2} = \left( r_{1} - r_{2} \right)^{2}
\end{array}
\end{equation}

and <math>P_l(1)=1</math>, <math>P_{l}(-1)=(-1)^{l}</math>.

Thus,

\begin{equation}
I_{l+1}\left(\frac{r_1r_2}{r_{12}}R^\prime\right) = \left( 2l+1 \right) I_{l}(R) - I_{l-1} \left( \frac{r_{1} r_{2}}{r_{12}} R^{\prime} \right)
\end{equation}

If

\begin{equation}
R = r_{1}^{a-1} r_{2}^{b-1} r_{12}^{c+2} e^{-\alpha r_{1} - \beta r_{2}}
\end{equation}

then [G.W.F. Drake, Phys. Rev. A 18, 820 (1978)]

\begin{equation}
I_{l+1}(r_1^ar_2^br_{12}^c) =
\frac{2l+1}{c+2} I_{l} \left( r_{1}^{a-1} r_{2}^{b-2} r_{12}^{c+2} \right)+I_{l-1} \left( r_{1}^{a} r_{2}^{b} r_{12}^{c} \right)
\end{equation}

For the case <math>c=-2</math>, take

\begin{equation}
R = r_{1}^{a-1} r_{2}^{b-1} \ln r_{12} e^{ -\alpha r_{1} -\beta r_{2}}
\end{equation}

Then

\begin{equation}
I_{l+1} \left( r_{1}^{a} r_{2}^{b} r_{12}^{c} \right) = I_{l} \left( r_{1}^{a-1} r_{2}^{b-1} \ln r_{12} \right) \left( 2l + 1 \right) - I_{l-1} \left( r_{1}^{a} r_{2}^{b} r_{12}^{-2} \right)
\end{equation}

This allows all <math>I_{l}</math> integrals to be calculated from tables of <math>I_0</math> and
<math>I_1</math> integrals.

===The General Integral===

The above results for the angular and radial integrals can now be combined into a general formula for integrals of the type

\begin{equation}
I = \int \int d{\bf r}_{1} d{\bf r}_{2} R_{1} \mathcal{Y}^{M^{\prime}}_{l^{\prime}_{1} l^{\prime}_{2} L^{\prime}} \left( \hat{r}_{1}, \hat{r}_{2} \right) T_{k_{1} k_{2} K}^{Q} \left( {\bf r}_{1},{\bf r}_{2} \right) R_{2} \mathcal{Y}_{l_{1} l_{2} L}^{M} \left( \hat{r}_{1}, \hat{r}_{2} \right)
\end{equation}

where

\begin{equation}
\mathcal{Y}^{M}_{l_{1} l_{2} L} \left( \hat{r}_{1}, \hat{r}_{2} \right) = \sum_{m_1,m_2} \langle l_{1} l_{2} m_{1} m_{2} | LM \rangle Y^{m_{1}}_{l_{1}} \left( \hat{r}_{1} \right) Y^{m_{2}}_{l_{2}} \left( \hat{r}_{2} \right)
\end{equation}

and

\begin{equation}
T^Q_{k_{1} k_{2} K} \left( {\bf r}_{1}, {\bf r}_{2} \right) = \sum_{q_{1}, q_{2}} \langle k_{1} k_{2} q_{1} q_{2} | KQ \rangle Y_{k_{1}}^{q_{1}} \left( \hat{r}_{1} \right) Y^{q_{2}}_{k_{2}} \left( \hat{r}_{2} \right)
\end{equation}

The basic idea is to make repeated use of the formula

\begin{equation}
Y^{m_{1}}_{l_{1}} \left( \hat{r}_{1} \right) Y^{m_{2}}_{l_{2}} \left( \hat{r}_{1} \right) = \sum_{l m} \left( \frac{\left( 2l_{1}+1 \right) \left(2l_{2}+1 \right) \left(2l+1 \right)}{4 \pi} \right)^{1/2}
\times \left( \begin{array}{ccc}
l_1 & l_2 & l \\
m_1 & m_2 & m
\end{array} \right) \left(\begin{array}{ccc}
l_1 & l_2 & l \\
0 & 0 & 0
\end{array} \right)
Y^{m*}_{l} \left( \hat{r}_{1} \right)
\end{equation}

where

\begin{equation}
Y^{m*}(\hat{r}) = (-1)^mY_l^{-m}(\hat{r})
\end{equation}

and

\begin{equation}
\left(\begin{array}{ccc}
l_1 & l_2 & l \\
m_1 & m_2 & m
\end{array} \right) = \frac{\left( -1 \right)^{l_{1} - l_{2} - m}}{\left( 2l + 1 \right)^{1/2}} \left( l_{1} l_{2} m_{1} m_{2} | l, -m \right)
\end{equation}

is a 3-<math>j</math> symbol. In particular, write

\begin{array}{cc}
Y^{m_1^\prime}_{l_1^\prime}(\hat{r}_1)^{*} \underbrace{Y_{k_{1}}^{q_{1}} \left( \hat{r}_{1} \right) Y^{m_{1}}_{l_{1}} \left( \hat{r}_{1} \right)} & = & \sum_{\Lambda M}(\ldots)Y^{M}_\Lambda(\hat{r}_1)\\
\ \ \ \ \ \ \ \sum_{\lambda_1\mu_1}Y_{\lambda_1}^{\mu_1}(\hat{r}_1) & & \\
& & \\
Y^{m_{2}^{\prime}}_{l_{2}^{\prime}} \left( \hat{r}_{2} \right) \underbrace{Y^{q_{2}}_{k_{2}} \left( \hat{r}_{2} \right) Y_{l_{2}}^{m_{2}} \left( \hat{r}_{2} \right)} & = & \sum_{\Lambda^{\prime} M^{\prime}} \left( \ldots \right)Y_{\Lambda^{\prime}}^{M^{\prime}*} \left( \hat{r}_{2} \right) \\
\ \ \ \ \ \ \ \ \ \sum_{\lambda_{2} \mu_{2}}Y_{\lambda_{2}}^{\mu_{2}} \left( \hat{r}_{2} \right) & &
\end{array}

The angular integral then gives a factor of <math>2 \pi \delta_{\Lambda, \Lambda^{\prime}} \delta_{M, M^{\prime}} P_{\Lambda} \left( \cos \theta_{12} \right)</math>. The total integral therefore reduces to the form

\begin{equation}
I = \sum_{\Lambda} C_{\Lambda} I_{\Lambda} \left( R_{1} R_{2} \right)
\end{equation}

where <math>C_{\Lambda} = \sum_{\lambda_{1}, \lambda_{2}} C_{ \lambda_{1}, \lambda_{2}, \Lambda}</math>. For further details and derivations, including graphical representations, see G.W.F. Drake, Phys. Rev. A <math>{\bf 18}</math>, 820 (1978).

==Graphical Representation==

[[File:GraphRep1.jpg|500px]]

==Matrix Elements of H==
Recall that

\begin{equation}
H = -\frac{1}{2} \nabla^{2}_{1} - \frac{1}{2} \nabla^{2}_{2} - \frac{1}{r_1} - \frac{1}{r_2} + \frac{Z^{-1}}{r_{12}}
\end{equation}

Consider matrix elements of

\begin{equation}
\nabla^2_1 = \frac{1}{r_1^2} \frac{\partial}{\partial r_1}\left(r_1^2\frac{\partial}{\partial r_1} \right) + \frac{1}{r^2}\frac{\partial}{\partial r}\left(r^2\frac{\partial}{\partial r}\right) - \frac{(\vec{l}_1^Y)^2}{r_1^2} + \frac{2 \left( r_1 - r_2 \cos \theta \right)}{r} \frac{\partial^2}{\partial r_1 \partial r} - 2 \left( \nabla^Y_1 \cdot {\bf r}_2 \right) \frac{1}{r} \frac{\partial}{\partial r}
\end{equation}

where <math>r \equiv r_{12}</math> and <math>\cos\theta \equiv \cos\theta_{12}</math>. Also in what follows, define <math>\hat{\nabla}^Y_1 \equiv r_1\nabla^Y_1</math> where <math>\nabla^Y</math> operates only on the spherical harmonic part of the wave function, and similarly for <math>\vec{l}_1^Y</math>.

A general matrix element is

\begin{equation}
\langle r_1^{a^\prime} r_2^{b^\prime} r_{12}^{c^\prime} e^{-\alpha^\prime r_1 - \beta^\prime r_2} \mathcal{Y}^{M^\prime}_{l^\prime_1 l^\prime_2 L^\prime} \left( \hat{r}_1, \hat{r}_2 \right) \left| \nabla^2_1 \right| r_1^a r_2^b r_{12}^c e^{-\alpha r_1 - \beta r_2} \mathcal{Y}^M_{l_1 l_2 L} \left( \hat{r}_1, \hat{r}_2 \right) \rangle
\end{equation}

Since <math>\nabla^2_1</math> is rotationally invariant, this vanishes unless <math>L=L^\prime</math>, <math>M=M^\prime</math>. Also <math>\nabla_1^2</math> is Hermitian, so that the result must be the same whether it operates to the right or left, even though the results look very different. In fact, requiring the results to be the same yields some interesting and useful integral identities as follows:

Put

\begin{equation}
\mathcal{F} = F \mathcal{Y}^M_{l_1 l_2 L} \left( \hat{r}_1, \hat{r_2} \right)
\end{equation}

and

\begin{equation}
F = r_1^a r_2^b r_{12}^c e^{-\alpha r_1 - \beta r_2}
\end{equation}

Then

\begin{equation}
\nabla^2_1 \mathcal{F} = \left\{ \frac{1}{r_1^2} \left[ a \left( a + 1 \right) - l_1 \left( l_1 + 1 \right) \right] + \frac{c \left( c + 1 \right)}{r^2} + \alpha^2\frac{2\alpha \left( a + 1 \right)}{r_1} + \frac{2 \left(r_1 - r_2 \cos \theta \right) }{r_1r^2}c \left[ a - \alpha r_1 \right] - \frac{2c}{r^2} \left( \hat{\nabla}^Y_1 \cdot \hat{r}_2 \right) \frac{r_2}{r_1} \right\} \mathcal{F}
\end{equation}

and

\begin{align}
\langle \mathcal{F}^\prime \left| \nabla^2_1 \right| \mathcal{F} \rangle & = \sum_\Lambda \int d\tau_r F^\prime C_\Lambda \left( 1 \right) P_\Lambda \left( \cos \theta \right) \times \left\{ \frac{1}{r_1^2} \left[ a \left( a + 1 \right) - l_1 \left( l_1 + 1 \right) \right] -\frac{2 \alpha \left( a + 1 \right)}{r_1} + \frac{c \left( c + 1 \right)}{r^2} +\alpha^2 + \frac{2 \left( r_2 - r_2 \cos \theta \right)}{r_1r^2}c \left( a - \alpha r_1 \right) \right\} F \\
& + \sum_\Lambda \int d\tau_r F^\prime C_\Lambda \left( \hat{\nabla}^Y_1 \cdot \hat{r}_2 \right) P_\Lambda \left( \cos \theta \right) \left( \frac{-2c r_2}{r_1 r^2} \right) F
\end{align}

where

\begin{equation}
\int d\tau_r = \int^\infty_0 r_1 dr_1 \int^\infty_0 r_2 dr_2 \int^{r_1+r_2}_{\left| r_1 - r_2 \right|} r dr
\end{equation}

For brevity, let the sum over <math>\Lambda</math> and the radial integrations be understood, and let <math>\left( \nabla_1^2 \right)</math> stand for the terms which appear in the integrand. Then operating to the right gives

\begin{equation}
(\nabla_1^2)_R = \frac{1}{r_1^2} \left[ a \left( a + 1 \right) - l_1 \left( l_1 + 1 \right) \right] + \frac{c \left( c + 1 \right)}{r^2} + \alpha^2 - \frac{2 \alpha \left( a + 1 \right)}{r_1}
+ \frac{2 \left( r_1 - r_2 \cos \theta \right)}{r_1 r^2}c \left( a - \alpha r_1 \right) - \frac{2c r_2}{r_1 r^2} \left( \hat{\nabla}^Y_1 \cdot \hat{r}_2 \right)
\end{equation}

Operating to the left gives

\begin{equation}
(\nabla_1^2)_L = \frac{1}{r_1^2} \left[a^\prime \left( a^\prime + 1 \right) - l_1^\prime \left( l_1^\prime + 1 \right) \right] + \frac{c^\prime \left( c^\prime + 1 \right)}{r^2} + \alpha^{\prime 2} -\frac{2 \alpha^\prime \left( a^\prime + 1 \right)}{r_1}
+ \frac{2 \left( r_1 - r_2 \cos \theta \right)}{r_1 r^2} c^\prime \left( a^\prime - \alpha^\prime r_1 \right) - \frac{2 c^\prime r_2}{r_1 r^2} \left( \hat{\nabla}^{Y^\prime}_1 \cdot \hat{r}_2 \right)
\end{equation}

Now put

\begin{array}{cc}
a_+ = a+a^\prime, & \hat{\nabla}_1^+ = \hat{\nabla}_1^Y + \hat{\nabla}_1^{Y\prime} \\
a_{-} = a-a^\prime, & \hat{\nabla}_1^{-} = \hat{\nabla}_1^Y - \hat{\nabla}_1^{Y\prime}
\end{array}

etc., and substitute <math>a^\prime = a_+ - a</math>, <math>c^\prime = c_+ - c</math>, and <math>\alpha^\prime = \alpha_+ - \alpha</math> in <math>(\nabla^2_1)_L</math>. If <math>a_+</math>, <math>c_+</math> and <math>\alpha_+</math> are held fixed, then the equation

\begin{equation}
(\nabla^2_1)_R = (\nabla^2_1)_L
\end{equation}

must be true for arbitrary <math>a</math>, <math>c</math>, and <math>\alpha</math>. Their coefficients must thus vanish.

This yields the integral relations

\begin{equation}
\label{I}
\frac{\left( r_1 - r_2 \cos \theta \right)}{r_1 r^2} = \frac{1}{c_+} \left( \frac{- \left( a_+ + 1 \right)}{r_1^2} + \frac{\alpha_+}{r_1} \right) \tag{I}
\end{equation}

from the coefficient of <math>a</math> and

\begin{equation}
\label{II}
\frac{ \left( r_1 - r_2 \cos \theta \right) \left( a_+ - \alpha_+ r_1 \right)}{r_1 r^2} = \frac{r_2}{r_1 r^2}\left( \hat{r}_2 \cdot \hat{\nabla}_1^+ \right) - \frac{ \left( c_+ + 1 \right)}{r^2} \tag{II}
\end{equation}

from the coefficient of <math>c</math>. The coefficient of <math>\alpha</math> gives an equation equivalent to (I).

Furthermore, if can be show that (see problem)

\begin{align}
& \sum_\Lambda \int d\tau_r \frac{r^c}{r^2} C_\Lambda \left( \hat{r}_2 \cdot \hat{\nabla}_1^Y \right) P_\Lambda \left( \cos \theta \right) \\
= & \sum_\Lambda \int d\tau_r \frac{r^c}{c r_1 r_2}C_\Lambda \left( 1 \right)P_\Lambda \left( \cos \theta \right) \times \left( \frac{l_1^\prime \left( l_1^\prime + 1 \right) - l_1 \left( l_1 + 1 \right) -\Lambda \left( \Lambda + 1 \right)}{2} \right)
\end{align}

and similarly for <math>\left( \hat{r}_2 \cdot \hat{\nabla}^{Y\prime}_1 \right)</math> with <math>l_1</math> and <math>l_1^\prime</math> interchangeable, then it follows that

\begin{equation}
\frac{r^{c_+}}{r^2} \left( \hat{r}_2 \cdot \hat{\nabla}_1^+ \right) = - \frac{r^{c_+}}{c_+ r_1 r_2} \Lambda \left( \Lambda + 1 \right)
\end{equation}

and

\begin{equation}
\frac{r^{c_+}}{r^2} \left( \hat{r}_2 \cdot \hat{\nabla}_1^- \right) = \frac{r^{c_+}}{c_+ r_1 r_2} \left[ l_1^\prime \left( l_1^\prime + 1 \right) - l_1 \left( l_1 + 1 \right) \right]
\end{equation}

where equality applies after integration and summation over <math>\Lambda</math>.

Thus (II) becomes

\begin{equation}\label{IIa}
\frac{ \left( r_1 - r_2 \cos \theta \right) \left( a_+ - \alpha_+ r_1 \right)}{r_1 r^2} = - \frac{\Lambda \left( \Lambda + 1 \right)}{c_+ r^2_1} - \frac{ \left( c_+ + 1 \right)}{r^2} \tag{IIa}
\end{equation}

===Problem===

Prove the integral relation

\begin{align}
& \sum_\Lambda \int d\tau_r f \left( r_1, r_2 \right) \frac{1}{r} \left( \frac{d}{dr} q(r) \right) C_\Lambda \left( \hat{r}_2 \cdot \hat{\nabla}^Y_1 \right) P_\Lambda \left( \cos \theta \right) \\
= & \sum_\Lambda \int d\tau_r \frac{ f \left( r_1, r_2 \right)}{r_1, r_2} q(r) C_\Lambda \left( 1 \right) P_\Lambda \left( \cos \theta \right) \left(\frac{l_1^\prime \left( l_1^\prime + 1 \right) - l_1 \left( l_1 + 1 \right) - \Lambda \left( \Lambda + 1 \right)}{2}\right)
\end{align}

where the coefficients <math>C_\Lambda \left( 1 \right)</math> are the angular coefficients from the overlap integral

\begin{align}
\int & d\Omega\, \mathcal{Y}^{M*}_{l_1^\prime l_2^\prime L}(\hat{r}_1,\hat{r}_2)\, \mathcal{Y}^{M}_{l_1 l_2 L}(\hat{r}_1,\hat{r}_2)\\
= & \sum_\Lambda C_\Lambda(1)P_\Lambda(\cos\theta_{12}).
\end{align}

Hint: Use the fact that <math>l_1^2</math> is Hermitian so that
<math>
\int d\tau (l_1^2\mathcal{Y}^\prime)^*q(r)\mathcal{Y} = \int d\tau
\mathcal{Y}^{\prime *}l_1^2(q(r)\mathcal{Y})
</math>
with <math>\ \vec{l}_1= \frac{1}{i}\vec{r}_1\times\nabla_1</math>.

It is also useful to use
<math>\begin{eqnarray}\ \
\cos\theta\, P_L(\cos\theta) &=& \frac{1}{2L+1}[L\,P_{L-1}(\cos\theta) + (L+1)P_{L+1}(\cos\theta)]
\\
(\cos^2\theta-1)P_L(\cos\theta) &=&
\frac{L(L-1)}{(2L-1)(2L+1)}P_{L-2}(\cos\theta)\nonumber
-\frac{2(L^2+L-1)}{(2L-1)(2L+3)}P_L(\cos\theta) + \frac{(L+1)(L+2)}{(2L+1)(2L+3)}P_{L+2}(\cos\theta)\nonumber\\
&=& \frac{L(L-1)}{(2L-1)(2L+1)}[P_{L-2}(\cos\theta) - P_L(\cos\theta)]
+ \frac{(L+1)(L+2)}{(2L+1)(2L+3)}[P_{L+2}(\cos\theta) - P_L(\cos\theta)]
\end{eqnarray}</math>

together with a double application of the integral recursion relation
<math>
I_{L+1}\left(\frac{1}{r}\frac{d}{dr}q(r)\right) =
(2L+1)I_L\left(\frac{1}{r_1r_2}q(r)\right)+I_{L-1}\left(\frac{1}{r}\frac{d}{dr}q(r)\right)
</math>

Of course <math>l_1^2\mathcal{Y}^M_{l_1l_2L}(\hat{r}_1,\hat{r}_2)=l_1(l_1+1)\mathcal{Y}^M_{l_1l_2L}(\hat{r}_1,\hat{r}_2)</math>. Begin by expanding
<math>
l_1^2(g\mathcal{Y}) = \mathcal{Y}l_1^2g + g\,l_1^2\mathcal{Y} + 2(\vec{l}_1g)\cdot(\vec{l}_1\mathcal{Y})
</math>, and show that

<math>
\begin{eqnarray}\ \
(\vec{l}_1g)\cdot(\vec{l}_1\mathcal{Y}) & = & \frac{r_1^2}{r}\frac{dg}{dr}\vec{r}_2\cdot \nabla_1\mathcal{Y}\,\,\,\,\mbox{ and}\\
l_1^2g(r) &=& 2r_1r_2\cos\theta\frac{1}{r}\frac{dg}{dr} + r_1^2r_2^2(\cos^2\theta - 1)
\frac{1}{r}\frac{d}{dr}\left(\frac{1}{r}\frac{dg}{dr}\right).
\end{eqnarray}
</math>

The proof amounts to showing that the term <math>l_1^2g(r)</math> can be replaced by
<math>\Lambda(\Lambda+1)g(r)</math> after multiplying by <math>P_\Lambda(\cos\theta)</math>
and integrating by parts with respect to the radial integrations over <math>r_1</math>, <math>r_2</math> and
<math>r \equiv |\vec{r}_1 - \vec{r}_2|</math>. Remember that <math>\cos\theta</math> is just a short-hand notation for the radial function <math>(r_1^2 + r_2^2 - r^2)/(2r_1r_2)</math>.

==General Hermitian Property==

Each combination of terms of the form
\begin{equation}
<f>_R = a^2f_1+af_2+abf_3+b^2f_4+bf_5+af_6\nabla_1^Y+bf_7\nabla^Y_2+f_8(Y)
\end{equation}
can be rewritten
\begin{equation}
<f>_L= (a_+-a)^2f_1+(a_+-a)f_2 + (a_+-a)(b_+-b)f_3+(b_+-b)^2f_4
+(b_+-b)f_5+(a_+-a)f_6\nabla^{Y\prime}_1+(b_+-b)f_7\nabla^{Y\prime}_2+f_8)Y^\prime
\end{equation}
where as usual <math>\nabla_1^Y</math> acts only on the spherical harmonic part of the wave function denoted for short by <math>Y</math>.
Since these must be equal for arbitrary <math>a</math> and <math>b</math>,
\begin{equation}
a_+^2f_1+a_+f_2+_+b_+f_3+b_+f_4+b_+f_5+a_+f_6\nabla^{Y\prime}_1 +
b_+f_7\nabla^{Y\prime}_2+f_8(Y^\prime)-f_8(Y)=0
\end{equation}

Adding the corresponding expression with <math>Y</math> and <math>Y^\prime</math> interchanged
yields
<math>\label{III}
a_+^2f_1+a_+f_2+a_+b_+f_3+b_+^2f_4+b_+f_5+\frac{1}{2}a_+f_6\nabla^+_1+\frac{1}{2}b_+f_7\nabla^+_2=0\tag{III}
</math>

Subtracting gives
<math>\label{IV}
f_8(Y)-f_8(Y^\prime)=-\frac{1}{2}[a_+f_6\nabla^-_1+b_+f_7\nabla_2^-]\tag{IV}
</math>
<math>\label{V}
a[-2a_+f_1-2f_2-b_+f_3-f_6\nabla^+_1]=0\tag{V}
</math>
<math>\label{VI}
b[-2b_+f_4-2f_5-a_+f_3-f_7\nabla^+_2]=0\tag{VI}
</math>

Adding the two forms gives

\begin{eqnarray}
<f>_R+<f>_L=\frac{1}{2}(a_+^2+a_-^2)f_1+a_+f_2+\frac{1}{2}(a_+b_++a_-b_-)f_3\nonumber\\
\ +\frac{1}{2}(b_+^2+b_-^2)f_4+b_+f_5+\frac{1}{2}f_6(a_+\nabla_1^++a_-\nabla^-_1)\nonumber\\
\ +\frac{1}{2}(b_+\nabla_2^++b_-\nabla^-_2)+f_8(Y)+f_8(Y^\prime)\nonumber
\end{eqnarray}

Subtracting <math>\frac{x}{2}\times({\rm I})</math> gives

\begin{eqnarray}\ \
<f>_R+<f>_L = \frac{1}{2}[(1-x)a_+^2 + a_-^2]f_1 +
(1-\frac{x}{2})a_+f_2+\frac{1}{2}[(1-x)a_+b_++a_-b_-]f_3\nonumber\\
+\frac{1}{2}[(1-x)b_+^2+b_-^2]f_4+(1-\frac{x}{2})f_5b_++\frac{1}{2}f_6[(1-\frac{x}{2})a_+\nabla_1^++a_-\nabla^-_1]\nonumber\\
+\frac{1}{2}[(1-\frac{x}{2})b_+\nabla^+_2+b_-\nabla^-_2]+f_8(Y)+f_8(Y^\prime)\nonumber
\end{eqnarray}

If <math>x=1</math>,
\begin{eqnarray}
<f>_R+<f>_L &=&
\frac{1}{2}[a_-^2f_1+a_+f_2+a_-b_-f_3+b_-^2f_4+b_+f_5\nonumber\\
&+&\left(\frac{1}{2}a_+\nabla^+_1+a_-\nabla^-_1\right)f_6+\left(\frac{1}{2}b_+\nabla^+_2+b_-\nabla^-_2\right)f_7]+f_8(Y)+f_8(Y^\prime)\nonumber
\end{eqnarray}

The General Hermitian Property for arbitrary <math>x</math> gives
\begin{eqnarray}\ \
\nabla_1^2=
\frac{1}{4}\{\frac{1}{r_1^2}[(1-x)a_+^2+a_-^2+2(1-\frac{x}{2})a_+-2[l_1(l_1+1)+l_1^\prime(l_1^\prime+1)]\nonumber\\
-\frac{2}{r_1}[(1-x)\alpha_+a_++\alpha_-a_-+2(1-\frac{x}{2})\alpha_+]+(1-x)\alpha_+^2
+ \alpha^2_-]\nonumber\\
+\frac{2(r_1-r_2\cos\theta)}{r_1r^2}[(1-x)(a_+-\alpha_+r_1)c_++(a_--\alpha_-r_2)c_-]\nonumber\\
-\frac{2r_2}{r_1r^2}[(1-\frac{x}{2})c_+\hat{r}_2\cdot\hat{\nabla}_1^++c_-\hat{r}_2\hat{\nabla}_1^-]+\frac{(1-x)c_+^2+c_-^2+2(1-\frac{x}{2})c_+}{r^2}\}\nonumber
\end{eqnarray}.

Use
<math>\begin{eqnarray}\ \
\frac{-2r_2}{r_1r^2}[(1-\frac{x}{2})c_+\hat{r}_2\hat{\nabla}_1^++c_-\hat{r}_2\hat{\nabla}_1^-]=\frac{2}{r_1^2}\left[(1-\frac{x}{2})\Lambda(\Lambda+1)-\frac{c_-}{c_+}[l_1^\prime(l_1^\prime+1)+l_1(l_1+1)]\right]\end{eqnarray}</math>,

<math>\begin{eqnarray}
\frac{2(r_1-r_2\cos\theta)}{r_1r^2}(a_--\alpha_-r_1)c_- = 2\frac{c_-}{c_+}\left[\frac{-1}{r_1^2}[a_-(a_++1)]+\frac{1}{r_1}[a_-\alpha_++\alpha_-(a_++2)]-\alpha_-\alpha_+\right]\nonumber
\end{eqnarray}</math>,

and
<math>\begin{eqnarray}\ \
\frac{2(r_1-r_2\cos\theta)}{r_1r^2}(a_+-\alpha_+r_1)c_+ = 2\left[-\frac{1}{r_1^2}[a_+(a_++1)]+\frac{1}{r_1}[a_+\alpha_++\alpha_+(a_++2)]-\alpha_+^2\right]
\end{eqnarray}</math>.

Substituting into <math>\nabla_1^2</math> gives

\begin{eqnarray}
\nabla^2_1 = \frac{1}{4}\{\frac{1}{r_1^2}[-(1-x)a_+^2 + a_-^2 +xa_+ +
2(1-\frac{x}{2})\Lambda(\Lambda+1)\nonumber\\
-2l_1(l_1+1)(1-\frac{c_-}{c_+}) -
2l_1^\prime(l_1^\prime+1)(1+\frac{c_-}{c_+})-\frac{2c_-a_-}{c_+}(a_++1)]\nonumber\\
- \frac{2}{r_1}\left[-(1-x)\alpha_+(a_++2)+\alpha_-a_-+2(1-\frac{x}{2})\alpha_+
-\frac{c_-}{c_+}[a_-\alpha_++\alpha_-(a_++2)]\right]\nonumber\\
-(1-x)\alpha_+^2+\alpha_-^2-2\frac{c_-}{c_+}\alpha_-\alpha_+ +\frac{1}{r^2}\left[(1-x)c_+^2+c_-^2+2(1-\frac{x}{2})c_+\right]\}\nonumber
\end{eqnarray}

This has the form
<math>
\nabla_1^2 =
\frac{1}{4}\left[\frac{A_1}{r_1^2}+\frac{B_1}{r_1}+\frac{C_1}{r^2}+D_1\right]
</math>
with

<math>\begin{array}{ccc}
A_1=-(1-x)a_+^2+a_-^2
+xa_++2(1-\frac{x}{2})\Lambda(\Lambda+1)-2l_1(l_1+1)(1-\frac{c_-}{c_+})\
-2l_1^\prime(l_1^\prime+1)(1+\frac{c_-}{c_+})-\frac{2c_-a_-}{c_+}(a_++1) \ \
\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \
\end{array}
</math>

<math>
B_1 =
2\left[(1-x)\alpha_+(a_++2)-\alpha_-a_--2(1-\frac{x}{2})\alpha_++\frac{c_-}{c_+}[a_-\alpha_++\alpha_-(a_++2)]\right]
</math>

<math>
C_1 = (1-x)c_+^2+ c_-^2 +2(1-\frac{x}{2})c_+
</math>

<math>
D_1 = -1(1-x)\alpha_+^2 +\alpha_-^2 - 2\frac{c_-}{c_+}\alpha_-\alpha_+
</math>

The complete Hamiltonian is then (in Z-scaled a.u.)

<math>\begin{eqnarray}
H =-\frac{1}{2}\nabla^2_1 -\frac{1}{2}\nabla^2_2 -\frac{1}{r_1} -\frac{1}{r_2} +\frac{Z^{-1}}{r}\nonumber\\
= -\frac{1}{2}\nabla^2_1 -\frac{1}{2}\nabla^2_2 -\frac{1}{r_1} -\frac{Z-1}{Zr_2}+Z^{-1}\left(\frac{1}{r}-\frac{1}{r_2}\right)\nonumber\\
\ =-\frac{1}{8}\left[\frac{A_1}{r_1^2}+\frac{B_1+8}{r_1}+\frac{C_1}{r^2}+D_1+D_2 +\frac{A_2}{r_2^2} + \frac{B_2 + 8(Z-1)/Z}{r_2}+\frac{C_2}{r^2}\right]\nonumber
+ Z^{-1}\left(\frac{1}{r}-\frac{1}{r_2}\right)\nonumber
\end{eqnarray}</math>

The screened hydrogenic energy is
<math>\begin{eqnarray}
E_{SH}=-\frac{1}{2}\left[\frac{1}{n_1^2}+ \left(\frac{Z-1}{Z}\right)^2\frac{1}{n_2^2}\right]
\end{eqnarray}</math>
so that

<math>\begin{eqnarray}
H-E_{SH} &=& -\frac{1}{8}[
\frac{A_1}{r_1^2}+\frac{B_1+8}{r_1}+\frac{C_1}{r^2} + D_1 + D_2
-\frac{4}{n_1^2} -\left(\frac{Z-1}{Z}\right)^2\frac{4}{n_2^2}\nonumber
+\frac{A_2}{r_2}+\frac{B_2+8(Z-1)/Z}{r_2}+\frac{C_2}{r^2}] +
Z^{-1}\left(\frac{1}{r} -\frac{1}{r_2}\right)\nonumber
\end{eqnarray}</math>

==Optimization of Nonlinear Parameters==
- The traditional method of performing Hylleraas calculations is to write the
basis set in the form

<math>
\Psi = \sum_{i,j,k}c_{i,j,k}\varphi_{i,j,k}(\alpha,\beta)\pm exchange
</math>

with

<math>
\varphi_{i,j,k}(\alpha,\beta) = r_1^ir_2^jr_{12}^ke^{-\alpha r_1 - \beta
r_2}\mathcal{Y}^M_{l_1l_2L}.
</math>

The usual procedure is to set <math>\alpha = Z</math> so that it represents the inner
<math>1s</math> electron, and then to vary <math>\beta</math> so as to minimize the energy.

[[File:OptNon.jpg|500px]]

Since <math>\beta</math> appears in <math>\Psi</math> as a non-linear parameter, the entire
calculation must be repeated for each value of <math>\beta</math>. However, the minimum
becomes progressively smaller as the basis set is enlarged.

[[File:OptNon02.jpg|500px]]

*Difficulties

1. If the basis set is constructed so that <math> i +j +j \leq N</math>, the the
number of terms is
<math>
(N+1)(N+2)(N+3)/6
</math>
N=14 already gives 680 terms and an accuracy of about 1 part in <math>10^{10}</math> for
low-lying states. A substantial improvement in accuracy would require much
larger basis sets, together with multiple precision arithmetic to avoid loss
of significant figures when high powers are included.

2. The accuracy rapidly deteriorates as one goes to more highly excited
states - about 1 significant figure is lost each time the principle quantum
number is increased.

*Cure

We have found that writing basis sets in the form
\begin{eqnarray}
\Psi &=& \sum_{i,j,k}\left(c^{(1)}_{i,j,k}\varphi_{ijk}(\alpha_1\beta_1) +
c_{ijk}^{(2)}\varphi_{ijk}(\alpha_2\beta_2)\right) \pm exchange \nonumber\\
&=& \psi({\bf r}_1,{\bf r}_2) \pm \psi({\bf r}_2,{\bf r}_1)\nonumber
\end{eqnarray}
so that each combination of powers is included twice with different nonlinear
parameters gives a dramatic improvement in accuracy for basis sets of about
the same total size.

However, the optimization of the nonlinear parameters is now much more
difficult, and an automated procedure is needed.

If
<math>
E = \frac{<\Psi|H|\Psi>}{<\Psi|\Psi>}</math> and we assume <math><\Psi|\Psi> = 1</math>, then

<math>
\frac{\partial E}{\partial\alpha_t} = -2 <\Psi|H-E|r_1\psi({\bf r}_1,{\bf
r}_2;\alpha_t)\pm r_21\psi({\bf r}_2,{\bf
r}_1;\alpha_t)>
</math>
where
<math>
\psi({\bf r}_1,{\bf r}_2;\alpha_t) =
\sum_{i,j,k}c_{ijk}^{(t)}\varphi_{ijk}(\alpha_t\beta_t)
</math>

*Problem

1. Prove the above.

2. Prove that there is no contribution from the implicit dependence of
<math>c_{ijk}^{(t)}</math> on <math>\alpha_t</math> if the linear parameters have been optimized.

==The Screened Hydrogenic Term==

If the Hamiltonian is written in the form (in Z-scaled a.u.)
<math>\begin{eqnarray}\
H = H_0({\bf r}_1,Z) + H_0({\bf r}_2,Z-1) +
Z^{-1}\left(\frac{1}{r}-\frac{1}{r_2}\right)
\end{eqnarray}</math>
with
<math>\
H_0({\bf r}_1,Z) = -\frac{1}{2}\nabla_1^2 - \frac{1}{r_1}\nonumber
</math> and
<math>
H_0({\bf r}_2,Z-1) = -\frac{1}{2}\nabla_2^2 - \frac{Z-1}Z{r_2}\nonumber
</math>,
then the eigenvectors of <math>H_0({\bf r},Z) +H_0({\bf r}_2,Z-1)</math> are products of
hydrogenic orbitals
<math>
\Psi_0=\psi_0(1s,Z)\psi_0(nl,Z-1)
</math>
and the eigenvalue is
<math>\begin{eqnarray}
E_{SH} = \left[-\frac{1}{2} - \left(\frac{Z-1}{Z}\right)^2\frac{1}{2n^2}\right]Z^2 a.u.
\end{eqnarray}</math>
called the screened hydrogenic eigenvalue. From highly excited states,
<math>E_{SH}</math> and <math>\Psi_0</math>are already excellent approximations.

For example, for the 1s8d states, the energies are

<math>\begin{eqnarray}\\
E(1s8d^1D)&=& -2.007\,816\,512\,563\,81 {\rm\ a.u.}\nonumber\\
E(1s8d^3D)&=& - 2.007\,817\,934\,711\,71 {\rm\ a.u.}\nonumber\\
E_{SH} &=& -2.007\,812\,5 {\rm\ a.u.} \nonumber
\end{eqnarray}</math>

It is therefore advantageous to include the screened hydrogenic terms in the
basis set so that the complete trial function becomes
<math>
\Psi = c_0\Psi_0 +
\sum_{ijk}\left[c_{ijk}^{(1)}\varphi_{ijk}(\alpha_1,\beta_1)+c_{ijk}^{(2)}\varphi(\alpha_2,\beta_2)\right]\pm
{\rm exchange}
</math>

Also, the variational principal can be re-expressed in the form
<math>\begin{eqnarray}
E= E_{SH}+\frac{<\Psi|H-E_{SH}|\Psi>}{<\Psi|\Psi>}
\end{eqnarray}</math>
so that the <math>E_{SH}</math> term can be cancelled analytically from the matrix
elements.

For example
<math>
<\Psi_0|H-E_{SH}|\Psi_0> = 0
</math>

and
<math>
<\varphi_{ijk}|H-E_{SH}|\Psi_0> =
Z^{-1}<\varphi_{ijk}|\frac{1}{r}-\frac{1}{r_2}|\Psi_0>
</math>

Recall that
<math>\begin{eqnarray}\ \
I_0(a,b,c) = \frac{2}{c+2}\sum^{[c+1]/2}_{i=0}\left(\begin{array}{c}
c+2\\
2i+1
\end{array}\right) \left[f(p,q;\beta) + f(p^\prime,q^\prime;\alpha)\right]
\end{eqnarray}</math>
where

<math>\begin{eqnarray}
f(p,q;x) =
\frac{q!}{x^{q+1}(\alpha+\beta)^{p+1}}\sum^p_{j=0}\frac{(p+j)!}{j!}\left(\frac{x}{\alpha+\beta}\right)^j
\end{eqnarray}</math>

<math>
\begin{array}{cc}
p=a+2i+2 & p^\prime = b+2i+2\\
q= b+c-2i+2 & q^\prime =a+c-2i+2
\end{array}
</math>

If <math>\beta\ll\alpha</math>, the <math>f(p,q;\beta)\gg f(p^\prime,q^\prime;\alpha)</math> and the
<math>i=0</math> term is the dominant contribution to <math>f(p,q;\beta)</math>. However, since
this term depends only on the sum of powers <math>b+c</math> for <math>r_2</math>, it cancels exactly
from the matrix element of
<math>
\frac{1}{r_{12}} - \frac{1}{r_2}
</math>
and can therefore
be omitted in the calculations of integrands, thereby saving many significant
figures. This is especially valuable when <math>b+c</math> is large and <math>a</math> is small.

==Small Corrections==
*Mass Polarization
If the nuclear mass is not taken to be infinite, then the Hamiltonian is
<math>\begin{eqnarray}\ \
H = \frac{P_N^2}{2M} + \sum_{i=1}\frac{p_i^2}{2m} + V
\end{eqnarray}</math>

[[File:SmallCor.jpg|500px]]

where <math>M =M_A-nm</math> is the nuclear mass and <math>m</math> is the electron mass.

Change variables to the C of M
<math>\begin{eqnarray}
{\bf R} = \frac{1}{M+nm}\left[M{\bf r}_N + m({\bf r}_1 + {\bf r}_2 +
\ldots)\right]
\end{eqnarray}</math>

and relative variables

<math>
{\bf s}_i = {\bf r}_i - {\bf r}_N.
</math>

Then

<math>\begin{eqnarray}\ \ \ \ \
H = \frac{1}{2(M+nm)}{\bf p}_R^2+\frac{1}{2m}\sum_{i=1}^n{\bf p}_{s_i}^2
+\frac{1}{2M}\sum_{i,k}{\bf p}_{s_i}\cdot {\bf p}_{s_k} + V({\bf s}_1, {\bf
s}_2, \ldots)\nonumber
= \frac{1}{M+nm}{\bf p}_R^2+\frac{1}{2\mu}\sum_{i=1}^n{\bf
p}_{s_i}^2+\frac{1}{2M}\sum_{i\neq k}{\bf p}_{s_i}\cdot {\bf p}_{s_k}+V({\bf s}_1, {\bf
s}_2, \ldots)\nonumber
\end{eqnarray}</math>

where
<math>\begin{eqnarray}
\mu = \left(\frac{1}{M} + \frac{1}{m}\right)^{-1}
\end{eqnarray}</math>

If <math>n=2</math>, the Schrödinger equation is
<math>\begin{eqnarray}\ \ \
\left[ -\frac{\hbar^2}{2\mu}(\nabla^2_{s_1}+ \nabla^2_{s_2}) -
\frac{\hbar^2}{M}\nabla_{s_1}\nabla_{s_2}-\frac{Ze^2}{s_1}-\frac{Ze^2}{s_2}+\frac{e^2}{s_{12}}\right]\psi
= E\psi
\end{eqnarray}</math>.

Define the reduced mass Bohr radius
<math>\begin{eqnarray}
a_\mu = \frac{\hbar^2}{\mu e^2} = \frac{m}{\mu}a_0
\end{eqnarray}</math>
and

<math>\begin{eqnarray}
\rho_i = \frac{Zs_i}{a_\mu}\nonumber\\
\varepsilon = \frac{E}{Z^2(e^2/a_\mu)}\nonumber
\end{eqnarray}</math>

The Schrödinger equation then becomes
<math>\begin{eqnarray}\ \ \
\left[
-\frac{1}{2}(\nabla^2_{\rho_1}+\nabla^2_{\rho_2})-\frac{\mu}{M}\vec{\nabla}_{\rho_1}\cdot\vec{\nabla}_{\rho_2} - \frac{1}{\rho_1}-\frac{1}{\rho_2}+\frac{Z^{-1}}{\rho_{12}}\right]\psi = \varepsilon \psi
\end{eqnarray}</math>

The effects of finite mass are thus

1. A reduced mass correction to all energy levels of <math>(e^2/a_\mu)/(e^2/a_0)
= \mu/m</math>, i.e.
<math>\begin{eqnarray}
E_M=\frac{\mu}{m}E_\infty
\end{eqnarray}</math>

2. A specific mass shift given in first order perturbation theory by
<math>\begin{eqnarray}
\Delta\varepsilon =
-\frac{\mu}{M}<\psi|\nabla_{\rho_1}\cdot\nabla_{\rho_2}|\psi>
\end{eqnarray}</math>
or

<math>\begin{eqnarray}
\Delta\varepsilon =
-\frac{\mu}{M}<\psi|\nabla_{\rho_1}\cdot\nabla_{\rho_2}|\psi>\frac{e^2}{a_\mu}\nonumber\\
= -\frac{\mu^2}{mM}<\psi|\nabla_{\rho_1}\cdot\nabla_{\rho_2}|\psi>\frac{e^2}{a_0}\nonumber
\end{eqnarray}</math>

Since <math>\nabla_{\rho_1}\cdot\nabla_{\rho_2}</math> has the same angular properties as
<math>{\bf\rho}_1 \cdot {\bf\rho}_2</math>, the operator is like the product of two
dipole operators. For product type wave functions of the form
$
\psi = \psi(1s)\psi(nl) \pm \mbox{exchange},
$
the matrix element vanishes for all but P-states.

=Radiative Transitions=

==Transition Integrals and the Line Strength==

For an electric dipole transition <math>\gamma'\,L'\,S'\,J'\,M' \rightarrow \gamma\,L\,S\,J\,M</math>, the line strength is defined by
\begin{equation}
S = \sum_{M',\mu,M}|\langle\gamma'\,L'\,S'\,J'\,M'|r_\mu|\gamma\,L\,S\,J\,M\rangle|^2
\end{equation}
Using the Wigner-Eckart theorem, this is related to the reduced matrix element by
\begin{eqnarray}
S(J',J) = |\langle\gamma'\,L'\,S'\,J'||r||\gamma\,L\,S\,J\,M\rangle|^2
\sum_{M',\mu,M}\left|\left(\matrix{J' & 1 & J \cr
-M'& \mu &M \cr}\right)\right|^2
\end{eqnarray}
Using the sum rule for the 3-<math>j</math> symbols
\begin{equation}
\sum_{M',\mu,M}\left|\left(\matrix{J' & 1 & J \cr
-M'& \mu &M \cr}\right)\right|^2 = 1
\end{equation}
we see that the line strength is identical to the reduced matrix element
\begin{equation}
S(J',J) = |\langle\gamma'\,L'\,S'\,J'||r||\gamma\,L\,S\,J\,M\rangle|^2
\end{equation}
For a nonrelativistic calculation in <math>LS</math> coupling, <math>S = S'</math> and we can next strip out the dependence on <math>J</math> and <math>J'</math> by use of Eq.\ (7.1.7) of Edmonds { Angular Momentum in Quantum Mechanics} with <math>k = 1</math> to obtain
\begin{equation}
S(J',J) = (-1)^{L'+S+J+1}[(2J+1)(2J'+1)]^{1/2}\left\{
\matrix{L' & J'& S\cr
J & L & 1\cr}\right\}\langle\gamma' L'|| r || \gamma L\rangle
\end{equation}
where <math>\langle\gamma' L'|| r || \gamma L\rangle</math> is the reduced matrix element if spin is neglected. Note that if spin is neglected, then the line-strength is simply
\begin{equation}
S(L',L) = |\langle\gamma' L'|| r || \gamma L\rangle|^2
\end{equation}
To calculate the reduced matrix element, use a particular example along with the Wigner-Eckart theorem
such as
\begin{equation}
\langle\gamma'\, L'|| r || \gamma\, L\rangle = (-1)^{L'}\langle\gamma'\, L'\, 0| z | \gamma\, L\, 0\rangle{\Large /}\left(\matrix{L' & 1 & L \cr
0 & 0 & 0 \cr}\right)
\end{equation}

===Correlated Two-electron Integrals===
For the case of two-electron integrals, <math>z = z_1 + z_2</math> and <math>r = r_1 + r_2</math>. As discussed in G.W.F. Drake, Phys. Rev. A 18, 820 (1978), the general integral in Hylleraas coordinates for the integral over the <math>z_1</math> term can be written in the form
\begin{equation}
\label{int1}
\langle\gamma'\, L'\, 0| z_1 | \gamma\, L\, 0\rangle = \sum_\Lambda C_\Lambda I_\Lambda(\gamma',\gamma)
\end{equation}
where <math>I_\Lambda(\gamma',\gamma)</math> stands for the purely radial integral
\begin{equation}
\label{radial}
I_\Lambda(\gamma',\gamma) = \int_0^\infty r_1\,dr_1\int_0^\infty r_2\,dr_2\int_{|r_1-r_2|}^{r_1+r_2}r_{12}\, dr_{12}
R_{\gamma'}^*\,r_1\, R_\gamma P_\Lambda(\cos\theta_{12})
\end{equation}
and <math>\cos\theta_{12} = (r_1^2 + r_2^2 - r_{12}^2)/(2r_1r_2)</math> is a short-hand notation for a radial function of <math>r_1</math>, <math>r_2</math> and <math>r_{12}</math>. Also, <math>P_\Lambda(\cos\theta_{12})</math> is a Legendre polynomial,
and, <math>R_\gamma = R_\gamma(r_1,r_2,r_{12})</math> is the radial part of the Hylleraas wave function for state <math>\gamma</math> such
that the total wave function is <math>\psi({\bf r}_1,{\bf r}_2) = R_\gamma(r_1,r_2,r_{12}){\cal Y}_{l_1,l_2,L}^M(\hat{\bf r}_1,\hat{\bf r}_2)\pm {\rm exchange}</math>, normalized to unity, and <math>{\cal Y}_{l_1,l_2,L}^M(\hat{\bf r}_1,\hat{\bf r}_2)</math> is a vector coupled product of spherical harmonics. The coefficents <math>C_\Lambda</math> in Eq.\ (\ref{int1}) can be calculated from Eq.\ (32) of the above 1978 paper. They need to be calculated separately for he direct-direct and the direct-exchange contributions. For the reduced matrix element of <math>z_1</math>, the leading 3-<math>j</math> symbol can be dropped, leaving (for the case <math>k = 1</math>)
\begin{eqnarray}
C_\Lambda^{\rm reduced} &=& \frac{(-1)^{l_2'+L'+L+1}}{2} (l_1,l_1',l_2,l_2',L,L')^{1/2}
\sum_{\lambda={\rm max}(l_1-1,0)}^{l_1+1} (\lambda,\Lambda,1)
\left(\matrix{l_1 & 1 & \lambda \cr
0 & 0 & 0 \cr}\right)
\left(\matrix{l_1'& \lambda & \Lambda \cr
0 & 0 & 0 \cr}\right)\nonumber\\
&\times&\left(\matrix{l_2' & 1_2 & \Lambda \cr
0 & 0 & 0 \cr}\right)
\left\{\matrix{L' & \lambda & l_2 \cr
\Lambda& l_2' & l_1' \cr}\right\}
\left\{\matrix{L' & \lambda & l_2 \cr
l_1 & L & 1 \cr}\right\}
\end{eqnarray}
where the notation <math>(a,b,c,\cdots)</math> means <math>(2a+1)(2b+1)(2c+1)\cdots</math>.
A similar expression can be obtained for the angular coefficients for the <math>z_2</math> term by interchanging the subscripts 1 and 2 throughout and multiplying by <math>(-1)^{L+L'+1}</math>.

===Example===
As an example, for the <math>1s^2\;^1S - 1s2p\;^1P</math> transition of helium, the only nonvanishing <math>C_\Lambda</math> angular coefficients are <math>C_1 = 1/2</math> for the direct-direct term (i.e. <math>P_1(\cos\theta_{12}) = \cos\theta_{12}</math> is included in the radial integral (\ref{radial})) and <math>C_0 = 1/2</math> for the direct-exchange term (i.e. <math>P_0(\cos\theta_{12}) = 1</math> is included in the radial integral (\ref{radial})). The results for infinite nuclear mass are
\begin{equation}
S(1\;^1S-2\;^1P) = (0.781\,127\,158\,667)^2
\end{equation}
\begin{equation}f(1\;^1S-2\;^1P) = -0.276\,164\,703\,736
\end{equation}
where
\begin{equation}
f(1\;^1S-2\;^1P) = \frac{2}{3}[E(1\;^1S) - E(2\;^1P)]S(1\;^1S-2\;^1P)
\end{equation}
is the absorption oscillator strength.

===Test===
This is a test edit

Theory Notes

2020-12-14T05:19:06Z

Drake: /* General Hermitian Property */

<center>
<H1>Notes on solving the Schrödinger equation in Hylleraas coordinates for heliumlike atoms
Gordon W.F. Drake
Department of Physics, University of Windsor
Windsor, Ontario, Canada N9B 3P4
(Transcribed from hand-written notes and edited by Lauren Moffatt. Last revised November 22, 2017 to correct a missing subscript in the expression for $B_1$ in the Hamiltonian in Sect. 11 General Hermitian Property.)
</H1>
</center>

==Introduction==
These notes describe some of the technical details involved in solving the Schrödinger equation for a heliumlike atom of nuclear charge Z in correlated Hylleraas coordinates. The standard for computational purposes is to express all quantities in a form that is valid for electronic states of arbitrary angular momentum, and to express matrix elements of the Hamiltonian in a manifestly Hermitian symmetrized form. Further information and derivations of the mathematical techniques for angular integrals and radial recursion relations can be found in the paper
G.W.F. Drake, Phys. Rev. A <math>{\bf 18}</math>, 820 (1978).

The starting point is the two-electron Schrödinger equation for infinite nuclear mass
\begin{equation} \left[-\frac{\hbar^2}{2m}(\nabla^2_1 +\nabla^2_2) - \frac{Ze^2}{r_1} -
\frac{Ze^2}{r_2}+\frac{e^2}{r_{12}} \right]\psi = E\psi\nonumber
\end{equation}
where <math>m</math> is the electron mass, and <math>r_{12} = |{\bf r}_1 - {\bf r}_2|</math> (see diagram below).

[[File:Pic1.jpg|250px]]

Begin by rescaling distances and energies so that the Schrödinger equation can be expressed in a dimensionless form. The dimensionless Z-scaled distance is defined by <math>\begin{eqnarray} \rho = \frac{Zr}{a_0} \end{eqnarray} </math> where <math> \begin{eqnarray} a_0 = \frac{\hbar^2}{me^2} \end{eqnarray} </math> is the atomic unit (a.u.) of distance equal to the Bohr radius <math>0.529\, 177\, 210\, 92(17) \times 10^{-10}</math> m. Then

\begin{equation}[-\frac{\hbar^2}{2m}Z^2(\frac{me^2}{\hbar^2})^2(\nabla^2_{\rho_1}+\nabla^2_{\rho_2})
- Z^2\frac{e^2}{a_0}\rho^{-1}_1 - Z^2\frac{e^2}{a_0}\rho^{-1}_2 +
\frac{e^2}{a_0}Z\rho^{-1}_{12}]\psi= E\psi\nonumber\end{equation}

But <math>\begin{eqnarray} \frac{\hbar^2}{m}(\frac{me^2}{\hbar^2})^2 = \frac{e^2}{a_0} \end{eqnarray} </math> is the hartree atomic unit of of energy <math>(E_h = 27.211\,385\,05(60)</math> eV, or equivalently <math>E_h/(hc) = 2194.746\,313\,708(11)</math> m<math>^{-1})</math>. Therefore, after multiplying through by <math>a_0/(Ze)^2</math>, the problem to be solved in Z-scaled dimensionless units becomes

\begin{equation} \left[-\frac{1}{2}(\nabla^2_{\rho_1}+\nabla^2_{\rho_2}) -
\frac{1}{\rho_1} - \frac{1}{\rho_2} + \frac{Z^{-1}}{\rho_{12}}\right]\psi =
\varepsilon\psi\nonumber \end{equation}

where <math>\begin{eqnarray}\varepsilon = \frac{Ea_0}{(Ze)^2} \end{eqnarray}</math> is the energy in Z-scaled atomic units. For convenience, rewrite this in the conventional form
\begin{equation}
H\psi = \varepsilon\psi\nonumber \end{equation}
where (using <math>r</math> in place of <math>\rho</math> for the Z-scaled distance)
\begin{equation}
H = -\frac{1}{2}(\nabla^2_1+\nabla^2_2) - \frac{1}{r_1}-\frac{1}{r_2} + \frac{Z^{-1}}{r_{12}}
\nonumber\end{equation}
is the atomic Hamiltonian for infinite nuclear mass.

==The Hartree Fock Method==

For purposes of comparison, and to define the correlation energy, assume that <math>\psi({\bf r}_1,{\bf r}_2)</math> can be written in the separable product form
\begin{equation}
\psi({\bf r}_1,{\bf r}_2) = \frac{1}{\sqrt{2}} \left[ u_1(r_1) u_2(r_2) \pm u_2(r_1) u_1(r_2) \right] \nonumber
\end{equation}
for the <math>1s^2\ ^1\!S</math> ground state. Because of the <math>\frac{1}{r_{12}}</math> term in the Schrödinger equation, the exact solution connot be expressed in this form as a separable product. However, the Hartree-Fock approximation corresponds to finding the best possible solution to the Schrödinger equation
\begin{equation}
H\psi({\bf r}_1,{\bf r}_2) = E\psi({\bf r}_1,{\bf r}_2)\nonumber
\end{equation}
that can nevertheless be expressed in this separable product form, where as before
\begin{equation}
H = -\frac{1}{2}(\nabla^2_1+\nabla^2_2) - \frac{1}{r_1}- \frac{1}{r_2} + \frac{Z^{-1}}{r_{12}}\nonumber
\end{equation}
is the full two-electron Hamiltonian. To find the best solution, substitute into <math>\left\langle\psi \lvert H-E \rvert \psi\right\rangle</math> and require this expression to be stationary
with respect to arbitrary infinitesimal variations <math>\delta u_1</math> and <math>\delta
u_2</math> in <math>u_1</math> and <math>u_2</math> respectively; i.e.

\begin{equation}
\frac{1}{2} \left\langle \delta u_1(r_1)u_2(r_2) \pm u_2(r_1)\delta
u_1(r_2) \lvert H-E \rvert u_1(r_1) u_2(r_2)\pm u_2(r_1) u_1(r_2)\right\rangle\nonumber
\end{equation}

\begin{equation}
=\int\delta u_1(r_1) d{\bf r}_1 \left\{ \int d{\bf r}_2 u_2(r_2) \left( H - E \right) \left[ u_1(r_1)u_2(r_2) \pm u_2(r_1) u_1(r_2) \right] \right\}\nonumber
\end{equation}

\begin{equation}
= 0 \nonumber
\end{equation}

for arbitrary <math> \delta u_1(r_1).\nonumber </math> Therefore <math>\{\int d{\bf r}_2 \ldots \} = 0</math>.

Similarly, the coefficient of <math>\delta u_2</math> would give

\begin{equation}
\int d{\bf r}_1 u_1(r_1) \left( H-E \right) \left[ u_1(r_1) u_2(r_2) \pm
u_2(r_1) u_1(r_2) \right] = 0\nonumber
\end{equation}

Define

\begin{equation}
I_{12} = I_{21} = \int d{\bf r}\, u_1(r)u_2(r), \nonumber
\end{equation}

\begin{equation}
H_{ij} = \int d{\bf r}\, u_i(r)(-\frac{1}{2}\nabla - \frac{1}{r})u_j(r), \nonumber
\end{equation}

\begin{equation}
G_{ij}(r) = \int d{\bf r}^\prime u_i(r^\prime)\frac{1}{|{\bf r} - {\bf r}\prime|}u_j(r^\prime)\nonumber
\end{equation}

Then the above equations become the pair of integro-differential equations

\begin{equation}
[ H_0 - E + H_{22}+G_{22}(r)]u_1(r) = \mp [ I_{12}(H_0-E) + H_{12}+G_{12}(r)]u_2(r)\nonumber
\end{equation}

\begin{equation}
[H_0-E+H_{11}+G_{11}(r)]u_2(r) = \mp [I_{12}(H_0-E) + H_{12}+G_{12}(r)]u_1(r)\nonumber
\end{equation}

These must be solved self-consistently for the "constants" <math>I_{12}</math> and
<math>H_{ij}</math> and the function <math>G_{ij}(r)</math>.

The Hartree Fock energy is <math>E \simeq -2.87\ldots </math>a.u. while the exact energy is <math>E = -2.903724\ldots </math>a.u.

The difference is called the "correlation energy" because it arises from the
way in which the motion of one electron is correlated to the other. The
Hartree Fock equations only describe how one electron moves in the average field
provided by the other (mean-field theory).

==Configuration Interaction==
Expand
\begin{equation}
\psi({\bf r}_1,{\bf r}_2)= C_0u^{(s)}_1(r_1)u^{(s)}_1(r_2) + C_1u^{(P)}_1({\bf r}_1)u^{(P)}_1({\bf r}_2){\cal Y}^0_{1,1,0}(\hat{\bf r}_1, \hat{\bf r}_2)+C_2u^{(d)}_1({\bf r}_1)u^{(d)}_2({\bf r}_2){\cal Y}^0_{2,2,0}(\hat{\bf r}_1, \hat{\bf r}_2)+... \pm {\rm\ exchange}
\end{equation}
where
\begin{equation}
{\cal Y}^M_{l_1,l_2,L}(\hat{\bf r}_1, \hat{\bf r}_2)=\sum_{m_1,m_2}Y^{m_1}_{l_1}({\bf r}_1)Y^{m_2}_{l_2}({\bf r}_2)\times <l_1l_2m_1m_2\mid LM>
\end{equation}

This works, but is slowly convergent, and very laborious. The best CI calculations are accurate to <math> ~10^{-7}</math> a.u.

==Hylleraas Coordinates==
[E.A. Hylleraas, Z. Phys. <math>{\bf 48}, 469(1928)</math> and <math>{\bf 54}, 347(1929)</math>]
suggested using the co-ordinates <math>r_1</math>, <math>r_2</math> and <math>r_{12}</math> or equivalently

\begin{eqnarray}
s &=& r_1 + r_2, \nonumber\\
t &=& r_1-r_2, \nonumber\\
u &=& r_{12}\nonumber
\end{eqnarray}

and writing the trial functions in the form

\begin{equation}
\Psi({\bf r}_1,{\bf r}_2) = \sum^{i+j+k\leq
N}_{i,j,k}c_{i,j,k}r_1^{i+l_1}r_2^{j+l_2}r_{12}^ke^{-\alpha r_1 - \beta r_2}
\mathcal{Y}^M_{l_1,l_2,L}(\hat{r}_1,\hat{r}_2)\pm \text{exchange}\nonumber
\end{equation}

Diagonalizing H in this non-orthogonal basis set is equivalent to solving
<math> \begin{eqnarray}
\frac{\partial E}{\partial c_{i,j,k}} = 0\nonumber
\end{eqnarray} </math>
for fixed <math>\alpha</math> and <math>\beta</math>.

The diagonalization must be repeated for different values of <math>\alpha</math> and
<math>\beta</math> in order to optimize the non-linear parameters.

==Completeness==
The completeness of the above basis set can be shown by first writing
\begin{equation}
r_{12}^2 = r_1^2 + r_2^2 - 2r_1r_2\cos\theta_{12}
\end{equation}
and, from the spherical harmonic addition theorem,
\begin{equation}
\cos(\theta_{12})=\frac{4\pi}{3}\sum^1_{m=-1}Y^{m*}_1(\theta_1,\varphi_1)Y^m_1(\theta_2,\varphi_2)
\end{equation}
Consider first the S-states. The <math>r_{12}^0</math> terms are like the ss terms in a CI calculation. The
<math>r_{12}^2</math> terms bring in p-p type contributions, and the higher powers bring
in d-d, f-f etc. type terms. In general

\begin{equation}
P_l(\cos\theta_{12}) = \frac{4\pi}{2l+1}\sum^l_{m=-l}{Y^{m}_l}^*(\theta_1,\varphi_1)Y^m_l(\theta_2,\varphi_2)\nonumber
\end{equation}

For P-states, one would have similarly

\begin{array}{lr}
r_{12}^0 & (sp)P\nonumber\\
r_{12}^2 & (pd)P\nonumber\\
r_{12}^4 & (df)P\nonumber\\
\vdots & \vdots\nonumber
\end{array}

For D-states

\begin{array}{lr}
r_{12}^0 & (sd)D & (pp^\prime)D\nonumber\\
r_{12}^2 & (pf)D & (dd^\prime)D\nonumber\\
r_{12}^4 & (dg)D & (ff^\prime)D\nonumber\\
\vdots & \vdots & \vdots\nonumber
\end{array}

In this case, since there are two "lowest-order" couplings to form a
D-state, both must be present in the basis set, i.e.

\begin{equation}
\Psi(r_2,r_2) = \sum c_{ijk}r_1^ir_2^{j+2}r_{12}^ke^{-\alpha r_1-\beta
r_2}\mathcal{Y}^M_{022}(\hat{r}_1,\hat{r}_2)
+\sum d_{ijk}r_1^{i+1}r_2^{j+1}r_{12^k}e^{-\alpha^\prime r_1 - \beta^\prime
r_2}\mathcal{Y}^M_{112}(\hat{r}_1,\hat{r}_2)\nonumber
\end{equation}

For F-states, one would need <math>(sf)F</math> and <math>(pd)F</math> terms.

For G-states, one would need <math>(sg)G</math>, <math>(pf)G</math> and <math>(dd^\prime)G</math> terms.

Completeness of the radial functions can be proven by considering the
Sturm-Liouville problem

\begin{equation}
\left(-\frac{1}{2}\nabla^2-\frac{\lambda}{r_s}-E\right)\psi({\bf r}) = 0\nonumber
\end{equation}

or

\begin{equation}
\left(-\frac{1}{2}\frac{1}{r^2}\frac{\partial}{\partial r}\left({r^2}\frac{\partial}{\partial r}\right) -
\frac{l\left(l+1\right)}{2r^2} - \frac{\lambda}{r} - E\right)u(r) = 0.\nonumber
\end{equation}

For fixed E and variable <math>\lambda</math> (nuclear charge).

The eigenvalues are <math>\lambda_n = (E/E_n)^{1/2}</math>, where <math>E_n =- \frac{1}{2n^2}</math>

[[File:Pic2.jpg|500px]]

and the eigenfunctions are

\begin{equation}
u_{nl}(r) =
\frac{1}{(2l+1)!}\left(\frac{(n+l)!}{(n-l-1)2!}\right)^{1/2}(2\alpha)^{3/2}e^{-\alpha
r}\nonumber
\times \ \left( 2 \alpha r \right)^l\,_{1}F_{1}\left(-n+l+1,2l+2; 2 \alpha r \right)
\end{equation}

with <math>\alpha = (-2E)^{1/2}</math> and <math>n\geq l+1</math>. The confluent hypergeometric function <math> _1F_1(a,b,;z)</math> then denotes a finite polynomial since <math>a=-n+l+1</math> is a negative integer or zero.

Unlike the hydrogen spectrum, which has both a discrete part for <math>E<0</math> and a
continuous part for <math>E>0</math>, this forms an entirely discrete set of finite
polynomials, called Sturmian functions. They are orthogonal with respect to
the potential, i.e.

\begin{equation}
\int^\infty_0 r^2dr \left( u_{n^\prime l}(r)\frac{1}{r}u_{nl}(r) \right) = \delta_{n,n^\prime}\nonumber
\end{equation}

Since they become complete in the limit <math>n\rightarrow\infty</math>, this assures the
completeness of the variational basis set.

[See also B Klahn and W.A. Bingel Theo. Chim. Acta (Berlin) 44, 9 and 27
(1977)].

==Solutions of the Eigenvalue Problem==

For convenience, write

\[
\Psi({\bf r}_1,{\bf r}_2) = \sum^N_{m=1}c_m\varphi_m\nonumber
\]

where <math>m</math> represents the <math>m</math>'th combination of <math>i,j,k</math> and

\begin{equation}
\varphi_{ijk}=r_1^ir_2^jr_{12}^ke^{-\alpha r_1 - \beta
r_2}\mathcal{Y}^M_{l_1,l_2,L}(\hat{r}_1,\hat{r}_2) \pm \mbox{exchange}\nonumber
\end{equation}

\begin{align}

&
\left( \begin{array}{lr}
\cos(\theta) & \sin(\theta)\\
-\sin(\theta) & \cos(\theta)
\end{array} \right)
\left( \begin{array}{lr}
H_{11} & H_{12} \\
H_{12} & H_{22}
\end{array}\right)
\left( \begin{array}{lr}
\cos(\theta) & -\sin(\theta)\\
\sin(\theta) & \cos(\theta)
\end{array} \right)
\\[5pt]
= & \left(\begin{array}{lr}
cH_{11}+sH_{12} & cH_{12} + sH_{22}\\
-sH_{11} + cH_{12} & -sH_{12} + cH_{22}
\end{array}\right)
\left( \begin{array}{cc}
c & -s \\
s & c
\end{array}\right)
\\[5pt]
= & \left(\begin{array}{lr}
c^2H_{11}+s^2H_{22} + 2csH_{12} & (c^2-s^2)H_{12}+cs(H_{22}-H_{11})\\
(c^2-s^2)H_{12}+cs(H_{22}-H_{11}) & s^2H_{11}+c^2H_{22}-2csH_{12}
\end{array}\right)
\end{align}

Therefore

\begin{equation}
\left[\cos^2(\theta)-\sin^2(\theta)\right]H_{12} =
\cos(\theta)\sin(\theta)(H_{11}-H_{22})\nonumber
\end{equation}

and

\begin{equation}
\tan(2\theta) = \frac{2H_{12}}{H_{11}-H_{22}}\nonumber
\end{equation}

i.e.

\begin{equation}
\cos(\theta)=\left(\frac{r+\omega}{2r}\right)^{1/2}\nonumber
\end{equation}

\begin{equation}
\sin(\theta)=-\mbox{sgn}(H_{12})\left(\frac{r-\omega}{2r}\right)^{1/2}\nonumber
\end{equation}

where

\begin{equation}
\omega = H_{22}-H_{11}\nonumber
\end{equation}

\begin{equation}
r=\left(\omega^2+4H_{12}^2\right)^{1/2}\nonumber
\end{equation}

\begin{equation}
E_1=\frac{1}{2}\left(H_{11}+H_{22}-r\right)\nonumber
\end{equation}

\begin{equation}
E_2=\frac{1}{2}\left(H_{11}+H_{22}+r\right)\nonumber
\end{equation}

===Brute Force Method===

- Gives all the eigenvalues and eigenvectors, but it is slow

- First orthonormalize the <math>N</math>-dimensional basis set; i.e. form linear combinations

\begin{equation}
\Phi_m = \sum_{n=1}^N\varphi_nR_{nm}\nonumber
\end{equation}

such that
\begin{equation}
\langle \Phi_m | \Phi_n \rangle = \delta_{m,n}
\end{equation}

This can be done by finding an orthogonal tranformation T such that

\begin{equation}
T^TOT = I = \left(
\begin{array}{ccccc}
I_1 & 0 & 0 & \ldots & 0 \\
0 & I_2 & 0 & \ldots & 0 \\
0 & 0 & I_3 & \ldots & 0 \\
\vdots & \vdots & \vdots & \ddots & \vdots \\
0 & 0 & 0 & \ldots & I_N
\end{array}\right);
\end{equation}

\begin{equation}
O_{mn} = \langle \varphi_m | \varphi_n \rangle \nonumber
\end{equation}

and then applying a scale change matrix

\begin{equation}
S = \left(\begin{array}{ccccc}
\frac{1}{I_1^{1/2}} & 0 & 0 & \ldots & 0\\
0 & \frac{1}{I_2^{1/2}} & 0 & \ldots & 0 \\
0 & 0 & \frac{1}{I_3^{1/2}} & \ldots & 0 \\
\vdots & \vdots & \vdots & \ddots & \vdots \\
0 & 0 & 0 & \ldots & I_N^{1/2}
\end{array}\right)= S^T\nonumber
\end{equation}

Then

\begin{equation}
S^TT^TOTS = 1
\end{equation}

i.e.

\begin{equation}
R^TOR = 1
\end{equation}

with <math>R=TS</math>.

If H is the matrix with elements <math>H_{mn} = \langle \varphi_m | \varphi_n \rangle </math>, then H
expressed in the <math>\Phi_m</math> basis set is

\begin{equation}
H^\prime = R^THR.\nonumber
\end{equation}

We next diagonalize <math>H^\prime</math> by finding an orthogonal transformation W such
that

\begin{equation}
W^T H^\prime W = \lambda = \left(
\begin{array}{cccc}
\lambda_1 & 0 & \ldots & 0 \\
0 & \lambda_2 & \ldots & 0 \\
\vdots & \vdots & \ddots & \vdots \\
0 & 0 & \ldots & \lambda_N
\end{array}\right)\nonumber
\end{equation}

The q'th eigenvector is

\begin{equation}
\Psi^{(q)} = \sum_{n=1}^N\Phi_n W_{n,q} = \sum_{n,n^\prime} \varphi_{n^\prime} R_{n^\prime ,n} W_{n,q} \nonumber
\end{equation}

i.e.

\begin{equation}
c_{n^\prime}^{(q)} = \sum_{n=1}^N R_{n^\prime n} W_{n,q}
\end{equation}

===The Power Method===

- Based on the observation that if H has one eigenvalue, <math>\lambda_M</math>, much bigger than all the rest, and <math>\chi = \left(
\begin{array}{c}a_1\\a_2\\\vdots\end{array}\right)</math>
is an arbitrary starting vector, then <math>\chi = \sum_{q=1}^N x_q\Psi^{(q)}</math>.

\begin{equation}
(H)^n\chi = \sum_{q=1}^N x_q \lambda^n_q\Psi^{(q)}\nonumber\rightarrow x_M\lambda_M^n\Psi^{(M)}\nonumber
\end{equation}

provided <math>x_M\neq 0</math>.

To pick out the eigenvector corresponding to any eigenvalue, write the original problem in the form

\begin{align}
H\Psi & = \lambda O \Psi \nonumber \\
\left( H - \lambda_q O \right) \Psi & = \left( \lambda - \lambda_q \right) O \Psi \nonumber
\end{align}

Therefore,

\begin{equation}
G \Psi = \frac{1}{\lambda - \lambda_q } \Psi \nonumber \\
\end{equation}

where <math>G=(H-\lambda_qO)^{-1}O</math> with eigenvalues <math>\frac{1}{\lambda_n - \lambda_q}</math>.

By picking <math>\lambda_q</math> close to any one of the <math>\lambda_n</math>, say
<math>\lambda_{n^\prime}</math>, then <math>\frac{1}{\lambda_n-\lambda_q}</math> is much larger for
<math>n=n^\prime</math> than for any other value. The sequence is then

\begin{array}{c}
\chi_1=G\chi\nonumber\\
\chi_2=G\chi_1\nonumber\\
\chi_3=G\chi_2\nonumber\\
\vdots\nonumber
\end{array}

until the ratios of components in <math>\chi_n</math> stop changing.

- To avoid matrix inversion and multiplication, note that the sequence is equivalent to
\begin{equation}
F \chi_n = \left( \lambda - \lambda_q \right) O \chi_{n-1} \nonumber \\
\end{equation}
where <math>F = H-\lambda_qO</math>. The factor of <math>\left( \lambda - \lambda_q \right)</math> can be
dropped because this only affects the normalization of <math>\chi_n</math>. To find <math>\chi_n</math>, solve
\begin{equation}
F \chi_n = O \chi_{n-1} \nonumber
\end{equation}
(N equations in N unknowns). Then
\begin{equation}
\lambda = \frac{ \langle \chi_n | H | \chi_n \rangle}{\langle \chi_n | \chi_n \rangle} \nonumber
\end{equation}

==Matrix Elements of H==

\begin{equation}
H=-\frac{1}{2}\nabla^2_1 -\frac{1}{2}\nabla^2_2 - \frac{1}{r_1} - \frac{1}{r_2} +\frac{Z^{-1}}{r_{12}}\nonumber
\end{equation}

Taking <math>r_1, r_2</math> and <math>r_{12}</math> as independent variables,

\begin{equation}
\nabla_1^2 = \frac{1}{r_1^2}\frac{\partial}{\partial r_1}\left(r_1^2\frac{\partial}{\partial r_1}\right)+ \frac{1}{r_{12}^2}\frac{\partial}{\partial r_{12}}
\left(r_{12}^2\frac{\partial}{\partial r_{12}}\right)\nonumber -\frac{l_1(l_1+1)}{r_1^2}+2(r_1 - r_2\cos(\theta_{12}))\frac{1}{r_{12}}\frac{\partial^2}{\partial r_1 \partial r_{12}} \nonumber - 2(\nabla_1^Y \cdot {\bf r}_2)\frac{1}{r_{12}}\frac{\partial}{\partial r_{12}}\nonumber
\end{equation}

where <math>\nabla_1^Y</math> acts only on the spherical harmonic part of the wave function and the diagram

[[File:Pic3.jpg|400px]]

defines the complete set of 6 independent variables is <math>r_1</math>, <math>r_2</math>, <math>r_{12}</math>, <math>\theta_1</math>, <math>\varphi_1</math>, <math>\chi</math>.

If <math>r_{12}</math> were not an independent variable, then one could take the volume element to be

\begin{equation}
d\tau = r_1^2dr_1\sin(\theta_1)d\theta_1d\varphi_1r_2^2dr_2\sin(\theta_2)d\theta_2d\varphi_2.\nonumber
\end{equation}

However, <math>\theta_2</math> and <math>\varphi_2</math> are no longer independent variables. To eliminate them, take the point <math>{\bf r}_1</math> as the
origin of a new polar co-ordinate system, and write

\begin{equation}
d\tau=-r_1^2dr_1\sin(\theta_1)d\theta_1d\varphi_1r_{12}^2dr_{12}\sin(\psi)d\psi d\chi\nonumber
\end{equation}

and use

\begin{equation}
r_2^2=r_1^2+r_{12}^2 +2r_1r_{12}\cos(\psi)
\end{equation}

Then for fixed <math>r_1</math> and <math>r_{12}</math>,

\begin{equation}
2r_2dr_2 = -2r_1r_{12}\sin(\psi)d\psi
\end{equation}

Thus

\begin{equation}
d\tau= r_1dr_1r_2dr_2r_{12}dr_{12}\sin(\theta_1)d\theta_1d\varphi_1d\chi
\end{equation}

The basic type of integral to be calculated is

\begin{align}
I \left( l_1, m_1, l_2, m_2; R \right) & = \int \sin \left( \theta_1 \right) d\theta_1 d\varphi_1 d\chi Y^{m_1}_{l_1} \left( \theta_1, \varphi_1 \right)^{*} Y^{m_2}_{l_2} \left( \theta_2, \varphi_2 \right) \\
& \times \int r_1dr_1r_2dr_2r_{12}dr_{12}R \left( r_1, r_2, r_{12} \right)
\end{align}

Consider first the angular integral. <math>Y^{m_2}_{l_2} \left( \theta_2, \varphi_2 \right)</math> can be expressed in terms of the independent variables <math>\theta_1, \varphi_1, \chi</math> by use of the rotation matrix relation
\begin{equation}
Y^{m_2}_{l_2}( \theta_2, \varphi_2) = \sum_m \mathcal{D}^{(l_2)}_{m_2,m}(\varphi_1,\theta_1,\chi)^*Y^m_{l_2}( \theta_{12}, \varphi)
\end{equation}
where <math>\theta, \varphi</math> are the polar angles of <math>{\bf r}_2</math> relative to <math>{\bf r}_1</math>. The angular integral is then

\begin{align}
I_{ang} & = \int^{2 \pi}_{0} d\chi \int^{2\pi}_{0} d\varphi_1 \int^{\pi}_{0} \sin \left( \theta_1 \right) d\theta_{1} Y^{m_1}_{l_1} \left( \theta_1, \varphi_1 \right)^{*} \\
& \times \sum_m \mathcal{D}^{(l_2)}_{m_2,m} \left( \varphi_1,\theta_1,\chi \right)^{*} Y^{m}_{l_2} \left( \theta_{12},\varphi \right) \\
\end{align}

Use

\begin{equation}
Y^{m_1}_{l_1} \left( \theta_1, \varphi_1 \right)^{*} = \sqrt{\frac{2l_1+1}{4\pi}}\mathcal{D}^{(l_1)}_{m_1,0} \left( \varphi_1, \theta_1, \chi \right)
\end{equation}

together with the orthogonality property of the rotation matrices (Brink and Satchler, p 147)

\begin{align}
& \int \mathcal{D}^{(j)*}_{m,m^\prime}\mathcal{D}^{(J)}_{M,M^\prime}\sin(\theta_1)d\theta_1d\varphi_1d\chi \\
& = \frac{8\pi^2}{2j+1}\delta_{jJ}\delta_{mM}\delta_{m^\prime M^\prime}\nonumber
\end{align}

to obtain

\begin{align}
I_{ang} & = \sqrt{ \frac{2l_1+1}{4\pi}} \frac{8\pi^2}{2l_1+1} \delta_{l_1,l_2} \delta_{m_1,m_2} Y^{0}_{l_2} \left( \theta_{12}, \varphi \right) \\
& = 2 \pi \delta_{l_1,l_2} \delta_{m_1,m_2} P_{l_2} \left( \cos \theta_{12} \right)
\end{align}

since

\begin{equation}
Y^0_{l_2} \left( \theta_{12}, \varphi \right) = \sqrt{\frac{2l_1+1}{4\pi}} P_{l_2} \left( \cos \left( \theta_{12} \right) \right)
\end{equation}

Note that <math>P_{l_2} \left( \cos \theta \right)</math> is just a short hand expression for a radial function because

\begin{equation}
\cos \theta_{12} = \frac{r_1^2 + r_2^2 - r_{12}^2}{2r_1r_2}
\end{equation}

The original integral is thus

\begin{equation}
I \left( l_1, m_1, l_2, m_2; R \right) = 2 \pi \delta_{l_1,l_2} \delta_{m_1,m_2} \int^{\infty}_{0} r_{1} dr_{1} \int^{\infty}_{0} r_{2} dr_{2} \int^{r_{1} + r_{2}}_{|r_{1} - r_{2}|} r_{12} dr_{12} R \left( r_{1}, r_{2}, r_{12} \right) P_{l_2} \left( \cos \theta_{12} \right)
\end{equation}

where again

\begin{equation}
\cos \theta_{12} = \frac{r_{1}^{2} +r^{2}_{2} - r_{12}^{2}}{2 r_{1} r_{2}}
\end{equation}

is a purely radial function.

The above would become quite complicated for large <math>l_2</math> because <math>P_{l_2}(\cos \theta_{12})</math> contains terms up to <math>\left( \cos \theta_{12} \right)^{l_2}</math>. However, recursion relations exist which allow any integral containing <math>P_l \left( \cos \theta_{12} \right)</math> in terms of those containing just <math>P_0 \left( \cos \theta_{12} \right) = 1</math> and <math>P_1 \left( \cos \theta_{12} \right) = \cos \theta_{12}</math>.

==Radial Integrals and Recursion Relations==

The basic radial integral is [see G.W.F. Drake, Phys. Rev. A 18, 820 (1978)]

\begin{align}
I_0 \left( a, b, c \right) & =
\int^{\infty}_{0} r_{1} dr_{1} \int^{\infty}_{r_{1}} r_{2} dr_{2} \int^{r_{1}+r_{2}}_{r_{2} - r_{1}} r_{12} dr_{12} r_{1}^{a} r_{2}^{b} r_{12}^{c} e^{-\alpha r_{1} - \beta r_{2}} + \int^{\infty}_{0} r_{2} dr_{2} \int^{\infty}_{r_{2}} r_{1} dr_{1} \int^{r_{1} + r_{2}}_{r_{1} - r_{2}} r_{12} dr_{12} r_{1}^{a} r_{2}^{b} r_{12}^{c} e^{- \alpha r_{1} - \beta r_{2}} \\
& = \frac{2}{c+2} \sum^{[(c+1)/2]}_{i=0}
\left(\begin{array}{c}
c+2 \\
2i+1
\end{array} \right)
\left\{ \frac{q!}{\beta^{q+1} \left( \alpha + \beta \right)^{p+1}} \sum_{j=0}^{q} \frac{\left( p + j \right)!}{j!} \left( \frac{\beta}{\alpha + \beta} \right)^j + \frac{q^{\prime}!}{\alpha^{q^{\prime}+1} \left( \alpha + \beta \right)^{p^{\prime}+1}} \sum^{q^{\prime}}_{j=0} \frac{\left( p^{\prime} + j \right)!}{j!} \left( \frac{\alpha}{\alpha + \beta} \right)^j \right\}
\end{align}

where

\begin{array}{ll}
p = a + 2i + 2 & p^{\prime} = b + 2i + 2 \\
q = b + c - 2i + 2 & q^{\prime} = a + c - 2i + 2
\end{array}

The above applies for <math> a, b \geq -2, c\geq -1</math>.
<math>[x]</math> means "greatest integer in" <math>x</math>.

Then

\begin{align}
I_1 \left( a, b, c \right) & = \int d\tau_{r} r_{1}^{a} r_{2}^{b} r_{12}^{c} e^{ -\alpha r_{1} - \beta r_{2}} P_{1}\left( \cos \theta \right) \\
& = \frac{1}{2} \left( I_{0} \left( a + 1, b - 1, c \right) + I_{0} \left( a - 1, b + 1, c \right) - I_{0} \left( a - 1, b - 1, c + 2 \right) \right)
\end{align}

===The Radial Recursion Relation===

Recall that the full integral is

\begin{equation}
I(l_1m_1,l_2m_2;R)=2\pi\delta_{l_1,l_2}\delta_{m_1,m_2}I_{l_2}(R)
\end{equation}

where

\begin{equation}
I_{l_2}(R) = \int d\tau_r R(r_1,r_2,r_{12})P_{l_2}(\cos \theta_{12})
\end{equation}

To obtain the recursion relation, use

\begin{equation}
P_l(x) = \frac{[P^{\prime}_{l+1}(x)-P^{\prime}_{l-1}(x)]}{2l+1}
\end{equation}

with

\begin{equation}
P_{l+1}^\prime (x) = \frac{d}{dx}P_{l+1}(x)
\end{equation}

Here <math>x=\cos\theta_{12}</math> and

\begin{align}
\frac{d}{d \cos \theta_{12}} & = \frac{dr_{12}}{d \cos \theta_{12}} \frac{d}{dr_{12}} \\
& = -\frac{r_{1} r_{2}}{r_{12}} \frac{d}{dr_{12}}
\end{align}

Then,

\begin{equation}
I_{l}\left( R \right) = - \int d\tau_r R \frac{r_{1} r_{2}}{r_{12}} \frac{d}{dr_{12}} \frac{ \left[ P_{l+1} \left( \cos \theta_{12} \right) - P_{l-1} \left( \cos \theta_{12} \right) \right]}{2l+1}
\end{equation}

The <math>r_{12}</math> part of the integral is

\begin{align}
& \int^{r_{1} + r_{2}}_{|r_{1} - r_{2}|} r_{12} dr_{12} R \frac{r_{1} r_{2}}{r_{12}} \frac{d}{dr_{12}} \left[P_{l+1}-P_{l-1} \right] \\
& = R r_{1} r_{2} \left[ P_{l+1} - P_{l-1} \right]^{r_{1} + r_{2}}_{|r_{1} - r_{2}|} - \int^{r_{1} + r_{2}}_{|r_{1} - r_{2}|} r_{12} dr_{12} \left( \frac{d}{dr_{12}} R \right) \frac{r_{1} r_{2}}{r_{12}} \frac{ \left[ P_{l+1} - P_{l-1} \right]}{2l+1}
\end{align}

The integrated term vanishes because

\begin{equation}
\cos \theta_{12} = \frac{r_{1}^{2} + r_{2}^{2} - r_{12}^{2}}{2 r_{1} r_{2}}
\begin{array}{lll}
& = -1 & {\rm when} & r_{12}^{2} = \left( r_{1} + r_{2} \right)^{2} \\
& = 1 & {\rm when} & r_{12}^{2} = \left( r_{1} - r_{2} \right)^{2}
\end{array}
\end{equation}

and <math>P_l(1)=1</math>, <math>P_{l}(-1)=(-1)^{l}</math>.

Thus,

\begin{equation}
I_{l+1}\left(\frac{r_1r_2}{r_{12}}R^\prime\right) = \left( 2l+1 \right) I_{l}(R) - I_{l-1} \left( \frac{r_{1} r_{2}}{r_{12}} R^{\prime} \right)
\end{equation}

If

\begin{equation}
R = r_{1}^{a-1} r_{2}^{b-1} r_{12}^{c+2} e^{-\alpha r_{1} - \beta r_{2}}
\end{equation}

then [G.W.F. Drake, Phys. Rev. A 18, 820 (1978)]

\begin{equation}
I_{l+1}(r_1^ar_2^br_{12}^c) =
\frac{2l+1}{c+2} I_{l} \left( r_{1}^{a-1} r_{2}^{b-2} r_{12}^{c+2} \right)+I_{l-1} \left( r_{1}^{a} r_{2}^{b} r_{12}^{c} \right)
\end{equation}

For the case <math>c=-2</math>, take

\begin{equation}
R = r_{1}^{a-1} r_{2}^{b-1} \ln r_{12} e^{ -\alpha r_{1} -\beta r_{2}}
\end{equation}

Then

\begin{equation}
I_{l+1} \left( r_{1}^{a} r_{2}^{b} r_{12}^{c} \right) = I_{l} \left( r_{1}^{a-1} r_{2}^{b-1} \ln r_{12} \right) \left( 2l + 1 \right) - I_{l-1} \left( r_{1}^{a} r_{2}^{b} r_{12}^{-2} \right)
\end{equation}

This allows all <math>I_{l}</math> integrals to be calculated from tables of <math>I_0</math> and
<math>I_1</math> integrals.

===The General Integral===

The above results for the angular and radial integrals can now be combined into a general formula for integrals of the type

\begin{equation}
I = \int \int d{\bf r}_{1} d{\bf r}_{2} R_{1} \mathcal{Y}^{M^{\prime}}_{l^{\prime}_{1} l^{\prime}_{2} L^{\prime}} \left( \hat{r}_{1}, \hat{r}_{2} \right) T_{k_{1} k_{2} K}^{Q} \left( {\bf r}_{1},{\bf r}_{2} \right) R_{2} \mathcal{Y}_{l_{1} l_{2} L}^{M} \left( \hat{r}_{1}, \hat{r}_{2} \right)
\end{equation}

where

\begin{equation}
\mathcal{Y}^{M}_{l_{1} l_{2} L} \left( \hat{r}_{1}, \hat{r}_{2} \right) = \sum_{m_1,m_2} \langle l_{1} l_{2} m_{1} m_{2} | LM \rangle Y^{m_{1}}_{l_{1}} \left( \hat{r}_{1} \right) Y^{m_{2}}_{l_{2}} \left( \hat{r}_{2} \right)
\end{equation}

and

\begin{equation}
T^Q_{k_{1} k_{2} K} \left( {\bf r}_{1}, {\bf r}_{2} \right) = \sum_{q_{1}, q_{2}} \langle k_{1} k_{2} q_{1} q_{2} | KQ \rangle Y_{k_{1}}^{q_{1}} \left( \hat{r}_{1} \right) Y^{q_{2}}_{k_{2}} \left( \hat{r}_{2} \right)
\end{equation}

The basic idea is to make repeated use of the formula

\begin{equation}
Y^{m_{1}}_{l_{1}} \left( \hat{r}_{1} \right) Y^{m_{2}}_{l_{2}} \left( \hat{r}_{1} \right) = \sum_{l m} \left( \frac{\left( 2l_{1}+1 \right) \left(2l_{2}+1 \right) \left(2l+1 \right)}{4 \pi} \right)^{1/2}
\times \left( \begin{array}{ccc}
l_1 & l_2 & l \\
m_1 & m_2 & m
\end{array} \right) \left(\begin{array}{ccc}
l_1 & l_2 & l \\
0 & 0 & 0
\end{array} \right)
Y^{m*}_{l} \left( \hat{r}_{1} \right)
\end{equation}

where

\begin{equation}
Y^{m*}(\hat{r}) = (-1)^mY_l^{-m}(\hat{r})
\end{equation}

and

\begin{equation}
\left(\begin{array}{ccc}
l_1 & l_2 & l \\
m_1 & m_2 & m
\end{array} \right) = \frac{\left( -1 \right)^{l_{1} - l_{2} - m}}{\left( 2l + 1 \right)^{1/2}} \left( l_{1} l_{2} m_{1} m_{2} | l, -m \right)
\end{equation}

is a 3-<math>j</math> symbol. In particular, write

\begin{array}{cc}
Y^{m_1^\prime}_{l_1^\prime}(\hat{r}_1)^{*} \underbrace{Y_{k_{1}}^{q_{1}} \left( \hat{r}_{1} \right) Y^{m_{1}}_{l_{1}} \left( \hat{r}_{1} \right)} & = & \sum_{\Lambda M}(\ldots)Y^{M}_\Lambda(\hat{r}_1)\\
\ \ \ \ \ \ \ \sum_{\lambda_1\mu_1}Y_{\lambda_1}^{\mu_1}(\hat{r}_1) & & \\
& & \\
Y^{m_{2}^{\prime}}_{l_{2}^{\prime}} \left( \hat{r}_{2} \right) \underbrace{Y^{q_{2}}_{k_{2}} \left( \hat{r}_{2} \right) Y_{l_{2}}^{m_{2}} \left( \hat{r}_{2} \right)} & = & \sum_{\Lambda^{\prime} M^{\prime}} \left( \ldots \right)Y_{\Lambda^{\prime}}^{M^{\prime}*} \left( \hat{r}_{2} \right) \\
\ \ \ \ \ \ \ \ \ \sum_{\lambda_{2} \mu_{2}}Y_{\lambda_{2}}^{\mu_{2}} \left( \hat{r}_{2} \right) & &
\end{array}

The angular integral then gives a factor of <math>2 \pi \delta_{\Lambda, \Lambda^{\prime}} \delta_{M, M^{\prime}} P_{\Lambda} \left( \cos \theta_{12} \right)</math>. The total integral therefore reduces to the form

\begin{equation}
I = \sum_{\Lambda} C_{\Lambda} I_{\Lambda} \left( R_{1} R_{2} \right)
\end{equation}

where <math>C_{\Lambda} = \sum_{\lambda_{1}, \lambda_{2}} C_{ \lambda_{1}, \lambda_{2}, \Lambda}</math>. For further details and derivations, including graphical representations, see G.W.F. Drake, Phys. Rev. A <math>{\bf 18}</math>, 820 (1978).

==Graphical Representation==

[[File:GraphRep1.jpg|500px]]

==Matrix Elements of H==
Recall that

\begin{equation}
H = -\frac{1}{2} \nabla^{2}_{1} - \frac{1}{2} \nabla^{2}_{2} - \frac{1}{r_1} - \frac{1}{r_2} + \frac{Z^{-1}}{r_{12}}
\end{equation}

Consider matrix elements of

\begin{equation}
\nabla^2_1 = \frac{1}{r_1^2} \frac{\partial}{\partial r_1}\left(r_1^2\frac{\partial}{\partial r_1} \right) + \frac{1}{r^2}\frac{\partial}{\partial r}\left(r^2\frac{\partial}{\partial r}\right) - \frac{(\vec{l}_1^Y)^2}{r_1^2} + \frac{2 \left( r_1 - r_2 \cos \theta \right)}{r} \frac{\partial^2}{\partial r_1 \partial r} - 2 \left( \nabla^Y_1 \cdot {\bf r}_2 \right) \frac{1}{r} \frac{\partial}{\partial r}
\end{equation}

where <math>r \equiv r_{12}</math> and <math>\cos\theta \equiv \cos\theta_{12}</math>. Also in what follows, define <math>\hat{\nabla}^Y_1 \equiv r_1\nabla^Y_1</math> where <math>\nabla^Y</math> operates only on the spherical harmonic part of the wave function, and similarly for <math>\vec{l}_1^Y</math>.

A general matrix element is

\begin{equation}
\langle r_1^{a^\prime} r_2^{b^\prime} r_{12}^{c^\prime} e^{-\alpha^\prime r_1 - \beta^\prime r_2} \mathcal{Y}^{M^\prime}_{l^\prime_1 l^\prime_2 L^\prime} \left( \hat{r}_1, \hat{r}_2 \right) \left| \nabla^2_1 \right| r_1^a r_2^b r_{12}^c e^{-\alpha r_1 - \beta r_2} \mathcal{Y}^M_{l_1 l_2 L} \left( \hat{r}_1, \hat{r}_2 \right) \rangle
\end{equation}

Since <math>\nabla^2_1</math> is rotationally invariant, this vanishes unless <math>L=L^\prime</math>, <math>M=M^\prime</math>. Also <math>\nabla_1^2</math> is Hermitian, so that the result must be the same whether it operates to the right or left, even though the results look very different. In fact, requiring the results to be the same yields some interesting and useful integral identities as follows:

Put

\begin{equation}
\mathcal{F} = F \mathcal{Y}^M_{l_1 l_2 L} \left( \hat{r}_1, \hat{r_2} \right)
\end{equation}

and

\begin{equation}
F = r_1^a r_2^b r_{12}^c e^{-\alpha r_1 - \beta r_2}
\end{equation}

Then

\begin{equation}
\nabla^2_1 \mathcal{F} = \left\{ \frac{1}{r_1^2} \left[ a \left( a + 1 \right) - l_1 \left( l_1 + 1 \right) \right] + \frac{c \left( c + 1 \right)}{r^2} + \alpha^2\frac{2\alpha \left( a + 1 \right)}{r_1} + \frac{2 \left(r_1 - r_2 \cos \theta \right) }{r_1r^2}c \left[ a - \alpha r_1 \right] - \frac{2c}{r^2} \left( \hat{\nabla}^Y_1 \cdot \hat{r}_2 \right) \frac{r_2}{r_1} \right\} \mathcal{F}
\end{equation}

and

\begin{align}
\langle \mathcal{F}^\prime \left| \nabla^2_1 \right| \mathcal{F} \rangle & = \sum_\Lambda \int d\tau_r F^\prime C_\Lambda \left( 1 \right) P_\Lambda \left( \cos \theta \right) \times \left\{ \frac{1}{r_1^2} \left[ a \left( a + 1 \right) - l_1 \left( l_1 + 1 \right) \right] -\frac{2 \alpha \left( a + 1 \right)}{r_1} + \frac{c \left( c + 1 \right)}{r^2} +\alpha^2 + \frac{2 \left( r_2 - r_2 \cos \theta \right)}{r_1r^2}c \left( a - \alpha r_1 \right) \right\} F \\
& + \sum_\Lambda \int d\tau_r F^\prime C_\Lambda \left( \hat{\nabla}^Y_1 \cdot \hat{r}_2 \right) P_\Lambda \left( \cos \theta \right) \left( \frac{-2c r_2}{r_1 r^2} \right) F
\end{align}

where

\begin{equation}
\int d\tau_r = \int^\infty_0 r_1 dr_1 \int^\infty_0 r_2 dr_2 \int^{r_1+r_2}_{\left| r_1 - r_2 \right|} r dr
\end{equation}

For brevity, let the sum over <math>\Lambda</math> and the radial integrations be understood, and let <math>\left( \nabla_1^2 \right)</math> stand for the terms which appear in the integrand. Then operating to the right gives

\begin{equation}
(\nabla_1^2)_R = \frac{1}{r_1^2} \left[ a \left( a + 1 \right) - l_1 \left( l_1 + 1 \right) \right] + \frac{c \left( c + 1 \right)}{r^2} + \alpha^2 - \frac{2 \alpha \left( a + 1 \right)}{r_1}
+ \frac{2 \left( r_1 - r_2 \cos \theta \right)}{r_1 r^2}c \left( a - \alpha r_1 \right) - \frac{2c r_2}{r_1 r^2} \left( \hat{\nabla}^Y_1 \cdot \hat{r}_2 \right)
\end{equation}

Operating to the left gives

\begin{equation}
(\nabla_1^2)_L = \frac{1}{r_1^2} \left[a^\prime \left( a^\prime + 1 \right) - l_1^\prime \left( l_1^\prime + 1 \right) \right] + \frac{c^\prime \left( c^\prime + 1 \right)}{r^2} + \alpha^{\prime 2} -\frac{2 \alpha^\prime \left( a^\prime + 1 \right)}{r_1}
+ \frac{2 \left( r_1 - r_2 \cos \theta \right)}{r_1 r^2} c^\prime \left( a^\prime - \alpha^\prime r_1 \right) - \frac{2 c^\prime r_2}{r_1 r^2} \left( \hat{\nabla}^{Y^\prime}_1 \cdot \hat{r}_2 \right)
\end{equation}

Now put

\begin{array}{cc}
a_+ = a+a^\prime, & \hat{\nabla}_1^+ = \hat{\nabla}_1^Y + \hat{\nabla}_1^{Y\prime} \\
a_{-} = a-a^\prime, & \hat{\nabla}_1^{-} = \hat{\nabla}_1^Y - \hat{\nabla}_1^{Y\prime}
\end{array}

etc., and substitute <math>a^\prime = a_+ - a</math>, <math>c^\prime = c_+ - c</math>, and <math>\alpha^\prime = \alpha_+ - \alpha</math> in <math>(\nabla^2_1)_L</math>. If <math>a_+</math>, <math>c_+</math> and <math>\alpha_+</math> are held fixed, then the equation

\begin{equation}
(\nabla^2_1)_R = (\nabla^2_1)_L
\end{equation}

must be true for arbitrary <math>a</math>, <math>c</math>, and <math>\alpha</math>. Their coefficients must thus vanish.

This yields the integral relations

\begin{equation}
\label{I}
\frac{\left( r_1 - r_2 \cos \theta \right)}{r_1 r^2} = \frac{1}{c_+} \left( \frac{- \left( a_+ + 1 \right)}{r_1^2} + \frac{\alpha_+}{r_1} \right) \tag{I}
\end{equation}

from the coefficient of <math>a</math> and

\begin{equation}
\label{II}
\frac{ \left( r_1 - r_2 \cos \theta \right) \left( a_+ - \alpha_+ r_1 \right)}{r_1 r^2} = \frac{r_2}{r_1 r^2}\left( \hat{r}_2 \cdot \hat{\nabla}_1^+ \right) - \frac{ \left( c_+ + 1 \right)}{r^2} \tag{II}
\end{equation}

from the coefficient of <math>c</math>. The coefficient of <math>\alpha</math> gives an equation equivalent to (I).

Furthermore, if can be show that (see problem)

\begin{align}
& \sum_\Lambda \int d\tau_r \frac{r^c}{r^2} C_\Lambda \left( \hat{r}_2 \cdot \hat{\nabla}_1^Y \right) P_\Lambda \left( \cos \theta \right) \\
= & \sum_\Lambda \int d\tau_r \frac{r^c}{c r_1 r_2}C_\Lambda \left( 1 \right)P_\Lambda \left( \cos \theta \right) \times \left( \frac{l_1^\prime \left( l_1^\prime + 1 \right) - l_1 \left( l_1 + 1 \right) -\Lambda \left( \Lambda + 1 \right)}{2} \right)
\end{align}

and similarly for <math>\left( \hat{r}_2 \cdot \hat{\nabla}^{Y\prime}_1 \right)</math> with <math>l_1</math> and <math>l_1^\prime</math> interchangeable, then it follows that

\begin{equation}
\frac{r^{c_+}}{r^2} \left( \hat{r}_2 \cdot \hat{\nabla}_1^+ \right) = - \frac{r^{c_+}}{c_+ r_1 r_2} \Lambda \left( \Lambda + 1 \right)
\end{equation}

and

\begin{equation}
\frac{r^{c_+}}{r^2} \left( \hat{r}_2 \cdot \hat{\nabla}_1^- \right) = \frac{r^{c_+}}{c_+ r_1 r_2} \left[ l_1^\prime \left( l_1^\prime + 1 \right) - l_1 \left( l_1 + 1 \right) \right]
\end{equation}

where equality applies after integration and summation over <math>\Lambda</math>.

Thus (II) becomes

\begin{equation}\label{IIa}
\frac{ \left( r_1 - r_2 \cos \theta \right) \left( a_+ - \alpha_+ r_1 \right)}{r_1 r^2} = - \frac{\Lambda \left( \Lambda + 1 \right)}{c_+ r^2_1} - \frac{ \left( c_+ + 1 \right)}{r^2} \tag{IIa}
\end{equation}

===Problem===

Prove the integral relation

\begin{align}
& \sum_\Lambda \int d\tau_r f \left( r_1, r_2 \right) \frac{1}{r} \left( \frac{d}{dr} q(r) \right) C_\Lambda \left( \hat{r}_2 \cdot \hat{\nabla}^Y_1 \right) P_\Lambda \left( \cos \theta \right) \\
= & \sum_\Lambda \int d\tau_r \frac{ f \left( r_1, r_2 \right)}{r_1, r_2} q(r) C_\Lambda \left( 1 \right) P_\Lambda \left( \cos \theta \right) \left(\frac{l_1^\prime \left( l_1^\prime + 1 \right) - l_1 \left( l_1 + 1 \right) - \Lambda \left( \Lambda + 1 \right)}{2}\right)
\end{align}

where the coefficients <math>C_\Lambda \left( 1 \right)</math> are the angular coefficients from the overlap integral

\begin{align}
\int & d\Omega\, \mathcal{Y}^{M*}_{l_1^\prime l_2^\prime L}(\hat{r}_1,\hat{r}_2)\, \mathcal{Y}^{M}_{l_1 l_2 L}(\hat{r}_1,\hat{r}_2)\\
= & \sum_\Lambda C_\Lambda(1)P_\Lambda(\cos\theta_{12}).
\end{align}

Hint: Use the fact that <math>l_1^2</math> is Hermitian so that
<math>
\int d\tau (l_1^2\mathcal{Y}^\prime)^*q(r)\mathcal{Y} = \int d\tau
\mathcal{Y}^{\prime *}l_1^2(q(r)\mathcal{Y})
</math>
with <math>\ \vec{l}_1= \frac{1}{i}\vec{r}_1\times\nabla_1</math>.

It is also useful to use
<math>\begin{eqnarray}\ \
\cos\theta\, P_L(\cos\theta) &=& \frac{1}{2L+1}[L\,P_{L-1}(\cos\theta) + (L+1)P_{L+1}(\cos\theta)]
\\
(\cos^2\theta-1)P_L(\cos\theta) &=&
\frac{L(L-1)}{(2L-1)(2L+1)}P_{L-2}(\cos\theta)\nonumber
-\frac{2(L^2+L-1)}{(2L-1)(2L+3)}P_L(\cos\theta) + \frac{(L+1)(L+2)}{(2L+1)(2L+3)}P_{L+2}(\cos\theta)\nonumber\\
&=& \frac{L(L-1)}{(2L-1)(2L+1)}[P_{L-2}(\cos\theta) - P_L(\cos\theta)]
+ \frac{(L+1)(L+2)}{(2L+1)(2L+3)}[P_{L+2}(\cos\theta) - P_L(\cos\theta)]
\end{eqnarray}</math>

together with a double application of the integral recursion relation
<math>
I_{L+1}\left(\frac{1}{r}\frac{d}{dr}q(r)\right) =
(2L+1)I_L\left(\frac{1}{r_1r_2}q(r)\right)+I_{L-1}\left(\frac{1}{r}\frac{d}{dr}q(r)\right)
</math>

Of course <math>l_1^2\mathcal{Y}^M_{l_1l_2L}(\hat{r}_1,\hat{r}_2)=l_1(l_1+1)\mathcal{Y}^M_{l_1l_2L}(\hat{r}_1,\hat{r}_2)</math>. Begin by expanding
<math>
l_1^2(g\mathcal{Y}) = \mathcal{Y}l_1^2g + g\,l_1^2\mathcal{Y} + 2(\vec{l}_1g)\cdot(\vec{l}_1\mathcal{Y})
</math>, and show that

<math>
\begin{eqnarray}\ \
(\vec{l}_1g)\cdot(\vec{l}_1\mathcal{Y}) & = & \frac{r_1^2}{r}\frac{dg}{dr}\vec{r}_2\cdot \nabla_1\mathcal{Y}\,\,\,\,\mbox{ and}\\
l_1^2g(r) &=& 2r_1r_2\cos\theta\frac{1}{r}\frac{dg}{dr} + r_1^2r_2^2(\cos^2\theta - 1)
\frac{1}{r}\frac{d}{dr}\left(\frac{1}{r}\frac{dg}{dr}\right).
\end{eqnarray}
</math>

The proof amounts to showing that the term <math>l_1^2g(r)</math> can be replaced by
<math>\Lambda(\Lambda+1)g(r)</math> after multiplying by <math>P_\Lambda(\cos\theta)</math>
and integrating by parts with respect to the radial integrations over <math>r_1</math>, <math>r_2</math> and
<math>r \equiv |\vec{r}_1 - \vec{r}_2|</math>. Remember that <math>\cos\theta</math> is just a short-hand notation for the radial function <math>(r_1^2 + r_2^2 - r^2)/(2r_1r_2)</math>.

==General Hermitian Property==

Each combination of terms of the form
\begin{equation}
<f>_R = a^2f_1+af_2+abf_3+b^2f_4+bf_5+af_6\nabla_1^Y+bf_7\nabla^Y_2+f_8(Y)
\end{equation}
can be rewritten
\begin{equation}
<f>_L= (a_+-a)^2f_1+(a_+-a)f_2 + (a_+-a)(b_+-b)f_3+(b_+-b)^2f_4
+(b_+-b)f_5+(a_+-a)f_6\nabla^{Y\prime}_1+(b_+-b)f_7\nabla^{Y\prime}_2+f_8)Y^\prime
\end{equation}
where as usual <math>\nabla_1^Y</math> acts only on the spherical harmonic part of the wave function denoted for short by <math>Y</math>.
Since these must be equal for arbitrary <math>a</math> and <math>b</math>,
\begin{equation}
a_+^2f_1+a_+f_2+_+b_+f_3+b_+f_4+b_+f_5+a_+f_6\nabla^{Y\prime}_1 +
b_+f_7\nabla^{Y\prime}_2+f_8(Y^\prime)-f_8(Y)=0
\end{equation}

Adding the corresponding expression with <math>Y</math> and <math>Y^\prime</math> interchanged
yields
<math>\label{III}
a_+^2f_1+a_+f_2+a_+b_+f_3+b_+^2f_4+b_+f_5+\frac{1}{2}a_+f_6\nabla^+_1+\frac{1}{2}b_+f_7\nabla^+_2=0\tag{III}
</math>

Subtracting gives
\[\label{IV}
f_8(Y)-f_8(Y^\prime)=-\frac{1}{2}[a_+f_6\nabla^-_1+b_+f_7\nabla_2^-]\tag{IV}
\]
\[\label{V}
a[-2a_+f_1-2f_2-b_+f_3-f_6\nabla^+_1]=0\tag{V}
\]
\[\label{VI}
b[-2b_+f_4-2f_5-a_+f_3-f_7\nabla^+_2]=0\tag{VI}
\]

Adding the two forms gives

\begin{eqnarray}
<f>_R+<f>_L=\frac{1}{2}(a_+^2+a_-^2)f_1+a_+f_2+\frac{1}{2}(a_+b_++a_-b_-)f_3\nonumber\\
\ +\frac{1}{2}(b_+^2+b_-^2)f_4+b_+f_5+\frac{1}{2}f_6(a_+\nabla_1^++a_-\nabla^-_1)\nonumber\\
\ +\frac{1}{2}(b_+\nabla_2^++b_-\nabla^-_2)+f_8(Y)+f_8(Y^\prime)\nonumber
\end{eqnarray}

Subtracting <math>\frac{x}{2}\times({\rm I})</math> gives

\begin{eqnarray}\ \
<f>_R+<f>_L = \frac{1}{2}[(1-x)a_+^2 + a_-^2]f_1 +
(1-\frac{x}{2})a_+f_2+\frac{1}{2}[(1-x)a_+b_++a_-b_-]f_3\nonumber\\
+\frac{1}{2}[(1-x)b_+^2+b_-^2]f_4+(1-\frac{x}{2})f_5b_++\frac{1}{2}f_6[(1-\frac{x}{2})a_+\nabla_1^++a_-\nabla^-_1]\nonumber\\
+\frac{1}{2}[(1-\frac{x}{2})b_+\nabla^+_2+b_-\nabla^-_2]+f_8(Y)+f_8(Y^\prime)\nonumber
\end{eqnarray}

If <math>x=1</math>,
\begin{eqnarray}
<f>_R+<f>_L &=&
\frac{1}{2}[a_-^2f_1+a_+f_2+a_-b_-f_3+b_-^2f_4+b_+f_5\nonumber\\
&+&\left(\frac{1}{2}a_+\nabla^+_1+a_-\nabla^-_1\right)f_6+\left(\frac{1}{2}b_+\nabla^+_2+b_-\nabla^-_2\right)f_7]+f_8(Y)+f_8(Y^\prime)\nonumber
\end{eqnarray}

The General Hermitian Property for arbitrary <math>x</math> gives
\begin{eqnarray}\ \
\nabla_1^2=
\frac{1}{4}\{\frac{1}{r_1^2}[(1-x)a_+^2+a_-^2+2(1-\frac{x}{2})a_+-2[l_1(l_1+1)+l_1^\prime(l_1^\prime+1)]\nonumber\\
-\frac{2}{r_1}[(1-x)\alpha_+a_++\alpha_-a_-+2(1-\frac{x}{2})\alpha_+]+(1-x)\alpha_+^2
+ \alpha^2_-]\nonumber\\
+\frac{2(r_1-r_2\cos\theta)}{r_1r^2}[(1-x)(a_+-\alpha_+r_1)c_++(a_--\alpha_-r_2)c_-]\nonumber\\
-\frac{2r_2}{r_1r^2}[(1-\frac{x}{2})c_+\hat{r}_2\cdot\hat{\nabla}_1^++c_-\hat{r}_2\hat{\nabla}_1^-]+\frac{(1-x)c_+^2+c_-^2+2(1-\frac{x}{2})c_+}{r^2}\}\nonumber
\end{eqnarray}.

Use
<math>\begin{eqnarray}\ \
\frac{-2r_2}{r_1r^2}[(1-\frac{x}{2})c_+\hat{r}_2\hat{\nabla}_1^++c_-\hat{r}_2\hat{\nabla}_1^-]=\frac{2}{r_1^2}\left[(1-\frac{x}{2})\Lambda(\Lambda+1)-\frac{c_-}{c_+}[l_1^\prime(l_1^\prime+1)+l_1(l_1+1)]\right]\end{eqnarray}</math>,

<math>\begin{eqnarray}
\frac{2(r_1-r_2\cos\theta)}{r_1r^2}(a_--\alpha_-r_1)c_- = 2\frac{c_-}{c_+}\left[\frac{-1}{r_1^2}[a_-(a_++1)]+\frac{1}{r_1}[a_-\alpha_++\alpha_-(a_++2)]-\alpha_-\alpha_+\right]\nonumber
\end{eqnarray}</math>,

and
<math>\begin{eqnarray}\ \
\frac{2(r_1-r_2\cos\theta)}{r_1r^2}(a_+-\alpha_+r_1)c_+ = 2\left[-\frac{1}{r_1^2}[a_+(a_++1)]+\frac{1}{r_1}[a_+\alpha_++\alpha_+(a_++2)]-\alpha_+^2\right]
\end{eqnarray}</math>.

Substituting into <math>\nabla_1^2</math> gives

\begin{eqnarray}
\nabla^2_1 = \frac{1}{4}\{\frac{1}{r_1^2}[-(1-x)a_+^2 + a_-^2 +xa_+ +
2(1-\frac{x}{2})\Lambda(\Lambda+1)\nonumber\\
-2l_1(l_1+1)(1-\frac{c_-}{c_+}) -
2l_1^\prime(l_1^\prime+1)(1+\frac{c_-}{c_+})-\frac{2c_-a_-}{c_+}(a_++1)]\nonumber\\
- \frac{2}{r_1}\left[-(1-x)\alpha_+(a_++2)+\alpha_-a_-+2(1-\frac{x}{2})\alpha_+
-\frac{c_-}{c_+}[a_-\alpha_++\alpha_-(a_++2)]\right]\nonumber\\
-(1-x)\alpha_+^2+\alpha_-^2-2\frac{c_-}{c_+}\alpha_-\alpha_+ +\frac{1}{r^2}\left[(1-x)c_+^2+c_-^2+2(1-\frac{x}{2})c_+\right]\}\nonumber
\end{eqnarray}

This has the form
<math>
\nabla_1^2 =
\frac{1}{4}\left[\frac{A_1}{r_1^2}+\frac{B_1}{r_1}+\frac{C_1}{r^2}+D_1\right]
</math>
with

<math>\begin{array}{ccc}
A_1=-(1-x)a_+^2+a_-^2
+xa_++2(1-\frac{x}{2})\Lambda(\Lambda+1)-2l_1(l_1+1)(1-\frac{c_-}{c_+})\
-2l_1^\prime(l_1^\prime+1)(1+\frac{c_-}{c_+})-\frac{2c_-a_-}{c_+}(a_++1) \ \
\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \
\end{array}
</math>

<math>
B_1 =
2\left[(1-x)\alpha_+(a_++2)-\alpha_-a_--2(1-\frac{x}{2})\alpha_++\frac{c_-}{c_+}[a_-\alpha_++\alpha_-(a_++2)]\right]
</math>

<math>
C_1 = (1-x)c_+^2+ c_-^2 +2(1-\frac{x}{2})c_+
</math>

<math>
D_1 = -1(1-x)\alpha_+^2 +\alpha_-^2 - 2\frac{c_-}{c_+}\alpha_-\alpha_+
</math>

The complete Hamiltonian is then (in Z-scaled a.u.)

<math>\begin{eqnarray}
H =-\frac{1}{2}\nabla^2_1 -\frac{1}{2}\nabla^2_2 -\frac{1}{r_1} -\frac{1}{r_2} +\frac{Z^{-1}}{r}\nonumber\\
= -\frac{1}{2}\nabla^2_1 -\frac{1}{2}\nabla^2_2 -\frac{1}{r_1} -\frac{Z-1}{Zr_2}+Z^{-1}\left(\frac{1}{r}-\frac{1}{r_2}\right)\nonumber\\
\ =-\frac{1}{8}\left[\frac{A_1}{r_1^2}+\frac{B_1+8}{r_1}+\frac{C_1}{r^2}+D_1+D_2 +\frac{A_2}{r_2^2} + \frac{B_2 + 8(Z-1)/Z}{r_2}+\frac{C_2}{r^2}\right]\nonumber
+ Z^{-1}\left(\frac{1}{r}-\frac{1}{r_2}\right)\nonumber
\end{eqnarray}</math>

The screened hydrogenic energy is
<math>\begin{eqnarray}
E_{SH}=-\frac{1}{2}\left[\frac{1}{n_1^2}+ \left(\frac{Z-1}{Z}\right)^2\frac{1}{n_2^2}\right]
\end{eqnarray}</math>
so that

<math>\begin{eqnarray}
H-E_{SH} &=& -\frac{1}{8}[
\frac{A_1}{r_1^2}+\frac{B_1+8}{r_1}+\frac{C_1}{r^2} + D_1 + D_2
-\frac{4}{n_1^2} -\left(\frac{Z-1}{Z}\right)^2\frac{4}{n_2^2}\nonumber
+\frac{A_2}{r_2}+\frac{B_2+8(Z-1)/Z}{r_2}+\frac{C_2}{r^2}] +
Z^{-1}\left(\frac{1}{r} -\frac{1}{r_2}\right)\nonumber
\end{eqnarray}</math>

==Optimization of Nonlinear Parameters==
- The traditional method of performing Hylleraas calculations is to write the
basis set in the form

<math>
\Psi = \sum_{i,j,k}c_{i,j,k}\varphi_{i,j,k}(\alpha,\beta)\pm exchange
</math>

with

<math>
\varphi_{i,j,k}(\alpha,\beta) = r_1^ir_2^jr_{12}^ke^{-\alpha r_1 - \beta
r_2}\mathcal{Y}^M_{l_1l_2L}.
</math>

The usual procedure is to set <math>\alpha = Z</math> so that it represents the inner
<math>1s</math> electron, and then to vary <math>\beta</math> so as to minimize the energy.

[[File:OptNon.jpg|500px]]

Since <math>\beta</math> appears in <math>\Psi</math> as a non-linear parameter, the entire
calculation must be repeated for each value of <math>\beta</math>. However, the minimum
becomes progressively smaller as the basis set is enlarged.

[[File:OptNon02.jpg|500px]]

*Difficulties

1. If the basis set is constructed so that <math> i +j +j \leq N</math>, the the
number of terms is
<math>
(N+1)(N+2)(N+3)/6
</math>
N=14 already gives 680 terms and an accuracy of about 1 part in <math>10^{10}</math> for
low-lying states. A substantial improvement in accuracy would require much
larger basis sets, together with multiple precision arithmetic to avoid loss
of significant figures when high powers are included.

2. The accuracy rapidly deteriorates as one goes to more highly excited
states - about 1 significant figure is lost each time the principle quantum
number is increased.

*Cure

We have found that writing basis sets in the form
\begin{eqnarray}
\Psi &=& \sum_{i,j,k}\left(c^{(1)}_{i,j,k}\varphi_{ijk}(\alpha_1\beta_1) +
c_{ijk}^{(2)}\varphi_{ijk}(\alpha_2\beta_2)\right) \pm exchange \nonumber\\
&=& \psi({\bf r}_1,{\bf r}_2) \pm \psi({\bf r}_2,{\bf r}_1)\nonumber
\end{eqnarray}
so that each combination of powers is included twice with different nonlinear
parameters gives a dramatic improvement in accuracy for basis sets of about
the same total size.

However, the optimization of the nonlinear parameters is now much more
difficult, and an automated procedure is needed.

If
<math>
E = \frac{<\Psi|H|\Psi>}{<\Psi|\Psi>}</math> and we assume <math><\Psi|\Psi> = 1</math>, then

<math>
\frac{\partial E}{\partial\alpha_t} = -2 <\Psi|H-E|r_1\psi({\bf r}_1,{\bf
r}_2;\alpha_t)\pm r_21\psi({\bf r}_2,{\bf
r}_1;\alpha_t)>
</math>
where
<math>
\psi({\bf r}_1,{\bf r}_2;\alpha_t) =
\sum_{i,j,k}c_{ijk}^{(t)}\varphi_{ijk}(\alpha_t\beta_t)
</math>

*Problem

1. Prove the above.

2. Prove that there is no contribution from the implicit dependence of
<math>c_{ijk}^{(t)}</math> on <math>\alpha_t</math> if the linear parameters have been optimized.

==The Screened Hydrogenic Term==

If the Hamiltonian is written in the form (in Z-scaled a.u.)
<math>\begin{eqnarray}\
H = H_0({\bf r}_1,Z) + H_0({\bf r}_2,Z-1) +
Z^{-1}\left(\frac{1}{r}-\frac{1}{r_2}\right)
\end{eqnarray}</math>
with
<math>\
H_0({\bf r}_1,Z) = -\frac{1}{2}\nabla_1^2 - \frac{1}{r_1}\nonumber
</math> and
<math>
H_0({\bf r}_2,Z-1) = -\frac{1}{2}\nabla_2^2 - \frac{Z-1}Z{r_2}\nonumber
</math>,
then the eigenvectors of <math>H_0({\bf r},Z) +H_0({\bf r}_2,Z-1)</math> are products of
hydrogenic orbitals
<math>
\Psi_0=\psi_0(1s,Z)\psi_0(nl,Z-1)
</math>
and the eigenvalue is
<math>\begin{eqnarray}
E_{SH} = \left[-\frac{1}{2} - \left(\frac{Z-1}{Z}\right)^2\frac{1}{2n^2}\right]Z^2 a.u.
\end{eqnarray}</math>
called the screened hydrogenic eigenvalue. From highly excited states,
<math>E_{SH}</math> and <math>\Psi_0</math>are already excellent approximations.

For example, for the 1s8d states, the energies are

<math>\begin{eqnarray}\\
E(1s8d^1D)&=& -2.007\,816\,512\,563\,81 {\rm\ a.u.}\nonumber\\
E(1s8d^3D)&=& - 2.007\,817\,934\,711\,71 {\rm\ a.u.}\nonumber\\
E_{SH} &=& -2.007\,812\,5 {\rm\ a.u.} \nonumber
\end{eqnarray}</math>

It is therefore advantageous to include the screened hydrogenic terms in the
basis set so that the complete trial function becomes
<math>
\Psi = c_0\Psi_0 +
\sum_{ijk}\left[c_{ijk}^{(1)}\varphi_{ijk}(\alpha_1,\beta_1)+c_{ijk}^{(2)}\varphi(\alpha_2,\beta_2)\right]\pm
{\rm exchange}
</math>

Also, the variational principal can be re-expressed in the form
<math>\begin{eqnarray}
E= E_{SH}+\frac{<\Psi|H-E_{SH}|\Psi>}{<\Psi|\Psi>}
\end{eqnarray}</math>
so that the <math>E_{SH}</math> term can be cancelled analytically from the matrix
elements.

For example
<math>
<\Psi_0|H-E_{SH}|\Psi_0> = 0
</math>

and
<math>
<\varphi_{ijk}|H-E_{SH}|\Psi_0> =
Z^{-1}<\varphi_{ijk}|\frac{1}{r}-\frac{1}{r_2}|\Psi_0>
</math>

Recall that
<math>\begin{eqnarray}\ \
I_0(a,b,c) = \frac{2}{c+2}\sum^{[c+1]/2}_{i=0}\left(\begin{array}{c}
c+2\\
2i+1
\end{array}\right) \left[f(p,q;\beta) + f(p^\prime,q^\prime;\alpha)\right]
\end{eqnarray}</math>
where

<math>\begin{eqnarray}
f(p,q;x) =
\frac{q!}{x^{q+1}(\alpha+\beta)^{p+1}}\sum^p_{j=0}\frac{(p+j)!}{j!}\left(\frac{x}{\alpha+\beta}\right)^j
\end{eqnarray}</math>

<math>
\begin{array}{cc}
p=a+2i+2 & p^\prime = b+2i+2\\
q= b+c-2i+2 & q^\prime =a+c-2i+2
\end{array}
</math>

If <math>\beta\ll\alpha</math>, the <math>f(p,q;\beta)\gg f(p^\prime,q^\prime;\alpha)</math> and the
<math>i=0</math> term is the dominant contribution to <math>f(p,q;\beta)</math>. However, since
this term depends only on the sum of powers <math>b+c</math> for <math>r_2</math>, it cancels exactly
from the matrix element of
<math>
\frac{1}{r_{12}} - \frac{1}{r_2}
</math>
and can therefore
be omitted in the calculations of integrands, thereby saving many significant
figures. This is especially valuable when <math>b+c</math> is large and <math>a</math> is small.

==Small Corrections==
*Mass Polarization
If the nuclear mass is not taken to be infinite, then the Hamiltonian is
<math>\begin{eqnarray}\ \
H = \frac{P_N^2}{2M} + \sum_{i=1}\frac{p_i^2}{2m} + V
\end{eqnarray}</math>

[[File:SmallCor.jpg|500px]]

where <math>M =M_A-nm</math> is the nuclear mass and <math>m</math> is the electron mass.

Change variables to the C of M
<math>\begin{eqnarray}
{\bf R} = \frac{1}{M+nm}\left[M{\bf r}_N + m({\bf r}_1 + {\bf r}_2 +
\ldots)\right]
\end{eqnarray}</math>

and relative variables

<math>
{\bf s}_i = {\bf r}_i - {\bf r}_N.
</math>

Then

<math>\begin{eqnarray}\ \ \ \ \
H = \frac{1}{2(M+nm)}{\bf p}_R^2+\frac{1}{2m}\sum_{i=1}^n{\bf p}_{s_i}^2
+\frac{1}{2M}\sum_{i,k}{\bf p}_{s_i}\cdot {\bf p}_{s_k} + V({\bf s}_1, {\bf
s}_2, \ldots)\nonumber
= \frac{1}{M+nm}{\bf p}_R^2+\frac{1}{2\mu}\sum_{i=1}^n{\bf
p}_{s_i}^2+\frac{1}{2M}\sum_{i\neq k}{\bf p}_{s_i}\cdot {\bf p}_{s_k}+V({\bf s}_1, {\bf
s}_2, \ldots)\nonumber
\end{eqnarray}</math>

where
<math>\begin{eqnarray}
\mu = \left(\frac{1}{M} + \frac{1}{m}\right)^{-1}
\end{eqnarray}</math>

If <math>n=2</math>, the Schrödinger equation is
<math>\begin{eqnarray}\ \ \
\left[ -\frac{\hbar^2}{2\mu}(\nabla^2_{s_1}+ \nabla^2_{s_2}) -
\frac{\hbar^2}{M}\nabla_{s_1}\nabla_{s_2}-\frac{Ze^2}{s_1}-\frac{Ze^2}{s_2}+\frac{e^2}{s_{12}}\right]\psi
= E\psi
\end{eqnarray}</math>.

Define the reduced mass Bohr radius
<math>\begin{eqnarray}
a_\mu = \frac{\hbar^2}{\mu e^2} = \frac{m}{\mu}a_0
\end{eqnarray}</math>
and

<math>\begin{eqnarray}
\rho_i = \frac{Zs_i}{a_\mu}\nonumber\\
\varepsilon = \frac{E}{Z^2(e^2/a_\mu)}\nonumber
\end{eqnarray}</math>

The Schrödinger equation then becomes
<math>\begin{eqnarray}\ \ \
\left[
-\frac{1}{2}(\nabla^2_{\rho_1}+\nabla^2_{\rho_2})-\frac{\mu}{M}\vec{\nabla}_{\rho_1}\cdot\vec{\nabla}_{\rho_2} - \frac{1}{\rho_1}-\frac{1}{\rho_2}+\frac{Z^{-1}}{\rho_{12}}\right]\psi = \varepsilon \psi
\end{eqnarray}</math>

The effects of finite mass are thus

1. A reduced mass correction to all energy levels of <math>(e^2/a_\mu)/(e^2/a_0)
= \mu/m</math>, i.e.
<math>\begin{eqnarray}
E_M=\frac{\mu}{m}E_\infty
\end{eqnarray}</math>

2. A specific mass shift given in first order perturbation theory by
<math>\begin{eqnarray}
\Delta\varepsilon =
-\frac{\mu}{M}<\psi|\nabla_{\rho_1}\cdot\nabla_{\rho_2}|\psi>
\end{eqnarray}</math>
or

<math>\begin{eqnarray}
\Delta\varepsilon =
-\frac{\mu}{M}<\psi|\nabla_{\rho_1}\cdot\nabla_{\rho_2}|\psi>\frac{e^2}{a_\mu}\nonumber\\
= -\frac{\mu^2}{mM}<\psi|\nabla_{\rho_1}\cdot\nabla_{\rho_2}|\psi>\frac{e^2}{a_0}\nonumber
\end{eqnarray}</math>

Since <math>\nabla_{\rho_1}\cdot\nabla_{\rho_2}</math> has the same angular properties as
<math>{\bf\rho}_1 \cdot {\bf\rho}_2</math>, the operator is like the product of two
dipole operators. For product type wave functions of the form
$
\psi = \psi(1s)\psi(nl) \pm \mbox{exchange},
$
the matrix element vanishes for all but P-states.

=Radiative Transitions=

==Transition Integrals and the Line Strength==

For an electric dipole transition <math>\gamma'\,L'\,S'\,J'\,M' \rightarrow \gamma\,L\,S\,J\,M</math>, the line strength is defined by
\begin{equation}
S = \sum_{M',\mu,M}|\langle\gamma'\,L'\,S'\,J'\,M'|r_\mu|\gamma\,L\,S\,J\,M\rangle|^2
\end{equation}
Using the Wigner-Eckart theorem, this is related to the reduced matrix element by
\begin{eqnarray}
S(J',J) = |\langle\gamma'\,L'\,S'\,J'||r||\gamma\,L\,S\,J\,M\rangle|^2
\sum_{M',\mu,M}\left|\left(\matrix{J' & 1 & J \cr
-M'& \mu &M \cr}\right)\right|^2
\end{eqnarray}
Using the sum rule for the 3-<math>j</math> symbols
\begin{equation}
\sum_{M',\mu,M}\left|\left(\matrix{J' & 1 & J \cr
-M'& \mu &M \cr}\right)\right|^2 = 1
\end{equation}
we see that the line strength is identical to the reduced matrix element
\begin{equation}
S(J',J) = |\langle\gamma'\,L'\,S'\,J'||r||\gamma\,L\,S\,J\,M\rangle|^2
\end{equation}
For a nonrelativistic calculation in <math>LS</math> coupling, <math>S = S'</math> and we can next strip out the dependence on <math>J</math> and <math>J'</math> by use of Eq.\ (7.1.7) of Edmonds { Angular Momentum in Quantum Mechanics} with <math>k = 1</math> to obtain
\begin{equation}
S(J',J) = (-1)^{L'+S+J+1}[(2J+1)(2J'+1)]^{1/2}\left\{
\matrix{L' & J'& S\cr
J & L & 1\cr}\right\}\langle\gamma' L'|| r || \gamma L\rangle
\end{equation}
where <math>\langle\gamma' L'|| r || \gamma L\rangle</math> is the reduced matrix element if spin is neglected. Note that if spin is neglected, then the line-strength is simply
\begin{equation}
S(L',L) = |\langle\gamma' L'|| r || \gamma L\rangle|^2
\end{equation}
To calculate the reduced matrix element, use a particular example along with the Wigner-Eckart theorem
such as
\begin{equation}
\langle\gamma'\, L'|| r || \gamma\, L\rangle = (-1)^{L'}\langle\gamma'\, L'\, 0| z | \gamma\, L\, 0\rangle{\Large /}\left(\matrix{L' & 1 & L \cr
0 & 0 & 0 \cr}\right)
\end{equation}

===Correlated Two-electron Integrals===
For the case of two-electron integrals, <math>z = z_1 + z_2</math> and <math>r = r_1 + r_2</math>. As discussed in G.W.F. Drake, Phys. Rev. A 18, 820 (1978), the general integral in Hylleraas coordinates for the integral over the <math>z_1</math> term can be written in the form
\begin{equation}
\label{int1}
\langle\gamma'\, L'\, 0| z_1 | \gamma\, L\, 0\rangle = \sum_\Lambda C_\Lambda I_\Lambda(\gamma',\gamma)
\end{equation}
where <math>I_\Lambda(\gamma',\gamma)</math> stands for the purely radial integral
\begin{equation}
\label{radial}
I_\Lambda(\gamma',\gamma) = \int_0^\infty r_1\,dr_1\int_0^\infty r_2\,dr_2\int_{|r_1-r_2|}^{r_1+r_2}r_{12}\, dr_{12}
R_{\gamma'}^*\,r_1\, R_\gamma P_\Lambda(\cos\theta_{12})
\end{equation}
and <math>\cos\theta_{12} = (r_1^2 + r_2^2 - r_{12}^2)/(2r_1r_2)</math> is a short-hand notation for a radial function of <math>r_1</math>, <math>r_2</math> and <math>r_{12}</math>. Also, <math>P_\Lambda(\cos\theta_{12})</math> is a Legendre polynomial,
and, <math>R_\gamma = R_\gamma(r_1,r_2,r_{12})</math> is the radial part of the Hylleraas wave function for state <math>\gamma</math> such
that the total wave function is <math>\psi({\bf r}_1,{\bf r}_2) = R_\gamma(r_1,r_2,r_{12}){\cal Y}_{l_1,l_2,L}^M(\hat{\bf r}_1,\hat{\bf r}_2)\pm {\rm exchange}</math>, normalized to unity, and <math>{\cal Y}_{l_1,l_2,L}^M(\hat{\bf r}_1,\hat{\bf r}_2)</math> is a vector coupled product of spherical harmonics. The coefficents <math>C_\Lambda</math> in Eq.\ (\ref{int1}) can be calculated from Eq.\ (32) of the above 1978 paper. They need to be calculated separately for he direct-direct and the direct-exchange contributions. For the reduced matrix element of <math>z_1</math>, the leading 3-<math>j</math> symbol can be dropped, leaving (for the case <math>k = 1</math>)
\begin{eqnarray}
C_\Lambda^{\rm reduced} &=& \frac{(-1)^{l_2'+L'+L+1}}{2} (l_1,l_1',l_2,l_2',L,L')^{1/2}
\sum_{\lambda={\rm max}(l_1-1,0)}^{l_1+1} (\lambda,\Lambda,1)
\left(\matrix{l_1 & 1 & \lambda \cr
0 & 0 & 0 \cr}\right)
\left(\matrix{l_1'& \lambda & \Lambda \cr
0 & 0 & 0 \cr}\right)\nonumber\\
&\times&\left(\matrix{l_2' & 1_2 & \Lambda \cr
0 & 0 & 0 \cr}\right)
\left\{\matrix{L' & \lambda & l_2 \cr
\Lambda& l_2' & l_1' \cr}\right\}
\left\{\matrix{L' & \lambda & l_2 \cr
l_1 & L & 1 \cr}\right\}
\end{eqnarray}
where the notation <math>(a,b,c,\cdots)</math> means <math>(2a+1)(2b+1)(2c+1)\cdots</math>.
A similar expression can be obtained for the angular coefficients for the <math>z_2</math> term by interchanging the subscripts 1 and 2 throughout and multiplying by <math>(-1)^{L+L'+1}</math>.

===Example===
As an example, for the <math>1s^2\;^1S - 1s2p\;^1P</math> transition of helium, the only nonvanishing <math>C_\Lambda</math> angular coefficients are <math>C_1 = 1/2</math> for the direct-direct term (i.e. <math>P_1(\cos\theta_{12}) = \cos\theta_{12}</math> is included in the radial integral (\ref{radial})) and <math>C_0 = 1/2</math> for the direct-exchange term (i.e. <math>P_0(\cos\theta_{12}) = 1</math> is included in the radial integral (\ref{radial})). The results for infinite nuclear mass are
\begin{equation}
S(1\;^1S-2\;^1P) = (0.781\,127\,158\,667)^2
\end{equation}
\begin{equation}f(1\;^1S-2\;^1P) = -0.276\,164\,703\,736
\end{equation}
where
\begin{equation}
f(1\;^1S-2\;^1P) = \frac{2}{3}[E(1\;^1S) - E(2\;^1P)]S(1\;^1S-2\;^1P)
\end{equation}
is the absorption oscillator strength.

===Test===
This is a test edit

Theory Notes

2020-12-14T04:30:26Z

Drake: /* Example */

<center>
<H1>Notes on solving the Schrödinger equation in Hylleraas coordinates for heliumlike atoms
Gordon W.F. Drake
Department of Physics, University of Windsor
Windsor, Ontario, Canada N9B 3P4
(Transcribed from hand-written notes and edited by Lauren Moffatt. Last revised November 22, 2017 to correct a missing subscript in the expression for $B_1$ in the Hamiltonian in Sect. 11 General Hermitian Property.)
</H1>
</center>

==Introduction==
These notes describe some of the technical details involved in solving the Schrödinger equation for a heliumlike atom of nuclear charge Z in correlated Hylleraas coordinates. The standard for computational purposes is to express all quantities in a form that is valid for electronic states of arbitrary angular momentum, and to express matrix elements of the Hamiltonian in a manifestly Hermitian symmetrized form. Further information and derivations of the mathematical techniques for angular integrals and radial recursion relations can be found in the paper
G.W.F. Drake, Phys. Rev. A <math>{\bf 18}</math>, 820 (1978).

The starting point is the two-electron Schrödinger equation for infinite nuclear mass
\begin{equation} \left[-\frac{\hbar^2}{2m}(\nabla^2_1 +\nabla^2_2) - \frac{Ze^2}{r_1} -
\frac{Ze^2}{r_2}+\frac{e^2}{r_{12}} \right]\psi = E\psi\nonumber
\end{equation}
where <math>m</math> is the electron mass, and <math>r_{12} = |{\bf r}_1 - {\bf r}_2|</math> (see diagram below).

[[File:Pic1.jpg|250px]]

Begin by rescaling distances and energies so that the Schrödinger equation can be expressed in a dimensionless form. The dimensionless Z-scaled distance is defined by <math>\begin{eqnarray} \rho = \frac{Zr}{a_0} \end{eqnarray} </math> where <math> \begin{eqnarray} a_0 = \frac{\hbar^2}{me^2} \end{eqnarray} </math> is the atomic unit (a.u.) of distance equal to the Bohr radius <math>0.529\, 177\, 210\, 92(17) \times 10^{-10}</math> m. Then

\begin{equation}[-\frac{\hbar^2}{2m}Z^2(\frac{me^2}{\hbar^2})^2(\nabla^2_{\rho_1}+\nabla^2_{\rho_2})
- Z^2\frac{e^2}{a_0}\rho^{-1}_1 - Z^2\frac{e^2}{a_0}\rho^{-1}_2 +
\frac{e^2}{a_0}Z\rho^{-1}_{12}]\psi= E\psi\nonumber\end{equation}

But <math>\begin{eqnarray} \frac{\hbar^2}{m}(\frac{me^2}{\hbar^2})^2 = \frac{e^2}{a_0} \end{eqnarray} </math> is the hartree atomic unit of of energy <math>(E_h = 27.211\,385\,05(60)</math> eV, or equivalently <math>E_h/(hc) = 2194.746\,313\,708(11)</math> m<math>^{-1})</math>. Therefore, after multiplying through by <math>a_0/(Ze)^2</math>, the problem to be solved in Z-scaled dimensionless units becomes

\begin{equation} \left[-\frac{1}{2}(\nabla^2_{\rho_1}+\nabla^2_{\rho_2}) -
\frac{1}{\rho_1} - \frac{1}{\rho_2} + \frac{Z^{-1}}{\rho_{12}}\right]\psi =
\varepsilon\psi\nonumber \end{equation}

where <math>\begin{eqnarray}\varepsilon = \frac{Ea_0}{(Ze)^2} \end{eqnarray}</math> is the energy in Z-scaled atomic units. For convenience, rewrite this in the conventional form
\begin{equation}
H\psi = \varepsilon\psi\nonumber \end{equation}
where (using <math>r</math> in place of <math>\rho</math> for the Z-scaled distance)
\begin{equation}
H = -\frac{1}{2}(\nabla^2_1+\nabla^2_2) - \frac{1}{r_1}-\frac{1}{r_2} + \frac{Z^{-1}}{r_{12}}
\nonumber\end{equation}
is the atomic Hamiltonian for infinite nuclear mass.

==The Hartree Fock Method==

For purposes of comparison, and to define the correlation energy, assume that <math>\psi({\bf r}_1,{\bf r}_2)</math> can be written in the separable product form
\begin{equation}
\psi({\bf r}_1,{\bf r}_2) = \frac{1}{\sqrt{2}} \left[ u_1(r_1) u_2(r_2) \pm u_2(r_1) u_1(r_2) \right] \nonumber
\end{equation}
for the <math>1s^2\ ^1\!S</math> ground state. Because of the <math>\frac{1}{r_{12}}</math> term in the Schrödinger equation, the exact solution connot be expressed in this form as a separable product. However, the Hartree-Fock approximation corresponds to finding the best possible solution to the Schrödinger equation
\begin{equation}
H\psi({\bf r}_1,{\bf r}_2) = E\psi({\bf r}_1,{\bf r}_2)\nonumber
\end{equation}
that can nevertheless be expressed in this separable product form, where as before
\begin{equation}
H = -\frac{1}{2}(\nabla^2_1+\nabla^2_2) - \frac{1}{r_1}- \frac{1}{r_2} + \frac{Z^{-1}}{r_{12}}\nonumber
\end{equation}
is the full two-electron Hamiltonian. To find the best solution, substitute into <math>\left\langle\psi \lvert H-E \rvert \psi\right\rangle</math> and require this expression to be stationary
with respect to arbitrary infinitesimal variations <math>\delta u_1</math> and <math>\delta
u_2</math> in <math>u_1</math> and <math>u_2</math> respectively; i.e.

\begin{equation}
\frac{1}{2} \left\langle \delta u_1(r_1)u_2(r_2) \pm u_2(r_1)\delta
u_1(r_2) \lvert H-E \rvert u_1(r_1) u_2(r_2)\pm u_2(r_1) u_1(r_2)\right\rangle\nonumber
\end{equation}

\begin{equation}
=\int\delta u_1(r_1) d{\bf r}_1 \left\{ \int d{\bf r}_2 u_2(r_2) \left( H - E \right) \left[ u_1(r_1)u_2(r_2) \pm u_2(r_1) u_1(r_2) \right] \right\}\nonumber
\end{equation}

\begin{equation}
= 0 \nonumber
\end{equation}

for arbitrary <math> \delta u_1(r_1).\nonumber </math> Therefore <math>\{\int d{\bf r}_2 \ldots \} = 0</math>.

Similarly, the coefficient of <math>\delta u_2</math> would give

\begin{equation}
\int d{\bf r}_1 u_1(r_1) \left( H-E \right) \left[ u_1(r_1) u_2(r_2) \pm
u_2(r_1) u_1(r_2) \right] = 0\nonumber
\end{equation}

Define

\begin{equation}
I_{12} = I_{21} = \int d{\bf r}\, u_1(r)u_2(r), \nonumber
\end{equation}

\begin{equation}
H_{ij} = \int d{\bf r}\, u_i(r)(-\frac{1}{2}\nabla - \frac{1}{r})u_j(r), \nonumber
\end{equation}

\begin{equation}
G_{ij}(r) = \int d{\bf r}^\prime u_i(r^\prime)\frac{1}{|{\bf r} - {\bf r}\prime|}u_j(r^\prime)\nonumber
\end{equation}

Then the above equations become the pair of integro-differential equations

\begin{equation}
[ H_0 - E + H_{22}+G_{22}(r)]u_1(r) = \mp [ I_{12}(H_0-E) + H_{12}+G_{12}(r)]u_2(r)\nonumber
\end{equation}

\begin{equation}
[H_0-E+H_{11}+G_{11}(r)]u_2(r) = \mp [I_{12}(H_0-E) + H_{12}+G_{12}(r)]u_1(r)\nonumber
\end{equation}

These must be solved self-consistently for the "constants" <math>I_{12}</math> and
<math>H_{ij}</math> and the function <math>G_{ij}(r)</math>.

The Hartree Fock energy is <math>E \simeq -2.87\ldots </math>a.u. while the exact energy is <math>E = -2.903724\ldots </math>a.u.

The difference is called the "correlation energy" because it arises from the
way in which the motion of one electron is correlated to the other. The
Hartree Fock equations only describe how one electron moves in the average field
provided by the other (mean-field theory).

==Configuration Interaction==
Expand
\begin{equation}
\psi({\bf r}_1,{\bf r}_2)= C_0u^{(s)}_1(r_1)u^{(s)}_1(r_2) + C_1u^{(P)}_1({\bf r}_1)u^{(P)}_1({\bf r}_2){\cal Y}^0_{1,1,0}(\hat{\bf r}_1, \hat{\bf r}_2)+C_2u^{(d)}_1({\bf r}_1)u^{(d)}_2({\bf r}_2){\cal Y}^0_{2,2,0}(\hat{\bf r}_1, \hat{\bf r}_2)+... \pm {\rm\ exchange}
\end{equation}
where
\begin{equation}
{\cal Y}^M_{l_1,l_2,L}(\hat{\bf r}_1, \hat{\bf r}_2)=\sum_{m_1,m_2}Y^{m_1}_{l_1}({\bf r}_1)Y^{m_2}_{l_2}({\bf r}_2)\times <l_1l_2m_1m_2\mid LM>
\end{equation}

This works, but is slowly convergent, and very laborious. The best CI calculations are accurate to <math> ~10^{-7}</math> a.u.

==Hylleraas Coordinates==
[E.A. Hylleraas, Z. Phys. <math>{\bf 48}, 469(1928)</math> and <math>{\bf 54}, 347(1929)</math>]
suggested using the co-ordinates <math>r_1</math>, <math>r_2</math> and <math>r_{12}</math> or equivalently

\begin{eqnarray}
s &=& r_1 + r_2, \nonumber\\
t &=& r_1-r_2, \nonumber\\
u &=& r_{12}\nonumber
\end{eqnarray}

and writing the trial functions in the form

\begin{equation}
\Psi({\bf r}_1,{\bf r}_2) = \sum^{i+j+k\leq
N}_{i,j,k}c_{i,j,k}r_1^{i+l_1}r_2^{j+l_2}r_{12}^ke^{-\alpha r_1 - \beta r_2}
\mathcal{Y}^M_{l_1,l_2,L}(\hat{r}_1,\hat{r}_2)\pm \text{exchange}\nonumber
\end{equation}

Diagonalizing H in this non-orthogonal basis set is equivalent to solving
<math> \begin{eqnarray}
\frac{\partial E}{\partial c_{i,j,k}} = 0\nonumber
\end{eqnarray} </math>
for fixed <math>\alpha</math> and <math>\beta</math>.

The diagonalization must be repeated for different values of <math>\alpha</math> and
<math>\beta</math> in order to optimize the non-linear parameters.

==Completeness==
The completeness of the above basis set can be shown by first writing
\begin{equation}
r_{12}^2 = r_1^2 + r_2^2 - 2r_1r_2\cos\theta_{12}
\end{equation}
and, from the spherical harmonic addition theorem,
\begin{equation}
\cos(\theta_{12})=\frac{4\pi}{3}\sum^1_{m=-1}Y^{m*}_1(\theta_1,\varphi_1)Y^m_1(\theta_2,\varphi_2)
\end{equation}
Consider first the S-states. The <math>r_{12}^0</math> terms are like the ss terms in a CI calculation. The
<math>r_{12}^2</math> terms bring in p-p type contributions, and the higher powers bring
in d-d, f-f etc. type terms. In general

\begin{equation}
P_l(\cos\theta_{12}) = \frac{4\pi}{2l+1}\sum^l_{m=-l}{Y^{m}_l}^*(\theta_1,\varphi_1)Y^m_l(\theta_2,\varphi_2)\nonumber
\end{equation}

For P-states, one would have similarly

\begin{array}{lr}
r_{12}^0 & (sp)P\nonumber\\
r_{12}^2 & (pd)P\nonumber\\
r_{12}^4 & (df)P\nonumber\\
\vdots & \vdots\nonumber
\end{array}

For D-states

\begin{array}{lr}
r_{12}^0 & (sd)D & (pp^\prime)D\nonumber\\
r_{12}^2 & (pf)D & (dd^\prime)D\nonumber\\
r_{12}^4 & (dg)D & (ff^\prime)D\nonumber\\
\vdots & \vdots & \vdots\nonumber
\end{array}

In this case, since there are two "lowest-order" couplings to form a
D-state, both must be present in the basis set, i.e.

\begin{equation}
\Psi(r_2,r_2) = \sum c_{ijk}r_1^ir_2^{j+2}r_{12}^ke^{-\alpha r_1-\beta
r_2}\mathcal{Y}^M_{022}(\hat{r}_1,\hat{r}_2)
+\sum d_{ijk}r_1^{i+1}r_2^{j+1}r_{12^k}e^{-\alpha^\prime r_1 - \beta^\prime
r_2}\mathcal{Y}^M_{112}(\hat{r}_1,\hat{r}_2)\nonumber
\end{equation}

For F-states, one would need <math>(sf)F</math> and <math>(pd)F</math> terms.

For G-states, one would need <math>(sg)G</math>, <math>(pf)G</math> and <math>(dd^\prime)G</math> terms.

Completeness of the radial functions can be proven by considering the
Sturm-Liouville problem

\begin{equation}
\left(-\frac{1}{2}\nabla^2-\frac{\lambda}{r_s}-E\right)\psi({\bf r}) = 0\nonumber
\end{equation}

or

\begin{equation}
\left(-\frac{1}{2}\frac{1}{r^2}\frac{\partial}{\partial r}\left({r^2}\frac{\partial}{\partial r}\right) -
\frac{l\left(l+1\right)}{2r^2} - \frac{\lambda}{r} - E\right)u(r) = 0.\nonumber
\end{equation}

For fixed E and variable <math>\lambda</math> (nuclear charge).

The eigenvalues are <math>\lambda_n = (E/E_n)^{1/2}</math>, where <math>E_n =- \frac{1}{2n^2}</math>

[[File:Pic2.jpg|500px]]

and the eigenfunctions are

\begin{equation}
u_{nl}(r) =
\frac{1}{(2l+1)!}\left(\frac{(n+l)!}{(n-l-1)2!}\right)^{1/2}(2\alpha)^{3/2}e^{-\alpha
r}\nonumber
\times \ \left( 2 \alpha r \right)^l\,_{1}F_{1}\left(-n+l+1,2l+2; 2 \alpha r \right)
\end{equation}

with <math>\alpha = (-2E)^{1/2}</math> and <math>n\geq l+1</math>. The confluent hypergeometric function <math> _1F_1(a,b,;z)</math> then denotes a finite polynomial since <math>a=-n+l+1</math> is a negative integer or zero.

Unlike the hydrogen spectrum, which has both a discrete part for <math>E<0</math> and a
continuous part for <math>E>0</math>, this forms an entirely discrete set of finite
polynomials, called Sturmian functions. They are orthogonal with respect to
the potential, i.e.

\begin{equation}
\int^\infty_0 r^2dr \left( u_{n^\prime l}(r)\frac{1}{r}u_{nl}(r) \right) = \delta_{n,n^\prime}\nonumber
\end{equation}

Since they become complete in the limit <math>n\rightarrow\infty</math>, this assures the
completeness of the variational basis set.

[See also B Klahn and W.A. Bingel Theo. Chim. Acta (Berlin) 44, 9 and 27
(1977)].

==Solutions of the Eigenvalue Problem==

For convenience, write

\[
\Psi({\bf r}_1,{\bf r}_2) = \sum^N_{m=1}c_m\varphi_m\nonumber
\]

where <math>m</math> represents the <math>m</math>'th combination of <math>i,j,k</math> and

\begin{equation}
\varphi_{ijk}=r_1^ir_2^jr_{12}^ke^{-\alpha r_1 - \beta
r_2}\mathcal{Y}^M_{l_1,l_2,L}(\hat{r}_1,\hat{r}_2) \pm \mbox{exchange}\nonumber
\end{equation}

\begin{align}

&
\left( \begin{array}{lr}
\cos(\theta) & \sin(\theta)\\
-\sin(\theta) & \cos(\theta)
\end{array} \right)
\left( \begin{array}{lr}
H_{11} & H_{12} \\
H_{12} & H_{22}
\end{array}\right)
\left( \begin{array}{lr}
\cos(\theta) & -\sin(\theta)\\
\sin(\theta) & \cos(\theta)
\end{array} \right)
\\[5pt]
= & \left(\begin{array}{lr}
cH_{11}+sH_{12} & cH_{12} + sH_{22}\\
-sH_{11} + cH_{12} & -sH_{12} + cH_{22}
\end{array}\right)
\left( \begin{array}{cc}
c & -s \\
s & c
\end{array}\right)
\\[5pt]
= & \left(\begin{array}{lr}
c^2H_{11}+s^2H_{22} + 2csH_{12} & (c^2-s^2)H_{12}+cs(H_{22}-H_{11})\\
(c^2-s^2)H_{12}+cs(H_{22}-H_{11}) & s^2H_{11}+c^2H_{22}-2csH_{12}
\end{array}\right)
\end{align}

Therefore

\begin{equation}
\left[\cos^2(\theta)-\sin^2(\theta)\right]H_{12} =
\cos(\theta)\sin(\theta)(H_{11}-H_{22})\nonumber
\end{equation}

and

\begin{equation}
\tan(2\theta) = \frac{2H_{12}}{H_{11}-H_{22}}\nonumber
\end{equation}

i.e.

\begin{equation}
\cos(\theta)=\left(\frac{r+\omega}{2r}\right)^{1/2}\nonumber
\end{equation}

\begin{equation}
\sin(\theta)=-\mbox{sgn}(H_{12})\left(\frac{r-\omega}{2r}\right)^{1/2}\nonumber
\end{equation}

where

\begin{equation}
\omega = H_{22}-H_{11}\nonumber
\end{equation}

\begin{equation}
r=\left(\omega^2+4H_{12}^2\right)^{1/2}\nonumber
\end{equation}

\begin{equation}
E_1=\frac{1}{2}\left(H_{11}+H_{22}-r\right)\nonumber
\end{equation}

\begin{equation}
E_2=\frac{1}{2}\left(H_{11}+H_{22}+r\right)\nonumber
\end{equation}

===Brute Force Method===

- Gives all the eigenvalues and eigenvectors, but it is slow

- First orthonormalize the <math>N</math>-dimensional basis set; i.e. form linear combinations

\begin{equation}
\Phi_m = \sum_{n=1}^N\varphi_nR_{nm}\nonumber
\end{equation}

such that
\begin{equation}
\langle \Phi_m | \Phi_n \rangle = \delta_{m,n}
\end{equation}

This can be done by finding an orthogonal tranformation T such that

\begin{equation}
T^TOT = I = \left(
\begin{array}{ccccc}
I_1 & 0 & 0 & \ldots & 0 \\
0 & I_2 & 0 & \ldots & 0 \\
0 & 0 & I_3 & \ldots & 0 \\
\vdots & \vdots & \vdots & \ddots & \vdots \\
0 & 0 & 0 & \ldots & I_N
\end{array}\right);
\end{equation}

\begin{equation}
O_{mn} = \langle \varphi_m | \varphi_n \rangle \nonumber
\end{equation}

and then applying a scale change matrix

\begin{equation}
S = \left(\begin{array}{ccccc}
\frac{1}{I_1^{1/2}} & 0 & 0 & \ldots & 0\\
0 & \frac{1}{I_2^{1/2}} & 0 & \ldots & 0 \\
0 & 0 & \frac{1}{I_3^{1/2}} & \ldots & 0 \\
\vdots & \vdots & \vdots & \ddots & \vdots \\
0 & 0 & 0 & \ldots & I_N^{1/2}
\end{array}\right)= S^T\nonumber
\end{equation}

Then

\begin{equation}
S^TT^TOTS = 1
\end{equation}

i.e.

\begin{equation}
R^TOR = 1
\end{equation}

with <math>R=TS</math>.

If H is the matrix with elements <math>H_{mn} = \langle \varphi_m | \varphi_n \rangle </math>, then H
expressed in the <math>\Phi_m</math> basis set is

\begin{equation}
H^\prime = R^THR.\nonumber
\end{equation}

We next diagonalize <math>H^\prime</math> by finding an orthogonal transformation W such
that

\begin{equation}
W^T H^\prime W = \lambda = \left(
\begin{array}{cccc}
\lambda_1 & 0 & \ldots & 0 \\
0 & \lambda_2 & \ldots & 0 \\
\vdots & \vdots & \ddots & \vdots \\
0 & 0 & \ldots & \lambda_N
\end{array}\right)\nonumber
\end{equation}

The q'th eigenvector is

\begin{equation}
\Psi^{(q)} = \sum_{n=1}^N\Phi_n W_{n,q} = \sum_{n,n^\prime} \varphi_{n^\prime} R_{n^\prime ,n} W_{n,q} \nonumber
\end{equation}

i.e.

\begin{equation}
c_{n^\prime}^{(q)} = \sum_{n=1}^N R_{n^\prime n} W_{n,q}
\end{equation}

===The Power Method===

- Based on the observation that if H has one eigenvalue, <math>\lambda_M</math>, much bigger than all the rest, and <math>\chi = \left(
\begin{array}{c}a_1\\a_2\\\vdots\end{array}\right)</math>
is an arbitrary starting vector, then <math>\chi = \sum_{q=1}^N x_q\Psi^{(q)}</math>.

\begin{equation}
(H)^n\chi = \sum_{q=1}^N x_q \lambda^n_q\Psi^{(q)}\nonumber\rightarrow x_M\lambda_M^n\Psi^{(M)}\nonumber
\end{equation}

provided <math>x_M\neq 0</math>.

To pick out the eigenvector corresponding to any eigenvalue, write the original problem in the form

\begin{align}
H\Psi & = \lambda O \Psi \nonumber \\
\left( H - \lambda_q O \right) \Psi & = \left( \lambda - \lambda_q \right) O \Psi \nonumber
\end{align}

Therefore,

\begin{equation}
G \Psi = \frac{1}{\lambda - \lambda_q } \Psi \nonumber \\
\end{equation}

where <math>G=(H-\lambda_qO)^{-1}O</math> with eigenvalues <math>\frac{1}{\lambda_n - \lambda_q}</math>.

By picking <math>\lambda_q</math> close to any one of the <math>\lambda_n</math>, say
<math>\lambda_{n^\prime}</math>, then <math>\frac{1}{\lambda_n-\lambda_q}</math> is much larger for
<math>n=n^\prime</math> than for any other value. The sequence is then

\begin{array}{c}
\chi_1=G\chi\nonumber\\
\chi_2=G\chi_1\nonumber\\
\chi_3=G\chi_2\nonumber\\
\vdots\nonumber
\end{array}

until the ratios of components in <math>\chi_n</math> stop changing.

- To avoid matrix inversion and multiplication, note that the sequence is equivalent to
\begin{equation}
F \chi_n = \left( \lambda - \lambda_q \right) O \chi_{n-1} \nonumber \\
\end{equation}
where <math>F = H-\lambda_qO</math>. The factor of <math>\left( \lambda - \lambda_q \right)</math> can be
dropped because this only affects the normalization of <math>\chi_n</math>. To find <math>\chi_n</math>, solve
\begin{equation}
F \chi_n = O \chi_{n-1} \nonumber
\end{equation}
(N equations in N unknowns). Then
\begin{equation}
\lambda = \frac{ \langle \chi_n | H | \chi_n \rangle}{\langle \chi_n | \chi_n \rangle} \nonumber
\end{equation}

==Matrix Elements of H==

\begin{equation}
H=-\frac{1}{2}\nabla^2_1 -\frac{1}{2}\nabla^2_2 - \frac{1}{r_1} - \frac{1}{r_2} +\frac{Z^{-1}}{r_{12}}\nonumber
\end{equation}

Taking <math>r_1, r_2</math> and <math>r_{12}</math> as independent variables,

\begin{equation}
\nabla_1^2 = \frac{1}{r_1^2}\frac{\partial}{\partial r_1}\left(r_1^2\frac{\partial}{\partial r_1}\right)+ \frac{1}{r_{12}^2}\frac{\partial}{\partial r_{12}}
\left(r_{12}^2\frac{\partial}{\partial r_{12}}\right)\nonumber -\frac{l_1(l_1+1)}{r_1^2}+2(r_1 - r_2\cos(\theta_{12}))\frac{1}{r_{12}}\frac{\partial^2}{\partial r_1 \partial r_{12}} \nonumber - 2(\nabla_1^Y \cdot {\bf r}_2)\frac{1}{r_{12}}\frac{\partial}{\partial r_{12}}\nonumber
\end{equation}

where <math>\nabla_1^Y</math> acts only on the spherical harmonic part of the wave function and the diagram

[[File:Pic3.jpg|400px]]

defines the complete set of 6 independent variables is <math>r_1</math>, <math>r_2</math>, <math>r_{12}</math>, <math>\theta_1</math>, <math>\varphi_1</math>, <math>\chi</math>.

If <math>r_{12}</math> were not an independent variable, then one could take the volume element to be

\begin{equation}
d\tau = r_1^2dr_1\sin(\theta_1)d\theta_1d\varphi_1r_2^2dr_2\sin(\theta_2)d\theta_2d\varphi_2.\nonumber
\end{equation}

However, <math>\theta_2</math> and <math>\varphi_2</math> are no longer independent variables. To eliminate them, take the point <math>{\bf r}_1</math> as the
origin of a new polar co-ordinate system, and write

\begin{equation}
d\tau=-r_1^2dr_1\sin(\theta_1)d\theta_1d\varphi_1r_{12}^2dr_{12}\sin(\psi)d\psi d\chi\nonumber
\end{equation}

and use

\begin{equation}
r_2^2=r_1^2+r_{12}^2 +2r_1r_{12}\cos(\psi)
\end{equation}

Then for fixed <math>r_1</math> and <math>r_{12}</math>,

\begin{equation}
2r_2dr_2 = -2r_1r_{12}\sin(\psi)d\psi
\end{equation}

Thus

\begin{equation}
d\tau= r_1dr_1r_2dr_2r_{12}dr_{12}\sin(\theta_1)d\theta_1d\varphi_1d\chi
\end{equation}

The basic type of integral to be calculated is

\begin{align}
I \left( l_1, m_1, l_2, m_2; R \right) & = \int \sin \left( \theta_1 \right) d\theta_1 d\varphi_1 d\chi Y^{m_1}_{l_1} \left( \theta_1, \varphi_1 \right)^{*} Y^{m_2}_{l_2} \left( \theta_2, \varphi_2 \right) \\
& \times \int r_1dr_1r_2dr_2r_{12}dr_{12}R \left( r_1, r_2, r_{12} \right)
\end{align}

Consider first the angular integral. <math>Y^{m_2}_{l_2} \left( \theta_2, \varphi_2 \right)</math> can be expressed in terms of the independent variables <math>\theta_1, \varphi_1, \chi</math> by use of the rotation matrix relation
\begin{equation}
Y^{m_2}_{l_2}( \theta_2, \varphi_2) = \sum_m \mathcal{D}^{(l_2)}_{m_2,m}(\varphi_1,\theta_1,\chi)^*Y^m_{l_2}( \theta_{12}, \varphi)
\end{equation}
where <math>\theta, \varphi</math> are the polar angles of <math>{\bf r}_2</math> relative to <math>{\bf r}_1</math>. The angular integral is then

\begin{align}
I_{ang} & = \int^{2 \pi}_{0} d\chi \int^{2\pi}_{0} d\varphi_1 \int^{\pi}_{0} \sin \left( \theta_1 \right) d\theta_{1} Y^{m_1}_{l_1} \left( \theta_1, \varphi_1 \right)^{*} \\
& \times \sum_m \mathcal{D}^{(l_2)}_{m_2,m} \left( \varphi_1,\theta_1,\chi \right)^{*} Y^{m}_{l_2} \left( \theta_{12},\varphi \right) \\
\end{align}

Use

\begin{equation}
Y^{m_1}_{l_1} \left( \theta_1, \varphi_1 \right)^{*} = \sqrt{\frac{2l_1+1}{4\pi}}\mathcal{D}^{(l_1)}_{m_1,0} \left( \varphi_1, \theta_1, \chi \right)
\end{equation}

together with the orthogonality property of the rotation matrices (Brink and Satchler, p 147)

\begin{align}
& \int \mathcal{D}^{(j)*}_{m,m^\prime}\mathcal{D}^{(J)}_{M,M^\prime}\sin(\theta_1)d\theta_1d\varphi_1d\chi \\
& = \frac{8\pi^2}{2j+1}\delta_{jJ}\delta_{mM}\delta_{m^\prime M^\prime}\nonumber
\end{align}

to obtain

\begin{align}
I_{ang} & = \sqrt{ \frac{2l_1+1}{4\pi}} \frac{8\pi^2}{2l_1+1} \delta_{l_1,l_2} \delta_{m_1,m_2} Y^{0}_{l_2} \left( \theta_{12}, \varphi \right) \\
& = 2 \pi \delta_{l_1,l_2} \delta_{m_1,m_2} P_{l_2} \left( \cos \theta_{12} \right)
\end{align}

since

\begin{equation}
Y^0_{l_2} \left( \theta_{12}, \varphi \right) = \sqrt{\frac{2l_1+1}{4\pi}} P_{l_2} \left( \cos \left( \theta_{12} \right) \right)
\end{equation}

Note that <math>P_{l_2} \left( \cos \theta \right)</math> is just a short hand expression for a radial function because

\begin{equation}
\cos \theta_{12} = \frac{r_1^2 + r_2^2 - r_{12}^2}{2r_1r_2}
\end{equation}

The original integral is thus

\begin{equation}
I \left( l_1, m_1, l_2, m_2; R \right) = 2 \pi \delta_{l_1,l_2} \delta_{m_1,m_2} \int^{\infty}_{0} r_{1} dr_{1} \int^{\infty}_{0} r_{2} dr_{2} \int^{r_{1} + r_{2}}_{|r_{1} - r_{2}|} r_{12} dr_{12} R \left( r_{1}, r_{2}, r_{12} \right) P_{l_2} \left( \cos \theta_{12} \right)
\end{equation}

where again

\begin{equation}
\cos \theta_{12} = \frac{r_{1}^{2} +r^{2}_{2} - r_{12}^{2}}{2 r_{1} r_{2}}
\end{equation}

is a purely radial function.

The above would become quite complicated for large <math>l_2</math> because <math>P_{l_2}(\cos \theta_{12})</math> contains terms up to <math>\left( \cos \theta_{12} \right)^{l_2}</math>. However, recursion relations exist which allow any integral containing <math>P_l \left( \cos \theta_{12} \right)</math> in terms of those containing just <math>P_0 \left( \cos \theta_{12} \right) = 1</math> and <math>P_1 \left( \cos \theta_{12} \right) = \cos \theta_{12}</math>.

==Radial Integrals and Recursion Relations==

The basic radial integral is [see G.W.F. Drake, Phys. Rev. A 18, 820 (1978)]

\begin{align}
I_0 \left( a, b, c \right) & =
\int^{\infty}_{0} r_{1} dr_{1} \int^{\infty}_{r_{1}} r_{2} dr_{2} \int^{r_{1}+r_{2}}_{r_{2} - r_{1}} r_{12} dr_{12} r_{1}^{a} r_{2}^{b} r_{12}^{c} e^{-\alpha r_{1} - \beta r_{2}} + \int^{\infty}_{0} r_{2} dr_{2} \int^{\infty}_{r_{2}} r_{1} dr_{1} \int^{r_{1} + r_{2}}_{r_{1} - r_{2}} r_{12} dr_{12} r_{1}^{a} r_{2}^{b} r_{12}^{c} e^{- \alpha r_{1} - \beta r_{2}} \\
& = \frac{2}{c+2} \sum^{[(c+1)/2]}_{i=0}
\left(\begin{array}{c}
c+2 \\
2i+1
\end{array} \right)
\left\{ \frac{q!}{\beta^{q+1} \left( \alpha + \beta \right)^{p+1}} \sum_{j=0}^{q} \frac{\left( p + j \right)!}{j!} \left( \frac{\beta}{\alpha + \beta} \right)^j + \frac{q^{\prime}!}{\alpha^{q^{\prime}+1} \left( \alpha + \beta \right)^{p^{\prime}+1}} \sum^{q^{\prime}}_{j=0} \frac{\left( p^{\prime} + j \right)!}{j!} \left( \frac{\alpha}{\alpha + \beta} \right)^j \right\}
\end{align}

where

\begin{array}{ll}
p = a + 2i + 2 & p^{\prime} = b + 2i + 2 \\
q = b + c - 2i + 2 & q^{\prime} = a + c - 2i + 2
\end{array}

The above applies for <math> a, b \geq -2, c\geq -1</math>.
<math>[x]</math> means "greatest integer in" <math>x</math>.

Then

\begin{align}
I_1 \left( a, b, c \right) & = \int d\tau_{r} r_{1}^{a} r_{2}^{b} r_{12}^{c} e^{ -\alpha r_{1} - \beta r_{2}} P_{1}\left( \cos \theta \right) \\
& = \frac{1}{2} \left( I_{0} \left( a + 1, b - 1, c \right) + I_{0} \left( a - 1, b + 1, c \right) - I_{0} \left( a - 1, b - 1, c + 2 \right) \right)
\end{align}

===The Radial Recursion Relation===

Recall that the full integral is

\begin{equation}
I(l_1m_1,l_2m_2;R)=2\pi\delta_{l_1,l_2}\delta_{m_1,m_2}I_{l_2}(R)
\end{equation}

where

\begin{equation}
I_{l_2}(R) = \int d\tau_r R(r_1,r_2,r_{12})P_{l_2}(\cos \theta_{12})
\end{equation}

To obtain the recursion relation, use

\begin{equation}
P_l(x) = \frac{[P^{\prime}_{l+1}(x)-P^{\prime}_{l-1}(x)]}{2l+1}
\end{equation}

with

\begin{equation}
P_{l+1}^\prime (x) = \frac{d}{dx}P_{l+1}(x)
\end{equation}

Here <math>x=\cos\theta_{12}</math> and

\begin{align}
\frac{d}{d \cos \theta_{12}} & = \frac{dr_{12}}{d \cos \theta_{12}} \frac{d}{dr_{12}} \\
& = -\frac{r_{1} r_{2}}{r_{12}} \frac{d}{dr_{12}}
\end{align}

Then,

\begin{equation}
I_{l}\left( R \right) = - \int d\tau_r R \frac{r_{1} r_{2}}{r_{12}} \frac{d}{dr_{12}} \frac{ \left[ P_{l+1} \left( \cos \theta_{12} \right) - P_{l-1} \left( \cos \theta_{12} \right) \right]}{2l+1}
\end{equation}

The <math>r_{12}</math> part of the integral is

\begin{align}
& \int^{r_{1} + r_{2}}_{|r_{1} - r_{2}|} r_{12} dr_{12} R \frac{r_{1} r_{2}}{r_{12}} \frac{d}{dr_{12}} \left[P_{l+1}-P_{l-1} \right] \\
& = R r_{1} r_{2} \left[ P_{l+1} - P_{l-1} \right]^{r_{1} + r_{2}}_{|r_{1} - r_{2}|} - \int^{r_{1} + r_{2}}_{|r_{1} - r_{2}|} r_{12} dr_{12} \left( \frac{d}{dr_{12}} R \right) \frac{r_{1} r_{2}}{r_{12}} \frac{ \left[ P_{l+1} - P_{l-1} \right]}{2l+1}
\end{align}

The integrated term vanishes because

\begin{equation}
\cos \theta_{12} = \frac{r_{1}^{2} + r_{2}^{2} - r_{12}^{2}}{2 r_{1} r_{2}}
\begin{array}{lll}
& = -1 & {\rm when} & r_{12}^{2} = \left( r_{1} + r_{2} \right)^{2} \\
& = 1 & {\rm when} & r_{12}^{2} = \left( r_{1} - r_{2} \right)^{2}
\end{array}
\end{equation}

and <math>P_l(1)=1</math>, <math>P_{l}(-1)=(-1)^{l}</math>.

Thus,

\begin{equation}
I_{l+1}\left(\frac{r_1r_2}{r_{12}}R^\prime\right) = \left( 2l+1 \right) I_{l}(R) - I_{l-1} \left( \frac{r_{1} r_{2}}{r_{12}} R^{\prime} \right)
\end{equation}

If

\begin{equation}
R = r_{1}^{a-1} r_{2}^{b-1} r_{12}^{c+2} e^{-\alpha r_{1} - \beta r_{2}}
\end{equation}

then [G.W.F. Drake, Phys. Rev. A 18, 820 (1978)]

\begin{equation}
I_{l+1}(r_1^ar_2^br_{12}^c) =
\frac{2l+1}{c+2} I_{l} \left( r_{1}^{a-1} r_{2}^{b-2} r_{12}^{c+2} \right)+I_{l-1} \left( r_{1}^{a} r_{2}^{b} r_{12}^{c} \right)
\end{equation}

For the case <math>c=-2</math>, take

\begin{equation}
R = r_{1}^{a-1} r_{2}^{b-1} \ln r_{12} e^{ -\alpha r_{1} -\beta r_{2}}
\end{equation}

Then

\begin{equation}
I_{l+1} \left( r_{1}^{a} r_{2}^{b} r_{12}^{c} \right) = I_{l} \left( r_{1}^{a-1} r_{2}^{b-1} \ln r_{12} \right) \left( 2l + 1 \right) - I_{l-1} \left( r_{1}^{a} r_{2}^{b} r_{12}^{-2} \right)
\end{equation}

This allows all <math>I_{l}</math> integrals to be calculated from tables of <math>I_0</math> and
<math>I_1</math> integrals.

===The General Integral===

The above results for the angular and radial integrals can now be combined into a general formula for integrals of the type

\begin{equation}
I = \int \int d{\bf r}_{1} d{\bf r}_{2} R_{1} \mathcal{Y}^{M^{\prime}}_{l^{\prime}_{1} l^{\prime}_{2} L^{\prime}} \left( \hat{r}_{1}, \hat{r}_{2} \right) T_{k_{1} k_{2} K}^{Q} \left( {\bf r}_{1},{\bf r}_{2} \right) R_{2} \mathcal{Y}_{l_{1} l_{2} L}^{M} \left( \hat{r}_{1}, \hat{r}_{2} \right)
\end{equation}

where

\begin{equation}
\mathcal{Y}^{M}_{l_{1} l_{2} L} \left( \hat{r}_{1}, \hat{r}_{2} \right) = \sum_{m_1,m_2} \langle l_{1} l_{2} m_{1} m_{2} | LM \rangle Y^{m_{1}}_{l_{1}} \left( \hat{r}_{1} \right) Y^{m_{2}}_{l_{2}} \left( \hat{r}_{2} \right)
\end{equation}

and

\begin{equation}
T^Q_{k_{1} k_{2} K} \left( {\bf r}_{1}, {\bf r}_{2} \right) = \sum_{q_{1}, q_{2}} \langle k_{1} k_{2} q_{1} q_{2} | KQ \rangle Y_{k_{1}}^{q_{1}} \left( \hat{r}_{1} \right) Y^{q_{2}}_{k_{2}} \left( \hat{r}_{2} \right)
\end{equation}

The basic idea is to make repeated use of the formula

\begin{equation}
Y^{m_{1}}_{l_{1}} \left( \hat{r}_{1} \right) Y^{m_{2}}_{l_{2}} \left( \hat{r}_{1} \right) = \sum_{l m} \left( \frac{\left( 2l_{1}+1 \right) \left(2l_{2}+1 \right) \left(2l+1 \right)}{4 \pi} \right)^{1/2}
\times \left( \begin{array}{ccc}
l_1 & l_2 & l \\
m_1 & m_2 & m
\end{array} \right) \left(\begin{array}{ccc}
l_1 & l_2 & l \\
0 & 0 & 0
\end{array} \right)
Y^{m*}_{l} \left( \hat{r}_{1} \right)
\end{equation}

where

\begin{equation}
Y^{m*}(\hat{r}) = (-1)^mY_l^{-m}(\hat{r})
\end{equation}

and

\begin{equation}
\left(\begin{array}{ccc}
l_1 & l_2 & l \\
m_1 & m_2 & m
\end{array} \right) = \frac{\left( -1 \right)^{l_{1} - l_{2} - m}}{\left( 2l + 1 \right)^{1/2}} \left( l_{1} l_{2} m_{1} m_{2} | l, -m \right)
\end{equation}

is a 3-<math>j</math> symbol. In particular, write

\begin{array}{cc}
Y^{m_1^\prime}_{l_1^\prime}(\hat{r}_1)^{*} \underbrace{Y_{k_{1}}^{q_{1}} \left( \hat{r}_{1} \right) Y^{m_{1}}_{l_{1}} \left( \hat{r}_{1} \right)} & = & \sum_{\Lambda M}(\ldots)Y^{M}_\Lambda(\hat{r}_1)\\
\ \ \ \ \ \ \ \sum_{\lambda_1\mu_1}Y_{\lambda_1}^{\mu_1}(\hat{r}_1) & & \\
& & \\
Y^{m_{2}^{\prime}}_{l_{2}^{\prime}} \left( \hat{r}_{2} \right) \underbrace{Y^{q_{2}}_{k_{2}} \left( \hat{r}_{2} \right) Y_{l_{2}}^{m_{2}} \left( \hat{r}_{2} \right)} & = & \sum_{\Lambda^{\prime} M^{\prime}} \left( \ldots \right)Y_{\Lambda^{\prime}}^{M^{\prime}*} \left( \hat{r}_{2} \right) \\
\ \ \ \ \ \ \ \ \ \sum_{\lambda_{2} \mu_{2}}Y_{\lambda_{2}}^{\mu_{2}} \left( \hat{r}_{2} \right) & &
\end{array}

The angular integral then gives a factor of <math>2 \pi \delta_{\Lambda, \Lambda^{\prime}} \delta_{M, M^{\prime}} P_{\Lambda} \left( \cos \theta_{12} \right)</math>. The total integral therefore reduces to the form

\begin{equation}
I = \sum_{\Lambda} C_{\Lambda} I_{\Lambda} \left( R_{1} R_{2} \right)
\end{equation}

where <math>C_{\Lambda} = \sum_{\lambda_{1}, \lambda_{2}} C_{ \lambda_{1}, \lambda_{2}, \Lambda}</math>. For further details and derivations, including graphical representations, see G.W.F. Drake, Phys. Rev. A <math>{\bf 18}</math>, 820 (1978).

==Graphical Representation==

[[File:GraphRep1.jpg|500px]]

==Matrix Elements of H==
Recall that

\begin{equation}
H = -\frac{1}{2} \nabla^{2}_{1} - \frac{1}{2} \nabla^{2}_{2} - \frac{1}{r_1} - \frac{1}{r_2} + \frac{Z^{-1}}{r_{12}}
\end{equation}

Consider matrix elements of

\begin{equation}
\nabla^2_1 = \frac{1}{r_1^2} \frac{\partial}{\partial r_1}\left(r_1^2\frac{\partial}{\partial r_1} \right) + \frac{1}{r^2}\frac{\partial}{\partial r}\left(r^2\frac{\partial}{\partial r}\right) - \frac{(\vec{l}_1^Y)^2}{r_1^2} + \frac{2 \left( r_1 - r_2 \cos \theta \right)}{r} \frac{\partial^2}{\partial r_1 \partial r} - 2 \left( \nabla^Y_1 \cdot {\bf r}_2 \right) \frac{1}{r} \frac{\partial}{\partial r}
\end{equation}

where <math>r \equiv r_{12}</math> and <math>\cos\theta \equiv \cos\theta_{12}</math>. Also in what follows, define <math>\hat{\nabla}^Y_1 \equiv r_1\nabla^Y_1</math> where <math>\nabla^Y</math> operates only on the spherical harmonic part of the wave function, and similarly for <math>\vec{l}_1^Y</math>.

A general matrix element is

\begin{equation}
\langle r_1^{a^\prime} r_2^{b^\prime} r_{12}^{c^\prime} e^{-\alpha^\prime r_1 - \beta^\prime r_2} \mathcal{Y}^{M^\prime}_{l^\prime_1 l^\prime_2 L^\prime} \left( \hat{r}_1, \hat{r}_2 \right) \left| \nabla^2_1 \right| r_1^a r_2^b r_{12}^c e^{-\alpha r_1 - \beta r_2} \mathcal{Y}^M_{l_1 l_2 L} \left( \hat{r}_1, \hat{r}_2 \right) \rangle
\end{equation}

Since <math>\nabla^2_1</math> is rotationally invariant, this vanishes unless <math>L=L^\prime</math>, <math>M=M^\prime</math>. Also <math>\nabla_1^2</math> is Hermitian, so that the result must be the same whether it operates to the right or left, even though the results look very different. In fact, requiring the results to be the same yields some interesting and useful integral identities as follows:

Put

\begin{equation}
\mathcal{F} = F \mathcal{Y}^M_{l_1 l_2 L} \left( \hat{r}_1, \hat{r_2} \right)
\end{equation}

and

\begin{equation}
F = r_1^a r_2^b r_{12}^c e^{-\alpha r_1 - \beta r_2}
\end{equation}

Then

\begin{equation}
\nabla^2_1 \mathcal{F} = \left\{ \frac{1}{r_1^2} \left[ a \left( a + 1 \right) - l_1 \left( l_1 + 1 \right) \right] + \frac{c \left( c + 1 \right)}{r^2} + \alpha^2\frac{2\alpha \left( a + 1 \right)}{r_1} + \frac{2 \left(r_1 - r_2 \cos \theta \right) }{r_1r^2}c \left[ a - \alpha r_1 \right] - \frac{2c}{r^2} \left( \hat{\nabla}^Y_1 \cdot \hat{r}_2 \right) \frac{r_2}{r_1} \right\} \mathcal{F}
\end{equation}

and

\begin{align}
\langle \mathcal{F}^\prime \left| \nabla^2_1 \right| \mathcal{F} \rangle & = \sum_\Lambda \int d\tau_r F^\prime C_\Lambda \left( 1 \right) P_\Lambda \left( \cos \theta \right) \times \left\{ \frac{1}{r_1^2} \left[ a \left( a + 1 \right) - l_1 \left( l_1 + 1 \right) \right] -\frac{2 \alpha \left( a + 1 \right)}{r_1} + \frac{c \left( c + 1 \right)}{r^2} +\alpha^2 + \frac{2 \left( r_2 - r_2 \cos \theta \right)}{r_1r^2}c \left( a - \alpha r_1 \right) \right\} F \\
& + \sum_\Lambda \int d\tau_r F^\prime C_\Lambda \left( \hat{\nabla}^Y_1 \cdot \hat{r}_2 \right) P_\Lambda \left( \cos \theta \right) \left( \frac{-2c r_2}{r_1 r^2} \right) F
\end{align}

where

\begin{equation}
\int d\tau_r = \int^\infty_0 r_1 dr_1 \int^\infty_0 r_2 dr_2 \int^{r_1+r_2}_{\left| r_1 - r_2 \right|} r dr
\end{equation}

For brevity, let the sum over <math>\Lambda</math> and the radial integrations be understood, and let <math>\left( \nabla_1^2 \right)</math> stand for the terms which appear in the integrand. Then operating to the right gives

\begin{equation}
(\nabla_1^2)_R = \frac{1}{r_1^2} \left[ a \left( a + 1 \right) - l_1 \left( l_1 + 1 \right) \right] + \frac{c \left( c + 1 \right)}{r^2} + \alpha^2 - \frac{2 \alpha \left( a + 1 \right)}{r_1}
+ \frac{2 \left( r_1 - r_2 \cos \theta \right)}{r_1 r^2}c \left( a - \alpha r_1 \right) - \frac{2c r_2}{r_1 r^2} \left( \hat{\nabla}^Y_1 \cdot \hat{r}_2 \right)
\end{equation}

Operating to the left gives

\begin{equation}
(\nabla_1^2)_L = \frac{1}{r_1^2} \left[a^\prime \left( a^\prime + 1 \right) - l_1^\prime \left( l_1^\prime + 1 \right) \right] + \frac{c^\prime \left( c^\prime + 1 \right)}{r^2} + \alpha^{\prime 2} -\frac{2 \alpha^\prime \left( a^\prime + 1 \right)}{r_1}
+ \frac{2 \left( r_1 - r_2 \cos \theta \right)}{r_1 r^2} c^\prime \left( a^\prime - \alpha^\prime r_1 \right) - \frac{2 c^\prime r_2}{r_1 r^2} \left( \hat{\nabla}^{Y^\prime}_1 \cdot \hat{r}_2 \right)
\end{equation}

Now put

\begin{array}{cc}
a_+ = a+a^\prime, & \hat{\nabla}_1^+ = \hat{\nabla}_1^Y + \hat{\nabla}_1^{Y\prime} \\
a_{-} = a-a^\prime, & \hat{\nabla}_1^{-} = \hat{\nabla}_1^Y - \hat{\nabla}_1^{Y\prime}
\end{array}

etc., and substitute <math>a^\prime = a_+ - a</math>, <math>c^\prime = c_+ - c</math>, and <math>\alpha^\prime = \alpha_+ - \alpha</math> in <math>(\nabla^2_1)_L</math>. If <math>a_+</math>, <math>c_+</math> and <math>\alpha_+</math> are held fixed, then the equation

\begin{equation}
(\nabla^2_1)_R = (\nabla^2_1)_L
\end{equation}

must be true for arbitrary <math>a</math>, <math>c</math>, and <math>\alpha</math>. Their coefficients must thus vanish.

This yields the integral relations

\begin{equation}
\label{I}
\frac{\left( r_1 - r_2 \cos \theta \right)}{r_1 r^2} = \frac{1}{c_+} \left( \frac{- \left( a_+ + 1 \right)}{r_1^2} + \frac{\alpha_+}{r_1} \right) \tag{I}
\end{equation}

from the coefficient of <math>a</math> and

\begin{equation}
\label{II}
\frac{ \left( r_1 - r_2 \cos \theta \right) \left( a_+ - \alpha_+ r_1 \right)}{r_1 r^2} = \frac{r_2}{r_1 r^2}\left( \hat{r}_2 \cdot \hat{\nabla}_1^+ \right) - \frac{ \left( c_+ + 1 \right)}{r^2} \tag{II}
\end{equation}

from the coefficient of <math>c</math>. The coefficient of <math>\alpha</math> gives an equation equivalent to (I).

Furthermore, if can be show that (see problem)

\begin{align}
& \sum_\Lambda \int d\tau_r \frac{r^c}{r^2} C_\Lambda \left( \hat{r}_2 \cdot \hat{\nabla}_1^Y \right) P_\Lambda \left( \cos \theta \right) \\
= & \sum_\Lambda \int d\tau_r \frac{r^c}{c r_1 r_2}C_\Lambda \left( 1 \right)P_\Lambda \left( \cos \theta \right) \times \left( \frac{l_1^\prime \left( l_1^\prime + 1 \right) - l_1 \left( l_1 + 1 \right) -\Lambda \left( \Lambda + 1 \right)}{2} \right)
\end{align}

and similarly for <math>\left( \hat{r}_2 \cdot \hat{\nabla}^{Y\prime}_1 \right)</math> with <math>l_1</math> and <math>l_1^\prime</math> interchangeable, then it follows that

\begin{equation}
\frac{r^{c_+}}{r^2} \left( \hat{r}_2 \cdot \hat{\nabla}_1^+ \right) = - \frac{r^{c_+}}{c_+ r_1 r_2} \Lambda \left( \Lambda + 1 \right)
\end{equation}

and

\begin{equation}
\frac{r^{c_+}}{r^2} \left( \hat{r}_2 \cdot \hat{\nabla}_1^- \right) = \frac{r^{c_+}}{c_+ r_1 r_2} \left[ l_1^\prime \left( l_1^\prime + 1 \right) - l_1 \left( l_1 + 1 \right) \right]
\end{equation}

where equality applies after integration and summation over <math>\Lambda</math>.

Thus (II) becomes

\begin{equation}\label{IIa}
\frac{ \left( r_1 - r_2 \cos \theta \right) \left( a_+ - \alpha_+ r_1 \right)}{r_1 r^2} = - \frac{\Lambda \left( \Lambda + 1 \right)}{c_+ r^2_1} - \frac{ \left( c_+ + 1 \right)}{r^2} \tag{IIa}
\end{equation}

===Problem===

Prove the integral relation

\begin{align}
& \sum_\Lambda \int d\tau_r f \left( r_1, r_2 \right) \frac{1}{r} \left( \frac{d}{dr} q(r) \right) C_\Lambda \left( \hat{r}_2 \cdot \hat{\nabla}^Y_1 \right) P_\Lambda \left( \cos \theta \right) \\
= & \sum_\Lambda \int d\tau_r \frac{ f \left( r_1, r_2 \right)}{r_1, r_2} q(r) C_\Lambda \left( 1 \right) P_\Lambda \left( \cos \theta \right) \left(\frac{l_1^\prime \left( l_1^\prime + 1 \right) - l_1 \left( l_1 + 1 \right) - \Lambda \left( \Lambda + 1 \right)}{2}\right)
\end{align}

where the coefficients <math>C_\Lambda \left( 1 \right)</math> are the angular coefficients from the overlap integral

\begin{align}
\int & d\Omega\, \mathcal{Y}^{M*}_{l_1^\prime l_2^\prime L}(\hat{r}_1,\hat{r}_2)\, \mathcal{Y}^{M}_{l_1 l_2 L}(\hat{r}_1,\hat{r}_2)\\
= & \sum_\Lambda C_\Lambda(1)P_\Lambda(\cos\theta_{12}).
\end{align}

Hint: Use the fact that <math>l_1^2</math> is Hermitian so that
<math>
\int d\tau (l_1^2\mathcal{Y}^\prime)^*q(r)\mathcal{Y} = \int d\tau
\mathcal{Y}^{\prime *}l_1^2(q(r)\mathcal{Y})
</math>
with <math>\ \vec{l}_1= \frac{1}{i}\vec{r}_1\times\nabla_1</math>.

It is also useful to use
<math>\begin{eqnarray}\ \
\cos\theta\, P_L(\cos\theta) &=& \frac{1}{2L+1}[L\,P_{L-1}(\cos\theta) + (L+1)P_{L+1}(\cos\theta)]
\\
(\cos^2\theta-1)P_L(\cos\theta) &=&
\frac{L(L-1)}{(2L-1)(2L+1)}P_{L-2}(\cos\theta)\nonumber
-\frac{2(L^2+L-1)}{(2L-1)(2L+3)}P_L(\cos\theta) + \frac{(L+1)(L+2)}{(2L+1)(2L+3)}P_{L+2}(\cos\theta)\nonumber\\
&=& \frac{L(L-1)}{(2L-1)(2L+1)}[P_{L-2}(\cos\theta) - P_L(\cos\theta)]
+ \frac{(L+1)(L+2)}{(2L+1)(2L+3)}[P_{L+2}(\cos\theta) - P_L(\cos\theta)]
\end{eqnarray}</math>

together with a double application of the integral recursion relation
<math>
I_{L+1}\left(\frac{1}{r}\frac{d}{dr}q(r)\right) =
(2L+1)I_L\left(\frac{1}{r_1r_2}q(r)\right)+I_{L-1}\left(\frac{1}{r}\frac{d}{dr}q(r)\right)
</math>

Of course <math>l_1^2\mathcal{Y}^M_{l_1l_2L}(\hat{r}_1,\hat{r}_2)=l_1(l_1+1)\mathcal{Y}^M_{l_1l_2L}(\hat{r}_1,\hat{r}_2)</math>. Begin by expanding
<math>
l_1^2(g\mathcal{Y}) = \mathcal{Y}l_1^2g + g\,l_1^2\mathcal{Y} + 2(\vec{l}_1g)\cdot(\vec{l}_1\mathcal{Y})
</math>, and show that

<math>
\begin{eqnarray}\ \
(\vec{l}_1g)\cdot(\vec{l}_1\mathcal{Y}) & = & \frac{r_1^2}{r}\frac{dg}{dr}\vec{r}_2\cdot \nabla_1\mathcal{Y}\,\,\,\,\mbox{ and}\\
l_1^2g(r) &=& 2r_1r_2\cos\theta\frac{1}{r}\frac{dg}{dr} + r_1^2r_2^2(\cos^2\theta - 1)
\frac{1}{r}\frac{d}{dr}\left(\frac{1}{r}\frac{dg}{dr}\right).
\end{eqnarray}
</math>

The proof amounts to showing that the term <math>l_1^2g(r)</math> can be replaced by
<math>\Lambda(\Lambda+1)g(r)</math> after multiplying by <math>P_\Lambda(\cos\theta)</math>
and integrating by parts with respect to the radial integrations over <math>r_1</math>, <math>r_2</math> and
<math>r \equiv |\vec{r}_1 - \vec{r}_2|</math>. Remember that <math>\cos\theta</math> is just a short-hand notation for the radial function <math>(r_1^2 + r_2^2 - r^2)/(2r_1r_2)</math>.

==General Hermitian Property==

Each combination of terms of the form
\begin{equation}
<f>_R = a^2f_1+af_2+abf_3+b^2f_4+bf_5+af_6\nabla_1^Y+bf_7\nabla^Y_2+f_8(Y)
\end{equation}
can be rewritten
\begin{equation}
<f>_L= (a_+-a)^2f_1+(a_+-a)f_2 + (a_+-a)(b_+-b)f_3+(b_+-b)^2f_4
+(b_+-b)f_5+(a_+-a)f_6\nabla^{Y\prime}_1+(b_+-b)f_7\nabla^{Y\prime}_2+f_8)Y^\prime
\end{equation}
where as usual <math>\nabla_1^Y</math> acts only on the spherical harmonic part of the wave function denoted for short by <math>Y</math>.
Since these must be equal for arbitrary <math>a</math> and <math>b</math>,
\begin{equation}
a_+^2f_1+a_+f_2+_+b_+f_3+b_+f_4+b_+f_5+a_+f_6\nabla^{Y\prime}_1 +
b_+f_7\nabla^{Y\prime}_2+f_8(Y^\prime)-f_8(Y)=0
\end{equation}

Adding the corresponding expression with <math>Y</math> and <math>Y^\prime</math> interchanged
yields
<math>\label{III}
a_+^2f_1+a_+f_2+a_+b_+f_3+b_+^2f_4+b_+f_5+\frac{1}{2}a_+f_6\nabla^+_1+\frac{1}{2}b_+f_7\nabla^+_2=0\tag{III}
</math>

Subtracting gives
<math>\label{IV}
f_8(Y)-f_8(Y^\prime)=-\frac{1}{2}[a_+f_6\nabla^-_1+b_+f_7\nabla_2^-]\tag{IV}
</math>
<math>\label{V}
a[-2a_+f_1-2f_2-b_+f_3-f_6\nabla^+_1]=0\tag{V}
</math>
<math>\label{VI}
b[-2b_+f_4-2f_5-a_+f_3-f_7\nabla^+_2]=0\tag{VI}
</math>

Adding the two forms gives

\begin{eqnarray}
<f>_R+<f>_L=\frac{1}{2}(a_+^2+a_-^2)f_1+a_+f_2+\frac{1}{2}(a_+b_++a_-b_-)f_3\nonumber\\
\ +\frac{1}{2}(b_+^2+b_-^2)f_4+b_+f_5+\frac{1}{2}f_6(a_+\nabla_1^++a_-\nabla^-_1)\nonumber\\
\ +\frac{1}{2}(b_+\nabla_2^++b_-\nabla^-_2)+f_8(Y)+f_8(Y^\prime)\nonumber
\end{eqnarray}

Subtracting <math>\frac{x}{2}\times({\rm I})</math> gives

\begin{eqnarray}\ \
<f>_R+<f>_L = \frac{1}{2}[(1-x)a_+^2 + a_-^2]f_1 +
(1-\frac{x}{2})a_+f_2+\frac{1}{2}[(1-x)a_+b_++a_-b_-]f_3\nonumber\\
+\frac{1}{2}[(1-x)b_+^2+b_-^2]f_4+(1-\frac{x}{2})f_5b_++\frac{1}{2}f_6[(1-\frac{x}{2})a_+\nabla_1^++a_-\nabla^-_1]\nonumber\\
+\frac{1}{2}[(1-\frac{x}{2})b_+\nabla^+_2+b_-\nabla^-_2]+f_8(Y)+f_8(Y^\prime)\nonumber
\end{eqnarray}

If <math>x=1</math>,
\begin{eqnarray}
<f>_R+<f>_L &=&
\frac{1}{2}[a_-^2f_1+a_+f_2+a_-b_-f_3+b_-^2f_4+b_+f_5\nonumber\\
&+&\left(\frac{1}{2}a_+\nabla^+_1+a_-\nabla^-_1\right)f_6+\left(\frac{1}{2}b_+\nabla^+_2+b_-\nabla^-_2\right)f_7]+f_8(Y)+f_8(Y^\prime)\nonumber
\end{eqnarray}

The General Hermitian Property for arbitrary <math>x</math> gives
\begin{eqnarray}\ \
\nabla_1^2=
\frac{1}{4}\{\frac{1}{r_1^2}[(1-x)a_+^2+a_-^2+2(1-\frac{x}{2})a_+-2[l_1(l_1+1)+l_1^\prime(l_1^\prime+1)]\nonumber\\
-\frac{2}{r_1}[(1-x)\alpha_+a_++\alpha_-a_-+2(1-\frac{x}{2})\alpha_+]+(1-x)\alpha_+^2
+ \alpha^2_-]\nonumber\\
+\frac{2(r_1-r_2\cos\theta)}{r_1r^2}[(1-x)(a_+-\alpha_+r_1)c_++(a_--\alpha_-r_2)c_-]\nonumber\\
-\frac{2r_2}{r_1r^2}[(1-\frac{x}{2})c_+\hat{r}_2\cdot\hat{\nabla}_1^++c_-\hat{r}_2\hat{\nabla}_1^-]+\frac{(1-x)c_+^2+c_-^2+2(1-\frac{x}{2})c_+}{r^2}\}\nonumber
\end{eqnarray}.

Use
<math>\begin{eqnarray}\ \
\frac{-2r_2}{r_1r^2}[(1-\frac{x}{2})c_+\hat{r}_2\hat{\nabla}_1^++c_-\hat{r}_2\hat{\nabla}_1^-]=\frac{2}{r_1^2}\left[(1-\frac{x}{2})\Lambda(\Lambda+1)-\frac{c_-}{c_+}[l_1^\prime(l_1^\prime+1)+l_1(l_1+1)]\right]\end{eqnarray}</math>,

<math>\begin{eqnarray}
\frac{2(r_1-r_2\cos\theta)}{r_1r^2}(a_--\alpha_-r_1)c_- = 2\frac{c_-}{c_+}\left[\frac{-1}{r_1^2}[a_-(a_++1)]+\frac{1}{r_1}[a_-\alpha_++\alpha_-(a_++2)]-\alpha_-\alpha_+\right]\nonumber
\end{eqnarray}</math>,

and
<math>\begin{eqnarray}\ \
\frac{2(r_1-r_2\cos\theta)}{r_1r^2}(a_+-\alpha_+r_1)c_+ = 2\left[-\frac{1}{r_1^2}[a_+(a_++1)]+\frac{1}{r_1}[a_+\alpha_++\alpha_+(a_++2)]-\alpha_+^2\right]
\end{eqnarray}</math>.

Substituting into <math>\nabla_1^2</math> gives

\begin{eqnarray}
\nabla^2_1 = \frac{1}{4}\{\frac{1}{r_1^2}[-(1-x)a_+^2 + a_-^2 +xa_+ +
2(1-\frac{x}{2})\Lambda(\Lambda+1)\nonumber\\
-2l_1(l_1+1)(1-\frac{c_-}{c_+}) -
2l_1^\prime(l_1^\prime+1)(1+\frac{c_-}{c_+})-\frac{2c_-a_-}{c_+}(a_++1)]\nonumber\\
- \frac{2}{r_1}\left[-(1-x)\alpha_+(a_++2)+\alpha_-a_-+2(1-\frac{x}{2})\alpha_+
-\frac{c_-}{c_+}[a_-\alpha_++\alpha_-(a_++2)]\right]\nonumber\\
-(1-x)\alpha_+^2+\alpha_-^2-2\frac{c_-}{c_+}\alpha_-\alpha_+ +\frac{1}{r^2}\left[(1-x)c_+^2+c_-^2+2(1-\frac{x}{2})c_+\right]\}\nonumber
\end{eqnarray}

This has the form
<math>
\nabla_1^2 =
\frac{1}{4}\left[\frac{A_1}{r_1^2}+\frac{B_1}{r_1}+\frac{C_1}{r^2}+D_1\right]
</math>
with

<math>\begin{array}{ccc}
A_1=-(1-x)a_+^2+a_-^2
+xa_++2(1-\frac{x}{2})\Lambda(\Lambda+1)-2l_1(l_1+1)(1-\frac{c_-}{c_+})\
-2l_1^\prime(l_1^\prime+1)(1+\frac{c_-}{c_+})-\frac{2c_-a_-}{c_+}(a_++1) \ \
\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \
\end{array}
</math>

<math>
B_1 =
2\left[(1-x)\alpha_+(a_++2)-\alpha_-a_--2(1-\frac{x}{2})\alpha_++\frac{c_-}{c_+}[a_-\alpha_++\alpha_-(a_++2)]\right]
</math>

<math>
C_1 = (1-x)c_+^2+ c_-^2 +2(1-\frac{x}{2})c_+
</math>

<math>
D_1 = -1(1-x)\alpha_+^2 +\alpha_-^2 - 2\frac{c_-}{c_+}\alpha_-\alpha_+
</math>

The complete Hamiltonian is then (in Z-scaled a.u.)

<math>\begin{eqnarray}
H =-\frac{1}{2}\nabla^2_1 -\frac{1}{2}\nabla^2_2 -\frac{1}{r_1} -\frac{1}{r_2} +\frac{Z^{-1}}{r}\nonumber\\
= -\frac{1}{2}\nabla^2_1 -\frac{1}{2}\nabla^2_2 -\frac{1}{r_1} -\frac{Z-1}{Zr_2}+Z^{-1}\left(\frac{1}{r}-\frac{1}{r_2}\right)\nonumber\\
\ =-\frac{1}{8}\left[\frac{A_1}{r_1^2}+\frac{B_1+8}{r_1}+\frac{C_1}{r^2}+D_1+D_2 +\frac{A_2}{r_2^2} + \frac{B_2 + 8(Z-1)/Z}{r_2}+\frac{C_2}{r^2}\right]\nonumber
+ Z^{-1}\left(\frac{1}{r}-\frac{1}{r_2}\right)\nonumber
\end{eqnarray}</math>

The screened hydrogenic energy is
<math>\begin{eqnarray}
E_{SH}=-\frac{1}{2}\left[\frac{1}{n_1^2}+ \left(\frac{Z-1}{Z}\right)^2\frac{1}{n_2^2}\right]
\end{eqnarray}</math>
so that

<math>\begin{eqnarray}
H-E_{SH} &=& -\frac{1}{8}[
\frac{A_1}{r_1^2}+\frac{B_1+8}{r_1}+\frac{C_1}{r^2} + D_1 + D_2
-\frac{4}{n_1^2} -\left(\frac{Z-1}{Z}\right)^2\frac{4}{n_2^2}\nonumber
+\frac{A_2}{r_2}+\frac{B_2+8(Z-1)/Z}{r_2}+\frac{C_2}{r^2}] +
Z^{-1}\left(\frac{1}{r} -\frac{1}{r_2}\right)\nonumber
\end{eqnarray}</math>

==Optimization of Nonlinear Parameters==
- The traditional method of performing Hylleraas calculations is to write the
basis set in the form

<math>
\Psi = \sum_{i,j,k}c_{i,j,k}\varphi_{i,j,k}(\alpha,\beta)\pm exchange
</math>

with

<math>
\varphi_{i,j,k}(\alpha,\beta) = r_1^ir_2^jr_{12}^ke^{-\alpha r_1 - \beta
r_2}\mathcal{Y}^M_{l_1l_2L}.
</math>

The usual procedure is to set <math>\alpha = Z</math> so that it represents the inner
<math>1s</math> electron, and then to vary <math>\beta</math> so as to minimize the energy.

[[File:OptNon.jpg|500px]]

Since <math>\beta</math> appears in <math>\Psi</math> as a non-linear parameter, the entire
calculation must be repeated for each value of <math>\beta</math>. However, the minimum
becomes progressively smaller as the basis set is enlarged.

[[File:OptNon02.jpg|500px]]

*Difficulties

1. If the basis set is constructed so that <math> i +j +j \leq N</math>, the the
number of terms is
<math>
(N+1)(N+2)(N+3)/6
</math>
N=14 already gives 680 terms and an accuracy of about 1 part in <math>10^{10}</math> for
low-lying states. A substantial improvement in accuracy would require much
larger basis sets, together with multiple precision arithmetic to avoid loss
of significant figures when high powers are included.

2. The accuracy rapidly deteriorates as one goes to more highly excited
states - about 1 significant figure is lost each time the principle quantum
number is increased.

*Cure

We have found that writing basis sets in the form
\begin{eqnarray}
\Psi &=& \sum_{i,j,k}\left(c^{(1)}_{i,j,k}\varphi_{ijk}(\alpha_1\beta_1) +
c_{ijk}^{(2)}\varphi_{ijk}(\alpha_2\beta_2)\right) \pm exchange \nonumber\\
&=& \psi({\bf r}_1,{\bf r}_2) \pm \psi({\bf r}_2,{\bf r}_1)\nonumber
\end{eqnarray}
so that each combination of powers is included twice with different nonlinear
parameters gives a dramatic improvement in accuracy for basis sets of about
the same total size.

However, the optimization of the nonlinear parameters is now much more
difficult, and an automated procedure is needed.

If
<math>
E = \frac{<\Psi|H|\Psi>}{<\Psi|\Psi>}</math> and we assume <math><\Psi|\Psi> = 1</math>, then

<math>
\frac{\partial E}{\partial\alpha_t} = -2 <\Psi|H-E|r_1\psi({\bf r}_1,{\bf
r}_2;\alpha_t)\pm r_21\psi({\bf r}_2,{\bf
r}_1;\alpha_t)>
</math>
where
<math>
\psi({\bf r}_1,{\bf r}_2;\alpha_t) =
\sum_{i,j,k}c_{ijk}^{(t)}\varphi_{ijk}(\alpha_t\beta_t)
</math>

*Problem

1. Prove the above.

2. Prove that there is no contribution from the implicit dependence of
<math>c_{ijk}^{(t)}</math> on <math>\alpha_t</math> if the linear parameters have been optimized.

==The Screened Hydrogenic Term==

If the Hamiltonian is written in the form (in Z-scaled a.u.)
<math>\begin{eqnarray}\
H = H_0({\bf r}_1,Z) + H_0({\bf r}_2,Z-1) +
Z^{-1}\left(\frac{1}{r}-\frac{1}{r_2}\right)
\end{eqnarray}</math>
with
<math>\
H_0({\bf r}_1,Z) = -\frac{1}{2}\nabla_1^2 - \frac{1}{r_1}\nonumber
</math> and
<math>
H_0({\bf r}_2,Z-1) = -\frac{1}{2}\nabla_2^2 - \frac{Z-1}Z{r_2}\nonumber
</math>,
then the eigenvectors of <math>H_0({\bf r},Z) +H_0({\bf r}_2,Z-1)</math> are products of
hydrogenic orbitals
<math>
\Psi_0=\psi_0(1s,Z)\psi_0(nl,Z-1)
</math>
and the eigenvalue is
<math>\begin{eqnarray}
E_{SH} = \left[-\frac{1}{2} - \left(\frac{Z-1}{Z}\right)^2\frac{1}{2n^2}\right]Z^2 a.u.
\end{eqnarray}</math>
called the screened hydrogenic eigenvalue. From highly excited states,
<math>E_{SH}</math> and <math>\Psi_0</math>are already excellent approximations.

For example, for the 1s8d states, the energies are

<math>\begin{eqnarray}\\
E(1s8d^1D)&=& -2.007\,816\,512\,563\,81 {\rm\ a.u.}\nonumber\\
E(1s8d^3D)&=& - 2.007\,817\,934\,711\,71 {\rm\ a.u.}\nonumber\\
E_{SH} &=& -2.007\,812\,5 {\rm\ a.u.} \nonumber
\end{eqnarray}</math>

It is therefore advantageous to include the screened hydrogenic terms in the
basis set so that the complete trial function becomes
<math>
\Psi = c_0\Psi_0 +
\sum_{ijk}\left[c_{ijk}^{(1)}\varphi_{ijk}(\alpha_1,\beta_1)+c_{ijk}^{(2)}\varphi(\alpha_2,\beta_2)\right]\pm
{\rm exchange}
</math>

Also, the variational principal can be re-expressed in the form
<math>\begin{eqnarray}
E= E_{SH}+\frac{<\Psi|H-E_{SH}|\Psi>}{<\Psi|\Psi>}
\end{eqnarray}</math>
so that the <math>E_{SH}</math> term can be cancelled analytically from the matrix
elements.

For example
<math>
<\Psi_0|H-E_{SH}|\Psi_0> = 0
</math>

and
<math>
<\varphi_{ijk}|H-E_{SH}|\Psi_0> =
Z^{-1}<\varphi_{ijk}|\frac{1}{r}-\frac{1}{r_2}|\Psi_0>
</math>

Recall that
<math>\begin{eqnarray}\ \
I_0(a,b,c) = \frac{2}{c+2}\sum^{[c+1]/2}_{i=0}\left(\begin{array}{c}
c+2\\
2i+1
\end{array}\right) \left[f(p,q;\beta) + f(p^\prime,q^\prime;\alpha)\right]
\end{eqnarray}</math>
where

<math>\begin{eqnarray}
f(p,q;x) =
\frac{q!}{x^{q+1}(\alpha+\beta)^{p+1}}\sum^p_{j=0}\frac{(p+j)!}{j!}\left(\frac{x}{\alpha+\beta}\right)^j
\end{eqnarray}</math>

<math>
\begin{array}{cc}
p=a+2i+2 & p^\prime = b+2i+2\\
q= b+c-2i+2 & q^\prime =a+c-2i+2
\end{array}
</math>

If <math>\beta\ll\alpha</math>, the <math>f(p,q;\beta)\gg f(p^\prime,q^\prime;\alpha)</math> and the
<math>i=0</math> term is the dominant contribution to <math>f(p,q;\beta)</math>. However, since
this term depends only on the sum of powers <math>b+c</math> for <math>r_2</math>, it cancels exactly
from the matrix element of
<math>
\frac{1}{r_{12}} - \frac{1}{r_2}
</math>
and can therefore
be omitted in the calculations of integrands, thereby saving many significant
figures. This is especially valuable when <math>b+c</math> is large and <math>a</math> is small.

==Small Corrections==
*Mass Polarization
If the nuclear mass is not taken to be infinite, then the Hamiltonian is
<math>\begin{eqnarray}\ \
H = \frac{P_N^2}{2M} + \sum_{i=1}\frac{p_i^2}{2m} + V
\end{eqnarray}</math>

[[File:SmallCor.jpg|500px]]

where <math>M =M_A-nm</math> is the nuclear mass and <math>m</math> is the electron mass.

Change variables to the C of M
<math>\begin{eqnarray}
{\bf R} = \frac{1}{M+nm}\left[M{\bf r}_N + m({\bf r}_1 + {\bf r}_2 +
\ldots)\right]
\end{eqnarray}</math>

and relative variables

<math>
{\bf s}_i = {\bf r}_i - {\bf r}_N.
</math>

Then

<math>\begin{eqnarray}\ \ \ \ \
H = \frac{1}{2(M+nm)}{\bf p}_R^2+\frac{1}{2m}\sum_{i=1}^n{\bf p}_{s_i}^2
+\frac{1}{2M}\sum_{i,k}{\bf p}_{s_i}\cdot {\bf p}_{s_k} + V({\bf s}_1, {\bf
s}_2, \ldots)\nonumber
= \frac{1}{M+nm}{\bf p}_R^2+\frac{1}{2\mu}\sum_{i=1}^n{\bf
p}_{s_i}^2+\frac{1}{2M}\sum_{i\neq k}{\bf p}_{s_i}\cdot {\bf p}_{s_k}+V({\bf s}_1, {\bf
s}_2, \ldots)\nonumber
\end{eqnarray}</math>

where
<math>\begin{eqnarray}
\mu = \left(\frac{1}{M} + \frac{1}{m}\right)^{-1}
\end{eqnarray}</math>

If <math>n=2</math>, the Schrödinger equation is
<math>\begin{eqnarray}\ \ \
\left[ -\frac{\hbar^2}{2\mu}(\nabla^2_{s_1}+ \nabla^2_{s_2}) -
\frac{\hbar^2}{M}\nabla_{s_1}\nabla_{s_2}-\frac{Ze^2}{s_1}-\frac{Ze^2}{s_2}+\frac{e^2}{s_{12}}\right]\psi
= E\psi
\end{eqnarray}</math>.

Define the reduced mass Bohr radius
<math>\begin{eqnarray}
a_\mu = \frac{\hbar^2}{\mu e^2} = \frac{m}{\mu}a_0
\end{eqnarray}</math>
and

<math>\begin{eqnarray}
\rho_i = \frac{Zs_i}{a_\mu}\nonumber\\
\varepsilon = \frac{E}{Z^2(e^2/a_\mu)}\nonumber
\end{eqnarray}</math>

The Schrödinger equation then becomes
<math>\begin{eqnarray}\ \ \
\left[
-\frac{1}{2}(\nabla^2_{\rho_1}+\nabla^2_{\rho_2})-\frac{\mu}{M}\vec{\nabla}_{\rho_1}\cdot\vec{\nabla}_{\rho_2} - \frac{1}{\rho_1}-\frac{1}{\rho_2}+\frac{Z^{-1}}{\rho_{12}}\right]\psi = \varepsilon \psi
\end{eqnarray}</math>

The effects of finite mass are thus

1. A reduced mass correction to all energy levels of <math>(e^2/a_\mu)/(e^2/a_0)
= \mu/m</math>, i.e.
<math>\begin{eqnarray}
E_M=\frac{\mu}{m}E_\infty
\end{eqnarray}</math>

2. A specific mass shift given in first order perturbation theory by
<math>\begin{eqnarray}
\Delta\varepsilon =
-\frac{\mu}{M}<\psi|\nabla_{\rho_1}\cdot\nabla_{\rho_2}|\psi>
\end{eqnarray}</math>
or

<math>\begin{eqnarray}
\Delta\varepsilon =
-\frac{\mu}{M}<\psi|\nabla_{\rho_1}\cdot\nabla_{\rho_2}|\psi>\frac{e^2}{a_\mu}\nonumber\\
= -\frac{\mu^2}{mM}<\psi|\nabla_{\rho_1}\cdot\nabla_{\rho_2}|\psi>\frac{e^2}{a_0}\nonumber
\end{eqnarray}</math>

Since <math>\nabla_{\rho_1}\cdot\nabla_{\rho_2}</math> has the same angular properties as
<math>{\bf\rho}_1 \cdot {\bf\rho}_2</math>, the operator is like the product of two
dipole operators. For product type wave functions of the form
$
\psi = \psi(1s)\psi(nl) \pm \mbox{exchange},
$
the matrix element vanishes for all but P-states.

=Radiative Transitions=

==Transition Integrals and the Line Strength==

For an electric dipole transition <math>\gamma'\,L'\,S'\,J'\,M' \rightarrow \gamma\,L\,S\,J\,M</math>, the line strength is defined by
\begin{equation}
S = \sum_{M',\mu,M}|\langle\gamma'\,L'\,S'\,J'\,M'|r_\mu|\gamma\,L\,S\,J\,M\rangle|^2
\end{equation}
Using the Wigner-Eckart theorem, this is related to the reduced matrix element by
\begin{eqnarray}
S(J',J) = |\langle\gamma'\,L'\,S'\,J'||r||\gamma\,L\,S\,J\,M\rangle|^2
\sum_{M',\mu,M}\left|\left(\matrix{J' & 1 & J \cr
-M'& \mu &M \cr}\right)\right|^2
\end{eqnarray}
Using the sum rule for the 3-<math>j</math> symbols
\begin{equation}
\sum_{M',\mu,M}\left|\left(\matrix{J' & 1 & J \cr
-M'& \mu &M \cr}\right)\right|^2 = 1
\end{equation}
we see that the line strength is identical to the reduced matrix element
\begin{equation}
S(J',J) = |\langle\gamma'\,L'\,S'\,J'||r||\gamma\,L\,S\,J\,M\rangle|^2
\end{equation}
For a nonrelativistic calculation in <math>LS</math> coupling, <math>S = S'</math> and we can next strip out the dependence on <math>J</math> and <math>J'</math> by use of Eq.\ (7.1.7) of Edmonds { Angular Momentum in Quantum Mechanics} with <math>k = 1</math> to obtain
\begin{equation}
S(J',J) = (-1)^{L'+S+J+1}[(2J+1)(2J'+1)]^{1/2}\left\{
\matrix{L' & J'& S\cr
J & L & 1\cr}\right\}\langle\gamma' L'|| r || \gamma L\rangle
\end{equation}
where <math>\langle\gamma' L'|| r || \gamma L\rangle</math> is the reduced matrix element if spin is neglected. Note that if spin is neglected, then the line-strength is simply
\begin{equation}
S(L',L) = |\langle\gamma' L'|| r || \gamma L\rangle|^2
\end{equation}
To calculate the reduced matrix element, use a particular example along with the Wigner-Eckart theorem
such as
\begin{equation}
\langle\gamma'\, L'|| r || \gamma\, L\rangle = (-1)^{L'}\langle\gamma'\, L'\, 0| z | \gamma\, L\, 0\rangle{\Large /}\left(\matrix{L' & 1 & L \cr
0 & 0 & 0 \cr}\right)
\end{equation}

===Correlated Two-electron Integrals===
For the case of two-electron integrals, <math>z = z_1 + z_2</math> and <math>r = r_1 + r_2</math>. As discussed in G.W.F. Drake, Phys. Rev. A 18, 820 (1978), the general integral in Hylleraas coordinates for the integral over the <math>z_1</math> term can be written in the form
\begin{equation}
\label{int1}
\langle\gamma'\, L'\, 0| z_1 | \gamma\, L\, 0\rangle = \sum_\Lambda C_\Lambda I_\Lambda(\gamma',\gamma)
\end{equation}
where <math>I_\Lambda(\gamma',\gamma)</math> stands for the purely radial integral
\begin{equation}
\label{radial}
I_\Lambda(\gamma',\gamma) = \int_0^\infty r_1\,dr_1\int_0^\infty r_2\,dr_2\int_{|r_1-r_2|}^{r_1+r_2}r_{12}\, dr_{12}
R_{\gamma'}^*\,r_1\, R_\gamma P_\Lambda(\cos\theta_{12})
\end{equation}
and <math>\cos\theta_{12} = (r_1^2 + r_2^2 - r_{12}^2)/(2r_1r_2)</math> is a short-hand notation for a radial function of <math>r_1</math>, <math>r_2</math> and <math>r_{12}</math>. Also, <math>P_\Lambda(\cos\theta_{12})</math> is a Legendre polynomial,
and, <math>R_\gamma = R_\gamma(r_1,r_2,r_{12})</math> is the radial part of the Hylleraas wave function for state <math>\gamma</math> such
that the total wave function is <math>\psi({\bf r}_1,{\bf r}_2) = R_\gamma(r_1,r_2,r_{12}){\cal Y}_{l_1,l_2,L}^M(\hat{\bf r}_1,\hat{\bf r}_2)\pm {\rm exchange}</math>, normalized to unity, and <math>{\cal Y}_{l_1,l_2,L}^M(\hat{\bf r}_1,\hat{\bf r}_2)</math> is a vector coupled product of spherical harmonics. The coefficents <math>C_\Lambda</math> in Eq.\ (\ref{int1}) can be calculated from Eq.\ (32) of the above 1978 paper. They need to be calculated separately for he direct-direct and the direct-exchange contributions. For the reduced matrix element of <math>z_1</math>, the leading 3-<math>j</math> symbol can be dropped, leaving (for the case <math>k = 1</math>)
\begin{eqnarray}
C_\Lambda^{\rm reduced} &=& \frac{(-1)^{l_2'+L'+L+1}}{2} (l_1,l_1',l_2,l_2',L,L')^{1/2}
\sum_{\lambda={\rm max}(l_1-1,0)}^{l_1+1} (\lambda,\Lambda,1)
\left(\matrix{l_1 & 1 & \lambda \cr
0 & 0 & 0 \cr}\right)
\left(\matrix{l_1'& \lambda & \Lambda \cr
0 & 0 & 0 \cr}\right)\nonumber\\
&\times&\left(\matrix{l_2' & 1_2 & \Lambda \cr
0 & 0 & 0 \cr}\right)
\left\{\matrix{L' & \lambda & l_2 \cr
\Lambda& l_2' & l_1' \cr}\right\}
\left\{\matrix{L' & \lambda & l_2 \cr
l_1 & L & 1 \cr}\right\}
\end{eqnarray}
where the notation <math>(a,b,c,\cdots)</math> means <math>(2a+1)(2b+1)(2c+1)\cdots</math>.
A similar expression can be obtained for the angular coefficients for the <math>z_2</math> term by interchanging the subscripts 1 and 2 throughout and multiplying by <math>(-1)^{L+L'+1}</math>.

===Example===
As an example, for the <math>1s^2\;^1S - 1s2p\;^1P</math> transition of helium, the only nonvanishing <math>C_\Lambda</math> angular coefficients are <math>C_1 = 1/2</math> for the direct-direct term (i.e. <math>P_1(\cos\theta_{12}) = \cos\theta_{12}</math> is included in the radial integral (\ref{radial})) and <math>C_0 = 1/2</math> for the direct-exchange term (i.e. <math>P_0(\cos\theta_{12}) = 1</math> is included in the radial integral (\ref{radial})). The results for infinite nuclear mass are
\begin{equation}
S(1\;^1S-2\;^1P) = (0.781\,127\,158\,667)^2
\end{equation}
\begin{equation}f(1\;^1S-2\;^1P) = -0.276\,164\,703\,736
\end{equation}
where
\begin{equation}
f(1\;^1S-2\;^1P) = \frac{2}{3}[E(1\;^1S) - E(2\;^1P)]S(1\;^1S-2\;^1P)
\end{equation}
is the absorption oscillator strength.

===Test===
This is a test edit

Theory Notes

2020-12-14T04:29:21Z

Drake: /* Example */

<center>
<H1>Notes on solving the Schrödinger equation in Hylleraas coordinates for heliumlike atoms
Gordon W.F. Drake
Department of Physics, University of Windsor
Windsor, Ontario, Canada N9B 3P4
(Transcribed from hand-written notes and edited by Lauren Moffatt. Last revised November 22, 2017 to correct a missing subscript in the expression for $B_1$ in the Hamiltonian in Sect. 11 General Hermitian Property.)
</H1>
</center>

==Introduction==
These notes describe some of the technical details involved in solving the Schrödinger equation for a heliumlike atom of nuclear charge Z in correlated Hylleraas coordinates. The standard for computational purposes is to express all quantities in a form that is valid for electronic states of arbitrary angular momentum, and to express matrix elements of the Hamiltonian in a manifestly Hermitian symmetrized form. Further information and derivations of the mathematical techniques for angular integrals and radial recursion relations can be found in the paper
G.W.F. Drake, Phys. Rev. A <math>{\bf 18}</math>, 820 (1978).

The starting point is the two-electron Schrödinger equation for infinite nuclear mass
\begin{equation} \left[-\frac{\hbar^2}{2m}(\nabla^2_1 +\nabla^2_2) - \frac{Ze^2}{r_1} -
\frac{Ze^2}{r_2}+\frac{e^2}{r_{12}} \right]\psi = E\psi\nonumber
\end{equation}
where <math>m</math> is the electron mass, and <math>r_{12} = |{\bf r}_1 - {\bf r}_2|</math> (see diagram below).

[[File:Pic1.jpg|250px]]

Begin by rescaling distances and energies so that the Schrödinger equation can be expressed in a dimensionless form. The dimensionless Z-scaled distance is defined by <math>\begin{eqnarray} \rho = \frac{Zr}{a_0} \end{eqnarray} </math> where <math> \begin{eqnarray} a_0 = \frac{\hbar^2}{me^2} \end{eqnarray} </math> is the atomic unit (a.u.) of distance equal to the Bohr radius <math>0.529\, 177\, 210\, 92(17) \times 10^{-10}</math> m. Then

\begin{equation}[-\frac{\hbar^2}{2m}Z^2(\frac{me^2}{\hbar^2})^2(\nabla^2_{\rho_1}+\nabla^2_{\rho_2})
- Z^2\frac{e^2}{a_0}\rho^{-1}_1 - Z^2\frac{e^2}{a_0}\rho^{-1}_2 +
\frac{e^2}{a_0}Z\rho^{-1}_{12}]\psi= E\psi\nonumber\end{equation}

But <math>\begin{eqnarray} \frac{\hbar^2}{m}(\frac{me^2}{\hbar^2})^2 = \frac{e^2}{a_0} \end{eqnarray} </math> is the hartree atomic unit of of energy <math>(E_h = 27.211\,385\,05(60)</math> eV, or equivalently <math>E_h/(hc) = 2194.746\,313\,708(11)</math> m<math>^{-1})</math>. Therefore, after multiplying through by <math>a_0/(Ze)^2</math>, the problem to be solved in Z-scaled dimensionless units becomes

\begin{equation} \left[-\frac{1}{2}(\nabla^2_{\rho_1}+\nabla^2_{\rho_2}) -
\frac{1}{\rho_1} - \frac{1}{\rho_2} + \frac{Z^{-1}}{\rho_{12}}\right]\psi =
\varepsilon\psi\nonumber \end{equation}

where <math>\begin{eqnarray}\varepsilon = \frac{Ea_0}{(Ze)^2} \end{eqnarray}</math> is the energy in Z-scaled atomic units. For convenience, rewrite this in the conventional form
\begin{equation}
H\psi = \varepsilon\psi\nonumber \end{equation}
where (using <math>r</math> in place of <math>\rho</math> for the Z-scaled distance)
\begin{equation}
H = -\frac{1}{2}(\nabla^2_1+\nabla^2_2) - \frac{1}{r_1}-\frac{1}{r_2} + \frac{Z^{-1}}{r_{12}}
\nonumber\end{equation}
is the atomic Hamiltonian for infinite nuclear mass.

==The Hartree Fock Method==

For purposes of comparison, and to define the correlation energy, assume that <math>\psi({\bf r}_1,{\bf r}_2)</math> can be written in the separable product form
\begin{equation}
\psi({\bf r}_1,{\bf r}_2) = \frac{1}{\sqrt{2}} \left[ u_1(r_1) u_2(r_2) \pm u_2(r_1) u_1(r_2) \right] \nonumber
\end{equation}
for the <math>1s^2\ ^1\!S</math> ground state. Because of the <math>\frac{1}{r_{12}}</math> term in the Schrödinger equation, the exact solution connot be expressed in this form as a separable product. However, the Hartree-Fock approximation corresponds to finding the best possible solution to the Schrödinger equation
\begin{equation}
H\psi({\bf r}_1,{\bf r}_2) = E\psi({\bf r}_1,{\bf r}_2)\nonumber
\end{equation}
that can nevertheless be expressed in this separable product form, where as before
\begin{equation}
H = -\frac{1}{2}(\nabla^2_1+\nabla^2_2) - \frac{1}{r_1}- \frac{1}{r_2} + \frac{Z^{-1}}{r_{12}}\nonumber
\end{equation}
is the full two-electron Hamiltonian. To find the best solution, substitute into <math>\left\langle\psi \lvert H-E \rvert \psi\right\rangle</math> and require this expression to be stationary
with respect to arbitrary infinitesimal variations <math>\delta u_1</math> and <math>\delta
u_2</math> in <math>u_1</math> and <math>u_2</math> respectively; i.e.

\begin{equation}
\frac{1}{2} \left\langle \delta u_1(r_1)u_2(r_2) \pm u_2(r_1)\delta
u_1(r_2) \lvert H-E \rvert u_1(r_1) u_2(r_2)\pm u_2(r_1) u_1(r_2)\right\rangle\nonumber
\end{equation}

\begin{equation}
=\int\delta u_1(r_1) d{\bf r}_1 \left\{ \int d{\bf r}_2 u_2(r_2) \left( H - E \right) \left[ u_1(r_1)u_2(r_2) \pm u_2(r_1) u_1(r_2) \right] \right\}\nonumber
\end{equation}

\begin{equation}
= 0 \nonumber
\end{equation}

for arbitrary <math> \delta u_1(r_1).\nonumber </math> Therefore <math>\{\int d{\bf r}_2 \ldots \} = 0</math>.

Similarly, the coefficient of <math>\delta u_2</math> would give

\begin{equation}
\int d{\bf r}_1 u_1(r_1) \left( H-E \right) \left[ u_1(r_1) u_2(r_2) \pm
u_2(r_1) u_1(r_2) \right] = 0\nonumber
\end{equation}

Define

\begin{equation}
I_{12} = I_{21} = \int d{\bf r}\, u_1(r)u_2(r), \nonumber
\end{equation}

\begin{equation}
H_{ij} = \int d{\bf r}\, u_i(r)(-\frac{1}{2}\nabla - \frac{1}{r})u_j(r), \nonumber
\end{equation}

\begin{equation}
G_{ij}(r) = \int d{\bf r}^\prime u_i(r^\prime)\frac{1}{|{\bf r} - {\bf r}\prime|}u_j(r^\prime)\nonumber
\end{equation}

Then the above equations become the pair of integro-differential equations

\begin{equation}
[ H_0 - E + H_{22}+G_{22}(r)]u_1(r) = \mp [ I_{12}(H_0-E) + H_{12}+G_{12}(r)]u_2(r)\nonumber
\end{equation}

\begin{equation}
[H_0-E+H_{11}+G_{11}(r)]u_2(r) = \mp [I_{12}(H_0-E) + H_{12}+G_{12}(r)]u_1(r)\nonumber
\end{equation}

These must be solved self-consistently for the "constants" <math>I_{12}</math> and
<math>H_{ij}</math> and the function <math>G_{ij}(r)</math>.

The Hartree Fock energy is <math>E \simeq -2.87\ldots </math>a.u. while the exact energy is <math>E = -2.903724\ldots </math>a.u.

The difference is called the "correlation energy" because it arises from the
way in which the motion of one electron is correlated to the other. The
Hartree Fock equations only describe how one electron moves in the average field
provided by the other (mean-field theory).

==Configuration Interaction==
Expand
\begin{equation}
\psi({\bf r}_1,{\bf r}_2)= C_0u^{(s)}_1(r_1)u^{(s)}_1(r_2) + C_1u^{(P)}_1({\bf r}_1)u^{(P)}_1({\bf r}_2){\cal Y}^0_{1,1,0}(\hat{\bf r}_1, \hat{\bf r}_2)+C_2u^{(d)}_1({\bf r}_1)u^{(d)}_2({\bf r}_2){\cal Y}^0_{2,2,0}(\hat{\bf r}_1, \hat{\bf r}_2)+... \pm {\rm\ exchange}
\end{equation}
where
\begin{equation}
{\cal Y}^M_{l_1,l_2,L}(\hat{\bf r}_1, \hat{\bf r}_2)=\sum_{m_1,m_2}Y^{m_1}_{l_1}({\bf r}_1)Y^{m_2}_{l_2}({\bf r}_2)\times <l_1l_2m_1m_2\mid LM>
\end{equation}

This works, but is slowly convergent, and very laborious. The best CI calculations are accurate to <math> ~10^{-7}</math> a.u.

==Hylleraas Coordinates==
[E.A. Hylleraas, Z. Phys. <math>{\bf 48}, 469(1928)</math> and <math>{\bf 54}, 347(1929)</math>]
suggested using the co-ordinates <math>r_1</math>, <math>r_2</math> and <math>r_{12}</math> or equivalently

\begin{eqnarray}
s &=& r_1 + r_2, \nonumber\\
t &=& r_1-r_2, \nonumber\\
u &=& r_{12}\nonumber
\end{eqnarray}

and writing the trial functions in the form

\begin{equation}
\Psi({\bf r}_1,{\bf r}_2) = \sum^{i+j+k\leq
N}_{i,j,k}c_{i,j,k}r_1^{i+l_1}r_2^{j+l_2}r_{12}^ke^{-\alpha r_1 - \beta r_2}
\mathcal{Y}^M_{l_1,l_2,L}(\hat{r}_1,\hat{r}_2)\pm \text{exchange}\nonumber
\end{equation}

Diagonalizing H in this non-orthogonal basis set is equivalent to solving
<math> \begin{eqnarray}
\frac{\partial E}{\partial c_{i,j,k}} = 0\nonumber
\end{eqnarray} </math>
for fixed <math>\alpha</math> and <math>\beta</math>.

The diagonalization must be repeated for different values of <math>\alpha</math> and
<math>\beta</math> in order to optimize the non-linear parameters.

==Completeness==
The completeness of the above basis set can be shown by first writing
\begin{equation}
r_{12}^2 = r_1^2 + r_2^2 - 2r_1r_2\cos\theta_{12}
\end{equation}
and, from the spherical harmonic addition theorem,
\begin{equation}
\cos(\theta_{12})=\frac{4\pi}{3}\sum^1_{m=-1}Y^{m*}_1(\theta_1,\varphi_1)Y^m_1(\theta_2,\varphi_2)
\end{equation}
Consider first the S-states. The <math>r_{12}^0</math> terms are like the ss terms in a CI calculation. The
<math>r_{12}^2</math> terms bring in p-p type contributions, and the higher powers bring
in d-d, f-f etc. type terms. In general

\begin{equation}
P_l(\cos\theta_{12}) = \frac{4\pi}{2l+1}\sum^l_{m=-l}{Y^{m}_l}^*(\theta_1,\varphi_1)Y^m_l(\theta_2,\varphi_2)\nonumber
\end{equation}

For P-states, one would have similarly

\begin{array}{lr}
r_{12}^0 & (sp)P\nonumber\\
r_{12}^2 & (pd)P\nonumber\\
r_{12}^4 & (df)P\nonumber\\
\vdots & \vdots\nonumber
\end{array}

For D-states

\begin{array}{lr}
r_{12}^0 & (sd)D & (pp^\prime)D\nonumber\\
r_{12}^2 & (pf)D & (dd^\prime)D\nonumber\\
r_{12}^4 & (dg)D & (ff^\prime)D\nonumber\\
\vdots & \vdots & \vdots\nonumber
\end{array}

In this case, since there are two "lowest-order" couplings to form a
D-state, both must be present in the basis set, i.e.

\begin{equation}
\Psi(r_2,r_2) = \sum c_{ijk}r_1^ir_2^{j+2}r_{12}^ke^{-\alpha r_1-\beta
r_2}\mathcal{Y}^M_{022}(\hat{r}_1,\hat{r}_2)
+\sum d_{ijk}r_1^{i+1}r_2^{j+1}r_{12^k}e^{-\alpha^\prime r_1 - \beta^\prime
r_2}\mathcal{Y}^M_{112}(\hat{r}_1,\hat{r}_2)\nonumber
\end{equation}

For F-states, one would need <math>(sf)F</math> and <math>(pd)F</math> terms.

For G-states, one would need <math>(sg)G</math>, <math>(pf)G</math> and <math>(dd^\prime)G</math> terms.

Completeness of the radial functions can be proven by considering the
Sturm-Liouville problem

\begin{equation}
\left(-\frac{1}{2}\nabla^2-\frac{\lambda}{r_s}-E\right)\psi({\bf r}) = 0\nonumber
\end{equation}

or

\begin{equation}
\left(-\frac{1}{2}\frac{1}{r^2}\frac{\partial}{\partial r}\left({r^2}\frac{\partial}{\partial r}\right) -
\frac{l\left(l+1\right)}{2r^2} - \frac{\lambda}{r} - E\right)u(r) = 0.\nonumber
\end{equation}

For fixed E and variable <math>\lambda</math> (nuclear charge).

The eigenvalues are <math>\lambda_n = (E/E_n)^{1/2}</math>, where <math>E_n =- \frac{1}{2n^2}</math>

[[File:Pic2.jpg|500px]]

and the eigenfunctions are

\begin{equation}
u_{nl}(r) =
\frac{1}{(2l+1)!}\left(\frac{(n+l)!}{(n-l-1)2!}\right)^{1/2}(2\alpha)^{3/2}e^{-\alpha
r}\nonumber
\times \ \left( 2 \alpha r \right)^l\,_{1}F_{1}\left(-n+l+1,2l+2; 2 \alpha r \right)
\end{equation}

with <math>\alpha = (-2E)^{1/2}</math> and <math>n\geq l+1</math>. The confluent hypergeometric function <math> _1F_1(a,b,;z)</math> then denotes a finite polynomial since <math>a=-n+l+1</math> is a negative integer or zero.

Unlike the hydrogen spectrum, which has both a discrete part for <math>E<0</math> and a
continuous part for <math>E>0</math>, this forms an entirely discrete set of finite
polynomials, called Sturmian functions. They are orthogonal with respect to
the potential, i.e.

\begin{equation}
\int^\infty_0 r^2dr \left( u_{n^\prime l}(r)\frac{1}{r}u_{nl}(r) \right) = \delta_{n,n^\prime}\nonumber
\end{equation}

Since they become complete in the limit <math>n\rightarrow\infty</math>, this assures the
completeness of the variational basis set.

[See also B Klahn and W.A. Bingel Theo. Chim. Acta (Berlin) 44, 9 and 27
(1977)].

==Solutions of the Eigenvalue Problem==

For convenience, write

\[
\Psi({\bf r}_1,{\bf r}_2) = \sum^N_{m=1}c_m\varphi_m\nonumber
\]

where <math>m</math> represents the <math>m</math>'th combination of <math>i,j,k</math> and

\begin{equation}
\varphi_{ijk}=r_1^ir_2^jr_{12}^ke^{-\alpha r_1 - \beta
r_2}\mathcal{Y}^M_{l_1,l_2,L}(\hat{r}_1,\hat{r}_2) \pm \mbox{exchange}\nonumber
\end{equation}

\begin{align}

&
\left( \begin{array}{lr}
\cos(\theta) & \sin(\theta)\\
-\sin(\theta) & \cos(\theta)
\end{array} \right)
\left( \begin{array}{lr}
H_{11} & H_{12} \\
H_{12} & H_{22}
\end{array}\right)
\left( \begin{array}{lr}
\cos(\theta) & -\sin(\theta)\\
\sin(\theta) & \cos(\theta)
\end{array} \right)
\\[5pt]
= & \left(\begin{array}{lr}
cH_{11}+sH_{12} & cH_{12} + sH_{22}\\
-sH_{11} + cH_{12} & -sH_{12} + cH_{22}
\end{array}\right)
\left( \begin{array}{cc}
c & -s \\
s & c
\end{array}\right)
\\[5pt]
= & \left(\begin{array}{lr}
c^2H_{11}+s^2H_{22} + 2csH_{12} & (c^2-s^2)H_{12}+cs(H_{22}-H_{11})\\
(c^2-s^2)H_{12}+cs(H_{22}-H_{11}) & s^2H_{11}+c^2H_{22}-2csH_{12}
\end{array}\right)
\end{align}

Therefore

\begin{equation}
\left[\cos^2(\theta)-\sin^2(\theta)\right]H_{12} =
\cos(\theta)\sin(\theta)(H_{11}-H_{22})\nonumber
\end{equation}

and

\begin{equation}
\tan(2\theta) = \frac{2H_{12}}{H_{11}-H_{22}}\nonumber
\end{equation}

i.e.

\begin{equation}
\cos(\theta)=\left(\frac{r+\omega}{2r}\right)^{1/2}\nonumber
\end{equation}

\begin{equation}
\sin(\theta)=-\mbox{sgn}(H_{12})\left(\frac{r-\omega}{2r}\right)^{1/2}\nonumber
\end{equation}

where

\begin{equation}
\omega = H_{22}-H_{11}\nonumber
\end{equation}

\begin{equation}
r=\left(\omega^2+4H_{12}^2\right)^{1/2}\nonumber
\end{equation}

\begin{equation}
E_1=\frac{1}{2}\left(H_{11}+H_{22}-r\right)\nonumber
\end{equation}

\begin{equation}
E_2=\frac{1}{2}\left(H_{11}+H_{22}+r\right)\nonumber
\end{equation}

===Brute Force Method===

- Gives all the eigenvalues and eigenvectors, but it is slow

- First orthonormalize the <math>N</math>-dimensional basis set; i.e. form linear combinations

\begin{equation}
\Phi_m = \sum_{n=1}^N\varphi_nR_{nm}\nonumber
\end{equation}

such that
\begin{equation}
\langle \Phi_m | \Phi_n \rangle = \delta_{m,n}
\end{equation}

This can be done by finding an orthogonal tranformation T such that

\begin{equation}
T^TOT = I = \left(
\begin{array}{ccccc}
I_1 & 0 & 0 & \ldots & 0 \\
0 & I_2 & 0 & \ldots & 0 \\
0 & 0 & I_3 & \ldots & 0 \\
\vdots & \vdots & \vdots & \ddots & \vdots \\
0 & 0 & 0 & \ldots & I_N
\end{array}\right);
\end{equation}

\begin{equation}
O_{mn} = \langle \varphi_m | \varphi_n \rangle \nonumber
\end{equation}

and then applying a scale change matrix

\begin{equation}
S = \left(\begin{array}{ccccc}
\frac{1}{I_1^{1/2}} & 0 & 0 & \ldots & 0\\
0 & \frac{1}{I_2^{1/2}} & 0 & \ldots & 0 \\
0 & 0 & \frac{1}{I_3^{1/2}} & \ldots & 0 \\
\vdots & \vdots & \vdots & \ddots & \vdots \\
0 & 0 & 0 & \ldots & I_N^{1/2}
\end{array}\right)= S^T\nonumber
\end{equation}

Then

\begin{equation}
S^TT^TOTS = 1
\end{equation}

i.e.

\begin{equation}
R^TOR = 1
\end{equation}

with <math>R=TS</math>.

If H is the matrix with elements <math>H_{mn} = \langle \varphi_m | \varphi_n \rangle </math>, then H
expressed in the <math>\Phi_m</math> basis set is

\begin{equation}
H^\prime = R^THR.\nonumber
\end{equation}

We next diagonalize <math>H^\prime</math> by finding an orthogonal transformation W such
that

\begin{equation}
W^T H^\prime W = \lambda = \left(
\begin{array}{cccc}
\lambda_1 & 0 & \ldots & 0 \\
0 & \lambda_2 & \ldots & 0 \\
\vdots & \vdots & \ddots & \vdots \\
0 & 0 & \ldots & \lambda_N
\end{array}\right)\nonumber
\end{equation}

The q'th eigenvector is

\begin{equation}
\Psi^{(q)} = \sum_{n=1}^N\Phi_n W_{n,q} = \sum_{n,n^\prime} \varphi_{n^\prime} R_{n^\prime ,n} W_{n,q} \nonumber
\end{equation}

i.e.

\begin{equation}
c_{n^\prime}^{(q)} = \sum_{n=1}^N R_{n^\prime n} W_{n,q}
\end{equation}

===The Power Method===

- Based on the observation that if H has one eigenvalue, <math>\lambda_M</math>, much bigger than all the rest, and <math>\chi = \left(
\begin{array}{c}a_1\\a_2\\\vdots\end{array}\right)</math>
is an arbitrary starting vector, then <math>\chi = \sum_{q=1}^N x_q\Psi^{(q)}</math>.

\begin{equation}
(H)^n\chi = \sum_{q=1}^N x_q \lambda^n_q\Psi^{(q)}\nonumber\rightarrow x_M\lambda_M^n\Psi^{(M)}\nonumber
\end{equation}

provided <math>x_M\neq 0</math>.

To pick out the eigenvector corresponding to any eigenvalue, write the original problem in the form

\begin{align}
H\Psi & = \lambda O \Psi \nonumber \\
\left( H - \lambda_q O \right) \Psi & = \left( \lambda - \lambda_q \right) O \Psi \nonumber
\end{align}

Therefore,

\begin{equation}
G \Psi = \frac{1}{\lambda - \lambda_q } \Psi \nonumber \\
\end{equation}

where <math>G=(H-\lambda_qO)^{-1}O</math> with eigenvalues <math>\frac{1}{\lambda_n - \lambda_q}</math>.

By picking <math>\lambda_q</math> close to any one of the <math>\lambda_n</math>, say
<math>\lambda_{n^\prime}</math>, then <math>\frac{1}{\lambda_n-\lambda_q}</math> is much larger for
<math>n=n^\prime</math> than for any other value. The sequence is then

\begin{array}{c}
\chi_1=G\chi\nonumber\\
\chi_2=G\chi_1\nonumber\\
\chi_3=G\chi_2\nonumber\\
\vdots\nonumber
\end{array}

until the ratios of components in <math>\chi_n</math> stop changing.

- To avoid matrix inversion and multiplication, note that the sequence is equivalent to
\begin{equation}
F \chi_n = \left( \lambda - \lambda_q \right) O \chi_{n-1} \nonumber \\
\end{equation}
where <math>F = H-\lambda_qO</math>. The factor of <math>\left( \lambda - \lambda_q \right)</math> can be
dropped because this only affects the normalization of <math>\chi_n</math>. To find <math>\chi_n</math>, solve
\begin{equation}
F \chi_n = O \chi_{n-1} \nonumber
\end{equation}
(N equations in N unknowns). Then
\begin{equation}
\lambda = \frac{ \langle \chi_n | H | \chi_n \rangle}{\langle \chi_n | \chi_n \rangle} \nonumber
\end{equation}

==Matrix Elements of H==

\begin{equation}
H=-\frac{1}{2}\nabla^2_1 -\frac{1}{2}\nabla^2_2 - \frac{1}{r_1} - \frac{1}{r_2} +\frac{Z^{-1}}{r_{12}}\nonumber
\end{equation}

Taking <math>r_1, r_2</math> and <math>r_{12}</math> as independent variables,

\begin{equation}
\nabla_1^2 = \frac{1}{r_1^2}\frac{\partial}{\partial r_1}\left(r_1^2\frac{\partial}{\partial r_1}\right)+ \frac{1}{r_{12}^2}\frac{\partial}{\partial r_{12}}
\left(r_{12}^2\frac{\partial}{\partial r_{12}}\right)\nonumber -\frac{l_1(l_1+1)}{r_1^2}+2(r_1 - r_2\cos(\theta_{12}))\frac{1}{r_{12}}\frac{\partial^2}{\partial r_1 \partial r_{12}} \nonumber - 2(\nabla_1^Y \cdot {\bf r}_2)\frac{1}{r_{12}}\frac{\partial}{\partial r_{12}}\nonumber
\end{equation}

where <math>\nabla_1^Y</math> acts only on the spherical harmonic part of the wave function and the diagram

[[File:Pic3.jpg|400px]]

defines the complete set of 6 independent variables is <math>r_1</math>, <math>r_2</math>, <math>r_{12}</math>, <math>\theta_1</math>, <math>\varphi_1</math>, <math>\chi</math>.

If <math>r_{12}</math> were not an independent variable, then one could take the volume element to be

\begin{equation}
d\tau = r_1^2dr_1\sin(\theta_1)d\theta_1d\varphi_1r_2^2dr_2\sin(\theta_2)d\theta_2d\varphi_2.\nonumber
\end{equation}

However, <math>\theta_2</math> and <math>\varphi_2</math> are no longer independent variables. To eliminate them, take the point <math>{\bf r}_1</math> as the
origin of a new polar co-ordinate system, and write

\begin{equation}
d\tau=-r_1^2dr_1\sin(\theta_1)d\theta_1d\varphi_1r_{12}^2dr_{12}\sin(\psi)d\psi d\chi\nonumber
\end{equation}

and use

\begin{equation}
r_2^2=r_1^2+r_{12}^2 +2r_1r_{12}\cos(\psi)
\end{equation}

Then for fixed <math>r_1</math> and <math>r_{12}</math>,

\begin{equation}
2r_2dr_2 = -2r_1r_{12}\sin(\psi)d\psi
\end{equation}

Thus

\begin{equation}
d\tau= r_1dr_1r_2dr_2r_{12}dr_{12}\sin(\theta_1)d\theta_1d\varphi_1d\chi
\end{equation}

The basic type of integral to be calculated is

\begin{align}
I \left( l_1, m_1, l_2, m_2; R \right) & = \int \sin \left( \theta_1 \right) d\theta_1 d\varphi_1 d\chi Y^{m_1}_{l_1} \left( \theta_1, \varphi_1 \right)^{*} Y^{m_2}_{l_2} \left( \theta_2, \varphi_2 \right) \\
& \times \int r_1dr_1r_2dr_2r_{12}dr_{12}R \left( r_1, r_2, r_{12} \right)
\end{align}

Consider first the angular integral. <math>Y^{m_2}_{l_2} \left( \theta_2, \varphi_2 \right)</math> can be expressed in terms of the independent variables <math>\theta_1, \varphi_1, \chi</math> by use of the rotation matrix relation
\begin{equation}
Y^{m_2}_{l_2}( \theta_2, \varphi_2) = \sum_m \mathcal{D}^{(l_2)}_{m_2,m}(\varphi_1,\theta_1,\chi)^*Y^m_{l_2}( \theta_{12}, \varphi)
\end{equation}
where <math>\theta, \varphi</math> are the polar angles of <math>{\bf r}_2</math> relative to <math>{\bf r}_1</math>. The angular integral is then

\begin{align}
I_{ang} & = \int^{2 \pi}_{0} d\chi \int^{2\pi}_{0} d\varphi_1 \int^{\pi}_{0} \sin \left( \theta_1 \right) d\theta_{1} Y^{m_1}_{l_1} \left( \theta_1, \varphi_1 \right)^{*} \\
& \times \sum_m \mathcal{D}^{(l_2)}_{m_2,m} \left( \varphi_1,\theta_1,\chi \right)^{*} Y^{m}_{l_2} \left( \theta_{12},\varphi \right) \\
\end{align}

Use

\begin{equation}
Y^{m_1}_{l_1} \left( \theta_1, \varphi_1 \right)^{*} = \sqrt{\frac{2l_1+1}{4\pi}}\mathcal{D}^{(l_1)}_{m_1,0} \left( \varphi_1, \theta_1, \chi \right)
\end{equation}

together with the orthogonality property of the rotation matrices (Brink and Satchler, p 147)

\begin{align}
& \int \mathcal{D}^{(j)*}_{m,m^\prime}\mathcal{D}^{(J)}_{M,M^\prime}\sin(\theta_1)d\theta_1d\varphi_1d\chi \\
& = \frac{8\pi^2}{2j+1}\delta_{jJ}\delta_{mM}\delta_{m^\prime M^\prime}\nonumber
\end{align}

to obtain

\begin{align}
I_{ang} & = \sqrt{ \frac{2l_1+1}{4\pi}} \frac{8\pi^2}{2l_1+1} \delta_{l_1,l_2} \delta_{m_1,m_2} Y^{0}_{l_2} \left( \theta_{12}, \varphi \right) \\
& = 2 \pi \delta_{l_1,l_2} \delta_{m_1,m_2} P_{l_2} \left( \cos \theta_{12} \right)
\end{align}

since

\begin{equation}
Y^0_{l_2} \left( \theta_{12}, \varphi \right) = \sqrt{\frac{2l_1+1}{4\pi}} P_{l_2} \left( \cos \left( \theta_{12} \right) \right)
\end{equation}

Note that <math>P_{l_2} \left( \cos \theta \right)</math> is just a short hand expression for a radial function because

\begin{equation}
\cos \theta_{12} = \frac{r_1^2 + r_2^2 - r_{12}^2}{2r_1r_2}
\end{equation}

The original integral is thus

\begin{equation}
I \left( l_1, m_1, l_2, m_2; R \right) = 2 \pi \delta_{l_1,l_2} \delta_{m_1,m_2} \int^{\infty}_{0} r_{1} dr_{1} \int^{\infty}_{0} r_{2} dr_{2} \int^{r_{1} + r_{2}}_{|r_{1} - r_{2}|} r_{12} dr_{12} R \left( r_{1}, r_{2}, r_{12} \right) P_{l_2} \left( \cos \theta_{12} \right)
\end{equation}

where again

\begin{equation}
\cos \theta_{12} = \frac{r_{1}^{2} +r^{2}_{2} - r_{12}^{2}}{2 r_{1} r_{2}}
\end{equation}

is a purely radial function.

The above would become quite complicated for large <math>l_2</math> because <math>P_{l_2}(\cos \theta_{12})</math> contains terms up to <math>\left( \cos \theta_{12} \right)^{l_2}</math>. However, recursion relations exist which allow any integral containing <math>P_l \left( \cos \theta_{12} \right)</math> in terms of those containing just <math>P_0 \left( \cos \theta_{12} \right) = 1</math> and <math>P_1 \left( \cos \theta_{12} \right) = \cos \theta_{12}</math>.

==Radial Integrals and Recursion Relations==

The basic radial integral is [see G.W.F. Drake, Phys. Rev. A 18, 820 (1978)]

\begin{align}
I_0 \left( a, b, c \right) & =
\int^{\infty}_{0} r_{1} dr_{1} \int^{\infty}_{r_{1}} r_{2} dr_{2} \int^{r_{1}+r_{2}}_{r_{2} - r_{1}} r_{12} dr_{12} r_{1}^{a} r_{2}^{b} r_{12}^{c} e^{-\alpha r_{1} - \beta r_{2}} + \int^{\infty}_{0} r_{2} dr_{2} \int^{\infty}_{r_{2}} r_{1} dr_{1} \int^{r_{1} + r_{2}}_{r_{1} - r_{2}} r_{12} dr_{12} r_{1}^{a} r_{2}^{b} r_{12}^{c} e^{- \alpha r_{1} - \beta r_{2}} \\
& = \frac{2}{c+2} \sum^{[(c+1)/2]}_{i=0}
\left(\begin{array}{c}
c+2 \\
2i+1
\end{array} \right)
\left\{ \frac{q!}{\beta^{q+1} \left( \alpha + \beta \right)^{p+1}} \sum_{j=0}^{q} \frac{\left( p + j \right)!}{j!} \left( \frac{\beta}{\alpha + \beta} \right)^j + \frac{q^{\prime}!}{\alpha^{q^{\prime}+1} \left( \alpha + \beta \right)^{p^{\prime}+1}} \sum^{q^{\prime}}_{j=0} \frac{\left( p^{\prime} + j \right)!}{j!} \left( \frac{\alpha}{\alpha + \beta} \right)^j \right\}
\end{align}

where

\begin{array}{ll}
p = a + 2i + 2 & p^{\prime} = b + 2i + 2 \\
q = b + c - 2i + 2 & q^{\prime} = a + c - 2i + 2
\end{array}

The above applies for <math> a, b \geq -2, c\geq -1</math>.
<math>[x]</math> means "greatest integer in" <math>x</math>.

Then

\begin{align}
I_1 \left( a, b, c \right) & = \int d\tau_{r} r_{1}^{a} r_{2}^{b} r_{12}^{c} e^{ -\alpha r_{1} - \beta r_{2}} P_{1}\left( \cos \theta \right) \\
& = \frac{1}{2} \left( I_{0} \left( a + 1, b - 1, c \right) + I_{0} \left( a - 1, b + 1, c \right) - I_{0} \left( a - 1, b - 1, c + 2 \right) \right)
\end{align}

===The Radial Recursion Relation===

Recall that the full integral is

\begin{equation}
I(l_1m_1,l_2m_2;R)=2\pi\delta_{l_1,l_2}\delta_{m_1,m_2}I_{l_2}(R)
\end{equation}

where

\begin{equation}
I_{l_2}(R) = \int d\tau_r R(r_1,r_2,r_{12})P_{l_2}(\cos \theta_{12})
\end{equation}

To obtain the recursion relation, use

\begin{equation}
P_l(x) = \frac{[P^{\prime}_{l+1}(x)-P^{\prime}_{l-1}(x)]}{2l+1}
\end{equation}

with

\begin{equation}
P_{l+1}^\prime (x) = \frac{d}{dx}P_{l+1}(x)
\end{equation}

Here <math>x=\cos\theta_{12}</math> and

\begin{align}
\frac{d}{d \cos \theta_{12}} & = \frac{dr_{12}}{d \cos \theta_{12}} \frac{d}{dr_{12}} \\
& = -\frac{r_{1} r_{2}}{r_{12}} \frac{d}{dr_{12}}
\end{align}

Then,

\begin{equation}
I_{l}\left( R \right) = - \int d\tau_r R \frac{r_{1} r_{2}}{r_{12}} \frac{d}{dr_{12}} \frac{ \left[ P_{l+1} \left( \cos \theta_{12} \right) - P_{l-1} \left( \cos \theta_{12} \right) \right]}{2l+1}
\end{equation}

The <math>r_{12}</math> part of the integral is

\begin{align}
& \int^{r_{1} + r_{2}}_{|r_{1} - r_{2}|} r_{12} dr_{12} R \frac{r_{1} r_{2}}{r_{12}} \frac{d}{dr_{12}} \left[P_{l+1}-P_{l-1} \right] \\
& = R r_{1} r_{2} \left[ P_{l+1} - P_{l-1} \right]^{r_{1} + r_{2}}_{|r_{1} - r_{2}|} - \int^{r_{1} + r_{2}}_{|r_{1} - r_{2}|} r_{12} dr_{12} \left( \frac{d}{dr_{12}} R \right) \frac{r_{1} r_{2}}{r_{12}} \frac{ \left[ P_{l+1} - P_{l-1} \right]}{2l+1}
\end{align}

The integrated term vanishes because

\begin{equation}
\cos \theta_{12} = \frac{r_{1}^{2} + r_{2}^{2} - r_{12}^{2}}{2 r_{1} r_{2}}
\begin{array}{lll}
& = -1 & {\rm when} & r_{12}^{2} = \left( r_{1} + r_{2} \right)^{2} \\
& = 1 & {\rm when} & r_{12}^{2} = \left( r_{1} - r_{2} \right)^{2}
\end{array}
\end{equation}

and <math>P_l(1)=1</math>, <math>P_{l}(-1)=(-1)^{l}</math>.

Thus,

\begin{equation}
I_{l+1}\left(\frac{r_1r_2}{r_{12}}R^\prime\right) = \left( 2l+1 \right) I_{l}(R) - I_{l-1} \left( \frac{r_{1} r_{2}}{r_{12}} R^{\prime} \right)
\end{equation}

If

\begin{equation}
R = r_{1}^{a-1} r_{2}^{b-1} r_{12}^{c+2} e^{-\alpha r_{1} - \beta r_{2}}
\end{equation}

then [G.W.F. Drake, Phys. Rev. A 18, 820 (1978)]

\begin{equation}
I_{l+1}(r_1^ar_2^br_{12}^c) =
\frac{2l+1}{c+2} I_{l} \left( r_{1}^{a-1} r_{2}^{b-2} r_{12}^{c+2} \right)+I_{l-1} \left( r_{1}^{a} r_{2}^{b} r_{12}^{c} \right)
\end{equation}

For the case <math>c=-2</math>, take

\begin{equation}
R = r_{1}^{a-1} r_{2}^{b-1} \ln r_{12} e^{ -\alpha r_{1} -\beta r_{2}}
\end{equation}

Then

\begin{equation}
I_{l+1} \left( r_{1}^{a} r_{2}^{b} r_{12}^{c} \right) = I_{l} \left( r_{1}^{a-1} r_{2}^{b-1} \ln r_{12} \right) \left( 2l + 1 \right) - I_{l-1} \left( r_{1}^{a} r_{2}^{b} r_{12}^{-2} \right)
\end{equation}

This allows all <math>I_{l}</math> integrals to be calculated from tables of <math>I_0</math> and
<math>I_1</math> integrals.

===The General Integral===

The above results for the angular and radial integrals can now be combined into a general formula for integrals of the type

\begin{equation}
I = \int \int d{\bf r}_{1} d{\bf r}_{2} R_{1} \mathcal{Y}^{M^{\prime}}_{l^{\prime}_{1} l^{\prime}_{2} L^{\prime}} \left( \hat{r}_{1}, \hat{r}_{2} \right) T_{k_{1} k_{2} K}^{Q} \left( {\bf r}_{1},{\bf r}_{2} \right) R_{2} \mathcal{Y}_{l_{1} l_{2} L}^{M} \left( \hat{r}_{1}, \hat{r}_{2} \right)
\end{equation}

where

\begin{equation}
\mathcal{Y}^{M}_{l_{1} l_{2} L} \left( \hat{r}_{1}, \hat{r}_{2} \right) = \sum_{m_1,m_2} \langle l_{1} l_{2} m_{1} m_{2} | LM \rangle Y^{m_{1}}_{l_{1}} \left( \hat{r}_{1} \right) Y^{m_{2}}_{l_{2}} \left( \hat{r}_{2} \right)
\end{equation}

and

\begin{equation}
T^Q_{k_{1} k_{2} K} \left( {\bf r}_{1}, {\bf r}_{2} \right) = \sum_{q_{1}, q_{2}} \langle k_{1} k_{2} q_{1} q_{2} | KQ \rangle Y_{k_{1}}^{q_{1}} \left( \hat{r}_{1} \right) Y^{q_{2}}_{k_{2}} \left( \hat{r}_{2} \right)
\end{equation}

The basic idea is to make repeated use of the formula

\begin{equation}
Y^{m_{1}}_{l_{1}} \left( \hat{r}_{1} \right) Y^{m_{2}}_{l_{2}} \left( \hat{r}_{1} \right) = \sum_{l m} \left( \frac{\left( 2l_{1}+1 \right) \left(2l_{2}+1 \right) \left(2l+1 \right)}{4 \pi} \right)^{1/2}
\times \left( \begin{array}{ccc}
l_1 & l_2 & l \\
m_1 & m_2 & m
\end{array} \right) \left(\begin{array}{ccc}
l_1 & l_2 & l \\
0 & 0 & 0
\end{array} \right)
Y^{m*}_{l} \left( \hat{r}_{1} \right)
\end{equation}

where

\begin{equation}
Y^{m*}(\hat{r}) = (-1)^mY_l^{-m}(\hat{r})
\end{equation}

and

\begin{equation}
\left(\begin{array}{ccc}
l_1 & l_2 & l \\
m_1 & m_2 & m
\end{array} \right) = \frac{\left( -1 \right)^{l_{1} - l_{2} - m}}{\left( 2l + 1 \right)^{1/2}} \left( l_{1} l_{2} m_{1} m_{2} | l, -m \right)
\end{equation}

is a 3-<math>j</math> symbol. In particular, write

\begin{array}{cc}
Y^{m_1^\prime}_{l_1^\prime}(\hat{r}_1)^{*} \underbrace{Y_{k_{1}}^{q_{1}} \left( \hat{r}_{1} \right) Y^{m_{1}}_{l_{1}} \left( \hat{r}_{1} \right)} & = & \sum_{\Lambda M}(\ldots)Y^{M}_\Lambda(\hat{r}_1)\\
\ \ \ \ \ \ \ \sum_{\lambda_1\mu_1}Y_{\lambda_1}^{\mu_1}(\hat{r}_1) & & \\
& & \\
Y^{m_{2}^{\prime}}_{l_{2}^{\prime}} \left( \hat{r}_{2} \right) \underbrace{Y^{q_{2}}_{k_{2}} \left( \hat{r}_{2} \right) Y_{l_{2}}^{m_{2}} \left( \hat{r}_{2} \right)} & = & \sum_{\Lambda^{\prime} M^{\prime}} \left( \ldots \right)Y_{\Lambda^{\prime}}^{M^{\prime}*} \left( \hat{r}_{2} \right) \\
\ \ \ \ \ \ \ \ \ \sum_{\lambda_{2} \mu_{2}}Y_{\lambda_{2}}^{\mu_{2}} \left( \hat{r}_{2} \right) & &
\end{array}

The angular integral then gives a factor of <math>2 \pi \delta_{\Lambda, \Lambda^{\prime}} \delta_{M, M^{\prime}} P_{\Lambda} \left( \cos \theta_{12} \right)</math>. The total integral therefore reduces to the form

\begin{equation}
I = \sum_{\Lambda} C_{\Lambda} I_{\Lambda} \left( R_{1} R_{2} \right)
\end{equation}

where <math>C_{\Lambda} = \sum_{\lambda_{1}, \lambda_{2}} C_{ \lambda_{1}, \lambda_{2}, \Lambda}</math>. For further details and derivations, including graphical representations, see G.W.F. Drake, Phys. Rev. A <math>{\bf 18}</math>, 820 (1978).

==Graphical Representation==

[[File:GraphRep1.jpg|500px]]

==Matrix Elements of H==
Recall that

\begin{equation}
H = -\frac{1}{2} \nabla^{2}_{1} - \frac{1}{2} \nabla^{2}_{2} - \frac{1}{r_1} - \frac{1}{r_2} + \frac{Z^{-1}}{r_{12}}
\end{equation}

Consider matrix elements of

\begin{equation}
\nabla^2_1 = \frac{1}{r_1^2} \frac{\partial}{\partial r_1}\left(r_1^2\frac{\partial}{\partial r_1} \right) + \frac{1}{r^2}\frac{\partial}{\partial r}\left(r^2\frac{\partial}{\partial r}\right) - \frac{(\vec{l}_1^Y)^2}{r_1^2} + \frac{2 \left( r_1 - r_2 \cos \theta \right)}{r} \frac{\partial^2}{\partial r_1 \partial r} - 2 \left( \nabla^Y_1 \cdot {\bf r}_2 \right) \frac{1}{r} \frac{\partial}{\partial r}
\end{equation}

where <math>r \equiv r_{12}</math> and <math>\cos\theta \equiv \cos\theta_{12}</math>. Also in what follows, define <math>\hat{\nabla}^Y_1 \equiv r_1\nabla^Y_1</math> where <math>\nabla^Y</math> operates only on the spherical harmonic part of the wave function, and similarly for <math>\vec{l}_1^Y</math>.

A general matrix element is

\begin{equation}
\langle r_1^{a^\prime} r_2^{b^\prime} r_{12}^{c^\prime} e^{-\alpha^\prime r_1 - \beta^\prime r_2} \mathcal{Y}^{M^\prime}_{l^\prime_1 l^\prime_2 L^\prime} \left( \hat{r}_1, \hat{r}_2 \right) \left| \nabla^2_1 \right| r_1^a r_2^b r_{12}^c e^{-\alpha r_1 - \beta r_2} \mathcal{Y}^M_{l_1 l_2 L} \left( \hat{r}_1, \hat{r}_2 \right) \rangle
\end{equation}

Since <math>\nabla^2_1</math> is rotationally invariant, this vanishes unless <math>L=L^\prime</math>, <math>M=M^\prime</math>. Also <math>\nabla_1^2</math> is Hermitian, so that the result must be the same whether it operates to the right or left, even though the results look very different. In fact, requiring the results to be the same yields some interesting and useful integral identities as follows:

Put

\begin{equation}
\mathcal{F} = F \mathcal{Y}^M_{l_1 l_2 L} \left( \hat{r}_1, \hat{r_2} \right)
\end{equation}

and

\begin{equation}
F = r_1^a r_2^b r_{12}^c e^{-\alpha r_1 - \beta r_2}
\end{equation}

Then

\begin{equation}
\nabla^2_1 \mathcal{F} = \left\{ \frac{1}{r_1^2} \left[ a \left( a + 1 \right) - l_1 \left( l_1 + 1 \right) \right] + \frac{c \left( c + 1 \right)}{r^2} + \alpha^2\frac{2\alpha \left( a + 1 \right)}{r_1} + \frac{2 \left(r_1 - r_2 \cos \theta \right) }{r_1r^2}c \left[ a - \alpha r_1 \right] - \frac{2c}{r^2} \left( \hat{\nabla}^Y_1 \cdot \hat{r}_2 \right) \frac{r_2}{r_1} \right\} \mathcal{F}
\end{equation}

and

\begin{align}
\langle \mathcal{F}^\prime \left| \nabla^2_1 \right| \mathcal{F} \rangle & = \sum_\Lambda \int d\tau_r F^\prime C_\Lambda \left( 1 \right) P_\Lambda \left( \cos \theta \right) \times \left\{ \frac{1}{r_1^2} \left[ a \left( a + 1 \right) - l_1 \left( l_1 + 1 \right) \right] -\frac{2 \alpha \left( a + 1 \right)}{r_1} + \frac{c \left( c + 1 \right)}{r^2} +\alpha^2 + \frac{2 \left( r_2 - r_2 \cos \theta \right)}{r_1r^2}c \left( a - \alpha r_1 \right) \right\} F \\
& + \sum_\Lambda \int d\tau_r F^\prime C_\Lambda \left( \hat{\nabla}^Y_1 \cdot \hat{r}_2 \right) P_\Lambda \left( \cos \theta \right) \left( \frac{-2c r_2}{r_1 r^2} \right) F
\end{align}

where

\begin{equation}
\int d\tau_r = \int^\infty_0 r_1 dr_1 \int^\infty_0 r_2 dr_2 \int^{r_1+r_2}_{\left| r_1 - r_2 \right|} r dr
\end{equation}

For brevity, let the sum over <math>\Lambda</math> and the radial integrations be understood, and let <math>\left( \nabla_1^2 \right)</math> stand for the terms which appear in the integrand. Then operating to the right gives

\begin{equation}
(\nabla_1^2)_R = \frac{1}{r_1^2} \left[ a \left( a + 1 \right) - l_1 \left( l_1 + 1 \right) \right] + \frac{c \left( c + 1 \right)}{r^2} + \alpha^2 - \frac{2 \alpha \left( a + 1 \right)}{r_1}
+ \frac{2 \left( r_1 - r_2 \cos \theta \right)}{r_1 r^2}c \left( a - \alpha r_1 \right) - \frac{2c r_2}{r_1 r^2} \left( \hat{\nabla}^Y_1 \cdot \hat{r}_2 \right)
\end{equation}

Operating to the left gives

\begin{equation}
(\nabla_1^2)_L = \frac{1}{r_1^2} \left[a^\prime \left( a^\prime + 1 \right) - l_1^\prime \left( l_1^\prime + 1 \right) \right] + \frac{c^\prime \left( c^\prime + 1 \right)}{r^2} + \alpha^{\prime 2} -\frac{2 \alpha^\prime \left( a^\prime + 1 \right)}{r_1}
+ \frac{2 \left( r_1 - r_2 \cos \theta \right)}{r_1 r^2} c^\prime \left( a^\prime - \alpha^\prime r_1 \right) - \frac{2 c^\prime r_2}{r_1 r^2} \left( \hat{\nabla}^{Y^\prime}_1 \cdot \hat{r}_2 \right)
\end{equation}

Now put

\begin{array}{cc}
a_+ = a+a^\prime, & \hat{\nabla}_1^+ = \hat{\nabla}_1^Y + \hat{\nabla}_1^{Y\prime} \\
a_{-} = a-a^\prime, & \hat{\nabla}_1^{-} = \hat{\nabla}_1^Y - \hat{\nabla}_1^{Y\prime}
\end{array}

etc., and substitute <math>a^\prime = a_+ - a</math>, <math>c^\prime = c_+ - c</math>, and <math>\alpha^\prime = \alpha_+ - \alpha</math> in <math>(\nabla^2_1)_L</math>. If <math>a_+</math>, <math>c_+</math> and <math>\alpha_+</math> are held fixed, then the equation

\begin{equation}
(\nabla^2_1)_R = (\nabla^2_1)_L
\end{equation}

must be true for arbitrary <math>a</math>, <math>c</math>, and <math>\alpha</math>. Their coefficients must thus vanish.

This yields the integral relations

\begin{equation}
\label{I}
\frac{\left( r_1 - r_2 \cos \theta \right)}{r_1 r^2} = \frac{1}{c_+} \left( \frac{- \left( a_+ + 1 \right)}{r_1^2} + \frac{\alpha_+}{r_1} \right) \tag{I}
\end{equation}

from the coefficient of <math>a</math> and

\begin{equation}
\label{II}
\frac{ \left( r_1 - r_2 \cos \theta \right) \left( a_+ - \alpha_+ r_1 \right)}{r_1 r^2} = \frac{r_2}{r_1 r^2}\left( \hat{r}_2 \cdot \hat{\nabla}_1^+ \right) - \frac{ \left( c_+ + 1 \right)}{r^2} \tag{II}
\end{equation}

from the coefficient of <math>c</math>. The coefficient of <math>\alpha</math> gives an equation equivalent to (I).

Furthermore, if can be show that (see problem)

\begin{align}
& \sum_\Lambda \int d\tau_r \frac{r^c}{r^2} C_\Lambda \left( \hat{r}_2 \cdot \hat{\nabla}_1^Y \right) P_\Lambda \left( \cos \theta \right) \\
= & \sum_\Lambda \int d\tau_r \frac{r^c}{c r_1 r_2}C_\Lambda \left( 1 \right)P_\Lambda \left( \cos \theta \right) \times \left( \frac{l_1^\prime \left( l_1^\prime + 1 \right) - l_1 \left( l_1 + 1 \right) -\Lambda \left( \Lambda + 1 \right)}{2} \right)
\end{align}

and similarly for <math>\left( \hat{r}_2 \cdot \hat{\nabla}^{Y\prime}_1 \right)</math> with <math>l_1</math> and <math>l_1^\prime</math> interchangeable, then it follows that

\begin{equation}
\frac{r^{c_+}}{r^2} \left( \hat{r}_2 \cdot \hat{\nabla}_1^+ \right) = - \frac{r^{c_+}}{c_+ r_1 r_2} \Lambda \left( \Lambda + 1 \right)
\end{equation}

and

\begin{equation}
\frac{r^{c_+}}{r^2} \left( \hat{r}_2 \cdot \hat{\nabla}_1^- \right) = \frac{r^{c_+}}{c_+ r_1 r_2} \left[ l_1^\prime \left( l_1^\prime + 1 \right) - l_1 \left( l_1 + 1 \right) \right]
\end{equation}

where equality applies after integration and summation over <math>\Lambda</math>.

Thus (II) becomes

\begin{equation}\label{IIa}
\frac{ \left( r_1 - r_2 \cos \theta \right) \left( a_+ - \alpha_+ r_1 \right)}{r_1 r^2} = - \frac{\Lambda \left( \Lambda + 1 \right)}{c_+ r^2_1} - \frac{ \left( c_+ + 1 \right)}{r^2} \tag{IIa}
\end{equation}

===Problem===

Prove the integral relation

\begin{align}
& \sum_\Lambda \int d\tau_r f \left( r_1, r_2 \right) \frac{1}{r} \left( \frac{d}{dr} q(r) \right) C_\Lambda \left( \hat{r}_2 \cdot \hat{\nabla}^Y_1 \right) P_\Lambda \left( \cos \theta \right) \\
= & \sum_\Lambda \int d\tau_r \frac{ f \left( r_1, r_2 \right)}{r_1, r_2} q(r) C_\Lambda \left( 1 \right) P_\Lambda \left( \cos \theta \right) \left(\frac{l_1^\prime \left( l_1^\prime + 1 \right) - l_1 \left( l_1 + 1 \right) - \Lambda \left( \Lambda + 1 \right)}{2}\right)
\end{align}

where the coefficients <math>C_\Lambda \left( 1 \right)</math> are the angular coefficients from the overlap integral

\begin{align}
\int & d\Omega\, \mathcal{Y}^{M*}_{l_1^\prime l_2^\prime L}(\hat{r}_1,\hat{r}_2)\, \mathcal{Y}^{M}_{l_1 l_2 L}(\hat{r}_1,\hat{r}_2)\\
= & \sum_\Lambda C_\Lambda(1)P_\Lambda(\cos\theta_{12}).
\end{align}

Hint: Use the fact that <math>l_1^2</math> is Hermitian so that
<math>
\int d\tau (l_1^2\mathcal{Y}^\prime)^*q(r)\mathcal{Y} = \int d\tau
\mathcal{Y}^{\prime *}l_1^2(q(r)\mathcal{Y})
</math>
with <math>\ \vec{l}_1= \frac{1}{i}\vec{r}_1\times\nabla_1</math>.

It is also useful to use
<math>\begin{eqnarray}\ \
\cos\theta\, P_L(\cos\theta) &=& \frac{1}{2L+1}[L\,P_{L-1}(\cos\theta) + (L+1)P_{L+1}(\cos\theta)]
\\
(\cos^2\theta-1)P_L(\cos\theta) &=&
\frac{L(L-1)}{(2L-1)(2L+1)}P_{L-2}(\cos\theta)\nonumber
-\frac{2(L^2+L-1)}{(2L-1)(2L+3)}P_L(\cos\theta) + \frac{(L+1)(L+2)}{(2L+1)(2L+3)}P_{L+2}(\cos\theta)\nonumber\\
&=& \frac{L(L-1)}{(2L-1)(2L+1)}[P_{L-2}(\cos\theta) - P_L(\cos\theta)]
+ \frac{(L+1)(L+2)}{(2L+1)(2L+3)}[P_{L+2}(\cos\theta) - P_L(\cos\theta)]
\end{eqnarray}</math>

together with a double application of the integral recursion relation
<math>
I_{L+1}\left(\frac{1}{r}\frac{d}{dr}q(r)\right) =
(2L+1)I_L\left(\frac{1}{r_1r_2}q(r)\right)+I_{L-1}\left(\frac{1}{r}\frac{d}{dr}q(r)\right)
</math>

Of course <math>l_1^2\mathcal{Y}^M_{l_1l_2L}(\hat{r}_1,\hat{r}_2)=l_1(l_1+1)\mathcal{Y}^M_{l_1l_2L}(\hat{r}_1,\hat{r}_2)</math>. Begin by expanding
<math>
l_1^2(g\mathcal{Y}) = \mathcal{Y}l_1^2g + g\,l_1^2\mathcal{Y} + 2(\vec{l}_1g)\cdot(\vec{l}_1\mathcal{Y})
</math>, and show that

<math>
\begin{eqnarray}\ \
(\vec{l}_1g)\cdot(\vec{l}_1\mathcal{Y}) & = & \frac{r_1^2}{r}\frac{dg}{dr}\vec{r}_2\cdot \nabla_1\mathcal{Y}\,\,\,\,\mbox{ and}\\
l_1^2g(r) &=& 2r_1r_2\cos\theta\frac{1}{r}\frac{dg}{dr} + r_1^2r_2^2(\cos^2\theta - 1)
\frac{1}{r}\frac{d}{dr}\left(\frac{1}{r}\frac{dg}{dr}\right).
\end{eqnarray}
</math>

The proof amounts to showing that the term <math>l_1^2g(r)</math> can be replaced by
<math>\Lambda(\Lambda+1)g(r)</math> after multiplying by <math>P_\Lambda(\cos\theta)</math>
and integrating by parts with respect to the radial integrations over <math>r_1</math>, <math>r_2</math> and
<math>r \equiv |\vec{r}_1 - \vec{r}_2|</math>. Remember that <math>\cos\theta</math> is just a short-hand notation for the radial function <math>(r_1^2 + r_2^2 - r^2)/(2r_1r_2)</math>.

==General Hermitian Property==

Each combination of terms of the form
\begin{equation}
<f>_R = a^2f_1+af_2+abf_3+b^2f_4+bf_5+af_6\nabla_1^Y+bf_7\nabla^Y_2+f_8(Y)
\end{equation}
can be rewritten
\begin{equation}
<f>_L= (a_+-a)^2f_1+(a_+-a)f_2 + (a_+-a)(b_+-b)f_3+(b_+-b)^2f_4
+(b_+-b)f_5+(a_+-a)f_6\nabla^{Y\prime}_1+(b_+-b)f_7\nabla^{Y\prime}_2+f_8)Y^\prime
\end{equation}
where as usual <math>\nabla_1^Y</math> acts only on the spherical harmonic part of the wave function denoted for short by <math>Y</math>.
Since these must be equal for arbitrary <math>a</math> and <math>b</math>,
\begin{equation}
a_+^2f_1+a_+f_2+_+b_+f_3+b_+f_4+b_+f_5+a_+f_6\nabla^{Y\prime}_1 +
b_+f_7\nabla^{Y\prime}_2+f_8(Y^\prime)-f_8(Y)=0
\end{equation}

Adding the corresponding expression with <math>Y</math> and <math>Y^\prime</math> interchanged
yields
<math>\label{III}
a_+^2f_1+a_+f_2+a_+b_+f_3+b_+^2f_4+b_+f_5+\frac{1}{2}a_+f_6\nabla^+_1+\frac{1}{2}b_+f_7\nabla^+_2=0\tag{III}
</math>

Subtracting gives
<math>\label{IV}
f_8(Y)-f_8(Y^\prime)=-\frac{1}{2}[a_+f_6\nabla^-_1+b_+f_7\nabla_2^-]\tag{IV}
</math>
<math>\label{V}
a[-2a_+f_1-2f_2-b_+f_3-f_6\nabla^+_1]=0\tag{V}
</math>
<math>\label{VI}
b[-2b_+f_4-2f_5-a_+f_3-f_7\nabla^+_2]=0\tag{VI}
</math>

Adding the two forms gives

\begin{eqnarray}
<f>_R+<f>_L=\frac{1}{2}(a_+^2+a_-^2)f_1+a_+f_2+\frac{1}{2}(a_+b_++a_-b_-)f_3\nonumber\\
\ +\frac{1}{2}(b_+^2+b_-^2)f_4+b_+f_5+\frac{1}{2}f_6(a_+\nabla_1^++a_-\nabla^-_1)\nonumber\\
\ +\frac{1}{2}(b_+\nabla_2^++b_-\nabla^-_2)+f_8(Y)+f_8(Y^\prime)\nonumber
\end{eqnarray}

Subtracting <math>\frac{x}{2}\times({\rm I})</math> gives

\begin{eqnarray}\ \
<f>_R+<f>_L = \frac{1}{2}[(1-x)a_+^2 + a_-^2]f_1 +
(1-\frac{x}{2})a_+f_2+\frac{1}{2}[(1-x)a_+b_++a_-b_-]f_3\nonumber\\
+\frac{1}{2}[(1-x)b_+^2+b_-^2]f_4+(1-\frac{x}{2})f_5b_++\frac{1}{2}f_6[(1-\frac{x}{2})a_+\nabla_1^++a_-\nabla^-_1]\nonumber\\
+\frac{1}{2}[(1-\frac{x}{2})b_+\nabla^+_2+b_-\nabla^-_2]+f_8(Y)+f_8(Y^\prime)\nonumber
\end{eqnarray}

If <math>x=1</math>,
\begin{eqnarray}
<f>_R+<f>_L &=&
\frac{1}{2}[a_-^2f_1+a_+f_2+a_-b_-f_3+b_-^2f_4+b_+f_5\nonumber\\
&+&\left(\frac{1}{2}a_+\nabla^+_1+a_-\nabla^-_1\right)f_6+\left(\frac{1}{2}b_+\nabla^+_2+b_-\nabla^-_2\right)f_7]+f_8(Y)+f_8(Y^\prime)\nonumber
\end{eqnarray}

The General Hermitian Property for arbitrary <math>x</math> gives
\begin{eqnarray}\ \
\nabla_1^2=
\frac{1}{4}\{\frac{1}{r_1^2}[(1-x)a_+^2+a_-^2+2(1-\frac{x}{2})a_+-2[l_1(l_1+1)+l_1^\prime(l_1^\prime+1)]\nonumber\\
-\frac{2}{r_1}[(1-x)\alpha_+a_++\alpha_-a_-+2(1-\frac{x}{2})\alpha_+]+(1-x)\alpha_+^2
+ \alpha^2_-]\nonumber\\
+\frac{2(r_1-r_2\cos\theta)}{r_1r^2}[(1-x)(a_+-\alpha_+r_1)c_++(a_--\alpha_-r_2)c_-]\nonumber\\
-\frac{2r_2}{r_1r^2}[(1-\frac{x}{2})c_+\hat{r}_2\cdot\hat{\nabla}_1^++c_-\hat{r}_2\hat{\nabla}_1^-]+\frac{(1-x)c_+^2+c_-^2+2(1-\frac{x}{2})c_+}{r^2}\}\nonumber
\end{eqnarray}.

Use
<math>\begin{eqnarray}\ \
\frac{-2r_2}{r_1r^2}[(1-\frac{x}{2})c_+\hat{r}_2\hat{\nabla}_1^++c_-\hat{r}_2\hat{\nabla}_1^-]=\frac{2}{r_1^2}\left[(1-\frac{x}{2})\Lambda(\Lambda+1)-\frac{c_-}{c_+}[l_1^\prime(l_1^\prime+1)+l_1(l_1+1)]\right]\end{eqnarray}</math>,

<math>\begin{eqnarray}
\frac{2(r_1-r_2\cos\theta)}{r_1r^2}(a_--\alpha_-r_1)c_- = 2\frac{c_-}{c_+}\left[\frac{-1}{r_1^2}[a_-(a_++1)]+\frac{1}{r_1}[a_-\alpha_++\alpha_-(a_++2)]-\alpha_-\alpha_+\right]\nonumber
\end{eqnarray}</math>,

and
<math>\begin{eqnarray}\ \
\frac{2(r_1-r_2\cos\theta)}{r_1r^2}(a_+-\alpha_+r_1)c_+ = 2\left[-\frac{1}{r_1^2}[a_+(a_++1)]+\frac{1}{r_1}[a_+\alpha_++\alpha_+(a_++2)]-\alpha_+^2\right]
\end{eqnarray}</math>.

Substituting into <math>\nabla_1^2</math> gives

\begin{eqnarray}
\nabla^2_1 = \frac{1}{4}\{\frac{1}{r_1^2}[-(1-x)a_+^2 + a_-^2 +xa_+ +
2(1-\frac{x}{2})\Lambda(\Lambda+1)\nonumber\\
-2l_1(l_1+1)(1-\frac{c_-}{c_+}) -
2l_1^\prime(l_1^\prime+1)(1+\frac{c_-}{c_+})-\frac{2c_-a_-}{c_+}(a_++1)]\nonumber\\
- \frac{2}{r_1}\left[-(1-x)\alpha_+(a_++2)+\alpha_-a_-+2(1-\frac{x}{2})\alpha_+
-\frac{c_-}{c_+}[a_-\alpha_++\alpha_-(a_++2)]\right]\nonumber\\
-(1-x)\alpha_+^2+\alpha_-^2-2\frac{c_-}{c_+}\alpha_-\alpha_+ +\frac{1}{r^2}\left[(1-x)c_+^2+c_-^2+2(1-\frac{x}{2})c_+\right]\}\nonumber
\end{eqnarray}

This has the form
<math>
\nabla_1^2 =
\frac{1}{4}\left[\frac{A_1}{r_1^2}+\frac{B_1}{r_1}+\frac{C_1}{r^2}+D_1\right]
</math>
with

<math>\begin{array}{ccc}
A_1=-(1-x)a_+^2+a_-^2
+xa_++2(1-\frac{x}{2})\Lambda(\Lambda+1)-2l_1(l_1+1)(1-\frac{c_-}{c_+})\
-2l_1^\prime(l_1^\prime+1)(1+\frac{c_-}{c_+})-\frac{2c_-a_-}{c_+}(a_++1) \ \
\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \
\end{array}
</math>

<math>
B_1 =
2\left[(1-x)\alpha_+(a_++2)-\alpha_-a_--2(1-\frac{x}{2})\alpha_++\frac{c_-}{c_+}[a_-\alpha_++\alpha_-(a_++2)]\right]
</math>

<math>
C_1 = (1-x)c_+^2+ c_-^2 +2(1-\frac{x}{2})c_+
</math>

<math>
D_1 = -1(1-x)\alpha_+^2 +\alpha_-^2 - 2\frac{c_-}{c_+}\alpha_-\alpha_+
</math>

The complete Hamiltonian is then (in Z-scaled a.u.)

<math>\begin{eqnarray}
H =-\frac{1}{2}\nabla^2_1 -\frac{1}{2}\nabla^2_2 -\frac{1}{r_1} -\frac{1}{r_2} +\frac{Z^{-1}}{r}\nonumber\\
= -\frac{1}{2}\nabla^2_1 -\frac{1}{2}\nabla^2_2 -\frac{1}{r_1} -\frac{Z-1}{Zr_2}+Z^{-1}\left(\frac{1}{r}-\frac{1}{r_2}\right)\nonumber\\
\ =-\frac{1}{8}\left[\frac{A_1}{r_1^2}+\frac{B_1+8}{r_1}+\frac{C_1}{r^2}+D_1+D_2 +\frac{A_2}{r_2^2} + \frac{B_2 + 8(Z-1)/Z}{r_2}+\frac{C_2}{r^2}\right]\nonumber
+ Z^{-1}\left(\frac{1}{r}-\frac{1}{r_2}\right)\nonumber
\end{eqnarray}</math>

The screened hydrogenic energy is
<math>\begin{eqnarray}
E_{SH}=-\frac{1}{2}\left[\frac{1}{n_1^2}+ \left(\frac{Z-1}{Z}\right)^2\frac{1}{n_2^2}\right]
\end{eqnarray}</math>
so that

<math>\begin{eqnarray}
H-E_{SH} &=& -\frac{1}{8}[
\frac{A_1}{r_1^2}+\frac{B_1+8}{r_1}+\frac{C_1}{r^2} + D_1 + D_2
-\frac{4}{n_1^2} -\left(\frac{Z-1}{Z}\right)^2\frac{4}{n_2^2}\nonumber
+\frac{A_2}{r_2}+\frac{B_2+8(Z-1)/Z}{r_2}+\frac{C_2}{r^2}] +
Z^{-1}\left(\frac{1}{r} -\frac{1}{r_2}\right)\nonumber
\end{eqnarray}</math>

==Optimization of Nonlinear Parameters==
- The traditional method of performing Hylleraas calculations is to write the
basis set in the form

<math>
\Psi = \sum_{i,j,k}c_{i,j,k}\varphi_{i,j,k}(\alpha,\beta)\pm exchange
</math>

with

<math>
\varphi_{i,j,k}(\alpha,\beta) = r_1^ir_2^jr_{12}^ke^{-\alpha r_1 - \beta
r_2}\mathcal{Y}^M_{l_1l_2L}.
</math>

The usual procedure is to set <math>\alpha = Z</math> so that it represents the inner
<math>1s</math> electron, and then to vary <math>\beta</math> so as to minimize the energy.

[[File:OptNon.jpg|500px]]

Since <math>\beta</math> appears in <math>\Psi</math> as a non-linear parameter, the entire
calculation must be repeated for each value of <math>\beta</math>. However, the minimum
becomes progressively smaller as the basis set is enlarged.

[[File:OptNon02.jpg|500px]]

*Difficulties

1. If the basis set is constructed so that <math> i +j +j \leq N</math>, the the
number of terms is
<math>
(N+1)(N+2)(N+3)/6
</math>
N=14 already gives 680 terms and an accuracy of about 1 part in <math>10^{10}</math> for
low-lying states. A substantial improvement in accuracy would require much
larger basis sets, together with multiple precision arithmetic to avoid loss
of significant figures when high powers are included.

2. The accuracy rapidly deteriorates as one goes to more highly excited
states - about 1 significant figure is lost each time the principle quantum
number is increased.

*Cure

We have found that writing basis sets in the form
\begin{eqnarray}
\Psi &=& \sum_{i,j,k}\left(c^{(1)}_{i,j,k}\varphi_{ijk}(\alpha_1\beta_1) +
c_{ijk}^{(2)}\varphi_{ijk}(\alpha_2\beta_2)\right) \pm exchange \nonumber\\
&=& \psi({\bf r}_1,{\bf r}_2) \pm \psi({\bf r}_2,{\bf r}_1)\nonumber
\end{eqnarray}
so that each combination of powers is included twice with different nonlinear
parameters gives a dramatic improvement in accuracy for basis sets of about
the same total size.

However, the optimization of the nonlinear parameters is now much more
difficult, and an automated procedure is needed.

If
<math>
E = \frac{<\Psi|H|\Psi>}{<\Psi|\Psi>}</math> and we assume <math><\Psi|\Psi> = 1</math>, then

<math>
\frac{\partial E}{\partial\alpha_t} = -2 <\Psi|H-E|r_1\psi({\bf r}_1,{\bf
r}_2;\alpha_t)\pm r_21\psi({\bf r}_2,{\bf
r}_1;\alpha_t)>
</math>
where
<math>
\psi({\bf r}_1,{\bf r}_2;\alpha_t) =
\sum_{i,j,k}c_{ijk}^{(t)}\varphi_{ijk}(\alpha_t\beta_t)
</math>

*Problem

1. Prove the above.

2. Prove that there is no contribution from the implicit dependence of
<math>c_{ijk}^{(t)}</math> on <math>\alpha_t</math> if the linear parameters have been optimized.

==The Screened Hydrogenic Term==

If the Hamiltonian is written in the form (in Z-scaled a.u.)
<math>\begin{eqnarray}\
H = H_0({\bf r}_1,Z) + H_0({\bf r}_2,Z-1) +
Z^{-1}\left(\frac{1}{r}-\frac{1}{r_2}\right)
\end{eqnarray}</math>
with
<math>\
H_0({\bf r}_1,Z) = -\frac{1}{2}\nabla_1^2 - \frac{1}{r_1}\nonumber
</math> and
<math>
H_0({\bf r}_2,Z-1) = -\frac{1}{2}\nabla_2^2 - \frac{Z-1}Z{r_2}\nonumber
</math>,
then the eigenvectors of <math>H_0({\bf r},Z) +H_0({\bf r}_2,Z-1)</math> are products of
hydrogenic orbitals
<math>
\Psi_0=\psi_0(1s,Z)\psi_0(nl,Z-1)
</math>
and the eigenvalue is
<math>\begin{eqnarray}
E_{SH} = \left[-\frac{1}{2} - \left(\frac{Z-1}{Z}\right)^2\frac{1}{2n^2}\right]Z^2 a.u.
\end{eqnarray}</math>
called the screened hydrogenic eigenvalue. From highly excited states,
<math>E_{SH}</math> and <math>\Psi_0</math>are already excellent approximations.

For example, for the 1s8d states, the energies are

<math>\begin{eqnarray}\\
E(1s8d^1D)&=& -2.007\,816\,512\,563\,81 {\rm\ a.u.}\nonumber\\
E(1s8d^3D)&=& - 2.007\,817\,934\,711\,71 {\rm\ a.u.}\nonumber\\
E_{SH} &=& -2.007\,812\,5 {\rm\ a.u.} \nonumber
\end{eqnarray}</math>

It is therefore advantageous to include the screened hydrogenic terms in the
basis set so that the complete trial function becomes
<math>
\Psi = c_0\Psi_0 +
\sum_{ijk}\left[c_{ijk}^{(1)}\varphi_{ijk}(\alpha_1,\beta_1)+c_{ijk}^{(2)}\varphi(\alpha_2,\beta_2)\right]\pm
{\rm exchange}
</math>

Also, the variational principal can be re-expressed in the form
<math>\begin{eqnarray}
E= E_{SH}+\frac{<\Psi|H-E_{SH}|\Psi>}{<\Psi|\Psi>}
\end{eqnarray}</math>
so that the <math>E_{SH}</math> term can be cancelled analytically from the matrix
elements.

For example
<math>
<\Psi_0|H-E_{SH}|\Psi_0> = 0
</math>

and
<math>
<\varphi_{ijk}|H-E_{SH}|\Psi_0> =
Z^{-1}<\varphi_{ijk}|\frac{1}{r}-\frac{1}{r_2}|\Psi_0>
</math>

Recall that
<math>\begin{eqnarray}\ \
I_0(a,b,c) = \frac{2}{c+2}\sum^{[c+1]/2}_{i=0}\left(\begin{array}{c}
c+2\\
2i+1
\end{array}\right) \left[f(p,q;\beta) + f(p^\prime,q^\prime;\alpha)\right]
\end{eqnarray}</math>
where

<math>\begin{eqnarray}
f(p,q;x) =
\frac{q!}{x^{q+1}(\alpha+\beta)^{p+1}}\sum^p_{j=0}\frac{(p+j)!}{j!}\left(\frac{x}{\alpha+\beta}\right)^j
\end{eqnarray}</math>

<math>
\begin{array}{cc}
p=a+2i+2 & p^\prime = b+2i+2\\
q= b+c-2i+2 & q^\prime =a+c-2i+2
\end{array}
</math>

If <math>\beta\ll\alpha</math>, the <math>f(p,q;\beta)\gg f(p^\prime,q^\prime;\alpha)</math> and the
<math>i=0</math> term is the dominant contribution to <math>f(p,q;\beta)</math>. However, since
this term depends only on the sum of powers <math>b+c</math> for <math>r_2</math>, it cancels exactly
from the matrix element of
<math>
\frac{1}{r_{12}} - \frac{1}{r_2}
</math>
and can therefore
be omitted in the calculations of integrands, thereby saving many significant
figures. This is especially valuable when <math>b+c</math> is large and <math>a</math> is small.

==Small Corrections==
*Mass Polarization
If the nuclear mass is not taken to be infinite, then the Hamiltonian is
<math>\begin{eqnarray}\ \
H = \frac{P_N^2}{2M} + \sum_{i=1}\frac{p_i^2}{2m} + V
\end{eqnarray}</math>

[[File:SmallCor.jpg|500px]]

where <math>M =M_A-nm</math> is the nuclear mass and <math>m</math> is the electron mass.

Change variables to the C of M
<math>\begin{eqnarray}
{\bf R} = \frac{1}{M+nm}\left[M{\bf r}_N + m({\bf r}_1 + {\bf r}_2 +
\ldots)\right]
\end{eqnarray}</math>

and relative variables

<math>
{\bf s}_i = {\bf r}_i - {\bf r}_N.
</math>

Then

<math>\begin{eqnarray}\ \ \ \ \
H = \frac{1}{2(M+nm)}{\bf p}_R^2+\frac{1}{2m}\sum_{i=1}^n{\bf p}_{s_i}^2
+\frac{1}{2M}\sum_{i,k}{\bf p}_{s_i}\cdot {\bf p}_{s_k} + V({\bf s}_1, {\bf
s}_2, \ldots)\nonumber
= \frac{1}{M+nm}{\bf p}_R^2+\frac{1}{2\mu}\sum_{i=1}^n{\bf
p}_{s_i}^2+\frac{1}{2M}\sum_{i\neq k}{\bf p}_{s_i}\cdot {\bf p}_{s_k}+V({\bf s}_1, {\bf
s}_2, \ldots)\nonumber
\end{eqnarray}</math>

where
<math>\begin{eqnarray}
\mu = \left(\frac{1}{M} + \frac{1}{m}\right)^{-1}
\end{eqnarray}</math>

If <math>n=2</math>, the Schrödinger equation is
<math>\begin{eqnarray}\ \ \
\left[ -\frac{\hbar^2}{2\mu}(\nabla^2_{s_1}+ \nabla^2_{s_2}) -
\frac{\hbar^2}{M}\nabla_{s_1}\nabla_{s_2}-\frac{Ze^2}{s_1}-\frac{Ze^2}{s_2}+\frac{e^2}{s_{12}}\right]\psi
= E\psi
\end{eqnarray}</math>.

Define the reduced mass Bohr radius
<math>\begin{eqnarray}
a_\mu = \frac{\hbar^2}{\mu e^2} = \frac{m}{\mu}a_0
\end{eqnarray}</math>
and

<math>\begin{eqnarray}
\rho_i = \frac{Zs_i}{a_\mu}\nonumber\\
\varepsilon = \frac{E}{Z^2(e^2/a_\mu)}\nonumber
\end{eqnarray}</math>

The Schrödinger equation then becomes
<math>\begin{eqnarray}\ \ \
\left[
-\frac{1}{2}(\nabla^2_{\rho_1}+\nabla^2_{\rho_2})-\frac{\mu}{M}\vec{\nabla}_{\rho_1}\cdot\vec{\nabla}_{\rho_2} - \frac{1}{\rho_1}-\frac{1}{\rho_2}+\frac{Z^{-1}}{\rho_{12}}\right]\psi = \varepsilon \psi
\end{eqnarray}</math>

The effects of finite mass are thus

1. A reduced mass correction to all energy levels of <math>(e^2/a_\mu)/(e^2/a_0)
= \mu/m</math>, i.e.
<math>\begin{eqnarray}
E_M=\frac{\mu}{m}E_\infty
\end{eqnarray}</math>

2. A specific mass shift given in first order perturbation theory by
<math>\begin{eqnarray}
\Delta\varepsilon =
-\frac{\mu}{M}<\psi|\nabla_{\rho_1}\cdot\nabla_{\rho_2}|\psi>
\end{eqnarray}</math>
or

<math>\begin{eqnarray}
\Delta\varepsilon =
-\frac{\mu}{M}<\psi|\nabla_{\rho_1}\cdot\nabla_{\rho_2}|\psi>\frac{e^2}{a_\mu}\nonumber\\
= -\frac{\mu^2}{mM}<\psi|\nabla_{\rho_1}\cdot\nabla_{\rho_2}|\psi>\frac{e^2}{a_0}\nonumber
\end{eqnarray}</math>

Since <math>\nabla_{\rho_1}\cdot\nabla_{\rho_2}</math> has the same angular properties as
<math>{\bf\rho}_1 \cdot {\bf\rho}_2</math>, the operator is like the product of two
dipole operators. For product type wave functions of the form
$
\psi = \psi(1s)\psi(nl) \pm \mbox{exchange},
$
the matrix element vanishes for all but P-states.

=Radiative Transitions=

==Transition Integrals and the Line Strength==

For an electric dipole transition <math>\gamma'\,L'\,S'\,J'\,M' \rightarrow \gamma\,L\,S\,J\,M</math>, the line strength is defined by
\begin{equation}
S = \sum_{M',\mu,M}|\langle\gamma'\,L'\,S'\,J'\,M'|r_\mu|\gamma\,L\,S\,J\,M\rangle|^2
\end{equation}
Using the Wigner-Eckart theorem, this is related to the reduced matrix element by
\begin{eqnarray}
S(J',J) = |\langle\gamma'\,L'\,S'\,J'||r||\gamma\,L\,S\,J\,M\rangle|^2
\sum_{M',\mu,M}\left|\left(\matrix{J' & 1 & J \cr
-M'& \mu &M \cr}\right)\right|^2
\end{eqnarray}
Using the sum rule for the 3-<math>j</math> symbols
\begin{equation}
\sum_{M',\mu,M}\left|\left(\matrix{J' & 1 & J \cr
-M'& \mu &M \cr}\right)\right|^2 = 1
\end{equation}
we see that the line strength is identical to the reduced matrix element
\begin{equation}
S(J',J) = |\langle\gamma'\,L'\,S'\,J'||r||\gamma\,L\,S\,J\,M\rangle|^2
\end{equation}
For a nonrelativistic calculation in <math>LS</math> coupling, <math>S = S'</math> and we can next strip out the dependence on <math>J</math> and <math>J'</math> by use of Eq.\ (7.1.7) of Edmonds { Angular Momentum in Quantum Mechanics} with <math>k = 1</math> to obtain
\begin{equation}
S(J',J) = (-1)^{L'+S+J+1}[(2J+1)(2J'+1)]^{1/2}\left\{
\matrix{L' & J'& S\cr
J & L & 1\cr}\right\}\langle\gamma' L'|| r || \gamma L\rangle
\end{equation}
where <math>\langle\gamma' L'|| r || \gamma L\rangle</math> is the reduced matrix element if spin is neglected. Note that if spin is neglected, then the line-strength is simply
\begin{equation}
S(L',L) = |\langle\gamma' L'|| r || \gamma L\rangle|^2
\end{equation}
To calculate the reduced matrix element, use a particular example along with the Wigner-Eckart theorem
such as
\begin{equation}
\langle\gamma'\, L'|| r || \gamma\, L\rangle = (-1)^{L'}\langle\gamma'\, L'\, 0| z | \gamma\, L\, 0\rangle{\Large /}\left(\matrix{L' & 1 & L \cr
0 & 0 & 0 \cr}\right)
\end{equation}

===Correlated Two-electron Integrals===
For the case of two-electron integrals, <math>z = z_1 + z_2</math> and <math>r = r_1 + r_2</math>. As discussed in G.W.F. Drake, Phys. Rev. A 18, 820 (1978), the general integral in Hylleraas coordinates for the integral over the <math>z_1</math> term can be written in the form
\begin{equation}
\label{int1}
\langle\gamma'\, L'\, 0| z_1 | \gamma\, L\, 0\rangle = \sum_\Lambda C_\Lambda I_\Lambda(\gamma',\gamma)
\end{equation}
where <math>I_\Lambda(\gamma',\gamma)</math> stands for the purely radial integral
\begin{equation}
\label{radial}
I_\Lambda(\gamma',\gamma) = \int_0^\infty r_1\,dr_1\int_0^\infty r_2\,dr_2\int_{|r_1-r_2|}^{r_1+r_2}r_{12}\, dr_{12}
R_{\gamma'}^*\,r_1\, R_\gamma P_\Lambda(\cos\theta_{12})
\end{equation}
and <math>\cos\theta_{12} = (r_1^2 + r_2^2 - r_{12}^2)/(2r_1r_2)</math> is a short-hand notation for a radial function of <math>r_1</math>, <math>r_2</math> and <math>r_{12}</math>. Also, <math>P_\Lambda(\cos\theta_{12})</math> is a Legendre polynomial,
and, <math>R_\gamma = R_\gamma(r_1,r_2,r_{12})</math> is the radial part of the Hylleraas wave function for state <math>\gamma</math> such
that the total wave function is <math>\psi({\bf r}_1,{\bf r}_2) = R_\gamma(r_1,r_2,r_{12}){\cal Y}_{l_1,l_2,L}^M(\hat{\bf r}_1,\hat{\bf r}_2)\pm {\rm exchange}</math>, normalized to unity, and <math>{\cal Y}_{l_1,l_2,L}^M(\hat{\bf r}_1,\hat{\bf r}_2)</math> is a vector coupled product of spherical harmonics. The coefficents <math>C_\Lambda</math> in Eq.\ (\ref{int1}) can be calculated from Eq.\ (32) of the above 1978 paper. They need to be calculated separately for he direct-direct and the direct-exchange contributions. For the reduced matrix element of <math>z_1</math>, the leading 3-<math>j</math> symbol can be dropped, leaving (for the case <math>k = 1</math>)
\begin{eqnarray}
C_\Lambda^{\rm reduced} &=& \frac{(-1)^{l_2'+L'+L+1}}{2} (l_1,l_1',l_2,l_2',L,L')^{1/2}
\sum_{\lambda={\rm max}(l_1-1,0)}^{l_1+1} (\lambda,\Lambda,1)
\left(\matrix{l_1 & 1 & \lambda \cr
0 & 0 & 0 \cr}\right)
\left(\matrix{l_1'& \lambda & \Lambda \cr
0 & 0 & 0 \cr}\right)\nonumber\\
&\times&\left(\matrix{l_2' & 1_2 & \Lambda \cr
0 & 0 & 0 \cr}\right)
\left\{\matrix{L' & \lambda & l_2 \cr
\Lambda& l_2' & l_1' \cr}\right\}
\left\{\matrix{L' & \lambda & l_2 \cr
l_1 & L & 1 \cr}\right\}
\end{eqnarray}
where the notation <math>(a,b,c,\cdots)</math> means <math>(2a+1)(2b+1)(2c+1)\cdots</math>.
A similar expression can be obtained for the angular coefficients for the <math>z_2</math> term by interchanging the subscripts 1 and 2 throughout and multiplying by <math>(-1)^{L+L'+1}</math>.

===Example===
As an example, for the <math>1s^2\;^1S - 1s2p\;^1P</math> transition of helium, the only nonvanishing <math>C_\Lambda</math> angular coefficients are <math>C_1 = 1/2</math> for the direct-direct term (i.e. <math>P_1(\cos\theta_{12}) = \cos\theta_{12}</math> is included in the radial integral (\ref{radial})) and <math>C_0 = 1/2</math> for the direct-exchange term (i.e. <math>P_0(\cos\theta_{12}) = 1</math> is included in the radial integral (\ref{radial})). The results for infinite nuclear mass are
\begin{equation}
S(1\;^1S-2\;^1P) = (0.781\,127\,158\,667)^2
\end{equation}
\begin{equation}f(1\;^1S-2\;^1P) = -0.276\,164\,703\,736
\end{equation}
where
\begin{equation}
f(1\;^1S-2\;^1P) = \frac{2}{3}[E(1\;^1S) - E(2\;^1P)]S(1\;^1S-2\;^1P)
\end{equation}
is the absorption oscillator strength.

As an example, for the <math>1s^2\;^1S - 1s2p\;^1P</math> transition of helium, the only nonvanishing <math>C_\Lambda</math> angular coefficients are <math>C_1 = 1/2</math> for the direct-direct term (i.e. <math>P_1(\cos\theta_{12}) = \cos\theta_{12}</math> is included in the radial integral (\ref{radial})) and <math>C_0 = 1/2</math> for the direct-exchange term (i.e. <math>P_0(\cos\theta_{12}) = 1</math> is included in the radial integral (\ref{radial})). The results for infinite nuclear mass are
\begin{equation}
S(1\;^1S-2\;^1P) = (0.781\,127\,158\,667)^2
\end{equation}
\begin{equation}f(1\;^1S-2\;^1P) = -0.276\,164\,703\,736
\end{equation}
where
\begin{equation}
f(1\;^1S-2\;^1P) = \frac{2}{3}[E(1\;^1S) - E(2\;^1P)]S(1\;^1S-2\;^1P)
\end{equation}
is the absorption oscillator strength.

As an example, for the <math>1s^2\;^1S - 1s2p\;^1P</math> transition of helium, the only nonvanishing <math>C_\Lambda</math> angular coefficients are <math>C_1 = 1/2</math> for the direct-direct term (i.e. <math>P_1(\cos\theta_{12}) = \cos\theta_{12}</math> is included in the radial integral (\ref{radial})) and <math>C_0 = 1/2</math> for the direct-exchange term (i.e. <math>P_0(\cos\theta_{12}) = 1</math> is included in the radial integral (\ref{radial})). The results for infinite nuclear mass are
\begin{equation}
S(1\;^1S-2\;^1P) = (0.781\,127\,158\,667)^2
\end{equation}
\begin{equation}f(1\;^1S-2\;^1P) = -0.276\,164\,703\,736
\end{equation}
where
\begin{equation}
f(1\;^1S-2\;^1P) = \frac{2}{3}[E(1\;^1S) - E(2\;^1P)]S(1\;^1S-2\;^1P)
\end{equation}
is the absorption oscillator strength.

As an example, for the <math>1s^2\;^1S - 1s2p\;^1P</math> transition of helium, the only nonvanishing <math>C_\Lambda</math> angular coefficients are <math>C_1 = 1/2</math> for the direct-direct term (i.e. <math>P_1(\cos\theta_{12}) = \cos\theta_{12}</math> is included in the radial integral (\ref{radial})) and <math>C_0 = 1/2</math> for the direct-exchange term (i.e. <math>P_0(\cos\theta_{12}) = 1</math> is included in the radial integral (\ref{radial})). The results for infinite nuclear mass are
\begin{equation}
S(1\;^1S-2\;^1P) = (0.781\,127\,158\,667)^2
\end{equation}
\begin{equation}f(1\;^1S-2\;^1P) = -0.276\,164\,703\,736
\end{equation}
where
\begin{equation}
f(1\;^1S-2\;^1P) = \frac{2}{3}[E(1\;^1S) - E(2\;^1P)]S(1\;^1S-2\;^1P)
\end{equation}
is the absorption oscillator strength.

As an example, for the <math>1s^2\;^1S - 1s2p\;^1P</math> transition of helium, the only nonvanishing <math>C_\Lambda</math> angular coefficients are <math>C_1 = 1/2</math> for the direct-direct term (i.e. <math>P_1(\cos\theta_{12}) = \cos\theta_{12}</math> is included in the radial integral (\ref{radial})) and <math>C_0 = 1/2</math> for the direct-exchange term (i.e. <math>P_0(\cos\theta_{12}) = 1</math> is included in the radial integral (\ref{radial})). The results for infinite nuclear mass are
\begin{equation}
S(1\;^1S-2\;^1P) = (0.781\,127\,158\,667)^2
\end{equation}
\begin{equation}f(1\;^1S-2\;^1P) = -0.276\,164\,703\,736
\end{equation}
where
\begin{equation}
f(1\;^1S-2\;^1P) = \frac{2}{3}[E(1\;^1S) - E(2\;^1P)]S(1\;^1S-2\;^1P)
\end{equation}
is the absorption oscillator strength.

As an example, for the <math>1s^2\;^1S - 1s2p\;^1P</math> transition of helium, the only nonvanishing <math>C_\Lambda</math> angular coefficients are <math>C_1 = 1/2</math> for the direct-direct term (i.e. <math>P_1(\cos\theta_{12}) = \cos\theta_{12}</math> is included in the radial integral (\ref{radial})) and <math>C_0 = 1/2</math> for the direct-exchange term (i.e. <math>P_0(\cos\theta_{12}) = 1</math> is included in the radial integral (\ref{radial})). The results for infinite nuclear mass are
\begin{equation}
S(1\;^1S-2\;^1P) = (0.781\,127\,158\,667)^2
\end{equation}
\begin{equation}f(1\;^1S-2\;^1P) = -0.276\,164\,703\,736
\end{equation}
where
\begin{equation}
f(1\;^1S-2\;^1P) = \frac{2}{3}[E(1\;^1S) - E(2\;^1P)]S(1\;^1S-2\;^1P)
\end{equation}
is the absorption oscillator strength.

===Test===
This is a test edit

Theory Notes

2020-12-14T04:28:10Z

Drake: /* Example */

<center>
<H1>Notes on solving the Schrödinger equation in Hylleraas coordinates for heliumlike atoms
Gordon W.F. Drake
Department of Physics, University of Windsor
Windsor, Ontario, Canada N9B 3P4
(Transcribed from hand-written notes and edited by Lauren Moffatt. Last revised November 22, 2017 to correct a missing subscript in the expression for $B_1$ in the Hamiltonian in Sect. 11 General Hermitian Property.)
</H1>
</center>

==Introduction==
These notes describe some of the technical details involved in solving the Schrödinger equation for a heliumlike atom of nuclear charge Z in correlated Hylleraas coordinates. The standard for computational purposes is to express all quantities in a form that is valid for electronic states of arbitrary angular momentum, and to express matrix elements of the Hamiltonian in a manifestly Hermitian symmetrized form. Further information and derivations of the mathematical techniques for angular integrals and radial recursion relations can be found in the paper
G.W.F. Drake, Phys. Rev. A <math>{\bf 18}</math>, 820 (1978).

The starting point is the two-electron Schrödinger equation for infinite nuclear mass
\begin{equation} \left[-\frac{\hbar^2}{2m}(\nabla^2_1 +\nabla^2_2) - \frac{Ze^2}{r_1} -
\frac{Ze^2}{r_2}+\frac{e^2}{r_{12}} \right]\psi = E\psi\nonumber
\end{equation}
where <math>m</math> is the electron mass, and <math>r_{12} = |{\bf r}_1 - {\bf r}_2|</math> (see diagram below).

[[File:Pic1.jpg|250px]]

Begin by rescaling distances and energies so that the Schrödinger equation can be expressed in a dimensionless form. The dimensionless Z-scaled distance is defined by <math>\begin{eqnarray} \rho = \frac{Zr}{a_0} \end{eqnarray} </math> where <math> \begin{eqnarray} a_0 = \frac{\hbar^2}{me^2} \end{eqnarray} </math> is the atomic unit (a.u.) of distance equal to the Bohr radius <math>0.529\, 177\, 210\, 92(17) \times 10^{-10}</math> m. Then

\begin{equation}[-\frac{\hbar^2}{2m}Z^2(\frac{me^2}{\hbar^2})^2(\nabla^2_{\rho_1}+\nabla^2_{\rho_2})
- Z^2\frac{e^2}{a_0}\rho^{-1}_1 - Z^2\frac{e^2}{a_0}\rho^{-1}_2 +
\frac{e^2}{a_0}Z\rho^{-1}_{12}]\psi= E\psi\nonumber\end{equation}

But <math>\begin{eqnarray} \frac{\hbar^2}{m}(\frac{me^2}{\hbar^2})^2 = \frac{e^2}{a_0} \end{eqnarray} </math> is the hartree atomic unit of of energy <math>(E_h = 27.211\,385\,05(60)</math> eV, or equivalently <math>E_h/(hc) = 2194.746\,313\,708(11)</math> m<math>^{-1})</math>. Therefore, after multiplying through by <math>a_0/(Ze)^2</math>, the problem to be solved in Z-scaled dimensionless units becomes

\begin{equation} \left[-\frac{1}{2}(\nabla^2_{\rho_1}+\nabla^2_{\rho_2}) -
\frac{1}{\rho_1} - \frac{1}{\rho_2} + \frac{Z^{-1}}{\rho_{12}}\right]\psi =
\varepsilon\psi\nonumber \end{equation}

where <math>\begin{eqnarray}\varepsilon = \frac{Ea_0}{(Ze)^2} \end{eqnarray}</math> is the energy in Z-scaled atomic units. For convenience, rewrite this in the conventional form
\begin{equation}
H\psi = \varepsilon\psi\nonumber \end{equation}
where (using <math>r</math> in place of <math>\rho</math> for the Z-scaled distance)
\begin{equation}
H = -\frac{1}{2}(\nabla^2_1+\nabla^2_2) - \frac{1}{r_1}-\frac{1}{r_2} + \frac{Z^{-1}}{r_{12}}
\nonumber\end{equation}
is the atomic Hamiltonian for infinite nuclear mass.

==The Hartree Fock Method==

For purposes of comparison, and to define the correlation energy, assume that <math>\psi({\bf r}_1,{\bf r}_2)</math> can be written in the separable product form
\begin{equation}
\psi({\bf r}_1,{\bf r}_2) = \frac{1}{\sqrt{2}} \left[ u_1(r_1) u_2(r_2) \pm u_2(r_1) u_1(r_2) \right] \nonumber
\end{equation}
for the <math>1s^2\ ^1\!S</math> ground state. Because of the <math>\frac{1}{r_{12}}</math> term in the Schrödinger equation, the exact solution connot be expressed in this form as a separable product. However, the Hartree-Fock approximation corresponds to finding the best possible solution to the Schrödinger equation
\begin{equation}
H\psi({\bf r}_1,{\bf r}_2) = E\psi({\bf r}_1,{\bf r}_2)\nonumber
\end{equation}
that can nevertheless be expressed in this separable product form, where as before
\begin{equation}
H = -\frac{1}{2}(\nabla^2_1+\nabla^2_2) - \frac{1}{r_1}- \frac{1}{r_2} + \frac{Z^{-1}}{r_{12}}\nonumber
\end{equation}
is the full two-electron Hamiltonian. To find the best solution, substitute into <math>\left\langle\psi \lvert H-E \rvert \psi\right\rangle</math> and require this expression to be stationary
with respect to arbitrary infinitesimal variations <math>\delta u_1</math> and <math>\delta
u_2</math> in <math>u_1</math> and <math>u_2</math> respectively; i.e.

\begin{equation}
\frac{1}{2} \left\langle \delta u_1(r_1)u_2(r_2) \pm u_2(r_1)\delta
u_1(r_2) \lvert H-E \rvert u_1(r_1) u_2(r_2)\pm u_2(r_1) u_1(r_2)\right\rangle\nonumber
\end{equation}

\begin{equation}
=\int\delta u_1(r_1) d{\bf r}_1 \left\{ \int d{\bf r}_2 u_2(r_2) \left( H - E \right) \left[ u_1(r_1)u_2(r_2) \pm u_2(r_1) u_1(r_2) \right] \right\}\nonumber
\end{equation}

\begin{equation}
= 0 \nonumber
\end{equation}

for arbitrary <math> \delta u_1(r_1).\nonumber </math> Therefore <math>\{\int d{\bf r}_2 \ldots \} = 0</math>.

Similarly, the coefficient of <math>\delta u_2</math> would give

\begin{equation}
\int d{\bf r}_1 u_1(r_1) \left( H-E \right) \left[ u_1(r_1) u_2(r_2) \pm
u_2(r_1) u_1(r_2) \right] = 0\nonumber
\end{equation}

Define

\begin{equation}
I_{12} = I_{21} = \int d{\bf r}\, u_1(r)u_2(r), \nonumber
\end{equation}

\begin{equation}
H_{ij} = \int d{\bf r}\, u_i(r)(-\frac{1}{2}\nabla - \frac{1}{r})u_j(r), \nonumber
\end{equation}

\begin{equation}
G_{ij}(r) = \int d{\bf r}^\prime u_i(r^\prime)\frac{1}{|{\bf r} - {\bf r}\prime|}u_j(r^\prime)\nonumber
\end{equation}

Then the above equations become the pair of integro-differential equations

\begin{equation}
[ H_0 - E + H_{22}+G_{22}(r)]u_1(r) = \mp [ I_{12}(H_0-E) + H_{12}+G_{12}(r)]u_2(r)\nonumber
\end{equation}

\begin{equation}
[H_0-E+H_{11}+G_{11}(r)]u_2(r) = \mp [I_{12}(H_0-E) + H_{12}+G_{12}(r)]u_1(r)\nonumber
\end{equation}

These must be solved self-consistently for the "constants" <math>I_{12}</math> and
<math>H_{ij}</math> and the function <math>G_{ij}(r)</math>.

The Hartree Fock energy is <math>E \simeq -2.87\ldots </math>a.u. while the exact energy is <math>E = -2.903724\ldots </math>a.u.

The difference is called the "correlation energy" because it arises from the
way in which the motion of one electron is correlated to the other. The
Hartree Fock equations only describe how one electron moves in the average field
provided by the other (mean-field theory).

==Configuration Interaction==
Expand
\begin{equation}
\psi({\bf r}_1,{\bf r}_2)= C_0u^{(s)}_1(r_1)u^{(s)}_1(r_2) + C_1u^{(P)}_1({\bf r}_1)u^{(P)}_1({\bf r}_2){\cal Y}^0_{1,1,0}(\hat{\bf r}_1, \hat{\bf r}_2)+C_2u^{(d)}_1({\bf r}_1)u^{(d)}_2({\bf r}_2){\cal Y}^0_{2,2,0}(\hat{\bf r}_1, \hat{\bf r}_2)+... \pm {\rm\ exchange}
\end{equation}
where
\begin{equation}
{\cal Y}^M_{l_1,l_2,L}(\hat{\bf r}_1, \hat{\bf r}_2)=\sum_{m_1,m_2}Y^{m_1}_{l_1}({\bf r}_1)Y^{m_2}_{l_2}({\bf r}_2)\times <l_1l_2m_1m_2\mid LM>
\end{equation}

This works, but is slowly convergent, and very laborious. The best CI calculations are accurate to <math> ~10^{-7}</math> a.u.

==Hylleraas Coordinates==
[E.A. Hylleraas, Z. Phys. <math>{\bf 48}, 469(1928)</math> and <math>{\bf 54}, 347(1929)</math>]
suggested using the co-ordinates <math>r_1</math>, <math>r_2</math> and <math>r_{12}</math> or equivalently

\begin{eqnarray}
s &=& r_1 + r_2, \nonumber\\
t &=& r_1-r_2, \nonumber\\
u &=& r_{12}\nonumber
\end{eqnarray}

and writing the trial functions in the form

\begin{equation}
\Psi({\bf r}_1,{\bf r}_2) = \sum^{i+j+k\leq
N}_{i,j,k}c_{i,j,k}r_1^{i+l_1}r_2^{j+l_2}r_{12}^ke^{-\alpha r_1 - \beta r_2}
\mathcal{Y}^M_{l_1,l_2,L}(\hat{r}_1,\hat{r}_2)\pm \text{exchange}\nonumber
\end{equation}

Diagonalizing H in this non-orthogonal basis set is equivalent to solving
<math> \begin{eqnarray}
\frac{\partial E}{\partial c_{i,j,k}} = 0\nonumber
\end{eqnarray} </math>
for fixed <math>\alpha</math> and <math>\beta</math>.

The diagonalization must be repeated for different values of <math>\alpha</math> and
<math>\beta</math> in order to optimize the non-linear parameters.

==Completeness==
The completeness of the above basis set can be shown by first writing
\begin{equation}
r_{12}^2 = r_1^2 + r_2^2 - 2r_1r_2\cos\theta_{12}
\end{equation}
and, from the spherical harmonic addition theorem,
\begin{equation}
\cos(\theta_{12})=\frac{4\pi}{3}\sum^1_{m=-1}Y^{m*}_1(\theta_1,\varphi_1)Y^m_1(\theta_2,\varphi_2)
\end{equation}
Consider first the S-states. The <math>r_{12}^0</math> terms are like the ss terms in a CI calculation. The
<math>r_{12}^2</math> terms bring in p-p type contributions, and the higher powers bring
in d-d, f-f etc. type terms. In general

\begin{equation}
P_l(\cos\theta_{12}) = \frac{4\pi}{2l+1}\sum^l_{m=-l}{Y^{m}_l}^*(\theta_1,\varphi_1)Y^m_l(\theta_2,\varphi_2)\nonumber
\end{equation}

For P-states, one would have similarly

\begin{array}{lr}
r_{12}^0 & (sp)P\nonumber\\
r_{12}^2 & (pd)P\nonumber\\
r_{12}^4 & (df)P\nonumber\\
\vdots & \vdots\nonumber
\end{array}

For D-states

\begin{array}{lr}
r_{12}^0 & (sd)D & (pp^\prime)D\nonumber\\
r_{12}^2 & (pf)D & (dd^\prime)D\nonumber\\
r_{12}^4 & (dg)D & (ff^\prime)D\nonumber\\
\vdots & \vdots & \vdots\nonumber
\end{array}

In this case, since there are two "lowest-order" couplings to form a
D-state, both must be present in the basis set, i.e.

\begin{equation}
\Psi(r_2,r_2) = \sum c_{ijk}r_1^ir_2^{j+2}r_{12}^ke^{-\alpha r_1-\beta
r_2}\mathcal{Y}^M_{022}(\hat{r}_1,\hat{r}_2)
+\sum d_{ijk}r_1^{i+1}r_2^{j+1}r_{12^k}e^{-\alpha^\prime r_1 - \beta^\prime
r_2}\mathcal{Y}^M_{112}(\hat{r}_1,\hat{r}_2)\nonumber
\end{equation}

For F-states, one would need <math>(sf)F</math> and <math>(pd)F</math> terms.

For G-states, one would need <math>(sg)G</math>, <math>(pf)G</math> and <math>(dd^\prime)G</math> terms.

Completeness of the radial functions can be proven by considering the
Sturm-Liouville problem

\begin{equation}
\left(-\frac{1}{2}\nabla^2-\frac{\lambda}{r_s}-E\right)\psi({\bf r}) = 0\nonumber
\end{equation}

or

\begin{equation}
\left(-\frac{1}{2}\frac{1}{r^2}\frac{\partial}{\partial r}\left({r^2}\frac{\partial}{\partial r}\right) -
\frac{l\left(l+1\right)}{2r^2} - \frac{\lambda}{r} - E\right)u(r) = 0.\nonumber
\end{equation}

For fixed E and variable <math>\lambda</math> (nuclear charge).

The eigenvalues are <math>\lambda_n = (E/E_n)^{1/2}</math>, where <math>E_n =- \frac{1}{2n^2}</math>

[[File:Pic2.jpg|500px]]

and the eigenfunctions are

\begin{equation}
u_{nl}(r) =
\frac{1}{(2l+1)!}\left(\frac{(n+l)!}{(n-l-1)2!}\right)^{1/2}(2\alpha)^{3/2}e^{-\alpha
r}\nonumber
\times \ \left( 2 \alpha r \right)^l\,_{1}F_{1}\left(-n+l+1,2l+2; 2 \alpha r \right)
\end{equation}

with <math>\alpha = (-2E)^{1/2}</math> and <math>n\geq l+1</math>. The confluent hypergeometric function <math> _1F_1(a,b,;z)</math> then denotes a finite polynomial since <math>a=-n+l+1</math> is a negative integer or zero.

Unlike the hydrogen spectrum, which has both a discrete part for <math>E<0</math> and a
continuous part for <math>E>0</math>, this forms an entirely discrete set of finite
polynomials, called Sturmian functions. They are orthogonal with respect to
the potential, i.e.

\begin{equation}
\int^\infty_0 r^2dr \left( u_{n^\prime l}(r)\frac{1}{r}u_{nl}(r) \right) = \delta_{n,n^\prime}\nonumber
\end{equation}

Since they become complete in the limit <math>n\rightarrow\infty</math>, this assures the
completeness of the variational basis set.

[See also B Klahn and W.A. Bingel Theo. Chim. Acta (Berlin) 44, 9 and 27
(1977)].

==Solutions of the Eigenvalue Problem==

For convenience, write

\[
\Psi({\bf r}_1,{\bf r}_2) = \sum^N_{m=1}c_m\varphi_m\nonumber
\]

where <math>m</math> represents the <math>m</math>'th combination of <math>i,j,k</math> and

\begin{equation}
\varphi_{ijk}=r_1^ir_2^jr_{12}^ke^{-\alpha r_1 - \beta
r_2}\mathcal{Y}^M_{l_1,l_2,L}(\hat{r}_1,\hat{r}_2) \pm \mbox{exchange}\nonumber
\end{equation}

\begin{align}

&
\left( \begin{array}{lr}
\cos(\theta) & \sin(\theta)\\
-\sin(\theta) & \cos(\theta)
\end{array} \right)
\left( \begin{array}{lr}
H_{11} & H_{12} \\
H_{12} & H_{22}
\end{array}\right)
\left( \begin{array}{lr}
\cos(\theta) & -\sin(\theta)\\
\sin(\theta) & \cos(\theta)
\end{array} \right)
\\[5pt]
= & \left(\begin{array}{lr}
cH_{11}+sH_{12} & cH_{12} + sH_{22}\\
-sH_{11} + cH_{12} & -sH_{12} + cH_{22}
\end{array}\right)
\left( \begin{array}{cc}
c & -s \\
s & c
\end{array}\right)
\\[5pt]
= & \left(\begin{array}{lr}
c^2H_{11}+s^2H_{22} + 2csH_{12} & (c^2-s^2)H_{12}+cs(H_{22}-H_{11})\\
(c^2-s^2)H_{12}+cs(H_{22}-H_{11}) & s^2H_{11}+c^2H_{22}-2csH_{12}
\end{array}\right)
\end{align}

Therefore

\begin{equation}
\left[\cos^2(\theta)-\sin^2(\theta)\right]H_{12} =
\cos(\theta)\sin(\theta)(H_{11}-H_{22})\nonumber
\end{equation}

and

\begin{equation}
\tan(2\theta) = \frac{2H_{12}}{H_{11}-H_{22}}\nonumber
\end{equation}

i.e.

\begin{equation}
\cos(\theta)=\left(\frac{r+\omega}{2r}\right)^{1/2}\nonumber
\end{equation}

\begin{equation}
\sin(\theta)=-\mbox{sgn}(H_{12})\left(\frac{r-\omega}{2r}\right)^{1/2}\nonumber
\end{equation}

where

\begin{equation}
\omega = H_{22}-H_{11}\nonumber
\end{equation}

\begin{equation}
r=\left(\omega^2+4H_{12}^2\right)^{1/2}\nonumber
\end{equation}

\begin{equation}
E_1=\frac{1}{2}\left(H_{11}+H_{22}-r\right)\nonumber
\end{equation}

\begin{equation}
E_2=\frac{1}{2}\left(H_{11}+H_{22}+r\right)\nonumber
\end{equation}

===Brute Force Method===

- Gives all the eigenvalues and eigenvectors, but it is slow

- First orthonormalize the <math>N</math>-dimensional basis set; i.e. form linear combinations

\begin{equation}
\Phi_m = \sum_{n=1}^N\varphi_nR_{nm}\nonumber
\end{equation}

such that
\begin{equation}
\langle \Phi_m | \Phi_n \rangle = \delta_{m,n}
\end{equation}

This can be done by finding an orthogonal tranformation T such that

\begin{equation}
T^TOT = I = \left(
\begin{array}{ccccc}
I_1 & 0 & 0 & \ldots & 0 \\
0 & I_2 & 0 & \ldots & 0 \\
0 & 0 & I_3 & \ldots & 0 \\
\vdots & \vdots & \vdots & \ddots & \vdots \\
0 & 0 & 0 & \ldots & I_N
\end{array}\right);
\end{equation}

\begin{equation}
O_{mn} = \langle \varphi_m | \varphi_n \rangle \nonumber
\end{equation}

and then applying a scale change matrix

\begin{equation}
S = \left(\begin{array}{ccccc}
\frac{1}{I_1^{1/2}} & 0 & 0 & \ldots & 0\\
0 & \frac{1}{I_2^{1/2}} & 0 & \ldots & 0 \\
0 & 0 & \frac{1}{I_3^{1/2}} & \ldots & 0 \\
\vdots & \vdots & \vdots & \ddots & \vdots \\
0 & 0 & 0 & \ldots & I_N^{1/2}
\end{array}\right)= S^T\nonumber
\end{equation}

Then

\begin{equation}
S^TT^TOTS = 1
\end{equation}

i.e.

\begin{equation}
R^TOR = 1
\end{equation}

with <math>R=TS</math>.

If H is the matrix with elements <math>H_{mn} = \langle \varphi_m | \varphi_n \rangle </math>, then H
expressed in the <math>\Phi_m</math> basis set is

\begin{equation}
H^\prime = R^THR.\nonumber
\end{equation}

We next diagonalize <math>H^\prime</math> by finding an orthogonal transformation W such
that

\begin{equation}
W^T H^\prime W = \lambda = \left(
\begin{array}{cccc}
\lambda_1 & 0 & \ldots & 0 \\
0 & \lambda_2 & \ldots & 0 \\
\vdots & \vdots & \ddots & \vdots \\
0 & 0 & \ldots & \lambda_N
\end{array}\right)\nonumber
\end{equation}

The q'th eigenvector is

\begin{equation}
\Psi^{(q)} = \sum_{n=1}^N\Phi_n W_{n,q} = \sum_{n,n^\prime} \varphi_{n^\prime} R_{n^\prime ,n} W_{n,q} \nonumber
\end{equation}

i.e.

\begin{equation}
c_{n^\prime}^{(q)} = \sum_{n=1}^N R_{n^\prime n} W_{n,q}
\end{equation}

===The Power Method===

- Based on the observation that if H has one eigenvalue, <math>\lambda_M</math>, much bigger than all the rest, and <math>\chi = \left(
\begin{array}{c}a_1\\a_2\\\vdots\end{array}\right)</math>
is an arbitrary starting vector, then <math>\chi = \sum_{q=1}^N x_q\Psi^{(q)}</math>.

\begin{equation}
(H)^n\chi = \sum_{q=1}^N x_q \lambda^n_q\Psi^{(q)}\nonumber\rightarrow x_M\lambda_M^n\Psi^{(M)}\nonumber
\end{equation}

provided <math>x_M\neq 0</math>.

To pick out the eigenvector corresponding to any eigenvalue, write the original problem in the form

\begin{align}
H\Psi & = \lambda O \Psi \nonumber \\
\left( H - \lambda_q O \right) \Psi & = \left( \lambda - \lambda_q \right) O \Psi \nonumber
\end{align}

Therefore,

\begin{equation}
G \Psi = \frac{1}{\lambda - \lambda_q } \Psi \nonumber \\
\end{equation}

where <math>G=(H-\lambda_qO)^{-1}O</math> with eigenvalues <math>\frac{1}{\lambda_n - \lambda_q}</math>.

By picking <math>\lambda_q</math> close to any one of the <math>\lambda_n</math>, say
<math>\lambda_{n^\prime}</math>, then <math>\frac{1}{\lambda_n-\lambda_q}</math> is much larger for
<math>n=n^\prime</math> than for any other value. The sequence is then

\begin{array}{c}
\chi_1=G\chi\nonumber\\
\chi_2=G\chi_1\nonumber\\
\chi_3=G\chi_2\nonumber\\
\vdots\nonumber
\end{array}

until the ratios of components in <math>\chi_n</math> stop changing.

- To avoid matrix inversion and multiplication, note that the sequence is equivalent to
\begin{equation}
F \chi_n = \left( \lambda - \lambda_q \right) O \chi_{n-1} \nonumber \\
\end{equation}
where <math>F = H-\lambda_qO</math>. The factor of <math>\left( \lambda - \lambda_q \right)</math> can be
dropped because this only affects the normalization of <math>\chi_n</math>. To find <math>\chi_n</math>, solve
\begin{equation}
F \chi_n = O \chi_{n-1} \nonumber
\end{equation}
(N equations in N unknowns). Then
\begin{equation}
\lambda = \frac{ \langle \chi_n | H | \chi_n \rangle}{\langle \chi_n | \chi_n \rangle} \nonumber
\end{equation}

==Matrix Elements of H==

\begin{equation}
H=-\frac{1}{2}\nabla^2_1 -\frac{1}{2}\nabla^2_2 - \frac{1}{r_1} - \frac{1}{r_2} +\frac{Z^{-1}}{r_{12}}\nonumber
\end{equation}

Taking <math>r_1, r_2</math> and <math>r_{12}</math> as independent variables,

\begin{equation}
\nabla_1^2 = \frac{1}{r_1^2}\frac{\partial}{\partial r_1}\left(r_1^2\frac{\partial}{\partial r_1}\right)+ \frac{1}{r_{12}^2}\frac{\partial}{\partial r_{12}}
\left(r_{12}^2\frac{\partial}{\partial r_{12}}\right)\nonumber -\frac{l_1(l_1+1)}{r_1^2}+2(r_1 - r_2\cos(\theta_{12}))\frac{1}{r_{12}}\frac{\partial^2}{\partial r_1 \partial r_{12}} \nonumber - 2(\nabla_1^Y \cdot {\bf r}_2)\frac{1}{r_{12}}\frac{\partial}{\partial r_{12}}\nonumber
\end{equation}

where <math>\nabla_1^Y</math> acts only on the spherical harmonic part of the wave function and the diagram

[[File:Pic3.jpg|400px]]

defines the complete set of 6 independent variables is <math>r_1</math>, <math>r_2</math>, <math>r_{12}</math>, <math>\theta_1</math>, <math>\varphi_1</math>, <math>\chi</math>.

If <math>r_{12}</math> were not an independent variable, then one could take the volume element to be

\begin{equation}
d\tau = r_1^2dr_1\sin(\theta_1)d\theta_1d\varphi_1r_2^2dr_2\sin(\theta_2)d\theta_2d\varphi_2.\nonumber
\end{equation}

However, <math>\theta_2</math> and <math>\varphi_2</math> are no longer independent variables. To eliminate them, take the point <math>{\bf r}_1</math> as the
origin of a new polar co-ordinate system, and write

\begin{equation}
d\tau=-r_1^2dr_1\sin(\theta_1)d\theta_1d\varphi_1r_{12}^2dr_{12}\sin(\psi)d\psi d\chi\nonumber
\end{equation}

and use

\begin{equation}
r_2^2=r_1^2+r_{12}^2 +2r_1r_{12}\cos(\psi)
\end{equation}

Then for fixed <math>r_1</math> and <math>r_{12}</math>,

\begin{equation}
2r_2dr_2 = -2r_1r_{12}\sin(\psi)d\psi
\end{equation}

Thus

\begin{equation}
d\tau= r_1dr_1r_2dr_2r_{12}dr_{12}\sin(\theta_1)d\theta_1d\varphi_1d\chi
\end{equation}

The basic type of integral to be calculated is

\begin{align}
I \left( l_1, m_1, l_2, m_2; R \right) & = \int \sin \left( \theta_1 \right) d\theta_1 d\varphi_1 d\chi Y^{m_1}_{l_1} \left( \theta_1, \varphi_1 \right)^{*} Y^{m_2}_{l_2} \left( \theta_2, \varphi_2 \right) \\
& \times \int r_1dr_1r_2dr_2r_{12}dr_{12}R \left( r_1, r_2, r_{12} \right)
\end{align}

Consider first the angular integral. <math>Y^{m_2}_{l_2} \left( \theta_2, \varphi_2 \right)</math> can be expressed in terms of the independent variables <math>\theta_1, \varphi_1, \chi</math> by use of the rotation matrix relation
\begin{equation}
Y^{m_2}_{l_2}( \theta_2, \varphi_2) = \sum_m \mathcal{D}^{(l_2)}_{m_2,m}(\varphi_1,\theta_1,\chi)^*Y^m_{l_2}( \theta_{12}, \varphi)
\end{equation}
where <math>\theta, \varphi</math> are the polar angles of <math>{\bf r}_2</math> relative to <math>{\bf r}_1</math>. The angular integral is then

\begin{align}
I_{ang} & = \int^{2 \pi}_{0} d\chi \int^{2\pi}_{0} d\varphi_1 \int^{\pi}_{0} \sin \left( \theta_1 \right) d\theta_{1} Y^{m_1}_{l_1} \left( \theta_1, \varphi_1 \right)^{*} \\
& \times \sum_m \mathcal{D}^{(l_2)}_{m_2,m} \left( \varphi_1,\theta_1,\chi \right)^{*} Y^{m}_{l_2} \left( \theta_{12},\varphi \right) \\
\end{align}

Use

\begin{equation}
Y^{m_1}_{l_1} \left( \theta_1, \varphi_1 \right)^{*} = \sqrt{\frac{2l_1+1}{4\pi}}\mathcal{D}^{(l_1)}_{m_1,0} \left( \varphi_1, \theta_1, \chi \right)
\end{equation}

together with the orthogonality property of the rotation matrices (Brink and Satchler, p 147)

\begin{align}
& \int \mathcal{D}^{(j)*}_{m,m^\prime}\mathcal{D}^{(J)}_{M,M^\prime}\sin(\theta_1)d\theta_1d\varphi_1d\chi \\
& = \frac{8\pi^2}{2j+1}\delta_{jJ}\delta_{mM}\delta_{m^\prime M^\prime}\nonumber
\end{align}

to obtain

\begin{align}
I_{ang} & = \sqrt{ \frac{2l_1+1}{4\pi}} \frac{8\pi^2}{2l_1+1} \delta_{l_1,l_2} \delta_{m_1,m_2} Y^{0}_{l_2} \left( \theta_{12}, \varphi \right) \\
& = 2 \pi \delta_{l_1,l_2} \delta_{m_1,m_2} P_{l_2} \left( \cos \theta_{12} \right)
\end{align}

since

\begin{equation}
Y^0_{l_2} \left( \theta_{12}, \varphi \right) = \sqrt{\frac{2l_1+1}{4\pi}} P_{l_2} \left( \cos \left( \theta_{12} \right) \right)
\end{equation}

Note that <math>P_{l_2} \left( \cos \theta \right)</math> is just a short hand expression for a radial function because

\begin{equation}
\cos \theta_{12} = \frac{r_1^2 + r_2^2 - r_{12}^2}{2r_1r_2}
\end{equation}

The original integral is thus

\begin{equation}
I \left( l_1, m_1, l_2, m_2; R \right) = 2 \pi \delta_{l_1,l_2} \delta_{m_1,m_2} \int^{\infty}_{0} r_{1} dr_{1} \int^{\infty}_{0} r_{2} dr_{2} \int^{r_{1} + r_{2}}_{|r_{1} - r_{2}|} r_{12} dr_{12} R \left( r_{1}, r_{2}, r_{12} \right) P_{l_2} \left( \cos \theta_{12} \right)
\end{equation}

where again

\begin{equation}
\cos \theta_{12} = \frac{r_{1}^{2} +r^{2}_{2} - r_{12}^{2}}{2 r_{1} r_{2}}
\end{equation}

is a purely radial function.

The above would become quite complicated for large <math>l_2</math> because <math>P_{l_2}(\cos \theta_{12})</math> contains terms up to <math>\left( \cos \theta_{12} \right)^{l_2}</math>. However, recursion relations exist which allow any integral containing <math>P_l \left( \cos \theta_{12} \right)</math> in terms of those containing just <math>P_0 \left( \cos \theta_{12} \right) = 1</math> and <math>P_1 \left( \cos \theta_{12} \right) = \cos \theta_{12}</math>.

==Radial Integrals and Recursion Relations==

The basic radial integral is [see G.W.F. Drake, Phys. Rev. A 18, 820 (1978)]

\begin{align}
I_0 \left( a, b, c \right) & =
\int^{\infty}_{0} r_{1} dr_{1} \int^{\infty}_{r_{1}} r_{2} dr_{2} \int^{r_{1}+r_{2}}_{r_{2} - r_{1}} r_{12} dr_{12} r_{1}^{a} r_{2}^{b} r_{12}^{c} e^{-\alpha r_{1} - \beta r_{2}} + \int^{\infty}_{0} r_{2} dr_{2} \int^{\infty}_{r_{2}} r_{1} dr_{1} \int^{r_{1} + r_{2}}_{r_{1} - r_{2}} r_{12} dr_{12} r_{1}^{a} r_{2}^{b} r_{12}^{c} e^{- \alpha r_{1} - \beta r_{2}} \\
& = \frac{2}{c+2} \sum^{[(c+1)/2]}_{i=0}
\left(\begin{array}{c}
c+2 \\
2i+1
\end{array} \right)
\left\{ \frac{q!}{\beta^{q+1} \left( \alpha + \beta \right)^{p+1}} \sum_{j=0}^{q} \frac{\left( p + j \right)!}{j!} \left( \frac{\beta}{\alpha + \beta} \right)^j + \frac{q^{\prime}!}{\alpha^{q^{\prime}+1} \left( \alpha + \beta \right)^{p^{\prime}+1}} \sum^{q^{\prime}}_{j=0} \frac{\left( p^{\prime} + j \right)!}{j!} \left( \frac{\alpha}{\alpha + \beta} \right)^j \right\}
\end{align}

where

\begin{array}{ll}
p = a + 2i + 2 & p^{\prime} = b + 2i + 2 \\
q = b + c - 2i + 2 & q^{\prime} = a + c - 2i + 2
\end{array}

The above applies for <math> a, b \geq -2, c\geq -1</math>.
<math>[x]</math> means "greatest integer in" <math>x</math>.

Then

\begin{align}
I_1 \left( a, b, c \right) & = \int d\tau_{r} r_{1}^{a} r_{2}^{b} r_{12}^{c} e^{ -\alpha r_{1} - \beta r_{2}} P_{1}\left( \cos \theta \right) \\
& = \frac{1}{2} \left( I_{0} \left( a + 1, b - 1, c \right) + I_{0} \left( a - 1, b + 1, c \right) - I_{0} \left( a - 1, b - 1, c + 2 \right) \right)
\end{align}

===The Radial Recursion Relation===

Recall that the full integral is

\begin{equation}
I(l_1m_1,l_2m_2;R)=2\pi\delta_{l_1,l_2}\delta_{m_1,m_2}I_{l_2}(R)
\end{equation}

where

\begin{equation}
I_{l_2}(R) = \int d\tau_r R(r_1,r_2,r_{12})P_{l_2}(\cos \theta_{12})
\end{equation}

To obtain the recursion relation, use

\begin{equation}
P_l(x) = \frac{[P^{\prime}_{l+1}(x)-P^{\prime}_{l-1}(x)]}{2l+1}
\end{equation}

with

\begin{equation}
P_{l+1}^\prime (x) = \frac{d}{dx}P_{l+1}(x)
\end{equation}

Here <math>x=\cos\theta_{12}</math> and

\begin{align}
\frac{d}{d \cos \theta_{12}} & = \frac{dr_{12}}{d \cos \theta_{12}} \frac{d}{dr_{12}} \\
& = -\frac{r_{1} r_{2}}{r_{12}} \frac{d}{dr_{12}}
\end{align}

Then,

\begin{equation}
I_{l}\left( R \right) = - \int d\tau_r R \frac{r_{1} r_{2}}{r_{12}} \frac{d}{dr_{12}} \frac{ \left[ P_{l+1} \left( \cos \theta_{12} \right) - P_{l-1} \left( \cos \theta_{12} \right) \right]}{2l+1}
\end{equation}

The <math>r_{12}</math> part of the integral is

\begin{align}
& \int^{r_{1} + r_{2}}_{|r_{1} - r_{2}|} r_{12} dr_{12} R \frac{r_{1} r_{2}}{r_{12}} \frac{d}{dr_{12}} \left[P_{l+1}-P_{l-1} \right] \\
& = R r_{1} r_{2} \left[ P_{l+1} - P_{l-1} \right]^{r_{1} + r_{2}}_{|r_{1} - r_{2}|} - \int^{r_{1} + r_{2}}_{|r_{1} - r_{2}|} r_{12} dr_{12} \left( \frac{d}{dr_{12}} R \right) \frac{r_{1} r_{2}}{r_{12}} \frac{ \left[ P_{l+1} - P_{l-1} \right]}{2l+1}
\end{align}

The integrated term vanishes because

\begin{equation}
\cos \theta_{12} = \frac{r_{1}^{2} + r_{2}^{2} - r_{12}^{2}}{2 r_{1} r_{2}}
\begin{array}{lll}
& = -1 & {\rm when} & r_{12}^{2} = \left( r_{1} + r_{2} \right)^{2} \\
& = 1 & {\rm when} & r_{12}^{2} = \left( r_{1} - r_{2} \right)^{2}
\end{array}
\end{equation}

and <math>P_l(1)=1</math>, <math>P_{l}(-1)=(-1)^{l}</math>.

Thus,

\begin{equation}
I_{l+1}\left(\frac{r_1r_2}{r_{12}}R^\prime\right) = \left( 2l+1 \right) I_{l}(R) - I_{l-1} \left( \frac{r_{1} r_{2}}{r_{12}} R^{\prime} \right)
\end{equation}

If

\begin{equation}
R = r_{1}^{a-1} r_{2}^{b-1} r_{12}^{c+2} e^{-\alpha r_{1} - \beta r_{2}}
\end{equation}

then [G.W.F. Drake, Phys. Rev. A 18, 820 (1978)]

\begin{equation}
I_{l+1}(r_1^ar_2^br_{12}^c) =
\frac{2l+1}{c+2} I_{l} \left( r_{1}^{a-1} r_{2}^{b-2} r_{12}^{c+2} \right)+I_{l-1} \left( r_{1}^{a} r_{2}^{b} r_{12}^{c} \right)
\end{equation}

For the case <math>c=-2</math>, take

\begin{equation}
R = r_{1}^{a-1} r_{2}^{b-1} \ln r_{12} e^{ -\alpha r_{1} -\beta r_{2}}
\end{equation}

Then

\begin{equation}
I_{l+1} \left( r_{1}^{a} r_{2}^{b} r_{12}^{c} \right) = I_{l} \left( r_{1}^{a-1} r_{2}^{b-1} \ln r_{12} \right) \left( 2l + 1 \right) - I_{l-1} \left( r_{1}^{a} r_{2}^{b} r_{12}^{-2} \right)
\end{equation}

This allows all <math>I_{l}</math> integrals to be calculated from tables of <math>I_0</math> and
<math>I_1</math> integrals.

===The General Integral===

The above results for the angular and radial integrals can now be combined into a general formula for integrals of the type

\begin{equation}
I = \int \int d{\bf r}_{1} d{\bf r}_{2} R_{1} \mathcal{Y}^{M^{\prime}}_{l^{\prime}_{1} l^{\prime}_{2} L^{\prime}} \left( \hat{r}_{1}, \hat{r}_{2} \right) T_{k_{1} k_{2} K}^{Q} \left( {\bf r}_{1},{\bf r}_{2} \right) R_{2} \mathcal{Y}_{l_{1} l_{2} L}^{M} \left( \hat{r}_{1}, \hat{r}_{2} \right)
\end{equation}

where

\begin{equation}
\mathcal{Y}^{M}_{l_{1} l_{2} L} \left( \hat{r}_{1}, \hat{r}_{2} \right) = \sum_{m_1,m_2} \langle l_{1} l_{2} m_{1} m_{2} | LM \rangle Y^{m_{1}}_{l_{1}} \left( \hat{r}_{1} \right) Y^{m_{2}}_{l_{2}} \left( \hat{r}_{2} \right)
\end{equation}

and

\begin{equation}
T^Q_{k_{1} k_{2} K} \left( {\bf r}_{1}, {\bf r}_{2} \right) = \sum_{q_{1}, q_{2}} \langle k_{1} k_{2} q_{1} q_{2} | KQ \rangle Y_{k_{1}}^{q_{1}} \left( \hat{r}_{1} \right) Y^{q_{2}}_{k_{2}} \left( \hat{r}_{2} \right)
\end{equation}

The basic idea is to make repeated use of the formula

\begin{equation}
Y^{m_{1}}_{l_{1}} \left( \hat{r}_{1} \right) Y^{m_{2}}_{l_{2}} \left( \hat{r}_{1} \right) = \sum_{l m} \left( \frac{\left( 2l_{1}+1 \right) \left(2l_{2}+1 \right) \left(2l+1 \right)}{4 \pi} \right)^{1/2}
\times \left( \begin{array}{ccc}
l_1 & l_2 & l \\
m_1 & m_2 & m
\end{array} \right) \left(\begin{array}{ccc}
l_1 & l_2 & l \\
0 & 0 & 0
\end{array} \right)
Y^{m*}_{l} \left( \hat{r}_{1} \right)
\end{equation}

where

\begin{equation}
Y^{m*}(\hat{r}) = (-1)^mY_l^{-m}(\hat{r})
\end{equation}

and

\begin{equation}
\left(\begin{array}{ccc}
l_1 & l_2 & l \\
m_1 & m_2 & m
\end{array} \right) = \frac{\left( -1 \right)^{l_{1} - l_{2} - m}}{\left( 2l + 1 \right)^{1/2}} \left( l_{1} l_{2} m_{1} m_{2} | l, -m \right)
\end{equation}

is a 3-<math>j</math> symbol. In particular, write

\begin{array}{cc}
Y^{m_1^\prime}_{l_1^\prime}(\hat{r}_1)^{*} \underbrace{Y_{k_{1}}^{q_{1}} \left( \hat{r}_{1} \right) Y^{m_{1}}_{l_{1}} \left( \hat{r}_{1} \right)} & = & \sum_{\Lambda M}(\ldots)Y^{M}_\Lambda(\hat{r}_1)\\
\ \ \ \ \ \ \ \sum_{\lambda_1\mu_1}Y_{\lambda_1}^{\mu_1}(\hat{r}_1) & & \\
& & \\
Y^{m_{2}^{\prime}}_{l_{2}^{\prime}} \left( \hat{r}_{2} \right) \underbrace{Y^{q_{2}}_{k_{2}} \left( \hat{r}_{2} \right) Y_{l_{2}}^{m_{2}} \left( \hat{r}_{2} \right)} & = & \sum_{\Lambda^{\prime} M^{\prime}} \left( \ldots \right)Y_{\Lambda^{\prime}}^{M^{\prime}*} \left( \hat{r}_{2} \right) \\
\ \ \ \ \ \ \ \ \ \sum_{\lambda_{2} \mu_{2}}Y_{\lambda_{2}}^{\mu_{2}} \left( \hat{r}_{2} \right) & &
\end{array}

The angular integral then gives a factor of <math>2 \pi \delta_{\Lambda, \Lambda^{\prime}} \delta_{M, M^{\prime}} P_{\Lambda} \left( \cos \theta_{12} \right)</math>. The total integral therefore reduces to the form

\begin{equation}
I = \sum_{\Lambda} C_{\Lambda} I_{\Lambda} \left( R_{1} R_{2} \right)
\end{equation}

where <math>C_{\Lambda} = \sum_{\lambda_{1}, \lambda_{2}} C_{ \lambda_{1}, \lambda_{2}, \Lambda}</math>. For further details and derivations, including graphical representations, see G.W.F. Drake, Phys. Rev. A <math>{\bf 18}</math>, 820 (1978).

==Graphical Representation==

[[File:GraphRep1.jpg|500px]]

==Matrix Elements of H==
Recall that

\begin{equation}
H = -\frac{1}{2} \nabla^{2}_{1} - \frac{1}{2} \nabla^{2}_{2} - \frac{1}{r_1} - \frac{1}{r_2} + \frac{Z^{-1}}{r_{12}}
\end{equation}

Consider matrix elements of

\begin{equation}
\nabla^2_1 = \frac{1}{r_1^2} \frac{\partial}{\partial r_1}\left(r_1^2\frac{\partial}{\partial r_1} \right) + \frac{1}{r^2}\frac{\partial}{\partial r}\left(r^2\frac{\partial}{\partial r}\right) - \frac{(\vec{l}_1^Y)^2}{r_1^2} + \frac{2 \left( r_1 - r_2 \cos \theta \right)}{r} \frac{\partial^2}{\partial r_1 \partial r} - 2 \left( \nabla^Y_1 \cdot {\bf r}_2 \right) \frac{1}{r} \frac{\partial}{\partial r}
\end{equation}

where <math>r \equiv r_{12}</math> and <math>\cos\theta \equiv \cos\theta_{12}</math>. Also in what follows, define <math>\hat{\nabla}^Y_1 \equiv r_1\nabla^Y_1</math> where <math>\nabla^Y</math> operates only on the spherical harmonic part of the wave function, and similarly for <math>\vec{l}_1^Y</math>.

A general matrix element is

\begin{equation}
\langle r_1^{a^\prime} r_2^{b^\prime} r_{12}^{c^\prime} e^{-\alpha^\prime r_1 - \beta^\prime r_2} \mathcal{Y}^{M^\prime}_{l^\prime_1 l^\prime_2 L^\prime} \left( \hat{r}_1, \hat{r}_2 \right) \left| \nabla^2_1 \right| r_1^a r_2^b r_{12}^c e^{-\alpha r_1 - \beta r_2} \mathcal{Y}^M_{l_1 l_2 L} \left( \hat{r}_1, \hat{r}_2 \right) \rangle
\end{equation}

Since <math>\nabla^2_1</math> is rotationally invariant, this vanishes unless <math>L=L^\prime</math>, <math>M=M^\prime</math>. Also <math>\nabla_1^2</math> is Hermitian, so that the result must be the same whether it operates to the right or left, even though the results look very different. In fact, requiring the results to be the same yields some interesting and useful integral identities as follows:

Put

\begin{equation}
\mathcal{F} = F \mathcal{Y}^M_{l_1 l_2 L} \left( \hat{r}_1, \hat{r_2} \right)
\end{equation}

and

\begin{equation}
F = r_1^a r_2^b r_{12}^c e^{-\alpha r_1 - \beta r_2}
\end{equation}

Then

\begin{equation}
\nabla^2_1 \mathcal{F} = \left\{ \frac{1}{r_1^2} \left[ a \left( a + 1 \right) - l_1 \left( l_1 + 1 \right) \right] + \frac{c \left( c + 1 \right)}{r^2} + \alpha^2\frac{2\alpha \left( a + 1 \right)}{r_1} + \frac{2 \left(r_1 - r_2 \cos \theta \right) }{r_1r^2}c \left[ a - \alpha r_1 \right] - \frac{2c}{r^2} \left( \hat{\nabla}^Y_1 \cdot \hat{r}_2 \right) \frac{r_2}{r_1} \right\} \mathcal{F}
\end{equation}

and

\begin{align}
\langle \mathcal{F}^\prime \left| \nabla^2_1 \right| \mathcal{F} \rangle & = \sum_\Lambda \int d\tau_r F^\prime C_\Lambda \left( 1 \right) P_\Lambda \left( \cos \theta \right) \times \left\{ \frac{1}{r_1^2} \left[ a \left( a + 1 \right) - l_1 \left( l_1 + 1 \right) \right] -\frac{2 \alpha \left( a + 1 \right)}{r_1} + \frac{c \left( c + 1 \right)}{r^2} +\alpha^2 + \frac{2 \left( r_2 - r_2 \cos \theta \right)}{r_1r^2}c \left( a - \alpha r_1 \right) \right\} F \\
& + \sum_\Lambda \int d\tau_r F^\prime C_\Lambda \left( \hat{\nabla}^Y_1 \cdot \hat{r}_2 \right) P_\Lambda \left( \cos \theta \right) \left( \frac{-2c r_2}{r_1 r^2} \right) F
\end{align}

where

\begin{equation}
\int d\tau_r = \int^\infty_0 r_1 dr_1 \int^\infty_0 r_2 dr_2 \int^{r_1+r_2}_{\left| r_1 - r_2 \right|} r dr
\end{equation}

For brevity, let the sum over <math>\Lambda</math> and the radial integrations be understood, and let <math>\left( \nabla_1^2 \right)</math> stand for the terms which appear in the integrand. Then operating to the right gives

\begin{equation}
(\nabla_1^2)_R = \frac{1}{r_1^2} \left[ a \left( a + 1 \right) - l_1 \left( l_1 + 1 \right) \right] + \frac{c \left( c + 1 \right)}{r^2} + \alpha^2 - \frac{2 \alpha \left( a + 1 \right)}{r_1}
+ \frac{2 \left( r_1 - r_2 \cos \theta \right)}{r_1 r^2}c \left( a - \alpha r_1 \right) - \frac{2c r_2}{r_1 r^2} \left( \hat{\nabla}^Y_1 \cdot \hat{r}_2 \right)
\end{equation}

Operating to the left gives

\begin{equation}
(\nabla_1^2)_L = \frac{1}{r_1^2} \left[a^\prime \left( a^\prime + 1 \right) - l_1^\prime \left( l_1^\prime + 1 \right) \right] + \frac{c^\prime \left( c^\prime + 1 \right)}{r^2} + \alpha^{\prime 2} -\frac{2 \alpha^\prime \left( a^\prime + 1 \right)}{r_1}
+ \frac{2 \left( r_1 - r_2 \cos \theta \right)}{r_1 r^2} c^\prime \left( a^\prime - \alpha^\prime r_1 \right) - \frac{2 c^\prime r_2}{r_1 r^2} \left( \hat{\nabla}^{Y^\prime}_1 \cdot \hat{r}_2 \right)
\end{equation}

Now put

\begin{array}{cc}
a_+ = a+a^\prime, & \hat{\nabla}_1^+ = \hat{\nabla}_1^Y + \hat{\nabla}_1^{Y\prime} \\
a_{-} = a-a^\prime, & \hat{\nabla}_1^{-} = \hat{\nabla}_1^Y - \hat{\nabla}_1^{Y\prime}
\end{array}

etc., and substitute <math>a^\prime = a_+ - a</math>, <math>c^\prime = c_+ - c</math>, and <math>\alpha^\prime = \alpha_+ - \alpha</math> in <math>(\nabla^2_1)_L</math>. If <math>a_+</math>, <math>c_+</math> and <math>\alpha_+</math> are held fixed, then the equation

\begin{equation}
(\nabla^2_1)_R = (\nabla^2_1)_L
\end{equation}

must be true for arbitrary <math>a</math>, <math>c</math>, and <math>\alpha</math>. Their coefficients must thus vanish.

This yields the integral relations

\begin{equation}
\label{I}
\frac{\left( r_1 - r_2 \cos \theta \right)}{r_1 r^2} = \frac{1}{c_+} \left( \frac{- \left( a_+ + 1 \right)}{r_1^2} + \frac{\alpha_+}{r_1} \right) \tag{I}
\end{equation}

from the coefficient of <math>a</math> and

\begin{equation}
\label{II}
\frac{ \left( r_1 - r_2 \cos \theta \right) \left( a_+ - \alpha_+ r_1 \right)}{r_1 r^2} = \frac{r_2}{r_1 r^2}\left( \hat{r}_2 \cdot \hat{\nabla}_1^+ \right) - \frac{ \left( c_+ + 1 \right)}{r^2} \tag{II}
\end{equation}

from the coefficient of <math>c</math>. The coefficient of <math>\alpha</math> gives an equation equivalent to (I).

Furthermore, if can be show that (see problem)

\begin{align}
& \sum_\Lambda \int d\tau_r \frac{r^c}{r^2} C_\Lambda \left( \hat{r}_2 \cdot \hat{\nabla}_1^Y \right) P_\Lambda \left( \cos \theta \right) \\
= & \sum_\Lambda \int d\tau_r \frac{r^c}{c r_1 r_2}C_\Lambda \left( 1 \right)P_\Lambda \left( \cos \theta \right) \times \left( \frac{l_1^\prime \left( l_1^\prime + 1 \right) - l_1 \left( l_1 + 1 \right) -\Lambda \left( \Lambda + 1 \right)}{2} \right)
\end{align}

and similarly for <math>\left( \hat{r}_2 \cdot \hat{\nabla}^{Y\prime}_1 \right)</math> with <math>l_1</math> and <math>l_1^\prime</math> interchangeable, then it follows that

\begin{equation}
\frac{r^{c_+}}{r^2} \left( \hat{r}_2 \cdot \hat{\nabla}_1^+ \right) = - \frac{r^{c_+}}{c_+ r_1 r_2} \Lambda \left( \Lambda + 1 \right)
\end{equation}

and

\begin{equation}
\frac{r^{c_+}}{r^2} \left( \hat{r}_2 \cdot \hat{\nabla}_1^- \right) = \frac{r^{c_+}}{c_+ r_1 r_2} \left[ l_1^\prime \left( l_1^\prime + 1 \right) - l_1 \left( l_1 + 1 \right) \right]
\end{equation}

where equality applies after integration and summation over <math>\Lambda</math>.

Thus (II) becomes

\begin{equation}\label{IIa}
\frac{ \left( r_1 - r_2 \cos \theta \right) \left( a_+ - \alpha_+ r_1 \right)}{r_1 r^2} = - \frac{\Lambda \left( \Lambda + 1 \right)}{c_+ r^2_1} - \frac{ \left( c_+ + 1 \right)}{r^2} \tag{IIa}
\end{equation}

===Problem===

Prove the integral relation

\begin{align}
& \sum_\Lambda \int d\tau_r f \left( r_1, r_2 \right) \frac{1}{r} \left( \frac{d}{dr} q(r) \right) C_\Lambda \left( \hat{r}_2 \cdot \hat{\nabla}^Y_1 \right) P_\Lambda \left( \cos \theta \right) \\
= & \sum_\Lambda \int d\tau_r \frac{ f \left( r_1, r_2 \right)}{r_1, r_2} q(r) C_\Lambda \left( 1 \right) P_\Lambda \left( \cos \theta \right) \left(\frac{l_1^\prime \left( l_1^\prime + 1 \right) - l_1 \left( l_1 + 1 \right) - \Lambda \left( \Lambda + 1 \right)}{2}\right)
\end{align}

where the coefficients <math>C_\Lambda \left( 1 \right)</math> are the angular coefficients from the overlap integral

\begin{align}
\int & d\Omega\, \mathcal{Y}^{M*}_{l_1^\prime l_2^\prime L}(\hat{r}_1,\hat{r}_2)\, \mathcal{Y}^{M}_{l_1 l_2 L}(\hat{r}_1,\hat{r}_2)\\
= & \sum_\Lambda C_\Lambda(1)P_\Lambda(\cos\theta_{12}).
\end{align}

Hint: Use the fact that <math>l_1^2</math> is Hermitian so that
<math>
\int d\tau (l_1^2\mathcal{Y}^\prime)^*q(r)\mathcal{Y} = \int d\tau
\mathcal{Y}^{\prime *}l_1^2(q(r)\mathcal{Y})
</math>
with <math>\ \vec{l}_1= \frac{1}{i}\vec{r}_1\times\nabla_1</math>.

It is also useful to use
<math>\begin{eqnarray}\ \
\cos\theta\, P_L(\cos\theta) &=& \frac{1}{2L+1}[L\,P_{L-1}(\cos\theta) + (L+1)P_{L+1}(\cos\theta)]
\\
(\cos^2\theta-1)P_L(\cos\theta) &=&
\frac{L(L-1)}{(2L-1)(2L+1)}P_{L-2}(\cos\theta)\nonumber
-\frac{2(L^2+L-1)}{(2L-1)(2L+3)}P_L(\cos\theta) + \frac{(L+1)(L+2)}{(2L+1)(2L+3)}P_{L+2}(\cos\theta)\nonumber\\
&=& \frac{L(L-1)}{(2L-1)(2L+1)}[P_{L-2}(\cos\theta) - P_L(\cos\theta)]
+ \frac{(L+1)(L+2)}{(2L+1)(2L+3)}[P_{L+2}(\cos\theta) - P_L(\cos\theta)]
\end{eqnarray}</math>

together with a double application of the integral recursion relation
<math>
I_{L+1}\left(\frac{1}{r}\frac{d}{dr}q(r)\right) =
(2L+1)I_L\left(\frac{1}{r_1r_2}q(r)\right)+I_{L-1}\left(\frac{1}{r}\frac{d}{dr}q(r)\right)
</math>

Of course <math>l_1^2\mathcal{Y}^M_{l_1l_2L}(\hat{r}_1,\hat{r}_2)=l_1(l_1+1)\mathcal{Y}^M_{l_1l_2L}(\hat{r}_1,\hat{r}_2)</math>. Begin by expanding
<math>
l_1^2(g\mathcal{Y}) = \mathcal{Y}l_1^2g + g\,l_1^2\mathcal{Y} + 2(\vec{l}_1g)\cdot(\vec{l}_1\mathcal{Y})
</math>, and show that

<math>
\begin{eqnarray}\ \
(\vec{l}_1g)\cdot(\vec{l}_1\mathcal{Y}) & = & \frac{r_1^2}{r}\frac{dg}{dr}\vec{r}_2\cdot \nabla_1\mathcal{Y}\,\,\,\,\mbox{ and}\\
l_1^2g(r) &=& 2r_1r_2\cos\theta\frac{1}{r}\frac{dg}{dr} + r_1^2r_2^2(\cos^2\theta - 1)
\frac{1}{r}\frac{d}{dr}\left(\frac{1}{r}\frac{dg}{dr}\right).
\end{eqnarray}
</math>

The proof amounts to showing that the term <math>l_1^2g(r)</math> can be replaced by
<math>\Lambda(\Lambda+1)g(r)</math> after multiplying by <math>P_\Lambda(\cos\theta)</math>
and integrating by parts with respect to the radial integrations over <math>r_1</math>, <math>r_2</math> and
<math>r \equiv |\vec{r}_1 - \vec{r}_2|</math>. Remember that <math>\cos\theta</math> is just a short-hand notation for the radial function <math>(r_1^2 + r_2^2 - r^2)/(2r_1r_2)</math>.

==General Hermitian Property==

Each combination of terms of the form
\begin{equation}
<f>_R = a^2f_1+af_2+abf_3+b^2f_4+bf_5+af_6\nabla_1^Y+bf_7\nabla^Y_2+f_8(Y)
\end{equation}
can be rewritten
\begin{equation}
<f>_L= (a_+-a)^2f_1+(a_+-a)f_2 + (a_+-a)(b_+-b)f_3+(b_+-b)^2f_4
+(b_+-b)f_5+(a_+-a)f_6\nabla^{Y\prime}_1+(b_+-b)f_7\nabla^{Y\prime}_2+f_8)Y^\prime
\end{equation}
where as usual <math>\nabla_1^Y</math> acts only on the spherical harmonic part of the wave function denoted for short by <math>Y</math>.
Since these must be equal for arbitrary <math>a</math> and <math>b</math>,
\begin{equation}
a_+^2f_1+a_+f_2+_+b_+f_3+b_+f_4+b_+f_5+a_+f_6\nabla^{Y\prime}_1 +
b_+f_7\nabla^{Y\prime}_2+f_8(Y^\prime)-f_8(Y)=0
\end{equation}

Adding the corresponding expression with <math>Y</math> and <math>Y^\prime</math> interchanged
yields
<math>\label{III}
a_+^2f_1+a_+f_2+a_+b_+f_3+b_+^2f_4+b_+f_5+\frac{1}{2}a_+f_6\nabla^+_1+\frac{1}{2}b_+f_7\nabla^+_2=0\tag{III}
</math>

Subtracting gives
<math>\label{IV}
f_8(Y)-f_8(Y^\prime)=-\frac{1}{2}[a_+f_6\nabla^-_1+b_+f_7\nabla_2^-]\tag{IV}
</math>
<math>\label{V}
a[-2a_+f_1-2f_2-b_+f_3-f_6\nabla^+_1]=0\tag{V}
</math>
<math>\label{VI}
b[-2b_+f_4-2f_5-a_+f_3-f_7\nabla^+_2]=0\tag{VI}
</math>

Adding the two forms gives

\begin{eqnarray}
<f>_R+<f>_L=\frac{1}{2}(a_+^2+a_-^2)f_1+a_+f_2+\frac{1}{2}(a_+b_++a_-b_-)f_3\nonumber\\
\ +\frac{1}{2}(b_+^2+b_-^2)f_4+b_+f_5+\frac{1}{2}f_6(a_+\nabla_1^++a_-\nabla^-_1)\nonumber\\
\ +\frac{1}{2}(b_+\nabla_2^++b_-\nabla^-_2)+f_8(Y)+f_8(Y^\prime)\nonumber
\end{eqnarray}

Subtracting <math>\frac{x}{2}\times({\rm I})</math> gives

\begin{eqnarray}\ \
<f>_R+<f>_L = \frac{1}{2}[(1-x)a_+^2 + a_-^2]f_1 +
(1-\frac{x}{2})a_+f_2+\frac{1}{2}[(1-x)a_+b_++a_-b_-]f_3\nonumber\\
+\frac{1}{2}[(1-x)b_+^2+b_-^2]f_4+(1-\frac{x}{2})f_5b_++\frac{1}{2}f_6[(1-\frac{x}{2})a_+\nabla_1^++a_-\nabla^-_1]\nonumber\\
+\frac{1}{2}[(1-\frac{x}{2})b_+\nabla^+_2+b_-\nabla^-_2]+f_8(Y)+f_8(Y^\prime)\nonumber
\end{eqnarray}

If <math>x=1</math>,
\begin{eqnarray}
<f>_R+<f>_L &=&
\frac{1}{2}[a_-^2f_1+a_+f_2+a_-b_-f_3+b_-^2f_4+b_+f_5\nonumber\\
&+&\left(\frac{1}{2}a_+\nabla^+_1+a_-\nabla^-_1\right)f_6+\left(\frac{1}{2}b_+\nabla^+_2+b_-\nabla^-_2\right)f_7]+f_8(Y)+f_8(Y^\prime)\nonumber
\end{eqnarray}

The General Hermitian Property for arbitrary <math>x</math> gives
\begin{eqnarray}\ \
\nabla_1^2=
\frac{1}{4}\{\frac{1}{r_1^2}[(1-x)a_+^2+a_-^2+2(1-\frac{x}{2})a_+-2[l_1(l_1+1)+l_1^\prime(l_1^\prime+1)]\nonumber\\
-\frac{2}{r_1}[(1-x)\alpha_+a_++\alpha_-a_-+2(1-\frac{x}{2})\alpha_+]+(1-x)\alpha_+^2
+ \alpha^2_-]\nonumber\\
+\frac{2(r_1-r_2\cos\theta)}{r_1r^2}[(1-x)(a_+-\alpha_+r_1)c_++(a_--\alpha_-r_2)c_-]\nonumber\\
-\frac{2r_2}{r_1r^2}[(1-\frac{x}{2})c_+\hat{r}_2\cdot\hat{\nabla}_1^++c_-\hat{r}_2\hat{\nabla}_1^-]+\frac{(1-x)c_+^2+c_-^2+2(1-\frac{x}{2})c_+}{r^2}\}\nonumber
\end{eqnarray}.

Use
<math>\begin{eqnarray}\ \
\frac{-2r_2}{r_1r^2}[(1-\frac{x}{2})c_+\hat{r}_2\hat{\nabla}_1^++c_-\hat{r}_2\hat{\nabla}_1^-]=\frac{2}{r_1^2}\left[(1-\frac{x}{2})\Lambda(\Lambda+1)-\frac{c_-}{c_+}[l_1^\prime(l_1^\prime+1)+l_1(l_1+1)]\right]\end{eqnarray}</math>,

<math>\begin{eqnarray}
\frac{2(r_1-r_2\cos\theta)}{r_1r^2}(a_--\alpha_-r_1)c_- = 2\frac{c_-}{c_+}\left[\frac{-1}{r_1^2}[a_-(a_++1)]+\frac{1}{r_1}[a_-\alpha_++\alpha_-(a_++2)]-\alpha_-\alpha_+\right]\nonumber
\end{eqnarray}</math>,

and
<math>\begin{eqnarray}\ \
\frac{2(r_1-r_2\cos\theta)}{r_1r^2}(a_+-\alpha_+r_1)c_+ = 2\left[-\frac{1}{r_1^2}[a_+(a_++1)]+\frac{1}{r_1}[a_+\alpha_++\alpha_+(a_++2)]-\alpha_+^2\right]
\end{eqnarray}</math>.

Substituting into <math>\nabla_1^2</math> gives

\begin{eqnarray}
\nabla^2_1 = \frac{1}{4}\{\frac{1}{r_1^2}[-(1-x)a_+^2 + a_-^2 +xa_+ +
2(1-\frac{x}{2})\Lambda(\Lambda+1)\nonumber\\
-2l_1(l_1+1)(1-\frac{c_-}{c_+}) -
2l_1^\prime(l_1^\prime+1)(1+\frac{c_-}{c_+})-\frac{2c_-a_-}{c_+}(a_++1)]\nonumber\\
- \frac{2}{r_1}\left[-(1-x)\alpha_+(a_++2)+\alpha_-a_-+2(1-\frac{x}{2})\alpha_+
-\frac{c_-}{c_+}[a_-\alpha_++\alpha_-(a_++2)]\right]\nonumber\\
-(1-x)\alpha_+^2+\alpha_-^2-2\frac{c_-}{c_+}\alpha_-\alpha_+ +\frac{1}{r^2}\left[(1-x)c_+^2+c_-^2+2(1-\frac{x}{2})c_+\right]\}\nonumber
\end{eqnarray}

This has the form
<math>
\nabla_1^2 =
\frac{1}{4}\left[\frac{A_1}{r_1^2}+\frac{B_1}{r_1}+\frac{C_1}{r^2}+D_1\right]
</math>
with

<math>\begin{array}{ccc}
A_1=-(1-x)a_+^2+a_-^2
+xa_++2(1-\frac{x}{2})\Lambda(\Lambda+1)-2l_1(l_1+1)(1-\frac{c_-}{c_+})\
-2l_1^\prime(l_1^\prime+1)(1+\frac{c_-}{c_+})-\frac{2c_-a_-}{c_+}(a_++1) \ \
\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \
\end{array}
</math>

<math>
B_1 =
2\left[(1-x)\alpha_+(a_++2)-\alpha_-a_--2(1-\frac{x}{2})\alpha_++\frac{c_-}{c_+}[a_-\alpha_++\alpha_-(a_++2)]\right]
</math>

<math>
C_1 = (1-x)c_+^2+ c_-^2 +2(1-\frac{x}{2})c_+
</math>

<math>
D_1 = -1(1-x)\alpha_+^2 +\alpha_-^2 - 2\frac{c_-}{c_+}\alpha_-\alpha_+
</math>

The complete Hamiltonian is then (in Z-scaled a.u.)

<math>\begin{eqnarray}
H =-\frac{1}{2}\nabla^2_1 -\frac{1}{2}\nabla^2_2 -\frac{1}{r_1} -\frac{1}{r_2} +\frac{Z^{-1}}{r}\nonumber\\
= -\frac{1}{2}\nabla^2_1 -\frac{1}{2}\nabla^2_2 -\frac{1}{r_1} -\frac{Z-1}{Zr_2}+Z^{-1}\left(\frac{1}{r}-\frac{1}{r_2}\right)\nonumber\\
\ =-\frac{1}{8}\left[\frac{A_1}{r_1^2}+\frac{B_1+8}{r_1}+\frac{C_1}{r^2}+D_1+D_2 +\frac{A_2}{r_2^2} + \frac{B_2 + 8(Z-1)/Z}{r_2}+\frac{C_2}{r^2}\right]\nonumber
+ Z^{-1}\left(\frac{1}{r}-\frac{1}{r_2}\right)\nonumber
\end{eqnarray}</math>

The screened hydrogenic energy is
<math>\begin{eqnarray}
E_{SH}=-\frac{1}{2}\left[\frac{1}{n_1^2}+ \left(\frac{Z-1}{Z}\right)^2\frac{1}{n_2^2}\right]
\end{eqnarray}</math>
so that

<math>\begin{eqnarray}
H-E_{SH} &=& -\frac{1}{8}[
\frac{A_1}{r_1^2}+\frac{B_1+8}{r_1}+\frac{C_1}{r^2} + D_1 + D_2
-\frac{4}{n_1^2} -\left(\frac{Z-1}{Z}\right)^2\frac{4}{n_2^2}\nonumber
+\frac{A_2}{r_2}+\frac{B_2+8(Z-1)/Z}{r_2}+\frac{C_2}{r^2}] +
Z^{-1}\left(\frac{1}{r} -\frac{1}{r_2}\right)\nonumber
\end{eqnarray}</math>

==Optimization of Nonlinear Parameters==
- The traditional method of performing Hylleraas calculations is to write the
basis set in the form

<math>
\Psi = \sum_{i,j,k}c_{i,j,k}\varphi_{i,j,k}(\alpha,\beta)\pm exchange
</math>

with

<math>
\varphi_{i,j,k}(\alpha,\beta) = r_1^ir_2^jr_{12}^ke^{-\alpha r_1 - \beta
r_2}\mathcal{Y}^M_{l_1l_2L}.
</math>

The usual procedure is to set <math>\alpha = Z</math> so that it represents the inner
<math>1s</math> electron, and then to vary <math>\beta</math> so as to minimize the energy.

[[File:OptNon.jpg|500px]]

Since <math>\beta</math> appears in <math>\Psi</math> as a non-linear parameter, the entire
calculation must be repeated for each value of <math>\beta</math>. However, the minimum
becomes progressively smaller as the basis set is enlarged.

[[File:OptNon02.jpg|500px]]

*Difficulties

1. If the basis set is constructed so that <math> i +j +j \leq N</math>, the the
number of terms is
<math>
(N+1)(N+2)(N+3)/6
</math>
N=14 already gives 680 terms and an accuracy of about 1 part in <math>10^{10}</math> for
low-lying states. A substantial improvement in accuracy would require much
larger basis sets, together with multiple precision arithmetic to avoid loss
of significant figures when high powers are included.

2. The accuracy rapidly deteriorates as one goes to more highly excited
states - about 1 significant figure is lost each time the principle quantum
number is increased.

*Cure

We have found that writing basis sets in the form
\begin{eqnarray}
\Psi &=& \sum_{i,j,k}\left(c^{(1)}_{i,j,k}\varphi_{ijk}(\alpha_1\beta_1) +
c_{ijk}^{(2)}\varphi_{ijk}(\alpha_2\beta_2)\right) \pm exchange \nonumber\\
&=& \psi({\bf r}_1,{\bf r}_2) \pm \psi({\bf r}_2,{\bf r}_1)\nonumber
\end{eqnarray}
so that each combination of powers is included twice with different nonlinear
parameters gives a dramatic improvement in accuracy for basis sets of about
the same total size.

However, the optimization of the nonlinear parameters is now much more
difficult, and an automated procedure is needed.

If
<math>
E = \frac{<\Psi|H|\Psi>}{<\Psi|\Psi>}</math> and we assume <math><\Psi|\Psi> = 1</math>, then

<math>
\frac{\partial E}{\partial\alpha_t} = -2 <\Psi|H-E|r_1\psi({\bf r}_1,{\bf
r}_2;\alpha_t)\pm r_21\psi({\bf r}_2,{\bf
r}_1;\alpha_t)>
</math>
where
<math>
\psi({\bf r}_1,{\bf r}_2;\alpha_t) =
\sum_{i,j,k}c_{ijk}^{(t)}\varphi_{ijk}(\alpha_t\beta_t)
</math>

*Problem

1. Prove the above.

2. Prove that there is no contribution from the implicit dependence of
<math>c_{ijk}^{(t)}</math> on <math>\alpha_t</math> if the linear parameters have been optimized.

==The Screened Hydrogenic Term==

If the Hamiltonian is written in the form (in Z-scaled a.u.)
<math>\begin{eqnarray}\
H = H_0({\bf r}_1,Z) + H_0({\bf r}_2,Z-1) +
Z^{-1}\left(\frac{1}{r}-\frac{1}{r_2}\right)
\end{eqnarray}</math>
with
<math>\
H_0({\bf r}_1,Z) = -\frac{1}{2}\nabla_1^2 - \frac{1}{r_1}\nonumber
</math> and
<math>
H_0({\bf r}_2,Z-1) = -\frac{1}{2}\nabla_2^2 - \frac{Z-1}Z{r_2}\nonumber
</math>,
then the eigenvectors of <math>H_0({\bf r},Z) +H_0({\bf r}_2,Z-1)</math> are products of
hydrogenic orbitals
<math>
\Psi_0=\psi_0(1s,Z)\psi_0(nl,Z-1)
</math>
and the eigenvalue is
<math>\begin{eqnarray}
E_{SH} = \left[-\frac{1}{2} - \left(\frac{Z-1}{Z}\right)^2\frac{1}{2n^2}\right]Z^2 a.u.
\end{eqnarray}</math>
called the screened hydrogenic eigenvalue. From highly excited states,
<math>E_{SH}</math> and <math>\Psi_0</math>are already excellent approximations.

For example, for the 1s8d states, the energies are

<math>\begin{eqnarray}\\
E(1s8d^1D)&=& -2.007\,816\,512\,563\,81 {\rm\ a.u.}\nonumber\\
E(1s8d^3D)&=& - 2.007\,817\,934\,711\,71 {\rm\ a.u.}\nonumber\\
E_{SH} &=& -2.007\,812\,5 {\rm\ a.u.} \nonumber
\end{eqnarray}</math>

It is therefore advantageous to include the screened hydrogenic terms in the
basis set so that the complete trial function becomes
<math>
\Psi = c_0\Psi_0 +
\sum_{ijk}\left[c_{ijk}^{(1)}\varphi_{ijk}(\alpha_1,\beta_1)+c_{ijk}^{(2)}\varphi(\alpha_2,\beta_2)\right]\pm
{\rm exchange}
</math>

Also, the variational principal can be re-expressed in the form
<math>\begin{eqnarray}
E= E_{SH}+\frac{<\Psi|H-E_{SH}|\Psi>}{<\Psi|\Psi>}
\end{eqnarray}</math>
so that the <math>E_{SH}</math> term can be cancelled analytically from the matrix
elements.

For example
<math>
<\Psi_0|H-E_{SH}|\Psi_0> = 0
</math>

and
<math>
<\varphi_{ijk}|H-E_{SH}|\Psi_0> =
Z^{-1}<\varphi_{ijk}|\frac{1}{r}-\frac{1}{r_2}|\Psi_0>
</math>

Recall that
<math>\begin{eqnarray}\ \
I_0(a,b,c) = \frac{2}{c+2}\sum^{[c+1]/2}_{i=0}\left(\begin{array}{c}
c+2\\
2i+1
\end{array}\right) \left[f(p,q;\beta) + f(p^\prime,q^\prime;\alpha)\right]
\end{eqnarray}</math>
where

<math>\begin{eqnarray}
f(p,q;x) =
\frac{q!}{x^{q+1}(\alpha+\beta)^{p+1}}\sum^p_{j=0}\frac{(p+j)!}{j!}\left(\frac{x}{\alpha+\beta}\right)^j
\end{eqnarray}</math>

<math>
\begin{array}{cc}
p=a+2i+2 & p^\prime = b+2i+2\\
q= b+c-2i+2 & q^\prime =a+c-2i+2
\end{array}
</math>

If <math>\beta\ll\alpha</math>, the <math>f(p,q;\beta)\gg f(p^\prime,q^\prime;\alpha)</math> and the
<math>i=0</math> term is the dominant contribution to <math>f(p,q;\beta)</math>. However, since
this term depends only on the sum of powers <math>b+c</math> for <math>r_2</math>, it cancels exactly
from the matrix element of
<math>
\frac{1}{r_{12}} - \frac{1}{r_2}
</math>
and can therefore
be omitted in the calculations of integrands, thereby saving many significant
figures. This is especially valuable when <math>b+c</math> is large and <math>a</math> is small.

==Small Corrections==
*Mass Polarization
If the nuclear mass is not taken to be infinite, then the Hamiltonian is
<math>\begin{eqnarray}\ \
H = \frac{P_N^2}{2M} + \sum_{i=1}\frac{p_i^2}{2m} + V
\end{eqnarray}</math>

[[File:SmallCor.jpg|500px]]

where <math>M =M_A-nm</math> is the nuclear mass and <math>m</math> is the electron mass.

Change variables to the C of M
<math>\begin{eqnarray}
{\bf R} = \frac{1}{M+nm}\left[M{\bf r}_N + m({\bf r}_1 + {\bf r}_2 +
\ldots)\right]
\end{eqnarray}</math>

and relative variables

<math>
{\bf s}_i = {\bf r}_i - {\bf r}_N.
</math>

Then

<math>\begin{eqnarray}\ \ \ \ \
H = \frac{1}{2(M+nm)}{\bf p}_R^2+\frac{1}{2m}\sum_{i=1}^n{\bf p}_{s_i}^2
+\frac{1}{2M}\sum_{i,k}{\bf p}_{s_i}\cdot {\bf p}_{s_k} + V({\bf s}_1, {\bf
s}_2, \ldots)\nonumber
= \frac{1}{M+nm}{\bf p}_R^2+\frac{1}{2\mu}\sum_{i=1}^n{\bf
p}_{s_i}^2+\frac{1}{2M}\sum_{i\neq k}{\bf p}_{s_i}\cdot {\bf p}_{s_k}+V({\bf s}_1, {\bf
s}_2, \ldots)\nonumber
\end{eqnarray}</math>

where
<math>\begin{eqnarray}
\mu = \left(\frac{1}{M} + \frac{1}{m}\right)^{-1}
\end{eqnarray}</math>

If <math>n=2</math>, the Schrödinger equation is
<math>\begin{eqnarray}\ \ \
\left[ -\frac{\hbar^2}{2\mu}(\nabla^2_{s_1}+ \nabla^2_{s_2}) -
\frac{\hbar^2}{M}\nabla_{s_1}\nabla_{s_2}-\frac{Ze^2}{s_1}-\frac{Ze^2}{s_2}+\frac{e^2}{s_{12}}\right]\psi
= E\psi
\end{eqnarray}</math>.

Define the reduced mass Bohr radius
<math>\begin{eqnarray}
a_\mu = \frac{\hbar^2}{\mu e^2} = \frac{m}{\mu}a_0
\end{eqnarray}</math>
and

<math>\begin{eqnarray}
\rho_i = \frac{Zs_i}{a_\mu}\nonumber\\
\varepsilon = \frac{E}{Z^2(e^2/a_\mu)}\nonumber
\end{eqnarray}</math>

The Schrödinger equation then becomes
<math>\begin{eqnarray}\ \ \
\left[
-\frac{1}{2}(\nabla^2_{\rho_1}+\nabla^2_{\rho_2})-\frac{\mu}{M}\vec{\nabla}_{\rho_1}\cdot\vec{\nabla}_{\rho_2} - \frac{1}{\rho_1}-\frac{1}{\rho_2}+\frac{Z^{-1}}{\rho_{12}}\right]\psi = \varepsilon \psi
\end{eqnarray}</math>

The effects of finite mass are thus

1. A reduced mass correction to all energy levels of <math>(e^2/a_\mu)/(e^2/a_0)
= \mu/m</math>, i.e.
<math>\begin{eqnarray}
E_M=\frac{\mu}{m}E_\infty
\end{eqnarray}</math>

2. A specific mass shift given in first order perturbation theory by
<math>\begin{eqnarray}
\Delta\varepsilon =
-\frac{\mu}{M}<\psi|\nabla_{\rho_1}\cdot\nabla_{\rho_2}|\psi>
\end{eqnarray}</math>
or

<math>\begin{eqnarray}
\Delta\varepsilon =
-\frac{\mu}{M}<\psi|\nabla_{\rho_1}\cdot\nabla_{\rho_2}|\psi>\frac{e^2}{a_\mu}\nonumber\\
= -\frac{\mu^2}{mM}<\psi|\nabla_{\rho_1}\cdot\nabla_{\rho_2}|\psi>\frac{e^2}{a_0}\nonumber
\end{eqnarray}</math>

Since <math>\nabla_{\rho_1}\cdot\nabla_{\rho_2}</math> has the same angular properties as
<math>{\bf\rho}_1 \cdot {\bf\rho}_2</math>, the operator is like the product of two
dipole operators. For product type wave functions of the form
$
\psi = \psi(1s)\psi(nl) \pm \mbox{exchange},
$
the matrix element vanishes for all but P-states.

=Radiative Transitions=

==Transition Integrals and the Line Strength==

For an electric dipole transition <math>\gamma'\,L'\,S'\,J'\,M' \rightarrow \gamma\,L\,S\,J\,M</math>, the line strength is defined by
\begin{equation}
S = \sum_{M',\mu,M}|\langle\gamma'\,L'\,S'\,J'\,M'|r_\mu|\gamma\,L\,S\,J\,M\rangle|^2
\end{equation}
Using the Wigner-Eckart theorem, this is related to the reduced matrix element by
\begin{eqnarray}
S(J',J) = |\langle\gamma'\,L'\,S'\,J'||r||\gamma\,L\,S\,J\,M\rangle|^2
\sum_{M',\mu,M}\left|\left(\matrix{J' & 1 & J \cr
-M'& \mu &M \cr}\right)\right|^2
\end{eqnarray}
Using the sum rule for the 3-<math>j</math> symbols
\begin{equation}
\sum_{M',\mu,M}\left|\left(\matrix{J' & 1 & J \cr
-M'& \mu &M \cr}\right)\right|^2 = 1
\end{equation}
we see that the line strength is identical to the reduced matrix element
\begin{equation}
S(J',J) = |\langle\gamma'\,L'\,S'\,J'||r||\gamma\,L\,S\,J\,M\rangle|^2
\end{equation}
For a nonrelativistic calculation in <math>LS</math> coupling, <math>S = S'</math> and we can next strip out the dependence on <math>J</math> and <math>J'</math> by use of Eq.\ (7.1.7) of Edmonds { Angular Momentum in Quantum Mechanics} with <math>k = 1</math> to obtain
\begin{equation}
S(J',J) = (-1)^{L'+S+J+1}[(2J+1)(2J'+1)]^{1/2}\left\{
\matrix{L' & J'& S\cr
J & L & 1\cr}\right\}\langle\gamma' L'|| r || \gamma L\rangle
\end{equation}
where <math>\langle\gamma' L'|| r || \gamma L\rangle</math> is the reduced matrix element if spin is neglected. Note that if spin is neglected, then the line-strength is simply
\begin{equation}
S(L',L) = |\langle\gamma' L'|| r || \gamma L\rangle|^2
\end{equation}
To calculate the reduced matrix element, use a particular example along with the Wigner-Eckart theorem
such as
\begin{equation}
\langle\gamma'\, L'|| r || \gamma\, L\rangle = (-1)^{L'}\langle\gamma'\, L'\, 0| z | \gamma\, L\, 0\rangle{\Large /}\left(\matrix{L' & 1 & L \cr
0 & 0 & 0 \cr}\right)
\end{equation}

===Correlated Two-electron Integrals===
For the case of two-electron integrals, <math>z = z_1 + z_2</math> and <math>r = r_1 + r_2</math>. As discussed in G.W.F. Drake, Phys. Rev. A 18, 820 (1978), the general integral in Hylleraas coordinates for the integral over the <math>z_1</math> term can be written in the form
\begin{equation}
\label{int1}
\langle\gamma'\, L'\, 0| z_1 | \gamma\, L\, 0\rangle = \sum_\Lambda C_\Lambda I_\Lambda(\gamma',\gamma)
\end{equation}
where <math>I_\Lambda(\gamma',\gamma)</math> stands for the purely radial integral
\begin{equation}
\label{radial}
I_\Lambda(\gamma',\gamma) = \int_0^\infty r_1\,dr_1\int_0^\infty r_2\,dr_2\int_{|r_1-r_2|}^{r_1+r_2}r_{12}\, dr_{12}
R_{\gamma'}^*\,r_1\, R_\gamma P_\Lambda(\cos\theta_{12})
\end{equation}
and <math>\cos\theta_{12} = (r_1^2 + r_2^2 - r_{12}^2)/(2r_1r_2)</math> is a short-hand notation for a radial function of <math>r_1</math>, <math>r_2</math> and <math>r_{12}</math>. Also, <math>P_\Lambda(\cos\theta_{12})</math> is a Legendre polynomial,
and, <math>R_\gamma = R_\gamma(r_1,r_2,r_{12})</math> is the radial part of the Hylleraas wave function for state <math>\gamma</math> such
that the total wave function is <math>\psi({\bf r}_1,{\bf r}_2) = R_\gamma(r_1,r_2,r_{12}){\cal Y}_{l_1,l_2,L}^M(\hat{\bf r}_1,\hat{\bf r}_2)\pm {\rm exchange}</math>, normalized to unity, and <math>{\cal Y}_{l_1,l_2,L}^M(\hat{\bf r}_1,\hat{\bf r}_2)</math> is a vector coupled product of spherical harmonics. The coefficents <math>C_\Lambda</math> in Eq.\ (\ref{int1}) can be calculated from Eq.\ (32) of the above 1978 paper. They need to be calculated separately for he direct-direct and the direct-exchange contributions. For the reduced matrix element of <math>z_1</math>, the leading 3-<math>j</math> symbol can be dropped, leaving (for the case <math>k = 1</math>)
\begin{eqnarray}
C_\Lambda^{\rm reduced} &=& \frac{(-1)^{l_2'+L'+L+1}}{2} (l_1,l_1',l_2,l_2',L,L')^{1/2}
\sum_{\lambda={\rm max}(l_1-1,0)}^{l_1+1} (\lambda,\Lambda,1)
\left(\matrix{l_1 & 1 & \lambda \cr
0 & 0 & 0 \cr}\right)
\left(\matrix{l_1'& \lambda & \Lambda \cr
0 & 0 & 0 \cr}\right)\nonumber\\
&\times&\left(\matrix{l_2' & 1_2 & \Lambda \cr
0 & 0 & 0 \cr}\right)
\left\{\matrix{L' & \lambda & l_2 \cr
\Lambda& l_2' & l_1' \cr}\right\}
\left\{\matrix{L' & \lambda & l_2 \cr
l_1 & L & 1 \cr}\right\}
\end{eqnarray}
where the notation <math>(a,b,c,\cdots)</math> means <math>(2a+1)(2b+1)(2c+1)\cdots</math>.
A similar expression can be obtained for the angular coefficients for the <math>z_2</math> term by interchanging the subscripts 1 and 2 throughout and multiplying by <math>(-1)^{L+L'+1}</math>.

===Example===
As an example, for the <math>1s^2\;^1S - 1s2p\;^1P</math> transition of helium, the only nonvanishing <math>C_\Lambda</math> angular coefficients are <math>C_1 = 1/2</math> for the direct-direct term (i.e. <math>P_1(\cos\theta_{12}) = \cos\theta_{12}</math> is included in the radial integral (\ref{radial})) and <math>C_0 = 1/2</math> for the direct-exchange term (i.e. <math>P_0(\cos\theta_{12}) = 1</math> is included in the radial integral (\ref{radial})). The results for infinite nuclear mass are
\begin{equation}
S(1\;^1S-2\;^1P) = (0.781\,127\,158\,667)^2
\end{equation}
\begin{equation}f(1\;^1S-2\;^1P) = -0.276\,164\,703\,736
\end{equation}
where
\begin{equation}
f(1\;^1S-2\;^1P) = \frac{2}{3}[E(1\;^1S) - E(2\;^1P)]S(1\;^1S-2\;^1P)
\end{equation}
is the absorption oscillator strength.

As an example, for the <math>1s^2\;^1S - 1s2p\;^1P</math> transition of helium, the only nonvanishing <math>C_\Lambda</math> angular coefficients are <math>C_1 = 1/2</math> for the direct-direct term (i.e. <math>P_1(\cos\theta_{12}) = \cos\theta_{12}</math> is included in the radial integral (\ref{radial})) and <math>C_0 = 1/2</math> for the direct-exchange term (i.e. <math>P_0(\cos\theta_{12}) = 1</math> is included in the radial integral (\ref{radial})). The results for infinite nuclear mass are
\begin{equation}
S(1\;^1S-2\;^1P) = (0.781\,127\,158\,667)^2
\end{equation}
\begin{equation}f(1\;^1S-2\;^1P) = -0.276\,164\,703\,736
\end{equation}
where
\begin{equation}
f(1\;^1S-2\;^1P) = \frac{2}{3}[E(1\;^1S) - E(2\;^1P)]S(1\;^1S-2\;^1P)
\end{equation}
is the absorption oscillator strength.

===Test===
This is a test edit

Theory Notes

2020-12-14T04:24:56Z

Drake: Undo revision 793 by Drake (talk)

<center>
<H1>Notes on solving the Schrödinger equation in Hylleraas coordinates for heliumlike atoms
Gordon W.F. Drake
Department of Physics, University of Windsor
Windsor, Ontario, Canada N9B 3P4
(Transcribed from hand-written notes and edited by Lauren Moffatt. Last revised November 22, 2017 to correct a missing subscript in the expression for $B_1$ in the Hamiltonian in Sect. 11 General Hermitian Property.)
</H1>
</center>

==Introduction==
These notes describe some of the technical details involved in solving the Schrödinger equation for a heliumlike atom of nuclear charge Z in correlated Hylleraas coordinates. The standard for computational purposes is to express all quantities in a form that is valid for electronic states of arbitrary angular momentum, and to express matrix elements of the Hamiltonian in a manifestly Hermitian symmetrized form. Further information and derivations of the mathematical techniques for angular integrals and radial recursion relations can be found in the paper
G.W.F. Drake, Phys. Rev. A <math>{\bf 18}</math>, 820 (1978).

The starting point is the two-electron Schrödinger equation for infinite nuclear mass
\begin{equation} \left[-\frac{\hbar^2}{2m}(\nabla^2_1 +\nabla^2_2) - \frac{Ze^2}{r_1} -
\frac{Ze^2}{r_2}+\frac{e^2}{r_{12}} \right]\psi = E\psi\nonumber
\end{equation}
where <math>m</math> is the electron mass, and <math>r_{12} = |{\bf r}_1 - {\bf r}_2|</math> (see diagram below).

[[File:Pic1.jpg|250px]]

Begin by rescaling distances and energies so that the Schrödinger equation can be expressed in a dimensionless form. The dimensionless Z-scaled distance is defined by <math>\begin{eqnarray} \rho = \frac{Zr}{a_0} \end{eqnarray} </math> where <math> \begin{eqnarray} a_0 = \frac{\hbar^2}{me^2} \end{eqnarray} </math> is the atomic unit (a.u.) of distance equal to the Bohr radius <math>0.529\, 177\, 210\, 92(17) \times 10^{-10}</math> m. Then

\begin{equation}[-\frac{\hbar^2}{2m}Z^2(\frac{me^2}{\hbar^2})^2(\nabla^2_{\rho_1}+\nabla^2_{\rho_2})
- Z^2\frac{e^2}{a_0}\rho^{-1}_1 - Z^2\frac{e^2}{a_0}\rho^{-1}_2 +
\frac{e^2}{a_0}Z\rho^{-1}_{12}]\psi= E\psi\nonumber\end{equation}

But <math>\begin{eqnarray} \frac{\hbar^2}{m}(\frac{me^2}{\hbar^2})^2 = \frac{e^2}{a_0} \end{eqnarray} </math> is the hartree atomic unit of of energy <math>(E_h = 27.211\,385\,05(60)</math> eV, or equivalently <math>E_h/(hc) = 2194.746\,313\,708(11)</math> m<math>^{-1})</math>. Therefore, after multiplying through by <math>a_0/(Ze)^2</math>, the problem to be solved in Z-scaled dimensionless units becomes

\begin{equation} \left[-\frac{1}{2}(\nabla^2_{\rho_1}+\nabla^2_{\rho_2}) -
\frac{1}{\rho_1} - \frac{1}{\rho_2} + \frac{Z^{-1}}{\rho_{12}}\right]\psi =
\varepsilon\psi\nonumber \end{equation}

where <math>\begin{eqnarray}\varepsilon = \frac{Ea_0}{(Ze)^2} \end{eqnarray}</math> is the energy in Z-scaled atomic units. For convenience, rewrite this in the conventional form
\begin{equation}
H\psi = \varepsilon\psi\nonumber \end{equation}
where (using <math>r</math> in place of <math>\rho</math> for the Z-scaled distance)
\begin{equation}
H = -\frac{1}{2}(\nabla^2_1+\nabla^2_2) - \frac{1}{r_1}-\frac{1}{r_2} + \frac{Z^{-1}}{r_{12}}
\nonumber\end{equation}
is the atomic Hamiltonian for infinite nuclear mass.

==The Hartree Fock Method==

For purposes of comparison, and to define the correlation energy, assume that <math>\psi({\bf r}_1,{\bf r}_2)</math> can be written in the separable product form
\begin{equation}
\psi({\bf r}_1,{\bf r}_2) = \frac{1}{\sqrt{2}} \left[ u_1(r_1) u_2(r_2) \pm u_2(r_1) u_1(r_2) \right] \nonumber
\end{equation}
for the <math>1s^2\ ^1\!S</math> ground state. Because of the <math>\frac{1}{r_{12}}</math> term in the Schrödinger equation, the exact solution connot be expressed in this form as a separable product. However, the Hartree-Fock approximation corresponds to finding the best possible solution to the Schrödinger equation
\begin{equation}
H\psi({\bf r}_1,{\bf r}_2) = E\psi({\bf r}_1,{\bf r}_2)\nonumber
\end{equation}
that can nevertheless be expressed in this separable product form, where as before
\begin{equation}
H = -\frac{1}{2}(\nabla^2_1+\nabla^2_2) - \frac{1}{r_1}- \frac{1}{r_2} + \frac{Z^{-1}}{r_{12}}\nonumber
\end{equation}
is the full two-electron Hamiltonian. To find the best solution, substitute into <math>\left\langle\psi \lvert H-E \rvert \psi\right\rangle</math> and require this expression to be stationary
with respect to arbitrary infinitesimal variations <math>\delta u_1</math> and <math>\delta
u_2</math> in <math>u_1</math> and <math>u_2</math> respectively; i.e.

\begin{equation}
\frac{1}{2} \left\langle \delta u_1(r_1)u_2(r_2) \pm u_2(r_1)\delta
u_1(r_2) \lvert H-E \rvert u_1(r_1) u_2(r_2)\pm u_2(r_1) u_1(r_2)\right\rangle\nonumber
\end{equation}

\begin{equation}
=\int\delta u_1(r_1) d{\bf r}_1 \left\{ \int d{\bf r}_2 u_2(r_2) \left( H - E \right) \left[ u_1(r_1)u_2(r_2) \pm u_2(r_1) u_1(r_2) \right] \right\}\nonumber
\end{equation}

\begin{equation}
= 0 \nonumber
\end{equation}

for arbitrary <math> \delta u_1(r_1).\nonumber </math> Therefore <math>\{\int d{\bf r}_2 \ldots \} = 0</math>.

Similarly, the coefficient of <math>\delta u_2</math> would give

\begin{equation}
\int d{\bf r}_1 u_1(r_1) \left( H-E \right) \left[ u_1(r_1) u_2(r_2) \pm
u_2(r_1) u_1(r_2) \right] = 0\nonumber
\end{equation}

Define

\begin{equation}
I_{12} = I_{21} = \int d{\bf r}\, u_1(r)u_2(r), \nonumber
\end{equation}

\begin{equation}
H_{ij} = \int d{\bf r}\, u_i(r)(-\frac{1}{2}\nabla - \frac{1}{r})u_j(r), \nonumber
\end{equation}

\begin{equation}
G_{ij}(r) = \int d{\bf r}^\prime u_i(r^\prime)\frac{1}{|{\bf r} - {\bf r}\prime|}u_j(r^\prime)\nonumber
\end{equation}

Then the above equations become the pair of integro-differential equations

\begin{equation}
[ H_0 - E + H_{22}+G_{22}(r)]u_1(r) = \mp [ I_{12}(H_0-E) + H_{12}+G_{12}(r)]u_2(r)\nonumber
\end{equation}

\begin{equation}
[H_0-E+H_{11}+G_{11}(r)]u_2(r) = \mp [I_{12}(H_0-E) + H_{12}+G_{12}(r)]u_1(r)\nonumber
\end{equation}

These must be solved self-consistently for the "constants" <math>I_{12}</math> and
<math>H_{ij}</math> and the function <math>G_{ij}(r)</math>.

The Hartree Fock energy is <math>E \simeq -2.87\ldots </math>a.u. while the exact energy is <math>E = -2.903724\ldots </math>a.u.

The difference is called the "correlation energy" because it arises from the
way in which the motion of one electron is correlated to the other. The
Hartree Fock equations only describe how one electron moves in the average field
provided by the other (mean-field theory).

==Configuration Interaction==
Expand
\begin{equation}
\psi({\bf r}_1,{\bf r}_2)= C_0u^{(s)}_1(r_1)u^{(s)}_1(r_2) + C_1u^{(P)}_1({\bf r}_1)u^{(P)}_1({\bf r}_2){\cal Y}^0_{1,1,0}(\hat{\bf r}_1, \hat{\bf r}_2)+C_2u^{(d)}_1({\bf r}_1)u^{(d)}_2({\bf r}_2){\cal Y}^0_{2,2,0}(\hat{\bf r}_1, \hat{\bf r}_2)+... \pm {\rm\ exchange}
\end{equation}
where
\begin{equation}
{\cal Y}^M_{l_1,l_2,L}(\hat{\bf r}_1, \hat{\bf r}_2)=\sum_{m_1,m_2}Y^{m_1}_{l_1}({\bf r}_1)Y^{m_2}_{l_2}({\bf r}_2)\times <l_1l_2m_1m_2\mid LM>
\end{equation}

This works, but is slowly convergent, and very laborious. The best CI calculations are accurate to <math> ~10^{-7}</math> a.u.

==Hylleraas Coordinates==
[E.A. Hylleraas, Z. Phys. <math>{\bf 48}, 469(1928)</math> and <math>{\bf 54}, 347(1929)</math>]
suggested using the co-ordinates <math>r_1</math>, <math>r_2</math> and <math>r_{12}</math> or equivalently

\begin{eqnarray}
s &=& r_1 + r_2, \nonumber\\
t &=& r_1-r_2, \nonumber\\
u &=& r_{12}\nonumber
\end{eqnarray}

and writing the trial functions in the form

\begin{equation}
\Psi({\bf r}_1,{\bf r}_2) = \sum^{i+j+k\leq
N}_{i,j,k}c_{i,j,k}r_1^{i+l_1}r_2^{j+l_2}r_{12}^ke^{-\alpha r_1 - \beta r_2}
\mathcal{Y}^M_{l_1,l_2,L}(\hat{r}_1,\hat{r}_2)\pm \text{exchange}\nonumber
\end{equation}

Diagonalizing H in this non-orthogonal basis set is equivalent to solving
<math> \begin{eqnarray}
\frac{\partial E}{\partial c_{i,j,k}} = 0\nonumber
\end{eqnarray} </math>
for fixed <math>\alpha</math> and <math>\beta</math>.

The diagonalization must be repeated for different values of <math>\alpha</math> and
<math>\beta</math> in order to optimize the non-linear parameters.

==Completeness==
The completeness of the above basis set can be shown by first writing
\begin{equation}
r_{12}^2 = r_1^2 + r_2^2 - 2r_1r_2\cos\theta_{12}
\end{equation}
and, from the spherical harmonic addition theorem,
\begin{equation}
\cos(\theta_{12})=\frac{4\pi}{3}\sum^1_{m=-1}Y^{m*}_1(\theta_1,\varphi_1)Y^m_1(\theta_2,\varphi_2)
\end{equation}
Consider first the S-states. The <math>r_{12}^0</math> terms are like the ss terms in a CI calculation. The
<math>r_{12}^2</math> terms bring in p-p type contributions, and the higher powers bring
in d-d, f-f etc. type terms. In general

\begin{equation}
P_l(\cos\theta_{12}) = \frac{4\pi}{2l+1}\sum^l_{m=-l}{Y^{m}_l}^*(\theta_1,\varphi_1)Y^m_l(\theta_2,\varphi_2)\nonumber
\end{equation}

For P-states, one would have similarly

\begin{array}{lr}
r_{12}^0 & (sp)P\nonumber\\
r_{12}^2 & (pd)P\nonumber\\
r_{12}^4 & (df)P\nonumber\\
\vdots & \vdots\nonumber
\end{array}

For D-states

\begin{array}{lr}
r_{12}^0 & (sd)D & (pp^\prime)D\nonumber\\
r_{12}^2 & (pf)D & (dd^\prime)D\nonumber\\
r_{12}^4 & (dg)D & (ff^\prime)D\nonumber\\
\vdots & \vdots & \vdots\nonumber
\end{array}

In this case, since there are two "lowest-order" couplings to form a
D-state, both must be present in the basis set, i.e.

\begin{equation}
\Psi(r_2,r_2) = \sum c_{ijk}r_1^ir_2^{j+2}r_{12}^ke^{-\alpha r_1-\beta
r_2}\mathcal{Y}^M_{022}(\hat{r}_1,\hat{r}_2)
+\sum d_{ijk}r_1^{i+1}r_2^{j+1}r_{12^k}e^{-\alpha^\prime r_1 - \beta^\prime
r_2}\mathcal{Y}^M_{112}(\hat{r}_1,\hat{r}_2)\nonumber
\end{equation}

For F-states, one would need <math>(sf)F</math> and <math>(pd)F</math> terms.

For G-states, one would need <math>(sg)G</math>, <math>(pf)G</math> and <math>(dd^\prime)G</math> terms.

Completeness of the radial functions can be proven by considering the
Sturm-Liouville problem

\begin{equation}
\left(-\frac{1}{2}\nabla^2-\frac{\lambda}{r_s}-E\right)\psi({\bf r}) = 0\nonumber
\end{equation}

or

\begin{equation}
\left(-\frac{1}{2}\frac{1}{r^2}\frac{\partial}{\partial r}\left({r^2}\frac{\partial}{\partial r}\right) -
\frac{l\left(l+1\right)}{2r^2} - \frac{\lambda}{r} - E\right)u(r) = 0.\nonumber
\end{equation}

For fixed E and variable <math>\lambda</math> (nuclear charge).

The eigenvalues are <math>\lambda_n = (E/E_n)^{1/2}</math>, where <math>E_n =- \frac{1}{2n^2}</math>

[[File:Pic2.jpg|500px]]

and the eigenfunctions are

\begin{equation}
u_{nl}(r) =
\frac{1}{(2l+1)!}\left(\frac{(n+l)!}{(n-l-1)2!}\right)^{1/2}(2\alpha)^{3/2}e^{-\alpha
r}\nonumber
\times \ \left( 2 \alpha r \right)^l\,_{1}F_{1}\left(-n+l+1,2l+2; 2 \alpha r \right)
\end{equation}

with <math>\alpha = (-2E)^{1/2}</math> and <math>n\geq l+1</math>. The confluent hypergeometric function <math> _1F_1(a,b,;z)</math> then denotes a finite polynomial since <math>a=-n+l+1</math> is a negative integer or zero.

Unlike the hydrogen spectrum, which has both a discrete part for <math>E<0</math> and a
continuous part for <math>E>0</math>, this forms an entirely discrete set of finite
polynomials, called Sturmian functions. They are orthogonal with respect to
the potential, i.e.

\begin{equation}
\int^\infty_0 r^2dr \left( u_{n^\prime l}(r)\frac{1}{r}u_{nl}(r) \right) = \delta_{n,n^\prime}\nonumber
\end{equation}

Since they become complete in the limit <math>n\rightarrow\infty</math>, this assures the
completeness of the variational basis set.

[See also B Klahn and W.A. Bingel Theo. Chim. Acta (Berlin) 44, 9 and 27
(1977)].

==Solutions of the Eigenvalue Problem==

For convenience, write

\[
\Psi({\bf r}_1,{\bf r}_2) = \sum^N_{m=1}c_m\varphi_m\nonumber
\]

where <math>m</math> represents the <math>m</math>'th combination of <math>i,j,k</math> and

\begin{equation}
\varphi_{ijk}=r_1^ir_2^jr_{12}^ke^{-\alpha r_1 - \beta
r_2}\mathcal{Y}^M_{l_1,l_2,L}(\hat{r}_1,\hat{r}_2) \pm \mbox{exchange}\nonumber
\end{equation}

\begin{align}

&
\left( \begin{array}{lr}
\cos(\theta) & \sin(\theta)\\
-\sin(\theta) & \cos(\theta)
\end{array} \right)
\left( \begin{array}{lr}
H_{11} & H_{12} \\
H_{12} & H_{22}
\end{array}\right)
\left( \begin{array}{lr}
\cos(\theta) & -\sin(\theta)\\
\sin(\theta) & \cos(\theta)
\end{array} \right)
\\[5pt]
= & \left(\begin{array}{lr}
cH_{11}+sH_{12} & cH_{12} + sH_{22}\\
-sH_{11} + cH_{12} & -sH_{12} + cH_{22}
\end{array}\right)
\left( \begin{array}{cc}
c & -s \\
s & c
\end{array}\right)
\\[5pt]
= & \left(\begin{array}{lr}
c^2H_{11}+s^2H_{22} + 2csH_{12} & (c^2-s^2)H_{12}+cs(H_{22}-H_{11})\\
(c^2-s^2)H_{12}+cs(H_{22}-H_{11}) & s^2H_{11}+c^2H_{22}-2csH_{12}
\end{array}\right)
\end{align}

Therefore

\begin{equation}
\left[\cos^2(\theta)-\sin^2(\theta)\right]H_{12} =
\cos(\theta)\sin(\theta)(H_{11}-H_{22})\nonumber
\end{equation}

and

\begin{equation}
\tan(2\theta) = \frac{2H_{12}}{H_{11}-H_{22}}\nonumber
\end{equation}

i.e.

\begin{equation}
\cos(\theta)=\left(\frac{r+\omega}{2r}\right)^{1/2}\nonumber
\end{equation}

\begin{equation}
\sin(\theta)=-\mbox{sgn}(H_{12})\left(\frac{r-\omega}{2r}\right)^{1/2}\nonumber
\end{equation}

where

\begin{equation}
\omega = H_{22}-H_{11}\nonumber
\end{equation}

\begin{equation}
r=\left(\omega^2+4H_{12}^2\right)^{1/2}\nonumber
\end{equation}

\begin{equation}
E_1=\frac{1}{2}\left(H_{11}+H_{22}-r\right)\nonumber
\end{equation}

\begin{equation}
E_2=\frac{1}{2}\left(H_{11}+H_{22}+r\right)\nonumber
\end{equation}

===Brute Force Method===

- Gives all the eigenvalues and eigenvectors, but it is slow

- First orthonormalize the <math>N</math>-dimensional basis set; i.e. form linear combinations

\begin{equation}
\Phi_m = \sum_{n=1}^N\varphi_nR_{nm}\nonumber
\end{equation}

such that
\begin{equation}
\langle \Phi_m | \Phi_n \rangle = \delta_{m,n}
\end{equation}

This can be done by finding an orthogonal tranformation T such that

\begin{equation}
T^TOT = I = \left(
\begin{array}{ccccc}
I_1 & 0 & 0 & \ldots & 0 \\
0 & I_2 & 0 & \ldots & 0 \\
0 & 0 & I_3 & \ldots & 0 \\
\vdots & \vdots & \vdots & \ddots & \vdots \\
0 & 0 & 0 & \ldots & I_N
\end{array}\right);
\end{equation}

\begin{equation}
O_{mn} = \langle \varphi_m | \varphi_n \rangle \nonumber
\end{equation}

and then applying a scale change matrix

\begin{equation}
S = \left(\begin{array}{ccccc}
\frac{1}{I_1^{1/2}} & 0 & 0 & \ldots & 0\\
0 & \frac{1}{I_2^{1/2}} & 0 & \ldots & 0 \\
0 & 0 & \frac{1}{I_3^{1/2}} & \ldots & 0 \\
\vdots & \vdots & \vdots & \ddots & \vdots \\
0 & 0 & 0 & \ldots & I_N^{1/2}
\end{array}\right)= S^T\nonumber
\end{equation}

Then

\begin{equation}
S^TT^TOTS = 1
\end{equation}

i.e.

\begin{equation}
R^TOR = 1
\end{equation}

with <math>R=TS</math>.

If H is the matrix with elements <math>H_{mn} = \langle \varphi_m | \varphi_n \rangle </math>, then H
expressed in the <math>\Phi_m</math> basis set is

\begin{equation}
H^\prime = R^THR.\nonumber
\end{equation}

We next diagonalize <math>H^\prime</math> by finding an orthogonal transformation W such
that

\begin{equation}
W^T H^\prime W = \lambda = \left(
\begin{array}{cccc}
\lambda_1 & 0 & \ldots & 0 \\
0 & \lambda_2 & \ldots & 0 \\
\vdots & \vdots & \ddots & \vdots \\
0 & 0 & \ldots & \lambda_N
\end{array}\right)\nonumber
\end{equation}

The q'th eigenvector is

\begin{equation}
\Psi^{(q)} = \sum_{n=1}^N\Phi_n W_{n,q} = \sum_{n,n^\prime} \varphi_{n^\prime} R_{n^\prime ,n} W_{n,q} \nonumber
\end{equation}

i.e.

\begin{equation}
c_{n^\prime}^{(q)} = \sum_{n=1}^N R_{n^\prime n} W_{n,q}
\end{equation}

===The Power Method===

- Based on the observation that if H has one eigenvalue, <math>\lambda_M</math>, much bigger than all the rest, and <math>\chi = \left(
\begin{array}{c}a_1\\a_2\\\vdots\end{array}\right)</math>
is an arbitrary starting vector, then <math>\chi = \sum_{q=1}^N x_q\Psi^{(q)}</math>.

\begin{equation}
(H)^n\chi = \sum_{q=1}^N x_q \lambda^n_q\Psi^{(q)}\nonumber\rightarrow x_M\lambda_M^n\Psi^{(M)}\nonumber
\end{equation}

provided <math>x_M\neq 0</math>.

To pick out the eigenvector corresponding to any eigenvalue, write the original problem in the form

\begin{align}
H\Psi & = \lambda O \Psi \nonumber \\
\left( H - \lambda_q O \right) \Psi & = \left( \lambda - \lambda_q \right) O \Psi \nonumber
\end{align}

Therefore,

\begin{equation}
G \Psi = \frac{1}{\lambda - \lambda_q } \Psi \nonumber \\
\end{equation}

where <math>G=(H-\lambda_qO)^{-1}O</math> with eigenvalues <math>\frac{1}{\lambda_n - \lambda_q}</math>.

By picking <math>\lambda_q</math> close to any one of the <math>\lambda_n</math>, say
<math>\lambda_{n^\prime}</math>, then <math>\frac{1}{\lambda_n-\lambda_q}</math> is much larger for
<math>n=n^\prime</math> than for any other value. The sequence is then

\begin{array}{c}
\chi_1=G\chi\nonumber\\
\chi_2=G\chi_1\nonumber\\
\chi_3=G\chi_2\nonumber\\
\vdots\nonumber
\end{array}

until the ratios of components in <math>\chi_n</math> stop changing.

- To avoid matrix inversion and multiplication, note that the sequence is equivalent to
\begin{equation}
F \chi_n = \left( \lambda - \lambda_q \right) O \chi_{n-1} \nonumber \\
\end{equation}
where <math>F = H-\lambda_qO</math>. The factor of <math>\left( \lambda - \lambda_q \right)</math> can be
dropped because this only affects the normalization of <math>\chi_n</math>. To find <math>\chi_n</math>, solve
\begin{equation}
F \chi_n = O \chi_{n-1} \nonumber
\end{equation}
(N equations in N unknowns). Then
\begin{equation}
\lambda = \frac{ \langle \chi_n | H | \chi_n \rangle}{\langle \chi_n | \chi_n \rangle} \nonumber
\end{equation}

==Matrix Elements of H==

\begin{equation}
H=-\frac{1}{2}\nabla^2_1 -\frac{1}{2}\nabla^2_2 - \frac{1}{r_1} - \frac{1}{r_2} +\frac{Z^{-1}}{r_{12}}\nonumber
\end{equation}

Taking <math>r_1, r_2</math> and <math>r_{12}</math> as independent variables,

\begin{equation}
\nabla_1^2 = \frac{1}{r_1^2}\frac{\partial}{\partial r_1}\left(r_1^2\frac{\partial}{\partial r_1}\right)+ \frac{1}{r_{12}^2}\frac{\partial}{\partial r_{12}}
\left(r_{12}^2\frac{\partial}{\partial r_{12}}\right)\nonumber -\frac{l_1(l_1+1)}{r_1^2}+2(r_1 - r_2\cos(\theta_{12}))\frac{1}{r_{12}}\frac{\partial^2}{\partial r_1 \partial r_{12}} \nonumber - 2(\nabla_1^Y \cdot {\bf r}_2)\frac{1}{r_{12}}\frac{\partial}{\partial r_{12}}\nonumber
\end{equation}

where <math>\nabla_1^Y</math> acts only on the spherical harmonic part of the wave function and the diagram

[[File:Pic3.jpg|400px]]

defines the complete set of 6 independent variables is <math>r_1</math>, <math>r_2</math>, <math>r_{12}</math>, <math>\theta_1</math>, <math>\varphi_1</math>, <math>\chi</math>.

If <math>r_{12}</math> were not an independent variable, then one could take the volume element to be

\begin{equation}
d\tau = r_1^2dr_1\sin(\theta_1)d\theta_1d\varphi_1r_2^2dr_2\sin(\theta_2)d\theta_2d\varphi_2.\nonumber
\end{equation}

However, <math>\theta_2</math> and <math>\varphi_2</math> are no longer independent variables. To eliminate them, take the point <math>{\bf r}_1</math> as the
origin of a new polar co-ordinate system, and write

\begin{equation}
d\tau=-r_1^2dr_1\sin(\theta_1)d\theta_1d\varphi_1r_{12}^2dr_{12}\sin(\psi)d\psi d\chi\nonumber
\end{equation}

and use

\begin{equation}
r_2^2=r_1^2+r_{12}^2 +2r_1r_{12}\cos(\psi)
\end{equation}

Then for fixed <math>r_1</math> and <math>r_{12}</math>,

\begin{equation}
2r_2dr_2 = -2r_1r_{12}\sin(\psi)d\psi
\end{equation}

Thus

\begin{equation}
d\tau= r_1dr_1r_2dr_2r_{12}dr_{12}\sin(\theta_1)d\theta_1d\varphi_1d\chi
\end{equation}

The basic type of integral to be calculated is

\begin{align}
I \left( l_1, m_1, l_2, m_2; R \right) & = \int \sin \left( \theta_1 \right) d\theta_1 d\varphi_1 d\chi Y^{m_1}_{l_1} \left( \theta_1, \varphi_1 \right)^{*} Y^{m_2}_{l_2} \left( \theta_2, \varphi_2 \right) \\
& \times \int r_1dr_1r_2dr_2r_{12}dr_{12}R \left( r_1, r_2, r_{12} \right)
\end{align}

Consider first the angular integral. <math>Y^{m_2}_{l_2} \left( \theta_2, \varphi_2 \right)</math> can be expressed in terms of the independent variables <math>\theta_1, \varphi_1, \chi</math> by use of the rotation matrix relation
\begin{equation}
Y^{m_2}_{l_2}( \theta_2, \varphi_2) = \sum_m \mathcal{D}^{(l_2)}_{m_2,m}(\varphi_1,\theta_1,\chi)^*Y^m_{l_2}( \theta_{12}, \varphi)
\end{equation}
where <math>\theta, \varphi</math> are the polar angles of <math>{\bf r}_2</math> relative to <math>{\bf r}_1</math>. The angular integral is then

\begin{align}
I_{ang} & = \int^{2 \pi}_{0} d\chi \int^{2\pi}_{0} d\varphi_1 \int^{\pi}_{0} \sin \left( \theta_1 \right) d\theta_{1} Y^{m_1}_{l_1} \left( \theta_1, \varphi_1 \right)^{*} \\
& \times \sum_m \mathcal{D}^{(l_2)}_{m_2,m} \left( \varphi_1,\theta_1,\chi \right)^{*} Y^{m}_{l_2} \left( \theta_{12},\varphi \right) \\
\end{align}

Use

\begin{equation}
Y^{m_1}_{l_1} \left( \theta_1, \varphi_1 \right)^{*} = \sqrt{\frac{2l_1+1}{4\pi}}\mathcal{D}^{(l_1)}_{m_1,0} \left( \varphi_1, \theta_1, \chi \right)
\end{equation}

together with the orthogonality property of the rotation matrices (Brink and Satchler, p 147)

\begin{align}
& \int \mathcal{D}^{(j)*}_{m,m^\prime}\mathcal{D}^{(J)}_{M,M^\prime}\sin(\theta_1)d\theta_1d\varphi_1d\chi \\
& = \frac{8\pi^2}{2j+1}\delta_{jJ}\delta_{mM}\delta_{m^\prime M^\prime}\nonumber
\end{align}

to obtain

\begin{align}
I_{ang} & = \sqrt{ \frac{2l_1+1}{4\pi}} \frac{8\pi^2}{2l_1+1} \delta_{l_1,l_2} \delta_{m_1,m_2} Y^{0}_{l_2} \left( \theta_{12}, \varphi \right) \\
& = 2 \pi \delta_{l_1,l_2} \delta_{m_1,m_2} P_{l_2} \left( \cos \theta_{12} \right)
\end{align}

since

\begin{equation}
Y^0_{l_2} \left( \theta_{12}, \varphi \right) = \sqrt{\frac{2l_1+1}{4\pi}} P_{l_2} \left( \cos \left( \theta_{12} \right) \right)
\end{equation}

Note that <math>P_{l_2} \left( \cos \theta \right)</math> is just a short hand expression for a radial function because

\begin{equation}
\cos \theta_{12} = \frac{r_1^2 + r_2^2 - r_{12}^2}{2r_1r_2}
\end{equation}

The original integral is thus

\begin{equation}
I \left( l_1, m_1, l_2, m_2; R \right) = 2 \pi \delta_{l_1,l_2} \delta_{m_1,m_2} \int^{\infty}_{0} r_{1} dr_{1} \int^{\infty}_{0} r_{2} dr_{2} \int^{r_{1} + r_{2}}_{|r_{1} - r_{2}|} r_{12} dr_{12} R \left( r_{1}, r_{2}, r_{12} \right) P_{l_2} \left( \cos \theta_{12} \right)
\end{equation}

where again

\begin{equation}
\cos \theta_{12} = \frac{r_{1}^{2} +r^{2}_{2} - r_{12}^{2}}{2 r_{1} r_{2}}
\end{equation}

is a purely radial function.

The above would become quite complicated for large <math>l_2</math> because <math>P_{l_2}(\cos \theta_{12})</math> contains terms up to <math>\left( \cos \theta_{12} \right)^{l_2}</math>. However, recursion relations exist which allow any integral containing <math>P_l \left( \cos \theta_{12} \right)</math> in terms of those containing just <math>P_0 \left( \cos \theta_{12} \right) = 1</math> and <math>P_1 \left( \cos \theta_{12} \right) = \cos \theta_{12}</math>.

==Radial Integrals and Recursion Relations==

The basic radial integral is [see G.W.F. Drake, Phys. Rev. A 18, 820 (1978)]

\begin{align}
I_0 \left( a, b, c \right) & =
\int^{\infty}_{0} r_{1} dr_{1} \int^{\infty}_{r_{1}} r_{2} dr_{2} \int^{r_{1}+r_{2}}_{r_{2} - r_{1}} r_{12} dr_{12} r_{1}^{a} r_{2}^{b} r_{12}^{c} e^{-\alpha r_{1} - \beta r_{2}} + \int^{\infty}_{0} r_{2} dr_{2} \int^{\infty}_{r_{2}} r_{1} dr_{1} \int^{r_{1} + r_{2}}_{r_{1} - r_{2}} r_{12} dr_{12} r_{1}^{a} r_{2}^{b} r_{12}^{c} e^{- \alpha r_{1} - \beta r_{2}} \\
& = \frac{2}{c+2} \sum^{[(c+1)/2]}_{i=0}
\left(\begin{array}{c}
c+2 \\
2i+1
\end{array} \right)
\left\{ \frac{q!}{\beta^{q+1} \left( \alpha + \beta \right)^{p+1}} \sum_{j=0}^{q} \frac{\left( p + j \right)!}{j!} \left( \frac{\beta}{\alpha + \beta} \right)^j + \frac{q^{\prime}!}{\alpha^{q^{\prime}+1} \left( \alpha + \beta \right)^{p^{\prime}+1}} \sum^{q^{\prime}}_{j=0} \frac{\left( p^{\prime} + j \right)!}{j!} \left( \frac{\alpha}{\alpha + \beta} \right)^j \right\}
\end{align}

where

\begin{array}{ll}
p = a + 2i + 2 & p^{\prime} = b + 2i + 2 \\
q = b + c - 2i + 2 & q^{\prime} = a + c - 2i + 2
\end{array}

The above applies for <math> a, b \geq -2, c\geq -1</math>.
<math>[x]</math> means "greatest integer in" <math>x</math>.

Then

\begin{align}
I_1 \left( a, b, c \right) & = \int d\tau_{r} r_{1}^{a} r_{2}^{b} r_{12}^{c} e^{ -\alpha r_{1} - \beta r_{2}} P_{1}\left( \cos \theta \right) \\
& = \frac{1}{2} \left( I_{0} \left( a + 1, b - 1, c \right) + I_{0} \left( a - 1, b + 1, c \right) - I_{0} \left( a - 1, b - 1, c + 2 \right) \right)
\end{align}

===The Radial Recursion Relation===

Recall that the full integral is

\begin{equation}
I(l_1m_1,l_2m_2;R)=2\pi\delta_{l_1,l_2}\delta_{m_1,m_2}I_{l_2}(R)
\end{equation}

where

\begin{equation}
I_{l_2}(R) = \int d\tau_r R(r_1,r_2,r_{12})P_{l_2}(\cos \theta_{12})
\end{equation}

To obtain the recursion relation, use

\begin{equation}
P_l(x) = \frac{[P^{\prime}_{l+1}(x)-P^{\prime}_{l-1}(x)]}{2l+1}
\end{equation}

with

\begin{equation}
P_{l+1}^\prime (x) = \frac{d}{dx}P_{l+1}(x)
\end{equation}

Here <math>x=\cos\theta_{12}</math> and

\begin{align}
\frac{d}{d \cos \theta_{12}} & = \frac{dr_{12}}{d \cos \theta_{12}} \frac{d}{dr_{12}} \\
& = -\frac{r_{1} r_{2}}{r_{12}} \frac{d}{dr_{12}}
\end{align}

Then,

\begin{equation}
I_{l}\left( R \right) = - \int d\tau_r R \frac{r_{1} r_{2}}{r_{12}} \frac{d}{dr_{12}} \frac{ \left[ P_{l+1} \left( \cos \theta_{12} \right) - P_{l-1} \left( \cos \theta_{12} \right) \right]}{2l+1}
\end{equation}

The <math>r_{12}</math> part of the integral is

\begin{align}
& \int^{r_{1} + r_{2}}_{|r_{1} - r_{2}|} r_{12} dr_{12} R \frac{r_{1} r_{2}}{r_{12}} \frac{d}{dr_{12}} \left[P_{l+1}-P_{l-1} \right] \\
& = R r_{1} r_{2} \left[ P_{l+1} - P_{l-1} \right]^{r_{1} + r_{2}}_{|r_{1} - r_{2}|} - \int^{r_{1} + r_{2}}_{|r_{1} - r_{2}|} r_{12} dr_{12} \left( \frac{d}{dr_{12}} R \right) \frac{r_{1} r_{2}}{r_{12}} \frac{ \left[ P_{l+1} - P_{l-1} \right]}{2l+1}
\end{align}

The integrated term vanishes because

\begin{equation}
\cos \theta_{12} = \frac{r_{1}^{2} + r_{2}^{2} - r_{12}^{2}}{2 r_{1} r_{2}}
\begin{array}{lll}
& = -1 & {\rm when} & r_{12}^{2} = \left( r_{1} + r_{2} \right)^{2} \\
& = 1 & {\rm when} & r_{12}^{2} = \left( r_{1} - r_{2} \right)^{2}
\end{array}
\end{equation}

and <math>P_l(1)=1</math>, <math>P_{l}(-1)=(-1)^{l}</math>.

Thus,

\begin{equation}
I_{l+1}\left(\frac{r_1r_2}{r_{12}}R^\prime\right) = \left( 2l+1 \right) I_{l}(R) - I_{l-1} \left( \frac{r_{1} r_{2}}{r_{12}} R^{\prime} \right)
\end{equation}

If

\begin{equation}
R = r_{1}^{a-1} r_{2}^{b-1} r_{12}^{c+2} e^{-\alpha r_{1} - \beta r_{2}}
\end{equation}

then [G.W.F. Drake, Phys. Rev. A 18, 820 (1978)]

\begin{equation}
I_{l+1}(r_1^ar_2^br_{12}^c) =
\frac{2l+1}{c+2} I_{l} \left( r_{1}^{a-1} r_{2}^{b-2} r_{12}^{c+2} \right)+I_{l-1} \left( r_{1}^{a} r_{2}^{b} r_{12}^{c} \right)
\end{equation}

For the case <math>c=-2</math>, take

\begin{equation}
R = r_{1}^{a-1} r_{2}^{b-1} \ln r_{12} e^{ -\alpha r_{1} -\beta r_{2}}
\end{equation}

Then

\begin{equation}
I_{l+1} \left( r_{1}^{a} r_{2}^{b} r_{12}^{c} \right) = I_{l} \left( r_{1}^{a-1} r_{2}^{b-1} \ln r_{12} \right) \left( 2l + 1 \right) - I_{l-1} \left( r_{1}^{a} r_{2}^{b} r_{12}^{-2} \right)
\end{equation}

This allows all <math>I_{l}</math> integrals to be calculated from tables of <math>I_0</math> and
<math>I_1</math> integrals.

===The General Integral===

The above results for the angular and radial integrals can now be combined into a general formula for integrals of the type

\begin{equation}
I = \int \int d{\bf r}_{1} d{\bf r}_{2} R_{1} \mathcal{Y}^{M^{\prime}}_{l^{\prime}_{1} l^{\prime}_{2} L^{\prime}} \left( \hat{r}_{1}, \hat{r}_{2} \right) T_{k_{1} k_{2} K}^{Q} \left( {\bf r}_{1},{\bf r}_{2} \right) R_{2} \mathcal{Y}_{l_{1} l_{2} L}^{M} \left( \hat{r}_{1}, \hat{r}_{2} \right)
\end{equation}

where

\begin{equation}
\mathcal{Y}^{M}_{l_{1} l_{2} L} \left( \hat{r}_{1}, \hat{r}_{2} \right) = \sum_{m_1,m_2} \langle l_{1} l_{2} m_{1} m_{2} | LM \rangle Y^{m_{1}}_{l_{1}} \left( \hat{r}_{1} \right) Y^{m_{2}}_{l_{2}} \left( \hat{r}_{2} \right)
\end{equation}

and

\begin{equation}
T^Q_{k_{1} k_{2} K} \left( {\bf r}_{1}, {\bf r}_{2} \right) = \sum_{q_{1}, q_{2}} \langle k_{1} k_{2} q_{1} q_{2} | KQ \rangle Y_{k_{1}}^{q_{1}} \left( \hat{r}_{1} \right) Y^{q_{2}}_{k_{2}} \left( \hat{r}_{2} \right)
\end{equation}

The basic idea is to make repeated use of the formula

\begin{equation}
Y^{m_{1}}_{l_{1}} \left( \hat{r}_{1} \right) Y^{m_{2}}_{l_{2}} \left( \hat{r}_{1} \right) = \sum_{l m} \left( \frac{\left( 2l_{1}+1 \right) \left(2l_{2}+1 \right) \left(2l+1 \right)}{4 \pi} \right)^{1/2}
\times \left( \begin{array}{ccc}
l_1 & l_2 & l \\
m_1 & m_2 & m
\end{array} \right) \left(\begin{array}{ccc}
l_1 & l_2 & l \\
0 & 0 & 0
\end{array} \right)
Y^{m*}_{l} \left( \hat{r}_{1} \right)
\end{equation}

where

\begin{equation}
Y^{m*}(\hat{r}) = (-1)^mY_l^{-m}(\hat{r})
\end{equation}

and

\begin{equation}
\left(\begin{array}{ccc}
l_1 & l_2 & l \\
m_1 & m_2 & m
\end{array} \right) = \frac{\left( -1 \right)^{l_{1} - l_{2} - m}}{\left( 2l + 1 \right)^{1/2}} \left( l_{1} l_{2} m_{1} m_{2} | l, -m \right)
\end{equation}

is a 3-<math>j</math> symbol. In particular, write

\begin{array}{cc}
Y^{m_1^\prime}_{l_1^\prime}(\hat{r}_1)^{*} \underbrace{Y_{k_{1}}^{q_{1}} \left( \hat{r}_{1} \right) Y^{m_{1}}_{l_{1}} \left( \hat{r}_{1} \right)} & = & \sum_{\Lambda M}(\ldots)Y^{M}_\Lambda(\hat{r}_1)\\
\ \ \ \ \ \ \ \sum_{\lambda_1\mu_1}Y_{\lambda_1}^{\mu_1}(\hat{r}_1) & & \\
& & \\
Y^{m_{2}^{\prime}}_{l_{2}^{\prime}} \left( \hat{r}_{2} \right) \underbrace{Y^{q_{2}}_{k_{2}} \left( \hat{r}_{2} \right) Y_{l_{2}}^{m_{2}} \left( \hat{r}_{2} \right)} & = & \sum_{\Lambda^{\prime} M^{\prime}} \left( \ldots \right)Y_{\Lambda^{\prime}}^{M^{\prime}*} \left( \hat{r}_{2} \right) \\
\ \ \ \ \ \ \ \ \ \sum_{\lambda_{2} \mu_{2}}Y_{\lambda_{2}}^{\mu_{2}} \left( \hat{r}_{2} \right) & &
\end{array}

The angular integral then gives a factor of <math>2 \pi \delta_{\Lambda, \Lambda^{\prime}} \delta_{M, M^{\prime}} P_{\Lambda} \left( \cos \theta_{12} \right)</math>. The total integral therefore reduces to the form

\begin{equation}
I = \sum_{\Lambda} C_{\Lambda} I_{\Lambda} \left( R_{1} R_{2} \right)
\end{equation}

where <math>C_{\Lambda} = \sum_{\lambda_{1}, \lambda_{2}} C_{ \lambda_{1}, \lambda_{2}, \Lambda}</math>. For further details and derivations, including graphical representations, see G.W.F. Drake, Phys. Rev. A <math>{\bf 18}</math>, 820 (1978).

==Graphical Representation==

[[File:GraphRep1.jpg|500px]]

==Matrix Elements of H==
Recall that

\begin{equation}
H = -\frac{1}{2} \nabla^{2}_{1} - \frac{1}{2} \nabla^{2}_{2} - \frac{1}{r_1} - \frac{1}{r_2} + \frac{Z^{-1}}{r_{12}}
\end{equation}

Consider matrix elements of

\begin{equation}
\nabla^2_1 = \frac{1}{r_1^2} \frac{\partial}{\partial r_1}\left(r_1^2\frac{\partial}{\partial r_1} \right) + \frac{1}{r^2}\frac{\partial}{\partial r}\left(r^2\frac{\partial}{\partial r}\right) - \frac{(\vec{l}_1^Y)^2}{r_1^2} + \frac{2 \left( r_1 - r_2 \cos \theta \right)}{r} \frac{\partial^2}{\partial r_1 \partial r} - 2 \left( \nabla^Y_1 \cdot {\bf r}_2 \right) \frac{1}{r} \frac{\partial}{\partial r}
\end{equation}

where <math>r \equiv r_{12}</math> and <math>\cos\theta \equiv \cos\theta_{12}</math>. Also in what follows, define <math>\hat{\nabla}^Y_1 \equiv r_1\nabla^Y_1</math> where <math>\nabla^Y</math> operates only on the spherical harmonic part of the wave function, and similarly for <math>\vec{l}_1^Y</math>.

A general matrix element is

\begin{equation}
\langle r_1^{a^\prime} r_2^{b^\prime} r_{12}^{c^\prime} e^{-\alpha^\prime r_1 - \beta^\prime r_2} \mathcal{Y}^{M^\prime}_{l^\prime_1 l^\prime_2 L^\prime} \left( \hat{r}_1, \hat{r}_2 \right) \left| \nabla^2_1 \right| r_1^a r_2^b r_{12}^c e^{-\alpha r_1 - \beta r_2} \mathcal{Y}^M_{l_1 l_2 L} \left( \hat{r}_1, \hat{r}_2 \right) \rangle
\end{equation}

Since <math>\nabla^2_1</math> is rotationally invariant, this vanishes unless <math>L=L^\prime</math>, <math>M=M^\prime</math>. Also <math>\nabla_1^2</math> is Hermitian, so that the result must be the same whether it operates to the right or left, even though the results look very different. In fact, requiring the results to be the same yields some interesting and useful integral identities as follows:

Put

\begin{equation}
\mathcal{F} = F \mathcal{Y}^M_{l_1 l_2 L} \left( \hat{r}_1, \hat{r_2} \right)
\end{equation}

and

\begin{equation}
F = r_1^a r_2^b r_{12}^c e^{-\alpha r_1 - \beta r_2}
\end{equation}

Then

\begin{equation}
\nabla^2_1 \mathcal{F} = \left\{ \frac{1}{r_1^2} \left[ a \left( a + 1 \right) - l_1 \left( l_1 + 1 \right) \right] + \frac{c \left( c + 1 \right)}{r^2} + \alpha^2\frac{2\alpha \left( a + 1 \right)}{r_1} + \frac{2 \left(r_1 - r_2 \cos \theta \right) }{r_1r^2}c \left[ a - \alpha r_1 \right] - \frac{2c}{r^2} \left( \hat{\nabla}^Y_1 \cdot \hat{r}_2 \right) \frac{r_2}{r_1} \right\} \mathcal{F}
\end{equation}

and

\begin{align}
\langle \mathcal{F}^\prime \left| \nabla^2_1 \right| \mathcal{F} \rangle & = \sum_\Lambda \int d\tau_r F^\prime C_\Lambda \left( 1 \right) P_\Lambda \left( \cos \theta \right) \times \left\{ \frac{1}{r_1^2} \left[ a \left( a + 1 \right) - l_1 \left( l_1 + 1 \right) \right] -\frac{2 \alpha \left( a + 1 \right)}{r_1} + \frac{c \left( c + 1 \right)}{r^2} +\alpha^2 + \frac{2 \left( r_2 - r_2 \cos \theta \right)}{r_1r^2}c \left( a - \alpha r_1 \right) \right\} F \\
& + \sum_\Lambda \int d\tau_r F^\prime C_\Lambda \left( \hat{\nabla}^Y_1 \cdot \hat{r}_2 \right) P_\Lambda \left( \cos \theta \right) \left( \frac{-2c r_2}{r_1 r^2} \right) F
\end{align}

where

\begin{equation}
\int d\tau_r = \int^\infty_0 r_1 dr_1 \int^\infty_0 r_2 dr_2 \int^{r_1+r_2}_{\left| r_1 - r_2 \right|} r dr
\end{equation}

For brevity, let the sum over <math>\Lambda</math> and the radial integrations be understood, and let <math>\left( \nabla_1^2 \right)</math> stand for the terms which appear in the integrand. Then operating to the right gives

\begin{equation}
(\nabla_1^2)_R = \frac{1}{r_1^2} \left[ a \left( a + 1 \right) - l_1 \left( l_1 + 1 \right) \right] + \frac{c \left( c + 1 \right)}{r^2} + \alpha^2 - \frac{2 \alpha \left( a + 1 \right)}{r_1}
+ \frac{2 \left( r_1 - r_2 \cos \theta \right)}{r_1 r^2}c \left( a - \alpha r_1 \right) - \frac{2c r_2}{r_1 r^2} \left( \hat{\nabla}^Y_1 \cdot \hat{r}_2 \right)
\end{equation}

Operating to the left gives

\begin{equation}
(\nabla_1^2)_L = \frac{1}{r_1^2} \left[a^\prime \left( a^\prime + 1 \right) - l_1^\prime \left( l_1^\prime + 1 \right) \right] + \frac{c^\prime \left( c^\prime + 1 \right)}{r^2} + \alpha^{\prime 2} -\frac{2 \alpha^\prime \left( a^\prime + 1 \right)}{r_1}
+ \frac{2 \left( r_1 - r_2 \cos \theta \right)}{r_1 r^2} c^\prime \left( a^\prime - \alpha^\prime r_1 \right) - \frac{2 c^\prime r_2}{r_1 r^2} \left( \hat{\nabla}^{Y^\prime}_1 \cdot \hat{r}_2 \right)
\end{equation}

Now put

\begin{array}{cc}
a_+ = a+a^\prime, & \hat{\nabla}_1^+ = \hat{\nabla}_1^Y + \hat{\nabla}_1^{Y\prime} \\
a_{-} = a-a^\prime, & \hat{\nabla}_1^{-} = \hat{\nabla}_1^Y - \hat{\nabla}_1^{Y\prime}
\end{array}

etc., and substitute <math>a^\prime = a_+ - a</math>, <math>c^\prime = c_+ - c</math>, and <math>\alpha^\prime = \alpha_+ - \alpha</math> in <math>(\nabla^2_1)_L</math>. If <math>a_+</math>, <math>c_+</math> and <math>\alpha_+</math> are held fixed, then the equation

\begin{equation}
(\nabla^2_1)_R = (\nabla^2_1)_L
\end{equation}

must be true for arbitrary <math>a</math>, <math>c</math>, and <math>\alpha</math>. Their coefficients must thus vanish.

This yields the integral relations

\begin{equation}
\label{I}
\frac{\left( r_1 - r_2 \cos \theta \right)}{r_1 r^2} = \frac{1}{c_+} \left( \frac{- \left( a_+ + 1 \right)}{r_1^2} + \frac{\alpha_+}{r_1} \right) \tag{I}
\end{equation}

from the coefficient of <math>a</math> and

\begin{equation}
\label{II}
\frac{ \left( r_1 - r_2 \cos \theta \right) \left( a_+ - \alpha_+ r_1 \right)}{r_1 r^2} = \frac{r_2}{r_1 r^2}\left( \hat{r}_2 \cdot \hat{\nabla}_1^+ \right) - \frac{ \left( c_+ + 1 \right)}{r^2} \tag{II}
\end{equation}

from the coefficient of <math>c</math>. The coefficient of <math>\alpha</math> gives an equation equivalent to (I).

Furthermore, if can be show that (see problem)

\begin{align}
& \sum_\Lambda \int d\tau_r \frac{r^c}{r^2} C_\Lambda \left( \hat{r}_2 \cdot \hat{\nabla}_1^Y \right) P_\Lambda \left( \cos \theta \right) \\
= & \sum_\Lambda \int d\tau_r \frac{r^c}{c r_1 r_2}C_\Lambda \left( 1 \right)P_\Lambda \left( \cos \theta \right) \times \left( \frac{l_1^\prime \left( l_1^\prime + 1 \right) - l_1 \left( l_1 + 1 \right) -\Lambda \left( \Lambda + 1 \right)}{2} \right)
\end{align}

and similarly for <math>\left( \hat{r}_2 \cdot \hat{\nabla}^{Y\prime}_1 \right)</math> with <math>l_1</math> and <math>l_1^\prime</math> interchangeable, then it follows that

\begin{equation}
\frac{r^{c_+}}{r^2} \left( \hat{r}_2 \cdot \hat{\nabla}_1^+ \right) = - \frac{r^{c_+}}{c_+ r_1 r_2} \Lambda \left( \Lambda + 1 \right)
\end{equation}

and

\begin{equation}
\frac{r^{c_+}}{r^2} \left( \hat{r}_2 \cdot \hat{\nabla}_1^- \right) = \frac{r^{c_+}}{c_+ r_1 r_2} \left[ l_1^\prime \left( l_1^\prime + 1 \right) - l_1 \left( l_1 + 1 \right) \right]
\end{equation}

where equality applies after integration and summation over <math>\Lambda</math>.

Thus (II) becomes

\begin{equation}\label{IIa}
\frac{ \left( r_1 - r_2 \cos \theta \right) \left( a_+ - \alpha_+ r_1 \right)}{r_1 r^2} = - \frac{\Lambda \left( \Lambda + 1 \right)}{c_+ r^2_1} - \frac{ \left( c_+ + 1 \right)}{r^2} \tag{IIa}
\end{equation}

===Problem===

Prove the integral relation

\begin{align}
& \sum_\Lambda \int d\tau_r f \left( r_1, r_2 \right) \frac{1}{r} \left( \frac{d}{dr} q(r) \right) C_\Lambda \left( \hat{r}_2 \cdot \hat{\nabla}^Y_1 \right) P_\Lambda \left( \cos \theta \right) \\
= & \sum_\Lambda \int d\tau_r \frac{ f \left( r_1, r_2 \right)}{r_1, r_2} q(r) C_\Lambda \left( 1 \right) P_\Lambda \left( \cos \theta \right) \left(\frac{l_1^\prime \left( l_1^\prime + 1 \right) - l_1 \left( l_1 + 1 \right) - \Lambda \left( \Lambda + 1 \right)}{2}\right)
\end{align}

where the coefficients <math>C_\Lambda \left( 1 \right)</math> are the angular coefficients from the overlap integral

\begin{align}
\int & d\Omega\, \mathcal{Y}^{M*}_{l_1^\prime l_2^\prime L}(\hat{r}_1,\hat{r}_2)\, \mathcal{Y}^{M}_{l_1 l_2 L}(\hat{r}_1,\hat{r}_2)\\
= & \sum_\Lambda C_\Lambda(1)P_\Lambda(\cos\theta_{12}).
\end{align}

Hint: Use the fact that <math>l_1^2</math> is Hermitian so that
<math>
\int d\tau (l_1^2\mathcal{Y}^\prime)^*q(r)\mathcal{Y} = \int d\tau
\mathcal{Y}^{\prime *}l_1^2(q(r)\mathcal{Y})
</math>
with <math>\ \vec{l}_1= \frac{1}{i}\vec{r}_1\times\nabla_1</math>.

It is also useful to use
<math>\begin{eqnarray}\ \
\cos\theta\, P_L(\cos\theta) &=& \frac{1}{2L+1}[L\,P_{L-1}(\cos\theta) + (L+1)P_{L+1}(\cos\theta)]
\\
(\cos^2\theta-1)P_L(\cos\theta) &=&
\frac{L(L-1)}{(2L-1)(2L+1)}P_{L-2}(\cos\theta)\nonumber
-\frac{2(L^2+L-1)}{(2L-1)(2L+3)}P_L(\cos\theta) + \frac{(L+1)(L+2)}{(2L+1)(2L+3)}P_{L+2}(\cos\theta)\nonumber\\
&=& \frac{L(L-1)}{(2L-1)(2L+1)}[P_{L-2}(\cos\theta) - P_L(\cos\theta)]
+ \frac{(L+1)(L+2)}{(2L+1)(2L+3)}[P_{L+2}(\cos\theta) - P_L(\cos\theta)]
\end{eqnarray}</math>

together with a double application of the integral recursion relation
<math>
I_{L+1}\left(\frac{1}{r}\frac{d}{dr}q(r)\right) =
(2L+1)I_L\left(\frac{1}{r_1r_2}q(r)\right)+I_{L-1}\left(\frac{1}{r}\frac{d}{dr}q(r)\right)
</math>

Of course <math>l_1^2\mathcal{Y}^M_{l_1l_2L}(\hat{r}_1,\hat{r}_2)=l_1(l_1+1)\mathcal{Y}^M_{l_1l_2L}(\hat{r}_1,\hat{r}_2)</math>. Begin by expanding
<math>
l_1^2(g\mathcal{Y}) = \mathcal{Y}l_1^2g + g\,l_1^2\mathcal{Y} + 2(\vec{l}_1g)\cdot(\vec{l}_1\mathcal{Y})
</math>, and show that

<math>
\begin{eqnarray}\ \
(\vec{l}_1g)\cdot(\vec{l}_1\mathcal{Y}) & = & \frac{r_1^2}{r}\frac{dg}{dr}\vec{r}_2\cdot \nabla_1\mathcal{Y}\,\,\,\,\mbox{ and}\\
l_1^2g(r) &=& 2r_1r_2\cos\theta\frac{1}{r}\frac{dg}{dr} + r_1^2r_2^2(\cos^2\theta - 1)
\frac{1}{r}\frac{d}{dr}\left(\frac{1}{r}\frac{dg}{dr}\right).
\end{eqnarray}
</math>

The proof amounts to showing that the term <math>l_1^2g(r)</math> can be replaced by
<math>\Lambda(\Lambda+1)g(r)</math> after multiplying by <math>P_\Lambda(\cos\theta)</math>
and integrating by parts with respect to the radial integrations over <math>r_1</math>, <math>r_2</math> and
<math>r \equiv |\vec{r}_1 - \vec{r}_2|</math>. Remember that <math>\cos\theta</math> is just a short-hand notation for the radial function <math>(r_1^2 + r_2^2 - r^2)/(2r_1r_2)</math>.

==General Hermitian Property==

Each combination of terms of the form
\begin{equation}
<f>_R = a^2f_1+af_2+abf_3+b^2f_4+bf_5+af_6\nabla_1^Y+bf_7\nabla^Y_2+f_8(Y)
\end{equation}
can be rewritten
\begin{equation}
<f>_L= (a_+-a)^2f_1+(a_+-a)f_2 + (a_+-a)(b_+-b)f_3+(b_+-b)^2f_4
+(b_+-b)f_5+(a_+-a)f_6\nabla^{Y\prime}_1+(b_+-b)f_7\nabla^{Y\prime}_2+f_8)Y^\prime
\end{equation}
where as usual <math>\nabla_1^Y</math> acts only on the spherical harmonic part of the wave function denoted for short by <math>Y</math>.
Since these must be equal for arbitrary <math>a</math> and <math>b</math>,
\begin{equation}
a_+^2f_1+a_+f_2+_+b_+f_3+b_+f_4+b_+f_5+a_+f_6\nabla^{Y\prime}_1 +
b_+f_7\nabla^{Y\prime}_2+f_8(Y^\prime)-f_8(Y)=0
\end{equation}

Adding the corresponding expression with <math>Y</math> and <math>Y^\prime</math> interchanged
yields
<math>\label{III}
a_+^2f_1+a_+f_2+a_+b_+f_3+b_+^2f_4+b_+f_5+\frac{1}{2}a_+f_6\nabla^+_1+\frac{1}{2}b_+f_7\nabla^+_2=0\tag{III}
</math>

Subtracting gives
<math>\label{IV}
f_8(Y)-f_8(Y^\prime)=-\frac{1}{2}[a_+f_6\nabla^-_1+b_+f_7\nabla_2^-]\tag{IV}
</math>
<math>\label{V}
a[-2a_+f_1-2f_2-b_+f_3-f_6\nabla^+_1]=0\tag{V}
</math>
<math>\label{VI}
b[-2b_+f_4-2f_5-a_+f_3-f_7\nabla^+_2]=0\tag{VI}
</math>

Adding the two forms gives

\begin{eqnarray}
<f>_R+<f>_L=\frac{1}{2}(a_+^2+a_-^2)f_1+a_+f_2+\frac{1}{2}(a_+b_++a_-b_-)f_3\nonumber\\
\ +\frac{1}{2}(b_+^2+b_-^2)f_4+b_+f_5+\frac{1}{2}f_6(a_+\nabla_1^++a_-\nabla^-_1)\nonumber\\
\ +\frac{1}{2}(b_+\nabla_2^++b_-\nabla^-_2)+f_8(Y)+f_8(Y^\prime)\nonumber
\end{eqnarray}

Subtracting <math>\frac{x}{2}\times({\rm I})</math> gives

\begin{eqnarray}\ \
<f>_R+<f>_L = \frac{1}{2}[(1-x)a_+^2 + a_-^2]f_1 +
(1-\frac{x}{2})a_+f_2+\frac{1}{2}[(1-x)a_+b_++a_-b_-]f_3\nonumber\\
+\frac{1}{2}[(1-x)b_+^2+b_-^2]f_4+(1-\frac{x}{2})f_5b_++\frac{1}{2}f_6[(1-\frac{x}{2})a_+\nabla_1^++a_-\nabla^-_1]\nonumber\\
+\frac{1}{2}[(1-\frac{x}{2})b_+\nabla^+_2+b_-\nabla^-_2]+f_8(Y)+f_8(Y^\prime)\nonumber
\end{eqnarray}

If <math>x=1</math>,
\begin{eqnarray}
<f>_R+<f>_L &=&
\frac{1}{2}[a_-^2f_1+a_+f_2+a_-b_-f_3+b_-^2f_4+b_+f_5\nonumber\\
&+&\left(\frac{1}{2}a_+\nabla^+_1+a_-\nabla^-_1\right)f_6+\left(\frac{1}{2}b_+\nabla^+_2+b_-\nabla^-_2\right)f_7]+f_8(Y)+f_8(Y^\prime)\nonumber
\end{eqnarray}

The General Hermitian Property for arbitrary <math>x</math> gives
\begin{eqnarray}\ \
\nabla_1^2=
\frac{1}{4}\{\frac{1}{r_1^2}[(1-x)a_+^2+a_-^2+2(1-\frac{x}{2})a_+-2[l_1(l_1+1)+l_1^\prime(l_1^\prime+1)]\nonumber\\
-\frac{2}{r_1}[(1-x)\alpha_+a_++\alpha_-a_-+2(1-\frac{x}{2})\alpha_+]+(1-x)\alpha_+^2
+ \alpha^2_-]\nonumber\\
+\frac{2(r_1-r_2\cos\theta)}{r_1r^2}[(1-x)(a_+-\alpha_+r_1)c_++(a_--\alpha_-r_2)c_-]\nonumber\\
-\frac{2r_2}{r_1r^2}[(1-\frac{x}{2})c_+\hat{r}_2\cdot\hat{\nabla}_1^++c_-\hat{r}_2\hat{\nabla}_1^-]+\frac{(1-x)c_+^2+c_-^2+2(1-\frac{x}{2})c_+}{r^2}\}\nonumber
\end{eqnarray}.

Use
<math>\begin{eqnarray}\ \
\frac{-2r_2}{r_1r^2}[(1-\frac{x}{2})c_+\hat{r}_2\hat{\nabla}_1^++c_-\hat{r}_2\hat{\nabla}_1^-]=\frac{2}{r_1^2}\left[(1-\frac{x}{2})\Lambda(\Lambda+1)-\frac{c_-}{c_+}[l_1^\prime(l_1^\prime+1)+l_1(l_1+1)]\right]\end{eqnarray}</math>,

<math>\begin{eqnarray}
\frac{2(r_1-r_2\cos\theta)}{r_1r^2}(a_--\alpha_-r_1)c_- = 2\frac{c_-}{c_+}\left[\frac{-1}{r_1^2}[a_-(a_++1)]+\frac{1}{r_1}[a_-\alpha_++\alpha_-(a_++2)]-\alpha_-\alpha_+\right]\nonumber
\end{eqnarray}</math>,

and
<math>\begin{eqnarray}\ \
\frac{2(r_1-r_2\cos\theta)}{r_1r^2}(a_+-\alpha_+r_1)c_+ = 2\left[-\frac{1}{r_1^2}[a_+(a_++1)]+\frac{1}{r_1}[a_+\alpha_++\alpha_+(a_++2)]-\alpha_+^2\right]
\end{eqnarray}</math>.

Substituting into <math>\nabla_1^2</math> gives

\begin{eqnarray}
\nabla^2_1 = \frac{1}{4}\{\frac{1}{r_1^2}[-(1-x)a_+^2 + a_-^2 +xa_+ +
2(1-\frac{x}{2})\Lambda(\Lambda+1)\nonumber\\
-2l_1(l_1+1)(1-\frac{c_-}{c_+}) -
2l_1^\prime(l_1^\prime+1)(1+\frac{c_-}{c_+})-\frac{2c_-a_-}{c_+}(a_++1)]\nonumber\\
- \frac{2}{r_1}\left[-(1-x)\alpha_+(a_++2)+\alpha_-a_-+2(1-\frac{x}{2})\alpha_+
-\frac{c_-}{c_+}[a_-\alpha_++\alpha_-(a_++2)]\right]\nonumber\\
-(1-x)\alpha_+^2+\alpha_-^2-2\frac{c_-}{c_+}\alpha_-\alpha_+ +\frac{1}{r^2}\left[(1-x)c_+^2+c_-^2+2(1-\frac{x}{2})c_+\right]\}\nonumber
\end{eqnarray}

This has the form
<math>
\nabla_1^2 =
\frac{1}{4}\left[\frac{A_1}{r_1^2}+\frac{B_1}{r_1}+\frac{C_1}{r^2}+D_1\right]
</math>
with

<math>\begin{array}{ccc}
A_1=-(1-x)a_+^2+a_-^2
+xa_++2(1-\frac{x}{2})\Lambda(\Lambda+1)-2l_1(l_1+1)(1-\frac{c_-}{c_+})\
-2l_1^\prime(l_1^\prime+1)(1+\frac{c_-}{c_+})-\frac{2c_-a_-}{c_+}(a_++1) \ \
\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \
\end{array}
</math>

<math>
B_1 =
2\left[(1-x)\alpha_+(a_++2)-\alpha_-a_--2(1-\frac{x}{2})\alpha_++\frac{c_-}{c_+}[a_-\alpha_++\alpha_-(a_++2)]\right]
</math>

<math>
C_1 = (1-x)c_+^2+ c_-^2 +2(1-\frac{x}{2})c_+
</math>

<math>
D_1 = -1(1-x)\alpha_+^2 +\alpha_-^2 - 2\frac{c_-}{c_+}\alpha_-\alpha_+
</math>

The complete Hamiltonian is then (in Z-scaled a.u.)

<math>\begin{eqnarray}
H =-\frac{1}{2}\nabla^2_1 -\frac{1}{2}\nabla^2_2 -\frac{1}{r_1} -\frac{1}{r_2} +\frac{Z^{-1}}{r}\nonumber\\
= -\frac{1}{2}\nabla^2_1 -\frac{1}{2}\nabla^2_2 -\frac{1}{r_1} -\frac{Z-1}{Zr_2}+Z^{-1}\left(\frac{1}{r}-\frac{1}{r_2}\right)\nonumber\\
\ =-\frac{1}{8}\left[\frac{A_1}{r_1^2}+\frac{B_1+8}{r_1}+\frac{C_1}{r^2}+D_1+D_2 +\frac{A_2}{r_2^2} + \frac{B_2 + 8(Z-1)/Z}{r_2}+\frac{C_2}{r^2}\right]\nonumber
+ Z^{-1}\left(\frac{1}{r}-\frac{1}{r_2}\right)\nonumber
\end{eqnarray}</math>

The screened hydrogenic energy is
<math>\begin{eqnarray}
E_{SH}=-\frac{1}{2}\left[\frac{1}{n_1^2}+ \left(\frac{Z-1}{Z}\right)^2\frac{1}{n_2^2}\right]
\end{eqnarray}</math>
so that

<math>\begin{eqnarray}
H-E_{SH} &=& -\frac{1}{8}[
\frac{A_1}{r_1^2}+\frac{B_1+8}{r_1}+\frac{C_1}{r^2} + D_1 + D_2
-\frac{4}{n_1^2} -\left(\frac{Z-1}{Z}\right)^2\frac{4}{n_2^2}\nonumber
+\frac{A_2}{r_2}+\frac{B_2+8(Z-1)/Z}{r_2}+\frac{C_2}{r^2}] +
Z^{-1}\left(\frac{1}{r} -\frac{1}{r_2}\right)\nonumber
\end{eqnarray}</math>

==Optimization of Nonlinear Parameters==
- The traditional method of performing Hylleraas calculations is to write the
basis set in the form

<math>
\Psi = \sum_{i,j,k}c_{i,j,k}\varphi_{i,j,k}(\alpha,\beta)\pm exchange
</math>

with

<math>
\varphi_{i,j,k}(\alpha,\beta) = r_1^ir_2^jr_{12}^ke^{-\alpha r_1 - \beta
r_2}\mathcal{Y}^M_{l_1l_2L}.
</math>

The usual procedure is to set <math>\alpha = Z</math> so that it represents the inner
<math>1s</math> electron, and then to vary <math>\beta</math> so as to minimize the energy.

[[File:OptNon.jpg|500px]]

Since <math>\beta</math> appears in <math>\Psi</math> as a non-linear parameter, the entire
calculation must be repeated for each value of <math>\beta</math>. However, the minimum
becomes progressively smaller as the basis set is enlarged.

[[File:OptNon02.jpg|500px]]

*Difficulties

1. If the basis set is constructed so that <math> i +j +j \leq N</math>, the the
number of terms is
<math>
(N+1)(N+2)(N+3)/6
</math>
N=14 already gives 680 terms and an accuracy of about 1 part in <math>10^{10}</math> for
low-lying states. A substantial improvement in accuracy would require much
larger basis sets, together with multiple precision arithmetic to avoid loss
of significant figures when high powers are included.

2. The accuracy rapidly deteriorates as one goes to more highly excited
states - about 1 significant figure is lost each time the principle quantum
number is increased.

*Cure

We have found that writing basis sets in the form
\begin{eqnarray}
\Psi &=& \sum_{i,j,k}\left(c^{(1)}_{i,j,k}\varphi_{ijk}(\alpha_1\beta_1) +
c_{ijk}^{(2)}\varphi_{ijk}(\alpha_2\beta_2)\right) \pm exchange \nonumber\\
&=& \psi({\bf r}_1,{\bf r}_2) \pm \psi({\bf r}_2,{\bf r}_1)\nonumber
\end{eqnarray}
so that each combination of powers is included twice with different nonlinear
parameters gives a dramatic improvement in accuracy for basis sets of about
the same total size.

However, the optimization of the nonlinear parameters is now much more
difficult, and an automated procedure is needed.

If
<math>
E = \frac{<\Psi|H|\Psi>}{<\Psi|\Psi>}</math> and we assume <math><\Psi|\Psi> = 1</math>, then

<math>
\frac{\partial E}{\partial\alpha_t} = -2 <\Psi|H-E|r_1\psi({\bf r}_1,{\bf
r}_2;\alpha_t)\pm r_21\psi({\bf r}_2,{\bf
r}_1;\alpha_t)>
</math>
where
<math>
\psi({\bf r}_1,{\bf r}_2;\alpha_t) =
\sum_{i,j,k}c_{ijk}^{(t)}\varphi_{ijk}(\alpha_t\beta_t)
</math>

*Problem

1. Prove the above.

2. Prove that there is no contribution from the implicit dependence of
<math>c_{ijk}^{(t)}</math> on <math>\alpha_t</math> if the linear parameters have been optimized.

==The Screened Hydrogenic Term==

If the Hamiltonian is written in the form (in Z-scaled a.u.)
<math>\begin{eqnarray}\
H = H_0({\bf r}_1,Z) + H_0({\bf r}_2,Z-1) +
Z^{-1}\left(\frac{1}{r}-\frac{1}{r_2}\right)
\end{eqnarray}</math>
with
<math>\
H_0({\bf r}_1,Z) = -\frac{1}{2}\nabla_1^2 - \frac{1}{r_1}\nonumber
</math> and
<math>
H_0({\bf r}_2,Z-1) = -\frac{1}{2}\nabla_2^2 - \frac{Z-1}Z{r_2}\nonumber
</math>,
then the eigenvectors of <math>H_0({\bf r},Z) +H_0({\bf r}_2,Z-1)</math> are products of
hydrogenic orbitals
<math>
\Psi_0=\psi_0(1s,Z)\psi_0(nl,Z-1)
</math>
and the eigenvalue is
<math>\begin{eqnarray}
E_{SH} = \left[-\frac{1}{2} - \left(\frac{Z-1}{Z}\right)^2\frac{1}{2n^2}\right]Z^2 a.u.
\end{eqnarray}</math>
called the screened hydrogenic eigenvalue. From highly excited states,
<math>E_{SH}</math> and <math>\Psi_0</math>are already excellent approximations.

For example, for the 1s8d states, the energies are

<math>\begin{eqnarray}\\
E(1s8d^1D)&=& -2.007\,816\,512\,563\,81 {\rm\ a.u.}\nonumber\\
E(1s8d^3D)&=& - 2.007\,817\,934\,711\,71 {\rm\ a.u.}\nonumber\\
E_{SH} &=& -2.007\,812\,5 {\rm\ a.u.} \nonumber
\end{eqnarray}</math>

It is therefore advantageous to include the screened hydrogenic terms in the
basis set so that the complete trial function becomes
<math>
\Psi = c_0\Psi_0 +
\sum_{ijk}\left[c_{ijk}^{(1)}\varphi_{ijk}(\alpha_1,\beta_1)+c_{ijk}^{(2)}\varphi(\alpha_2,\beta_2)\right]\pm
{\rm exchange}
</math>

Also, the variational principal can be re-expressed in the form
<math>\begin{eqnarray}
E= E_{SH}+\frac{<\Psi|H-E_{SH}|\Psi>}{<\Psi|\Psi>}
\end{eqnarray}</math>
so that the <math>E_{SH}</math> term can be cancelled analytically from the matrix
elements.

For example
<math>
<\Psi_0|H-E_{SH}|\Psi_0> = 0
</math>

and
<math>
<\varphi_{ijk}|H-E_{SH}|\Psi_0> =
Z^{-1}<\varphi_{ijk}|\frac{1}{r}-\frac{1}{r_2}|\Psi_0>
</math>

Recall that
<math>\begin{eqnarray}\ \
I_0(a,b,c) = \frac{2}{c+2}\sum^{[c+1]/2}_{i=0}\left(\begin{array}{c}
c+2\\
2i+1
\end{array}\right) \left[f(p,q;\beta) + f(p^\prime,q^\prime;\alpha)\right]
\end{eqnarray}</math>
where

<math>\begin{eqnarray}
f(p,q;x) =
\frac{q!}{x^{q+1}(\alpha+\beta)^{p+1}}\sum^p_{j=0}\frac{(p+j)!}{j!}\left(\frac{x}{\alpha+\beta}\right)^j
\end{eqnarray}</math>

<math>
\begin{array}{cc}
p=a+2i+2 & p^\prime = b+2i+2\\
q= b+c-2i+2 & q^\prime =a+c-2i+2
\end{array}
</math>

If <math>\beta\ll\alpha</math>, the <math>f(p,q;\beta)\gg f(p^\prime,q^\prime;\alpha)</math> and the
<math>i=0</math> term is the dominant contribution to <math>f(p,q;\beta)</math>. However, since
this term depends only on the sum of powers <math>b+c</math> for <math>r_2</math>, it cancels exactly
from the matrix element of
<math>
\frac{1}{r_{12}} - \frac{1}{r_2}
</math>
and can therefore
be omitted in the calculations of integrands, thereby saving many significant
figures. This is especially valuable when <math>b+c</math> is large and <math>a</math> is small.

==Small Corrections==
*Mass Polarization
If the nuclear mass is not taken to be infinite, then the Hamiltonian is
<math>\begin{eqnarray}\ \
H = \frac{P_N^2}{2M} + \sum_{i=1}\frac{p_i^2}{2m} + V
\end{eqnarray}</math>

[[File:SmallCor.jpg|500px]]

where <math>M =M_A-nm</math> is the nuclear mass and <math>m</math> is the electron mass.

Change variables to the C of M
<math>\begin{eqnarray}
{\bf R} = \frac{1}{M+nm}\left[M{\bf r}_N + m({\bf r}_1 + {\bf r}_2 +
\ldots)\right]
\end{eqnarray}</math>

and relative variables

<math>
{\bf s}_i = {\bf r}_i - {\bf r}_N.
</math>

Then

<math>\begin{eqnarray}\ \ \ \ \
H = \frac{1}{2(M+nm)}{\bf p}_R^2+\frac{1}{2m}\sum_{i=1}^n{\bf p}_{s_i}^2
+\frac{1}{2M}\sum_{i,k}{\bf p}_{s_i}\cdot {\bf p}_{s_k} + V({\bf s}_1, {\bf
s}_2, \ldots)\nonumber
= \frac{1}{M+nm}{\bf p}_R^2+\frac{1}{2\mu}\sum_{i=1}^n{\bf
p}_{s_i}^2+\frac{1}{2M}\sum_{i\neq k}{\bf p}_{s_i}\cdot {\bf p}_{s_k}+V({\bf s}_1, {\bf
s}_2, \ldots)\nonumber
\end{eqnarray}</math>

where
<math>\begin{eqnarray}
\mu = \left(\frac{1}{M} + \frac{1}{m}\right)^{-1}
\end{eqnarray}</math>

If <math>n=2</math>, the Schrödinger equation is
<math>\begin{eqnarray}\ \ \
\left[ -\frac{\hbar^2}{2\mu}(\nabla^2_{s_1}+ \nabla^2_{s_2}) -
\frac{\hbar^2}{M}\nabla_{s_1}\nabla_{s_2}-\frac{Ze^2}{s_1}-\frac{Ze^2}{s_2}+\frac{e^2}{s_{12}}\right]\psi
= E\psi
\end{eqnarray}</math>.

Define the reduced mass Bohr radius
<math>\begin{eqnarray}
a_\mu = \frac{\hbar^2}{\mu e^2} = \frac{m}{\mu}a_0
\end{eqnarray}</math>
and

<math>\begin{eqnarray}
\rho_i = \frac{Zs_i}{a_\mu}\nonumber\\
\varepsilon = \frac{E}{Z^2(e^2/a_\mu)}\nonumber
\end{eqnarray}</math>

The Schrödinger equation then becomes
<math>\begin{eqnarray}\ \ \
\left[
-\frac{1}{2}(\nabla^2_{\rho_1}+\nabla^2_{\rho_2})-\frac{\mu}{M}\vec{\nabla}_{\rho_1}\cdot\vec{\nabla}_{\rho_2} - \frac{1}{\rho_1}-\frac{1}{\rho_2}+\frac{Z^{-1}}{\rho_{12}}\right]\psi = \varepsilon \psi
\end{eqnarray}</math>

The effects of finite mass are thus

1. A reduced mass correction to all energy levels of <math>(e^2/a_\mu)/(e^2/a_0)
= \mu/m</math>, i.e.
<math>\begin{eqnarray}
E_M=\frac{\mu}{m}E_\infty
\end{eqnarray}</math>

2. A specific mass shift given in first order perturbation theory by
<math>\begin{eqnarray}
\Delta\varepsilon =
-\frac{\mu}{M}<\psi|\nabla_{\rho_1}\cdot\nabla_{\rho_2}|\psi>
\end{eqnarray}</math>
or

<math>\begin{eqnarray}
\Delta\varepsilon =
-\frac{\mu}{M}<\psi|\nabla_{\rho_1}\cdot\nabla_{\rho_2}|\psi>\frac{e^2}{a_\mu}\nonumber\\
= -\frac{\mu^2}{mM}<\psi|\nabla_{\rho_1}\cdot\nabla_{\rho_2}|\psi>\frac{e^2}{a_0}\nonumber
\end{eqnarray}</math>

Since <math>\nabla_{\rho_1}\cdot\nabla_{\rho_2}</math> has the same angular properties as
<math>{\bf\rho}_1 \cdot {\bf\rho}_2</math>, the operator is like the product of two
dipole operators. For product type wave functions of the form
$
\psi = \psi(1s)\psi(nl) \pm \mbox{exchange},
$
the matrix element vanishes for all but P-states.

=Radiative Transitions=

==Transition Integrals and the Line Strength==

For an electric dipole transition <math>\gamma'\,L'\,S'\,J'\,M' \rightarrow \gamma\,L\,S\,J\,M</math>, the line strength is defined by
\begin{equation}
S = \sum_{M',\mu,M}|\langle\gamma'\,L'\,S'\,J'\,M'|r_\mu|\gamma\,L\,S\,J\,M\rangle|^2
\end{equation}
Using the Wigner-Eckart theorem, this is related to the reduced matrix element by
\begin{eqnarray}
S(J',J) = |\langle\gamma'\,L'\,S'\,J'||r||\gamma\,L\,S\,J\,M\rangle|^2
\sum_{M',\mu,M}\left|\left(\matrix{J' & 1 & J \cr
-M'& \mu &M \cr}\right)\right|^2
\end{eqnarray}
Using the sum rule for the 3-<math>j</math> symbols
\begin{equation}
\sum_{M',\mu,M}\left|\left(\matrix{J' & 1 & J \cr
-M'& \mu &M \cr}\right)\right|^2 = 1
\end{equation}
we see that the line strength is identical to the reduced matrix element
\begin{equation}
S(J',J) = |\langle\gamma'\,L'\,S'\,J'||r||\gamma\,L\,S\,J\,M\rangle|^2
\end{equation}
For a nonrelativistic calculation in <math>LS</math> coupling, <math>S = S'</math> and we can next strip out the dependence on <math>J</math> and <math>J'</math> by use of Eq.\ (7.1.7) of Edmonds { Angular Momentum in Quantum Mechanics} with <math>k = 1</math> to obtain
\begin{equation}
S(J',J) = (-1)^{L'+S+J+1}[(2J+1)(2J'+1)]^{1/2}\left\{
\matrix{L' & J'& S\cr
J & L & 1\cr}\right\}\langle\gamma' L'|| r || \gamma L\rangle
\end{equation}
where <math>\langle\gamma' L'|| r || \gamma L\rangle</math> is the reduced matrix element if spin is neglected. Note that if spin is neglected, then the line-strength is simply
\begin{equation}
S(L',L) = |\langle\gamma' L'|| r || \gamma L\rangle|^2
\end{equation}
To calculate the reduced matrix element, use a particular example along with the Wigner-Eckart theorem
such as
\begin{equation}
\langle\gamma'\, L'|| r || \gamma\, L\rangle = (-1)^{L'}\langle\gamma'\, L'\, 0| z | \gamma\, L\, 0\rangle{\Large /}\left(\matrix{L' & 1 & L \cr
0 & 0 & 0 \cr}\right)
\end{equation}

===Correlated Two-electron Integrals===
For the case of two-electron integrals, <math>z = z_1 + z_2</math> and <math>r = r_1 + r_2</math>. As discussed in G.W.F. Drake, Phys. Rev. A 18, 820 (1978), the general integral in Hylleraas coordinates for the integral over the <math>z_1</math> term can be written in the form
\begin{equation}
\label{int1}
\langle\gamma'\, L'\, 0| z_1 | \gamma\, L\, 0\rangle = \sum_\Lambda C_\Lambda I_\Lambda(\gamma',\gamma)
\end{equation}
where <math>I_\Lambda(\gamma',\gamma)</math> stands for the purely radial integral
\begin{equation}
\label{radial}
I_\Lambda(\gamma',\gamma) = \int_0^\infty r_1\,dr_1\int_0^\infty r_2\,dr_2\int_{|r_1-r_2|}^{r_1+r_2}r_{12}\, dr_{12}
R_{\gamma'}^*\,r_1\, R_\gamma P_\Lambda(\cos\theta_{12})
\end{equation}
and <math>\cos\theta_{12} = (r_1^2 + r_2^2 - r_{12}^2)/(2r_1r_2)</math> is a short-hand notation for a radial function of <math>r_1</math>, <math>r_2</math> and <math>r_{12}</math>. Also, <math>P_\Lambda(\cos\theta_{12})</math> is a Legendre polynomial,
and, <math>R_\gamma = R_\gamma(r_1,r_2,r_{12})</math> is the radial part of the Hylleraas wave function for state <math>\gamma</math> such
that the total wave function is <math>\psi({\bf r}_1,{\bf r}_2) = R_\gamma(r_1,r_2,r_{12}){\cal Y}_{l_1,l_2,L}^M(\hat{\bf r}_1,\hat{\bf r}_2)\pm {\rm exchange}</math>, normalized to unity, and <math>{\cal Y}_{l_1,l_2,L}^M(\hat{\bf r}_1,\hat{\bf r}_2)</math> is a vector coupled product of spherical harmonics. The coefficents <math>C_\Lambda</math> in Eq.\ (\ref{int1}) can be calculated from Eq.\ (32) of the above 1978 paper. They need to be calculated separately for he direct-direct and the direct-exchange contributions. For the reduced matrix element of <math>z_1</math>, the leading 3-<math>j</math> symbol can be dropped, leaving (for the case <math>k = 1</math>)
\begin{eqnarray}
C_\Lambda^{\rm reduced} &=& \frac{(-1)^{l_2'+L'+L+1}}{2} (l_1,l_1',l_2,l_2',L,L')^{1/2}
\sum_{\lambda={\rm max}(l_1-1,0)}^{l_1+1} (\lambda,\Lambda,1)
\left(\matrix{l_1 & 1 & \lambda \cr
0 & 0 & 0 \cr}\right)
\left(\matrix{l_1'& \lambda & \Lambda \cr
0 & 0 & 0 \cr}\right)\nonumber\\
&\times&\left(\matrix{l_2' & 1_2 & \Lambda \cr
0 & 0 & 0 \cr}\right)
\left\{\matrix{L' & \lambda & l_2 \cr
\Lambda& l_2' & l_1' \cr}\right\}
\left\{\matrix{L' & \lambda & l_2 \cr
l_1 & L & 1 \cr}\right\}
\end{eqnarray}
where the notation <math>(a,b,c,\cdots)</math> means <math>(2a+1)(2b+1)(2c+1)\cdots</math>.
A similar expression can be obtained for the angular coefficients for the <math>z_2</math> term by interchanging the subscripts 1 and 2 throughout and multiplying by <math>(-1)^{L+L'+1}</math>.

===Example===
As an example, for the <math>1s^2\;^1S - 1s2p\;^1P</math> transition of helium, the only nonvanishing <math>C_\Lambda</math> angular coefficients are <math>C_1 = 1/2</math> for the direct-direct term (i.e. <math>P_1(\cos\theta_{12}) = \cos\theta_{12}</math> is included in the radial integral (\ref{radial})) and <math>C_0 = 1/2</math> for the direct-exchange term (i.e. <math>P_0(\cos\theta_{12}) = 1</math> is included in the radial integral (\ref{radial})). The results for infinite nuclear mass are
\begin{equation}
S(1\;^1S-2\;^1P) = (0.781\,127\,158\,667)^2
\end{equation}
\begin{equation}f(1\;^1S-2\;^1P) = -0.276\,164\,703\,736
\end{equation}
where
\begin{equation}
f(1\;^1S-2\;^1P) = \frac{2}{3}[E(1\;^1S) - E(2\;^1P)]S(1\;^1S-2\;^1P)
\end{equation}
is the absorption oscillator strength.

===Test===
This is a test edit

Theory Notes

2020-12-14T04:23:01Z

Drake: /* General Hermitian Property */

<center>
<H1>Notes on solving the Schrödinger equation in Hylleraas coordinates for heliumlike atoms
Gordon W.F. Drake
Department of Physics, University of Windsor
Windsor, Ontario, Canada N9B 3P4
(Transcribed from hand-written notes and edited by Lauren Moffatt. Last revised November 22, 2017 to correct a missing subscript in the expression for $B_1$ in the Hamiltonian in Sect. 11 General Hermitian Property.)
</H1>
</center>

==Introduction==
These notes describe some of the technical details involved in solving the Schrödinger equation for a heliumlike atom of nuclear charge Z in correlated Hylleraas coordinates. The standard for computational purposes is to express all quantities in a form that is valid for electronic states of arbitrary angular momentum, and to express matrix elements of the Hamiltonian in a manifestly Hermitian symmetrized form. Further information and derivations of the mathematical techniques for angular integrals and radial recursion relations can be found in the paper
G.W.F. Drake, Phys. Rev. A <math>{\bf 18}</math>, 820 (1978).

The starting point is the two-electron Schrödinger equation for infinite nuclear mass
\begin{equation} \left[-\frac{\hbar^2}{2m}(\nabla^2_1 +\nabla^2_2) - \frac{Ze^2}{r_1} -
\frac{Ze^2}{r_2}+\frac{e^2}{r_{12}} \right]\psi = E\psi\nonumber
\end{equation}
where <math>m</math> is the electron mass, and <math>r_{12} = |{\bf r}_1 - {\bf r}_2|</math> (see diagram below).

[[File:Pic1.jpg|250px]]

Begin by rescaling distances and energies so that the Schrödinger equation can be expressed in a dimensionless form. The dimensionless Z-scaled distance is defined by <math>\begin{eqnarray} \rho = \frac{Zr}{a_0} \end{eqnarray} </math> where <math> \begin{eqnarray} a_0 = \frac{\hbar^2}{me^2} \end{eqnarray} </math> is the atomic unit (a.u.) of distance equal to the Bohr radius <math>0.529\, 177\, 210\, 92(17) \times 10^{-10}</math> m. Then

\begin{equation}[-\frac{\hbar^2}{2m}Z^2(\frac{me^2}{\hbar^2})^2(\nabla^2_{\rho_1}+\nabla^2_{\rho_2})
- Z^2\frac{e^2}{a_0}\rho^{-1}_1 - Z^2\frac{e^2}{a_0}\rho^{-1}_2 +
\frac{e^2}{a_0}Z\rho^{-1}_{12}]\psi= E\psi\nonumber\end{equation}

But <math>\begin{eqnarray} \frac{\hbar^2}{m}(\frac{me^2}{\hbar^2})^2 = \frac{e^2}{a_0} \end{eqnarray} </math> is the hartree atomic unit of of energy <math>(E_h = 27.211\,385\,05(60)</math> eV, or equivalently <math>E_h/(hc) = 2194.746\,313\,708(11)</math> m<math>^{-1})</math>. Therefore, after multiplying through by <math>a_0/(Ze)^2</math>, the problem to be solved in Z-scaled dimensionless units becomes

\begin{equation} \left[-\frac{1}{2}(\nabla^2_{\rho_1}+\nabla^2_{\rho_2}) -
\frac{1}{\rho_1} - \frac{1}{\rho_2} + \frac{Z^{-1}}{\rho_{12}}\right]\psi =
\varepsilon\psi\nonumber \end{equation}

where <math>\begin{eqnarray}\varepsilon = \frac{Ea_0}{(Ze)^2} \end{eqnarray}</math> is the energy in Z-scaled atomic units. For convenience, rewrite this in the conventional form
\begin{equation}
H\psi = \varepsilon\psi\nonumber \end{equation}
where (using <math>r</math> in place of <math>\rho</math> for the Z-scaled distance)
\begin{equation}
H = -\frac{1}{2}(\nabla^2_1+\nabla^2_2) - \frac{1}{r_1}-\frac{1}{r_2} + \frac{Z^{-1}}{r_{12}}
\nonumber\end{equation}
is the atomic Hamiltonian for infinite nuclear mass.

==The Hartree Fock Method==

For purposes of comparison, and to define the correlation energy, assume that <math>\psi({\bf r}_1,{\bf r}_2)</math> can be written in the separable product form
\begin{equation}
\psi({\bf r}_1,{\bf r}_2) = \frac{1}{\sqrt{2}} \left[ u_1(r_1) u_2(r_2) \pm u_2(r_1) u_1(r_2) \right] \nonumber
\end{equation}
for the <math>1s^2\ ^1\!S</math> ground state. Because of the <math>\frac{1}{r_{12}}</math> term in the Schrödinger equation, the exact solution connot be expressed in this form as a separable product. However, the Hartree-Fock approximation corresponds to finding the best possible solution to the Schrödinger equation
\begin{equation}
H\psi({\bf r}_1,{\bf r}_2) = E\psi({\bf r}_1,{\bf r}_2)\nonumber
\end{equation}
that can nevertheless be expressed in this separable product form, where as before
\begin{equation}
H = -\frac{1}{2}(\nabla^2_1+\nabla^2_2) - \frac{1}{r_1}- \frac{1}{r_2} + \frac{Z^{-1}}{r_{12}}\nonumber
\end{equation}
is the full two-electron Hamiltonian. To find the best solution, substitute into <math>\left\langle\psi \lvert H-E \rvert \psi\right\rangle</math> and require this expression to be stationary
with respect to arbitrary infinitesimal variations <math>\delta u_1</math> and <math>\delta
u_2</math> in <math>u_1</math> and <math>u_2</math> respectively; i.e.

\begin{equation}
\frac{1}{2} \left\langle \delta u_1(r_1)u_2(r_2) \pm u_2(r_1)\delta
u_1(r_2) \lvert H-E \rvert u_1(r_1) u_2(r_2)\pm u_2(r_1) u_1(r_2)\right\rangle\nonumber
\end{equation}

\begin{equation}
=\int\delta u_1(r_1) d{\bf r}_1 \left\{ \int d{\bf r}_2 u_2(r_2) \left( H - E \right) \left[ u_1(r_1)u_2(r_2) \pm u_2(r_1) u_1(r_2) \right] \right\}\nonumber
\end{equation}

\begin{equation}
= 0 \nonumber
\end{equation}

for arbitrary <math> \delta u_1(r_1).\nonumber </math> Therefore <math>\{\int d{\bf r}_2 \ldots \} = 0</math>.

Similarly, the coefficient of <math>\delta u_2</math> would give

\begin{equation}
\int d{\bf r}_1 u_1(r_1) \left( H-E \right) \left[ u_1(r_1) u_2(r_2) \pm
u_2(r_1) u_1(r_2) \right] = 0\nonumber
\end{equation}

Define

\begin{equation}
I_{12} = I_{21} = \int d{\bf r}\, u_1(r)u_2(r), \nonumber
\end{equation}

\begin{equation}
H_{ij} = \int d{\bf r}\, u_i(r)(-\frac{1}{2}\nabla - \frac{1}{r})u_j(r), \nonumber
\end{equation}

\begin{equation}
G_{ij}(r) = \int d{\bf r}^\prime u_i(r^\prime)\frac{1}{|{\bf r} - {\bf r}\prime|}u_j(r^\prime)\nonumber
\end{equation}

Then the above equations become the pair of integro-differential equations

\begin{equation}
[ H_0 - E + H_{22}+G_{22}(r)]u_1(r) = \mp [ I_{12}(H_0-E) + H_{12}+G_{12}(r)]u_2(r)\nonumber
\end{equation}

\begin{equation}
[H_0-E+H_{11}+G_{11}(r)]u_2(r) = \mp [I_{12}(H_0-E) + H_{12}+G_{12}(r)]u_1(r)\nonumber
\end{equation}

These must be solved self-consistently for the "constants" <math>I_{12}</math> and
<math>H_{ij}</math> and the function <math>G_{ij}(r)</math>.

The Hartree Fock energy is <math>E \simeq -2.87\ldots </math>a.u. while the exact energy is <math>E = -2.903724\ldots </math>a.u.

The difference is called the "correlation energy" because it arises from the
way in which the motion of one electron is correlated to the other. The
Hartree Fock equations only describe how one electron moves in the average field
provided by the other (mean-field theory).

==Configuration Interaction==
Expand
\begin{equation}
\psi({\bf r}_1,{\bf r}_2)= C_0u^{(s)}_1(r_1)u^{(s)}_1(r_2) + C_1u^{(P)}_1({\bf r}_1)u^{(P)}_1({\bf r}_2){\cal Y}^0_{1,1,0}(\hat{\bf r}_1, \hat{\bf r}_2)+C_2u^{(d)}_1({\bf r}_1)u^{(d)}_2({\bf r}_2){\cal Y}^0_{2,2,0}(\hat{\bf r}_1, \hat{\bf r}_2)+... \pm {\rm\ exchange}
\end{equation}
where
\begin{equation}
{\cal Y}^M_{l_1,l_2,L}(\hat{\bf r}_1, \hat{\bf r}_2)=\sum_{m_1,m_2}Y^{m_1}_{l_1}({\bf r}_1)Y^{m_2}_{l_2}({\bf r}_2)\times <l_1l_2m_1m_2\mid LM>
\end{equation}

This works, but is slowly convergent, and very laborious. The best CI calculations are accurate to <math> ~10^{-7}</math> a.u.

==Hylleraas Coordinates==
[E.A. Hylleraas, Z. Phys. <math>{\bf 48}, 469(1928)</math> and <math>{\bf 54}, 347(1929)</math>]
suggested using the co-ordinates <math>r_1</math>, <math>r_2</math> and <math>r_{12}</math> or equivalently

\begin{eqnarray}
s &=& r_1 + r_2, \nonumber\\
t &=& r_1-r_2, \nonumber\\
u &=& r_{12}\nonumber
\end{eqnarray}

and writing the trial functions in the form

\begin{equation}
\Psi({\bf r}_1,{\bf r}_2) = \sum^{i+j+k\leq
N}_{i,j,k}c_{i,j,k}r_1^{i+l_1}r_2^{j+l_2}r_{12}^ke^{-\alpha r_1 - \beta r_2}
\mathcal{Y}^M_{l_1,l_2,L}(\hat{r}_1,\hat{r}_2)\pm \text{exchange}\nonumber
\end{equation}

Diagonalizing H in this non-orthogonal basis set is equivalent to solving
<math> \begin{eqnarray}
\frac{\partial E}{\partial c_{i,j,k}} = 0\nonumber
\end{eqnarray} </math>
for fixed <math>\alpha</math> and <math>\beta</math>.

The diagonalization must be repeated for different values of <math>\alpha</math> and
<math>\beta</math> in order to optimize the non-linear parameters.

==Completeness==
The completeness of the above basis set can be shown by first writing
\begin{equation}
r_{12}^2 = r_1^2 + r_2^2 - 2r_1r_2\cos\theta_{12}
\end{equation}
and, from the spherical harmonic addition theorem,
\begin{equation}
\cos(\theta_{12})=\frac{4\pi}{3}\sum^1_{m=-1}Y^{m*}_1(\theta_1,\varphi_1)Y^m_1(\theta_2,\varphi_2)
\end{equation}
Consider first the S-states. The <math>r_{12}^0</math> terms are like the ss terms in a CI calculation. The
<math>r_{12}^2</math> terms bring in p-p type contributions, and the higher powers bring
in d-d, f-f etc. type terms. In general

\begin{equation}
P_l(\cos\theta_{12}) = \frac{4\pi}{2l+1}\sum^l_{m=-l}{Y^{m}_l}^*(\theta_1,\varphi_1)Y^m_l(\theta_2,\varphi_2)\nonumber
\end{equation}

For P-states, one would have similarly

\begin{array}{lr}
r_{12}^0 & (sp)P\nonumber\\
r_{12}^2 & (pd)P\nonumber\\
r_{12}^4 & (df)P\nonumber\\
\vdots & \vdots\nonumber
\end{array}

For D-states

\begin{array}{lr}
r_{12}^0 & (sd)D & (pp^\prime)D\nonumber\\
r_{12}^2 & (pf)D & (dd^\prime)D\nonumber\\
r_{12}^4 & (dg)D & (ff^\prime)D\nonumber\\
\vdots & \vdots & \vdots\nonumber
\end{array}

In this case, since there are two "lowest-order" couplings to form a
D-state, both must be present in the basis set, i.e.

\begin{equation}
\Psi(r_2,r_2) = \sum c_{ijk}r_1^ir_2^{j+2}r_{12}^ke^{-\alpha r_1-\beta
r_2}\mathcal{Y}^M_{022}(\hat{r}_1,\hat{r}_2)
+\sum d_{ijk}r_1^{i+1}r_2^{j+1}r_{12^k}e^{-\alpha^\prime r_1 - \beta^\prime
r_2}\mathcal{Y}^M_{112}(\hat{r}_1,\hat{r}_2)\nonumber
\end{equation}

For F-states, one would need <math>(sf)F</math> and <math>(pd)F</math> terms.

For G-states, one would need <math>(sg)G</math>, <math>(pf)G</math> and <math>(dd^\prime)G</math> terms.

Completeness of the radial functions can be proven by considering the
Sturm-Liouville problem

\begin{equation}
\left(-\frac{1}{2}\nabla^2-\frac{\lambda}{r_s}-E\right)\psi({\bf r}) = 0\nonumber
\end{equation}

or

\begin{equation}
\left(-\frac{1}{2}\frac{1}{r^2}\frac{\partial}{\partial r}\left({r^2}\frac{\partial}{\partial r}\right) -
\frac{l\left(l+1\right)}{2r^2} - \frac{\lambda}{r} - E\right)u(r) = 0.\nonumber
\end{equation}

For fixed E and variable <math>\lambda</math> (nuclear charge).

The eigenvalues are <math>\lambda_n = (E/E_n)^{1/2}</math>, where <math>E_n =- \frac{1}{2n^2}</math>

[[File:Pic2.jpg|500px]]

and the eigenfunctions are

\begin{equation}
u_{nl}(r) =
\frac{1}{(2l+1)!}\left(\frac{(n+l)!}{(n-l-1)2!}\right)^{1/2}(2\alpha)^{3/2}e^{-\alpha
r}\nonumber
\times \ \left( 2 \alpha r \right)^l\,_{1}F_{1}\left(-n+l+1,2l+2; 2 \alpha r \right)
\end{equation}

with <math>\alpha = (-2E)^{1/2}</math> and <math>n\geq l+1</math>. The confluent hypergeometric function <math> _1F_1(a,b,;z)</math> then denotes a finite polynomial since <math>a=-n+l+1</math> is a negative integer or zero.

Unlike the hydrogen spectrum, which has both a discrete part for <math>E<0</math> and a
continuous part for <math>E>0</math>, this forms an entirely discrete set of finite
polynomials, called Sturmian functions. They are orthogonal with respect to
the potential, i.e.

\begin{equation}
\int^\infty_0 r^2dr \left( u_{n^\prime l}(r)\frac{1}{r}u_{nl}(r) \right) = \delta_{n,n^\prime}\nonumber
\end{equation}

Since they become complete in the limit <math>n\rightarrow\infty</math>, this assures the
completeness of the variational basis set.

[See also B Klahn and W.A. Bingel Theo. Chim. Acta (Berlin) 44, 9 and 27
(1977)].

==Solutions of the Eigenvalue Problem==

For convenience, write

\[
\Psi({\bf r}_1,{\bf r}_2) = \sum^N_{m=1}c_m\varphi_m\nonumber
\]

where <math>m</math> represents the <math>m</math>'th combination of <math>i,j,k</math> and

\begin{equation}
\varphi_{ijk}=r_1^ir_2^jr_{12}^ke^{-\alpha r_1 - \beta
r_2}\mathcal{Y}^M_{l_1,l_2,L}(\hat{r}_1,\hat{r}_2) \pm \mbox{exchange}\nonumber
\end{equation}

\begin{align}

&
\left( \begin{array}{lr}
\cos(\theta) & \sin(\theta)\\
-\sin(\theta) & \cos(\theta)
\end{array} \right)
\left( \begin{array}{lr}
H_{11} & H_{12} \\
H_{12} & H_{22}
\end{array}\right)
\left( \begin{array}{lr}
\cos(\theta) & -\sin(\theta)\\
\sin(\theta) & \cos(\theta)
\end{array} \right)
\\[5pt]
= & \left(\begin{array}{lr}
cH_{11}+sH_{12} & cH_{12} + sH_{22}\\
-sH_{11} + cH_{12} & -sH_{12} + cH_{22}
\end{array}\right)
\left( \begin{array}{cc}
c & -s \\
s & c
\end{array}\right)
\\[5pt]
= & \left(\begin{array}{lr}
c^2H_{11}+s^2H_{22} + 2csH_{12} & (c^2-s^2)H_{12}+cs(H_{22}-H_{11})\\
(c^2-s^2)H_{12}+cs(H_{22}-H_{11}) & s^2H_{11}+c^2H_{22}-2csH_{12}
\end{array}\right)
\end{align}

Therefore

\begin{equation}
\left[\cos^2(\theta)-\sin^2(\theta)\right]H_{12} =
\cos(\theta)\sin(\theta)(H_{11}-H_{22})\nonumber
\end{equation}

and

\begin{equation}
\tan(2\theta) = \frac{2H_{12}}{H_{11}-H_{22}}\nonumber
\end{equation}

i.e.

\begin{equation}
\cos(\theta)=\left(\frac{r+\omega}{2r}\right)^{1/2}\nonumber
\end{equation}

\begin{equation}
\sin(\theta)=-\mbox{sgn}(H_{12})\left(\frac{r-\omega}{2r}\right)^{1/2}\nonumber
\end{equation}

where

\begin{equation}
\omega = H_{22}-H_{11}\nonumber
\end{equation}

\begin{equation}
r=\left(\omega^2+4H_{12}^2\right)^{1/2}\nonumber
\end{equation}

\begin{equation}
E_1=\frac{1}{2}\left(H_{11}+H_{22}-r\right)\nonumber
\end{equation}

\begin{equation}
E_2=\frac{1}{2}\left(H_{11}+H_{22}+r\right)\nonumber
\end{equation}

===Brute Force Method===

- Gives all the eigenvalues and eigenvectors, but it is slow

- First orthonormalize the <math>N</math>-dimensional basis set; i.e. form linear combinations

\begin{equation}
\Phi_m = \sum_{n=1}^N\varphi_nR_{nm}\nonumber
\end{equation}

such that
\begin{equation}
\langle \Phi_m | \Phi_n \rangle = \delta_{m,n}
\end{equation}

This can be done by finding an orthogonal tranformation T such that

\begin{equation}
T^TOT = I = \left(
\begin{array}{ccccc}
I_1 & 0 & 0 & \ldots & 0 \\
0 & I_2 & 0 & \ldots & 0 \\
0 & 0 & I_3 & \ldots & 0 \\
\vdots & \vdots & \vdots & \ddots & \vdots \\
0 & 0 & 0 & \ldots & I_N
\end{array}\right);
\end{equation}

\begin{equation}
O_{mn} = \langle \varphi_m | \varphi_n \rangle \nonumber
\end{equation}

and then applying a scale change matrix

\begin{equation}
S = \left(\begin{array}{ccccc}
\frac{1}{I_1^{1/2}} & 0 & 0 & \ldots & 0\\
0 & \frac{1}{I_2^{1/2}} & 0 & \ldots & 0 \\
0 & 0 & \frac{1}{I_3^{1/2}} & \ldots & 0 \\
\vdots & \vdots & \vdots & \ddots & \vdots \\
0 & 0 & 0 & \ldots & I_N^{1/2}
\end{array}\right)= S^T\nonumber
\end{equation}

Then

\begin{equation}
S^TT^TOTS = 1
\end{equation}

i.e.

\begin{equation}
R^TOR = 1
\end{equation}

with <math>R=TS</math>.

If H is the matrix with elements <math>H_{mn} = \langle \varphi_m | \varphi_n \rangle </math>, then H
expressed in the <math>\Phi_m</math> basis set is

\begin{equation}
H^\prime = R^THR.\nonumber
\end{equation}

We next diagonalize <math>H^\prime</math> by finding an orthogonal transformation W such
that

\begin{equation}
W^T H^\prime W = \lambda = \left(
\begin{array}{cccc}
\lambda_1 & 0 & \ldots & 0 \\
0 & \lambda_2 & \ldots & 0 \\
\vdots & \vdots & \ddots & \vdots \\
0 & 0 & \ldots & \lambda_N
\end{array}\right)\nonumber
\end{equation}

The q'th eigenvector is

\begin{equation}
\Psi^{(q)} = \sum_{n=1}^N\Phi_n W_{n,q} = \sum_{n,n^\prime} \varphi_{n^\prime} R_{n^\prime ,n} W_{n,q} \nonumber
\end{equation}

i.e.

\begin{equation}
c_{n^\prime}^{(q)} = \sum_{n=1}^N R_{n^\prime n} W_{n,q}
\end{equation}

===The Power Method===

- Based on the observation that if H has one eigenvalue, <math>\lambda_M</math>, much bigger than all the rest, and <math>\chi = \left(
\begin{array}{c}a_1\\a_2\\\vdots\end{array}\right)</math>
is an arbitrary starting vector, then <math>\chi = \sum_{q=1}^N x_q\Psi^{(q)}</math>.

\begin{equation}
(H)^n\chi = \sum_{q=1}^N x_q \lambda^n_q\Psi^{(q)}\nonumber\rightarrow x_M\lambda_M^n\Psi^{(M)}\nonumber
\end{equation}

provided <math>x_M\neq 0</math>.

To pick out the eigenvector corresponding to any eigenvalue, write the original problem in the form

\begin{align}
H\Psi & = \lambda O \Psi \nonumber \\
\left( H - \lambda_q O \right) \Psi & = \left( \lambda - \lambda_q \right) O \Psi \nonumber
\end{align}

Therefore,

\begin{equation}
G \Psi = \frac{1}{\lambda - \lambda_q } \Psi \nonumber \\
\end{equation}

where <math>G=(H-\lambda_qO)^{-1}O</math> with eigenvalues <math>\frac{1}{\lambda_n - \lambda_q}</math>.

By picking <math>\lambda_q</math> close to any one of the <math>\lambda_n</math>, say
<math>\lambda_{n^\prime}</math>, then <math>\frac{1}{\lambda_n-\lambda_q}</math> is much larger for
<math>n=n^\prime</math> than for any other value. The sequence is then

\begin{array}{c}
\chi_1=G\chi\nonumber\\
\chi_2=G\chi_1\nonumber\\
\chi_3=G\chi_2\nonumber\\
\vdots\nonumber
\end{array}

until the ratios of components in <math>\chi_n</math> stop changing.

- To avoid matrix inversion and multiplication, note that the sequence is equivalent to
\begin{equation}
F \chi_n = \left( \lambda - \lambda_q \right) O \chi_{n-1} \nonumber \\
\end{equation}
where <math>F = H-\lambda_qO</math>. The factor of <math>\left( \lambda - \lambda_q \right)</math> can be
dropped because this only affects the normalization of <math>\chi_n</math>. To find <math>\chi_n</math>, solve
\begin{equation}
F \chi_n = O \chi_{n-1} \nonumber
\end{equation}
(N equations in N unknowns). Then
\begin{equation}
\lambda = \frac{ \langle \chi_n | H | \chi_n \rangle}{\langle \chi_n | \chi_n \rangle} \nonumber
\end{equation}

==Matrix Elements of H==

\begin{equation}
H=-\frac{1}{2}\nabla^2_1 -\frac{1}{2}\nabla^2_2 - \frac{1}{r_1} - \frac{1}{r_2} +\frac{Z^{-1}}{r_{12}}\nonumber
\end{equation}

Taking <math>r_1, r_2</math> and <math>r_{12}</math> as independent variables,

\begin{equation}
\nabla_1^2 = \frac{1}{r_1^2}\frac{\partial}{\partial r_1}\left(r_1^2\frac{\partial}{\partial r_1}\right)+ \frac{1}{r_{12}^2}\frac{\partial}{\partial r_{12}}
\left(r_{12}^2\frac{\partial}{\partial r_{12}}\right)\nonumber -\frac{l_1(l_1+1)}{r_1^2}+2(r_1 - r_2\cos(\theta_{12}))\frac{1}{r_{12}}\frac{\partial^2}{\partial r_1 \partial r_{12}} \nonumber - 2(\nabla_1^Y \cdot {\bf r}_2)\frac{1}{r_{12}}\frac{\partial}{\partial r_{12}}\nonumber
\end{equation}

where <math>\nabla_1^Y</math> acts only on the spherical harmonic part of the wave function and the diagram

[[File:Pic3.jpg|400px]]

defines the complete set of 6 independent variables is <math>r_1</math>, <math>r_2</math>, <math>r_{12}</math>, <math>\theta_1</math>, <math>\varphi_1</math>, <math>\chi</math>.

If <math>r_{12}</math> were not an independent variable, then one could take the volume element to be

\begin{equation}
d\tau = r_1^2dr_1\sin(\theta_1)d\theta_1d\varphi_1r_2^2dr_2\sin(\theta_2)d\theta_2d\varphi_2.\nonumber
\end{equation}

However, <math>\theta_2</math> and <math>\varphi_2</math> are no longer independent variables. To eliminate them, take the point <math>{\bf r}_1</math> as the
origin of a new polar co-ordinate system, and write

\begin{equation}
d\tau=-r_1^2dr_1\sin(\theta_1)d\theta_1d\varphi_1r_{12}^2dr_{12}\sin(\psi)d\psi d\chi\nonumber
\end{equation}

and use

\begin{equation}
r_2^2=r_1^2+r_{12}^2 +2r_1r_{12}\cos(\psi)
\end{equation}

Then for fixed <math>r_1</math> and <math>r_{12}</math>,

\begin{equation}
2r_2dr_2 = -2r_1r_{12}\sin(\psi)d\psi
\end{equation}

Thus

\begin{equation}
d\tau= r_1dr_1r_2dr_2r_{12}dr_{12}\sin(\theta_1)d\theta_1d\varphi_1d\chi
\end{equation}

The basic type of integral to be calculated is

\begin{align}
I \left( l_1, m_1, l_2, m_2; R \right) & = \int \sin \left( \theta_1 \right) d\theta_1 d\varphi_1 d\chi Y^{m_1}_{l_1} \left( \theta_1, \varphi_1 \right)^{*} Y^{m_2}_{l_2} \left( \theta_2, \varphi_2 \right) \\
& \times \int r_1dr_1r_2dr_2r_{12}dr_{12}R \left( r_1, r_2, r_{12} \right)
\end{align}

Consider first the angular integral. <math>Y^{m_2}_{l_2} \left( \theta_2, \varphi_2 \right)</math> can be expressed in terms of the independent variables <math>\theta_1, \varphi_1, \chi</math> by use of the rotation matrix relation
\begin{equation}
Y^{m_2}_{l_2}( \theta_2, \varphi_2) = \sum_m \mathcal{D}^{(l_2)}_{m_2,m}(\varphi_1,\theta_1,\chi)^*Y^m_{l_2}( \theta_{12}, \varphi)
\end{equation}
where <math>\theta, \varphi</math> are the polar angles of <math>{\bf r}_2</math> relative to <math>{\bf r}_1</math>. The angular integral is then

\begin{align}
I_{ang} & = \int^{2 \pi}_{0} d\chi \int^{2\pi}_{0} d\varphi_1 \int^{\pi}_{0} \sin \left( \theta_1 \right) d\theta_{1} Y^{m_1}_{l_1} \left( \theta_1, \varphi_1 \right)^{*} \\
& \times \sum_m \mathcal{D}^{(l_2)}_{m_2,m} \left( \varphi_1,\theta_1,\chi \right)^{*} Y^{m}_{l_2} \left( \theta_{12},\varphi \right) \\
\end{align}

Use

\begin{equation}
Y^{m_1}_{l_1} \left( \theta_1, \varphi_1 \right)^{*} = \sqrt{\frac{2l_1+1}{4\pi}}\mathcal{D}^{(l_1)}_{m_1,0} \left( \varphi_1, \theta_1, \chi \right)
\end{equation}

together with the orthogonality property of the rotation matrices (Brink and Satchler, p 147)

\begin{align}
& \int \mathcal{D}^{(j)*}_{m,m^\prime}\mathcal{D}^{(J)}_{M,M^\prime}\sin(\theta_1)d\theta_1d\varphi_1d\chi \\
& = \frac{8\pi^2}{2j+1}\delta_{jJ}\delta_{mM}\delta_{m^\prime M^\prime}\nonumber
\end{align}

to obtain

\begin{align}
I_{ang} & = \sqrt{ \frac{2l_1+1}{4\pi}} \frac{8\pi^2}{2l_1+1} \delta_{l_1,l_2} \delta_{m_1,m_2} Y^{0}_{l_2} \left( \theta_{12}, \varphi \right) \\
& = 2 \pi \delta_{l_1,l_2} \delta_{m_1,m_2} P_{l_2} \left( \cos \theta_{12} \right)
\end{align}

since

\begin{equation}
Y^0_{l_2} \left( \theta_{12}, \varphi \right) = \sqrt{\frac{2l_1+1}{4\pi}} P_{l_2} \left( \cos \left( \theta_{12} \right) \right)
\end{equation}

Note that <math>P_{l_2} \left( \cos \theta \right)</math> is just a short hand expression for a radial function because

\begin{equation}
\cos \theta_{12} = \frac{r_1^2 + r_2^2 - r_{12}^2}{2r_1r_2}
\end{equation}

The original integral is thus

\begin{equation}
I \left( l_1, m_1, l_2, m_2; R \right) = 2 \pi \delta_{l_1,l_2} \delta_{m_1,m_2} \int^{\infty}_{0} r_{1} dr_{1} \int^{\infty}_{0} r_{2} dr_{2} \int^{r_{1} + r_{2}}_{|r_{1} - r_{2}|} r_{12} dr_{12} R \left( r_{1}, r_{2}, r_{12} \right) P_{l_2} \left( \cos \theta_{12} \right)
\end{equation}

where again

\begin{equation}
\cos \theta_{12} = \frac{r_{1}^{2} +r^{2}_{2} - r_{12}^{2}}{2 r_{1} r_{2}}
\end{equation}

is a purely radial function.

The above would become quite complicated for large <math>l_2</math> because <math>P_{l_2}(\cos \theta_{12})</math> contains terms up to <math>\left( \cos \theta_{12} \right)^{l_2}</math>. However, recursion relations exist which allow any integral containing <math>P_l \left( \cos \theta_{12} \right)</math> in terms of those containing just <math>P_0 \left( \cos \theta_{12} \right) = 1</math> and <math>P_1 \left( \cos \theta_{12} \right) = \cos \theta_{12}</math>.

==Radial Integrals and Recursion Relations==

The basic radial integral is [see G.W.F. Drake, Phys. Rev. A 18, 820 (1978)]

\begin{align}
I_0 \left( a, b, c \right) & =
\int^{\infty}_{0} r_{1} dr_{1} \int^{\infty}_{r_{1}} r_{2} dr_{2} \int^{r_{1}+r_{2}}_{r_{2} - r_{1}} r_{12} dr_{12} r_{1}^{a} r_{2}^{b} r_{12}^{c} e^{-\alpha r_{1} - \beta r_{2}} + \int^{\infty}_{0} r_{2} dr_{2} \int^{\infty}_{r_{2}} r_{1} dr_{1} \int^{r_{1} + r_{2}}_{r_{1} - r_{2}} r_{12} dr_{12} r_{1}^{a} r_{2}^{b} r_{12}^{c} e^{- \alpha r_{1} - \beta r_{2}} \\
& = \frac{2}{c+2} \sum^{[(c+1)/2]}_{i=0}
\left(\begin{array}{c}
c+2 \\
2i+1
\end{array} \right)
\left\{ \frac{q!}{\beta^{q+1} \left( \alpha + \beta \right)^{p+1}} \sum_{j=0}^{q} \frac{\left( p + j \right)!}{j!} \left( \frac{\beta}{\alpha + \beta} \right)^j + \frac{q^{\prime}!}{\alpha^{q^{\prime}+1} \left( \alpha + \beta \right)^{p^{\prime}+1}} \sum^{q^{\prime}}_{j=0} \frac{\left( p^{\prime} + j \right)!}{j!} \left( \frac{\alpha}{\alpha + \beta} \right)^j \right\}
\end{align}

where

\begin{array}{ll}
p = a + 2i + 2 & p^{\prime} = b + 2i + 2 \\
q = b + c - 2i + 2 & q^{\prime} = a + c - 2i + 2
\end{array}

The above applies for <math> a, b \geq -2, c\geq -1</math>.
<math>[x]</math> means "greatest integer in" <math>x</math>.

Then

\begin{align}
I_1 \left( a, b, c \right) & = \int d\tau_{r} r_{1}^{a} r_{2}^{b} r_{12}^{c} e^{ -\alpha r_{1} - \beta r_{2}} P_{1}\left( \cos \theta \right) \\
& = \frac{1}{2} \left( I_{0} \left( a + 1, b - 1, c \right) + I_{0} \left( a - 1, b + 1, c \right) - I_{0} \left( a - 1, b - 1, c + 2 \right) \right)
\end{align}

===The Radial Recursion Relation===

Recall that the full integral is

\begin{equation}
I(l_1m_1,l_2m_2;R)=2\pi\delta_{l_1,l_2}\delta_{m_1,m_2}I_{l_2}(R)
\end{equation}

where

\begin{equation}
I_{l_2}(R) = \int d\tau_r R(r_1,r_2,r_{12})P_{l_2}(\cos \theta_{12})
\end{equation}

To obtain the recursion relation, use

\begin{equation}
P_l(x) = \frac{[P^{\prime}_{l+1}(x)-P^{\prime}_{l-1}(x)]}{2l+1}
\end{equation}

with

\begin{equation}
P_{l+1}^\prime (x) = \frac{d}{dx}P_{l+1}(x)
\end{equation}

Here <math>x=\cos\theta_{12}</math> and

\begin{align}
\frac{d}{d \cos \theta_{12}} & = \frac{dr_{12}}{d \cos \theta_{12}} \frac{d}{dr_{12}} \\
& = -\frac{r_{1} r_{2}}{r_{12}} \frac{d}{dr_{12}}
\end{align}

Then,

\begin{equation}
I_{l}\left( R \right) = - \int d\tau_r R \frac{r_{1} r_{2}}{r_{12}} \frac{d}{dr_{12}} \frac{ \left[ P_{l+1} \left( \cos \theta_{12} \right) - P_{l-1} \left( \cos \theta_{12} \right) \right]}{2l+1}
\end{equation}

The <math>r_{12}</math> part of the integral is

\begin{align}
& \int^{r_{1} + r_{2}}_{|r_{1} - r_{2}|} r_{12} dr_{12} R \frac{r_{1} r_{2}}{r_{12}} \frac{d}{dr_{12}} \left[P_{l+1}-P_{l-1} \right] \\
& = R r_{1} r_{2} \left[ P_{l+1} - P_{l-1} \right]^{r_{1} + r_{2}}_{|r_{1} - r_{2}|} - \int^{r_{1} + r_{2}}_{|r_{1} - r_{2}|} r_{12} dr_{12} \left( \frac{d}{dr_{12}} R \right) \frac{r_{1} r_{2}}{r_{12}} \frac{ \left[ P_{l+1} - P_{l-1} \right]}{2l+1}
\end{align}

The integrated term vanishes because

\begin{equation}
\cos \theta_{12} = \frac{r_{1}^{2} + r_{2}^{2} - r_{12}^{2}}{2 r_{1} r_{2}}
\begin{array}{lll}
& = -1 & {\rm when} & r_{12}^{2} = \left( r_{1} + r_{2} \right)^{2} \\
& = 1 & {\rm when} & r_{12}^{2} = \left( r_{1} - r_{2} \right)^{2}
\end{array}
\end{equation}

and <math>P_l(1)=1</math>, <math>P_{l}(-1)=(-1)^{l}</math>.

Thus,

\begin{equation}
I_{l+1}\left(\frac{r_1r_2}{r_{12}}R^\prime\right) = \left( 2l+1 \right) I_{l}(R) - I_{l-1} \left( \frac{r_{1} r_{2}}{r_{12}} R^{\prime} \right)
\end{equation}

If

\begin{equation}
R = r_{1}^{a-1} r_{2}^{b-1} r_{12}^{c+2} e^{-\alpha r_{1} - \beta r_{2}}
\end{equation}

then [G.W.F. Drake, Phys. Rev. A 18, 820 (1978)]

\begin{equation}
I_{l+1}(r_1^ar_2^br_{12}^c) =
\frac{2l+1}{c+2} I_{l} \left( r_{1}^{a-1} r_{2}^{b-2} r_{12}^{c+2} \right)+I_{l-1} \left( r_{1}^{a} r_{2}^{b} r_{12}^{c} \right)
\end{equation}

For the case <math>c=-2</math>, take

\begin{equation}
R = r_{1}^{a-1} r_{2}^{b-1} \ln r_{12} e^{ -\alpha r_{1} -\beta r_{2}}
\end{equation}

Then

\begin{equation}
I_{l+1} \left( r_{1}^{a} r_{2}^{b} r_{12}^{c} \right) = I_{l} \left( r_{1}^{a-1} r_{2}^{b-1} \ln r_{12} \right) \left( 2l + 1 \right) - I_{l-1} \left( r_{1}^{a} r_{2}^{b} r_{12}^{-2} \right)
\end{equation}

This allows all <math>I_{l}</math> integrals to be calculated from tables of <math>I_0</math> and
<math>I_1</math> integrals.

===The General Integral===

The above results for the angular and radial integrals can now be combined into a general formula for integrals of the type

\begin{equation}
I = \int \int d{\bf r}_{1} d{\bf r}_{2} R_{1} \mathcal{Y}^{M^{\prime}}_{l^{\prime}_{1} l^{\prime}_{2} L^{\prime}} \left( \hat{r}_{1}, \hat{r}_{2} \right) T_{k_{1} k_{2} K}^{Q} \left( {\bf r}_{1},{\bf r}_{2} \right) R_{2} \mathcal{Y}_{l_{1} l_{2} L}^{M} \left( \hat{r}_{1}, \hat{r}_{2} \right)
\end{equation}

where

\begin{equation}
\mathcal{Y}^{M}_{l_{1} l_{2} L} \left( \hat{r}_{1}, \hat{r}_{2} \right) = \sum_{m_1,m_2} \langle l_{1} l_{2} m_{1} m_{2} | LM \rangle Y^{m_{1}}_{l_{1}} \left( \hat{r}_{1} \right) Y^{m_{2}}_{l_{2}} \left( \hat{r}_{2} \right)
\end{equation}

and

\begin{equation}
T^Q_{k_{1} k_{2} K} \left( {\bf r}_{1}, {\bf r}_{2} \right) = \sum_{q_{1}, q_{2}} \langle k_{1} k_{2} q_{1} q_{2} | KQ \rangle Y_{k_{1}}^{q_{1}} \left( \hat{r}_{1} \right) Y^{q_{2}}_{k_{2}} \left( \hat{r}_{2} \right)
\end{equation}

The basic idea is to make repeated use of the formula

\begin{equation}
Y^{m_{1}}_{l_{1}} \left( \hat{r}_{1} \right) Y^{m_{2}}_{l_{2}} \left( \hat{r}_{1} \right) = \sum_{l m} \left( \frac{\left( 2l_{1}+1 \right) \left(2l_{2}+1 \right) \left(2l+1 \right)}{4 \pi} \right)^{1/2}
\times \left( \begin{array}{ccc}
l_1 & l_2 & l \\
m_1 & m_2 & m
\end{array} \right) \left(\begin{array}{ccc}
l_1 & l_2 & l \\
0 & 0 & 0
\end{array} \right)
Y^{m*}_{l} \left( \hat{r}_{1} \right)
\end{equation}

where

\begin{equation}
Y^{m*}(\hat{r}) = (-1)^mY_l^{-m}(\hat{r})
\end{equation}

and

\begin{equation}
\left(\begin{array}{ccc}
l_1 & l_2 & l \\
m_1 & m_2 & m
\end{array} \right) = \frac{\left( -1 \right)^{l_{1} - l_{2} - m}}{\left( 2l + 1 \right)^{1/2}} \left( l_{1} l_{2} m_{1} m_{2} | l, -m \right)
\end{equation}

is a 3-<math>j</math> symbol. In particular, write

\begin{array}{cc}
Y^{m_1^\prime}_{l_1^\prime}(\hat{r}_1)^{*} \underbrace{Y_{k_{1}}^{q_{1}} \left( \hat{r}_{1} \right) Y^{m_{1}}_{l_{1}} \left( \hat{r}_{1} \right)} & = & \sum_{\Lambda M}(\ldots)Y^{M}_\Lambda(\hat{r}_1)\\
\ \ \ \ \ \ \ \sum_{\lambda_1\mu_1}Y_{\lambda_1}^{\mu_1}(\hat{r}_1) & & \\
& & \\
Y^{m_{2}^{\prime}}_{l_{2}^{\prime}} \left( \hat{r}_{2} \right) \underbrace{Y^{q_{2}}_{k_{2}} \left( \hat{r}_{2} \right) Y_{l_{2}}^{m_{2}} \left( \hat{r}_{2} \right)} & = & \sum_{\Lambda^{\prime} M^{\prime}} \left( \ldots \right)Y_{\Lambda^{\prime}}^{M^{\prime}*} \left( \hat{r}_{2} \right) \\
\ \ \ \ \ \ \ \ \ \sum_{\lambda_{2} \mu_{2}}Y_{\lambda_{2}}^{\mu_{2}} \left( \hat{r}_{2} \right) & &
\end{array}

The angular integral then gives a factor of <math>2 \pi \delta_{\Lambda, \Lambda^{\prime}} \delta_{M, M^{\prime}} P_{\Lambda} \left( \cos \theta_{12} \right)</math>. The total integral therefore reduces to the form

\begin{equation}
I = \sum_{\Lambda} C_{\Lambda} I_{\Lambda} \left( R_{1} R_{2} \right)
\end{equation}

where <math>C_{\Lambda} = \sum_{\lambda_{1}, \lambda_{2}} C_{ \lambda_{1}, \lambda_{2}, \Lambda}</math>. For further details and derivations, including graphical representations, see G.W.F. Drake, Phys. Rev. A <math>{\bf 18}</math>, 820 (1978).

==Graphical Representation==

[[File:GraphRep1.jpg|500px]]

==Matrix Elements of H==
Recall that

\begin{equation}
H = -\frac{1}{2} \nabla^{2}_{1} - \frac{1}{2} \nabla^{2}_{2} - \frac{1}{r_1} - \frac{1}{r_2} + \frac{Z^{-1}}{r_{12}}
\end{equation}

Consider matrix elements of

\begin{equation}
\nabla^2_1 = \frac{1}{r_1^2} \frac{\partial}{\partial r_1}\left(r_1^2\frac{\partial}{\partial r_1} \right) + \frac{1}{r^2}\frac{\partial}{\partial r}\left(r^2\frac{\partial}{\partial r}\right) - \frac{(\vec{l}_1^Y)^2}{r_1^2} + \frac{2 \left( r_1 - r_2 \cos \theta \right)}{r} \frac{\partial^2}{\partial r_1 \partial r} - 2 \left( \nabla^Y_1 \cdot {\bf r}_2 \right) \frac{1}{r} \frac{\partial}{\partial r}
\end{equation}

where <math>r \equiv r_{12}</math> and <math>\cos\theta \equiv \cos\theta_{12}</math>. Also in what follows, define <math>\hat{\nabla}^Y_1 \equiv r_1\nabla^Y_1</math> where <math>\nabla^Y</math> operates only on the spherical harmonic part of the wave function, and similarly for <math>\vec{l}_1^Y</math>.

A general matrix element is

\begin{equation}
\langle r_1^{a^\prime} r_2^{b^\prime} r_{12}^{c^\prime} e^{-\alpha^\prime r_1 - \beta^\prime r_2} \mathcal{Y}^{M^\prime}_{l^\prime_1 l^\prime_2 L^\prime} \left( \hat{r}_1, \hat{r}_2 \right) \left| \nabla^2_1 \right| r_1^a r_2^b r_{12}^c e^{-\alpha r_1 - \beta r_2} \mathcal{Y}^M_{l_1 l_2 L} \left( \hat{r}_1, \hat{r}_2 \right) \rangle
\end{equation}

Since <math>\nabla^2_1</math> is rotationally invariant, this vanishes unless <math>L=L^\prime</math>, <math>M=M^\prime</math>. Also <math>\nabla_1^2</math> is Hermitian, so that the result must be the same whether it operates to the right or left, even though the results look very different. In fact, requiring the results to be the same yields some interesting and useful integral identities as follows:

Put

\begin{equation}
\mathcal{F} = F \mathcal{Y}^M_{l_1 l_2 L} \left( \hat{r}_1, \hat{r_2} \right)
\end{equation}

and

\begin{equation}
F = r_1^a r_2^b r_{12}^c e^{-\alpha r_1 - \beta r_2}
\end{equation}

Then

\begin{equation}
\nabla^2_1 \mathcal{F} = \left\{ \frac{1}{r_1^2} \left[ a \left( a + 1 \right) - l_1 \left( l_1 + 1 \right) \right] + \frac{c \left( c + 1 \right)}{r^2} + \alpha^2\frac{2\alpha \left( a + 1 \right)}{r_1} + \frac{2 \left(r_1 - r_2 \cos \theta \right) }{r_1r^2}c \left[ a - \alpha r_1 \right] - \frac{2c}{r^2} \left( \hat{\nabla}^Y_1 \cdot \hat{r}_2 \right) \frac{r_2}{r_1} \right\} \mathcal{F}
\end{equation}

and

\begin{align}
\langle \mathcal{F}^\prime \left| \nabla^2_1 \right| \mathcal{F} \rangle & = \sum_\Lambda \int d\tau_r F^\prime C_\Lambda \left( 1 \right) P_\Lambda \left( \cos \theta \right) \times \left\{ \frac{1}{r_1^2} \left[ a \left( a + 1 \right) - l_1 \left( l_1 + 1 \right) \right] -\frac{2 \alpha \left( a + 1 \right)}{r_1} + \frac{c \left( c + 1 \right)}{r^2} +\alpha^2 + \frac{2 \left( r_2 - r_2 \cos \theta \right)}{r_1r^2}c \left( a - \alpha r_1 \right) \right\} F \\
& + \sum_\Lambda \int d\tau_r F^\prime C_\Lambda \left( \hat{\nabla}^Y_1 \cdot \hat{r}_2 \right) P_\Lambda \left( \cos \theta \right) \left( \frac{-2c r_2}{r_1 r^2} \right) F
\end{align}

where

\begin{equation}
\int d\tau_r = \int^\infty_0 r_1 dr_1 \int^\infty_0 r_2 dr_2 \int^{r_1+r_2}_{\left| r_1 - r_2 \right|} r dr
\end{equation}

For brevity, let the sum over <math>\Lambda</math> and the radial integrations be understood, and let <math>\left( \nabla_1^2 \right)</math> stand for the terms which appear in the integrand. Then operating to the right gives

\begin{equation}
(\nabla_1^2)_R = \frac{1}{r_1^2} \left[ a \left( a + 1 \right) - l_1 \left( l_1 + 1 \right) \right] + \frac{c \left( c + 1 \right)}{r^2} + \alpha^2 - \frac{2 \alpha \left( a + 1 \right)}{r_1}
+ \frac{2 \left( r_1 - r_2 \cos \theta \right)}{r_1 r^2}c \left( a - \alpha r_1 \right) - \frac{2c r_2}{r_1 r^2} \left( \hat{\nabla}^Y_1 \cdot \hat{r}_2 \right)
\end{equation}

Operating to the left gives

\begin{equation}
(\nabla_1^2)_L = \frac{1}{r_1^2} \left[a^\prime \left( a^\prime + 1 \right) - l_1^\prime \left( l_1^\prime + 1 \right) \right] + \frac{c^\prime \left( c^\prime + 1 \right)}{r^2} + \alpha^{\prime 2} -\frac{2 \alpha^\prime \left( a^\prime + 1 \right)}{r_1}
+ \frac{2 \left( r_1 - r_2 \cos \theta \right)}{r_1 r^2} c^\prime \left( a^\prime - \alpha^\prime r_1 \right) - \frac{2 c^\prime r_2}{r_1 r^2} \left( \hat{\nabla}^{Y^\prime}_1 \cdot \hat{r}_2 \right)
\end{equation}

Now put

\begin{array}{cc}
a_+ = a+a^\prime, & \hat{\nabla}_1^+ = \hat{\nabla}_1^Y + \hat{\nabla}_1^{Y\prime} \\
a_{-} = a-a^\prime, & \hat{\nabla}_1^{-} = \hat{\nabla}_1^Y - \hat{\nabla}_1^{Y\prime}
\end{array}

etc., and substitute <math>a^\prime = a_+ - a</math>, <math>c^\prime = c_+ - c</math>, and <math>\alpha^\prime = \alpha_+ - \alpha</math> in <math>(\nabla^2_1)_L</math>. If <math>a_+</math>, <math>c_+</math> and <math>\alpha_+</math> are held fixed, then the equation

\begin{equation}
(\nabla^2_1)_R = (\nabla^2_1)_L
\end{equation}

must be true for arbitrary <math>a</math>, <math>c</math>, and <math>\alpha</math>. Their coefficients must thus vanish.

This yields the integral relations

\begin{equation}
\label{I}
\frac{\left( r_1 - r_2 \cos \theta \right)}{r_1 r^2} = \frac{1}{c_+} \left( \frac{- \left( a_+ + 1 \right)}{r_1^2} + \frac{\alpha_+}{r_1} \right) \tag{I}
\end{equation}

from the coefficient of <math>a</math> and

\begin{equation}
\label{II}
\frac{ \left( r_1 - r_2 \cos \theta \right) \left( a_+ - \alpha_+ r_1 \right)}{r_1 r^2} = \frac{r_2}{r_1 r^2}\left( \hat{r}_2 \cdot \hat{\nabla}_1^+ \right) - \frac{ \left( c_+ + 1 \right)}{r^2} \tag{II}
\end{equation}

from the coefficient of <math>c</math>. The coefficient of <math>\alpha</math> gives an equation equivalent to (I).

Furthermore, if can be show that (see problem)

\begin{align}
& \sum_\Lambda \int d\tau_r \frac{r^c}{r^2} C_\Lambda \left( \hat{r}_2 \cdot \hat{\nabla}_1^Y \right) P_\Lambda \left( \cos \theta \right) \\
= & \sum_\Lambda \int d\tau_r \frac{r^c}{c r_1 r_2}C_\Lambda \left( 1 \right)P_\Lambda \left( \cos \theta \right) \times \left( \frac{l_1^\prime \left( l_1^\prime + 1 \right) - l_1 \left( l_1 + 1 \right) -\Lambda \left( \Lambda + 1 \right)}{2} \right)
\end{align}

and similarly for <math>\left( \hat{r}_2 \cdot \hat{\nabla}^{Y\prime}_1 \right)</math> with <math>l_1</math> and <math>l_1^\prime</math> interchangeable, then it follows that

\begin{equation}
\frac{r^{c_+}}{r^2} \left( \hat{r}_2 \cdot \hat{\nabla}_1^+ \right) = - \frac{r^{c_+}}{c_+ r_1 r_2} \Lambda \left( \Lambda + 1 \right)
\end{equation}

and

\begin{equation}
\frac{r^{c_+}}{r^2} \left( \hat{r}_2 \cdot \hat{\nabla}_1^- \right) = \frac{r^{c_+}}{c_+ r_1 r_2} \left[ l_1^\prime \left( l_1^\prime + 1 \right) - l_1 \left( l_1 + 1 \right) \right]
\end{equation}

where equality applies after integration and summation over <math>\Lambda</math>.

Thus (II) becomes

\begin{equation}\label{IIa}
\frac{ \left( r_1 - r_2 \cos \theta \right) \left( a_+ - \alpha_+ r_1 \right)}{r_1 r^2} = - \frac{\Lambda \left( \Lambda + 1 \right)}{c_+ r^2_1} - \frac{ \left( c_+ + 1 \right)}{r^2} \tag{IIa}
\end{equation}

===Problem===

Prove the integral relation

\begin{align}
& \sum_\Lambda \int d\tau_r f \left( r_1, r_2 \right) \frac{1}{r} \left( \frac{d}{dr} q(r) \right) C_\Lambda \left( \hat{r}_2 \cdot \hat{\nabla}^Y_1 \right) P_\Lambda \left( \cos \theta \right) \\
= & \sum_\Lambda \int d\tau_r \frac{ f \left( r_1, r_2 \right)}{r_1, r_2} q(r) C_\Lambda \left( 1 \right) P_\Lambda \left( \cos \theta \right) \left(\frac{l_1^\prime \left( l_1^\prime + 1 \right) - l_1 \left( l_1 + 1 \right) - \Lambda \left( \Lambda + 1 \right)}{2}\right)
\end{align}

where the coefficients <math>C_\Lambda \left( 1 \right)</math> are the angular coefficients from the overlap integral

\begin{align}
\int & d\Omega\, \mathcal{Y}^{M*}_{l_1^\prime l_2^\prime L}(\hat{r}_1,\hat{r}_2)\, \mathcal{Y}^{M}_{l_1 l_2 L}(\hat{r}_1,\hat{r}_2)\\
= & \sum_\Lambda C_\Lambda(1)P_\Lambda(\cos\theta_{12}).
\end{align}

Hint: Use the fact that <math>l_1^2</math> is Hermitian so that
<math>
\int d\tau (l_1^2\mathcal{Y}^\prime)^*q(r)\mathcal{Y} = \int d\tau
\mathcal{Y}^{\prime *}l_1^2(q(r)\mathcal{Y})
</math>
with <math>\ \vec{l}_1= \frac{1}{i}\vec{r}_1\times\nabla_1</math>.

It is also useful to use
<math>\begin{eqnarray}\ \
\cos\theta\, P_L(\cos\theta) &=& \frac{1}{2L+1}[L\,P_{L-1}(\cos\theta) + (L+1)P_{L+1}(\cos\theta)]
\\
(\cos^2\theta-1)P_L(\cos\theta) &=&
\frac{L(L-1)}{(2L-1)(2L+1)}P_{L-2}(\cos\theta)\nonumber
-\frac{2(L^2+L-1)}{(2L-1)(2L+3)}P_L(\cos\theta) + \frac{(L+1)(L+2)}{(2L+1)(2L+3)}P_{L+2}(\cos\theta)\nonumber\\
&=& \frac{L(L-1)}{(2L-1)(2L+1)}[P_{L-2}(\cos\theta) - P_L(\cos\theta)]
+ \frac{(L+1)(L+2)}{(2L+1)(2L+3)}[P_{L+2}(\cos\theta) - P_L(\cos\theta)]
\end{eqnarray}</math>

together with a double application of the integral recursion relation
<math>
I_{L+1}\left(\frac{1}{r}\frac{d}{dr}q(r)\right) =
(2L+1)I_L\left(\frac{1}{r_1r_2}q(r)\right)+I_{L-1}\left(\frac{1}{r}\frac{d}{dr}q(r)\right)
</math>

Of course <math>l_1^2\mathcal{Y}^M_{l_1l_2L}(\hat{r}_1,\hat{r}_2)=l_1(l_1+1)\mathcal{Y}^M_{l_1l_2L}(\hat{r}_1,\hat{r}_2)</math>. Begin by expanding
<math>
l_1^2(g\mathcal{Y}) = \mathcal{Y}l_1^2g + g\,l_1^2\mathcal{Y} + 2(\vec{l}_1g)\cdot(\vec{l}_1\mathcal{Y})
</math>, and show that

<math>
\begin{eqnarray}\ \
(\vec{l}_1g)\cdot(\vec{l}_1\mathcal{Y}) & = & \frac{r_1^2}{r}\frac{dg}{dr}\vec{r}_2\cdot \nabla_1\mathcal{Y}\,\,\,\,\mbox{ and}\\
l_1^2g(r) &=& 2r_1r_2\cos\theta\frac{1}{r}\frac{dg}{dr} + r_1^2r_2^2(\cos^2\theta - 1)
\frac{1}{r}\frac{d}{dr}\left(\frac{1}{r}\frac{dg}{dr}\right).
\end{eqnarray}
</math>

The proof amounts to showing that the term <math>l_1^2g(r)</math> can be replaced by
<math>\Lambda(\Lambda+1)g(r)</math> after multiplying by <math>P_\Lambda(\cos\theta)</math>
and integrating by parts with respect to the radial integrations over <math>r_1</math>, <math>r_2</math> and
<math>r \equiv |\vec{r}_1 - \vec{r}_2|</math>. Remember that <math>\cos\theta</math> is just a short-hand notation for the radial function <math>(r_1^2 + r_2^2 - r^2)/(2r_1r_2)</math>.

==General Hermitian Property==

Each combination of terms of the form
\begin{equation}
<f>_R = a^2f_1+af_2+abf_3+b^2f_4+bf_5+af_6\nabla_1^Y+bf_7\nabla^Y_2+f_8(Y)
\end{equation}
can be rewritten
\begin{equation}
<f>_L= (a_+-a)^2f_1+(a_+-a)f_2 + (a_+-a)(b_+-b)f_3+(b_+-b)^2f_4
+(b_+-b)f_5+(a_+-a)f_6\nabla^{Y\prime}_1+(b_+-b)f_7\nabla^{Y\prime}_2+f_8)Y^\prime
\end{equation}
where as usual <math>\nabla_1^Y</math> acts only on the spherical harmonic part of the wave function denoted for short by <math>Y</math>.
Since these must be equal for arbitrary <math>a</math> and <math>b</math>,
\begin{equation}
a_+^2f_1+a_+f_2+_+b_+f_3+b_+f_4+b_+f_5+a_+f_6\nabla^{Y\prime}_1 +
b_+f_7\nabla^{Y\prime}_2+f_8(Y^\prime)-f_8(Y)=0
\end{equation}

Adding the corresponding expression with <math>Y</math> and <math>Y^\prime</math> interchanged
yields
<math>\label{III}
a_+^2f_1+a_+f_2+a_+b_+f_3+b_+^2f_4+b_+f_5+\frac{1}{2}a_+f_6\nabla^+_1+\frac{1}{2}b_+f_7\nabla^+_2=0\tag{III}
</math>

Subtracting gives
\[\label{IV}
f_8(Y)-f_8(Y^\prime)=-\frac{1}{2}[a_+f_6\nabla^-_1+b_+f_7\nabla_2^-]\tag{IV}
\]
\[\label{V}
a[-2a_+f_1-2f_2-b_+f_3-f_6\nabla^+_1]=0\tag{V}
\]
\[\label{VI}
b[-2b_+f_4-2f_5-a_+f_3-f_7\nabla^+_2]=0\tag{VI}
\]

Adding the two forms gives

\begin{eqnarray}
<f>_R+<f>_L=\frac{1}{2}(a_+^2+a_-^2)f_1+a_+f_2+\frac{1}{2}(a_+b_++a_-b_-)f_3\nonumber\\
\ +\frac{1}{2}(b_+^2+b_-^2)f_4+b_+f_5+\frac{1}{2}f_6(a_+\nabla_1^++a_-\nabla^-_1)\nonumber\\
\ +\frac{1}{2}(b_+\nabla_2^++b_-\nabla^-_2)+f_8(Y)+f_8(Y^\prime)\nonumber
\end{eqnarray}

Subtracting <math>\frac{x}{2}\times({\rm I})</math> gives

\begin{eqnarray}\ \
<f>_R+<f>_L = \frac{1}{2}[(1-x)a_+^2 + a_-^2]f_1 +
(1-\frac{x}{2})a_+f_2+\frac{1}{2}[(1-x)a_+b_++a_-b_-]f_3\nonumber\\
+\frac{1}{2}[(1-x)b_+^2+b_-^2]f_4+(1-\frac{x}{2})f_5b_++\frac{1}{2}f_6[(1-\frac{x}{2})a_+\nabla_1^++a_-\nabla^-_1]\nonumber\\
+\frac{1}{2}[(1-\frac{x}{2})b_+\nabla^+_2+b_-\nabla^-_2]+f_8(Y)+f_8(Y^\prime)\nonumber
\end{eqnarray}

If <math>x=1</math>,
\begin{eqnarray}
<f>_R+<f>_L &=&
\frac{1}{2}[a_-^2f_1+a_+f_2+a_-b_-f_3+b_-^2f_4+b_+f_5\nonumber\\
&+&\left(\frac{1}{2}a_+\nabla^+_1+a_-\nabla^-_1\right)f_6+\left(\frac{1}{2}b_+\nabla^+_2+b_-\nabla^-_2\right)f_7]+f_8(Y)+f_8(Y^\prime)\nonumber
\end{eqnarray}

The General Hermitian Property for arbitrary <math>x</math> gives
\begin{eqnarray}\ \
\nabla_1^2=
\frac{1}{4}\{\frac{1}{r_1^2}[(1-x)a_+^2+a_-^2+2(1-\frac{x}{2})a_+-2[l_1(l_1+1)+l_1^\prime(l_1^\prime+1)]\nonumber\\
-\frac{2}{r_1}[(1-x)\alpha_+a_++\alpha_-a_-+2(1-\frac{x}{2})\alpha_+]+(1-x)\alpha_+^2
+ \alpha^2_-]\nonumber\\
+\frac{2(r_1-r_2\cos\theta)}{r_1r^2}[(1-x)(a_+-\alpha_+r_1)c_++(a_--\alpha_-r_2)c_-]\nonumber\\
-\frac{2r_2}{r_1r^2}[(1-\frac{x}{2})c_+\hat{r}_2\cdot\hat{\nabla}_1^++c_-\hat{r}_2\hat{\nabla}_1^-]+\frac{(1-x)c_+^2+c_-^2+2(1-\frac{x}{2})c_+}{r^2}\}\nonumber
\end{eqnarray}.

Use
<math>\begin{eqnarray}\ \
\frac{-2r_2}{r_1r^2}[(1-\frac{x}{2})c_+\hat{r}_2\hat{\nabla}_1^++c_-\hat{r}_2\hat{\nabla}_1^-]=\frac{2}{r_1^2}\left[(1-\frac{x}{2})\Lambda(\Lambda+1)-\frac{c_-}{c_+}[l_1^\prime(l_1^\prime+1)+l_1(l_1+1)]\right]\end{eqnarray}</math>,

<math>\begin{eqnarray}
\frac{2(r_1-r_2\cos\theta)}{r_1r^2}(a_--\alpha_-r_1)c_- = 2\frac{c_-}{c_+}\left[\frac{-1}{r_1^2}[a_-(a_++1)]+\frac{1}{r_1}[a_-\alpha_++\alpha_-(a_++2)]-\alpha_-\alpha_+\right]\nonumber
\end{eqnarray}</math>,

and
<math>\begin{eqnarray}\ \
\frac{2(r_1-r_2\cos\theta)}{r_1r^2}(a_+-\alpha_+r_1)c_+ = 2\left[-\frac{1}{r_1^2}[a_+(a_++1)]+\frac{1}{r_1}[a_+\alpha_++\alpha_+(a_++2)]-\alpha_+^2\right]
\end{eqnarray}</math>.

Substituting into <math>\nabla_1^2</math> gives

\begin{eqnarray}
\nabla^2_1 = \frac{1}{4}\{\frac{1}{r_1^2}[-(1-x)a_+^2 + a_-^2 +xa_+ +
2(1-\frac{x}{2})\Lambda(\Lambda+1)\nonumber\\
-2l_1(l_1+1)(1-\frac{c_-}{c_+}) -
2l_1^\prime(l_1^\prime+1)(1+\frac{c_-}{c_+})-\frac{2c_-a_-}{c_+}(a_++1)]\nonumber\\
- \frac{2}{r_1}\left[-(1-x)\alpha_+(a_++2)+\alpha_-a_-+2(1-\frac{x}{2})\alpha_+
-\frac{c_-}{c_+}[a_-\alpha_++\alpha_-(a_++2)]\right]\nonumber\\
-(1-x)\alpha_+^2+\alpha_-^2-2\frac{c_-}{c_+}\alpha_-\alpha_+ +\frac{1}{r^2}\left[(1-x)c_+^2+c_-^2+2(1-\frac{x}{2})c_+\right]\}\nonumber
\end{eqnarray}

This has the form
<math>
\nabla_1^2 =
\frac{1}{4}\left[\frac{A_1}{r_1^2}+\frac{B_1}{r_1}+\frac{C_1}{r^2}+D_1\right]
</math>
with

<math>\begin{array}{ccc}
A_1=-(1-x)a_+^2+a_-^2
+xa_++2(1-\frac{x}{2})\Lambda(\Lambda+1)-2l_1(l_1+1)(1-\frac{c_-}{c_+})\
-2l_1^\prime(l_1^\prime+1)(1+\frac{c_-}{c_+})-\frac{2c_-a_-}{c_+}(a_++1) \ \
\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \
\end{array}
</math>

<math>
B_1 =
2\left[(1-x)\alpha_+(a_++2)-\alpha_-a_--2(1-\frac{x}{2})\alpha_++\frac{c_-}{c_+}[a_-\alpha_++\alpha_-(a_++2)]\right]
</math>

<math>
C_1 = (1-x)c_+^2+ c_-^2 +2(1-\frac{x}{2})c_+
</math>

<math>
D_1 = -1(1-x)\alpha_+^2 +\alpha_-^2 - 2\frac{c_-}{c_+}\alpha_-\alpha_+
</math>

The complete Hamiltonian is then (in Z-scaled a.u.)

<math>\begin{eqnarray}
H =-\frac{1}{2}\nabla^2_1 -\frac{1}{2}\nabla^2_2 -\frac{1}{r_1} -\frac{1}{r_2} +\frac{Z^{-1}}{r}\nonumber\\
= -\frac{1}{2}\nabla^2_1 -\frac{1}{2}\nabla^2_2 -\frac{1}{r_1} -\frac{Z-1}{Zr_2}+Z^{-1}\left(\frac{1}{r}-\frac{1}{r_2}\right)\nonumber\\
\ =-\frac{1}{8}\left[\frac{A_1}{r_1^2}+\frac{B_1+8}{r_1}+\frac{C_1}{r^2}+D_1+D_2 +\frac{A_2}{r_2^2} + \frac{B_2 + 8(Z-1)/Z}{r_2}+\frac{C_2}{r^2}\right]\nonumber
+ Z^{-1}\left(\frac{1}{r}-\frac{1}{r_2}\right)\nonumber
\end{eqnarray}</math>

The screened hydrogenic energy is
<math>\begin{eqnarray}
E_{SH}=-\frac{1}{2}\left[\frac{1}{n_1^2}+ \left(\frac{Z-1}{Z}\right)^2\frac{1}{n_2^2}\right]
\end{eqnarray}</math>
so that

<math>\begin{eqnarray}
H-E_{SH} &=& -\frac{1}{8}[
\frac{A_1}{r_1^2}+\frac{B_1+8}{r_1}+\frac{C_1}{r^2} + D_1 + D_2
-\frac{4}{n_1^2} -\left(\frac{Z-1}{Z}\right)^2\frac{4}{n_2^2}\nonumber
+\frac{A_2}{r_2}+\frac{B_2+8(Z-1)/Z}{r_2}+\frac{C_2}{r^2}] +
Z^{-1}\left(\frac{1}{r} -\frac{1}{r_2}\right)\nonumber
\end{eqnarray}</math>

==Optimization of Nonlinear Parameters==
- The traditional method of performing Hylleraas calculations is to write the
basis set in the form

<math>
\Psi = \sum_{i,j,k}c_{i,j,k}\varphi_{i,j,k}(\alpha,\beta)\pm exchange
</math>

with

<math>
\varphi_{i,j,k}(\alpha,\beta) = r_1^ir_2^jr_{12}^ke^{-\alpha r_1 - \beta
r_2}\mathcal{Y}^M_{l_1l_2L}.
</math>

The usual procedure is to set <math>\alpha = Z</math> so that it represents the inner
<math>1s</math> electron, and then to vary <math>\beta</math> so as to minimize the energy.

[[File:OptNon.jpg|500px]]

Since <math>\beta</math> appears in <math>\Psi</math> as a non-linear parameter, the entire
calculation must be repeated for each value of <math>\beta</math>. However, the minimum
becomes progressively smaller as the basis set is enlarged.

[[File:OptNon02.jpg|500px]]

*Difficulties

1. If the basis set is constructed so that <math> i +j +j \leq N</math>, the the
number of terms is
<math>
(N+1)(N+2)(N+3)/6
</math>
N=14 already gives 680 terms and an accuracy of about 1 part in <math>10^{10}</math> for
low-lying states. A substantial improvement in accuracy would require much
larger basis sets, together with multiple precision arithmetic to avoid loss
of significant figures when high powers are included.

2. The accuracy rapidly deteriorates as one goes to more highly excited
states - about 1 significant figure is lost each time the principle quantum
number is increased.

*Cure

We have found that writing basis sets in the form
\begin{eqnarray}
\Psi &=& \sum_{i,j,k}\left(c^{(1)}_{i,j,k}\varphi_{ijk}(\alpha_1\beta_1) +
c_{ijk}^{(2)}\varphi_{ijk}(\alpha_2\beta_2)\right) \pm exchange \nonumber\\
&=& \psi({\bf r}_1,{\bf r}_2) \pm \psi({\bf r}_2,{\bf r}_1)\nonumber
\end{eqnarray}
so that each combination of powers is included twice with different nonlinear
parameters gives a dramatic improvement in accuracy for basis sets of about
the same total size.

However, the optimization of the nonlinear parameters is now much more
difficult, and an automated procedure is needed.

If
<math>
E = \frac{<\Psi|H|\Psi>}{<\Psi|\Psi>}</math> and we assume <math><\Psi|\Psi> = 1</math>, then

<math>
\frac{\partial E}{\partial\alpha_t} = -2 <\Psi|H-E|r_1\psi({\bf r}_1,{\bf
r}_2;\alpha_t)\pm r_21\psi({\bf r}_2,{\bf
r}_1;\alpha_t)>
</math>
where
<math>
\psi({\bf r}_1,{\bf r}_2;\alpha_t) =
\sum_{i,j,k}c_{ijk}^{(t)}\varphi_{ijk}(\alpha_t\beta_t)
</math>

*Problem

1. Prove the above.

2. Prove that there is no contribution from the implicit dependence of
<math>c_{ijk}^{(t)}</math> on <math>\alpha_t</math> if the linear parameters have been optimized.

==The Screened Hydrogenic Term==

If the Hamiltonian is written in the form (in Z-scaled a.u.)
<math>\begin{eqnarray}\
H = H_0({\bf r}_1,Z) + H_0({\bf r}_2,Z-1) +
Z^{-1}\left(\frac{1}{r}-\frac{1}{r_2}\right)
\end{eqnarray}</math>
with
<math>\
H_0({\bf r}_1,Z) = -\frac{1}{2}\nabla_1^2 - \frac{1}{r_1}\nonumber
</math> and
<math>
H_0({\bf r}_2,Z-1) = -\frac{1}{2}\nabla_2^2 - \frac{Z-1}Z{r_2}\nonumber
</math>,
then the eigenvectors of <math>H_0({\bf r},Z) +H_0({\bf r}_2,Z-1)</math> are products of
hydrogenic orbitals
<math>
\Psi_0=\psi_0(1s,Z)\psi_0(nl,Z-1)
</math>
and the eigenvalue is
<math>\begin{eqnarray}
E_{SH} = \left[-\frac{1}{2} - \left(\frac{Z-1}{Z}\right)^2\frac{1}{2n^2}\right]Z^2 a.u.
\end{eqnarray}</math>
called the screened hydrogenic eigenvalue. From highly excited states,
<math>E_{SH}</math> and <math>\Psi_0</math>are already excellent approximations.

For example, for the 1s8d states, the energies are

<math>\begin{eqnarray}\\
E(1s8d^1D)&=& -2.007\,816\,512\,563\,81 {\rm\ a.u.}\nonumber\\
E(1s8d^3D)&=& - 2.007\,817\,934\,711\,71 {\rm\ a.u.}\nonumber\\
E_{SH} &=& -2.007\,812\,5 {\rm\ a.u.} \nonumber
\end{eqnarray}</math>

It is therefore advantageous to include the screened hydrogenic terms in the
basis set so that the complete trial function becomes
<math>
\Psi = c_0\Psi_0 +
\sum_{ijk}\left[c_{ijk}^{(1)}\varphi_{ijk}(\alpha_1,\beta_1)+c_{ijk}^{(2)}\varphi(\alpha_2,\beta_2)\right]\pm
{\rm exchange}
</math>

Also, the variational principal can be re-expressed in the form
<math>\begin{eqnarray}
E= E_{SH}+\frac{<\Psi|H-E_{SH}|\Psi>}{<\Psi|\Psi>}
\end{eqnarray}</math>
so that the <math>E_{SH}</math> term can be cancelled analytically from the matrix
elements.

For example
<math>
<\Psi_0|H-E_{SH}|\Psi_0> = 0
</math>

and
<math>
<\varphi_{ijk}|H-E_{SH}|\Psi_0> =
Z^{-1}<\varphi_{ijk}|\frac{1}{r}-\frac{1}{r_2}|\Psi_0>
</math>

Recall that
<math>\begin{eqnarray}\ \
I_0(a,b,c) = \frac{2}{c+2}\sum^{[c+1]/2}_{i=0}\left(\begin{array}{c}
c+2\\
2i+1
\end{array}\right) \left[f(p,q;\beta) + f(p^\prime,q^\prime;\alpha)\right]
\end{eqnarray}</math>
where

<math>\begin{eqnarray}
f(p,q;x) =
\frac{q!}{x^{q+1}(\alpha+\beta)^{p+1}}\sum^p_{j=0}\frac{(p+j)!}{j!}\left(\frac{x}{\alpha+\beta}\right)^j
\end{eqnarray}</math>

<math>
\begin{array}{cc}
p=a+2i+2 & p^\prime = b+2i+2\\
q= b+c-2i+2 & q^\prime =a+c-2i+2
\end{array}
</math>

If <math>\beta\ll\alpha</math>, the <math>f(p,q;\beta)\gg f(p^\prime,q^\prime;\alpha)</math> and the
<math>i=0</math> term is the dominant contribution to <math>f(p,q;\beta)</math>. However, since
this term depends only on the sum of powers <math>b+c</math> for <math>r_2</math>, it cancels exactly
from the matrix element of
<math>
\frac{1}{r_{12}} - \frac{1}{r_2}
</math>
and can therefore
be omitted in the calculations of integrands, thereby saving many significant
figures. This is especially valuable when <math>b+c</math> is large and <math>a</math> is small.

==Small Corrections==
*Mass Polarization
If the nuclear mass is not taken to be infinite, then the Hamiltonian is
<math>\begin{eqnarray}\ \
H = \frac{P_N^2}{2M} + \sum_{i=1}\frac{p_i^2}{2m} + V
\end{eqnarray}</math>

[[File:SmallCor.jpg|500px]]

where <math>M =M_A-nm</math> is the nuclear mass and <math>m</math> is the electron mass.

Change variables to the C of M
<math>\begin{eqnarray}
{\bf R} = \frac{1}{M+nm}\left[M{\bf r}_N + m({\bf r}_1 + {\bf r}_2 +
\ldots)\right]
\end{eqnarray}</math>

and relative variables

<math>
{\bf s}_i = {\bf r}_i - {\bf r}_N.
</math>

Then

<math>\begin{eqnarray}\ \ \ \ \
H = \frac{1}{2(M+nm)}{\bf p}_R^2+\frac{1}{2m}\sum_{i=1}^n{\bf p}_{s_i}^2
+\frac{1}{2M}\sum_{i,k}{\bf p}_{s_i}\cdot {\bf p}_{s_k} + V({\bf s}_1, {\bf
s}_2, \ldots)\nonumber
= \frac{1}{M+nm}{\bf p}_R^2+\frac{1}{2\mu}\sum_{i=1}^n{\bf
p}_{s_i}^2+\frac{1}{2M}\sum_{i\neq k}{\bf p}_{s_i}\cdot {\bf p}_{s_k}+V({\bf s}_1, {\bf
s}_2, \ldots)\nonumber
\end{eqnarray}</math>

where
<math>\begin{eqnarray}
\mu = \left(\frac{1}{M} + \frac{1}{m}\right)^{-1}
\end{eqnarray}</math>

If <math>n=2</math>, the Schrödinger equation is
<math>\begin{eqnarray}\ \ \
\left[ -\frac{\hbar^2}{2\mu}(\nabla^2_{s_1}+ \nabla^2_{s_2}) -
\frac{\hbar^2}{M}\nabla_{s_1}\nabla_{s_2}-\frac{Ze^2}{s_1}-\frac{Ze^2}{s_2}+\frac{e^2}{s_{12}}\right]\psi
= E\psi
\end{eqnarray}</math>.

Define the reduced mass Bohr radius
<math>\begin{eqnarray}
a_\mu = \frac{\hbar^2}{\mu e^2} = \frac{m}{\mu}a_0
\end{eqnarray}</math>
and

<math>\begin{eqnarray}
\rho_i = \frac{Zs_i}{a_\mu}\nonumber\\
\varepsilon = \frac{E}{Z^2(e^2/a_\mu)}\nonumber
\end{eqnarray}</math>

The Schrödinger equation then becomes
<math>\begin{eqnarray}\ \ \
\left[
-\frac{1}{2}(\nabla^2_{\rho_1}+\nabla^2_{\rho_2})-\frac{\mu}{M}\vec{\nabla}_{\rho_1}\cdot\vec{\nabla}_{\rho_2} - \frac{1}{\rho_1}-\frac{1}{\rho_2}+\frac{Z^{-1}}{\rho_{12}}\right]\psi = \varepsilon \psi
\end{eqnarray}</math>

The effects of finite mass are thus

1. A reduced mass correction to all energy levels of <math>(e^2/a_\mu)/(e^2/a_0)
= \mu/m</math>, i.e.
<math>\begin{eqnarray}
E_M=\frac{\mu}{m}E_\infty
\end{eqnarray}</math>

2. A specific mass shift given in first order perturbation theory by
<math>\begin{eqnarray}
\Delta\varepsilon =
-\frac{\mu}{M}<\psi|\nabla_{\rho_1}\cdot\nabla_{\rho_2}|\psi>
\end{eqnarray}</math>
or

<math>\begin{eqnarray}
\Delta\varepsilon =
-\frac{\mu}{M}<\psi|\nabla_{\rho_1}\cdot\nabla_{\rho_2}|\psi>\frac{e^2}{a_\mu}\nonumber\\
= -\frac{\mu^2}{mM}<\psi|\nabla_{\rho_1}\cdot\nabla_{\rho_2}|\psi>\frac{e^2}{a_0}\nonumber
\end{eqnarray}</math>

Since <math>\nabla_{\rho_1}\cdot\nabla_{\rho_2}</math> has the same angular properties as
<math>{\bf\rho}_1 \cdot {\bf\rho}_2</math>, the operator is like the product of two
dipole operators. For product type wave functions of the form
$
\psi = \psi(1s)\psi(nl) \pm \mbox{exchange},
$
the matrix element vanishes for all but P-states.

=Radiative Transitions=

==Transition Integrals and the Line Strength==

For an electric dipole transition <math>\gamma'\,L'\,S'\,J'\,M' \rightarrow \gamma\,L\,S\,J\,M</math>, the line strength is defined by
\begin{equation}
S = \sum_{M',\mu,M}|\langle\gamma'\,L'\,S'\,J'\,M'|r_\mu|\gamma\,L\,S\,J\,M\rangle|^2
\end{equation}
Using the Wigner-Eckart theorem, this is related to the reduced matrix element by
\begin{eqnarray}
S(J',J) = |\langle\gamma'\,L'\,S'\,J'||r||\gamma\,L\,S\,J\,M\rangle|^2
\sum_{M',\mu,M}\left|\left(\matrix{J' & 1 & J \cr
-M'& \mu &M \cr}\right)\right|^2
\end{eqnarray}
Using the sum rule for the 3-<math>j</math> symbols
\begin{equation}
\sum_{M',\mu,M}\left|\left(\matrix{J' & 1 & J \cr
-M'& \mu &M \cr}\right)\right|^2 = 1
\end{equation}
we see that the line strength is identical to the reduced matrix element
\begin{equation}
S(J',J) = |\langle\gamma'\,L'\,S'\,J'||r||\gamma\,L\,S\,J\,M\rangle|^2
\end{equation}
For a nonrelativistic calculation in <math>LS</math> coupling, <math>S = S'</math> and we can next strip out the dependence on <math>J</math> and <math>J'</math> by use of Eq.\ (7.1.7) of Edmonds { Angular Momentum in Quantum Mechanics} with <math>k = 1</math> to obtain
\begin{equation}
S(J',J) = (-1)^{L'+S+J+1}[(2J+1)(2J'+1)]^{1/2}\left\{
\matrix{L' & J'& S\cr
J & L & 1\cr}\right\}\langle\gamma' L'|| r || \gamma L\rangle
\end{equation}
where <math>\langle\gamma' L'|| r || \gamma L\rangle</math> is the reduced matrix element if spin is neglected. Note that if spin is neglected, then the line-strength is simply
\begin{equation}
S(L',L) = |\langle\gamma' L'|| r || \gamma L\rangle|^2
\end{equation}
To calculate the reduced matrix element, use a particular example along with the Wigner-Eckart theorem
such as
\begin{equation}
\langle\gamma'\, L'|| r || \gamma\, L\rangle = (-1)^{L'}\langle\gamma'\, L'\, 0| z | \gamma\, L\, 0\rangle{\Large /}\left(\matrix{L' & 1 & L \cr
0 & 0 & 0 \cr}\right)
\end{equation}

===Correlated Two-electron Integrals===
For the case of two-electron integrals, <math>z = z_1 + z_2</math> and <math>r = r_1 + r_2</math>. As discussed in G.W.F. Drake, Phys. Rev. A 18, 820 (1978), the general integral in Hylleraas coordinates for the integral over the <math>z_1</math> term can be written in the form
\begin{equation}
\label{int1}
\langle\gamma'\, L'\, 0| z_1 | \gamma\, L\, 0\rangle = \sum_\Lambda C_\Lambda I_\Lambda(\gamma',\gamma)
\end{equation}
where <math>I_\Lambda(\gamma',\gamma)</math> stands for the purely radial integral
\begin{equation}
\label{radial}
I_\Lambda(\gamma',\gamma) = \int_0^\infty r_1\,dr_1\int_0^\infty r_2\,dr_2\int_{|r_1-r_2|}^{r_1+r_2}r_{12}\, dr_{12}
R_{\gamma'}^*\,r_1\, R_\gamma P_\Lambda(\cos\theta_{12})
\end{equation}
and <math>\cos\theta_{12} = (r_1^2 + r_2^2 - r_{12}^2)/(2r_1r_2)</math> is a short-hand notation for a radial function of <math>r_1</math>, <math>r_2</math> and <math>r_{12}</math>. Also, <math>P_\Lambda(\cos\theta_{12})</math> is a Legendre polynomial,
and, <math>R_\gamma = R_\gamma(r_1,r_2,r_{12})</math> is the radial part of the Hylleraas wave function for state <math>\gamma</math> such
that the total wave function is <math>\psi({\bf r}_1,{\bf r}_2) = R_\gamma(r_1,r_2,r_{12}){\cal Y}_{l_1,l_2,L}^M(\hat{\bf r}_1,\hat{\bf r}_2)\pm {\rm exchange}</math>, normalized to unity, and <math>{\cal Y}_{l_1,l_2,L}^M(\hat{\bf r}_1,\hat{\bf r}_2)</math> is a vector coupled product of spherical harmonics. The coefficents <math>C_\Lambda</math> in Eq.\ (\ref{int1}) can be calculated from Eq.\ (32) of the above 1978 paper. They need to be calculated separately for he direct-direct and the direct-exchange contributions. For the reduced matrix element of <math>z_1</math>, the leading 3-<math>j</math> symbol can be dropped, leaving (for the case <math>k = 1</math>)
\begin{eqnarray}
C_\Lambda^{\rm reduced} &=& \frac{(-1)^{l_2'+L'+L+1}}{2} (l_1,l_1',l_2,l_2',L,L')^{1/2}
\sum_{\lambda={\rm max}(l_1-1,0)}^{l_1+1} (\lambda,\Lambda,1)
\left(\matrix{l_1 & 1 & \lambda \cr
0 & 0 & 0 \cr}\right)
\left(\matrix{l_1'& \lambda & \Lambda \cr
0 & 0 & 0 \cr}\right)\nonumber\\
&\times&\left(\matrix{l_2' & 1_2 & \Lambda \cr
0 & 0 & 0 \cr}\right)
\left\{\matrix{L' & \lambda & l_2 \cr
\Lambda& l_2' & l_1' \cr}\right\}
\left\{\matrix{L' & \lambda & l_2 \cr
l_1 & L & 1 \cr}\right\}
\end{eqnarray}
where the notation <math>(a,b,c,\cdots)</math> means <math>(2a+1)(2b+1)(2c+1)\cdots</math>.
A similar expression can be obtained for the angular coefficients for the <math>z_2</math> term by interchanging the subscripts 1 and 2 throughout and multiplying by <math>(-1)^{L+L'+1}</math>.

===Example===
As an example, for the <math>1s^2\;^1S - 1s2p\;^1P</math> transition of helium, the only nonvanishing <math>C_\Lambda</math> angular coefficients are <math>C_1 = 1/2</math> for the direct-direct term (i.e. <math>P_1(\cos\theta_{12}) = \cos\theta_{12}</math> is included in the radial integral (\ref{radial})) and <math>C_0 = 1/2</math> for the direct-exchange term (i.e. <math>P_0(\cos\theta_{12}) = 1</math> is included in the radial integral (\ref{radial})). The results for infinite nuclear mass are
\begin{equation}
S(1\;^1S-2\;^1P) = (0.781\,127\,158\,667)^2
\end{equation}
\begin{equation}f(1\;^1S-2\;^1P) = -0.276\,164\,703\,736
\end{equation}
where
\begin{equation}
f(1\;^1S-2\;^1P) = \frac{2}{3}[E(1\;^1S) - E(2\;^1P)]S(1\;^1S-2\;^1P)
\end{equation}
is the absorption oscillator strength.

===Test===
This is a test edit

Theory Notes

2020-12-14T04:10:41Z

Drake: /* Example */

<center>
<H1>Notes on solving the Schrödinger equation in Hylleraas coordinates for heliumlike atoms
Gordon W.F. Drake
Department of Physics, University of Windsor
Windsor, Ontario, Canada N9B 3P4
(Transcribed from hand-written notes and edited by Lauren Moffatt. Last revised November 22, 2017 to correct a missing subscript in the expression for $B_1$ in the Hamiltonian in Sect. 11 General Hermitian Property.)
</H1>
</center>

==Introduction==
These notes describe some of the technical details involved in solving the Schrödinger equation for a heliumlike atom of nuclear charge Z in correlated Hylleraas coordinates. The standard for computational purposes is to express all quantities in a form that is valid for electronic states of arbitrary angular momentum, and to express matrix elements of the Hamiltonian in a manifestly Hermitian symmetrized form. Further information and derivations of the mathematical techniques for angular integrals and radial recursion relations can be found in the paper
G.W.F. Drake, Phys. Rev. A <math>{\bf 18}</math>, 820 (1978).

The starting point is the two-electron Schrödinger equation for infinite nuclear mass
\begin{equation} \left[-\frac{\hbar^2}{2m}(\nabla^2_1 +\nabla^2_2) - \frac{Ze^2}{r_1} -
\frac{Ze^2}{r_2}+\frac{e^2}{r_{12}} \right]\psi = E\psi\nonumber
\end{equation}
where <math>m</math> is the electron mass, and <math>r_{12} = |{\bf r}_1 - {\bf r}_2|</math> (see diagram below).

[[File:Pic1.jpg|250px]]

Begin by rescaling distances and energies so that the Schrödinger equation can be expressed in a dimensionless form. The dimensionless Z-scaled distance is defined by <math>\begin{eqnarray} \rho = \frac{Zr}{a_0} \end{eqnarray} </math> where <math> \begin{eqnarray} a_0 = \frac{\hbar^2}{me^2} \end{eqnarray} </math> is the atomic unit (a.u.) of distance equal to the Bohr radius <math>0.529\, 177\, 210\, 92(17) \times 10^{-10}</math> m. Then

\begin{equation}[-\frac{\hbar^2}{2m}Z^2(\frac{me^2}{\hbar^2})^2(\nabla^2_{\rho_1}+\nabla^2_{\rho_2})
- Z^2\frac{e^2}{a_0}\rho^{-1}_1 - Z^2\frac{e^2}{a_0}\rho^{-1}_2 +
\frac{e^2}{a_0}Z\rho^{-1}_{12}]\psi= E\psi\nonumber\end{equation}

But <math>\begin{eqnarray} \frac{\hbar^2}{m}(\frac{me^2}{\hbar^2})^2 = \frac{e^2}{a_0} \end{eqnarray} </math> is the hartree atomic unit of of energy <math>(E_h = 27.211\,385\,05(60)</math> eV, or equivalently <math>E_h/(hc) = 2194.746\,313\,708(11)</math> m<math>^{-1})</math>. Therefore, after multiplying through by <math>a_0/(Ze)^2</math>, the problem to be solved in Z-scaled dimensionless units becomes

\begin{equation} \left[-\frac{1}{2}(\nabla^2_{\rho_1}+\nabla^2_{\rho_2}) -
\frac{1}{\rho_1} - \frac{1}{\rho_2} + \frac{Z^{-1}}{\rho_{12}}\right]\psi =
\varepsilon\psi\nonumber \end{equation}

where <math>\begin{eqnarray}\varepsilon = \frac{Ea_0}{(Ze)^2} \end{eqnarray}</math> is the energy in Z-scaled atomic units. For convenience, rewrite this in the conventional form
\begin{equation}
H\psi = \varepsilon\psi\nonumber \end{equation}
where (using <math>r</math> in place of <math>\rho</math> for the Z-scaled distance)
\begin{equation}
H = -\frac{1}{2}(\nabla^2_1+\nabla^2_2) - \frac{1}{r_1}-\frac{1}{r_2} + \frac{Z^{-1}}{r_{12}}
\nonumber\end{equation}
is the atomic Hamiltonian for infinite nuclear mass.

==The Hartree Fock Method==

For purposes of comparison, and to define the correlation energy, assume that <math>\psi({\bf r}_1,{\bf r}_2)</math> can be written in the separable product form
\begin{equation}
\psi({\bf r}_1,{\bf r}_2) = \frac{1}{\sqrt{2}} \left[ u_1(r_1) u_2(r_2) \pm u_2(r_1) u_1(r_2) \right] \nonumber
\end{equation}
for the <math>1s^2\ ^1\!S</math> ground state. Because of the <math>\frac{1}{r_{12}}</math> term in the Schrödinger equation, the exact solution connot be expressed in this form as a separable product. However, the Hartree-Fock approximation corresponds to finding the best possible solution to the Schrödinger equation
\begin{equation}
H\psi({\bf r}_1,{\bf r}_2) = E\psi({\bf r}_1,{\bf r}_2)\nonumber
\end{equation}
that can nevertheless be expressed in this separable product form, where as before
\begin{equation}
H = -\frac{1}{2}(\nabla^2_1+\nabla^2_2) - \frac{1}{r_1}- \frac{1}{r_2} + \frac{Z^{-1}}{r_{12}}\nonumber
\end{equation}
is the full two-electron Hamiltonian. To find the best solution, substitute into <math>\left\langle\psi \lvert H-E \rvert \psi\right\rangle</math> and require this expression to be stationary
with respect to arbitrary infinitesimal variations <math>\delta u_1</math> and <math>\delta
u_2</math> in <math>u_1</math> and <math>u_2</math> respectively; i.e.

\begin{equation}
\frac{1}{2} \left\langle \delta u_1(r_1)u_2(r_2) \pm u_2(r_1)\delta
u_1(r_2) \lvert H-E \rvert u_1(r_1) u_2(r_2)\pm u_2(r_1) u_1(r_2)\right\rangle\nonumber
\end{equation}

\begin{equation}
=\int\delta u_1(r_1) d{\bf r}_1 \left\{ \int d{\bf r}_2 u_2(r_2) \left( H - E \right) \left[ u_1(r_1)u_2(r_2) \pm u_2(r_1) u_1(r_2) \right] \right\}\nonumber
\end{equation}

\begin{equation}
= 0 \nonumber
\end{equation}

for arbitrary <math> \delta u_1(r_1).\nonumber </math> Therefore <math>\{\int d{\bf r}_2 \ldots \} = 0</math>.

Similarly, the coefficient of <math>\delta u_2</math> would give

\begin{equation}
\int d{\bf r}_1 u_1(r_1) \left( H-E \right) \left[ u_1(r_1) u_2(r_2) \pm
u_2(r_1) u_1(r_2) \right] = 0\nonumber
\end{equation}

Define

\begin{equation}
I_{12} = I_{21} = \int d{\bf r}\, u_1(r)u_2(r), \nonumber
\end{equation}

\begin{equation}
H_{ij} = \int d{\bf r}\, u_i(r)(-\frac{1}{2}\nabla - \frac{1}{r})u_j(r), \nonumber
\end{equation}

\begin{equation}
G_{ij}(r) = \int d{\bf r}^\prime u_i(r^\prime)\frac{1}{|{\bf r} - {\bf r}\prime|}u_j(r^\prime)\nonumber
\end{equation}

Then the above equations become the pair of integro-differential equations

\begin{equation}
[ H_0 - E + H_{22}+G_{22}(r)]u_1(r) = \mp [ I_{12}(H_0-E) + H_{12}+G_{12}(r)]u_2(r)\nonumber
\end{equation}

\begin{equation}
[H_0-E+H_{11}+G_{11}(r)]u_2(r) = \mp [I_{12}(H_0-E) + H_{12}+G_{12}(r)]u_1(r)\nonumber
\end{equation}

These must be solved self-consistently for the "constants" <math>I_{12}</math> and
<math>H_{ij}</math> and the function <math>G_{ij}(r)</math>.

The Hartree Fock energy is <math>E \simeq -2.87\ldots </math>a.u. while the exact energy is <math>E = -2.903724\ldots </math>a.u.

The difference is called the "correlation energy" because it arises from the
way in which the motion of one electron is correlated to the other. The
Hartree Fock equations only describe how one electron moves in the average field
provided by the other (mean-field theory).

==Configuration Interaction==
Expand
\begin{equation}
\psi({\bf r}_1,{\bf r}_2)= C_0u^{(s)}_1(r_1)u^{(s)}_1(r_2) + C_1u^{(P)}_1({\bf r}_1)u^{(P)}_1({\bf r}_2){\cal Y}^0_{1,1,0}(\hat{\bf r}_1, \hat{\bf r}_2)+C_2u^{(d)}_1({\bf r}_1)u^{(d)}_2({\bf r}_2){\cal Y}^0_{2,2,0}(\hat{\bf r}_1, \hat{\bf r}_2)+... \pm {\rm\ exchange}
\end{equation}
where
\begin{equation}
{\cal Y}^M_{l_1,l_2,L}(\hat{\bf r}_1, \hat{\bf r}_2)=\sum_{m_1,m_2}Y^{m_1}_{l_1}({\bf r}_1)Y^{m_2}_{l_2}({\bf r}_2)\times <l_1l_2m_1m_2\mid LM>
\end{equation}

This works, but is slowly convergent, and very laborious. The best CI calculations are accurate to <math> ~10^{-7}</math> a.u.

==Hylleraas Coordinates==
[E.A. Hylleraas, Z. Phys. <math>{\bf 48}, 469(1928)</math> and <math>{\bf 54}, 347(1929)</math>]
suggested using the co-ordinates <math>r_1</math>, <math>r_2</math> and <math>r_{12}</math> or equivalently

\begin{eqnarray}
s &=& r_1 + r_2, \nonumber\\
t &=& r_1-r_2, \nonumber\\
u &=& r_{12}\nonumber
\end{eqnarray}

and writing the trial functions in the form

\begin{equation}
\Psi({\bf r}_1,{\bf r}_2) = \sum^{i+j+k\leq
N}_{i,j,k}c_{i,j,k}r_1^{i+l_1}r_2^{j+l_2}r_{12}^ke^{-\alpha r_1 - \beta r_2}
\mathcal{Y}^M_{l_1,l_2,L}(\hat{r}_1,\hat{r}_2)\pm \text{exchange}\nonumber
\end{equation}

Diagonalizing H in this non-orthogonal basis set is equivalent to solving
<math> \begin{eqnarray}
\frac{\partial E}{\partial c_{i,j,k}} = 0\nonumber
\end{eqnarray} </math>
for fixed <math>\alpha</math> and <math>\beta</math>.

The diagonalization must be repeated for different values of <math>\alpha</math> and
<math>\beta</math> in order to optimize the non-linear parameters.

==Completeness==
The completeness of the above basis set can be shown by first writing
\begin{equation}
r_{12}^2 = r_1^2 + r_2^2 - 2r_1r_2\cos\theta_{12}
\end{equation}
and, from the spherical harmonic addition theorem,
\begin{equation}
\cos(\theta_{12})=\frac{4\pi}{3}\sum^1_{m=-1}Y^{m*}_1(\theta_1,\varphi_1)Y^m_1(\theta_2,\varphi_2)
\end{equation}
Consider first the S-states. The <math>r_{12}^0</math> terms are like the ss terms in a CI calculation. The
<math>r_{12}^2</math> terms bring in p-p type contributions, and the higher powers bring
in d-d, f-f etc. type terms. In general

\begin{equation}
P_l(\cos\theta_{12}) = \frac{4\pi}{2l+1}\sum^l_{m=-l}{Y^{m}_l}^*(\theta_1,\varphi_1)Y^m_l(\theta_2,\varphi_2)\nonumber
\end{equation}

For P-states, one would have similarly

\begin{array}{lr}
r_{12}^0 & (sp)P\nonumber\\
r_{12}^2 & (pd)P\nonumber\\
r_{12}^4 & (df)P\nonumber\\
\vdots & \vdots\nonumber
\end{array}

For D-states

\begin{array}{lr}
r_{12}^0 & (sd)D & (pp^\prime)D\nonumber\\
r_{12}^2 & (pf)D & (dd^\prime)D\nonumber\\
r_{12}^4 & (dg)D & (ff^\prime)D\nonumber\\
\vdots & \vdots & \vdots\nonumber
\end{array}

In this case, since there are two "lowest-order" couplings to form a
D-state, both must be present in the basis set, i.e.

\begin{equation}
\Psi(r_2,r_2) = \sum c_{ijk}r_1^ir_2^{j+2}r_{12}^ke^{-\alpha r_1-\beta
r_2}\mathcal{Y}^M_{022}(\hat{r}_1,\hat{r}_2)
+\sum d_{ijk}r_1^{i+1}r_2^{j+1}r_{12^k}e^{-\alpha^\prime r_1 - \beta^\prime
r_2}\mathcal{Y}^M_{112}(\hat{r}_1,\hat{r}_2)\nonumber
\end{equation}

For F-states, one would need <math>(sf)F</math> and <math>(pd)F</math> terms.

For G-states, one would need <math>(sg)G</math>, <math>(pf)G</math> and <math>(dd^\prime)G</math> terms.

Completeness of the radial functions can be proven by considering the
Sturm-Liouville problem

\begin{equation}
\left(-\frac{1}{2}\nabla^2-\frac{\lambda}{r_s}-E\right)\psi({\bf r}) = 0\nonumber
\end{equation}

or

\begin{equation}
\left(-\frac{1}{2}\frac{1}{r^2}\frac{\partial}{\partial r}\left({r^2}\frac{\partial}{\partial r}\right) -
\frac{l\left(l+1\right)}{2r^2} - \frac{\lambda}{r} - E\right)u(r) = 0.\nonumber
\end{equation}

For fixed E and variable <math>\lambda</math> (nuclear charge).

The eigenvalues are <math>\lambda_n = (E/E_n)^{1/2}</math>, where <math>E_n =- \frac{1}{2n^2}</math>

[[File:Pic2.jpg|500px]]

and the eigenfunctions are

\begin{equation}
u_{nl}(r) =
\frac{1}{(2l+1)!}\left(\frac{(n+l)!}{(n-l-1)2!}\right)^{1/2}(2\alpha)^{3/2}e^{-\alpha
r}\nonumber
\times \ \left( 2 \alpha r \right)^l\,_{1}F_{1}\left(-n+l+1,2l+2; 2 \alpha r \right)
\end{equation}

with <math>\alpha = (-2E)^{1/2}</math> and <math>n\geq l+1</math>. The confluent hypergeometric function <math> _1F_1(a,b,;z)</math> then denotes a finite polynomial since <math>a=-n+l+1</math> is a negative integer or zero.

Unlike the hydrogen spectrum, which has both a discrete part for <math>E<0</math> and a
continuous part for <math>E>0</math>, this forms an entirely discrete set of finite
polynomials, called Sturmian functions. They are orthogonal with respect to
the potential, i.e.

\begin{equation}
\int^\infty_0 r^2dr \left( u_{n^\prime l}(r)\frac{1}{r}u_{nl}(r) \right) = \delta_{n,n^\prime}\nonumber
\end{equation}

Since they become complete in the limit <math>n\rightarrow\infty</math>, this assures the
completeness of the variational basis set.

[See also B Klahn and W.A. Bingel Theo. Chim. Acta (Berlin) 44, 9 and 27
(1977)].

==Solutions of the Eigenvalue Problem==

For convenience, write

\[
\Psi({\bf r}_1,{\bf r}_2) = \sum^N_{m=1}c_m\varphi_m\nonumber
\]

where <math>m</math> represents the <math>m</math>'th combination of <math>i,j,k</math> and

\begin{equation}
\varphi_{ijk}=r_1^ir_2^jr_{12}^ke^{-\alpha r_1 - \beta
r_2}\mathcal{Y}^M_{l_1,l_2,L}(\hat{r}_1,\hat{r}_2) \pm \mbox{exchange}\nonumber
\end{equation}

\begin{align}

&
\left( \begin{array}{lr}
\cos(\theta) & \sin(\theta)\\
-\sin(\theta) & \cos(\theta)
\end{array} \right)
\left( \begin{array}{lr}
H_{11} & H_{12} \\
H_{12} & H_{22}
\end{array}\right)
\left( \begin{array}{lr}
\cos(\theta) & -\sin(\theta)\\
\sin(\theta) & \cos(\theta)
\end{array} \right)
\\[5pt]
= & \left(\begin{array}{lr}
cH_{11}+sH_{12} & cH_{12} + sH_{22}\\
-sH_{11} + cH_{12} & -sH_{12} + cH_{22}
\end{array}\right)
\left( \begin{array}{cc}
c & -s \\
s & c
\end{array}\right)
\\[5pt]
= & \left(\begin{array}{lr}
c^2H_{11}+s^2H_{22} + 2csH_{12} & (c^2-s^2)H_{12}+cs(H_{22}-H_{11})\\
(c^2-s^2)H_{12}+cs(H_{22}-H_{11}) & s^2H_{11}+c^2H_{22}-2csH_{12}
\end{array}\right)
\end{align}

Therefore

\begin{equation}
\left[\cos^2(\theta)-\sin^2(\theta)\right]H_{12} =
\cos(\theta)\sin(\theta)(H_{11}-H_{22})\nonumber
\end{equation}

and

\begin{equation}
\tan(2\theta) = \frac{2H_{12}}{H_{11}-H_{22}}\nonumber
\end{equation}

i.e.

\begin{equation}
\cos(\theta)=\left(\frac{r+\omega}{2r}\right)^{1/2}\nonumber
\end{equation}

\begin{equation}
\sin(\theta)=-\mbox{sgn}(H_{12})\left(\frac{r-\omega}{2r}\right)^{1/2}\nonumber
\end{equation}

where

\begin{equation}
\omega = H_{22}-H_{11}\nonumber
\end{equation}

\begin{equation}
r=\left(\omega^2+4H_{12}^2\right)^{1/2}\nonumber
\end{equation}

\begin{equation}
E_1=\frac{1}{2}\left(H_{11}+H_{22}-r\right)\nonumber
\end{equation}

\begin{equation}
E_2=\frac{1}{2}\left(H_{11}+H_{22}+r\right)\nonumber
\end{equation}

===Brute Force Method===

- Gives all the eigenvalues and eigenvectors, but it is slow

- First orthonormalize the <math>N</math>-dimensional basis set; i.e. form linear combinations

\begin{equation}
\Phi_m = \sum_{n=1}^N\varphi_nR_{nm}\nonumber
\end{equation}

such that
\begin{equation}
\langle \Phi_m | \Phi_n \rangle = \delta_{m,n}
\end{equation}

This can be done by finding an orthogonal tranformation T such that

\begin{equation}
T^TOT = I = \left(
\begin{array}{ccccc}
I_1 & 0 & 0 & \ldots & 0 \\
0 & I_2 & 0 & \ldots & 0 \\
0 & 0 & I_3 & \ldots & 0 \\
\vdots & \vdots & \vdots & \ddots & \vdots \\
0 & 0 & 0 & \ldots & I_N
\end{array}\right);
\end{equation}

\begin{equation}
O_{mn} = \langle \varphi_m | \varphi_n \rangle \nonumber
\end{equation}

and then applying a scale change matrix

\begin{equation}
S = \left(\begin{array}{ccccc}
\frac{1}{I_1^{1/2}} & 0 & 0 & \ldots & 0\\
0 & \frac{1}{I_2^{1/2}} & 0 & \ldots & 0 \\
0 & 0 & \frac{1}{I_3^{1/2}} & \ldots & 0 \\
\vdots & \vdots & \vdots & \ddots & \vdots \\
0 & 0 & 0 & \ldots & I_N^{1/2}
\end{array}\right)= S^T\nonumber
\end{equation}

Then

\begin{equation}
S^TT^TOTS = 1
\end{equation}

i.e.

\begin{equation}
R^TOR = 1
\end{equation}

with <math>R=TS</math>.

If H is the matrix with elements <math>H_{mn} = \langle \varphi_m | \varphi_n \rangle </math>, then H
expressed in the <math>\Phi_m</math> basis set is

\begin{equation}
H^\prime = R^THR.\nonumber
\end{equation}

We next diagonalize <math>H^\prime</math> by finding an orthogonal transformation W such
that

\begin{equation}
W^T H^\prime W = \lambda = \left(
\begin{array}{cccc}
\lambda_1 & 0 & \ldots & 0 \\
0 & \lambda_2 & \ldots & 0 \\
\vdots & \vdots & \ddots & \vdots \\
0 & 0 & \ldots & \lambda_N
\end{array}\right)\nonumber
\end{equation}

The q'th eigenvector is

\begin{equation}
\Psi^{(q)} = \sum_{n=1}^N\Phi_n W_{n,q} = \sum_{n,n^\prime} \varphi_{n^\prime} R_{n^\prime ,n} W_{n,q} \nonumber
\end{equation}

i.e.

\begin{equation}
c_{n^\prime}^{(q)} = \sum_{n=1}^N R_{n^\prime n} W_{n,q}
\end{equation}

===The Power Method===

- Based on the observation that if H has one eigenvalue, <math>\lambda_M</math>, much bigger than all the rest, and <math>\chi = \left(
\begin{array}{c}a_1\\a_2\\\vdots\end{array}\right)</math>
is an arbitrary starting vector, then <math>\chi = \sum_{q=1}^N x_q\Psi^{(q)}</math>.

\begin{equation}
(H)^n\chi = \sum_{q=1}^N x_q \lambda^n_q\Psi^{(q)}\nonumber\rightarrow x_M\lambda_M^n\Psi^{(M)}\nonumber
\end{equation}

provided <math>x_M\neq 0</math>.

To pick out the eigenvector corresponding to any eigenvalue, write the original problem in the form

\begin{align}
H\Psi & = \lambda O \Psi \nonumber \\
\left( H - \lambda_q O \right) \Psi & = \left( \lambda - \lambda_q \right) O \Psi \nonumber
\end{align}

Therefore,

\begin{equation}
G \Psi = \frac{1}{\lambda - \lambda_q } \Psi \nonumber \\
\end{equation}

where <math>G=(H-\lambda_qO)^{-1}O</math> with eigenvalues <math>\frac{1}{\lambda_n - \lambda_q}</math>.

By picking <math>\lambda_q</math> close to any one of the <math>\lambda_n</math>, say
<math>\lambda_{n^\prime}</math>, then <math>\frac{1}{\lambda_n-\lambda_q}</math> is much larger for
<math>n=n^\prime</math> than for any other value. The sequence is then

\begin{array}{c}
\chi_1=G\chi\nonumber\\
\chi_2=G\chi_1\nonumber\\
\chi_3=G\chi_2\nonumber\\
\vdots\nonumber
\end{array}

until the ratios of components in <math>\chi_n</math> stop changing.

- To avoid matrix inversion and multiplication, note that the sequence is equivalent to
\begin{equation}
F \chi_n = \left( \lambda - \lambda_q \right) O \chi_{n-1} \nonumber \\
\end{equation}
where <math>F = H-\lambda_qO</math>. The factor of <math>\left( \lambda - \lambda_q \right)</math> can be
dropped because this only affects the normalization of <math>\chi_n</math>. To find <math>\chi_n</math>, solve
\begin{equation}
F \chi_n = O \chi_{n-1} \nonumber
\end{equation}
(N equations in N unknowns). Then
\begin{equation}
\lambda = \frac{ \langle \chi_n | H | \chi_n \rangle}{\langle \chi_n | \chi_n \rangle} \nonumber
\end{equation}

==Matrix Elements of H==

\begin{equation}
H=-\frac{1}{2}\nabla^2_1 -\frac{1}{2}\nabla^2_2 - \frac{1}{r_1} - \frac{1}{r_2} +\frac{Z^{-1}}{r_{12}}\nonumber
\end{equation}

Taking <math>r_1, r_2</math> and <math>r_{12}</math> as independent variables,

\begin{equation}
\nabla_1^2 = \frac{1}{r_1^2}\frac{\partial}{\partial r_1}\left(r_1^2\frac{\partial}{\partial r_1}\right)+ \frac{1}{r_{12}^2}\frac{\partial}{\partial r_{12}}
\left(r_{12}^2\frac{\partial}{\partial r_{12}}\right)\nonumber -\frac{l_1(l_1+1)}{r_1^2}+2(r_1 - r_2\cos(\theta_{12}))\frac{1}{r_{12}}\frac{\partial^2}{\partial r_1 \partial r_{12}} \nonumber - 2(\nabla_1^Y \cdot {\bf r}_2)\frac{1}{r_{12}}\frac{\partial}{\partial r_{12}}\nonumber
\end{equation}

where <math>\nabla_1^Y</math> acts only on the spherical harmonic part of the wave function and the diagram

[[File:Pic3.jpg|400px]]

defines the complete set of 6 independent variables is <math>r_1</math>, <math>r_2</math>, <math>r_{12}</math>, <math>\theta_1</math>, <math>\varphi_1</math>, <math>\chi</math>.

If <math>r_{12}</math> were not an independent variable, then one could take the volume element to be

\begin{equation}
d\tau = r_1^2dr_1\sin(\theta_1)d\theta_1d\varphi_1r_2^2dr_2\sin(\theta_2)d\theta_2d\varphi_2.\nonumber
\end{equation}

However, <math>\theta_2</math> and <math>\varphi_2</math> are no longer independent variables. To eliminate them, take the point <math>{\bf r}_1</math> as the
origin of a new polar co-ordinate system, and write

\begin{equation}
d\tau=-r_1^2dr_1\sin(\theta_1)d\theta_1d\varphi_1r_{12}^2dr_{12}\sin(\psi)d\psi d\chi\nonumber
\end{equation}

and use

\begin{equation}
r_2^2=r_1^2+r_{12}^2 +2r_1r_{12}\cos(\psi)
\end{equation}

Then for fixed <math>r_1</math> and <math>r_{12}</math>,

\begin{equation}
2r_2dr_2 = -2r_1r_{12}\sin(\psi)d\psi
\end{equation}

Thus

\begin{equation}
d\tau= r_1dr_1r_2dr_2r_{12}dr_{12}\sin(\theta_1)d\theta_1d\varphi_1d\chi
\end{equation}

The basic type of integral to be calculated is

\begin{align}
I \left( l_1, m_1, l_2, m_2; R \right) & = \int \sin \left( \theta_1 \right) d\theta_1 d\varphi_1 d\chi Y^{m_1}_{l_1} \left( \theta_1, \varphi_1 \right)^{*} Y^{m_2}_{l_2} \left( \theta_2, \varphi_2 \right) \\
& \times \int r_1dr_1r_2dr_2r_{12}dr_{12}R \left( r_1, r_2, r_{12} \right)
\end{align}

Consider first the angular integral. <math>Y^{m_2}_{l_2} \left( \theta_2, \varphi_2 \right)</math> can be expressed in terms of the independent variables <math>\theta_1, \varphi_1, \chi</math> by use of the rotation matrix relation
\begin{equation}
Y^{m_2}_{l_2}( \theta_2, \varphi_2) = \sum_m \mathcal{D}^{(l_2)}_{m_2,m}(\varphi_1,\theta_1,\chi)^*Y^m_{l_2}( \theta_{12}, \varphi)
\end{equation}
where <math>\theta, \varphi</math> are the polar angles of <math>{\bf r}_2</math> relative to <math>{\bf r}_1</math>. The angular integral is then

\begin{align}
I_{ang} & = \int^{2 \pi}_{0} d\chi \int^{2\pi}_{0} d\varphi_1 \int^{\pi}_{0} \sin \left( \theta_1 \right) d\theta_{1} Y^{m_1}_{l_1} \left( \theta_1, \varphi_1 \right)^{*} \\
& \times \sum_m \mathcal{D}^{(l_2)}_{m_2,m} \left( \varphi_1,\theta_1,\chi \right)^{*} Y^{m}_{l_2} \left( \theta_{12},\varphi \right) \\
\end{align}

Use

\begin{equation}
Y^{m_1}_{l_1} \left( \theta_1, \varphi_1 \right)^{*} = \sqrt{\frac{2l_1+1}{4\pi}}\mathcal{D}^{(l_1)}_{m_1,0} \left( \varphi_1, \theta_1, \chi \right)
\end{equation}

together with the orthogonality property of the rotation matrices (Brink and Satchler, p 147)

\begin{align}
& \int \mathcal{D}^{(j)*}_{m,m^\prime}\mathcal{D}^{(J)}_{M,M^\prime}\sin(\theta_1)d\theta_1d\varphi_1d\chi \\
& = \frac{8\pi^2}{2j+1}\delta_{jJ}\delta_{mM}\delta_{m^\prime M^\prime}\nonumber
\end{align}

to obtain

\begin{align}
I_{ang} & = \sqrt{ \frac{2l_1+1}{4\pi}} \frac{8\pi^2}{2l_1+1} \delta_{l_1,l_2} \delta_{m_1,m_2} Y^{0}_{l_2} \left( \theta_{12}, \varphi \right) \\
& = 2 \pi \delta_{l_1,l_2} \delta_{m_1,m_2} P_{l_2} \left( \cos \theta_{12} \right)
\end{align}

since

\begin{equation}
Y^0_{l_2} \left( \theta_{12}, \varphi \right) = \sqrt{\frac{2l_1+1}{4\pi}} P_{l_2} \left( \cos \left( \theta_{12} \right) \right)
\end{equation}

Note that <math>P_{l_2} \left( \cos \theta \right)</math> is just a short hand expression for a radial function because

\begin{equation}
\cos \theta_{12} = \frac{r_1^2 + r_2^2 - r_{12}^2}{2r_1r_2}
\end{equation}

The original integral is thus

\begin{equation}
I \left( l_1, m_1, l_2, m_2; R \right) = 2 \pi \delta_{l_1,l_2} \delta_{m_1,m_2} \int^{\infty}_{0} r_{1} dr_{1} \int^{\infty}_{0} r_{2} dr_{2} \int^{r_{1} + r_{2}}_{|r_{1} - r_{2}|} r_{12} dr_{12} R \left( r_{1}, r_{2}, r_{12} \right) P_{l_2} \left( \cos \theta_{12} \right)
\end{equation}

where again

\begin{equation}
\cos \theta_{12} = \frac{r_{1}^{2} +r^{2}_{2} - r_{12}^{2}}{2 r_{1} r_{2}}
\end{equation}

is a purely radial function.

The above would become quite complicated for large <math>l_2</math> because <math>P_{l_2}(\cos \theta_{12})</math> contains terms up to <math>\left( \cos \theta_{12} \right)^{l_2}</math>. However, recursion relations exist which allow any integral containing <math>P_l \left( \cos \theta_{12} \right)</math> in terms of those containing just <math>P_0 \left( \cos \theta_{12} \right) = 1</math> and <math>P_1 \left( \cos \theta_{12} \right) = \cos \theta_{12}</math>.

==Radial Integrals and Recursion Relations==

The basic radial integral is [see G.W.F. Drake, Phys. Rev. A 18, 820 (1978)]

\begin{align}
I_0 \left( a, b, c \right) & =
\int^{\infty}_{0} r_{1} dr_{1} \int^{\infty}_{r_{1}} r_{2} dr_{2} \int^{r_{1}+r_{2}}_{r_{2} - r_{1}} r_{12} dr_{12} r_{1}^{a} r_{2}^{b} r_{12}^{c} e^{-\alpha r_{1} - \beta r_{2}} + \int^{\infty}_{0} r_{2} dr_{2} \int^{\infty}_{r_{2}} r_{1} dr_{1} \int^{r_{1} + r_{2}}_{r_{1} - r_{2}} r_{12} dr_{12} r_{1}^{a} r_{2}^{b} r_{12}^{c} e^{- \alpha r_{1} - \beta r_{2}} \\
& = \frac{2}{c+2} \sum^{[(c+1)/2]}_{i=0}
\left(\begin{array}{c}
c+2 \\
2i+1
\end{array} \right)
\left\{ \frac{q!}{\beta^{q+1} \left( \alpha + \beta \right)^{p+1}} \sum_{j=0}^{q} \frac{\left( p + j \right)!}{j!} \left( \frac{\beta}{\alpha + \beta} \right)^j + \frac{q^{\prime}!}{\alpha^{q^{\prime}+1} \left( \alpha + \beta \right)^{p^{\prime}+1}} \sum^{q^{\prime}}_{j=0} \frac{\left( p^{\prime} + j \right)!}{j!} \left( \frac{\alpha}{\alpha + \beta} \right)^j \right\}
\end{align}

where

\begin{array}{ll}
p = a + 2i + 2 & p^{\prime} = b + 2i + 2 \\
q = b + c - 2i + 2 & q^{\prime} = a + c - 2i + 2
\end{array}

The above applies for <math> a, b \geq -2, c\geq -1</math>.
<math>[x]</math> means "greatest integer in" <math>x</math>.

Then

\begin{align}
I_1 \left( a, b, c \right) & = \int d\tau_{r} r_{1}^{a} r_{2}^{b} r_{12}^{c} e^{ -\alpha r_{1} - \beta r_{2}} P_{1}\left( \cos \theta \right) \\
& = \frac{1}{2} \left( I_{0} \left( a + 1, b - 1, c \right) + I_{0} \left( a - 1, b + 1, c \right) - I_{0} \left( a - 1, b - 1, c + 2 \right) \right)
\end{align}

===The Radial Recursion Relation===

Recall that the full integral is

\begin{equation}
I(l_1m_1,l_2m_2;R)=2\pi\delta_{l_1,l_2}\delta_{m_1,m_2}I_{l_2}(R)
\end{equation}

where

\begin{equation}
I_{l_2}(R) = \int d\tau_r R(r_1,r_2,r_{12})P_{l_2}(\cos \theta_{12})
\end{equation}

To obtain the recursion relation, use

\begin{equation}
P_l(x) = \frac{[P^{\prime}_{l+1}(x)-P^{\prime}_{l-1}(x)]}{2l+1}
\end{equation}

with

\begin{equation}
P_{l+1}^\prime (x) = \frac{d}{dx}P_{l+1}(x)
\end{equation}

Here <math>x=\cos\theta_{12}</math> and

\begin{align}
\frac{d}{d \cos \theta_{12}} & = \frac{dr_{12}}{d \cos \theta_{12}} \frac{d}{dr_{12}} \\
& = -\frac{r_{1} r_{2}}{r_{12}} \frac{d}{dr_{12}}
\end{align}

Then,

\begin{equation}
I_{l}\left( R \right) = - \int d\tau_r R \frac{r_{1} r_{2}}{r_{12}} \frac{d}{dr_{12}} \frac{ \left[ P_{l+1} \left( \cos \theta_{12} \right) - P_{l-1} \left( \cos \theta_{12} \right) \right]}{2l+1}
\end{equation}

The <math>r_{12}</math> part of the integral is

\begin{align}
& \int^{r_{1} + r_{2}}_{|r_{1} - r_{2}|} r_{12} dr_{12} R \frac{r_{1} r_{2}}{r_{12}} \frac{d}{dr_{12}} \left[P_{l+1}-P_{l-1} \right] \\
& = R r_{1} r_{2} \left[ P_{l+1} - P_{l-1} \right]^{r_{1} + r_{2}}_{|r_{1} - r_{2}|} - \int^{r_{1} + r_{2}}_{|r_{1} - r_{2}|} r_{12} dr_{12} \left( \frac{d}{dr_{12}} R \right) \frac{r_{1} r_{2}}{r_{12}} \frac{ \left[ P_{l+1} - P_{l-1} \right]}{2l+1}
\end{align}

The integrated term vanishes because

\begin{equation}
\cos \theta_{12} = \frac{r_{1}^{2} + r_{2}^{2} - r_{12}^{2}}{2 r_{1} r_{2}}
\begin{array}{lll}
& = -1 & {\rm when} & r_{12}^{2} = \left( r_{1} + r_{2} \right)^{2} \\
& = 1 & {\rm when} & r_{12}^{2} = \left( r_{1} - r_{2} \right)^{2}
\end{array}
\end{equation}

and <math>P_l(1)=1</math>, <math>P_{l}(-1)=(-1)^{l}</math>.

Thus,

\begin{equation}
I_{l+1}\left(\frac{r_1r_2}{r_{12}}R^\prime\right) = \left( 2l+1 \right) I_{l}(R) - I_{l-1} \left( \frac{r_{1} r_{2}}{r_{12}} R^{\prime} \right)
\end{equation}

If

\begin{equation}
R = r_{1}^{a-1} r_{2}^{b-1} r_{12}^{c+2} e^{-\alpha r_{1} - \beta r_{2}}
\end{equation}

then [G.W.F. Drake, Phys. Rev. A 18, 820 (1978)]

\begin{equation}
I_{l+1}(r_1^ar_2^br_{12}^c) =
\frac{2l+1}{c+2} I_{l} \left( r_{1}^{a-1} r_{2}^{b-2} r_{12}^{c+2} \right)+I_{l-1} \left( r_{1}^{a} r_{2}^{b} r_{12}^{c} \right)
\end{equation}

For the case <math>c=-2</math>, take

\begin{equation}
R = r_{1}^{a-1} r_{2}^{b-1} \ln r_{12} e^{ -\alpha r_{1} -\beta r_{2}}
\end{equation}

Then

\begin{equation}
I_{l+1} \left( r_{1}^{a} r_{2}^{b} r_{12}^{c} \right) = I_{l} \left( r_{1}^{a-1} r_{2}^{b-1} \ln r_{12} \right) \left( 2l + 1 \right) - I_{l-1} \left( r_{1}^{a} r_{2}^{b} r_{12}^{-2} \right)
\end{equation}

This allows all <math>I_{l}</math> integrals to be calculated from tables of <math>I_0</math> and
<math>I_1</math> integrals.

===The General Integral===

The above results for the angular and radial integrals can now be combined into a general formula for integrals of the type

\begin{equation}
I = \int \int d{\bf r}_{1} d{\bf r}_{2} R_{1} \mathcal{Y}^{M^{\prime}}_{l^{\prime}_{1} l^{\prime}_{2} L^{\prime}} \left( \hat{r}_{1}, \hat{r}_{2} \right) T_{k_{1} k_{2} K}^{Q} \left( {\bf r}_{1},{\bf r}_{2} \right) R_{2} \mathcal{Y}_{l_{1} l_{2} L}^{M} \left( \hat{r}_{1}, \hat{r}_{2} \right)
\end{equation}

where

\begin{equation}
\mathcal{Y}^{M}_{l_{1} l_{2} L} \left( \hat{r}_{1}, \hat{r}_{2} \right) = \sum_{m_1,m_2} \langle l_{1} l_{2} m_{1} m_{2} | LM \rangle Y^{m_{1}}_{l_{1}} \left( \hat{r}_{1} \right) Y^{m_{2}}_{l_{2}} \left( \hat{r}_{2} \right)
\end{equation}

and

\begin{equation}
T^Q_{k_{1} k_{2} K} \left( {\bf r}_{1}, {\bf r}_{2} \right) = \sum_{q_{1}, q_{2}} \langle k_{1} k_{2} q_{1} q_{2} | KQ \rangle Y_{k_{1}}^{q_{1}} \left( \hat{r}_{1} \right) Y^{q_{2}}_{k_{2}} \left( \hat{r}_{2} \right)
\end{equation}

The basic idea is to make repeated use of the formula

\begin{equation}
Y^{m_{1}}_{l_{1}} \left( \hat{r}_{1} \right) Y^{m_{2}}_{l_{2}} \left( \hat{r}_{1} \right) = \sum_{l m} \left( \frac{\left( 2l_{1}+1 \right) \left(2l_{2}+1 \right) \left(2l+1 \right)}{4 \pi} \right)^{1/2}
\times \left( \begin{array}{ccc}
l_1 & l_2 & l \\
m_1 & m_2 & m
\end{array} \right) \left(\begin{array}{ccc}
l_1 & l_2 & l \\
0 & 0 & 0
\end{array} \right)
Y^{m*}_{l} \left( \hat{r}_{1} \right)
\end{equation}

where

\begin{equation}
Y^{m*}(\hat{r}) = (-1)^mY_l^{-m}(\hat{r})
\end{equation}

and

\begin{equation}
\left(\begin{array}{ccc}
l_1 & l_2 & l \\
m_1 & m_2 & m
\end{array} \right) = \frac{\left( -1 \right)^{l_{1} - l_{2} - m}}{\left( 2l + 1 \right)^{1/2}} \left( l_{1} l_{2} m_{1} m_{2} | l, -m \right)
\end{equation}

is a 3-<math>j</math> symbol. In particular, write

\begin{array}{cc}
Y^{m_1^\prime}_{l_1^\prime}(\hat{r}_1)^{*} \underbrace{Y_{k_{1}}^{q_{1}} \left( \hat{r}_{1} \right) Y^{m_{1}}_{l_{1}} \left( \hat{r}_{1} \right)} & = & \sum_{\Lambda M}(\ldots)Y^{M}_\Lambda(\hat{r}_1)\\
\ \ \ \ \ \ \ \sum_{\lambda_1\mu_1}Y_{\lambda_1}^{\mu_1}(\hat{r}_1) & & \\
& & \\
Y^{m_{2}^{\prime}}_{l_{2}^{\prime}} \left( \hat{r}_{2} \right) \underbrace{Y^{q_{2}}_{k_{2}} \left( \hat{r}_{2} \right) Y_{l_{2}}^{m_{2}} \left( \hat{r}_{2} \right)} & = & \sum_{\Lambda^{\prime} M^{\prime}} \left( \ldots \right)Y_{\Lambda^{\prime}}^{M^{\prime}*} \left( \hat{r}_{2} \right) \\
\ \ \ \ \ \ \ \ \ \sum_{\lambda_{2} \mu_{2}}Y_{\lambda_{2}}^{\mu_{2}} \left( \hat{r}_{2} \right) & &
\end{array}

The angular integral then gives a factor of <math>2 \pi \delta_{\Lambda, \Lambda^{\prime}} \delta_{M, M^{\prime}} P_{\Lambda} \left( \cos \theta_{12} \right)</math>. The total integral therefore reduces to the form

\begin{equation}
I = \sum_{\Lambda} C_{\Lambda} I_{\Lambda} \left( R_{1} R_{2} \right)
\end{equation}

where <math>C_{\Lambda} = \sum_{\lambda_{1}, \lambda_{2}} C_{ \lambda_{1}, \lambda_{2}, \Lambda}</math>. For further details and derivations, including graphical representations, see G.W.F. Drake, Phys. Rev. A <math>{\bf 18}</math>, 820 (1978).

==Graphical Representation==

[[File:GraphRep1.jpg|500px]]

==Matrix Elements of H==
Recall that

\begin{equation}
H = -\frac{1}{2} \nabla^{2}_{1} - \frac{1}{2} \nabla^{2}_{2} - \frac{1}{r_1} - \frac{1}{r_2} + \frac{Z^{-1}}{r_{12}}
\end{equation}

Consider matrix elements of

\begin{equation}
\nabla^2_1 = \frac{1}{r_1^2} \frac{\partial}{\partial r_1}\left(r_1^2\frac{\partial}{\partial r_1} \right) + \frac{1}{r^2}\frac{\partial}{\partial r}\left(r^2\frac{\partial}{\partial r}\right) - \frac{(\vec{l}_1^Y)^2}{r_1^2} + \frac{2 \left( r_1 - r_2 \cos \theta \right)}{r} \frac{\partial^2}{\partial r_1 \partial r} - 2 \left( \nabla^Y_1 \cdot {\bf r}_2 \right) \frac{1}{r} \frac{\partial}{\partial r}
\end{equation}

where <math>r \equiv r_{12}</math> and <math>\cos\theta \equiv \cos\theta_{12}</math>. Also in what follows, define <math>\hat{\nabla}^Y_1 \equiv r_1\nabla^Y_1</math> where <math>\nabla^Y</math> operates only on the spherical harmonic part of the wave function, and similarly for <math>\vec{l}_1^Y</math>.

A general matrix element is

\begin{equation}
\langle r_1^{a^\prime} r_2^{b^\prime} r_{12}^{c^\prime} e^{-\alpha^\prime r_1 - \beta^\prime r_2} \mathcal{Y}^{M^\prime}_{l^\prime_1 l^\prime_2 L^\prime} \left( \hat{r}_1, \hat{r}_2 \right) \left| \nabla^2_1 \right| r_1^a r_2^b r_{12}^c e^{-\alpha r_1 - \beta r_2} \mathcal{Y}^M_{l_1 l_2 L} \left( \hat{r}_1, \hat{r}_2 \right) \rangle
\end{equation}

Since <math>\nabla^2_1</math> is rotationally invariant, this vanishes unless <math>L=L^\prime</math>, <math>M=M^\prime</math>. Also <math>\nabla_1^2</math> is Hermitian, so that the result must be the same whether it operates to the right or left, even though the results look very different. In fact, requiring the results to be the same yields some interesting and useful integral identities as follows:

Put

\begin{equation}
\mathcal{F} = F \mathcal{Y}^M_{l_1 l_2 L} \left( \hat{r}_1, \hat{r_2} \right)
\end{equation}

and

\begin{equation}
F = r_1^a r_2^b r_{12}^c e^{-\alpha r_1 - \beta r_2}
\end{equation}

Then

\begin{equation}
\nabla^2_1 \mathcal{F} = \left\{ \frac{1}{r_1^2} \left[ a \left( a + 1 \right) - l_1 \left( l_1 + 1 \right) \right] + \frac{c \left( c + 1 \right)}{r^2} + \alpha^2\frac{2\alpha \left( a + 1 \right)}{r_1} + \frac{2 \left(r_1 - r_2 \cos \theta \right) }{r_1r^2}c \left[ a - \alpha r_1 \right] - \frac{2c}{r^2} \left( \hat{\nabla}^Y_1 \cdot \hat{r}_2 \right) \frac{r_2}{r_1} \right\} \mathcal{F}
\end{equation}

and

\begin{align}
\langle \mathcal{F}^\prime \left| \nabla^2_1 \right| \mathcal{F} \rangle & = \sum_\Lambda \int d\tau_r F^\prime C_\Lambda \left( 1 \right) P_\Lambda \left( \cos \theta \right) \times \left\{ \frac{1}{r_1^2} \left[ a \left( a + 1 \right) - l_1 \left( l_1 + 1 \right) \right] -\frac{2 \alpha \left( a + 1 \right)}{r_1} + \frac{c \left( c + 1 \right)}{r^2} +\alpha^2 + \frac{2 \left( r_2 - r_2 \cos \theta \right)}{r_1r^2}c \left( a - \alpha r_1 \right) \right\} F \\
& + \sum_\Lambda \int d\tau_r F^\prime C_\Lambda \left( \hat{\nabla}^Y_1 \cdot \hat{r}_2 \right) P_\Lambda \left( \cos \theta \right) \left( \frac{-2c r_2}{r_1 r^2} \right) F
\end{align}

where

\begin{equation}
\int d\tau_r = \int^\infty_0 r_1 dr_1 \int^\infty_0 r_2 dr_2 \int^{r_1+r_2}_{\left| r_1 - r_2 \right|} r dr
\end{equation}

For brevity, let the sum over <math>\Lambda</math> and the radial integrations be understood, and let <math>\left( \nabla_1^2 \right)</math> stand for the terms which appear in the integrand. Then operating to the right gives

\begin{equation}
(\nabla_1^2)_R = \frac{1}{r_1^2} \left[ a \left( a + 1 \right) - l_1 \left( l_1 + 1 \right) \right] + \frac{c \left( c + 1 \right)}{r^2} + \alpha^2 - \frac{2 \alpha \left( a + 1 \right)}{r_1}
+ \frac{2 \left( r_1 - r_2 \cos \theta \right)}{r_1 r^2}c \left( a - \alpha r_1 \right) - \frac{2c r_2}{r_1 r^2} \left( \hat{\nabla}^Y_1 \cdot \hat{r}_2 \right)
\end{equation}

Operating to the left gives

\begin{equation}
(\nabla_1^2)_L = \frac{1}{r_1^2} \left[a^\prime \left( a^\prime + 1 \right) - l_1^\prime \left( l_1^\prime + 1 \right) \right] + \frac{c^\prime \left( c^\prime + 1 \right)}{r^2} + \alpha^{\prime 2} -\frac{2 \alpha^\prime \left( a^\prime + 1 \right)}{r_1}
+ \frac{2 \left( r_1 - r_2 \cos \theta \right)}{r_1 r^2} c^\prime \left( a^\prime - \alpha^\prime r_1 \right) - \frac{2 c^\prime r_2}{r_1 r^2} \left( \hat{\nabla}^{Y^\prime}_1 \cdot \hat{r}_2 \right)
\end{equation}

Now put

\begin{array}{cc}
a_+ = a+a^\prime, & \hat{\nabla}_1^+ = \hat{\nabla}_1^Y + \hat{\nabla}_1^{Y\prime} \\
a_{-} = a-a^\prime, & \hat{\nabla}_1^{-} = \hat{\nabla}_1^Y - \hat{\nabla}_1^{Y\prime}
\end{array}

etc., and substitute <math>a^\prime = a_+ - a</math>, <math>c^\prime = c_+ - c</math>, and <math>\alpha^\prime = \alpha_+ - \alpha</math> in <math>(\nabla^2_1)_L</math>. If <math>a_+</math>, <math>c_+</math> and <math>\alpha_+</math> are held fixed, then the equation

\begin{equation}
(\nabla^2_1)_R = (\nabla^2_1)_L
\end{equation}

must be true for arbitrary <math>a</math>, <math>c</math>, and <math>\alpha</math>. Their coefficients must thus vanish.

This yields the integral relations

\begin{equation}
\label{I}
\frac{\left( r_1 - r_2 \cos \theta \right)}{r_1 r^2} = \frac{1}{c_+} \left( \frac{- \left( a_+ + 1 \right)}{r_1^2} + \frac{\alpha_+}{r_1} \right) \tag{I}
\end{equation}

from the coefficient of <math>a</math> and

\begin{equation}
\label{II}
\frac{ \left( r_1 - r_2 \cos \theta \right) \left( a_+ - \alpha_+ r_1 \right)}{r_1 r^2} = \frac{r_2}{r_1 r^2}\left( \hat{r}_2 \cdot \hat{\nabla}_1^+ \right) - \frac{ \left( c_+ + 1 \right)}{r^2} \tag{II}
\end{equation}

from the coefficient of <math>c</math>. The coefficient of <math>\alpha</math> gives an equation equivalent to (I).

Furthermore, if can be show that (see problem)

\begin{align}
& \sum_\Lambda \int d\tau_r \frac{r^c}{r^2} C_\Lambda \left( \hat{r}_2 \cdot \hat{\nabla}_1^Y \right) P_\Lambda \left( \cos \theta \right) \\
= & \sum_\Lambda \int d\tau_r \frac{r^c}{c r_1 r_2}C_\Lambda \left( 1 \right)P_\Lambda \left( \cos \theta \right) \times \left( \frac{l_1^\prime \left( l_1^\prime + 1 \right) - l_1 \left( l_1 + 1 \right) -\Lambda \left( \Lambda + 1 \right)}{2} \right)
\end{align}

and similarly for <math>\left( \hat{r}_2 \cdot \hat{\nabla}^{Y\prime}_1 \right)</math> with <math>l_1</math> and <math>l_1^\prime</math> interchangeable, then it follows that

\begin{equation}
\frac{r^{c_+}}{r^2} \left( \hat{r}_2 \cdot \hat{\nabla}_1^+ \right) = - \frac{r^{c_+}}{c_+ r_1 r_2} \Lambda \left( \Lambda + 1 \right)
\end{equation}

and

\begin{equation}
\frac{r^{c_+}}{r^2} \left( \hat{r}_2 \cdot \hat{\nabla}_1^- \right) = \frac{r^{c_+}}{c_+ r_1 r_2} \left[ l_1^\prime \left( l_1^\prime + 1 \right) - l_1 \left( l_1 + 1 \right) \right]
\end{equation}

where equality applies after integration and summation over <math>\Lambda</math>.

Thus (II) becomes

\begin{equation}\label{IIa}
\frac{ \left( r_1 - r_2 \cos \theta \right) \left( a_+ - \alpha_+ r_1 \right)}{r_1 r^2} = - \frac{\Lambda \left( \Lambda + 1 \right)}{c_+ r^2_1} - \frac{ \left( c_+ + 1 \right)}{r^2} \tag{IIa}
\end{equation}

===Problem===

Prove the integral relation

\begin{align}
& \sum_\Lambda \int d\tau_r f \left( r_1, r_2 \right) \frac{1}{r} \left( \frac{d}{dr} q(r) \right) C_\Lambda \left( \hat{r}_2 \cdot \hat{\nabla}^Y_1 \right) P_\Lambda \left( \cos \theta \right) \\
= & \sum_\Lambda \int d\tau_r \frac{ f \left( r_1, r_2 \right)}{r_1, r_2} q(r) C_\Lambda \left( 1 \right) P_\Lambda \left( \cos \theta \right) \left(\frac{l_1^\prime \left( l_1^\prime + 1 \right) - l_1 \left( l_1 + 1 \right) - \Lambda \left( \Lambda + 1 \right)}{2}\right)
\end{align}

where the coefficients <math>C_\Lambda \left( 1 \right)</math> are the angular coefficients from the overlap integral

\begin{align}
\int & d\Omega\, \mathcal{Y}^{M*}_{l_1^\prime l_2^\prime L}(\hat{r}_1,\hat{r}_2)\, \mathcal{Y}^{M}_{l_1 l_2 L}(\hat{r}_1,\hat{r}_2)\\
= & \sum_\Lambda C_\Lambda(1)P_\Lambda(\cos\theta_{12}).
\end{align}

Hint: Use the fact that <math>l_1^2</math> is Hermitian so that
<math>
\int d\tau (l_1^2\mathcal{Y}^\prime)^*q(r)\mathcal{Y} = \int d\tau
\mathcal{Y}^{\prime *}l_1^2(q(r)\mathcal{Y})
</math>
with <math>\ \vec{l}_1= \frac{1}{i}\vec{r}_1\times\nabla_1</math>.

It is also useful to use
<math>\begin{eqnarray}\ \
\cos\theta\, P_L(\cos\theta) &=& \frac{1}{2L+1}[L\,P_{L-1}(\cos\theta) + (L+1)P_{L+1}(\cos\theta)]
\\
(\cos^2\theta-1)P_L(\cos\theta) &=&
\frac{L(L-1)}{(2L-1)(2L+1)}P_{L-2}(\cos\theta)\nonumber
-\frac{2(L^2+L-1)}{(2L-1)(2L+3)}P_L(\cos\theta) + \frac{(L+1)(L+2)}{(2L+1)(2L+3)}P_{L+2}(\cos\theta)\nonumber\\
&=& \frac{L(L-1)}{(2L-1)(2L+1)}[P_{L-2}(\cos\theta) - P_L(\cos\theta)]
+ \frac{(L+1)(L+2)}{(2L+1)(2L+3)}[P_{L+2}(\cos\theta) - P_L(\cos\theta)]
\end{eqnarray}</math>

together with a double application of the integral recursion relation
<math>
I_{L+1}\left(\frac{1}{r}\frac{d}{dr}q(r)\right) =
(2L+1)I_L\left(\frac{1}{r_1r_2}q(r)\right)+I_{L-1}\left(\frac{1}{r}\frac{d}{dr}q(r)\right)
</math>

Of course <math>l_1^2\mathcal{Y}^M_{l_1l_2L}(\hat{r}_1,\hat{r}_2)=l_1(l_1+1)\mathcal{Y}^M_{l_1l_2L}(\hat{r}_1,\hat{r}_2)</math>. Begin by expanding
<math>
l_1^2(g\mathcal{Y}) = \mathcal{Y}l_1^2g + g\,l_1^2\mathcal{Y} + 2(\vec{l}_1g)\cdot(\vec{l}_1\mathcal{Y})
</math>, and show that

<math>
\begin{eqnarray}\ \
(\vec{l}_1g)\cdot(\vec{l}_1\mathcal{Y}) & = & \frac{r_1^2}{r}\frac{dg}{dr}\vec{r}_2\cdot \nabla_1\mathcal{Y}\,\,\,\,\mbox{ and}\\
l_1^2g(r) &=& 2r_1r_2\cos\theta\frac{1}{r}\frac{dg}{dr} + r_1^2r_2^2(\cos^2\theta - 1)
\frac{1}{r}\frac{d}{dr}\left(\frac{1}{r}\frac{dg}{dr}\right).
\end{eqnarray}
</math>

The proof amounts to showing that the term <math>l_1^2g(r)</math> can be replaced by
<math>\Lambda(\Lambda+1)g(r)</math> after multiplying by <math>P_\Lambda(\cos\theta)</math>
and integrating by parts with respect to the radial integrations over <math>r_1</math>, <math>r_2</math> and
<math>r \equiv |\vec{r}_1 - \vec{r}_2|</math>. Remember that <math>\cos\theta</math> is just a short-hand notation for the radial function <math>(r_1^2 + r_2^2 - r^2)/(2r_1r_2)</math>.

==General Hermitian Property==

Each combination of terms of the form
\begin{equation}
<f>_R = a^2f_1+af_2+abf_3+b^2f_4+bf_5+af_6\nabla_1^Y+bf_7\nabla^Y_2+f_8(Y)
\end{equation}
can be rewritten
\begin{equation}
<f>_L= (a_+-a)^2f_1+(a_+-a)f_2 + (a_+-a)(b_+-b)f_3+(b_+-b)^2f_4
+(b_+-b)f_5+(a_+-a)f_6\nabla^{Y\prime}_1+(b_+-b)f_7\nabla^{Y\prime}_2+f_8)Y^\prime
\end{equation}
where as usual <math>\nabla_1^Y</math> acts only on the spherical harmonic part of the wave function denoted for short by <math>Y</math>.
Since these must be equal for arbitrary <math>a</math> and <math>b</math>,
\begin{equation}
a_+^2f_1+a_+f_2+_+b_+f_3+b_+f_4+b_+f_5+a_+f_6\nabla^{Y\prime}_1 +
b_+f_7\nabla^{Y\prime}_2+f_8(Y^\prime)-f_8(Y)=0
\end{equation}

Adding the corresponding expression with <math>Y</math> and <math>Y^\prime</math> interchanged
yields
<math>\label{III}
a_+^2f_1+a_+f_2+a_+b_+f_3+b_+^2f_4+b_+f_5+\frac{1}{2}a_+f_6\nabla^+_1+\frac{1}{2}b_+f_7\nabla^+_2=0\tag{III}
</math>

Subtracting gives
<math>\label{IV}
f_8(Y)-f_8(Y^\prime)=-\frac{1}{2}[a_+f_6\nabla^-_1+b_+f_7\nabla_2^-]\tag{IV}
</math>
<math>\label{V}
a[-2a_+f_1-2f_2-b_+f_3-f_6\nabla^+_1]=0\tag{V}
</math>
<math>\label{VI}
b[-2b_+f_4-2f_5-a_+f_3-f_7\nabla^+_2]=0\tag{VI}
</math>

Adding the two forms gives

\begin{eqnarray}
<f>_R+<f>_L=\frac{1}{2}(a_+^2+a_-^2)f_1+a_+f_2+\frac{1}{2}(a_+b_++a_-b_-)f_3\nonumber\\
\ +\frac{1}{2}(b_+^2+b_-^2)f_4+b_+f_5+\frac{1}{2}f_6(a_+\nabla_1^++a_-\nabla^-_1)\nonumber\\
\ +\frac{1}{2}(b_+\nabla_2^++b_-\nabla^-_2)+f_8(Y)+f_8(Y^\prime)\nonumber
\end{eqnarray}

Subtracting <math>\frac{x}{2}\times({\rm I})</math> gives

\begin{eqnarray}\ \
<f>_R+<f>_L = \frac{1}{2}[(1-x)a_+^2 + a_-^2]f_1 +
(1-\frac{x}{2})a_+f_2+\frac{1}{2}[(1-x)a_+b_++a_-b_-]f_3\nonumber\\
+\frac{1}{2}[(1-x)b_+^2+b_-^2]f_4+(1-\frac{x}{2})f_5b_++\frac{1}{2}f_6[(1-\frac{x}{2})a_+\nabla_1^++a_-\nabla^-_1]\nonumber\\
+\frac{1}{2}[(1-\frac{x}{2})b_+\nabla^+_2+b_-\nabla^-_2]+f_8(Y)+f_8(Y^\prime)\nonumber
\end{eqnarray}

If <math>x=1</math>,
\begin{eqnarray}
<f>_R+<f>_L &=&
\frac{1}{2}[a_-^2f_1+a_+f_2+a_-b_-f_3+b_-^2f_4+b_+f_5\nonumber\\
&+&\left(\frac{1}{2}a_+\nabla^+_1+a_-\nabla^-_1\right)f_6+\left(\frac{1}{2}b_+\nabla^+_2+b_-\nabla^-_2\right)f_7]+f_8(Y)+f_8(Y^\prime)\nonumber
\end{eqnarray}

The General Hermitian Property for arbitrary <math>x</math> gives
\begin{eqnarray}\ \
\nabla_1^2=
\frac{1}{4}\{\frac{1}{r_1^2}[(1-x)a_+^2+a_-^2+2(1-\frac{x}{2})a_+-2[l_1(l_1+1)+l_1^\prime(l_1^\prime+1)]\nonumber\\
-\frac{2}{r_1}[(1-x)\alpha_+a_++\alpha_-a_-+2(1-\frac{x}{2})\alpha_+]+(1-x)\alpha_+^2
+ \alpha^2_-]\nonumber\\
+\frac{2(r_1-r_2\cos\theta)}{r_1r^2}[(1-x)(a_+-\alpha_+r_1)c_++(a_--\alpha_-r_2)c_-]\nonumber\\
-\frac{2r_2}{r_1r^2}[(1-\frac{x}{2})c_+\hat{r}_2\cdot\hat{\nabla}_1^++c_-\hat{r}_2\hat{\nabla}_1^-]+\frac{(1-x)c_+^2+c_-^2+2(1-\frac{x}{2})c_+}{r^2}\}\nonumber
\end{eqnarray}.

Use
<math>\begin{eqnarray}\ \
\frac{-2r_2}{r_1r^2}[(1-\frac{x}{2})c_+\hat{r}_2\hat{\nabla}_1^++c_-\hat{r}_2\hat{\nabla}_1^-]=\frac{2}{r_1^2}\left[(1-\frac{x}{2})\Lambda(\Lambda+1)-\frac{c_-}{c_+}[l_1^\prime(l_1^\prime+1)+l_1(l_1+1)]\right]\end{eqnarray}</math>,

<math>\begin{eqnarray}
\frac{2(r_1-r_2\cos\theta)}{r_1r^2}(a_--\alpha_-r_1)c_- = 2\frac{c_-}{c_+}\left[\frac{-1}{r_1^2}[a_-(a_++1)]+\frac{1}{r_1}[a_-\alpha_++\alpha_-(a_++2)]-\alpha_-\alpha_+\right]\nonumber
\end{eqnarray}</math>,

and
<math>\begin{eqnarray}\ \
\frac{2(r_1-r_2\cos\theta)}{r_1r^2}(a_+-\alpha_+r_1)c_+ = 2\left[-\frac{1}{r_1^2}[a_+(a_++1)]+\frac{1}{r_1}[a_+\alpha_++\alpha_+(a_++2)]-\alpha_+^2\right]
\end{eqnarray}</math>.

Substituting into <math>\nabla_1^2</math> gives

\begin{eqnarray}
\nabla^2_1 = \frac{1}{4}\{\frac{1}{r_1^2}[-(1-x)a_+^2 + a_-^2 +xa_+ +
2(1-\frac{x}{2})\Lambda(\Lambda+1)\nonumber\\
-2l_1(l_1+1)(1-\frac{c_-}{c_+}) -
2l_1^\prime(l_1^\prime+1)(1+\frac{c_-}{c_+})-\frac{2c_-a_-}{c_+}(a_++1)]\nonumber\\
- \frac{2}{r_1}\left[-(1-x)\alpha_+(a_++2)+\alpha_-a_-+2(1-\frac{x}{2})\alpha_+
-\frac{c_-}{c_+}[a_-\alpha_++\alpha_-(a_++2)]\right]\nonumber\\
-(1-x)\alpha_+^2+\alpha_-^2-2\frac{c_-}{c_+}\alpha_-\alpha_+ +\frac{1}{r^2}\left[(1-x)c_+^2+c_-^2+2(1-\frac{x}{2})c_+\right]\}\nonumber
\end{eqnarray}

This has the form
<math>
\nabla_1^2 =
\frac{1}{4}\left[\frac{A_1}{r_1^2}+\frac{B_1}{r_1}+\frac{C_1}{r^2}+D_1\right]
</math>
with

<math>\begin{array}{ccc}
A_1=-(1-x)a_+^2+a_-^2
+xa_++2(1-\frac{x}{2})\Lambda(\Lambda+1)-2l_1(l_1+1)(1-\frac{c_-}{c_+})\
-2l_1^\prime(l_1^\prime+1)(1+\frac{c_-}{c_+})-\frac{2c_-a_-}{c_+}(a_++1) \ \
\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \
\end{array}
</math>

<math>
B_1 =
2\left[(1-x)\alpha_+(a_++2)-\alpha_-a_--2(1-\frac{x}{2})\alpha_++\frac{c_-}{c_+}[a_-\alpha_++\alpha_-(a_++2)]\right]
</math>

<math>
C_1 = (1-x)c_+^2+ c_-^2 +2(1-\frac{x}{2})c_+
</math>

<math>
D_1 = -1(1-x)\alpha_+^2 +\alpha_-^2 - 2\frac{c_-}{c_+}\alpha_-\alpha_+
</math>

The complete Hamiltonian is then (in Z-scaled a.u.)

<math>\begin{eqnarray}
H =-\frac{1}{2}\nabla^2_1 -\frac{1}{2}\nabla^2_2 -\frac{1}{r_1} -\frac{1}{r_2} +\frac{Z^{-1}}{r}\nonumber\\
= -\frac{1}{2}\nabla^2_1 -\frac{1}{2}\nabla^2_2 -\frac{1}{r_1} -\frac{Z-1}{Zr_2}+Z^{-1}\left(\frac{1}{r}-\frac{1}{r_2}\right)\nonumber\\
\ =-\frac{1}{8}\left[\frac{A_1}{r_1^2}+\frac{B_1+8}{r_1}+\frac{C_1}{r^2}+D_1+D_2 +\frac{A_2}{r_2^2} + \frac{B_2 + 8(Z-1)/Z}{r_2}+\frac{C_2}{r^2}\right]\nonumber
+ Z^{-1}\left(\frac{1}{r}-\frac{1}{r_2}\right)\nonumber
\end{eqnarray}</math>

The screened hydrogenic energy is
<math>\begin{eqnarray}
E_{SH}=-\frac{1}{2}\left[\frac{1}{n_1^2}+ \left(\frac{Z-1}{Z}\right)^2\frac{1}{n_2^2}\right]
\end{eqnarray}</math>
so that

<math>\begin{eqnarray}
H-E_{SH} &=& -\frac{1}{8}[
\frac{A_1}{r_1^2}+\frac{B_1+8}{r_1}+\frac{C_1}{r^2} + D_1 + D_2
-\frac{4}{n_1^2} -\left(\frac{Z-1}{Z}\right)^2\frac{4}{n_2^2}\nonumber
+\frac{A_2}{r_2}+\frac{B_2+8(Z-1)/Z}{r_2}+\frac{C_2}{r^2}] +
Z^{-1}\left(\frac{1}{r} -\frac{1}{r_2}\right)\nonumber
\end{eqnarray}</math>

==Optimization of Nonlinear Parameters==
- The traditional method of performing Hylleraas calculations is to write the
basis set in the form

<math>
\Psi = \sum_{i,j,k}c_{i,j,k}\varphi_{i,j,k}(\alpha,\beta)\pm exchange
</math>

with

<math>
\varphi_{i,j,k}(\alpha,\beta) = r_1^ir_2^jr_{12}^ke^{-\alpha r_1 - \beta
r_2}\mathcal{Y}^M_{l_1l_2L}.
</math>

The usual procedure is to set <math>\alpha = Z</math> so that it represents the inner
<math>1s</math> electron, and then to vary <math>\beta</math> so as to minimize the energy.

[[File:OptNon.jpg|500px]]

Since <math>\beta</math> appears in <math>\Psi</math> as a non-linear parameter, the entire
calculation must be repeated for each value of <math>\beta</math>. However, the minimum
becomes progressively smaller as the basis set is enlarged.

[[File:OptNon02.jpg|500px]]

*Difficulties

1. If the basis set is constructed so that <math> i +j +j \leq N</math>, the the
number of terms is
<math>
(N+1)(N+2)(N+3)/6
</math>
N=14 already gives 680 terms and an accuracy of about 1 part in <math>10^{10}</math> for
low-lying states. A substantial improvement in accuracy would require much
larger basis sets, together with multiple precision arithmetic to avoid loss
of significant figures when high powers are included.

2. The accuracy rapidly deteriorates as one goes to more highly excited
states - about 1 significant figure is lost each time the principle quantum
number is increased.

*Cure

We have found that writing basis sets in the form
\begin{eqnarray}
\Psi &=& \sum_{i,j,k}\left(c^{(1)}_{i,j,k}\varphi_{ijk}(\alpha_1\beta_1) +
c_{ijk}^{(2)}\varphi_{ijk}(\alpha_2\beta_2)\right) \pm exchange \nonumber\\
&=& \psi({\bf r}_1,{\bf r}_2) \pm \psi({\bf r}_2,{\bf r}_1)\nonumber
\end{eqnarray}
so that each combination of powers is included twice with different nonlinear
parameters gives a dramatic improvement in accuracy for basis sets of about
the same total size.

However, the optimization of the nonlinear parameters is now much more
difficult, and an automated procedure is needed.

If
<math>
E = \frac{<\Psi|H|\Psi>}{<\Psi|\Psi>}</math> and we assume <math><\Psi|\Psi> = 1</math>, then

<math>
\frac{\partial E}{\partial\alpha_t} = -2 <\Psi|H-E|r_1\psi({\bf r}_1,{\bf
r}_2;\alpha_t)\pm r_21\psi({\bf r}_2,{\bf
r}_1;\alpha_t)>
</math>
where
<math>
\psi({\bf r}_1,{\bf r}_2;\alpha_t) =
\sum_{i,j,k}c_{ijk}^{(t)}\varphi_{ijk}(\alpha_t\beta_t)
</math>

*Problem

1. Prove the above.

2. Prove that there is no contribution from the implicit dependence of
<math>c_{ijk}^{(t)}</math> on <math>\alpha_t</math> if the linear parameters have been optimized.

==The Screened Hydrogenic Term==

If the Hamiltonian is written in the form (in Z-scaled a.u.)
<math>\begin{eqnarray}\
H = H_0({\bf r}_1,Z) + H_0({\bf r}_2,Z-1) +
Z^{-1}\left(\frac{1}{r}-\frac{1}{r_2}\right)
\end{eqnarray}</math>
with
<math>\
H_0({\bf r}_1,Z) = -\frac{1}{2}\nabla_1^2 - \frac{1}{r_1}\nonumber
</math> and
<math>
H_0({\bf r}_2,Z-1) = -\frac{1}{2}\nabla_2^2 - \frac{Z-1}Z{r_2}\nonumber
</math>,
then the eigenvectors of <math>H_0({\bf r},Z) +H_0({\bf r}_2,Z-1)</math> are products of
hydrogenic orbitals
<math>
\Psi_0=\psi_0(1s,Z)\psi_0(nl,Z-1)
</math>
and the eigenvalue is
<math>\begin{eqnarray}
E_{SH} = \left[-\frac{1}{2} - \left(\frac{Z-1}{Z}\right)^2\frac{1}{2n^2}\right]Z^2 a.u.
\end{eqnarray}</math>
called the screened hydrogenic eigenvalue. From highly excited states,
<math>E_{SH}</math> and <math>\Psi_0</math>are already excellent approximations.

For example, for the 1s8d states, the energies are

<math>\begin{eqnarray}\\
E(1s8d^1D)&=& -2.007\,816\,512\,563\,81 {\rm\ a.u.}\nonumber\\
E(1s8d^3D)&=& - 2.007\,817\,934\,711\,71 {\rm\ a.u.}\nonumber\\
E_{SH} &=& -2.007\,812\,5 {\rm\ a.u.} \nonumber
\end{eqnarray}</math>

It is therefore advantageous to include the screened hydrogenic terms in the
basis set so that the complete trial function becomes
<math>
\Psi = c_0\Psi_0 +
\sum_{ijk}\left[c_{ijk}^{(1)}\varphi_{ijk}(\alpha_1,\beta_1)+c_{ijk}^{(2)}\varphi(\alpha_2,\beta_2)\right]\pm
{\rm exchange}
</math>

Also, the variational principal can be re-expressed in the form
<math>\begin{eqnarray}
E= E_{SH}+\frac{<\Psi|H-E_{SH}|\Psi>}{<\Psi|\Psi>}
\end{eqnarray}</math>
so that the <math>E_{SH}</math> term can be cancelled analytically from the matrix
elements.

For example
<math>
<\Psi_0|H-E_{SH}|\Psi_0> = 0
</math>

and
<math>
<\varphi_{ijk}|H-E_{SH}|\Psi_0> =
Z^{-1}<\varphi_{ijk}|\frac{1}{r}-\frac{1}{r_2}|\Psi_0>
</math>

Recall that
<math>\begin{eqnarray}\ \
I_0(a,b,c) = \frac{2}{c+2}\sum^{[c+1]/2}_{i=0}\left(\begin{array}{c}
c+2\\
2i+1
\end{array}\right) \left[f(p,q;\beta) + f(p^\prime,q^\prime;\alpha)\right]
\end{eqnarray}</math>
where

<math>\begin{eqnarray}
f(p,q;x) =
\frac{q!}{x^{q+1}(\alpha+\beta)^{p+1}}\sum^p_{j=0}\frac{(p+j)!}{j!}\left(\frac{x}{\alpha+\beta}\right)^j
\end{eqnarray}</math>

<math>
\begin{array}{cc}
p=a+2i+2 & p^\prime = b+2i+2\\
q= b+c-2i+2 & q^\prime =a+c-2i+2
\end{array}
</math>

If <math>\beta\ll\alpha</math>, the <math>f(p,q;\beta)\gg f(p^\prime,q^\prime;\alpha)</math> and the
<math>i=0</math> term is the dominant contribution to <math>f(p,q;\beta)</math>. However, since
this term depends only on the sum of powers <math>b+c</math> for <math>r_2</math>, it cancels exactly
from the matrix element of
<math>
\frac{1}{r_{12}} - \frac{1}{r_2}
</math>
and can therefore
be omitted in the calculations of integrands, thereby saving many significant
figures. This is especially valuable when <math>b+c</math> is large and <math>a</math> is small.

==Small Corrections==
*Mass Polarization
If the nuclear mass is not taken to be infinite, then the Hamiltonian is
<math>\begin{eqnarray}\ \
H = \frac{P_N^2}{2M} + \sum_{i=1}\frac{p_i^2}{2m} + V
\end{eqnarray}</math>

[[File:SmallCor.jpg|500px]]

where <math>M =M_A-nm</math> is the nuclear mass and <math>m</math> is the electron mass.

Change variables to the C of M
<math>\begin{eqnarray}
{\bf R} = \frac{1}{M+nm}\left[M{\bf r}_N + m({\bf r}_1 + {\bf r}_2 +
\ldots)\right]
\end{eqnarray}</math>

and relative variables

<math>
{\bf s}_i = {\bf r}_i - {\bf r}_N.
</math>

Then

<math>\begin{eqnarray}\ \ \ \ \
H = \frac{1}{2(M+nm)}{\bf p}_R^2+\frac{1}{2m}\sum_{i=1}^n{\bf p}_{s_i}^2
+\frac{1}{2M}\sum_{i,k}{\bf p}_{s_i}\cdot {\bf p}_{s_k} + V({\bf s}_1, {\bf
s}_2, \ldots)\nonumber
= \frac{1}{M+nm}{\bf p}_R^2+\frac{1}{2\mu}\sum_{i=1}^n{\bf
p}_{s_i}^2+\frac{1}{2M}\sum_{i\neq k}{\bf p}_{s_i}\cdot {\bf p}_{s_k}+V({\bf s}_1, {\bf
s}_2, \ldots)\nonumber
\end{eqnarray}</math>

where
<math>\begin{eqnarray}
\mu = \left(\frac{1}{M} + \frac{1}{m}\right)^{-1}
\end{eqnarray}</math>

If <math>n=2</math>, the Schrödinger equation is
<math>\begin{eqnarray}\ \ \
\left[ -\frac{\hbar^2}{2\mu}(\nabla^2_{s_1}+ \nabla^2_{s_2}) -
\frac{\hbar^2}{M}\nabla_{s_1}\nabla_{s_2}-\frac{Ze^2}{s_1}-\frac{Ze^2}{s_2}+\frac{e^2}{s_{12}}\right]\psi
= E\psi
\end{eqnarray}</math>.

Define the reduced mass Bohr radius
<math>\begin{eqnarray}
a_\mu = \frac{\hbar^2}{\mu e^2} = \frac{m}{\mu}a_0
\end{eqnarray}</math>
and

<math>\begin{eqnarray}
\rho_i = \frac{Zs_i}{a_\mu}\nonumber\\
\varepsilon = \frac{E}{Z^2(e^2/a_\mu)}\nonumber
\end{eqnarray}</math>

The Schrödinger equation then becomes
<math>\begin{eqnarray}\ \ \
\left[
-\frac{1}{2}(\nabla^2_{\rho_1}+\nabla^2_{\rho_2})-\frac{\mu}{M}\vec{\nabla}_{\rho_1}\cdot\vec{\nabla}_{\rho_2} - \frac{1}{\rho_1}-\frac{1}{\rho_2}+\frac{Z^{-1}}{\rho_{12}}\right]\psi = \varepsilon \psi
\end{eqnarray}</math>

The effects of finite mass are thus

1. A reduced mass correction to all energy levels of <math>(e^2/a_\mu)/(e^2/a_0)
= \mu/m</math>, i.e.
<math>\begin{eqnarray}
E_M=\frac{\mu}{m}E_\infty
\end{eqnarray}</math>

2. A specific mass shift given in first order perturbation theory by
<math>\begin{eqnarray}
\Delta\varepsilon =
-\frac{\mu}{M}<\psi|\nabla_{\rho_1}\cdot\nabla_{\rho_2}|\psi>
\end{eqnarray}</math>
or

<math>\begin{eqnarray}
\Delta\varepsilon =
-\frac{\mu}{M}<\psi|\nabla_{\rho_1}\cdot\nabla_{\rho_2}|\psi>\frac{e^2}{a_\mu}\nonumber\\
= -\frac{\mu^2}{mM}<\psi|\nabla_{\rho_1}\cdot\nabla_{\rho_2}|\psi>\frac{e^2}{a_0}\nonumber
\end{eqnarray}</math>

Since <math>\nabla_{\rho_1}\cdot\nabla_{\rho_2}</math> has the same angular properties as
<math>{\bf\rho}_1 \cdot {\bf\rho}_2</math>, the operator is like the product of two
dipole operators. For product type wave functions of the form
$
\psi = \psi(1s)\psi(nl) \pm \mbox{exchange},
$
the matrix element vanishes for all but P-states.

=Radiative Transitions=

==Transition Integrals and the Line Strength==

For an electric dipole transition <math>\gamma'\,L'\,S'\,J'\,M' \rightarrow \gamma\,L\,S\,J\,M</math>, the line strength is defined by
\begin{equation}
S = \sum_{M',\mu,M}|\langle\gamma'\,L'\,S'\,J'\,M'|r_\mu|\gamma\,L\,S\,J\,M\rangle|^2
\end{equation}
Using the Wigner-Eckart theorem, this is related to the reduced matrix element by
\begin{eqnarray}
S(J',J) = |\langle\gamma'\,L'\,S'\,J'||r||\gamma\,L\,S\,J\,M\rangle|^2
\sum_{M',\mu,M}\left|\left(\matrix{J' & 1 & J \cr
-M'& \mu &M \cr}\right)\right|^2
\end{eqnarray}
Using the sum rule for the 3-<math>j</math> symbols
\begin{equation}
\sum_{M',\mu,M}\left|\left(\matrix{J' & 1 & J \cr
-M'& \mu &M \cr}\right)\right|^2 = 1
\end{equation}
we see that the line strength is identical to the reduced matrix element
\begin{equation}
S(J',J) = |\langle\gamma'\,L'\,S'\,J'||r||\gamma\,L\,S\,J\,M\rangle|^2
\end{equation}
For a nonrelativistic calculation in <math>LS</math> coupling, <math>S = S'</math> and we can next strip out the dependence on <math>J</math> and <math>J'</math> by use of Eq.\ (7.1.7) of Edmonds { Angular Momentum in Quantum Mechanics} with <math>k = 1</math> to obtain
\begin{equation}
S(J',J) = (-1)^{L'+S+J+1}[(2J+1)(2J'+1)]^{1/2}\left\{
\matrix{L' & J'& S\cr
J & L & 1\cr}\right\}\langle\gamma' L'|| r || \gamma L\rangle
\end{equation}
where <math>\langle\gamma' L'|| r || \gamma L\rangle</math> is the reduced matrix element if spin is neglected. Note that if spin is neglected, then the line-strength is simply
\begin{equation}
S(L',L) = |\langle\gamma' L'|| r || \gamma L\rangle|^2
\end{equation}
To calculate the reduced matrix element, use a particular example along with the Wigner-Eckart theorem
such as
\begin{equation}
\langle\gamma'\, L'|| r || \gamma\, L\rangle = (-1)^{L'}\langle\gamma'\, L'\, 0| z | \gamma\, L\, 0\rangle{\Large /}\left(\matrix{L' & 1 & L \cr
0 & 0 & 0 \cr}\right)
\end{equation}

===Correlated Two-electron Integrals===
For the case of two-electron integrals, <math>z = z_1 + z_2</math> and <math>r = r_1 + r_2</math>. As discussed in G.W.F. Drake, Phys. Rev. A 18, 820 (1978), the general integral in Hylleraas coordinates for the integral over the <math>z_1</math> term can be written in the form
\begin{equation}
\label{int1}
\langle\gamma'\, L'\, 0| z_1 | \gamma\, L\, 0\rangle = \sum_\Lambda C_\Lambda I_\Lambda(\gamma',\gamma)
\end{equation}
where <math>I_\Lambda(\gamma',\gamma)</math> stands for the purely radial integral
\begin{equation}
\label{radial}
I_\Lambda(\gamma',\gamma) = \int_0^\infty r_1\,dr_1\int_0^\infty r_2\,dr_2\int_{|r_1-r_2|}^{r_1+r_2}r_{12}\, dr_{12}
R_{\gamma'}^*\,r_1\, R_\gamma P_\Lambda(\cos\theta_{12})
\end{equation}
and <math>\cos\theta_{12} = (r_1^2 + r_2^2 - r_{12}^2)/(2r_1r_2)</math> is a short-hand notation for a radial function of <math>r_1</math>, <math>r_2</math> and <math>r_{12}</math>. Also, <math>P_\Lambda(\cos\theta_{12})</math> is a Legendre polynomial,
and, <math>R_\gamma = R_\gamma(r_1,r_2,r_{12})</math> is the radial part of the Hylleraas wave function for state <math>\gamma</math> such
that the total wave function is <math>\psi({\bf r}_1,{\bf r}_2) = R_\gamma(r_1,r_2,r_{12}){\cal Y}_{l_1,l_2,L}^M(\hat{\bf r}_1,\hat{\bf r}_2)\pm {\rm exchange}</math>, normalized to unity, and <math>{\cal Y}_{l_1,l_2,L}^M(\hat{\bf r}_1,\hat{\bf r}_2)</math> is a vector coupled product of spherical harmonics. The coefficents <math>C_\Lambda</math> in Eq.\ (\ref{int1}) can be calculated from Eq.\ (32) of the above 1978 paper. They need to be calculated separately for he direct-direct and the direct-exchange contributions. For the reduced matrix element of <math>z_1</math>, the leading 3-<math>j</math> symbol can be dropped, leaving (for the case <math>k = 1</math>)
\begin{eqnarray}
C_\Lambda^{\rm reduced} &=& \frac{(-1)^{l_2'+L'+L+1}}{2} (l_1,l_1',l_2,l_2',L,L')^{1/2}
\sum_{\lambda={\rm max}(l_1-1,0)}^{l_1+1} (\lambda,\Lambda,1)
\left(\matrix{l_1 & 1 & \lambda \cr
0 & 0 & 0 \cr}\right)
\left(\matrix{l_1'& \lambda & \Lambda \cr
0 & 0 & 0 \cr}\right)\nonumber\\
&\times&\left(\matrix{l_2' & 1_2 & \Lambda \cr
0 & 0 & 0 \cr}\right)
\left\{\matrix{L' & \lambda & l_2 \cr
\Lambda& l_2' & l_1' \cr}\right\}
\left\{\matrix{L' & \lambda & l_2 \cr
l_1 & L & 1 \cr}\right\}
\end{eqnarray}
where the notation <math>(a,b,c,\cdots)</math> means <math>(2a+1)(2b+1)(2c+1)\cdots</math>.
A similar expression can be obtained for the angular coefficients for the <math>z_2</math> term by interchanging the subscripts 1 and 2 throughout and multiplying by <math>(-1)^{L+L'+1}</math>.

===Example===
As an example, for the <math>1s^2\;^1S - 1s2p\;^1P</math> transition of helium, the only nonvanishing <math>C_\Lambda</math> angular coefficients are <math>C_1 = 1/2</math> for the direct-direct term (i.e. <math>P_1(\cos\theta_{12}) = \cos\theta_{12}</math> is included in the radial integral (\ref{radial})) and <math>C_0 = 1/2</math> for the direct-exchange term (i.e. <math>P_0(\cos\theta_{12}) = 1</math> is included in the radial integral (\ref{radial})). The results for infinite nuclear mass are
\begin{equation}
S(1\;^1S-2\;^1P) = (0.781\,127\,158\,667)^2
\end{equation}
\begin{equation}f(1\;^1S-2\;^1P) = -0.276\,164\,703\,736
\end{equation}
where
\begin{equation}
f(1\;^1S-2\;^1P) = \frac{2}{3}[E(1\;^1S) - E(2\;^1P)]S(1\;^1S-2\;^1P)
\end{equation}
is the absorption oscillator strength.

===Test===
This is a test edit

Theory Notes

2020-12-14T03:46:04Z

Drake: just making sure editing works - Mike Busuttil

<center>
<H1>Notes on solving the Schrödinger equation in Hylleraas coordinates for heliumlike atoms
Gordon W.F. Drake
Department of Physics, University of Windsor
Windsor, Ontario, Canada N9B 3P4
(Transcribed from hand-written notes and edited by Lauren Moffatt. Last revised November 22, 2017 to correct a missing subscript in the expression for $B_1$ in the Hamiltonian in Sect. 11 General Hermitian Property.)
</H1>
</center>

==Introduction==
These notes describe some of the technical details involved in solving the Schrödinger equation for a heliumlike atom of nuclear charge Z in correlated Hylleraas coordinates. The standard for computational purposes is to express all quantities in a form that is valid for electronic states of arbitrary angular momentum, and to express matrix elements of the Hamiltonian in a manifestly Hermitian symmetrized form. Further information and derivations of the mathematical techniques for angular integrals and radial recursion relations can be found in the paper
G.W.F. Drake, Phys. Rev. A <math>{\bf 18}</math>, 820 (1978).

The starting point is the two-electron Schrödinger equation for infinite nuclear mass
\begin{equation} \left[-\frac{\hbar^2}{2m}(\nabla^2_1 +\nabla^2_2) - \frac{Ze^2}{r_1} -
\frac{Ze^2}{r_2}+\frac{e^2}{r_{12}} \right]\psi = E\psi\nonumber
\end{equation}
where <math>m</math> is the electron mass, and <math>r_{12} = |{\bf r}_1 - {\bf r}_2|</math> (see diagram below).

[[File:Pic1.jpg|250px]]

Begin by rescaling distances and energies so that the Schrödinger equation can be expressed in a dimensionless form. The dimensionless Z-scaled distance is defined by <math>\begin{eqnarray} \rho = \frac{Zr}{a_0} \end{eqnarray} </math> where <math> \begin{eqnarray} a_0 = \frac{\hbar^2}{me^2} \end{eqnarray} </math> is the atomic unit (a.u.) of distance equal to the Bohr radius <math>0.529\, 177\, 210\, 92(17) \times 10^{-10}</math> m. Then

\begin{equation}[-\frac{\hbar^2}{2m}Z^2(\frac{me^2}{\hbar^2})^2(\nabla^2_{\rho_1}+\nabla^2_{\rho_2})
- Z^2\frac{e^2}{a_0}\rho^{-1}_1 - Z^2\frac{e^2}{a_0}\rho^{-1}_2 +
\frac{e^2}{a_0}Z\rho^{-1}_{12}]\psi= E\psi\nonumber\end{equation}

But <math>\begin{eqnarray} \frac{\hbar^2}{m}(\frac{me^2}{\hbar^2})^2 = \frac{e^2}{a_0} \end{eqnarray} </math> is the hartree atomic unit of of energy <math>(E_h = 27.211\,385\,05(60)</math> eV, or equivalently <math>E_h/(hc) = 2194.746\,313\,708(11)</math> m<math>^{-1})</math>. Therefore, after multiplying through by <math>a_0/(Ze)^2</math>, the problem to be solved in Z-scaled dimensionless units becomes

\begin{equation} \left[-\frac{1}{2}(\nabla^2_{\rho_1}+\nabla^2_{\rho_2}) -
\frac{1}{\rho_1} - \frac{1}{\rho_2} + \frac{Z^{-1}}{\rho_{12}}\right]\psi =
\varepsilon\psi\nonumber \end{equation}

where <math>\begin{eqnarray}\varepsilon = \frac{Ea_0}{(Ze)^2} \end{eqnarray}</math> is the energy in Z-scaled atomic units. For convenience, rewrite this in the conventional form
\begin{equation}
H\psi = \varepsilon\psi\nonumber \end{equation}
where (using <math>r</math> in place of <math>\rho</math> for the Z-scaled distance)
\begin{equation}
H = -\frac{1}{2}(\nabla^2_1+\nabla^2_2) - \frac{1}{r_1}-\frac{1}{r_2} + \frac{Z^{-1}}{r_{12}}
\nonumber\end{equation}
is the atomic Hamiltonian for infinite nuclear mass.

==The Hartree Fock Method==

For purposes of comparison, and to define the correlation energy, assume that <math>\psi({\bf r}_1,{\bf r}_2)</math> can be written in the separable product form
\begin{equation}
\psi({\bf r}_1,{\bf r}_2) = \frac{1}{\sqrt{2}} \left[ u_1(r_1) u_2(r_2) \pm u_2(r_1) u_1(r_2) \right] \nonumber
\end{equation}
for the <math>1s^2\ ^1\!S</math> ground state. Because of the <math>\frac{1}{r_{12}}</math> term in the Schrödinger equation, the exact solution connot be expressed in this form as a separable product. However, the Hartree-Fock approximation corresponds to finding the best possible solution to the Schrödinger equation
\begin{equation}
H\psi({\bf r}_1,{\bf r}_2) = E\psi({\bf r}_1,{\bf r}_2)\nonumber
\end{equation}
that can nevertheless be expressed in this separable product form, where as before
\begin{equation}
H = -\frac{1}{2}(\nabla^2_1+\nabla^2_2) - \frac{1}{r_1}- \frac{1}{r_2} + \frac{Z^{-1}}{r_{12}}\nonumber
\end{equation}
is the full two-electron Hamiltonian. To find the best solution, substitute into <math>\left\langle\psi \lvert H-E \rvert \psi\right\rangle</math> and require this expression to be stationary
with respect to arbitrary infinitesimal variations <math>\delta u_1</math> and <math>\delta
u_2</math> in <math>u_1</math> and <math>u_2</math> respectively; i.e.

\begin{equation}
\frac{1}{2} \left\langle \delta u_1(r_1)u_2(r_2) \pm u_2(r_1)\delta
u_1(r_2) \lvert H-E \rvert u_1(r_1) u_2(r_2)\pm u_2(r_1) u_1(r_2)\right\rangle\nonumber
\end{equation}

\begin{equation}
=\int\delta u_1(r_1) d{\bf r}_1 \left\{ \int d{\bf r}_2 u_2(r_2) \left( H - E \right) \left[ u_1(r_1)u_2(r_2) \pm u_2(r_1) u_1(r_2) \right] \right\}\nonumber
\end{equation}

\begin{equation}
= 0 \nonumber
\end{equation}

for arbitrary <math> \delta u_1(r_1).\nonumber </math> Therefore <math>\{\int d{\bf r}_2 \ldots \} = 0</math>.

Similarly, the coefficient of <math>\delta u_2</math> would give

\begin{equation}
\int d{\bf r}_1 u_1(r_1) \left( H-E \right) \left[ u_1(r_1) u_2(r_2) \pm
u_2(r_1) u_1(r_2) \right] = 0\nonumber
\end{equation}

Define

\begin{equation}
I_{12} = I_{21} = \int d{\bf r}\, u_1(r)u_2(r), \nonumber
\end{equation}

\begin{equation}
H_{ij} = \int d{\bf r}\, u_i(r)(-\frac{1}{2}\nabla - \frac{1}{r})u_j(r), \nonumber
\end{equation}

\begin{equation}
G_{ij}(r) = \int d{\bf r}^\prime u_i(r^\prime)\frac{1}{|{\bf r} - {\bf r}\prime|}u_j(r^\prime)\nonumber
\end{equation}

Then the above equations become the pair of integro-differential equations

\begin{equation}
[ H_0 - E + H_{22}+G_{22}(r)]u_1(r) = \mp [ I_{12}(H_0-E) + H_{12}+G_{12}(r)]u_2(r)\nonumber
\end{equation}

\begin{equation}
[H_0-E+H_{11}+G_{11}(r)]u_2(r) = \mp [I_{12}(H_0-E) + H_{12}+G_{12}(r)]u_1(r)\nonumber
\end{equation}

These must be solved self-consistently for the "constants" <math>I_{12}</math> and
<math>H_{ij}</math> and the function <math>G_{ij}(r)</math>.

The Hartree Fock energy is <math>E \simeq -2.87\ldots </math>a.u. while the exact energy is <math>E = -2.903724\ldots </math>a.u.

The difference is called the "correlation energy" because it arises from the
way in which the motion of one electron is correlated to the other. The
Hartree Fock equations only describe how one electron moves in the average field
provided by the other (mean-field theory).

==Configuration Interaction==
Expand
\begin{equation}
\psi({\bf r}_1,{\bf r}_2)= C_0u^{(s)}_1(r_1)u^{(s)}_1(r_2) + C_1u^{(P)}_1({\bf r}_1)u^{(P)}_1({\bf r}_2){\cal Y}^0_{1,1,0}(\hat{\bf r}_1, \hat{\bf r}_2)+C_2u^{(d)}_1({\bf r}_1)u^{(d)}_2({\bf r}_2){\cal Y}^0_{2,2,0}(\hat{\bf r}_1, \hat{\bf r}_2)+... \pm {\rm\ exchange}
\end{equation}
where
\begin{equation}
{\cal Y}^M_{l_1,l_2,L}(\hat{\bf r}_1, \hat{\bf r}_2)=\sum_{m_1,m_2}Y^{m_1}_{l_1}({\bf r}_1)Y^{m_2}_{l_2}({\bf r}_2)\times <l_1l_2m_1m_2\mid LM>
\end{equation}

This works, but is slowly convergent, and very laborious. The best CI calculations are accurate to <math> ~10^{-7}</math> a.u.

==Hylleraas Coordinates==
[E.A. Hylleraas, Z. Phys. <math>{\bf 48}, 469(1928)</math> and <math>{\bf 54}, 347(1929)</math>]
suggested using the co-ordinates <math>r_1</math>, <math>r_2</math> and <math>r_{12}</math> or equivalently

\begin{eqnarray}
s &=& r_1 + r_2, \nonumber\\
t &=& r_1-r_2, \nonumber\\
u &=& r_{12}\nonumber
\end{eqnarray}

and writing the trial functions in the form

\begin{equation}
\Psi({\bf r}_1,{\bf r}_2) = \sum^{i+j+k\leq
N}_{i,j,k}c_{i,j,k}r_1^{i+l_1}r_2^{j+l_2}r_{12}^ke^{-\alpha r_1 - \beta r_2}
\mathcal{Y}^M_{l_1,l_2,L}(\hat{r}_1,\hat{r}_2)\pm \text{exchange}\nonumber
\end{equation}

Diagonalizing H in this non-orthogonal basis set is equivalent to solving
<math> \begin{eqnarray}
\frac{\partial E}{\partial c_{i,j,k}} = 0\nonumber
\end{eqnarray} </math>
for fixed <math>\alpha</math> and <math>\beta</math>.

The diagonalization must be repeated for different values of <math>\alpha</math> and
<math>\beta</math> in order to optimize the non-linear parameters.

==Completeness==
The completeness of the above basis set can be shown by first writing
\begin{equation}
r_{12}^2 = r_1^2 + r_2^2 - 2r_1r_2\cos\theta_{12}
\end{equation}
and, from the spherical harmonic addition theorem,
\begin{equation}
\cos(\theta_{12})=\frac{4\pi}{3}\sum^1_{m=-1}Y^{m*}_1(\theta_1,\varphi_1)Y^m_1(\theta_2,\varphi_2)
\end{equation}
Consider first the S-states. The <math>r_{12}^0</math> terms are like the ss terms in a CI calculation. The
<math>r_{12}^2</math> terms bring in p-p type contributions, and the higher powers bring
in d-d, f-f etc. type terms. In general

\begin{equation}
P_l(\cos\theta_{12}) = \frac{4\pi}{2l+1}\sum^l_{m=-l}{Y^{m}_l}^*(\theta_1,\varphi_1)Y^m_l(\theta_2,\varphi_2)\nonumber
\end{equation}

For P-states, one would have similarly

\begin{array}{lr}
r_{12}^0 & (sp)P\nonumber\\
r_{12}^2 & (pd)P\nonumber\\
r_{12}^4 & (df)P\nonumber\\
\vdots & \vdots\nonumber
\end{array}

For D-states

\begin{array}{lr}
r_{12}^0 & (sd)D & (pp^\prime)D\nonumber\\
r_{12}^2 & (pf)D & (dd^\prime)D\nonumber\\
r_{12}^4 & (dg)D & (ff^\prime)D\nonumber\\
\vdots & \vdots & \vdots\nonumber
\end{array}

In this case, since there are two "lowest-order" couplings to form a
D-state, both must be present in the basis set, i.e.

\begin{equation}
\Psi(r_2,r_2) = \sum c_{ijk}r_1^ir_2^{j+2}r_{12}^ke^{-\alpha r_1-\beta
r_2}\mathcal{Y}^M_{022}(\hat{r}_1,\hat{r}_2)
+\sum d_{ijk}r_1^{i+1}r_2^{j+1}r_{12^k}e^{-\alpha^\prime r_1 - \beta^\prime
r_2}\mathcal{Y}^M_{112}(\hat{r}_1,\hat{r}_2)\nonumber
\end{equation}

For F-states, one would need <math>(sf)F</math> and <math>(pd)F</math> terms.

For G-states, one would need <math>(sg)G</math>, <math>(pf)G</math> and <math>(dd^\prime)G</math> terms.

Completeness of the radial functions can be proven by considering the
Sturm-Liouville problem

\begin{equation}
\left(-\frac{1}{2}\nabla^2-\frac{\lambda}{r_s}-E\right)\psi({\bf r}) = 0\nonumber
\end{equation}

or

\begin{equation}
\left(-\frac{1}{2}\frac{1}{r^2}\frac{\partial}{\partial r}\left({r^2}\frac{\partial}{\partial r}\right) -
\frac{l\left(l+1\right)}{2r^2} - \frac{\lambda}{r} - E\right)u(r) = 0.\nonumber
\end{equation}

For fixed E and variable <math>\lambda</math> (nuclear charge).

The eigenvalues are <math>\lambda_n = (E/E_n)^{1/2}</math>, where <math>E_n =- \frac{1}{2n^2}</math>

[[File:Pic2.jpg|500px]]

and the eigenfunctions are

\begin{equation}
u_{nl}(r) =
\frac{1}{(2l+1)!}\left(\frac{(n+l)!}{(n-l-1)2!}\right)^{1/2}(2\alpha)^{3/2}e^{-\alpha
r}\nonumber
\times \ \left( 2 \alpha r \right)^l\,_{1}F_{1}\left(-n+l+1,2l+2; 2 \alpha r \right)
\end{equation}

with <math>\alpha = (-2E)^{1/2}</math> and <math>n\geq l+1</math>. The confluent hypergeometric function <math> _1F_1(a,b,;z)</math> then denotes a finite polynomial since <math>a=-n+l+1</math> is a negative integer or zero.

Unlike the hydrogen spectrum, which has both a discrete part for <math>E<0</math> and a
continuous part for <math>E>0</math>, this forms an entirely discrete set of finite
polynomials, called Sturmian functions. They are orthogonal with respect to
the potential, i.e.

\begin{equation}
\int^\infty_0 r^2dr \left( u_{n^\prime l}(r)\frac{1}{r}u_{nl}(r) \right) = \delta_{n,n^\prime}\nonumber
\end{equation}

Since they become complete in the limit <math>n\rightarrow\infty</math>, this assures the
completeness of the variational basis set.

[See also B Klahn and W.A. Bingel Theo. Chim. Acta (Berlin) 44, 9 and 27
(1977)].

==Solutions of the Eigenvalue Problem==

For convenience, write

\[
\Psi({\bf r}_1,{\bf r}_2) = \sum^N_{m=1}c_m\varphi_m\nonumber
\]

where <math>m</math> represents the <math>m</math>'th combination of <math>i,j,k</math> and

\begin{equation}
\varphi_{ijk}=r_1^ir_2^jr_{12}^ke^{-\alpha r_1 - \beta
r_2}\mathcal{Y}^M_{l_1,l_2,L}(\hat{r}_1,\hat{r}_2) \pm \mbox{exchange}\nonumber
\end{equation}

\begin{align}

&
\left( \begin{array}{lr}
\cos(\theta) & \sin(\theta)\\
-\sin(\theta) & \cos(\theta)
\end{array} \right)
\left( \begin{array}{lr}
H_{11} & H_{12} \\
H_{12} & H_{22}
\end{array}\right)
\left( \begin{array}{lr}
\cos(\theta) & -\sin(\theta)\\
\sin(\theta) & \cos(\theta)
\end{array} \right)
\\[5pt]
= & \left(\begin{array}{lr}
cH_{11}+sH_{12} & cH_{12} + sH_{22}\\
-sH_{11} + cH_{12} & -sH_{12} + cH_{22}
\end{array}\right)
\left( \begin{array}{cc}
c & -s \\
s & c
\end{array}\right)
\\[5pt]
= & \left(\begin{array}{lr}
c^2H_{11}+s^2H_{22} + 2csH_{12} & (c^2-s^2)H_{12}+cs(H_{22}-H_{11})\\
(c^2-s^2)H_{12}+cs(H_{22}-H_{11}) & s^2H_{11}+c^2H_{22}-2csH_{12}
\end{array}\right)
\end{align}

Therefore

\begin{equation}
\left[\cos^2(\theta)-\sin^2(\theta)\right]H_{12} =
\cos(\theta)\sin(\theta)(H_{11}-H_{22})\nonumber
\end{equation}

and

\begin{equation}
\tan(2\theta) = \frac{2H_{12}}{H_{11}-H_{22}}\nonumber
\end{equation}

i.e.

\begin{equation}
\cos(\theta)=\left(\frac{r+\omega}{2r}\right)^{1/2}\nonumber
\end{equation}

\begin{equation}
\sin(\theta)=-\mbox{sgn}(H_{12})\left(\frac{r-\omega}{2r}\right)^{1/2}\nonumber
\end{equation}

where

\begin{equation}
\omega = H_{22}-H_{11}\nonumber
\end{equation}

\begin{equation}
r=\left(\omega^2+4H_{12}^2\right)^{1/2}\nonumber
\end{equation}

\begin{equation}
E_1=\frac{1}{2}\left(H_{11}+H_{22}-r\right)\nonumber
\end{equation}

\begin{equation}
E_2=\frac{1}{2}\left(H_{11}+H_{22}+r\right)\nonumber
\end{equation}

===Brute Force Method===

- Gives all the eigenvalues and eigenvectors, but it is slow

- First orthonormalize the <math>N</math>-dimensional basis set; i.e. form linear combinations

\begin{equation}
\Phi_m = \sum_{n=1}^N\varphi_nR_{nm}\nonumber
\end{equation}

such that
\begin{equation}
\langle \Phi_m | \Phi_n \rangle = \delta_{m,n}
\end{equation}

This can be done by finding an orthogonal tranformation T such that

\begin{equation}
T^TOT = I = \left(
\begin{array}{ccccc}
I_1 & 0 & 0 & \ldots & 0 \\
0 & I_2 & 0 & \ldots & 0 \\
0 & 0 & I_3 & \ldots & 0 \\
\vdots & \vdots & \vdots & \ddots & \vdots \\
0 & 0 & 0 & \ldots & I_N
\end{array}\right);
\end{equation}

\begin{equation}
O_{mn} = \langle \varphi_m | \varphi_n \rangle \nonumber
\end{equation}

and then applying a scale change matrix

\begin{equation}
S = \left(\begin{array}{ccccc}
\frac{1}{I_1^{1/2}} & 0 & 0 & \ldots & 0\\
0 & \frac{1}{I_2^{1/2}} & 0 & \ldots & 0 \\
0 & 0 & \frac{1}{I_3^{1/2}} & \ldots & 0 \\
\vdots & \vdots & \vdots & \ddots & \vdots \\
0 & 0 & 0 & \ldots & I_N^{1/2}
\end{array}\right)= S^T\nonumber
\end{equation}

Then

\begin{equation}
S^TT^TOTS = 1
\end{equation}

i.e.

\begin{equation}
R^TOR = 1
\end{equation}

with <math>R=TS</math>.

If H is the matrix with elements <math>H_{mn} = \langle \varphi_m | \varphi_n \rangle </math>, then H
expressed in the <math>\Phi_m</math> basis set is

\begin{equation}
H^\prime = R^THR.\nonumber
\end{equation}

We next diagonalize <math>H^\prime</math> by finding an orthogonal transformation W such
that

\begin{equation}
W^T H^\prime W = \lambda = \left(
\begin{array}{cccc}
\lambda_1 & 0 & \ldots & 0 \\
0 & \lambda_2 & \ldots & 0 \\
\vdots & \vdots & \ddots & \vdots \\
0 & 0 & \ldots & \lambda_N
\end{array}\right)\nonumber
\end{equation}

The q'th eigenvector is

\begin{equation}
\Psi^{(q)} = \sum_{n=1}^N\Phi_n W_{n,q} = \sum_{n,n^\prime} \varphi_{n^\prime} R_{n^\prime ,n} W_{n,q} \nonumber
\end{equation}

i.e.

\begin{equation}
c_{n^\prime}^{(q)} = \sum_{n=1}^N R_{n^\prime n} W_{n,q}
\end{equation}

===The Power Method===

- Based on the observation that if H has one eigenvalue, <math>\lambda_M</math>, much bigger than all the rest, and <math>\chi = \left(
\begin{array}{c}a_1\\a_2\\\vdots\end{array}\right)</math>
is an arbitrary starting vector, then <math>\chi = \sum_{q=1}^N x_q\Psi^{(q)}</math>.

\begin{equation}
(H)^n\chi = \sum_{q=1}^N x_q \lambda^n_q\Psi^{(q)}\nonumber\rightarrow x_M\lambda_M^n\Psi^{(M)}\nonumber
\end{equation}

provided <math>x_M\neq 0</math>.

To pick out the eigenvector corresponding to any eigenvalue, write the original problem in the form

\begin{align}
H\Psi & = \lambda O \Psi \nonumber \\
\left( H - \lambda_q O \right) \Psi & = \left( \lambda - \lambda_q \right) O \Psi \nonumber
\end{align}

Therefore,

\begin{equation}
G \Psi = \frac{1}{\lambda - \lambda_q } \Psi \nonumber \\
\end{equation}

where <math>G=(H-\lambda_qO)^{-1}O</math> with eigenvalues <math>\frac{1}{\lambda_n - \lambda_q}</math>.

By picking <math>\lambda_q</math> close to any one of the <math>\lambda_n</math>, say
<math>\lambda_{n^\prime}</math>, then <math>\frac{1}{\lambda_n-\lambda_q}</math> is much larger for
<math>n=n^\prime</math> than for any other value. The sequence is then

\begin{array}{c}
\chi_1=G\chi\nonumber\\
\chi_2=G\chi_1\nonumber\\
\chi_3=G\chi_2\nonumber\\
\vdots\nonumber
\end{array}

until the ratios of components in <math>\chi_n</math> stop changing.

- To avoid matrix inversion and multiplication, note that the sequence is equivalent to
\begin{equation}
F \chi_n = \left( \lambda - \lambda_q \right) O \chi_{n-1} \nonumber \\
\end{equation}
where <math>F = H-\lambda_qO</math>. The factor of <math>\left( \lambda - \lambda_q \right)</math> can be
dropped because this only affects the normalization of <math>\chi_n</math>. To find <math>\chi_n</math>, solve
\begin{equation}
F \chi_n = O \chi_{n-1} \nonumber
\end{equation}
(N equations in N unknowns). Then
\begin{equation}
\lambda = \frac{ \langle \chi_n | H | \chi_n \rangle}{\langle \chi_n | \chi_n \rangle} \nonumber
\end{equation}

==Matrix Elements of H==

\begin{equation}
H=-\frac{1}{2}\nabla^2_1 -\frac{1}{2}\nabla^2_2 - \frac{1}{r_1} - \frac{1}{r_2} +\frac{Z^{-1}}{r_{12}}\nonumber
\end{equation}

Taking <math>r_1, r_2</math> and <math>r_{12}</math> as independent variables,

\begin{equation}
\nabla_1^2 = \frac{1}{r_1^2}\frac{\partial}{\partial r_1}\left(r_1^2\frac{\partial}{\partial r_1}\right)+ \frac{1}{r_{12}^2}\frac{\partial}{\partial r_{12}}
\left(r_{12}^2\frac{\partial}{\partial r_{12}}\right)\nonumber -\frac{l_1(l_1+1)}{r_1^2}+2(r_1 - r_2\cos(\theta_{12}))\frac{1}{r_{12}}\frac{\partial^2}{\partial r_1 \partial r_{12}} \nonumber - 2(\nabla_1^Y \cdot {\bf r}_2)\frac{1}{r_{12}}\frac{\partial}{\partial r_{12}}\nonumber
\end{equation}

where <math>\nabla_1^Y</math> acts only on the spherical harmonic part of the wave function and the diagram

[[File:Pic3.jpg|400px]]

defines the complete set of 6 independent variables is <math>r_1</math>, <math>r_2</math>, <math>r_{12}</math>, <math>\theta_1</math>, <math>\varphi_1</math>, <math>\chi</math>.

If <math>r_{12}</math> were not an independent variable, then one could take the volume element to be

\begin{equation}
d\tau = r_1^2dr_1\sin(\theta_1)d\theta_1d\varphi_1r_2^2dr_2\sin(\theta_2)d\theta_2d\varphi_2.\nonumber
\end{equation}

However, <math>\theta_2</math> and <math>\varphi_2</math> are no longer independent variables. To eliminate them, take the point <math>{\bf r}_1</math> as the
origin of a new polar co-ordinate system, and write

\begin{equation}
d\tau=-r_1^2dr_1\sin(\theta_1)d\theta_1d\varphi_1r_{12}^2dr_{12}\sin(\psi)d\psi d\chi\nonumber
\end{equation}

and use

\begin{equation}
r_2^2=r_1^2+r_{12}^2 +2r_1r_{12}\cos(\psi)
\end{equation}

Then for fixed <math>r_1</math> and <math>r_{12}</math>,

\begin{equation}
2r_2dr_2 = -2r_1r_{12}\sin(\psi)d\psi
\end{equation}

Thus

\begin{equation}
d\tau= r_1dr_1r_2dr_2r_{12}dr_{12}\sin(\theta_1)d\theta_1d\varphi_1d\chi
\end{equation}

The basic type of integral to be calculated is

\begin{align}
I \left( l_1, m_1, l_2, m_2; R \right) & = \int \sin \left( \theta_1 \right) d\theta_1 d\varphi_1 d\chi Y^{m_1}_{l_1} \left( \theta_1, \varphi_1 \right)^{*} Y^{m_2}_{l_2} \left( \theta_2, \varphi_2 \right) \\
& \times \int r_1dr_1r_2dr_2r_{12}dr_{12}R \left( r_1, r_2, r_{12} \right)
\end{align}

Consider first the angular integral. <math>Y^{m_2}_{l_2} \left( \theta_2, \varphi_2 \right)</math> can be expressed in terms of the independent variables <math>\theta_1, \varphi_1, \chi</math> by use of the rotation matrix relation
\begin{equation}
Y^{m_2}_{l_2}( \theta_2, \varphi_2) = \sum_m \mathcal{D}^{(l_2)}_{m_2,m}(\varphi_1,\theta_1,\chi)^*Y^m_{l_2}( \theta_{12}, \varphi)
\end{equation}
where <math>\theta, \varphi</math> are the polar angles of <math>{\bf r}_2</math> relative to <math>{\bf r}_1</math>. The angular integral is then

\begin{align}
I_{ang} & = \int^{2 \pi}_{0} d\chi \int^{2\pi}_{0} d\varphi_1 \int^{\pi}_{0} \sin \left( \theta_1 \right) d\theta_{1} Y^{m_1}_{l_1} \left( \theta_1, \varphi_1 \right)^{*} \\
& \times \sum_m \mathcal{D}^{(l_2)}_{m_2,m} \left( \varphi_1,\theta_1,\chi \right)^{*} Y^{m}_{l_2} \left( \theta_{12},\varphi \right) \\
\end{align}

Use

\begin{equation}
Y^{m_1}_{l_1} \left( \theta_1, \varphi_1 \right)^{*} = \sqrt{\frac{2l_1+1}{4\pi}}\mathcal{D}^{(l_1)}_{m_1,0} \left( \varphi_1, \theta_1, \chi \right)
\end{equation}

together with the orthogonality property of the rotation matrices (Brink and Satchler, p 147)

\begin{align}
& \int \mathcal{D}^{(j)*}_{m,m^\prime}\mathcal{D}^{(J)}_{M,M^\prime}\sin(\theta_1)d\theta_1d\varphi_1d\chi \\
& = \frac{8\pi^2}{2j+1}\delta_{jJ}\delta_{mM}\delta_{m^\prime M^\prime}\nonumber
\end{align}

to obtain

\begin{align}
I_{ang} & = \sqrt{ \frac{2l_1+1}{4\pi}} \frac{8\pi^2}{2l_1+1} \delta_{l_1,l_2} \delta_{m_1,m_2} Y^{0}_{l_2} \left( \theta_{12}, \varphi \right) \\
& = 2 \pi \delta_{l_1,l_2} \delta_{m_1,m_2} P_{l_2} \left( \cos \theta_{12} \right)
\end{align}

since

\begin{equation}
Y^0_{l_2} \left( \theta_{12}, \varphi \right) = \sqrt{\frac{2l_1+1}{4\pi}} P_{l_2} \left( \cos \left( \theta_{12} \right) \right)
\end{equation}

Note that <math>P_{l_2} \left( \cos \theta \right)</math> is just a short hand expression for a radial function because

\begin{equation}
\cos \theta_{12} = \frac{r_1^2 + r_2^2 - r_{12}^2}{2r_1r_2}
\end{equation}

The original integral is thus

\begin{equation}
I \left( l_1, m_1, l_2, m_2; R \right) = 2 \pi \delta_{l_1,l_2} \delta_{m_1,m_2} \int^{\infty}_{0} r_{1} dr_{1} \int^{\infty}_{0} r_{2} dr_{2} \int^{r_{1} + r_{2}}_{|r_{1} - r_{2}|} r_{12} dr_{12} R \left( r_{1}, r_{2}, r_{12} \right) P_{l_2} \left( \cos \theta_{12} \right)
\end{equation}

where again

\begin{equation}
\cos \theta_{12} = \frac{r_{1}^{2} +r^{2}_{2} - r_{12}^{2}}{2 r_{1} r_{2}}
\end{equation}

is a purely radial function.

The above would become quite complicated for large <math>l_2</math> because <math>P_{l_2}(\cos \theta_{12})</math> contains terms up to <math>\left( \cos \theta_{12} \right)^{l_2}</math>. However, recursion relations exist which allow any integral containing <math>P_l \left( \cos \theta_{12} \right)</math> in terms of those containing just <math>P_0 \left( \cos \theta_{12} \right) = 1</math> and <math>P_1 \left( \cos \theta_{12} \right) = \cos \theta_{12}</math>.

==Radial Integrals and Recursion Relations==

The basic radial integral is [see G.W.F. Drake, Phys. Rev. A 18, 820 (1978)]

\begin{align}
I_0 \left( a, b, c \right) & =
\int^{\infty}_{0} r_{1} dr_{1} \int^{\infty}_{r_{1}} r_{2} dr_{2} \int^{r_{1}+r_{2}}_{r_{2} - r_{1}} r_{12} dr_{12} r_{1}^{a} r_{2}^{b} r_{12}^{c} e^{-\alpha r_{1} - \beta r_{2}} + \int^{\infty}_{0} r_{2} dr_{2} \int^{\infty}_{r_{2}} r_{1} dr_{1} \int^{r_{1} + r_{2}}_{r_{1} - r_{2}} r_{12} dr_{12} r_{1}^{a} r_{2}^{b} r_{12}^{c} e^{- \alpha r_{1} - \beta r_{2}} \\
& = \frac{2}{c+2} \sum^{[(c+1)/2]}_{i=0}
\left(\begin{array}{c}
c+2 \\
2i+1
\end{array} \right)
\left\{ \frac{q!}{\beta^{q+1} \left( \alpha + \beta \right)^{p+1}} \sum_{j=0}^{q} \frac{\left( p + j \right)!}{j!} \left( \frac{\beta}{\alpha + \beta} \right)^j + \frac{q^{\prime}!}{\alpha^{q^{\prime}+1} \left( \alpha + \beta \right)^{p^{\prime}+1}} \sum^{q^{\prime}}_{j=0} \frac{\left( p^{\prime} + j \right)!}{j!} \left( \frac{\alpha}{\alpha + \beta} \right)^j \right\}
\end{align}

where

\begin{array}{ll}
p = a + 2i + 2 & p^{\prime} = b + 2i + 2 \\
q = b + c - 2i + 2 & q^{\prime} = a + c - 2i + 2
\end{array}

The above applies for <math> a, b \geq -2, c\geq -1</math>.
<math>[x]</math> means "greatest integer in" <math>x</math>.

Then

\begin{align}
I_1 \left( a, b, c \right) & = \int d\tau_{r} r_{1}^{a} r_{2}^{b} r_{12}^{c} e^{ -\alpha r_{1} - \beta r_{2}} P_{1}\left( \cos \theta \right) \\
& = \frac{1}{2} \left( I_{0} \left( a + 1, b - 1, c \right) + I_{0} \left( a - 1, b + 1, c \right) - I_{0} \left( a - 1, b - 1, c + 2 \right) \right)
\end{align}

===The Radial Recursion Relation===

Recall that the full integral is

\begin{equation}
I(l_1m_1,l_2m_2;R)=2\pi\delta_{l_1,l_2}\delta_{m_1,m_2}I_{l_2}(R)
\end{equation}

where

\begin{equation}
I_{l_2}(R) = \int d\tau_r R(r_1,r_2,r_{12})P_{l_2}(\cos \theta_{12})
\end{equation}

To obtain the recursion relation, use

\begin{equation}
P_l(x) = \frac{[P^{\prime}_{l+1}(x)-P^{\prime}_{l-1}(x)]}{2l+1}
\end{equation}

with

\begin{equation}
P_{l+1}^\prime (x) = \frac{d}{dx}P_{l+1}(x)
\end{equation}

Here <math>x=\cos\theta_{12}</math> and

\begin{align}
\frac{d}{d \cos \theta_{12}} & = \frac{dr_{12}}{d \cos \theta_{12}} \frac{d}{dr_{12}} \\
& = -\frac{r_{1} r_{2}}{r_{12}} \frac{d}{dr_{12}}
\end{align}

Then,

\begin{equation}
I_{l}\left( R \right) = - \int d\tau_r R \frac{r_{1} r_{2}}{r_{12}} \frac{d}{dr_{12}} \frac{ \left[ P_{l+1} \left( \cos \theta_{12} \right) - P_{l-1} \left( \cos \theta_{12} \right) \right]}{2l+1}
\end{equation}

The <math>r_{12}</math> part of the integral is

\begin{align}
& \int^{r_{1} + r_{2}}_{|r_{1} - r_{2}|} r_{12} dr_{12} R \frac{r_{1} r_{2}}{r_{12}} \frac{d}{dr_{12}} \left[P_{l+1}-P_{l-1} \right] \\
& = R r_{1} r_{2} \left[ P_{l+1} - P_{l-1} \right]^{r_{1} + r_{2}}_{|r_{1} - r_{2}|} - \int^{r_{1} + r_{2}}_{|r_{1} - r_{2}|} r_{12} dr_{12} \left( \frac{d}{dr_{12}} R \right) \frac{r_{1} r_{2}}{r_{12}} \frac{ \left[ P_{l+1} - P_{l-1} \right]}{2l+1}
\end{align}

The integrated term vanishes because

\begin{equation}
\cos \theta_{12} = \frac{r_{1}^{2} + r_{2}^{2} - r_{12}^{2}}{2 r_{1} r_{2}}
\begin{array}{lll}
& = -1 & {\rm when} & r_{12}^{2} = \left( r_{1} + r_{2} \right)^{2} \\
& = 1 & {\rm when} & r_{12}^{2} = \left( r_{1} - r_{2} \right)^{2}
\end{array}
\end{equation}

and <math>P_l(1)=1</math>, <math>P_{l}(-1)=(-1)^{l}</math>.

Thus,

\begin{equation}
I_{l+1}\left(\frac{r_1r_2}{r_{12}}R^\prime\right) = \left( 2l+1 \right) I_{l}(R) - I_{l-1} \left( \frac{r_{1} r_{2}}{r_{12}} R^{\prime} \right)
\end{equation}

If

\begin{equation}
R = r_{1}^{a-1} r_{2}^{b-1} r_{12}^{c+2} e^{-\alpha r_{1} - \beta r_{2}}
\end{equation}

then [G.W.F. Drake, Phys. Rev. A 18, 820 (1978)]

\begin{equation}
I_{l+1}(r_1^ar_2^br_{12}^c) =
\frac{2l+1}{c+2} I_{l} \left( r_{1}^{a-1} r_{2}^{b-2} r_{12}^{c+2} \right)+I_{l-1} \left( r_{1}^{a} r_{2}^{b} r_{12}^{c} \right)
\end{equation}

For the case <math>c=-2</math>, take

\begin{equation}
R = r_{1}^{a-1} r_{2}^{b-1} \ln r_{12} e^{ -\alpha r_{1} -\beta r_{2}}
\end{equation}

Then

\begin{equation}
I_{l+1} \left( r_{1}^{a} r_{2}^{b} r_{12}^{c} \right) = I_{l} \left( r_{1}^{a-1} r_{2}^{b-1} \ln r_{12} \right) \left( 2l + 1 \right) - I_{l-1} \left( r_{1}^{a} r_{2}^{b} r_{12}^{-2} \right)
\end{equation}

This allows all <math>I_{l}</math> integrals to be calculated from tables of <math>I_0</math> and
<math>I_1</math> integrals.

===The General Integral===

The above results for the angular and radial integrals can now be combined into a general formula for integrals of the type

\begin{equation}
I = \int \int d{\bf r}_{1} d{\bf r}_{2} R_{1} \mathcal{Y}^{M^{\prime}}_{l^{\prime}_{1} l^{\prime}_{2} L^{\prime}} \left( \hat{r}_{1}, \hat{r}_{2} \right) T_{k_{1} k_{2} K}^{Q} \left( {\bf r}_{1},{\bf r}_{2} \right) R_{2} \mathcal{Y}_{l_{1} l_{2} L}^{M} \left( \hat{r}_{1}, \hat{r}_{2} \right)
\end{equation}

where

\begin{equation}
\mathcal{Y}^{M}_{l_{1} l_{2} L} \left( \hat{r}_{1}, \hat{r}_{2} \right) = \sum_{m_1,m_2} \langle l_{1} l_{2} m_{1} m_{2} | LM \rangle Y^{m_{1}}_{l_{1}} \left( \hat{r}_{1} \right) Y^{m_{2}}_{l_{2}} \left( \hat{r}_{2} \right)
\end{equation}

and

\begin{equation}
T^Q_{k_{1} k_{2} K} \left( {\bf r}_{1}, {\bf r}_{2} \right) = \sum_{q_{1}, q_{2}} \langle k_{1} k_{2} q_{1} q_{2} | KQ \rangle Y_{k_{1}}^{q_{1}} \left( \hat{r}_{1} \right) Y^{q_{2}}_{k_{2}} \left( \hat{r}_{2} \right)
\end{equation}

The basic idea is to make repeated use of the formula

\begin{equation}
Y^{m_{1}}_{l_{1}} \left( \hat{r}_{1} \right) Y^{m_{2}}_{l_{2}} \left( \hat{r}_{1} \right) = \sum_{l m} \left( \frac{\left( 2l_{1}+1 \right) \left(2l_{2}+1 \right) \left(2l+1 \right)}{4 \pi} \right)^{1/2}
\times \left( \begin{array}{ccc}
l_1 & l_2 & l \\
m_1 & m_2 & m
\end{array} \right) \left(\begin{array}{ccc}
l_1 & l_2 & l \\
0 & 0 & 0
\end{array} \right)
Y^{m*}_{l} \left( \hat{r}_{1} \right)
\end{equation}

where

\begin{equation}
Y^{m*}(\hat{r}) = (-1)^mY_l^{-m}(\hat{r})
\end{equation}

and

\begin{equation}
\left(\begin{array}{ccc}
l_1 & l_2 & l \\
m_1 & m_2 & m
\end{array} \right) = \frac{\left( -1 \right)^{l_{1} - l_{2} - m}}{\left( 2l + 1 \right)^{1/2}} \left( l_{1} l_{2} m_{1} m_{2} | l, -m \right)
\end{equation}

is a 3-<math>j</math> symbol. In particular, write

\begin{array}{cc}
Y^{m_1^\prime}_{l_1^\prime}(\hat{r}_1)^{*} \underbrace{Y_{k_{1}}^{q_{1}} \left( \hat{r}_{1} \right) Y^{m_{1}}_{l_{1}} \left( \hat{r}_{1} \right)} & = & \sum_{\Lambda M}(\ldots)Y^{M}_\Lambda(\hat{r}_1)\\
\ \ \ \ \ \ \ \sum_{\lambda_1\mu_1}Y_{\lambda_1}^{\mu_1}(\hat{r}_1) & & \\
& & \\
Y^{m_{2}^{\prime}}_{l_{2}^{\prime}} \left( \hat{r}_{2} \right) \underbrace{Y^{q_{2}}_{k_{2}} \left( \hat{r}_{2} \right) Y_{l_{2}}^{m_{2}} \left( \hat{r}_{2} \right)} & = & \sum_{\Lambda^{\prime} M^{\prime}} \left( \ldots \right)Y_{\Lambda^{\prime}}^{M^{\prime}*} \left( \hat{r}_{2} \right) \\
\ \ \ \ \ \ \ \ \ \sum_{\lambda_{2} \mu_{2}}Y_{\lambda_{2}}^{\mu_{2}} \left( \hat{r}_{2} \right) & &
\end{array}

The angular integral then gives a factor of <math>2 \pi \delta_{\Lambda, \Lambda^{\prime}} \delta_{M, M^{\prime}} P_{\Lambda} \left( \cos \theta_{12} \right)</math>. The total integral therefore reduces to the form

\begin{equation}
I = \sum_{\Lambda} C_{\Lambda} I_{\Lambda} \left( R_{1} R_{2} \right)
\end{equation}

where <math>C_{\Lambda} = \sum_{\lambda_{1}, \lambda_{2}} C_{ \lambda_{1}, \lambda_{2}, \Lambda}</math>. For further details and derivations, including graphical representations, see G.W.F. Drake, Phys. Rev. A <math>{\bf 18}</math>, 820 (1978).

==Graphical Representation==

[[File:GraphRep1.jpg|500px]]

==Matrix Elements of H==
Recall that

\begin{equation}
H = -\frac{1}{2} \nabla^{2}_{1} - \frac{1}{2} \nabla^{2}_{2} - \frac{1}{r_1} - \frac{1}{r_2} + \frac{Z^{-1}}{r_{12}}
\end{equation}

Consider matrix elements of

\begin{equation}
\nabla^2_1 = \frac{1}{r_1^2} \frac{\partial}{\partial r_1}\left(r_1^2\frac{\partial}{\partial r_1} \right) + \frac{1}{r^2}\frac{\partial}{\partial r}\left(r^2\frac{\partial}{\partial r}\right) - \frac{(\vec{l}_1^Y)^2}{r_1^2} + \frac{2 \left( r_1 - r_2 \cos \theta \right)}{r} \frac{\partial^2}{\partial r_1 \partial r} - 2 \left( \nabla^Y_1 \cdot {\bf r}_2 \right) \frac{1}{r} \frac{\partial}{\partial r}
\end{equation}

where <math>r \equiv r_{12}</math> and <math>\cos\theta \equiv \cos\theta_{12}</math>. Also in what follows, define <math>\hat{\nabla}^Y_1 \equiv r_1\nabla^Y_1</math> where <math>\nabla^Y</math> operates only on the spherical harmonic part of the wave function, and similarly for <math>\vec{l}_1^Y</math>.

A general matrix element is

\begin{equation}
\langle r_1^{a^\prime} r_2^{b^\prime} r_{12}^{c^\prime} e^{-\alpha^\prime r_1 - \beta^\prime r_2} \mathcal{Y}^{M^\prime}_{l^\prime_1 l^\prime_2 L^\prime} \left( \hat{r}_1, \hat{r}_2 \right) \left| \nabla^2_1 \right| r_1^a r_2^b r_{12}^c e^{-\alpha r_1 - \beta r_2} \mathcal{Y}^M_{l_1 l_2 L} \left( \hat{r}_1, \hat{r}_2 \right) \rangle
\end{equation}

Since <math>\nabla^2_1</math> is rotationally invariant, this vanishes unless <math>L=L^\prime</math>, <math>M=M^\prime</math>. Also <math>\nabla_1^2</math> is Hermitian, so that the result must be the same whether it operates to the right or left, even though the results look very different. In fact, requiring the results to be the same yields some interesting and useful integral identities as follows:

Put

\begin{equation}
\mathcal{F} = F \mathcal{Y}^M_{l_1 l_2 L} \left( \hat{r}_1, \hat{r_2} \right)
\end{equation}

and

\begin{equation}
F = r_1^a r_2^b r_{12}^c e^{-\alpha r_1 - \beta r_2}
\end{equation}

Then

\begin{equation}
\nabla^2_1 \mathcal{F} = \left\{ \frac{1}{r_1^2} \left[ a \left( a + 1 \right) - l_1 \left( l_1 + 1 \right) \right] + \frac{c \left( c + 1 \right)}{r^2} + \alpha^2\frac{2\alpha \left( a + 1 \right)}{r_1} + \frac{2 \left(r_1 - r_2 \cos \theta \right) }{r_1r^2}c \left[ a - \alpha r_1 \right] - \frac{2c}{r^2} \left( \hat{\nabla}^Y_1 \cdot \hat{r}_2 \right) \frac{r_2}{r_1} \right\} \mathcal{F}
\end{equation}

and

\begin{align}
\langle \mathcal{F}^\prime \left| \nabla^2_1 \right| \mathcal{F} \rangle & = \sum_\Lambda \int d\tau_r F^\prime C_\Lambda \left( 1 \right) P_\Lambda \left( \cos \theta \right) \times \left\{ \frac{1}{r_1^2} \left[ a \left( a + 1 \right) - l_1 \left( l_1 + 1 \right) \right] -\frac{2 \alpha \left( a + 1 \right)}{r_1} + \frac{c \left( c + 1 \right)}{r^2} +\alpha^2 + \frac{2 \left( r_2 - r_2 \cos \theta \right)}{r_1r^2}c \left( a - \alpha r_1 \right) \right\} F \\
& + \sum_\Lambda \int d\tau_r F^\prime C_\Lambda \left( \hat{\nabla}^Y_1 \cdot \hat{r}_2 \right) P_\Lambda \left( \cos \theta \right) \left( \frac{-2c r_2}{r_1 r^2} \right) F
\end{align}

where

\begin{equation}
\int d\tau_r = \int^\infty_0 r_1 dr_1 \int^\infty_0 r_2 dr_2 \int^{r_1+r_2}_{\left| r_1 - r_2 \right|} r dr
\end{equation}

For brevity, let the sum over <math>\Lambda</math> and the radial integrations be understood, and let <math>\left( \nabla_1^2 \right)</math> stand for the terms which appear in the integrand. Then operating to the right gives

\begin{equation}
(\nabla_1^2)_R = \frac{1}{r_1^2} \left[ a \left( a + 1 \right) - l_1 \left( l_1 + 1 \right) \right] + \frac{c \left( c + 1 \right)}{r^2} + \alpha^2 - \frac{2 \alpha \left( a + 1 \right)}{r_1}
+ \frac{2 \left( r_1 - r_2 \cos \theta \right)}{r_1 r^2}c \left( a - \alpha r_1 \right) - \frac{2c r_2}{r_1 r^2} \left( \hat{\nabla}^Y_1 \cdot \hat{r}_2 \right)
\end{equation}

Operating to the left gives

\begin{equation}
(\nabla_1^2)_L = \frac{1}{r_1^2} \left[a^\prime \left( a^\prime + 1 \right) - l_1^\prime \left( l_1^\prime + 1 \right) \right] + \frac{c^\prime \left( c^\prime + 1 \right)}{r^2} + \alpha^{\prime 2} -\frac{2 \alpha^\prime \left( a^\prime + 1 \right)}{r_1}
+ \frac{2 \left( r_1 - r_2 \cos \theta \right)}{r_1 r^2} c^\prime \left( a^\prime - \alpha^\prime r_1 \right) - \frac{2 c^\prime r_2}{r_1 r^2} \left( \hat{\nabla}^{Y^\prime}_1 \cdot \hat{r}_2 \right)
\end{equation}

Now put

\begin{array}{cc}
a_+ = a+a^\prime, & \hat{\nabla}_1^+ = \hat{\nabla}_1^Y + \hat{\nabla}_1^{Y\prime} \\
a_{-} = a-a^\prime, & \hat{\nabla}_1^{-} = \hat{\nabla}_1^Y - \hat{\nabla}_1^{Y\prime}
\end{array}

etc., and substitute <math>a^\prime = a_+ - a</math>, <math>c^\prime = c_+ - c</math>, and <math>\alpha^\prime = \alpha_+ - \alpha</math> in <math>(\nabla^2_1)_L</math>. If <math>a_+</math>, <math>c_+</math> and <math>\alpha_+</math> are held fixed, then the equation

\begin{equation}
(\nabla^2_1)_R = (\nabla^2_1)_L
\end{equation}

must be true for arbitrary <math>a</math>, <math>c</math>, and <math>\alpha</math>. Their coefficients must thus vanish.

This yields the integral relations

\begin{equation}
\label{I}
\frac{\left( r_1 - r_2 \cos \theta \right)}{r_1 r^2} = \frac{1}{c_+} \left( \frac{- \left( a_+ + 1 \right)}{r_1^2} + \frac{\alpha_+}{r_1} \right) \tag{I}
\end{equation}

from the coefficient of <math>a</math> and

\begin{equation}
\label{II}
\frac{ \left( r_1 - r_2 \cos \theta \right) \left( a_+ - \alpha_+ r_1 \right)}{r_1 r^2} = \frac{r_2}{r_1 r^2}\left( \hat{r}_2 \cdot \hat{\nabla}_1^+ \right) - \frac{ \left( c_+ + 1 \right)}{r^2} \tag{II}
\end{equation}

from the coefficient of <math>c</math>. The coefficient of <math>\alpha</math> gives an equation equivalent to (I).

Furthermore, if can be show that (see problem)

\begin{align}
& \sum_\Lambda \int d\tau_r \frac{r^c}{r^2} C_\Lambda \left( \hat{r}_2 \cdot \hat{\nabla}_1^Y \right) P_\Lambda \left( \cos \theta \right) \\
= & \sum_\Lambda \int d\tau_r \frac{r^c}{c r_1 r_2}C_\Lambda \left( 1 \right)P_\Lambda \left( \cos \theta \right) \times \left( \frac{l_1^\prime \left( l_1^\prime + 1 \right) - l_1 \left( l_1 + 1 \right) -\Lambda \left( \Lambda + 1 \right)}{2} \right)
\end{align}

and similarly for <math>\left( \hat{r}_2 \cdot \hat{\nabla}^{Y\prime}_1 \right)</math> with <math>l_1</math> and <math>l_1^\prime</math> interchangeable, then it follows that

\begin{equation}
\frac{r^{c_+}}{r^2} \left( \hat{r}_2 \cdot \hat{\nabla}_1^+ \right) = - \frac{r^{c_+}}{c_+ r_1 r_2} \Lambda \left( \Lambda + 1 \right)
\end{equation}

and

\begin{equation}
\frac{r^{c_+}}{r^2} \left( \hat{r}_2 \cdot \hat{\nabla}_1^- \right) = \frac{r^{c_+}}{c_+ r_1 r_2} \left[ l_1^\prime \left( l_1^\prime + 1 \right) - l_1 \left( l_1 + 1 \right) \right]
\end{equation}

where equality applies after integration and summation over <math>\Lambda</math>.

Thus (II) becomes

\begin{equation}\label{IIa}
\frac{ \left( r_1 - r_2 \cos \theta \right) \left( a_+ - \alpha_+ r_1 \right)}{r_1 r^2} = - \frac{\Lambda \left( \Lambda + 1 \right)}{c_+ r^2_1} - \frac{ \left( c_+ + 1 \right)}{r^2} \tag{IIa}
\end{equation}

===Problem===

Prove the integral relation

\begin{align}
& \sum_\Lambda \int d\tau_r f \left( r_1, r_2 \right) \frac{1}{r} \left( \frac{d}{dr} q(r) \right) C_\Lambda \left( \hat{r}_2 \cdot \hat{\nabla}^Y_1 \right) P_\Lambda \left( \cos \theta \right) \\
= & \sum_\Lambda \int d\tau_r \frac{ f \left( r_1, r_2 \right)}{r_1, r_2} q(r) C_\Lambda \left( 1 \right) P_\Lambda \left( \cos \theta \right) \left(\frac{l_1^\prime \left( l_1^\prime + 1 \right) - l_1 \left( l_1 + 1 \right) - \Lambda \left( \Lambda + 1 \right)}{2}\right)
\end{align}

where the coefficients <math>C_\Lambda \left( 1 \right)</math> are the angular coefficients from the overlap integral

\begin{align}
\int & d\Omega\, \mathcal{Y}^{M*}_{l_1^\prime l_2^\prime L}(\hat{r}_1,\hat{r}_2)\, \mathcal{Y}^{M}_{l_1 l_2 L}(\hat{r}_1,\hat{r}_2)\\
= & \sum_\Lambda C_\Lambda(1)P_\Lambda(\cos\theta_{12}).
\end{align}

Hint: Use the fact that <math>l_1^2</math> is Hermitian so that
<math>
\int d\tau (l_1^2\mathcal{Y}^\prime)^*q(r)\mathcal{Y} = \int d\tau
\mathcal{Y}^{\prime *}l_1^2(q(r)\mathcal{Y})
</math>
with <math>\ \vec{l}_1= \frac{1}{i}\vec{r}_1\times\nabla_1</math>.

It is also useful to use
<math>\begin{eqnarray}\ \
\cos\theta\, P_L(\cos\theta) &=& \frac{1}{2L+1}[L\,P_{L-1}(\cos\theta) + (L+1)P_{L+1}(\cos\theta)]
\\
(\cos^2\theta-1)P_L(\cos\theta) &=&
\frac{L(L-1)}{(2L-1)(2L+1)}P_{L-2}(\cos\theta)\nonumber
-\frac{2(L^2+L-1)}{(2L-1)(2L+3)}P_L(\cos\theta) + \frac{(L+1)(L+2)}{(2L+1)(2L+3)}P_{L+2}(\cos\theta)\nonumber\\
&=& \frac{L(L-1)}{(2L-1)(2L+1)}[P_{L-2}(\cos\theta) - P_L(\cos\theta)]
+ \frac{(L+1)(L+2)}{(2L+1)(2L+3)}[P_{L+2}(\cos\theta) - P_L(\cos\theta)]
\end{eqnarray}</math>

together with a double application of the integral recursion relation
<math>
I_{L+1}\left(\frac{1}{r}\frac{d}{dr}q(r)\right) =
(2L+1)I_L\left(\frac{1}{r_1r_2}q(r)\right)+I_{L-1}\left(\frac{1}{r}\frac{d}{dr}q(r)\right)
</math>

Of course <math>l_1^2\mathcal{Y}^M_{l_1l_2L}(\hat{r}_1,\hat{r}_2)=l_1(l_1+1)\mathcal{Y}^M_{l_1l_2L}(\hat{r}_1,\hat{r}_2)</math>. Begin by expanding
<math>
l_1^2(g\mathcal{Y}) = \mathcal{Y}l_1^2g + g\,l_1^2\mathcal{Y} + 2(\vec{l}_1g)\cdot(\vec{l}_1\mathcal{Y})
</math>, and show that

<math>
\begin{eqnarray}\ \
(\vec{l}_1g)\cdot(\vec{l}_1\mathcal{Y}) & = & \frac{r_1^2}{r}\frac{dg}{dr}\vec{r}_2\cdot \nabla_1\mathcal{Y}\,\,\,\,\mbox{ and}\\
l_1^2g(r) &=& 2r_1r_2\cos\theta\frac{1}{r}\frac{dg}{dr} + r_1^2r_2^2(\cos^2\theta - 1)
\frac{1}{r}\frac{d}{dr}\left(\frac{1}{r}\frac{dg}{dr}\right).
\end{eqnarray}
</math>

The proof amounts to showing that the term <math>l_1^2g(r)</math> can be replaced by
<math>\Lambda(\Lambda+1)g(r)</math> after multiplying by <math>P_\Lambda(\cos\theta)</math>
and integrating by parts with respect to the radial integrations over <math>r_1</math>, <math>r_2</math> and
<math>r \equiv |\vec{r}_1 - \vec{r}_2|</math>. Remember that <math>\cos\theta</math> is just a short-hand notation for the radial function <math>(r_1^2 + r_2^2 - r^2)/(2r_1r_2)</math>.

==General Hermitian Property==

Each combination of terms of the form
\begin{equation}
<f>_R = a^2f_1+af_2+abf_3+b^2f_4+bf_5+af_6\nabla_1^Y+bf_7\nabla^Y_2+f_8(Y)
\end{equation}
can be rewritten
\begin{equation}
<f>_L= (a_+-a)^2f_1+(a_+-a)f_2 + (a_+-a)(b_+-b)f_3+(b_+-b)^2f_4
+(b_+-b)f_5+(a_+-a)f_6\nabla^{Y\prime}_1+(b_+-b)f_7\nabla^{Y\prime}_2+f_8)Y^\prime
\end{equation}
where as usual <math>\nabla_1^Y</math> acts only on the spherical harmonic part of the wave function denoted for short by <math>Y</math>.
Since these must be equal for arbitrary <math>a</math> and <math>b</math>,
\begin{equation}
a_+^2f_1+a_+f_2+_+b_+f_3+b_+f_4+b_+f_5+a_+f_6\nabla^{Y\prime}_1 +
b_+f_7\nabla^{Y\prime}_2+f_8(Y^\prime)-f_8(Y)=0
\end{equation}

Adding the corresponding expression with <math>Y</math> and <math>Y^\prime</math> interchanged
yields
<math>\label{III}
a_+^2f_1+a_+f_2+a_+b_+f_3+b_+^2f_4+b_+f_5+\frac{1}{2}a_+f_6\nabla^+_1+\frac{1}{2}b_+f_7\nabla^+_2=0\tag{III}
</math>

Subtracting gives
<math>\label{IV}
f_8(Y)-f_8(Y^\prime)=-\frac{1}{2}[a_+f_6\nabla^-_1+b_+f_7\nabla_2^-]\tag{IV}
</math>
<math>\label{V}
a[-2a_+f_1-2f_2-b_+f_3-f_6\nabla^+_1]=0\tag{V}
</math>
<math>\label{VI}
b[-2b_+f_4-2f_5-a_+f_3-f_7\nabla^+_2]=0\tag{VI}
</math>

Adding the two forms gives

\begin{eqnarray}
<f>_R+<f>_L=\frac{1}{2}(a_+^2+a_-^2)f_1+a_+f_2+\frac{1}{2}(a_+b_++a_-b_-)f_3\nonumber\\
\ +\frac{1}{2}(b_+^2+b_-^2)f_4+b_+f_5+\frac{1}{2}f_6(a_+\nabla_1^++a_-\nabla^-_1)\nonumber\\
\ +\frac{1}{2}(b_+\nabla_2^++b_-\nabla^-_2)+f_8(Y)+f_8(Y^\prime)\nonumber
\end{eqnarray}

Subtracting <math>\frac{x}{2}\times({\rm I})</math> gives

\begin{eqnarray}\ \
<f>_R+<f>_L = \frac{1}{2}[(1-x)a_+^2 + a_-^2]f_1 +
(1-\frac{x}{2})a_+f_2+\frac{1}{2}[(1-x)a_+b_++a_-b_-]f_3\nonumber\\
+\frac{1}{2}[(1-x)b_+^2+b_-^2]f_4+(1-\frac{x}{2})f_5b_++\frac{1}{2}f_6[(1-\frac{x}{2})a_+\nabla_1^++a_-\nabla^-_1]\nonumber\\
+\frac{1}{2}[(1-\frac{x}{2})b_+\nabla^+_2+b_-\nabla^-_2]+f_8(Y)+f_8(Y^\prime)\nonumber
\end{eqnarray}

If <math>x=1</math>,
\begin{eqnarray}
<f>_R+<f>_L &=&
\frac{1}{2}[a_-^2f_1+a_+f_2+a_-b_-f_3+b_-^2f_4+b_+f_5\nonumber\\
&+&\left(\frac{1}{2}a_+\nabla^+_1+a_-\nabla^-_1\right)f_6+\left(\frac{1}{2}b_+\nabla^+_2+b_-\nabla^-_2\right)f_7]+f_8(Y)+f_8(Y^\prime)\nonumber
\end{eqnarray}

The General Hermitian Property for arbitrary <math>x</math> gives
\begin{eqnarray}\ \
\nabla_1^2=
\frac{1}{4}\{\frac{1}{r_1^2}[(1-x)a_+^2+a_-^2+2(1-\frac{x}{2})a_+-2[l_1(l_1+1)+l_1^\prime(l_1^\prime+1)]\nonumber\\
-\frac{2}{r_1}[(1-x)\alpha_+a_++\alpha_-a_-+2(1-\frac{x}{2})\alpha_+]+(1-x)\alpha_+^2
+ \alpha^2_-]\nonumber\\
+\frac{2(r_1-r_2\cos\theta)}{r_1r^2}[(1-x)(a_+-\alpha_+r_1)c_++(a_--\alpha_-r_2)c_-]\nonumber\\
-\frac{2r_2}{r_1r^2}[(1-\frac{x}{2})c_+\hat{r}_2\cdot\hat{\nabla}_1^++c_-\hat{r}_2\hat{\nabla}_1^-]+\frac{(1-x)c_+^2+c_-^2+2(1-\frac{x}{2})c_+}{r^2}\}\nonumber
\end{eqnarray}.

Use
<math>\begin{eqnarray}\ \
\frac{-2r_2}{r_1r^2}[(1-\frac{x}{2})c_+\hat{r}_2\hat{\nabla}_1^++c_-\hat{r}_2\hat{\nabla}_1^-]=\frac{2}{r_1^2}\left[(1-\frac{x}{2})\Lambda(\Lambda+1)-\frac{c_-}{c_+}[l_1^\prime(l_1^\prime+1)+l_1(l_1+1)]\right]\end{eqnarray}</math>,

<math>\begin{eqnarray}
\frac{2(r_1-r_2\cos\theta)}{r_1r^2}(a_--\alpha_-r_1)c_- = 2\frac{c_-}{c_+}\left[\frac{-1}{r_1^2}[a_-(a_++1)]+\frac{1}{r_1}[a_-\alpha_++\alpha_-(a_++2)]-\alpha_-\alpha_+\right]\nonumber
\end{eqnarray}</math>,

and
<math>\begin{eqnarray}\ \
\frac{2(r_1-r_2\cos\theta)}{r_1r^2}(a_+-\alpha_+r_1)c_+ = 2\left[-\frac{1}{r_1^2}[a_+(a_++1)]+\frac{1}{r_1}[a_+\alpha_++\alpha_+(a_++2)]-\alpha_+^2\right]
\end{eqnarray}</math>.

Substituting into <math>\nabla_1^2</math> gives

\begin{eqnarray}
\nabla^2_1 = \frac{1}{4}\{\frac{1}{r_1^2}[-(1-x)a_+^2 + a_-^2 +xa_+ +
2(1-\frac{x}{2})\Lambda(\Lambda+1)\nonumber\\
-2l_1(l_1+1)(1-\frac{c_-}{c_+}) -
2l_1^\prime(l_1^\prime+1)(1+\frac{c_-}{c_+})-\frac{2c_-a_-}{c_+}(a_++1)]\nonumber\\
- \frac{2}{r_1}\left[-(1-x)\alpha_+(a_++2)+\alpha_-a_-+2(1-\frac{x}{2})\alpha_+
-\frac{c_-}{c_+}[a_-\alpha_++\alpha_-(a_++2)]\right]\nonumber\\
-(1-x)\alpha_+^2+\alpha_-^2-2\frac{c_-}{c_+}\alpha_-\alpha_+ +\frac{1}{r^2}\left[(1-x)c_+^2+c_-^2+2(1-\frac{x}{2})c_+\right]\}\nonumber
\end{eqnarray}

This has the form
<math>
\nabla_1^2 =
\frac{1}{4}\left[\frac{A_1}{r_1^2}+\frac{B_1}{r_1}+\frac{C_1}{r^2}+D_1\right]
</math>
with

<math>\begin{array}{ccc}
A_1=-(1-x)a_+^2+a_-^2
+xa_++2(1-\frac{x}{2})\Lambda(\Lambda+1)-2l_1(l_1+1)(1-\frac{c_-}{c_+})\
-2l_1^\prime(l_1^\prime+1)(1+\frac{c_-}{c_+})-\frac{2c_-a_-}{c_+}(a_++1) \ \
\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \
\end{array}
</math>

<math>
B_1 =
2\left[(1-x)\alpha_+(a_++2)-\alpha_-a_--2(1-\frac{x}{2})\alpha_++\frac{c_-}{c_+}[a_-\alpha_++\alpha_-(a_++2)]\right]
</math>

<math>
C_1 = (1-x)c_+^2+ c_-^2 +2(1-\frac{x}{2})c_+
</math>

<math>
D_1 = -1(1-x)\alpha_+^2 +\alpha_-^2 - 2\frac{c_-}{c_+}\alpha_-\alpha_+
</math>

The complete Hamiltonian is then (in Z-scaled a.u.)

<math>\begin{eqnarray}
H =-\frac{1}{2}\nabla^2_1 -\frac{1}{2}\nabla^2_2 -\frac{1}{r_1} -\frac{1}{r_2} +\frac{Z^{-1}}{r}\nonumber\\
= -\frac{1}{2}\nabla^2_1 -\frac{1}{2}\nabla^2_2 -\frac{1}{r_1} -\frac{Z-1}{Zr_2}+Z^{-1}\left(\frac{1}{r}-\frac{1}{r_2}\right)\nonumber\\
\ =-\frac{1}{8}\left[\frac{A_1}{r_1^2}+\frac{B_1+8}{r_1}+\frac{C_1}{r^2}+D_1+D_2 +\frac{A_2}{r_2^2} + \frac{B_2 + 8(Z-1)/Z}{r_2}+\frac{C_2}{r^2}\right]\nonumber
+ Z^{-1}\left(\frac{1}{r}-\frac{1}{r_2}\right)\nonumber
\end{eqnarray}</math>

The screened hydrogenic energy is
<math>\begin{eqnarray}
E_{SH}=-\frac{1}{2}\left[\frac{1}{n_1^2}+ \left(\frac{Z-1}{Z}\right)^2\frac{1}{n_2^2}\right]
\end{eqnarray}</math>
so that

<math>\begin{eqnarray}
H-E_{SH} &=& -\frac{1}{8}[
\frac{A_1}{r_1^2}+\frac{B_1+8}{r_1}+\frac{C_1}{r^2} + D_1 + D_2
-\frac{4}{n_1^2} -\left(\frac{Z-1}{Z}\right)^2\frac{4}{n_2^2}\nonumber
+\frac{A_2}{r_2}+\frac{B_2+8(Z-1)/Z}{r_2}+\frac{C_2}{r^2}] +
Z^{-1}\left(\frac{1}{r} -\frac{1}{r_2}\right)\nonumber
\end{eqnarray}</math>

==Optimization of Nonlinear Parameters==
- The traditional method of performing Hylleraas calculations is to write the
basis set in the form

<math>
\Psi = \sum_{i,j,k}c_{i,j,k}\varphi_{i,j,k}(\alpha,\beta)\pm exchange
</math>

with

<math>
\varphi_{i,j,k}(\alpha,\beta) = r_1^ir_2^jr_{12}^ke^{-\alpha r_1 - \beta
r_2}\mathcal{Y}^M_{l_1l_2L}.
</math>

The usual procedure is to set <math>\alpha = Z</math> so that it represents the inner
<math>1s</math> electron, and then to vary <math>\beta</math> so as to minimize the energy.

[[File:OptNon.jpg|500px]]

Since <math>\beta</math> appears in <math>\Psi</math> as a non-linear parameter, the entire
calculation must be repeated for each value of <math>\beta</math>. However, the minimum
becomes progressively smaller as the basis set is enlarged.

[[File:OptNon02.jpg|500px]]

*Difficulties

1. If the basis set is constructed so that <math> i +j +j \leq N</math>, the the
number of terms is
<math>
(N+1)(N+2)(N+3)/6
</math>
N=14 already gives 680 terms and an accuracy of about 1 part in <math>10^{10}</math> for
low-lying states. A substantial improvement in accuracy would require much
larger basis sets, together with multiple precision arithmetic to avoid loss
of significant figures when high powers are included.

2. The accuracy rapidly deteriorates as one goes to more highly excited
states - about 1 significant figure is lost each time the principle quantum
number is increased.

*Cure

We have found that writing basis sets in the form
\begin{eqnarray}
\Psi &=& \sum_{i,j,k}\left(c^{(1)}_{i,j,k}\varphi_{ijk}(\alpha_1\beta_1) +
c_{ijk}^{(2)}\varphi_{ijk}(\alpha_2\beta_2)\right) \pm exchange \nonumber\\
&=& \psi({\bf r}_1,{\bf r}_2) \pm \psi({\bf r}_2,{\bf r}_1)\nonumber
\end{eqnarray}
so that each combination of powers is included twice with different nonlinear
parameters gives a dramatic improvement in accuracy for basis sets of about
the same total size.

However, the optimization of the nonlinear parameters is now much more
difficult, and an automated procedure is needed.

If
<math>
E = \frac{<\Psi|H|\Psi>}{<\Psi|\Psi>}</math> and we assume <math><\Psi|\Psi> = 1</math>, then

<math>
\frac{\partial E}{\partial\alpha_t} = -2 <\Psi|H-E|r_1\psi({\bf r}_1,{\bf
r}_2;\alpha_t)\pm r_21\psi({\bf r}_2,{\bf
r}_1;\alpha_t)>
</math>
where
<math>
\psi({\bf r}_1,{\bf r}_2;\alpha_t) =
\sum_{i,j,k}c_{ijk}^{(t)}\varphi_{ijk}(\alpha_t\beta_t)
</math>

*Problem

1. Prove the above.

2. Prove that there is no contribution from the implicit dependence of
<math>c_{ijk}^{(t)}</math> on <math>\alpha_t</math> if the linear parameters have been optimized.

==The Screened Hydrogenic Term==

If the Hamiltonian is written in the form (in Z-scaled a.u.)
<math>\begin{eqnarray}\
H = H_0({\bf r}_1,Z) + H_0({\bf r}_2,Z-1) +
Z^{-1}\left(\frac{1}{r}-\frac{1}{r_2}\right)
\end{eqnarray}</math>
with
<math>\
H_0({\bf r}_1,Z) = -\frac{1}{2}\nabla_1^2 - \frac{1}{r_1}\nonumber
</math> and
<math>
H_0({\bf r}_2,Z-1) = -\frac{1}{2}\nabla_2^2 - \frac{Z-1}Z{r_2}\nonumber
</math>,
then the eigenvectors of <math>H_0({\bf r},Z) +H_0({\bf r}_2,Z-1)</math> are products of
hydrogenic orbitals
<math>
\Psi_0=\psi_0(1s,Z)\psi_0(nl,Z-1)
</math>
and the eigenvalue is
<math>\begin{eqnarray}
E_{SH} = \left[-\frac{1}{2} - \left(\frac{Z-1}{Z}\right)^2\frac{1}{2n^2}\right]Z^2 a.u.
\end{eqnarray}</math>
called the screened hydrogenic eigenvalue. From highly excited states,
<math>E_{SH}</math> and <math>\Psi_0</math>are already excellent approximations.

For example, for the 1s8d states, the energies are

<math>\begin{eqnarray}\\
E(1s8d^1D)&=& -2.007\,816\,512\,563\,81 {\rm\ a.u.}\nonumber\\
E(1s8d^3D)&=& - 2.007\,817\,934\,711\,71 {\rm\ a.u.}\nonumber\\
E_{SH} &=& -2.007\,812\,5 {\rm\ a.u.} \nonumber
\end{eqnarray}</math>

It is therefore advantageous to include the screened hydrogenic terms in the
basis set so that the complete trial function becomes
<math>
\Psi = c_0\Psi_0 +
\sum_{ijk}\left[c_{ijk}^{(1)}\varphi_{ijk}(\alpha_1,\beta_1)+c_{ijk}^{(2)}\varphi(\alpha_2,\beta_2)\right]\pm
{\rm exchange}
</math>

Also, the variational principal can be re-expressed in the form
<math>\begin{eqnarray}
E= E_{SH}+\frac{<\Psi|H-E_{SH}|\Psi>}{<\Psi|\Psi>}
\end{eqnarray}</math>
so that the <math>E_{SH}</math> term can be cancelled analytically from the matrix
elements.

For example
<math>
<\Psi_0|H-E_{SH}|\Psi_0> = 0
</math>

and
<math>
<\varphi_{ijk}|H-E_{SH}|\Psi_0> =
Z^{-1}<\varphi_{ijk}|\frac{1}{r}-\frac{1}{r_2}|\Psi_0>
</math>

Recall that
<math>\begin{eqnarray}\ \
I_0(a,b,c) = \frac{2}{c+2}\sum^{[c+1]/2}_{i=0}\left(\begin{array}{c}
c+2\\
2i+1
\end{array}\right) \left[f(p,q;\beta) + f(p^\prime,q^\prime;\alpha)\right]
\end{eqnarray}</math>
where

<math>\begin{eqnarray}
f(p,q;x) =
\frac{q!}{x^{q+1}(\alpha+\beta)^{p+1}}\sum^p_{j=0}\frac{(p+j)!}{j!}\left(\frac{x}{\alpha+\beta}\right)^j
\end{eqnarray}</math>

<math>
\begin{array}{cc}
p=a+2i+2 & p^\prime = b+2i+2\\
q= b+c-2i+2 & q^\prime =a+c-2i+2
\end{array}
</math>

If <math>\beta\ll\alpha</math>, the <math>f(p,q;\beta)\gg f(p^\prime,q^\prime;\alpha)</math> and the
<math>i=0</math> term is the dominant contribution to <math>f(p,q;\beta)</math>. However, since
this term depends only on the sum of powers <math>b+c</math> for <math>r_2</math>, it cancels exactly
from the matrix element of
<math>
\frac{1}{r_{12}} - \frac{1}{r_2}
</math>
and can therefore
be omitted in the calculations of integrands, thereby saving many significant
figures. This is especially valuable when <math>b+c</math> is large and <math>a</math> is small.

==Small Corrections==
*Mass Polarization
If the nuclear mass is not taken to be infinite, then the Hamiltonian is
<math>\begin{eqnarray}\ \
H = \frac{P_N^2}{2M} + \sum_{i=1}\frac{p_i^2}{2m} + V
\end{eqnarray}</math>

[[File:SmallCor.jpg|500px]]

where <math>M =M_A-nm</math> is the nuclear mass and <math>m</math> is the electron mass.

Change variables to the C of M
<math>\begin{eqnarray}
{\bf R} = \frac{1}{M+nm}\left[M{\bf r}_N + m({\bf r}_1 + {\bf r}_2 +
\ldots)\right]
\end{eqnarray}</math>

and relative variables

<math>
{\bf s}_i = {\bf r}_i - {\bf r}_N.
</math>

Then

<math>\begin{eqnarray}\ \ \ \ \
H = \frac{1}{2(M+nm)}{\bf p}_R^2+\frac{1}{2m}\sum_{i=1}^n{\bf p}_{s_i}^2
+\frac{1}{2M}\sum_{i,k}{\bf p}_{s_i}\cdot {\bf p}_{s_k} + V({\bf s}_1, {\bf
s}_2, \ldots)\nonumber
= \frac{1}{M+nm}{\bf p}_R^2+\frac{1}{2\mu}\sum_{i=1}^n{\bf
p}_{s_i}^2+\frac{1}{2M}\sum_{i\neq k}{\bf p}_{s_i}\cdot {\bf p}_{s_k}+V({\bf s}_1, {\bf
s}_2, \ldots)\nonumber
\end{eqnarray}</math>

where
<math>\begin{eqnarray}
\mu = \left(\frac{1}{M} + \frac{1}{m}\right)^{-1}
\end{eqnarray}</math>

If <math>n=2</math>, the Schrödinger equation is
<math>\begin{eqnarray}\ \ \
\left[ -\frac{\hbar^2}{2\mu}(\nabla^2_{s_1}+ \nabla^2_{s_2}) -
\frac{\hbar^2}{M}\nabla_{s_1}\nabla_{s_2}-\frac{Ze^2}{s_1}-\frac{Ze^2}{s_2}+\frac{e^2}{s_{12}}\right]\psi
= E\psi
\end{eqnarray}</math>.

Define the reduced mass Bohr radius
<math>\begin{eqnarray}
a_\mu = \frac{\hbar^2}{\mu e^2} = \frac{m}{\mu}a_0
\end{eqnarray}</math>
and

<math>\begin{eqnarray}
\rho_i = \frac{Zs_i}{a_\mu}\nonumber\\
\varepsilon = \frac{E}{Z^2(e^2/a_\mu)}\nonumber
\end{eqnarray}</math>

The Schrödinger equation then becomes
<math>\begin{eqnarray}\ \ \
\left[
-\frac{1}{2}(\nabla^2_{\rho_1}+\nabla^2_{\rho_2})-\frac{\mu}{M}\vec{\nabla}_{\rho_1}\cdot\vec{\nabla}_{\rho_2} - \frac{1}{\rho_1}-\frac{1}{\rho_2}+\frac{Z^{-1}}{\rho_{12}}\right]\psi = \varepsilon \psi
\end{eqnarray}</math>

The effects of finite mass are thus

1. A reduced mass correction to all energy levels of <math>(e^2/a_\mu)/(e^2/a_0)
= \mu/m</math>, i.e.
<math>\begin{eqnarray}
E_M=\frac{\mu}{m}E_\infty
\end{eqnarray}</math>

2. A specific mass shift given in first order perturbation theory by
<math>\begin{eqnarray}
\Delta\varepsilon =
-\frac{\mu}{M}<\psi|\nabla_{\rho_1}\cdot\nabla_{\rho_2}|\psi>
\end{eqnarray}</math>
or

<math>\begin{eqnarray}
\Delta\varepsilon =
-\frac{\mu}{M}<\psi|\nabla_{\rho_1}\cdot\nabla_{\rho_2}|\psi>\frac{e^2}{a_\mu}\nonumber\\
= -\frac{\mu^2}{mM}<\psi|\nabla_{\rho_1}\cdot\nabla_{\rho_2}|\psi>\frac{e^2}{a_0}\nonumber
\end{eqnarray}</math>

Since <math>\nabla_{\rho_1}\cdot\nabla_{\rho_2}</math> has the same angular properties as
<math>{\bf\rho}_1 \cdot {\bf\rho}_2</math>, the operator is like the product of two
dipole operators. For product type wave functions of the form
$
\psi = \psi(1s)\psi(nl) \pm \mbox{exchange},
$
the matrix element vanishes for all but P-states.

=Radiative Transitions=

==Transition Integrals and the Line Strength==

For an electric dipole transition <math>\gamma'\,L'\,S'\,J'\,M' \rightarrow \gamma\,L\,S\,J\,M</math>, the line strength is defined by
\begin{equation}
S = \sum_{M',\mu,M}|\langle\gamma'\,L'\,S'\,J'\,M'|r_\mu|\gamma\,L\,S\,J\,M\rangle|^2
\end{equation}
Using the Wigner-Eckart theorem, this is related to the reduced matrix element by
\begin{eqnarray}
S(J',J) = |\langle\gamma'\,L'\,S'\,J'||r||\gamma\,L\,S\,J\,M\rangle|^2
\sum_{M',\mu,M}\left|\left(\matrix{J' & 1 & J \cr
-M'& \mu &M \cr}\right)\right|^2
\end{eqnarray}
Using the sum rule for the 3-<math>j</math> symbols
\begin{equation}
\sum_{M',\mu,M}\left|\left(\matrix{J' & 1 & J \cr
-M'& \mu &M \cr}\right)\right|^2 = 1
\end{equation}
we see that the line strength is identical to the reduced matrix element
\begin{equation}
S(J',J) = |\langle\gamma'\,L'\,S'\,J'||r||\gamma\,L\,S\,J\,M\rangle|^2
\end{equation}
For a nonrelativistic calculation in <math>LS</math> coupling, <math>S = S'</math> and we can next strip out the dependence on <math>J</math> and <math>J'</math> by use of Eq.\ (7.1.7) of Edmonds { Angular Momentum in Quantum Mechanics} with <math>k = 1</math> to obtain
\begin{equation}
S(J',J) = (-1)^{L'+S+J+1}[(2J+1)(2J'+1)]^{1/2}\left\{
\matrix{L' & J'& S\cr
J & L & 1\cr}\right\}\langle\gamma' L'|| r || \gamma L\rangle
\end{equation}
where <math>\langle\gamma' L'|| r || \gamma L\rangle</math> is the reduced matrix element if spin is neglected. Note that if spin is neglected, then the line-strength is simply
\begin{equation}
S(L',L) = |\langle\gamma' L'|| r || \gamma L\rangle|^2
\end{equation}
To calculate the reduced matrix element, use a particular example along with the Wigner-Eckart theorem
such as
\begin{equation}
\langle\gamma'\, L'|| r || \gamma\, L\rangle = (-1)^{L'}\langle\gamma'\, L'\, 0| z | \gamma\, L\, 0\rangle{\Large /}\left(\matrix{L' & 1 & L \cr
0 & 0 & 0 \cr}\right)
\end{equation}

===Correlated Two-electron Integrals===
For the case of two-electron integrals, <math>z = z_1 + z_2</math> and <math>r = r_1 + r_2</math>. As discussed in G.W.F. Drake, Phys. Rev. A 18, 820 (1978), the general integral in Hylleraas coordinates for the integral over the <math>z_1</math> term can be written in the form
\begin{equation}
\label{int1}
\langle\gamma'\, L'\, 0| z_1 | \gamma\, L\, 0\rangle = \sum_\Lambda C_\Lambda I_\Lambda(\gamma',\gamma)
\end{equation}
where <math>I_\Lambda(\gamma',\gamma)</math> stands for the purely radial integral
\begin{equation}
\label{radial}
I_\Lambda(\gamma',\gamma) = \int_0^\infty r_1\,dr_1\int_0^\infty r_2\,dr_2\int_{|r_1-r_2|}^{r_1+r_2}r_{12}\, dr_{12}
R_{\gamma'}^*\,r_1\, R_\gamma P_\Lambda(\cos\theta_{12})
\end{equation}
and <math>\cos\theta_{12} = (r_1^2 + r_2^2 - r_{12}^2)/(2r_1r_2)</math> is a short-hand notation for a radial function of <math>r_1</math>, <math>r_2</math> and <math>r_{12}</math>. Also, <math>P_\Lambda(\cos\theta_{12})</math> is a Legendre polynomial,
and, <math>R_\gamma = R_\gamma(r_1,r_2,r_{12})</math> is the radial part of the Hylleraas wave function for state <math>\gamma</math> such
that the total wave function is <math>\psi({\bf r}_1,{\bf r}_2) = R_\gamma(r_1,r_2,r_{12}){\cal Y}_{l_1,l_2,L}^M(\hat{\bf r}_1,\hat{\bf r}_2)\pm {\rm exchange}</math>, normalized to unity, and <math>{\cal Y}_{l_1,l_2,L}^M(\hat{\bf r}_1,\hat{\bf r}_2)</math> is a vector coupled product of spherical harmonics. The coefficents <math>C_\Lambda</math> in Eq.\ (\ref{int1}) can be calculated from Eq.\ (32) of the above 1978 paper. They need to be calculated separately for he direct-direct and the direct-exchange contributions. For the reduced matrix element of <math>z_1</math>, the leading 3-<math>j</math> symbol can be dropped, leaving (for the case <math>k = 1</math>)
\begin{eqnarray}
C_\Lambda^{\rm reduced} &=& \frac{(-1)^{l_2'+L'+L+1}}{2} (l_1,l_1',l_2,l_2',L,L')^{1/2}
\sum_{\lambda={\rm max}(l_1-1,0)}^{l_1+1} (\lambda,\Lambda,1)
\left(\matrix{l_1 & 1 & \lambda \cr
0 & 0 & 0 \cr}\right)
\left(\matrix{l_1'& \lambda & \Lambda \cr
0 & 0 & 0 \cr}\right)\nonumber\\
&\times&\left(\matrix{l_2' & 1_2 & \Lambda \cr
0 & 0 & 0 \cr}\right)
\left\{\matrix{L' & \lambda & l_2 \cr
\Lambda& l_2' & l_1' \cr}\right\}
\left\{\matrix{L' & \lambda & l_2 \cr
l_1 & L & 1 \cr}\right\}
\end{eqnarray}
where the notation <math>(a,b,c,\cdots)</math> means <math>(2a+1)(2b+1)(2c+1)\cdots</math>.
A similar expression can be obtained for the angular coefficients for the <math>z_2</math> term by interchanging the subscripts 1 and 2 throughout and multiplying by <math>(-1)^{L+L'+1}</math>.

===Example===
As an example, for the <math>1s^2\;^1S - 1s2p\;^1P</math> transition of helium, the only nonvanishing <math>C_\Lambda</math> angular coefficients are <math>C_1 = 1/2</math> for the direct-direct term (i.e. <math>P_1(\cos\theta_{12}) = \cos\theta_{12}</math> is included in the radial integral (\ref{radial})) and <math>C_0 = 1/2</math> for the direct-exchange term (i.e. <math>P_0(\cos\theta_{12}) = 1</math> is included in the radial integral (\ref{radial})). The results for infinite nuclear mass are <math>S(1\;^1S-2\;^1P) = (0.781\,127\,158\,667)^2</math> and <math>f(1\;^1S-2\;^1P) = -0.276\,164\,703\,736</math> where
\begin{equation}
f(1\;^1S-2\;^1P) = \frac{2}{3}[E(1\;^1S) - E(2\;^1P)]S(1\;^1S-2\;^1P)
\end{equation}
is the absorption oscillator strength.

===Test===
This is a test edit

Theory Notes

2020-12-14T00:39:15Z

Drake: Undo revision 789 by Drake (talk)

<center>
<H1>Notes on solving the Schrödinger equation in Hylleraas coordinates for heliumlike atoms
Gordon W.F. Drake
Department of Physics, University of Windsor
Windsor, Ontario, Canada N9B 3P4
(Transcribed from hand-written notes and edited by Lauren Moffatt. Last revised November 22, 2017 to correct a missing subscript in the expression for $B_1$ in the Hamiltonian in Sect. 11 General Hermitian Property.)
</H1>
</center>

==Introduction==
These notes describe some of the technical details involved in solving the Schrödinger equation for a heliumlike atom of nuclear charge Z in correlated Hylleraas coordinates. The standard for computational purposes is to express all quantities in a form that is valid for electronic states of arbitrary angular momentum, and to express matrix elements of the Hamiltonian in a manifestly Hermitian symmetrized form. Further information and derivations of the mathematical techniques for angular integrals and radial recursion relations can be found in the paper
G.W.F. Drake, Phys. Rev. A <math>{\bf 18}</math>, 820 (1978).

The starting point is the two-electron Schrödinger equation for infinite nuclear mass
\begin{equation} \left[-\frac{\hbar^2}{2m}(\nabla^2_1 +\nabla^2_2) - \frac{Ze^2}{r_1} -
\frac{Ze^2}{r_2}+\frac{e^2}{r_{12}} \right]\psi = E\psi\nonumber
\end{equation}
where <math>m</math> is the electron mass, and <math>r_{12} = |{\bf r}_1 - {\bf r}_2|</math> (see diagram below).

[[File:Pic1.jpg|250px]]

Begin by rescaling distances and energies so that the Schrödinger equation can be expressed in a dimensionless form. The dimensionless Z-scaled distance is defined by <math>\begin{eqnarray} \rho = \frac{Zr}{a_0} \end{eqnarray} </math> where <math> \begin{eqnarray} a_0 = \frac{\hbar^2}{me^2} \end{eqnarray} </math> is the atomic unit (a.u.) of distance equal to the Bohr radius <math>0.529\, 177\, 210\, 92(17) \times 10^{-10}</math> m. Then

\begin{equation}[-\frac{\hbar^2}{2m}Z^2(\frac{me^2}{\hbar^2})^2(\nabla^2_{\rho_1}+\nabla^2_{\rho_2})
- Z^2\frac{e^2}{a_0}\rho^{-1}_1 - Z^2\frac{e^2}{a_0}\rho^{-1}_2 +
\frac{e^2}{a_0}Z\rho^{-1}_{12}]\psi= E\psi\nonumber\end{equation}

But <math>\begin{eqnarray} \frac{\hbar^2}{m}(\frac{me^2}{\hbar^2})^2 = \frac{e^2}{a_0} \end{eqnarray} </math> is the hartree atomic unit of of energy <math>(E_h = 27.211\,385\,05(60)</math> eV, or equivalently <math>E_h/(hc) = 2194.746\,313\,708(11)</math> m<math>^{-1})</math>. Therefore, after multiplying through by <math>a_0/(Ze)^2</math>, the problem to be solved in Z-scaled dimensionless units becomes

\begin{equation} \left[-\frac{1}{2}(\nabla^2_{\rho_1}+\nabla^2_{\rho_2}) -
\frac{1}{\rho_1} - \frac{1}{\rho_2} + \frac{Z^{-1}}{\rho_{12}}\right]\psi =
\varepsilon\psi\nonumber \end{equation}

where <math>\begin{eqnarray}\varepsilon = \frac{Ea_0}{(Ze)^2} \end{eqnarray}</math> is the energy in Z-scaled atomic units. For convenience, rewrite this in the conventional form
\begin{equation}
H\psi = \varepsilon\psi\nonumber \end{equation}
where (using <math>r</math> in place of <math>\rho</math> for the Z-scaled distance)
\begin{equation}
H = -\frac{1}{2}(\nabla^2_1+\nabla^2_2) - \frac{1}{r_1}-\frac{1}{r_2} + \frac{Z^{-1}}{r_{12}}
\nonumber\end{equation}
is the atomic Hamiltonian for infinite nuclear mass.

==The Hartree Fock Method==

For purposes of comparison, and to define the correlation energy, assume that <math>\psi({\bf r}_1,{\bf r}_2)</math> can be written in the separable product form
\begin{equation}
\psi({\bf r}_1,{\bf r}_2) = \frac{1}{\sqrt{2}} \left[ u_1(r_1) u_2(r_2) \pm u_2(r_1) u_1(r_2) \right] \nonumber
\end{equation}
for the <math>1s^2\ ^1\!S</math> ground state. Because of the <math>\frac{1}{r_{12}}</math> term in the Schrödinger equation, the exact solution connot be expressed in this form as a separable product. However, the Hartree-Fock approximation corresponds to finding the best possible solution to the Schrödinger equation
\begin{equation}
H\psi({\bf r}_1,{\bf r}_2) = E\psi({\bf r}_1,{\bf r}_2)\nonumber
\end{equation}
that can nevertheless be expressed in this separable product form, where as before
\begin{equation}
H = -\frac{1}{2}(\nabla^2_1+\nabla^2_2) - \frac{1}{r_1}- \frac{1}{r_2} + \frac{Z^{-1}}{r_{12}}\nonumber
\end{equation}
is the full two-electron Hamiltonian. To find the best solution, substitute into <math>\left\langle\psi \lvert H-E \rvert \psi\right\rangle</math> and require this expression to be stationary
with respect to arbitrary infinitesimal variations <math>\delta u_1</math> and <math>\delta
u_2</math> in <math>u_1</math> and <math>u_2</math> respectively; i.e.

\begin{equation}
\frac{1}{2} \left\langle \delta u_1(r_1)u_2(r_2) \pm u_2(r_1)\delta
u_1(r_2) \lvert H-E \rvert u_1(r_1) u_2(r_2)\pm u_2(r_1) u_1(r_2)\right\rangle\nonumber
\end{equation}

\begin{equation}
=\int\delta u_1(r_1) d{\bf r}_1 \left\{ \int d{\bf r}_2 u_2(r_2) \left( H - E \right) \left[ u_1(r_1)u_2(r_2) \pm u_2(r_1) u_1(r_2) \right] \right\}\nonumber
\end{equation}

\begin{equation}
= 0 \nonumber
\end{equation}

for arbitrary <math> \delta u_1(r_1).\nonumber </math> Therefore <math>\{\int d{\bf r}_2 \ldots \} = 0</math>.

Similarly, the coefficient of <math>\delta u_2</math> would give

\begin{equation}
\int d{\bf r}_1 u_1(r_1) \left( H-E \right) \left[ u_1(r_1) u_2(r_2) \pm
u_2(r_1) u_1(r_2) \right] = 0\nonumber
\end{equation}

Define

\begin{equation}
I_{12} = I_{21} = \int d{\bf r}\, u_1(r)u_2(r), \nonumber
\end{equation}

\begin{equation}
H_{ij} = \int d{\bf r}\, u_i(r)(-\frac{1}{2}\nabla - \frac{1}{r})u_j(r), \nonumber
\end{equation}

\begin{equation}
G_{ij}(r) = \int d{\bf r}^\prime u_i(r^\prime)\frac{1}{|{\bf r} - {\bf r}\prime|}u_j(r^\prime)\nonumber
\end{equation}

Then the above equations become the pair of integro-differential equations

\begin{equation}
[ H_0 - E + H_{22}+G_{22}(r)]u_1(r) = \mp [ I_{12}(H_0-E) + H_{12}+G_{12}(r)]u_2(r)\nonumber
\end{equation}

\begin{equation}
[H_0-E+H_{11}+G_{11}(r)]u_2(r) = \mp [I_{12}(H_0-E) + H_{12}+G_{12}(r)]u_1(r)\nonumber
\end{equation}

These must be solved self-consistently for the "constants" <math>I_{12}</math> and
<math>H_{ij}</math> and the function <math>G_{ij}(r)</math>.

The Hartree Fock energy is <math>E \simeq -2.87\ldots </math>a.u. while the exact energy is <math>E = -2.903724\ldots </math>a.u.

The difference is called the "correlation energy" because it arises from the
way in which the motion of one electron is correlated to the other. The
Hartree Fock equations only describe how one electron moves in the average field
provided by the other (mean-field theory).

==Configuration Interaction==
Expand
\begin{equation}
\psi({\bf r}_1,{\bf r}_2)= C_0u^{(s)}_1(r_1)u^{(s)}_1(r_2) + C_1u^{(P)}_1({\bf r}_1)u^{(P)}_1({\bf r}_2){\cal Y}^0_{1,1,0}(\hat{\bf r}_1, \hat{\bf r}_2)+C_2u^{(d)}_1({\bf r}_1)u^{(d)}_2({\bf r}_2){\cal Y}^0_{2,2,0}(\hat{\bf r}_1, \hat{\bf r}_2)+... \pm {\rm\ exchange}
\end{equation}
where
\begin{equation}
{\cal Y}^M_{l_1,l_2,L}(\hat{\bf r}_1, \hat{\bf r}_2)=\sum_{m_1,m_2}Y^{m_1}_{l_1}({\bf r}_1)Y^{m_2}_{l_2}({\bf r}_2)\times <l_1l_2m_1m_2\mid LM>
\end{equation}

This works, but is slowly convergent, and very laborious. The best CI calculations are accurate to <math> ~10^{-7}</math> a.u.

==Hylleraas Coordinates==
[E.A. Hylleraas, Z. Phys. <math>{\bf 48}, 469(1928)</math> and <math>{\bf 54}, 347(1929)</math>]
suggested using the co-ordinates <math>r_1</math>, <math>r_2</math> and <math>r_{12}</math> or equivalently

\begin{eqnarray}
s &=& r_1 + r_2, \nonumber\\
t &=& r_1-r_2, \nonumber\\
u &=& r_{12}\nonumber
\end{eqnarray}

and writing the trial functions in the form

\begin{equation}
\Psi({\bf r}_1,{\bf r}_2) = \sum^{i+j+k\leq
N}_{i,j,k}c_{i,j,k}r_1^{i+l_1}r_2^{j+l_2}r_{12}^ke^{-\alpha r_1 - \beta r_2}
\mathcal{Y}^M_{l_1,l_2,L}(\hat{r}_1,\hat{r}_2)\pm \text{exchange}\nonumber
\end{equation}

Diagonalizing H in this non-orthogonal basis set is equivalent to solving
<math> \begin{eqnarray}
\frac{\partial E}{\partial c_{i,j,k}} = 0\nonumber
\end{eqnarray} </math>
for fixed <math>\alpha</math> and <math>\beta</math>.

The diagonalization must be repeated for different values of <math>\alpha</math> and
<math>\beta</math> in order to optimize the non-linear parameters.

==Completeness==
The completeness of the above basis set can be shown by first writing
\begin{equation}
r_{12}^2 = r_1^2 + r_2^2 - 2r_1r_2\cos\theta_{12}
\end{equation}
and, from the spherical harmonic addition theorem,
\begin{equation}
\cos(\theta_{12})=\frac{4\pi}{3}\sum^1_{m=-1}Y^{m*}_1(\theta_1,\varphi_1)Y^m_1(\theta_2,\varphi_2)
\end{equation}
Consider first the S-states. The <math>r_{12}^0</math> terms are like the ss terms in a CI calculation. The
<math>r_{12}^2</math> terms bring in p-p type contributions, and the higher powers bring
in d-d, f-f etc. type terms. In general

\begin{equation}
P_l(\cos\theta_{12}) = \frac{4\pi}{2l+1}\sum^l_{m=-l}{Y^{m}_l}^*(\theta_1,\varphi_1)Y^m_l(\theta_2,\varphi_2)\nonumber
\end{equation}

For P-states, one would have similarly

\begin{array}{lr}
r_{12}^0 & (sp)P\nonumber\\
r_{12}^2 & (pd)P\nonumber\\
r_{12}^4 & (df)P\nonumber\\
\vdots & \vdots\nonumber
\end{array}

For D-states

\begin{array}{lr}
r_{12}^0 & (sd)D & (pp^\prime)D\nonumber\\
r_{12}^2 & (pf)D & (dd^\prime)D\nonumber\\
r_{12}^4 & (dg)D & (ff^\prime)D\nonumber\\
\vdots & \vdots & \vdots\nonumber
\end{array}

In this case, since there are two "lowest-order" couplings to form a
D-state, both must be present in the basis set, i.e.

\begin{equation}
\Psi(r_2,r_2) = \sum c_{ijk}r_1^ir_2^{j+2}r_{12}^ke^{-\alpha r_1-\beta
r_2}\mathcal{Y}^M_{022}(\hat{r}_1,\hat{r}_2)
+\sum d_{ijk}r_1^{i+1}r_2^{j+1}r_{12^k}e^{-\alpha^\prime r_1 - \beta^\prime
r_2}\mathcal{Y}^M_{112}(\hat{r}_1,\hat{r}_2)\nonumber
\end{equation}

For F-states, one would need <math>(sf)F</math> and <math>(pd)F</math> terms.

For G-states, one would need <math>(sg)G</math>, <math>(pf)G</math> and <math>(dd^\prime)G</math> terms.

Completeness of the radial functions can be proven by considering the
Sturm-Liouville problem

\begin{equation}
\left(-\frac{1}{2}\nabla^2-\frac{\lambda}{r_s}-E\right)\psi({\bf r}) = 0\nonumber
\end{equation}

or

\begin{equation}
\left(-\frac{1}{2}\frac{1}{r^2}\frac{\partial}{\partial r}\left({r^2}\frac{\partial}{\partial r}\right) -
\frac{l\left(l+1\right)}{2r^2} - \frac{\lambda}{r} - E\right)u(r) = 0.\nonumber
\end{equation}

For fixed E and variable <math>\lambda</math> (nuclear charge).

The eigenvalues are <math>\lambda_n = (E/E_n)^{1/2}</math>, where <math>E_n =- \frac{1}{2n^2}</math>

[[File:Pic2.jpg|500px]]

and the eigenfunctions are

\begin{equation}
u_{nl}(r) =
\frac{1}{(2l+1)!}\left(\frac{(n+l)!}{(n-l-1)2!}\right)^{1/2}(2\alpha)^{3/2}e^{-\alpha
r}\nonumber
\times \ \left( 2 \alpha r \right)^l\,_{1}F_{1}\left(-n+l+1,2l+2; 2 \alpha r \right)
\end{equation}

with <math>\alpha = (-2E)^{1/2}</math> and <math>n\geq l+1</math>. The confluent hypergeometric function <math> _1F_1(a,b,;z)</math> then denotes a finite polynomial since <math>a=-n+l+1</math> is a negative integer or zero.

Unlike the hydrogen spectrum, which has both a discrete part for <math>E<0</math> and a
continuous part for <math>E>0</math>, this forms an entirely discrete set of finite
polynomials, called Sturmian functions. They are orthogonal with respect to
the potential, i.e.

\begin{equation}
\int^\infty_0 r^2dr \left( u_{n^\prime l}(r)\frac{1}{r}u_{nl}(r) \right) = \delta_{n,n^\prime}\nonumber
\end{equation}

Since they become complete in the limit <math>n\rightarrow\infty</math>, this assures the
completeness of the variational basis set.

[See also B Klahn and W.A. Bingel Theo. Chim. Acta (Berlin) 44, 9 and 27
(1977)].

==Solutions of the Eigenvalue Problem==

For convenience, write

\[
\Psi({\bf r}_1,{\bf r}_2) = \sum^N_{m=1}c_m\varphi_m\nonumber
\]

where <math>m</math> represents the <math>m</math>'th combination of <math>i,j,k</math> and

\begin{equation}
\varphi_{ijk}=r_1^ir_2^jr_{12}^ke^{-\alpha r_1 - \beta
r_2}\mathcal{Y}^M_{l_1,l_2,L}(\hat{r}_1,\hat{r}_2) \pm \mbox{exchange}\nonumber
\end{equation}

\begin{align}

&
\left( \begin{array}{lr}
\cos(\theta) & \sin(\theta)\\
-\sin(\theta) & \cos(\theta)
\end{array} \right)
\left( \begin{array}{lr}
H_{11} & H_{12} \\
H_{12} & H_{22}
\end{array}\right)
\left( \begin{array}{lr}
\cos(\theta) & -\sin(\theta)\\
\sin(\theta) & \cos(\theta)
\end{array} \right)
\\[5pt]
= & \left(\begin{array}{lr}
cH_{11}+sH_{12} & cH_{12} + sH_{22}\\
-sH_{11} + cH_{12} & -sH_{12} + cH_{22}
\end{array}\right)
\left( \begin{array}{cc}
c & -s \\
s & c
\end{array}\right)
\\[5pt]
= & \left(\begin{array}{lr}
c^2H_{11}+s^2H_{22} + 2csH_{12} & (c^2-s^2)H_{12}+cs(H_{22}-H_{11})\\
(c^2-s^2)H_{12}+cs(H_{22}-H_{11}) & s^2H_{11}+c^2H_{22}-2csH_{12}
\end{array}\right)
\end{align}

Therefore

\begin{equation}
\left[\cos^2(\theta)-\sin^2(\theta)\right]H_{12} =
\cos(\theta)\sin(\theta)(H_{11}-H_{22})\nonumber
\end{equation}

and

\begin{equation}
\tan(2\theta) = \frac{2H_{12}}{H_{11}-H_{22}}\nonumber
\end{equation}

i.e.

\begin{equation}
\cos(\theta)=\left(\frac{r+\omega}{2r}\right)^{1/2}\nonumber
\end{equation}

\begin{equation}
\sin(\theta)=-\mbox{sgn}(H_{12})\left(\frac{r-\omega}{2r}\right)^{1/2}\nonumber
\end{equation}

where

\begin{equation}
\omega = H_{22}-H_{11}\nonumber
\end{equation}

\begin{equation}
r=\left(\omega^2+4H_{12}^2\right)^{1/2}\nonumber
\end{equation}

\begin{equation}
E_1=\frac{1}{2}\left(H_{11}+H_{22}-r\right)\nonumber
\end{equation}

\begin{equation}
E_2=\frac{1}{2}\left(H_{11}+H_{22}+r\right)\nonumber
\end{equation}

===Brute Force Method===

- Gives all the eigenvalues and eigenvectors, but it is slow

- First orthonormalize the <math>N</math>-dimensional basis set; i.e. form linear combinations

\begin{equation}
\Phi_m = \sum_{n=1}^N\varphi_nR_{nm}\nonumber
\end{equation}

such that
\begin{equation}
\langle \Phi_m | \Phi_n \rangle = \delta_{m,n}
\end{equation}

This can be done by finding an orthogonal tranformation T such that

\begin{equation}
T^TOT = I = \left(
\begin{array}{ccccc}
I_1 & 0 & 0 & \ldots & 0 \\
0 & I_2 & 0 & \ldots & 0 \\
0 & 0 & I_3 & \ldots & 0 \\
\vdots & \vdots & \vdots & \ddots & \vdots \\
0 & 0 & 0 & \ldots & I_N
\end{array}\right);
\end{equation}

\begin{equation}
O_{mn} = \langle \varphi_m | \varphi_n \rangle \nonumber
\end{equation}

and then applying a scale change matrix

\begin{equation}
S = \left(\begin{array}{ccccc}
\frac{1}{I_1^{1/2}} & 0 & 0 & \ldots & 0\\
0 & \frac{1}{I_2^{1/2}} & 0 & \ldots & 0 \\
0 & 0 & \frac{1}{I_3^{1/2}} & \ldots & 0 \\
\vdots & \vdots & \vdots & \ddots & \vdots \\
0 & 0 & 0 & \ldots & I_N^{1/2}
\end{array}\right)= S^T\nonumber
\end{equation}

Then

\begin{equation}
S^TT^TOTS = 1
\end{equation}

i.e.

\begin{equation}
R^TOR = 1
\end{equation}

with <math>R=TS</math>.

If H is the matrix with elements <math>H_{mn} = \langle \varphi_m | \varphi_n \rangle </math>, then H
expressed in the <math>\Phi_m</math> basis set is

\begin{equation}
H^\prime = R^THR.\nonumber
\end{equation}

We next diagonalize <math>H^\prime</math> by finding an orthogonal transformation W such
that

\begin{equation}
W^T H^\prime W = \lambda = \left(
\begin{array}{cccc}
\lambda_1 & 0 & \ldots & 0 \\
0 & \lambda_2 & \ldots & 0 \\
\vdots & \vdots & \ddots & \vdots \\
0 & 0 & \ldots & \lambda_N
\end{array}\right)\nonumber
\end{equation}

The q'th eigenvector is

\begin{equation}
\Psi^{(q)} = \sum_{n=1}^N\Phi_n W_{n,q} = \sum_{n,n^\prime} \varphi_{n^\prime} R_{n^\prime ,n} W_{n,q} \nonumber
\end{equation}

i.e.

\begin{equation}
c_{n^\prime}^{(q)} = \sum_{n=1}^N R_{n^\prime n} W_{n,q}
\end{equation}

===The Power Method===

- Based on the observation that if H has one eigenvalue, <math>\lambda_M</math>, much bigger than all the rest, and <math>\chi = \left(
\begin{array}{c}a_1\\a_2\\\vdots\end{array}\right)</math>
is an arbitrary starting vector, then <math>\chi = \sum_{q=1}^N x_q\Psi^{(q)}</math>.

\begin{equation}
(H)^n\chi = \sum_{q=1}^N x_q \lambda^n_q\Psi^{(q)}\nonumber\rightarrow x_M\lambda_M^n\Psi^{(M)}\nonumber
\end{equation}

provided <math>x_M\neq 0</math>.

To pick out the eigenvector corresponding to any eigenvalue, write the original problem in the form

\begin{align}
H\Psi & = \lambda O \Psi \nonumber \\
\left( H - \lambda_q O \right) \Psi & = \left( \lambda - \lambda_q \right) O \Psi \nonumber
\end{align}

Therefore,

\begin{equation}
G \Psi = \frac{1}{\lambda - \lambda_q } \Psi \nonumber \\
\end{equation}

where <math>G=(H-\lambda_qO)^{-1}O</math> with eigenvalues <math>\frac{1}{\lambda_n - \lambda_q}</math>.

By picking <math>\lambda_q</math> close to any one of the <math>\lambda_n</math>, say
<math>\lambda_{n^\prime}</math>, then <math>\frac{1}{\lambda_n-\lambda_q}</math> is much larger for
<math>n=n^\prime</math> than for any other value. The sequence is then

\begin{array}{c}
\chi_1=G\chi\nonumber\\
\chi_2=G\chi_1\nonumber\\
\chi_3=G\chi_2\nonumber\\
\vdots\nonumber
\end{array}

until the ratios of components in <math>\chi_n</math> stop changing.

- To avoid matrix inversion and multiplication, note that the sequence is equivalent to
\begin{equation}
F \chi_n = \left( \lambda - \lambda_q \right) O \chi_{n-1} \nonumber \\
\end{equation}
where <math>F = H-\lambda_qO</math>. The factor of <math>\left( \lambda - \lambda_q \right)</math> can be
dropped because this only affects the normalization of <math>\chi_n</math>. To find <math>\chi_n</math>, solve
\begin{equation}
F \chi_n = O \chi_{n-1} \nonumber
\end{equation}
(N equations in N unknowns). Then
\begin{equation}
\lambda = \frac{ \langle \chi_n | H | \chi_n \rangle}{\langle \chi_n | \chi_n \rangle} \nonumber
\end{equation}

==Matrix Elements of H==

\begin{equation}
H=-\frac{1}{2}\nabla^2_1 -\frac{1}{2}\nabla^2_2 - \frac{1}{r_1} - \frac{1}{r_2} +\frac{Z^{-1}}{r_{12}}\nonumber
\end{equation}

Taking <math>r_1, r_2</math> and <math>r_{12}</math> as independent variables,

\begin{equation}
\nabla_1^2 = \frac{1}{r_1^2}\frac{\partial}{\partial r_1}\left(r_1^2\frac{\partial}{\partial r_1}\right)+ \frac{1}{r_{12}^2}\frac{\partial}{\partial r_{12}}
\left(r_{12}^2\frac{\partial}{\partial r_{12}}\right)\nonumber -\frac{l_1(l_1+1)}{r_1^2}+2(r_1 - r_2\cos(\theta_{12}))\frac{1}{r_{12}}\frac{\partial^2}{\partial r_1 \partial r_{12}} \nonumber - 2(\nabla_1^Y \cdot {\bf r}_2)\frac{1}{r_{12}}\frac{\partial}{\partial r_{12}}\nonumber
\end{equation}

where <math>\nabla_1^Y</math> acts only on the spherical harmonic part of the wave function and the diagram

[[File:Pic3.jpg|400px]]

defines the complete set of 6 independent variables is <math>r_1</math>, <math>r_2</math>, <math>r_{12}</math>, <math>\theta_1</math>, <math>\varphi_1</math>, <math>\chi</math>.

If <math>r_{12}</math> were not an independent variable, then one could take the volume element to be

\begin{equation}
d\tau = r_1^2dr_1\sin(\theta_1)d\theta_1d\varphi_1r_2^2dr_2\sin(\theta_2)d\theta_2d\varphi_2.\nonumber
\end{equation}

However, <math>\theta_2</math> and <math>\varphi_2</math> are no longer independent variables. To eliminate them, take the point <math>{\bf r}_1</math> as the
origin of a new polar co-ordinate system, and write

\begin{equation}
d\tau=-r_1^2dr_1\sin(\theta_1)d\theta_1d\varphi_1r_{12}^2dr_{12}\sin(\psi)d\psi d\chi\nonumber
\end{equation}

and use

\begin{equation}
r_2^2=r_1^2+r_{12}^2 +2r_1r_{12}\cos(\psi)
\end{equation}

Then for fixed <math>r_1</math> and <math>r_{12}</math>,

\begin{equation}
2r_2dr_2 = -2r_1r_{12}\sin(\psi)d\psi
\end{equation}

Thus

\begin{equation}
d\tau= r_1dr_1r_2dr_2r_{12}dr_{12}\sin(\theta_1)d\theta_1d\varphi_1d\chi
\end{equation}

The basic type of integral to be calculated is

\begin{align}
I \left( l_1, m_1, l_2, m_2; R \right) & = \int \sin \left( \theta_1 \right) d\theta_1 d\varphi_1 d\chi Y^{m_1}_{l_1} \left( \theta_1, \varphi_1 \right)^{*} Y^{m_2}_{l_2} \left( \theta_2, \varphi_2 \right) \\
& \times \int r_1dr_1r_2dr_2r_{12}dr_{12}R \left( r_1, r_2, r_{12} \right)
\end{align}

Consider first the angular integral. <math>Y^{m_2}_{l_2} \left( \theta_2, \varphi_2 \right)</math> can be expressed in terms of the independent variables <math>\theta_1, \varphi_1, \chi</math> by use of the rotation matrix relation
\begin{equation}
Y^{m_2}_{l_2}( \theta_2, \varphi_2) = \sum_m \mathcal{D}^{(l_2)}_{m_2,m}(\varphi_1,\theta_1,\chi)^*Y^m_{l_2}( \theta_{12}, \varphi)
\end{equation}
where <math>\theta, \varphi</math> are the polar angles of <math>{\bf r}_2</math> relative to <math>{\bf r}_1</math>. The angular integral is then

\begin{align}
I_{ang} & = \int^{2 \pi}_{0} d\chi \int^{2\pi}_{0} d\varphi_1 \int^{\pi}_{0} \sin \left( \theta_1 \right) d\theta_{1} Y^{m_1}_{l_1} \left( \theta_1, \varphi_1 \right)^{*} \\
& \times \sum_m \mathcal{D}^{(l_2)}_{m_2,m} \left( \varphi_1,\theta_1,\chi \right)^{*} Y^{m}_{l_2} \left( \theta_{12},\varphi \right) \\
\end{align}

Use

\begin{equation}
Y^{m_1}_{l_1} \left( \theta_1, \varphi_1 \right)^{*} = \sqrt{\frac{2l_1+1}{4\pi}}\mathcal{D}^{(l_1)}_{m_1,0} \left( \varphi_1, \theta_1, \chi \right)
\end{equation}

together with the orthogonality property of the rotation matrices (Brink and Satchler, p 147)

\begin{align}
& \int \mathcal{D}^{(j)*}_{m,m^\prime}\mathcal{D}^{(J)}_{M,M^\prime}\sin(\theta_1)d\theta_1d\varphi_1d\chi \\
& = \frac{8\pi^2}{2j+1}\delta_{jJ}\delta_{mM}\delta_{m^\prime M^\prime}\nonumber
\end{align}

to obtain

\begin{align}
I_{ang} & = \sqrt{ \frac{2l_1+1}{4\pi}} \frac{8\pi^2}{2l_1+1} \delta_{l_1,l_2} \delta_{m_1,m_2} Y^{0}_{l_2} \left( \theta_{12}, \varphi \right) \\
& = 2 \pi \delta_{l_1,l_2} \delta_{m_1,m_2} P_{l_2} \left( \cos \theta_{12} \right)
\end{align}

since

\begin{equation}
Y^0_{l_2} \left( \theta_{12}, \varphi \right) = \sqrt{\frac{2l_1+1}{4\pi}} P_{l_2} \left( \cos \left( \theta_{12} \right) \right)
\end{equation}

Note that <math>P_{l_2} \left( \cos \theta \right)</math> is just a short hand expression for a radial function because

\begin{equation}
\cos \theta_{12} = \frac{r_1^2 + r_2^2 - r_{12}^2}{2r_1r_2}
\end{equation}

The original integral is thus

\begin{equation}
I \left( l_1, m_1, l_2, m_2; R \right) = 2 \pi \delta_{l_1,l_2} \delta_{m_1,m_2} \int^{\infty}_{0} r_{1} dr_{1} \int^{\infty}_{0} r_{2} dr_{2} \int^{r_{1} + r_{2}}_{|r_{1} - r_{2}|} r_{12} dr_{12} R \left( r_{1}, r_{2}, r_{12} \right) P_{l_2} \left( \cos \theta_{12} \right)
\end{equation}

where again

\begin{equation}
\cos \theta_{12} = \frac{r_{1}^{2} +r^{2}_{2} - r_{12}^{2}}{2 r_{1} r_{2}}
\end{equation}

is a purely radial function.

The above would become quite complicated for large <math>l_2</math> because <math>P_{l_2}(\cos \theta_{12})</math> contains terms up to <math>\left( \cos \theta_{12} \right)^{l_2}</math>. However, recursion relations exist which allow any integral containing <math>P_l \left( \cos \theta_{12} \right)</math> in terms of those containing just <math>P_0 \left( \cos \theta_{12} \right) = 1</math> and <math>P_1 \left( \cos \theta_{12} \right) = \cos \theta_{12}</math>.

==Radial Integrals and Recursion Relations==

The basic radial integral is [see G.W.F. Drake, Phys. Rev. A 18, 820 (1978)]

\begin{align}
I_0 \left( a, b, c \right) & =
\int^{\infty}_{0} r_{1} dr_{1} \int^{\infty}_{r_{1}} r_{2} dr_{2} \int^{r_{1}+r_{2}}_{r_{2} - r_{1}} r_{12} dr_{12} r_{1}^{a} r_{2}^{b} r_{12}^{c} e^{-\alpha r_{1} - \beta r_{2}} + \int^{\infty}_{0} r_{2} dr_{2} \int^{\infty}_{r_{2}} r_{1} dr_{1} \int^{r_{1} + r_{2}}_{r_{1} - r_{2}} r_{12} dr_{12} r_{1}^{a} r_{2}^{b} r_{12}^{c} e^{- \alpha r_{1} - \beta r_{2}} \\
& = \frac{2}{c+2} \sum^{[(c+1)/2]}_{i=0}
\left(\begin{array}{c}
c+2 \\
2i+1
\end{array} \right)
\left\{ \frac{q!}{\beta^{q+1} \left( \alpha + \beta \right)^{p+1}} \sum_{j=0}^{q} \frac{\left( p + j \right)!}{j!} \left( \frac{\beta}{\alpha + \beta} \right)^j + \frac{q^{\prime}!}{\alpha^{q^{\prime}+1} \left( \alpha + \beta \right)^{p^{\prime}+1}} \sum^{q^{\prime}}_{j=0} \frac{\left( p^{\prime} + j \right)!}{j!} \left( \frac{\alpha}{\alpha + \beta} \right)^j \right\}
\end{align}

where

\begin{array}{ll}
p = a + 2i + 2 & p^{\prime} = b + 2i + 2 \\
q = b + c - 2i + 2 & q^{\prime} = a + c - 2i + 2
\end{array}

The above applies for <math> a, b \geq -2, c\geq -1</math>.
<math>[x]</math> means "greatest integer in" <math>x</math>.

Then

\begin{align}
I_1 \left( a, b, c \right) & = \int d\tau_{r} r_{1}^{a} r_{2}^{b} r_{12}^{c} e^{ -\alpha r_{1} - \beta r_{2}} P_{1}\left( \cos \theta \right) \\
& = \frac{1}{2} \left( I_{0} \left( a + 1, b - 1, c \right) + I_{0} \left( a - 1, b + 1, c \right) - I_{0} \left( a - 1, b - 1, c + 2 \right) \right)
\end{align}

===The Radial Recursion Relation===

Recall that the full integral is

\begin{equation}
I(l_1m_1,l_2m_2;R)=2\pi\delta_{l_1,l_2}\delta_{m_1,m_2}I_{l_2}(R)
\end{equation}

where

\begin{equation}
I_{l_2}(R) = \int d\tau_r R(r_1,r_2,r_{12})P_{l_2}(\cos \theta_{12})
\end{equation}

To obtain the recursion relation, use

\begin{equation}
P_l(x) = \frac{[P^{\prime}_{l+1}(x)-P^{\prime}_{l-1}(x)]}{2l+1}
\end{equation}

with

\begin{equation}
P_{l+1}^\prime (x) = \frac{d}{dx}P_{l+1}(x)
\end{equation}

Here <math>x=\cos\theta_{12}</math> and

\begin{align}
\frac{d}{d \cos \theta_{12}} & = \frac{dr_{12}}{d \cos \theta_{12}} \frac{d}{dr_{12}} \\
& = -\frac{r_{1} r_{2}}{r_{12}} \frac{d}{dr_{12}}
\end{align}

Then,

\begin{equation}
I_{l}\left( R \right) = - \int d\tau_r R \frac{r_{1} r_{2}}{r_{12}} \frac{d}{dr_{12}} \frac{ \left[ P_{l+1} \left( \cos \theta_{12} \right) - P_{l-1} \left( \cos \theta_{12} \right) \right]}{2l+1}
\end{equation}

The <math>r_{12}</math> part of the integral is

\begin{align}
& \int^{r_{1} + r_{2}}_{|r_{1} - r_{2}|} r_{12} dr_{12} R \frac{r_{1} r_{2}}{r_{12}} \frac{d}{dr_{12}} \left[P_{l+1}-P_{l-1} \right] \\
& = R r_{1} r_{2} \left[ P_{l+1} - P_{l-1} \right]^{r_{1} + r_{2}}_{|r_{1} - r_{2}|} - \int^{r_{1} + r_{2}}_{|r_{1} - r_{2}|} r_{12} dr_{12} \left( \frac{d}{dr_{12}} R \right) \frac{r_{1} r_{2}}{r_{12}} \frac{ \left[ P_{l+1} - P_{l-1} \right]}{2l+1}
\end{align}

The integrated term vanishes because

\begin{equation}
\cos \theta_{12} = \frac{r_{1}^{2} + r_{2}^{2} - r_{12}^{2}}{2 r_{1} r_{2}}
\begin{array}{lll}
& = -1 & {\rm when} & r_{12}^{2} = \left( r_{1} + r_{2} \right)^{2} \\
& = 1 & {\rm when} & r_{12}^{2} = \left( r_{1} - r_{2} \right)^{2}
\end{array}
\end{equation}

and <math>P_l(1)=1</math>, <math>P_{l}(-1)=(-1)^{l}</math>.

Thus,

\begin{equation}
I_{l+1}\left(\frac{r_1r_2}{r_{12}}R^\prime\right) = \left( 2l+1 \right) I_{l}(R) - I_{l-1} \left( \frac{r_{1} r_{2}}{r_{12}} R^{\prime} \right)
\end{equation}

If

\begin{equation}
R = r_{1}^{a-1} r_{2}^{b-1} r_{12}^{c+2} e^{-\alpha r_{1} - \beta r_{2}}
\end{equation}

then [G.W.F. Drake, Phys. Rev. A 18, 820 (1978)]

\begin{equation}
I_{l+1}(r_1^ar_2^br_{12}^c) =
\frac{2l+1}{c+2} I_{l} \left( r_{1}^{a-1} r_{2}^{b-2} r_{12}^{c+2} \right)+I_{l-1} \left( r_{1}^{a} r_{2}^{b} r_{12}^{c} \right)
\end{equation}

For the case <math>c=-2</math>, take

\begin{equation}
R = r_{1}^{a-1} r_{2}^{b-1} \ln r_{12} e^{ -\alpha r_{1} -\beta r_{2}}
\end{equation}

Then

\begin{equation}
I_{l+1} \left( r_{1}^{a} r_{2}^{b} r_{12}^{c} \right) = I_{l} \left( r_{1}^{a-1} r_{2}^{b-1} \ln r_{12} \right) \left( 2l + 1 \right) - I_{l-1} \left( r_{1}^{a} r_{2}^{b} r_{12}^{-2} \right)
\end{equation}

This allows all <math>I_{l}</math> integrals to be calculated from tables of <math>I_0</math> and
<math>I_1</math> integrals.

===The General Integral===

The above results for the angular and radial integrals can now be combined into a general formula for integrals of the type

\begin{equation}
I = \int \int d{\bf r}_{1} d{\bf r}_{2} R_{1} \mathcal{Y}^{M^{\prime}}_{l^{\prime}_{1} l^{\prime}_{2} L^{\prime}} \left( \hat{r}_{1}, \hat{r}_{2} \right) T_{k_{1} k_{2} K}^{Q} \left( {\bf r}_{1},{\bf r}_{2} \right) R_{2} \mathcal{Y}_{l_{1} l_{2} L}^{M} \left( \hat{r}_{1}, \hat{r}_{2} \right)
\end{equation}

where

\begin{equation}
\mathcal{Y}^{M}_{l_{1} l_{2} L} \left( \hat{r}_{1}, \hat{r}_{2} \right) = \sum_{m_1,m_2} \langle l_{1} l_{2} m_{1} m_{2} | LM \rangle Y^{m_{1}}_{l_{1}} \left( \hat{r}_{1} \right) Y^{m_{2}}_{l_{2}} \left( \hat{r}_{2} \right)
\end{equation}

and

\begin{equation}
T^Q_{k_{1} k_{2} K} \left( {\bf r}_{1}, {\bf r}_{2} \right) = \sum_{q_{1}, q_{2}} \langle k_{1} k_{2} q_{1} q_{2} | KQ \rangle Y_{k_{1}}^{q_{1}} \left( \hat{r}_{1} \right) Y^{q_{2}}_{k_{2}} \left( \hat{r}_{2} \right)
\end{equation}

The basic idea is to make repeated use of the formula

\begin{equation}
Y^{m_{1}}_{l_{1}} \left( \hat{r}_{1} \right) Y^{m_{2}}_{l_{2}} \left( \hat{r}_{1} \right) = \sum_{l m} \left( \frac{\left( 2l_{1}+1 \right) \left(2l_{2}+1 \right) \left(2l+1 \right)}{4 \pi} \right)^{1/2}
\times \left( \begin{array}{ccc}
l_1 & l_2 & l \\
m_1 & m_2 & m
\end{array} \right) \left(\begin{array}{ccc}
l_1 & l_2 & l \\
0 & 0 & 0
\end{array} \right)
Y^{m*}_{l} \left( \hat{r}_{1} \right)
\end{equation}

where

\begin{equation}
Y^{m*}(\hat{r}) = (-1)^mY_l^{-m}(\hat{r})
\end{equation}

and

\begin{equation}
\left(\begin{array}{ccc}
l_1 & l_2 & l \\
m_1 & m_2 & m
\end{array} \right) = \frac{\left( -1 \right)^{l_{1} - l_{2} - m}}{\left( 2l + 1 \right)^{1/2}} \left( l_{1} l_{2} m_{1} m_{2} | l, -m \right)
\end{equation}

is a 3-<math>j</math> symbol. In particular, write

\begin{array}{cc}
Y^{m_1^\prime}_{l_1^\prime}(\hat{r}_1)^{*} \underbrace{Y_{k_{1}}^{q_{1}} \left( \hat{r}_{1} \right) Y^{m_{1}}_{l_{1}} \left( \hat{r}_{1} \right)} & = & \sum_{\Lambda M}(\ldots)Y^{M}_\Lambda(\hat{r}_1)\\
\ \ \ \ \ \ \ \sum_{\lambda_1\mu_1}Y_{\lambda_1}^{\mu_1}(\hat{r}_1) & & \\
& & \\
Y^{m_{2}^{\prime}}_{l_{2}^{\prime}} \left( \hat{r}_{2} \right) \underbrace{Y^{q_{2}}_{k_{2}} \left( \hat{r}_{2} \right) Y_{l_{2}}^{m_{2}} \left( \hat{r}_{2} \right)} & = & \sum_{\Lambda^{\prime} M^{\prime}} \left( \ldots \right)Y_{\Lambda^{\prime}}^{M^{\prime}*} \left( \hat{r}_{2} \right) \\
\ \ \ \ \ \ \ \ \ \sum_{\lambda_{2} \mu_{2}}Y_{\lambda_{2}}^{\mu_{2}} \left( \hat{r}_{2} \right) & &
\end{array}

The angular integral then gives a factor of <math>2 \pi \delta_{\Lambda, \Lambda^{\prime}} \delta_{M, M^{\prime}} P_{\Lambda} \left( \cos \theta_{12} \right)</math>. The total integral therefore reduces to the form

\begin{equation}
I = \sum_{\Lambda} C_{\Lambda} I_{\Lambda} \left( R_{1} R_{2} \right)
\end{equation}

where <math>C_{\Lambda} = \sum_{\lambda_{1}, \lambda_{2}} C_{ \lambda_{1}, \lambda_{2}, \Lambda}</math>. For further details and derivations, including graphical representations, see G.W.F. Drake, Phys. Rev. A <math>{\bf 18}</math>, 820 (1978).

==Graphical Representation==

[[File:GraphRep1.jpg|500px]]

==Matrix Elements of H==
Recall that

\begin{equation}
H = -\frac{1}{2} \nabla^{2}_{1} - \frac{1}{2} \nabla^{2}_{2} - \frac{1}{r_1} - \frac{1}{r_2} + \frac{Z^{-1}}{r_{12}}
\end{equation}

Consider matrix elements of

\begin{equation}
\nabla^2_1 = \frac{1}{r_1^2} \frac{\partial}{\partial r_1}\left(r_1^2\frac{\partial}{\partial r_1} \right) + \frac{1}{r^2}\frac{\partial}{\partial r}\left(r^2\frac{\partial}{\partial r}\right) - \frac{(\vec{l}_1^Y)^2}{r_1^2} + \frac{2 \left( r_1 - r_2 \cos \theta \right)}{r} \frac{\partial^2}{\partial r_1 \partial r} - 2 \left( \nabla^Y_1 \cdot {\bf r}_2 \right) \frac{1}{r} \frac{\partial}{\partial r}
\end{equation}

where <math>r \equiv r_{12}</math> and <math>\cos\theta \equiv \cos\theta_{12}</math>. Also in what follows, define <math>\hat{\nabla}^Y_1 \equiv r_1\nabla^Y_1</math> where <math>\nabla^Y</math> operates only on the spherical harmonic part of the wave function, and similarly for <math>\vec{l}_1^Y</math>.

A general matrix element is

\begin{equation}
\langle r_1^{a^\prime} r_2^{b^\prime} r_{12}^{c^\prime} e^{-\alpha^\prime r_1 - \beta^\prime r_2} \mathcal{Y}^{M^\prime}_{l^\prime_1 l^\prime_2 L^\prime} \left( \hat{r}_1, \hat{r}_2 \right) \left| \nabla^2_1 \right| r_1^a r_2^b r_{12}^c e^{-\alpha r_1 - \beta r_2} \mathcal{Y}^M_{l_1 l_2 L} \left( \hat{r}_1, \hat{r}_2 \right) \rangle
\end{equation}

Since <math>\nabla^2_1</math> is rotationally invariant, this vanishes unless <math>L=L^\prime</math>, <math>M=M^\prime</math>. Also <math>\nabla_1^2</math> is Hermitian, so that the result must be the same whether it operates to the right or left, even though the results look very different. In fact, requiring the results to be the same yields some interesting and useful integral identities as follows:

Put

\begin{equation}
\mathcal{F} = F \mathcal{Y}^M_{l_1 l_2 L} \left( \hat{r}_1, \hat{r_2} \right)
\end{equation}

and

\begin{equation}
F = r_1^a r_2^b r_{12}^c e^{-\alpha r_1 - \beta r_2}
\end{equation}

Then

\begin{equation}
\nabla^2_1 \mathcal{F} = \left\{ \frac{1}{r_1^2} \left[ a \left( a + 1 \right) - l_1 \left( l_1 + 1 \right) \right] + \frac{c \left( c + 1 \right)}{r^2} + \alpha^2\frac{2\alpha \left( a + 1 \right)}{r_1} + \frac{2 \left(r_1 - r_2 \cos \theta \right) }{r_1r^2}c \left[ a - \alpha r_1 \right] - \frac{2c}{r^2} \left( \hat{\nabla}^Y_1 \cdot \hat{r}_2 \right) \frac{r_2}{r_1} \right\} \mathcal{F}
\end{equation}

and

\begin{align}
\langle \mathcal{F}^\prime \left| \nabla^2_1 \right| \mathcal{F} \rangle & = \sum_\Lambda \int d\tau_r F^\prime C_\Lambda \left( 1 \right) P_\Lambda \left( \cos \theta \right) \times \left\{ \frac{1}{r_1^2} \left[ a \left( a + 1 \right) - l_1 \left( l_1 + 1 \right) \right] -\frac{2 \alpha \left( a + 1 \right)}{r_1} + \frac{c \left( c + 1 \right)}{r^2} +\alpha^2 + \frac{2 \left( r_2 - r_2 \cos \theta \right)}{r_1r^2}c \left( a - \alpha r_1 \right) \right\} F \\
& + \sum_\Lambda \int d\tau_r F^\prime C_\Lambda \left( \hat{\nabla}^Y_1 \cdot \hat{r}_2 \right) P_\Lambda \left( \cos \theta \right) \left( \frac{-2c r_2}{r_1 r^2} \right) F
\end{align}

where

\begin{equation}
\int d\tau_r = \int^\infty_0 r_1 dr_1 \int^\infty_0 r_2 dr_2 \int^{r_1+r_2}_{\left| r_1 - r_2 \right|} r dr
\end{equation}

For brevity, let the sum over <math>\Lambda</math> and the radial integrations be understood, and let <math>\left( \nabla_1^2 \right)</math> stand for the terms which appear in the integrand. Then operating to the right gives

\begin{equation}
(\nabla_1^2)_R = \frac{1}{r_1^2} \left[ a \left( a + 1 \right) - l_1 \left( l_1 + 1 \right) \right] + \frac{c \left( c + 1 \right)}{r^2} + \alpha^2 - \frac{2 \alpha \left( a + 1 \right)}{r_1}
+ \frac{2 \left( r_1 - r_2 \cos \theta \right)}{r_1 r^2}c \left( a - \alpha r_1 \right) - \frac{2c r_2}{r_1 r^2} \left( \hat{\nabla}^Y_1 \cdot \hat{r}_2 \right)
\end{equation}

Operating to the left gives

\begin{equation}
(\nabla_1^2)_L = \frac{1}{r_1^2} \left[a^\prime \left( a^\prime + 1 \right) - l_1^\prime \left( l_1^\prime + 1 \right) \right] + \frac{c^\prime \left( c^\prime + 1 \right)}{r^2} + \alpha^{\prime 2} -\frac{2 \alpha^\prime \left( a^\prime + 1 \right)}{r_1}
+ \frac{2 \left( r_1 - r_2 \cos \theta \right)}{r_1 r^2} c^\prime \left( a^\prime - \alpha^\prime r_1 \right) - \frac{2 c^\prime r_2}{r_1 r^2} \left( \hat{\nabla}^{Y^\prime}_1 \cdot \hat{r}_2 \right)
\end{equation}

Now put

\begin{array}{cc}
a_+ = a+a^\prime, & \hat{\nabla}_1^+ = \hat{\nabla}_1^Y + \hat{\nabla}_1^{Y\prime} \\
a_{-} = a-a^\prime, & \hat{\nabla}_1^{-} = \hat{\nabla}_1^Y - \hat{\nabla}_1^{Y\prime}
\end{array}

etc., and substitute <math>a^\prime = a_+ - a</math>, <math>c^\prime = c_+ - c</math>, and <math>\alpha^\prime = \alpha_+ - \alpha</math> in <math>(\nabla^2_1)_L</math>. If <math>a_+</math>, <math>c_+</math> and <math>\alpha_+</math> are held fixed, then the equation

\begin{equation}
(\nabla^2_1)_R = (\nabla^2_1)_L
\end{equation}

must be true for arbitrary <math>a</math>, <math>c</math>, and <math>\alpha</math>. Their coefficients must thus vanish.

This yields the integral relations

\begin{equation}
\label{I}
\frac{\left( r_1 - r_2 \cos \theta \right)}{r_1 r^2} = \frac{1}{c_+} \left( \frac{- \left( a_+ + 1 \right)}{r_1^2} + \frac{\alpha_+}{r_1} \right) \tag{I}
\end{equation}

from the coefficient of <math>a</math> and

\begin{equation}
\label{II}
\frac{ \left( r_1 - r_2 \cos \theta \right) \left( a_+ - \alpha_+ r_1 \right)}{r_1 r^2} = \frac{r_2}{r_1 r^2}\left( \hat{r}_2 \cdot \hat{\nabla}_1^+ \right) - \frac{ \left( c_+ + 1 \right)}{r^2} \tag{II}
\end{equation}

from the coefficient of <math>c</math>. The coefficient of <math>\alpha</math> gives an equation equivalent to (I).

Furthermore, if can be show that (see problem)

\begin{align}
& \sum_\Lambda \int d\tau_r \frac{r^c}{r^2} C_\Lambda \left( \hat{r}_2 \cdot \hat{\nabla}_1^Y \right) P_\Lambda \left( \cos \theta \right) \\
= & \sum_\Lambda \int d\tau_r \frac{r^c}{c r_1 r_2}C_\Lambda \left( 1 \right)P_\Lambda \left( \cos \theta \right) \times \left( \frac{l_1^\prime \left( l_1^\prime + 1 \right) - l_1 \left( l_1 + 1 \right) -\Lambda \left( \Lambda + 1 \right)}{2} \right)
\end{align}

and similarly for <math>\left( \hat{r}_2 \cdot \hat{\nabla}^{Y\prime}_1 \right)</math> with <math>l_1</math> and <math>l_1^\prime</math> interchangeable, then it follows that

\begin{equation}
\frac{r^{c_+}}{r^2} \left( \hat{r}_2 \cdot \hat{\nabla}_1^+ \right) = - \frac{r^{c_+}}{c_+ r_1 r_2} \Lambda \left( \Lambda + 1 \right)
\end{equation}

and

\begin{equation}
\frac{r^{c_+}}{r^2} \left( \hat{r}_2 \cdot \hat{\nabla}_1^- \right) = \frac{r^{c_+}}{c_+ r_1 r_2} \left[ l_1^\prime \left( l_1^\prime + 1 \right) - l_1 \left( l_1 + 1 \right) \right]
\end{equation}

where equality applies after integration and summation over <math>\Lambda</math>.

Thus (II) becomes

\begin{equation}\label{IIa}
\frac{ \left( r_1 - r_2 \cos \theta \right) \left( a_+ - \alpha_+ r_1 \right)}{r_1 r^2} = - \frac{\Lambda \left( \Lambda + 1 \right)}{c_+ r^2_1} - \frac{ \left( c_+ + 1 \right)}{r^2} \tag{IIa}
\end{equation}

===Problem===

Prove the integral relation

\begin{align}
& \sum_\Lambda \int d\tau_r f \left( r_1, r_2 \right) \frac{1}{r} \left( \frac{d}{dr} q(r) \right) C_\Lambda \left( \hat{r}_2 \cdot \hat{\nabla}^Y_1 \right) P_\Lambda \left( \cos \theta \right) \\
= & \sum_\Lambda \int d\tau_r \frac{ f \left( r_1, r_2 \right)}{r_1, r_2} q(r) C_\Lambda \left( 1 \right) P_\Lambda \left( \cos \theta \right) \left(\frac{l_1^\prime \left( l_1^\prime + 1 \right) - l_1 \left( l_1 + 1 \right) - \Lambda \left( \Lambda + 1 \right)}{2}\right)
\end{align}

where the coefficients <math>C_\Lambda \left( 1 \right)</math> are the angular coefficients from the overlap integral

\begin{align}
\int & d\Omega\, \mathcal{Y}^{M*}_{l_1^\prime l_2^\prime L}(\hat{r}_1,\hat{r}_2)\, \mathcal{Y}^{M}_{l_1 l_2 L}(\hat{r}_1,\hat{r}_2)\\
= & \sum_\Lambda C_\Lambda(1)P_\Lambda(\cos\theta_{12}).
\end{align}

Hint: Use the fact that <math>l_1^2</math> is Hermitian so that
<math>
\int d\tau (l_1^2\mathcal{Y}^\prime)^*q(r)\mathcal{Y} = \int d\tau
\mathcal{Y}^{\prime *}l_1^2(q(r)\mathcal{Y})
</math>
with <math>\ \vec{l}_1= \frac{1}{i}\vec{r}_1\times\nabla_1</math>.

It is also useful to use
<math>\begin{eqnarray}\ \
\cos\theta\, P_L(\cos\theta) &=& \frac{1}{2L+1}[L\,P_{L-1}(\cos\theta) + (L+1)P_{L+1}(\cos\theta)]
\\
(\cos^2\theta-1)P_L(\cos\theta) &=&
\frac{L(L-1)}{(2L-1)(2L+1)}P_{L-2}(\cos\theta)\nonumber
-\frac{2(L^2+L-1)}{(2L-1)(2L+3)}P_L(\cos\theta) + \frac{(L+1)(L+2)}{(2L+1)(2L+3)}P_{L+2}(\cos\theta)\nonumber\\
&=& \frac{L(L-1)}{(2L-1)(2L+1)}[P_{L-2}(\cos\theta) - P_L(\cos\theta)]
+ \frac{(L+1)(L+2)}{(2L+1)(2L+3)}[P_{L+2}(\cos\theta) - P_L(\cos\theta)]
\end{eqnarray}</math>

together with a double application of the integral recursion relation
<math>
I_{L+1}\left(\frac{1}{r}\frac{d}{dr}q(r)\right) =
(2L+1)I_L\left(\frac{1}{r_1r_2}q(r)\right)+I_{L-1}\left(\frac{1}{r}\frac{d}{dr}q(r)\right)
</math>

Of course <math>l_1^2\mathcal{Y}^M_{l_1l_2L}(\hat{r}_1,\hat{r}_2)=l_1(l_1+1)\mathcal{Y}^M_{l_1l_2L}(\hat{r}_1,\hat{r}_2)</math>. Begin by expanding
<math>
l_1^2(g\mathcal{Y}) = \mathcal{Y}l_1^2g + g\,l_1^2\mathcal{Y} + 2(\vec{l}_1g)\cdot(\vec{l}_1\mathcal{Y})
</math>, and show that

<math>
\begin{eqnarray}\ \
(\vec{l}_1g)\cdot(\vec{l}_1\mathcal{Y}) & = & \frac{r_1^2}{r}\frac{dg}{dr}\vec{r}_2\cdot \nabla_1\mathcal{Y}\,\,\,\,\mbox{ and}\\
l_1^2g(r) &=& 2r_1r_2\cos\theta\frac{1}{r}\frac{dg}{dr} + r_1^2r_2^2(\cos^2\theta - 1)
\frac{1}{r}\frac{d}{dr}\left(\frac{1}{r}\frac{dg}{dr}\right).
\end{eqnarray}
</math>

The proof amounts to showing that the term <math>l_1^2g(r)</math> can be replaced by
<math>\Lambda(\Lambda+1)g(r)</math> after multiplying by <math>P_\Lambda(\cos\theta)</math>
and integrating by parts with respect to the radial integrations over <math>r_1</math>, <math>r_2</math> and
<math>r \equiv |\vec{r}_1 - \vec{r}_2|</math>. Remember that <math>\cos\theta</math> is just a short-hand notation for the radial function <math>(r_1^2 + r_2^2 - r^2)/(2r_1r_2)</math>.

==General Hermitian Property==

Each combination of terms of the form
\begin{equation}
<f>_R = a^2f_1+af_2+abf_3+b^2f_4+bf_5+af_6\nabla_1^Y+bf_7\nabla^Y_2+f_8(Y)
\end{equation}
can be rewritten
\begin{equation}
<f>_L= (a_+-a)^2f_1+(a_+-a)f_2 + (a_+-a)(b_+-b)f_3+(b_+-b)^2f_4
+(b_+-b)f_5+(a_+-a)f_6\nabla^{Y\prime}_1+(b_+-b)f_7\nabla^{Y\prime}_2+f_8)Y^\prime
\end{equation}
where as usual <math>\nabla_1^Y</math> acts only on the spherical harmonic part of the wave function denoted for short by <math>Y</math>.
Since these must be equal for arbitrary <math>a</math> and <math>b</math>,
\begin{equation}
a_+^2f_1+a_+f_2+_+b_+f_3+b_+f_4+b_+f_5+a_+f_6\nabla^{Y\prime}_1 +
b_+f_7\nabla^{Y\prime}_2+f_8(Y^\prime)-f_8(Y)=0
\end{equation}

Adding the corresponding expression with <math>Y</math> and <math>Y^\prime</math> interchanged
yields
<math>\label{III}
a_+^2f_1+a_+f_2+a_+b_+f_3+b_+^2f_4+b_+f_5+\frac{1}{2}a_+f_6\nabla^+_1+\frac{1}{2}b_+f_7\nabla^+_2=0\tag{III}
</math>

Subtracting gives
<math>\label{IV}
f_8(Y)-f_8(Y^\prime)=-\frac{1}{2}[a_+f_6\nabla^-_1+b_+f_7\nabla_2^-]\tag{IV}
</math>
<math>\label{V}
a[-2a_+f_1-2f_2-b_+f_3-f_6\nabla^+_1]=0\tag{V}
</math>
<math>\label{VI}
b[-2b_+f_4-2f_5-a_+f_3-f_7\nabla^+_2]=0\tag{VI}
</math>

Adding the two forms gives

\begin{eqnarray}
<f>_R+<f>_L=\frac{1}{2}(a_+^2+a_-^2)f_1+a_+f_2+\frac{1}{2}(a_+b_++a_-b_-)f_3\nonumber\\
\ +\frac{1}{2}(b_+^2+b_-^2)f_4+b_+f_5+\frac{1}{2}f_6(a_+\nabla_1^++a_-\nabla^-_1)\nonumber\\
\ +\frac{1}{2}(b_+\nabla_2^++b_-\nabla^-_2)+f_8(Y)+f_8(Y^\prime)\nonumber
\end{eqnarray}

Subtracting <math>\frac{x}{2}\times({\rm I})</math> gives

\begin{eqnarray}\ \
<f>_R+<f>_L = \frac{1}{2}[(1-x)a_+^2 + a_-^2]f_1 +
(1-\frac{x}{2})a_+f_2+\frac{1}{2}[(1-x)a_+b_++a_-b_-]f_3\nonumber\\
+\frac{1}{2}[(1-x)b_+^2+b_-^2]f_4+(1-\frac{x}{2})f_5b_++\frac{1}{2}f_6[(1-\frac{x}{2})a_+\nabla_1^++a_-\nabla^-_1]\nonumber\\
+\frac{1}{2}[(1-\frac{x}{2})b_+\nabla^+_2+b_-\nabla^-_2]+f_8(Y)+f_8(Y^\prime)\nonumber
\end{eqnarray}

If <math>x=1</math>,
\begin{eqnarray}
<f>_R+<f>_L &=&
\frac{1}{2}[a_-^2f_1+a_+f_2+a_-b_-f_3+b_-^2f_4+b_+f_5\nonumber\\
&+&\left(\frac{1}{2}a_+\nabla^+_1+a_-\nabla^-_1\right)f_6+\left(\frac{1}{2}b_+\nabla^+_2+b_-\nabla^-_2\right)f_7]+f_8(Y)+f_8(Y^\prime)\nonumber
\end{eqnarray}

The General Hermitian Property for arbitrary <math>x</math> gives
\begin{eqnarray}\ \
\nabla_1^2=
\frac{1}{4}\{\frac{1}{r_1^2}[(1-x)a_+^2+a_-^2+2(1-\frac{x}{2})a_+-2[l_1(l_1+1)+l_1^\prime(l_1^\prime+1)]\nonumber\\
-\frac{2}{r_1}[(1-x)\alpha_+a_++\alpha_-a_-+2(1-\frac{x}{2})\alpha_+]+(1-x)\alpha_+^2
+ \alpha^2_-]\nonumber\\
+\frac{2(r_1-r_2\cos\theta)}{r_1r^2}[(1-x)(a_+-\alpha_+r_1)c_++(a_--\alpha_-r_2)c_-]\nonumber\\
-\frac{2r_2}{r_1r^2}[(1-\frac{x}{2})c_+\hat{r}_2\cdot\hat{\nabla}_1^++c_-\hat{r}_2\hat{\nabla}_1^-]+\frac{(1-x)c_+^2+c_-^2+2(1-\frac{x}{2})c_+}{r^2}\}\nonumber
\end{eqnarray}.

Use
<math>\begin{eqnarray}\ \
\frac{-2r_2}{r_1r^2}[(1-\frac{x}{2})c_+\hat{r}_2\hat{\nabla}_1^++c_-\hat{r}_2\hat{\nabla}_1^-]=\frac{2}{r_1^2}\left[(1-\frac{x}{2})\Lambda(\Lambda+1)-\frac{c_-}{c_+}[l_1^\prime(l_1^\prime+1)+l_1(l_1+1)]\right]\end{eqnarray}</math>,

<math>\begin{eqnarray}
\frac{2(r_1-r_2\cos\theta)}{r_1r^2}(a_--\alpha_-r_1)c_- = 2\frac{c_-}{c_+}\left[\frac{-1}{r_1^2}[a_-(a_++1)]+\frac{1}{r_1}[a_-\alpha_++\alpha_-(a_++2)]-\alpha_-\alpha_+\right]\nonumber
\end{eqnarray}</math>,

and
<math>\begin{eqnarray}\ \
\frac{2(r_1-r_2\cos\theta)}{r_1r^2}(a_+-\alpha_+r_1)c_+ = 2\left[-\frac{1}{r_1^2}[a_+(a_++1)]+\frac{1}{r_1}[a_+\alpha_++\alpha_+(a_++2)]-\alpha_+^2\right]
\end{eqnarray}</math>.

Substituting into <math>\nabla_1^2</math> gives

\begin{eqnarray}
\nabla^2_1 = \frac{1}{4}\{\frac{1}{r_1^2}[-(1-x)a_+^2 + a_-^2 +xa_+ +
2(1-\frac{x}{2})\Lambda(\Lambda+1)\nonumber\\
-2l_1(l_1+1)(1-\frac{c_-}{c_+}) -
2l_1^\prime(l_1^\prime+1)(1+\frac{c_-}{c_+})-\frac{2c_-a_-}{c_+}(a_++1)]\nonumber\\
- \frac{2}{r_1}\left[-(1-x)\alpha_+(a_++2)+\alpha_-a_-+2(1-\frac{x}{2})\alpha_+
-\frac{c_-}{c_+}[a_-\alpha_++\alpha_-(a_++2)]\right]\nonumber\\
-(1-x)\alpha_+^2+\alpha_-^2-2\frac{c_-}{c_+}\alpha_-\alpha_+ +\frac{1}{r^2}\left[(1-x)c_+^2+c_-^2+2(1-\frac{x}{2})c_+\right]\}\nonumber
\end{eqnarray}

This has the form
<math>
\nabla_1^2 =
\frac{1}{4}\left[\frac{A_1}{r_1^2}+\frac{B_1}{r_1}+\frac{C_1}{r^2}+D_1\right]
</math>
with

<math>\begin{array}{ccc}
A_1=-(1-x)a_+^2+a_-^2
+xa_++2(1-\frac{x}{2})\Lambda(\Lambda+1)-2l_1(l_1+1)(1-\frac{c_-}{c_+})\
-2l_1^\prime(l_1^\prime+1)(1+\frac{c_-}{c_+})-\frac{2c_-a_-}{c_+}(a_++1) \ \
\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \
\end{array}
</math>

<math>
B_1 =
2\left[(1-x)\alpha_+(a_++2)-\alpha_-a_--2(1-\frac{x}{2})\alpha_++\frac{c_-}{c_+}[a_-\alpha_++\alpha_-(a_++2)]\right]
</math>

<math>
C_1 = (1-x)c_+^2+ c_-^2 +2(1-\frac{x}{2})c_+
</math>

<math>
D_1 = -1(1-x)\alpha_+^2 +\alpha_-^2 - 2\frac{c_-}{c_+}\alpha_-\alpha_+
</math>

The complete Hamiltonian is then (in Z-scaled a.u.)

<math>\begin{eqnarray}
H =-\frac{1}{2}\nabla^2_1 -\frac{1}{2}\nabla^2_2 -\frac{1}{r_1} -\frac{1}{r_2} +\frac{Z^{-1}}{r}\nonumber\\
= -\frac{1}{2}\nabla^2_1 -\frac{1}{2}\nabla^2_2 -\frac{1}{r_1} -\frac{Z-1}{Zr_2}+Z^{-1}\left(\frac{1}{r}-\frac{1}{r_2}\right)\nonumber\\
\ =-\frac{1}{8}\left[\frac{A_1}{r_1^2}+\frac{B_1+8}{r_1}+\frac{C_1}{r^2}+D_1+D_2 +\frac{A_2}{r_2^2} + \frac{B_2 + 8(Z-1)/Z}{r_2}+\frac{C_2}{r^2}\right]\nonumber
+ Z^{-1}\left(\frac{1}{r}-\frac{1}{r_2}\right)\nonumber
\end{eqnarray}</math>

The screened hydrogenic energy is
<math>\begin{eqnarray}
E_{SH}=-\frac{1}{2}\left[\frac{1}{n_1^2}+ \left(\frac{Z-1}{Z}\right)^2\frac{1}{n_2^2}\right]
\end{eqnarray}</math>
so that

<math>\begin{eqnarray}
H-E_{SH} &=& -\frac{1}{8}[
\frac{A_1}{r_1^2}+\frac{B_1+8}{r_1}+\frac{C_1}{r^2} + D_1 + D_2
-\frac{4}{n_1^2} -\left(\frac{Z-1}{Z}\right)^2\frac{4}{n_2^2}\nonumber
+\frac{A_2}{r_2}+\frac{B_2+8(Z-1)/Z}{r_2}+\frac{C_2}{r^2}] +
Z^{-1}\left(\frac{1}{r} -\frac{1}{r_2}\right)\nonumber
\end{eqnarray}</math>

==Optimization of Nonlinear Parameters==
- The traditional method of performing Hylleraas calculations is to write the
basis set in the form

<math>
\Psi = \sum_{i,j,k}c_{i,j,k}\varphi_{i,j,k}(\alpha,\beta)\pm exchange
</math>

with

<math>
\varphi_{i,j,k}(\alpha,\beta) = r_1^ir_2^jr_{12}^ke^{-\alpha r_1 - \beta
r_2}\mathcal{Y}^M_{l_1l_2L}.
</math>

The usual procedure is to set <math>\alpha = Z</math> so that it represents the inner
<math>1s</math> electron, and then to vary <math>\beta</math> so as to minimize the energy.

[[File:OptNon.jpg|500px]]

Since <math>\beta</math> appears in <math>\Psi</math> as a non-linear parameter, the entire
calculation must be repeated for each value of <math>\beta</math>. However, the minimum
becomes progressively smaller as the basis set is enlarged.

[[File:OptNon02.jpg|500px]]

*Difficulties

1. If the basis set is constructed so that <math> i +j +j \leq N</math>, the the
number of terms is
<math>
(N+1)(N+2)(N+3)/6
</math>
N=14 already gives 680 terms and an accuracy of about 1 part in <math>10^{10}</math> for
low-lying states. A substantial improvement in accuracy would require much
larger basis sets, together with multiple precision arithmetic to avoid loss
of significant figures when high powers are included.

2. The accuracy rapidly deteriorates as one goes to more highly excited
states - about 1 significant figure is lost each time the principle quantum
number is increased.

*Cure

We have found that writing basis sets in the form
\begin{eqnarray}
\Psi &=& \sum_{i,j,k}\left(c^{(1)}_{i,j,k}\varphi_{ijk}(\alpha_1\beta_1) +
c_{ijk}^{(2)}\varphi_{ijk}(\alpha_2\beta_2)\right) \pm exchange \nonumber\\
&=& \psi({\bf r}_1,{\bf r}_2) \pm \psi({\bf r}_2,{\bf r}_1)\nonumber
\end{eqnarray}
so that each combination of powers is included twice with different nonlinear
parameters gives a dramatic improvement in accuracy for basis sets of about
the same total size.

However, the optimization of the nonlinear parameters is now much more
difficult, and an automated procedure is needed.

If
<math>
E = \frac{<\Psi|H|\Psi>}{<\Psi|\Psi>}</math> and we assume <math><\Psi|\Psi> = 1</math>, then

<math>
\frac{\partial E}{\partial\alpha_t} = -2 <\Psi|H-E|r_1\psi({\bf r}_1,{\bf
r}_2;\alpha_t)\pm r_21\psi({\bf r}_2,{\bf
r}_1;\alpha_t)>
</math>
where
<math>
\psi({\bf r}_1,{\bf r}_2;\alpha_t) =
\sum_{i,j,k}c_{ijk}^{(t)}\varphi_{ijk}(\alpha_t\beta_t)
</math>

*Problem

1. Prove the above.

2. Prove that there is no contribution from the implicit dependence of
<math>c_{ijk}^{(t)}</math> on <math>\alpha_t</math> if the linear parameters have been optimized.

==The Screened Hydrogenic Term==

If the Hamiltonian is written in the form (in Z-scaled a.u.)
<math>\begin{eqnarray}\
H = H_0({\bf r}_1,Z) + H_0({\bf r}_2,Z-1) +
Z^{-1}\left(\frac{1}{r}-\frac{1}{r_2}\right)
\end{eqnarray}</math>
with
<math>\
H_0({\bf r}_1,Z) = -\frac{1}{2}\nabla_1^2 - \frac{1}{r_1}\nonumber
</math> and
<math>
H_0({\bf r}_2,Z-1) = -\frac{1}{2}\nabla_2^2 - \frac{Z-1}Z{r_2}\nonumber
</math>,
then the eigenvectors of <math>H_0({\bf r},Z) +H_0({\bf r}_2,Z-1)</math> are products of
hydrogenic orbitals
<math>
\Psi_0=\psi_0(1s,Z)\psi_0(nl,Z-1)
</math>
and the eigenvalue is
<math>\begin{eqnarray}
E_{SH} = \left[-\frac{1}{2} - \left(\frac{Z-1}{Z}\right)^2\frac{1}{2n^2}\right]Z^2 a.u.
\end{eqnarray}</math>
called the screened hydrogenic eigenvalue. From highly excited states,
<math>E_{SH}</math> and <math>\Psi_0</math>are already excellent approximations.

For example, for the 1s8d states, the energies are

<math>\begin{eqnarray}\\
E(1s8d^1D)&=& -2.007\,816\,512\,563\,81 {\rm\ a.u.}\nonumber\\
E(1s8d^3D)&=& - 2.007\,817\,934\,711\,71 {\rm\ a.u.}\nonumber\\
E_{SH} &=& -2.007\,812\,5 {\rm\ a.u.} \nonumber
\end{eqnarray}</math>

It is therefore advantageous to include the screened hydrogenic terms in the
basis set so that the complete trial function becomes
<math>
\Psi = c_0\Psi_0 +
\sum_{ijk}\left[c_{ijk}^{(1)}\varphi_{ijk}(\alpha_1,\beta_1)+c_{ijk}^{(2)}\varphi(\alpha_2,\beta_2)\right]\pm
{\rm exchange}
</math>

Also, the variational principal can be re-expressed in the form
<math>\begin{eqnarray}
E= E_{SH}+\frac{<\Psi|H-E_{SH}|\Psi>}{<\Psi|\Psi>}
\end{eqnarray}</math>
so that the <math>E_{SH}</math> term can be cancelled analytically from the matrix
elements.

For example
<math>
<\Psi_0|H-E_{SH}|\Psi_0> = 0
</math>

and
<math>
<\varphi_{ijk}|H-E_{SH}|\Psi_0> =
Z^{-1}<\varphi_{ijk}|\frac{1}{r}-\frac{1}{r_2}|\Psi_0>
</math>

Recall that
<math>\begin{eqnarray}\ \
I_0(a,b,c) = \frac{2}{c+2}\sum^{[c+1]/2}_{i=0}\left(\begin{array}{c}
c+2\\
2i+1
\end{array}\right) \left[f(p,q;\beta) + f(p^\prime,q^\prime;\alpha)\right]
\end{eqnarray}</math>
where

<math>\begin{eqnarray}
f(p,q;x) =
\frac{q!}{x^{q+1}(\alpha+\beta)^{p+1}}\sum^p_{j=0}\frac{(p+j)!}{j!}\left(\frac{x}{\alpha+\beta}\right)^j
\end{eqnarray}</math>

<math>
\begin{array}{cc}
p=a+2i+2 & p^\prime = b+2i+2\\
q= b+c-2i+2 & q^\prime =a+c-2i+2
\end{array}
</math>

If <math>\beta\ll\alpha</math>, the <math>f(p,q;\beta)\gg f(p^\prime,q^\prime;\alpha)</math> and the
<math>i=0</math> term is the dominant contribution to <math>f(p,q;\beta)</math>. However, since
this term depends only on the sum of powers <math>b+c</math> for <math>r_2</math>, it cancels exactly
from the matrix element of
<math>
\frac{1}{r_{12}} - \frac{1}{r_2}
</math>
and can therefore
be omitted in the calculations of integrands, thereby saving many significant
figures. This is especially valuable when <math>b+c</math> is large and <math>a</math> is small.

==Small Corrections==
*Mass Polarization
If the nuclear mass is not taken to be infinite, then the Hamiltonian is
<math>\begin{eqnarray}\ \
H = \frac{P_N^2}{2M} + \sum_{i=1}\frac{p_i^2}{2m} + V
\end{eqnarray}</math>

[[File:SmallCor.jpg|500px]]

where <math>M =M_A-nm</math> is the nuclear mass and <math>m</math> is the electron mass.

Change variables to the C of M
<math>\begin{eqnarray}
{\bf R} = \frac{1}{M+nm}\left[M{\bf r}_N + m({\bf r}_1 + {\bf r}_2 +
\ldots)\right]
\end{eqnarray}</math>

and relative variables

<math>
{\bf s}_i = {\bf r}_i - {\bf r}_N.
</math>

Then

<math>\begin{eqnarray}\ \ \ \ \
H = \frac{1}{2(M+nm)}{\bf p}_R^2+\frac{1}{2m}\sum_{i=1}^n{\bf p}_{s_i}^2
+\frac{1}{2M}\sum_{i,k}{\bf p}_{s_i}\cdot {\bf p}_{s_k} + V({\bf s}_1, {\bf
s}_2, \ldots)\nonumber
= \frac{1}{M+nm}{\bf p}_R^2+\frac{1}{2\mu}\sum_{i=1}^n{\bf
p}_{s_i}^2+\frac{1}{2M}\sum_{i\neq k}{\bf p}_{s_i}\cdot {\bf p}_{s_k}+V({\bf s}_1, {\bf
s}_2, \ldots)\nonumber
\end{eqnarray}</math>

where
<math>\begin{eqnarray}
\mu = \left(\frac{1}{M} + \frac{1}{m}\right)^{-1}
\end{eqnarray}</math>

If <math>n=2</math>, the Schrödinger equation is
<math>\begin{eqnarray}\ \ \
\left[ -\frac{\hbar^2}{2\mu}(\nabla^2_{s_1}+ \nabla^2_{s_2}) -
\frac{\hbar^2}{M}\nabla_{s_1}\nabla_{s_2}-\frac{Ze^2}{s_1}-\frac{Ze^2}{s_2}+\frac{e^2}{s_{12}}\right]\psi
= E\psi
\end{eqnarray}</math>.

Define the reduced mass Bohr radius
<math>\begin{eqnarray}
a_\mu = \frac{\hbar^2}{\mu e^2} = \frac{m}{\mu}a_0
\end{eqnarray}</math>
and

<math>\begin{eqnarray}
\rho_i = \frac{Zs_i}{a_\mu}\nonumber\\
\varepsilon = \frac{E}{Z^2(e^2/a_\mu)}\nonumber
\end{eqnarray}</math>

The Schrödinger equation then becomes
<math>\begin{eqnarray}\ \ \
\left[
-\frac{1}{2}(\nabla^2_{\rho_1}+\nabla^2_{\rho_2})-\frac{\mu}{M}\vec{\nabla}_{\rho_1}\cdot\vec{\nabla}_{\rho_2} - \frac{1}{\rho_1}-\frac{1}{\rho_2}+\frac{Z^{-1}}{\rho_{12}}\right]\psi = \varepsilon \psi
\end{eqnarray}</math>

The effects of finite mass are thus

1. A reduced mass correction to all energy levels of <math>(e^2/a_\mu)/(e^2/a_0)
= \mu/m</math>, i.e.
<math>\begin{eqnarray}
E_M=\frac{\mu}{m}E_\infty
\end{eqnarray}</math>

2. A specific mass shift given in first order perturbation theory by
<math>\begin{eqnarray}
\Delta\varepsilon =
-\frac{\mu}{M}<\psi|\nabla_{\rho_1}\cdot\nabla_{\rho_2}|\psi>
\end{eqnarray}</math>
or

<math>\begin{eqnarray}
\Delta\varepsilon =
-\frac{\mu}{M}<\psi|\nabla_{\rho_1}\cdot\nabla_{\rho_2}|\psi>\frac{e^2}{a_\mu}\nonumber\\
= -\frac{\mu^2}{mM}<\psi|\nabla_{\rho_1}\cdot\nabla_{\rho_2}|\psi>\frac{e^2}{a_0}\nonumber
\end{eqnarray}</math>

Since <math>\nabla_{\rho_1}\cdot\nabla_{\rho_2}</math> has the same angular properties as
<math>{\bf\rho}_1 \cdot {\bf\rho}_2</math>, the operator is like the product of two
dipole operators. For product type wave functions of the form
$
\psi = \psi(1s)\psi(nl) \pm \mbox{exchange},
$
the matrix element vanishes for all but P-states.

=Radiative Transitions=

==Transition Integrals and the Line Strength==

For an electric dipole transition <math>\gamma'\,L'\,S'\,J'\,M' \rightarrow \gamma\,L\,S\,J\,M</math>, the line strength is defined by
\begin{equation}
S = \sum_{M',\mu,M}|\langle\gamma'\,L'\,S'\,J'\,M'|r_\mu|\gamma\,L\,S\,J\,M\rangle|^2
\end{equation}
Using the Wigner-Eckart theorem, this is related to the reduced matrix element by
\begin{eqnarray}
S(J',J) = |\langle\gamma'\,L'\,S'\,J'||r||\gamma\,L\,S\,J\,M\rangle|^2
\sum_{M',\mu,M}\left|\left(\matrix{J' & 1 & J \cr
-M'& \mu &M \cr}\right)\right|^2
\end{eqnarray}
Using the sum rule for the 3-<math>j</math> symbols
\begin{equation}
\sum_{M',\mu,M}\left|\left(\matrix{J' & 1 & J \cr
-M'& \mu &M \cr}\right)\right|^2 = 1
\end{equation}
we see that the line strength is identical to the reduced matrix element
\begin{equation}
S(J',J) = |\langle\gamma'\,L'\,S'\,J'||r||\gamma\,L\,S\,J\,M\rangle|^2
\end{equation}
For a nonrelativistic calculation in <math>LS</math> coupling, <math>S = S'</math> and we can next strip out the dependence on <math>J</math> and <math>J'</math> by use of Eq.\ (7.1.7) of Edmonds { Angular Momentum in Quantum Mechanics} with <math>k = 1</math> to obtain
\begin{equation}
S(J',J) = (-1)^{L'+S+J+1}[(2J+1)(2J'+1)]^{1/2}\left\{
\matrix{L' & J'& S\cr
J & L & 1\cr}\right\}\langle\gamma' L'|| r || \gamma L\rangle
\end{equation}
where <math>\langle\gamma' L'|| r || \gamma L\rangle</math> is the reduced matrix element if spin is neglected. Note that if spin is neglected, then the line-strength is simply
\begin{equation}
S(L',L) = |\langle\gamma' L'|| r || \gamma L\rangle|^2
\end{equation}
To calculate the reduced matrix element, use a particular example along with the Wigner-Eckart theorem
such as
\begin{equation}
\langle\gamma'\, L'|| r || \gamma\, L\rangle = (-1)^{L'}\langle\gamma'\, L'\, 0| z | \gamma\, L\, 0\rangle{\Large /}\left(\matrix{L' & 1 & L \cr
0 & 0 & 0 \cr}\right)
\end{equation}

===Correlated Two-electron Integrals===
For the case of two-electron integrals, <math>z = z_1 + z_2</math> and <math>r = r_1 + r_2</math>. As discussed in G.W.F. Drake, Phys. Rev. A 18, 820 (1978), the general integral in Hylleraas coordinates for the integral over the <math>z_1</math> term can be written in the form
\begin{equation}
\label{int1}
\langle\gamma'\, L'\, 0| z_1 | \gamma\, L\, 0\rangle = \sum_\Lambda C_\Lambda I_\Lambda(\gamma',\gamma)
\end{equation}
where <math>I_\Lambda(\gamma',\gamma)</math> stands for the purely radial integral
\begin{equation}
\label{radial}
I_\Lambda(\gamma',\gamma) = \int_0^\infty r_1\,dr_1\int_0^\infty r_2\,dr_2\int_{|r_1-r_2|}^{r_1+r_2}r_{12}\, dr_{12}
R_{\gamma'}^*\,r_1\, R_\gamma P_\Lambda(\cos\theta_{12})
\end{equation}
and <math>\cos\theta_{12} = (r_1^2 + r_2^2 - r_{12}^2)/(2r_1r_2)</math> is a short-hand notation for a radial function of <math>r_1</math>, <math>r_2</math> and <math>r_{12}</math>. Also, <math>P_\Lambda(\cos\theta_{12})</math> is a Legendre polynomial,
and, <math>R_\gamma = R_\gamma(r_1,r_2,r_{12})</math> is the radial part of the Hylleraas wave function for state <math>\gamma</math> such
that the total wave function is <math>\psi({\bf r}_1,{\bf r}_2) = R_\gamma(r_1,r_2,r_{12}){\cal Y}_{l_1,l_2,L}^M(\hat{\bf r}_1,\hat{\bf r}_2)\pm {\rm exchange}</math>, normalized to unity, and <math>{\cal Y}_{l_1,l_2,L}^M(\hat{\bf r}_1,\hat{\bf r}_2)</math> is a vector coupled product of spherical harmonics. The coefficents <math>C_\Lambda</math> in Eq.\ (\ref{int1}) can be calculated from Eq.\ (32) of the above 1978 paper. They need to be calculated separately for he direct-direct and the direct-exchange contributions. For the reduced matrix element of <math>z_1</math>, the leading 3-<math>j</math> symbol can be dropped, leaving (for the case <math>k = 1</math>)
\begin{eqnarray}
C_\Lambda^{\rm reduced} &=& \frac{(-1)^{l_2'+L'+L+1}}{2} (l_1,l_1',l_2,l_2',L,L')^{1/2}
\sum_{\lambda={\rm max}(l_1-1,0)}^{l_1+1} (\lambda,\Lambda,1)
\left(\matrix{l_1 & 1 & \lambda \cr
0 & 0 & 0 \cr}\right)
\left(\matrix{l_1'& \lambda & \Lambda \cr
0 & 0 & 0 \cr}\right)\nonumber\\
&\times&\left(\matrix{l_2' & 1_2 & \Lambda \cr
0 & 0 & 0 \cr}\right)
\left\{\matrix{L' & \lambda & l_2 \cr
\Lambda& l_2' & l_1' \cr}\right\}
\left\{\matrix{L' & \lambda & l_2 \cr
l_1 & L & 1 \cr}\right\}
\end{eqnarray}
where the notation <math>(a,b,c,\cdots)</math> means <math>(2a+1)(2b+1)(2c+1)\cdots</math>.
A similar expression can be obtained for the angular coefficients for the <math>z_2</math> term by interchanging the subscripts 1 and 2 throughout and multiplying by <math>(-1)^{L+L'+1}</math>.

===Example===
As an example, for the <math>1s^2\;^1S - 1s2p\;^1P</math> transition of helium, the only nonvanishing <math>C_\Lambda</math> angular coefficients are <math>C_1 = 1/2</math> for the direct-direct term (i.e. <math>P_1(\cos\theta_{12}) = \cos\theta_{12}</math> is included in the radial integral (\ref{radial})) and <math>C_0 = 1/2</math> for the direct-exchange term (i.e. <math>P_0(\cos\theta_{12}) = 1</math> is included in the radial integral (\ref{radial})). The results for infinite nuclear mass are <math>S(1\;^1S-2\;^1P) = (0.781\,127\,158\,667)^2</math> and <math>f(1\;^1S-2\;^1P) = -0.276\,164\,703\,736</math> where
\begin{equation}
f(1\;^1S-2\;^1P) = \frac{2}{3}[E(1\;^1S) - E(2\;^1P)]S(1\;^1S-2\;^1P)
\end{equation}
is the absorption oscillator strength.

Theory Notes

2020-12-14T00:38:37Z

Drake: /* General Hermitian Property */

<center>
<H1>Notes on solving the Schrödinger equation in Hylleraas coordinates for heliumlike atoms
Gordon W.F. Drake
Department of Physics, University of Windsor
Windsor, Ontario, Canada N9B 3P4
(Transcribed from hand-written notes and edited by Lauren Moffatt. Last revised November 22, 2017 to correct a missing subscript in the expression for $B_1$ in the Hamiltonian in Sect. 11 General Hermitian Property.)
</H1>
</center>

==Introduction==
These notes describe some of the technical details involved in solving the Schrödinger equation for a heliumlike atom of nuclear charge Z in correlated Hylleraas coordinates. The standard for computational purposes is to express all quantities in a form that is valid for electronic states of arbitrary angular momentum, and to express matrix elements of the Hamiltonian in a manifestly Hermitian symmetrized form. Further information and derivations of the mathematical techniques for angular integrals and radial recursion relations can be found in the paper
G.W.F. Drake, Phys. Rev. A <math>{\bf 18}</math>, 820 (1978).

The starting point is the two-electron Schrödinger equation for infinite nuclear mass
\begin{equation} \left[-\frac{\hbar^2}{2m}(\nabla^2_1 +\nabla^2_2) - \frac{Ze^2}{r_1} -
\frac{Ze^2}{r_2}+\frac{e^2}{r_{12}} \right]\psi = E\psi\nonumber
\end{equation}
where <math>m</math> is the electron mass, and <math>r_{12} = |{\bf r}_1 - {\bf r}_2|</math> (see diagram below).

[[File:Pic1.jpg|250px]]

Begin by rescaling distances and energies so that the Schrödinger equation can be expressed in a dimensionless form. The dimensionless Z-scaled distance is defined by <math>\begin{eqnarray} \rho = \frac{Zr}{a_0} \end{eqnarray} </math> where <math> \begin{eqnarray} a_0 = \frac{\hbar^2}{me^2} \end{eqnarray} </math> is the atomic unit (a.u.) of distance equal to the Bohr radius <math>0.529\, 177\, 210\, 92(17) \times 10^{-10}</math> m. Then

\begin{equation}[-\frac{\hbar^2}{2m}Z^2(\frac{me^2}{\hbar^2})^2(\nabla^2_{\rho_1}+\nabla^2_{\rho_2})
- Z^2\frac{e^2}{a_0}\rho^{-1}_1 - Z^2\frac{e^2}{a_0}\rho^{-1}_2 +
\frac{e^2}{a_0}Z\rho^{-1}_{12}]\psi= E\psi\nonumber\end{equation}

But <math>\begin{eqnarray} \frac{\hbar^2}{m}(\frac{me^2}{\hbar^2})^2 = \frac{e^2}{a_0} \end{eqnarray} </math> is the hartree atomic unit of of energy <math>(E_h = 27.211\,385\,05(60)</math> eV, or equivalently <math>E_h/(hc) = 2194.746\,313\,708(11)</math> m<math>^{-1})</math>. Therefore, after multiplying through by <math>a_0/(Ze)^2</math>, the problem to be solved in Z-scaled dimensionless units becomes

\begin{equation} \left[-\frac{1}{2}(\nabla^2_{\rho_1}+\nabla^2_{\rho_2}) -
\frac{1}{\rho_1} - \frac{1}{\rho_2} + \frac{Z^{-1}}{\rho_{12}}\right]\psi =
\varepsilon\psi\nonumber \end{equation}

where <math>\begin{eqnarray}\varepsilon = \frac{Ea_0}{(Ze)^2} \end{eqnarray}</math> is the energy in Z-scaled atomic units. For convenience, rewrite this in the conventional form
\begin{equation}
H\psi = \varepsilon\psi\nonumber \end{equation}
where (using <math>r</math> in place of <math>\rho</math> for the Z-scaled distance)
\begin{equation}
H = -\frac{1}{2}(\nabla^2_1+\nabla^2_2) - \frac{1}{r_1}-\frac{1}{r_2} + \frac{Z^{-1}}{r_{12}}
\nonumber\end{equation}
is the atomic Hamiltonian for infinite nuclear mass.

==The Hartree Fock Method==

For purposes of comparison, and to define the correlation energy, assume that <math>\psi({\bf r}_1,{\bf r}_2)</math> can be written in the separable product form
\begin{equation}
\psi({\bf r}_1,{\bf r}_2) = \frac{1}{\sqrt{2}} \left[ u_1(r_1) u_2(r_2) \pm u_2(r_1) u_1(r_2) \right] \nonumber
\end{equation}
for the <math>1s^2\ ^1\!S</math> ground state. Because of the <math>\frac{1}{r_{12}}</math> term in the Schrödinger equation, the exact solution connot be expressed in this form as a separable product. However, the Hartree-Fock approximation corresponds to finding the best possible solution to the Schrödinger equation
\begin{equation}
H\psi({\bf r}_1,{\bf r}_2) = E\psi({\bf r}_1,{\bf r}_2)\nonumber
\end{equation}
that can nevertheless be expressed in this separable product form, where as before
\begin{equation}
H = -\frac{1}{2}(\nabla^2_1+\nabla^2_2) - \frac{1}{r_1}- \frac{1}{r_2} + \frac{Z^{-1}}{r_{12}}\nonumber
\end{equation}
is the full two-electron Hamiltonian. To find the best solution, substitute into <math>\left\langle\psi \lvert H-E \rvert \psi\right\rangle</math> and require this expression to be stationary
with respect to arbitrary infinitesimal variations <math>\delta u_1</math> and <math>\delta
u_2</math> in <math>u_1</math> and <math>u_2</math> respectively; i.e.

\begin{equation}
\frac{1}{2} \left\langle \delta u_1(r_1)u_2(r_2) \pm u_2(r_1)\delta
u_1(r_2) \lvert H-E \rvert u_1(r_1) u_2(r_2)\pm u_2(r_1) u_1(r_2)\right\rangle\nonumber
\end{equation}

\begin{equation}
=\int\delta u_1(r_1) d{\bf r}_1 \left\{ \int d{\bf r}_2 u_2(r_2) \left( H - E \right) \left[ u_1(r_1)u_2(r_2) \pm u_2(r_1) u_1(r_2) \right] \right\}\nonumber
\end{equation}

\begin{equation}
= 0 \nonumber
\end{equation}

for arbitrary <math> \delta u_1(r_1).\nonumber </math> Therefore <math>\{\int d{\bf r}_2 \ldots \} = 0</math>.

Similarly, the coefficient of <math>\delta u_2</math> would give

\begin{equation}
\int d{\bf r}_1 u_1(r_1) \left( H-E \right) \left[ u_1(r_1) u_2(r_2) \pm
u_2(r_1) u_1(r_2) \right] = 0\nonumber
\end{equation}

Define

\begin{equation}
I_{12} = I_{21} = \int d{\bf r}\, u_1(r)u_2(r), \nonumber
\end{equation}

\begin{equation}
H_{ij} = \int d{\bf r}\, u_i(r)(-\frac{1}{2}\nabla - \frac{1}{r})u_j(r), \nonumber
\end{equation}

\begin{equation}
G_{ij}(r) = \int d{\bf r}^\prime u_i(r^\prime)\frac{1}{|{\bf r} - {\bf r}\prime|}u_j(r^\prime)\nonumber
\end{equation}

Then the above equations become the pair of integro-differential equations

\begin{equation}
[ H_0 - E + H_{22}+G_{22}(r)]u_1(r) = \mp [ I_{12}(H_0-E) + H_{12}+G_{12}(r)]u_2(r)\nonumber
\end{equation}

\begin{equation}
[H_0-E+H_{11}+G_{11}(r)]u_2(r) = \mp [I_{12}(H_0-E) + H_{12}+G_{12}(r)]u_1(r)\nonumber
\end{equation}

These must be solved self-consistently for the "constants" <math>I_{12}</math> and
<math>H_{ij}</math> and the function <math>G_{ij}(r)</math>.

The Hartree Fock energy is <math>E \simeq -2.87\ldots </math>a.u. while the exact energy is <math>E = -2.903724\ldots </math>a.u.

The difference is called the "correlation energy" because it arises from the
way in which the motion of one electron is correlated to the other. The
Hartree Fock equations only describe how one electron moves in the average field
provided by the other (mean-field theory).

==Configuration Interaction==
Expand
\begin{equation}
\psi({\bf r}_1,{\bf r}_2)= C_0u^{(s)}_1(r_1)u^{(s)}_1(r_2) + C_1u^{(P)}_1({\bf r}_1)u^{(P)}_1({\bf r}_2){\cal Y}^0_{1,1,0}(\hat{\bf r}_1, \hat{\bf r}_2)+C_2u^{(d)}_1({\bf r}_1)u^{(d)}_2({\bf r}_2){\cal Y}^0_{2,2,0}(\hat{\bf r}_1, \hat{\bf r}_2)+... \pm {\rm\ exchange}
\end{equation}
where
\begin{equation}
{\cal Y}^M_{l_1,l_2,L}(\hat{\bf r}_1, \hat{\bf r}_2)=\sum_{m_1,m_2}Y^{m_1}_{l_1}({\bf r}_1)Y^{m_2}_{l_2}({\bf r}_2)\times <l_1l_2m_1m_2\mid LM>
\end{equation}

This works, but is slowly convergent, and very laborious. The best CI calculations are accurate to <math> ~10^{-7}</math> a.u.

==Hylleraas Coordinates==
[E.A. Hylleraas, Z. Phys. <math>{\bf 48}, 469(1928)</math> and <math>{\bf 54}, 347(1929)</math>]
suggested using the co-ordinates <math>r_1</math>, <math>r_2</math> and <math>r_{12}</math> or equivalently

\begin{eqnarray}
s &=& r_1 + r_2, \nonumber\\
t &=& r_1-r_2, \nonumber\\
u &=& r_{12}\nonumber
\end{eqnarray}

and writing the trial functions in the form

\begin{equation}
\Psi({\bf r}_1,{\bf r}_2) = \sum^{i+j+k\leq
N}_{i,j,k}c_{i,j,k}r_1^{i+l_1}r_2^{j+l_2}r_{12}^ke^{-\alpha r_1 - \beta r_2}
\mathcal{Y}^M_{l_1,l_2,L}(\hat{r}_1,\hat{r}_2)\pm \text{exchange}\nonumber
\end{equation}

Diagonalizing H in this non-orthogonal basis set is equivalent to solving
<math> \begin{eqnarray}
\frac{\partial E}{\partial c_{i,j,k}} = 0\nonumber
\end{eqnarray} </math>
for fixed <math>\alpha</math> and <math>\beta</math>.

The diagonalization must be repeated for different values of <math>\alpha</math> and
<math>\beta</math> in order to optimize the non-linear parameters.

==Completeness==
The completeness of the above basis set can be shown by first writing
\begin{equation}
r_{12}^2 = r_1^2 + r_2^2 - 2r_1r_2\cos\theta_{12}
\end{equation}
and, from the spherical harmonic addition theorem,
\begin{equation}
\cos(\theta_{12})=\frac{4\pi}{3}\sum^1_{m=-1}Y^{m*}_1(\theta_1,\varphi_1)Y^m_1(\theta_2,\varphi_2)
\end{equation}
Consider first the S-states. The <math>r_{12}^0</math> terms are like the ss terms in a CI calculation. The
<math>r_{12}^2</math> terms bring in p-p type contributions, and the higher powers bring
in d-d, f-f etc. type terms. In general

\begin{equation}
P_l(\cos\theta_{12}) = \frac{4\pi}{2l+1}\sum^l_{m=-l}{Y^{m}_l}^*(\theta_1,\varphi_1)Y^m_l(\theta_2,\varphi_2)\nonumber
\end{equation}

For P-states, one would have similarly

\begin{array}{lr}
r_{12}^0 & (sp)P\nonumber\\
r_{12}^2 & (pd)P\nonumber\\
r_{12}^4 & (df)P\nonumber\\
\vdots & \vdots\nonumber
\end{array}

For D-states

\begin{array}{lr}
r_{12}^0 & (sd)D & (pp^\prime)D\nonumber\\
r_{12}^2 & (pf)D & (dd^\prime)D\nonumber\\
r_{12}^4 & (dg)D & (ff^\prime)D\nonumber\\
\vdots & \vdots & \vdots\nonumber
\end{array}

In this case, since there are two "lowest-order" couplings to form a
D-state, both must be present in the basis set, i.e.

\begin{equation}
\Psi(r_2,r_2) = \sum c_{ijk}r_1^ir_2^{j+2}r_{12}^ke^{-\alpha r_1-\beta
r_2}\mathcal{Y}^M_{022}(\hat{r}_1,\hat{r}_2)
+\sum d_{ijk}r_1^{i+1}r_2^{j+1}r_{12^k}e^{-\alpha^\prime r_1 - \beta^\prime
r_2}\mathcal{Y}^M_{112}(\hat{r}_1,\hat{r}_2)\nonumber
\end{equation}

For F-states, one would need <math>(sf)F</math> and <math>(pd)F</math> terms.

For G-states, one would need <math>(sg)G</math>, <math>(pf)G</math> and <math>(dd^\prime)G</math> terms.

Completeness of the radial functions can be proven by considering the
Sturm-Liouville problem

\begin{equation}
\left(-\frac{1}{2}\nabla^2-\frac{\lambda}{r_s}-E\right)\psi({\bf r}) = 0\nonumber
\end{equation}

or

\begin{equation}
\left(-\frac{1}{2}\frac{1}{r^2}\frac{\partial}{\partial r}\left({r^2}\frac{\partial}{\partial r}\right) -
\frac{l\left(l+1\right)}{2r^2} - \frac{\lambda}{r} - E\right)u(r) = 0.\nonumber
\end{equation}

For fixed E and variable <math>\lambda</math> (nuclear charge).

The eigenvalues are <math>\lambda_n = (E/E_n)^{1/2}</math>, where <math>E_n =- \frac{1}{2n^2}</math>

[[File:Pic2.jpg|500px]]

and the eigenfunctions are

\begin{equation}
u_{nl}(r) =
\frac{1}{(2l+1)!}\left(\frac{(n+l)!}{(n-l-1)2!}\right)^{1/2}(2\alpha)^{3/2}e^{-\alpha
r}\nonumber
\times \ \left( 2 \alpha r \right)^l\,_{1}F_{1}\left(-n+l+1,2l+2; 2 \alpha r \right)
\end{equation}

with <math>\alpha = (-2E)^{1/2}</math> and <math>n\geq l+1</math>. The confluent hypergeometric function <math> _1F_1(a,b,;z)</math> then denotes a finite polynomial since <math>a=-n+l+1</math> is a negative integer or zero.

Unlike the hydrogen spectrum, which has both a discrete part for <math>E<0</math> and a
continuous part for <math>E>0</math>, this forms an entirely discrete set of finite
polynomials, called Sturmian functions. They are orthogonal with respect to
the potential, i.e.

\begin{equation}
\int^\infty_0 r^2dr \left( u_{n^\prime l}(r)\frac{1}{r}u_{nl}(r) \right) = \delta_{n,n^\prime}\nonumber
\end{equation}

Since they become complete in the limit <math>n\rightarrow\infty</math>, this assures the
completeness of the variational basis set.

[See also B Klahn and W.A. Bingel Theo. Chim. Acta (Berlin) 44, 9 and 27
(1977)].

==Solutions of the Eigenvalue Problem==

For convenience, write

\[
\Psi({\bf r}_1,{\bf r}_2) = \sum^N_{m=1}c_m\varphi_m\nonumber
\]

where <math>m</math> represents the <math>m</math>'th combination of <math>i,j,k</math> and

\begin{equation}
\varphi_{ijk}=r_1^ir_2^jr_{12}^ke^{-\alpha r_1 - \beta
r_2}\mathcal{Y}^M_{l_1,l_2,L}(\hat{r}_1,\hat{r}_2) \pm \mbox{exchange}\nonumber
\end{equation}

\begin{align}

&
\left( \begin{array}{lr}
\cos(\theta) & \sin(\theta)\\
-\sin(\theta) & \cos(\theta)
\end{array} \right)
\left( \begin{array}{lr}
H_{11} & H_{12} \\
H_{12} & H_{22}
\end{array}\right)
\left( \begin{array}{lr}
\cos(\theta) & -\sin(\theta)\\
\sin(\theta) & \cos(\theta)
\end{array} \right)
\\[5pt]
= & \left(\begin{array}{lr}
cH_{11}+sH_{12} & cH_{12} + sH_{22}\\
-sH_{11} + cH_{12} & -sH_{12} + cH_{22}
\end{array}\right)
\left( \begin{array}{cc}
c & -s \\
s & c
\end{array}\right)
\\[5pt]
= & \left(\begin{array}{lr}
c^2H_{11}+s^2H_{22} + 2csH_{12} & (c^2-s^2)H_{12}+cs(H_{22}-H_{11})\\
(c^2-s^2)H_{12}+cs(H_{22}-H_{11}) & s^2H_{11}+c^2H_{22}-2csH_{12}
\end{array}\right)
\end{align}

Therefore

\begin{equation}
\left[\cos^2(\theta)-\sin^2(\theta)\right]H_{12} =
\cos(\theta)\sin(\theta)(H_{11}-H_{22})\nonumber
\end{equation}

and

\begin{equation}
\tan(2\theta) = \frac{2H_{12}}{H_{11}-H_{22}}\nonumber
\end{equation}

i.e.

\begin{equation}
\cos(\theta)=\left(\frac{r+\omega}{2r}\right)^{1/2}\nonumber
\end{equation}

\begin{equation}
\sin(\theta)=-\mbox{sgn}(H_{12})\left(\frac{r-\omega}{2r}\right)^{1/2}\nonumber
\end{equation}

where

\begin{equation}
\omega = H_{22}-H_{11}\nonumber
\end{equation}

\begin{equation}
r=\left(\omega^2+4H_{12}^2\right)^{1/2}\nonumber
\end{equation}

\begin{equation}
E_1=\frac{1}{2}\left(H_{11}+H_{22}-r\right)\nonumber
\end{equation}

\begin{equation}
E_2=\frac{1}{2}\left(H_{11}+H_{22}+r\right)\nonumber
\end{equation}

===Brute Force Method===

- Gives all the eigenvalues and eigenvectors, but it is slow

- First orthonormalize the <math>N</math>-dimensional basis set; i.e. form linear combinations

\begin{equation}
\Phi_m = \sum_{n=1}^N\varphi_nR_{nm}\nonumber
\end{equation}

such that
\begin{equation}
\langle \Phi_m | \Phi_n \rangle = \delta_{m,n}
\end{equation}

This can be done by finding an orthogonal tranformation T such that

\begin{equation}
T^TOT = I = \left(
\begin{array}{ccccc}
I_1 & 0 & 0 & \ldots & 0 \\
0 & I_2 & 0 & \ldots & 0 \\
0 & 0 & I_3 & \ldots & 0 \\
\vdots & \vdots & \vdots & \ddots & \vdots \\
0 & 0 & 0 & \ldots & I_N
\end{array}\right);
\end{equation}

\begin{equation}
O_{mn} = \langle \varphi_m | \varphi_n \rangle \nonumber
\end{equation}

and then applying a scale change matrix

\begin{equation}
S = \left(\begin{array}{ccccc}
\frac{1}{I_1^{1/2}} & 0 & 0 & \ldots & 0\\
0 & \frac{1}{I_2^{1/2}} & 0 & \ldots & 0 \\
0 & 0 & \frac{1}{I_3^{1/2}} & \ldots & 0 \\
\vdots & \vdots & \vdots & \ddots & \vdots \\
0 & 0 & 0 & \ldots & I_N^{1/2}
\end{array}\right)= S^T\nonumber
\end{equation}

Then

\begin{equation}
S^TT^TOTS = 1
\end{equation}

i.e.

\begin{equation}
R^TOR = 1
\end{equation}

with <math>R=TS</math>.

If H is the matrix with elements <math>H_{mn} = \langle \varphi_m | \varphi_n \rangle </math>, then H
expressed in the <math>\Phi_m</math> basis set is

\begin{equation}
H^\prime = R^THR.\nonumber
\end{equation}

We next diagonalize <math>H^\prime</math> by finding an orthogonal transformation W such
that

\begin{equation}
W^T H^\prime W = \lambda = \left(
\begin{array}{cccc}
\lambda_1 & 0 & \ldots & 0 \\
0 & \lambda_2 & \ldots & 0 \\
\vdots & \vdots & \ddots & \vdots \\
0 & 0 & \ldots & \lambda_N
\end{array}\right)\nonumber
\end{equation}

The q'th eigenvector is

\begin{equation}
\Psi^{(q)} = \sum_{n=1}^N\Phi_n W_{n,q} = \sum_{n,n^\prime} \varphi_{n^\prime} R_{n^\prime ,n} W_{n,q} \nonumber
\end{equation}

i.e.

\begin{equation}
c_{n^\prime}^{(q)} = \sum_{n=1}^N R_{n^\prime n} W_{n,q}
\end{equation}

===The Power Method===

- Based on the observation that if H has one eigenvalue, <math>\lambda_M</math>, much bigger than all the rest, and <math>\chi = \left(
\begin{array}{c}a_1\\a_2\\\vdots\end{array}\right)</math>
is an arbitrary starting vector, then <math>\chi = \sum_{q=1}^N x_q\Psi^{(q)}</math>.

\begin{equation}
(H)^n\chi = \sum_{q=1}^N x_q \lambda^n_q\Psi^{(q)}\nonumber\rightarrow x_M\lambda_M^n\Psi^{(M)}\nonumber
\end{equation}

provided <math>x_M\neq 0</math>.

To pick out the eigenvector corresponding to any eigenvalue, write the original problem in the form

\begin{align}
H\Psi & = \lambda O \Psi \nonumber \\
\left( H - \lambda_q O \right) \Psi & = \left( \lambda - \lambda_q \right) O \Psi \nonumber
\end{align}

Therefore,

\begin{equation}
G \Psi = \frac{1}{\lambda - \lambda_q } \Psi \nonumber \\
\end{equation}

where <math>G=(H-\lambda_qO)^{-1}O</math> with eigenvalues <math>\frac{1}{\lambda_n - \lambda_q}</math>.

By picking <math>\lambda_q</math> close to any one of the <math>\lambda_n</math>, say
<math>\lambda_{n^\prime}</math>, then <math>\frac{1}{\lambda_n-\lambda_q}</math> is much larger for
<math>n=n^\prime</math> than for any other value. The sequence is then

\begin{array}{c}
\chi_1=G\chi\nonumber\\
\chi_2=G\chi_1\nonumber\\
\chi_3=G\chi_2\nonumber\\
\vdots\nonumber
\end{array}

until the ratios of components in <math>\chi_n</math> stop changing.

- To avoid matrix inversion and multiplication, note that the sequence is equivalent to
\begin{equation}
F \chi_n = \left( \lambda - \lambda_q \right) O \chi_{n-1} \nonumber \\
\end{equation}
where <math>F = H-\lambda_qO</math>. The factor of <math>\left( \lambda - \lambda_q \right)</math> can be
dropped because this only affects the normalization of <math>\chi_n</math>. To find <math>\chi_n</math>, solve
\begin{equation}
F \chi_n = O \chi_{n-1} \nonumber
\end{equation}
(N equations in N unknowns). Then
\begin{equation}
\lambda = \frac{ \langle \chi_n | H | \chi_n \rangle}{\langle \chi_n | \chi_n \rangle} \nonumber
\end{equation}

==Matrix Elements of H==

\begin{equation}
H=-\frac{1}{2}\nabla^2_1 -\frac{1}{2}\nabla^2_2 - \frac{1}{r_1} - \frac{1}{r_2} +\frac{Z^{-1}}{r_{12}}\nonumber
\end{equation}

Taking <math>r_1, r_2</math> and <math>r_{12}</math> as independent variables,

\begin{equation}
\nabla_1^2 = \frac{1}{r_1^2}\frac{\partial}{\partial r_1}\left(r_1^2\frac{\partial}{\partial r_1}\right)+ \frac{1}{r_{12}^2}\frac{\partial}{\partial r_{12}}
\left(r_{12}^2\frac{\partial}{\partial r_{12}}\right)\nonumber -\frac{l_1(l_1+1)}{r_1^2}+2(r_1 - r_2\cos(\theta_{12}))\frac{1}{r_{12}}\frac{\partial^2}{\partial r_1 \partial r_{12}} \nonumber - 2(\nabla_1^Y \cdot {\bf r}_2)\frac{1}{r_{12}}\frac{\partial}{\partial r_{12}}\nonumber
\end{equation}

where <math>\nabla_1^Y</math> acts only on the spherical harmonic part of the wave function and the diagram

[[File:Pic3.jpg|400px]]

defines the complete set of 6 independent variables is <math>r_1</math>, <math>r_2</math>, <math>r_{12}</math>, <math>\theta_1</math>, <math>\varphi_1</math>, <math>\chi</math>.

If <math>r_{12}</math> were not an independent variable, then one could take the volume element to be

\begin{equation}
d\tau = r_1^2dr_1\sin(\theta_1)d\theta_1d\varphi_1r_2^2dr_2\sin(\theta_2)d\theta_2d\varphi_2.\nonumber
\end{equation}

However, <math>\theta_2</math> and <math>\varphi_2</math> are no longer independent variables. To eliminate them, take the point <math>{\bf r}_1</math> as the
origin of a new polar co-ordinate system, and write

\begin{equation}
d\tau=-r_1^2dr_1\sin(\theta_1)d\theta_1d\varphi_1r_{12}^2dr_{12}\sin(\psi)d\psi d\chi\nonumber
\end{equation}

and use

\begin{equation}
r_2^2=r_1^2+r_{12}^2 +2r_1r_{12}\cos(\psi)
\end{equation}

Then for fixed <math>r_1</math> and <math>r_{12}</math>,

\begin{equation}
2r_2dr_2 = -2r_1r_{12}\sin(\psi)d\psi
\end{equation}

Thus

\begin{equation}
d\tau= r_1dr_1r_2dr_2r_{12}dr_{12}\sin(\theta_1)d\theta_1d\varphi_1d\chi
\end{equation}

The basic type of integral to be calculated is

\begin{align}
I \left( l_1, m_1, l_2, m_2; R \right) & = \int \sin \left( \theta_1 \right) d\theta_1 d\varphi_1 d\chi Y^{m_1}_{l_1} \left( \theta_1, \varphi_1 \right)^{*} Y^{m_2}_{l_2} \left( \theta_2, \varphi_2 \right) \\
& \times \int r_1dr_1r_2dr_2r_{12}dr_{12}R \left( r_1, r_2, r_{12} \right)
\end{align}

Consider first the angular integral. <math>Y^{m_2}_{l_2} \left( \theta_2, \varphi_2 \right)</math> can be expressed in terms of the independent variables <math>\theta_1, \varphi_1, \chi</math> by use of the rotation matrix relation
\begin{equation}
Y^{m_2}_{l_2}( \theta_2, \varphi_2) = \sum_m \mathcal{D}^{(l_2)}_{m_2,m}(\varphi_1,\theta_1,\chi)^*Y^m_{l_2}( \theta_{12}, \varphi)
\end{equation}
where <math>\theta, \varphi</math> are the polar angles of <math>{\bf r}_2</math> relative to <math>{\bf r}_1</math>. The angular integral is then

\begin{align}
I_{ang} & = \int^{2 \pi}_{0} d\chi \int^{2\pi}_{0} d\varphi_1 \int^{\pi}_{0} \sin \left( \theta_1 \right) d\theta_{1} Y^{m_1}_{l_1} \left( \theta_1, \varphi_1 \right)^{*} \\
& \times \sum_m \mathcal{D}^{(l_2)}_{m_2,m} \left( \varphi_1,\theta_1,\chi \right)^{*} Y^{m}_{l_2} \left( \theta_{12},\varphi \right) \\
\end{align}

Use

\begin{equation}
Y^{m_1}_{l_1} \left( \theta_1, \varphi_1 \right)^{*} = \sqrt{\frac{2l_1+1}{4\pi}}\mathcal{D}^{(l_1)}_{m_1,0} \left( \varphi_1, \theta_1, \chi \right)
\end{equation}

together with the orthogonality property of the rotation matrices (Brink and Satchler, p 147)

\begin{align}
& \int \mathcal{D}^{(j)*}_{m,m^\prime}\mathcal{D}^{(J)}_{M,M^\prime}\sin(\theta_1)d\theta_1d\varphi_1d\chi \\
& = \frac{8\pi^2}{2j+1}\delta_{jJ}\delta_{mM}\delta_{m^\prime M^\prime}\nonumber
\end{align}

to obtain

\begin{align}
I_{ang} & = \sqrt{ \frac{2l_1+1}{4\pi}} \frac{8\pi^2}{2l_1+1} \delta_{l_1,l_2} \delta_{m_1,m_2} Y^{0}_{l_2} \left( \theta_{12}, \varphi \right) \\
& = 2 \pi \delta_{l_1,l_2} \delta_{m_1,m_2} P_{l_2} \left( \cos \theta_{12} \right)
\end{align}

since

\begin{equation}
Y^0_{l_2} \left( \theta_{12}, \varphi \right) = \sqrt{\frac{2l_1+1}{4\pi}} P_{l_2} \left( \cos \left( \theta_{12} \right) \right)
\end{equation}

Note that <math>P_{l_2} \left( \cos \theta \right)</math> is just a short hand expression for a radial function because

\begin{equation}
\cos \theta_{12} = \frac{r_1^2 + r_2^2 - r_{12}^2}{2r_1r_2}
\end{equation}

The original integral is thus

\begin{equation}
I \left( l_1, m_1, l_2, m_2; R \right) = 2 \pi \delta_{l_1,l_2} \delta_{m_1,m_2} \int^{\infty}_{0} r_{1} dr_{1} \int^{\infty}_{0} r_{2} dr_{2} \int^{r_{1} + r_{2}}_{|r_{1} - r_{2}|} r_{12} dr_{12} R \left( r_{1}, r_{2}, r_{12} \right) P_{l_2} \left( \cos \theta_{12} \right)
\end{equation}

where again

\begin{equation}
\cos \theta_{12} = \frac{r_{1}^{2} +r^{2}_{2} - r_{12}^{2}}{2 r_{1} r_{2}}
\end{equation}

is a purely radial function.

The above would become quite complicated for large <math>l_2</math> because <math>P_{l_2}(\cos \theta_{12})</math> contains terms up to <math>\left( \cos \theta_{12} \right)^{l_2}</math>. However, recursion relations exist which allow any integral containing <math>P_l \left( \cos \theta_{12} \right)</math> in terms of those containing just <math>P_0 \left( \cos \theta_{12} \right) = 1</math> and <math>P_1 \left( \cos \theta_{12} \right) = \cos \theta_{12}</math>.

==Radial Integrals and Recursion Relations==

The basic radial integral is [see G.W.F. Drake, Phys. Rev. A 18, 820 (1978)]

\begin{align}
I_0 \left( a, b, c \right) & =
\int^{\infty}_{0} r_{1} dr_{1} \int^{\infty}_{r_{1}} r_{2} dr_{2} \int^{r_{1}+r_{2}}_{r_{2} - r_{1}} r_{12} dr_{12} r_{1}^{a} r_{2}^{b} r_{12}^{c} e^{-\alpha r_{1} - \beta r_{2}} + \int^{\infty}_{0} r_{2} dr_{2} \int^{\infty}_{r_{2}} r_{1} dr_{1} \int^{r_{1} + r_{2}}_{r_{1} - r_{2}} r_{12} dr_{12} r_{1}^{a} r_{2}^{b} r_{12}^{c} e^{- \alpha r_{1} - \beta r_{2}} \\
& = \frac{2}{c+2} \sum^{[(c+1)/2]}_{i=0}
\left(\begin{array}{c}
c+2 \\
2i+1
\end{array} \right)
\left\{ \frac{q!}{\beta^{q+1} \left( \alpha + \beta \right)^{p+1}} \sum_{j=0}^{q} \frac{\left( p + j \right)!}{j!} \left( \frac{\beta}{\alpha + \beta} \right)^j + \frac{q^{\prime}!}{\alpha^{q^{\prime}+1} \left( \alpha + \beta \right)^{p^{\prime}+1}} \sum^{q^{\prime}}_{j=0} \frac{\left( p^{\prime} + j \right)!}{j!} \left( \frac{\alpha}{\alpha + \beta} \right)^j \right\}
\end{align}

where

\begin{array}{ll}
p = a + 2i + 2 & p^{\prime} = b + 2i + 2 \\
q = b + c - 2i + 2 & q^{\prime} = a + c - 2i + 2
\end{array}

The above applies for <math> a, b \geq -2, c\geq -1</math>.
<math>[x]</math> means "greatest integer in" <math>x</math>.

Then

\begin{align}
I_1 \left( a, b, c \right) & = \int d\tau_{r} r_{1}^{a} r_{2}^{b} r_{12}^{c} e^{ -\alpha r_{1} - \beta r_{2}} P_{1}\left( \cos \theta \right) \\
& = \frac{1}{2} \left( I_{0} \left( a + 1, b - 1, c \right) + I_{0} \left( a - 1, b + 1, c \right) - I_{0} \left( a - 1, b - 1, c + 2 \right) \right)
\end{align}

===The Radial Recursion Relation===

Recall that the full integral is

\begin{equation}
I(l_1m_1,l_2m_2;R)=2\pi\delta_{l_1,l_2}\delta_{m_1,m_2}I_{l_2}(R)
\end{equation}

where

\begin{equation}
I_{l_2}(R) = \int d\tau_r R(r_1,r_2,r_{12})P_{l_2}(\cos \theta_{12})
\end{equation}

To obtain the recursion relation, use

\begin{equation}
P_l(x) = \frac{[P^{\prime}_{l+1}(x)-P^{\prime}_{l-1}(x)]}{2l+1}
\end{equation}

with

\begin{equation}
P_{l+1}^\prime (x) = \frac{d}{dx}P_{l+1}(x)
\end{equation}

Here <math>x=\cos\theta_{12}</math> and

\begin{align}
\frac{d}{d \cos \theta_{12}} & = \frac{dr_{12}}{d \cos \theta_{12}} \frac{d}{dr_{12}} \\
& = -\frac{r_{1} r_{2}}{r_{12}} \frac{d}{dr_{12}}
\end{align}

Then,

\begin{equation}
I_{l}\left( R \right) = - \int d\tau_r R \frac{r_{1} r_{2}}{r_{12}} \frac{d}{dr_{12}} \frac{ \left[ P_{l+1} \left( \cos \theta_{12} \right) - P_{l-1} \left( \cos \theta_{12} \right) \right]}{2l+1}
\end{equation}

The <math>r_{12}</math> part of the integral is

\begin{align}
& \int^{r_{1} + r_{2}}_{|r_{1} - r_{2}|} r_{12} dr_{12} R \frac{r_{1} r_{2}}{r_{12}} \frac{d}{dr_{12}} \left[P_{l+1}-P_{l-1} \right] \\
& = R r_{1} r_{2} \left[ P_{l+1} - P_{l-1} \right]^{r_{1} + r_{2}}_{|r_{1} - r_{2}|} - \int^{r_{1} + r_{2}}_{|r_{1} - r_{2}|} r_{12} dr_{12} \left( \frac{d}{dr_{12}} R \right) \frac{r_{1} r_{2}}{r_{12}} \frac{ \left[ P_{l+1} - P_{l-1} \right]}{2l+1}
\end{align}

The integrated term vanishes because

\begin{equation}
\cos \theta_{12} = \frac{r_{1}^{2} + r_{2}^{2} - r_{12}^{2}}{2 r_{1} r_{2}}
\begin{array}{lll}
& = -1 & {\rm when} & r_{12}^{2} = \left( r_{1} + r_{2} \right)^{2} \\
& = 1 & {\rm when} & r_{12}^{2} = \left( r_{1} - r_{2} \right)^{2}
\end{array}
\end{equation}

and <math>P_l(1)=1</math>, <math>P_{l}(-1)=(-1)^{l}</math>.

Thus,

\begin{equation}
I_{l+1}\left(\frac{r_1r_2}{r_{12}}R^\prime\right) = \left( 2l+1 \right) I_{l}(R) - I_{l-1} \left( \frac{r_{1} r_{2}}{r_{12}} R^{\prime} \right)
\end{equation}

If

\begin{equation}
R = r_{1}^{a-1} r_{2}^{b-1} r_{12}^{c+2} e^{-\alpha r_{1} - \beta r_{2}}
\end{equation}

then [G.W.F. Drake, Phys. Rev. A 18, 820 (1978)]

\begin{equation}
I_{l+1}(r_1^ar_2^br_{12}^c) =
\frac{2l+1}{c+2} I_{l} \left( r_{1}^{a-1} r_{2}^{b-2} r_{12}^{c+2} \right)+I_{l-1} \left( r_{1}^{a} r_{2}^{b} r_{12}^{c} \right)
\end{equation}

For the case <math>c=-2</math>, take

\begin{equation}
R = r_{1}^{a-1} r_{2}^{b-1} \ln r_{12} e^{ -\alpha r_{1} -\beta r_{2}}
\end{equation}

Then

\begin{equation}
I_{l+1} \left( r_{1}^{a} r_{2}^{b} r_{12}^{c} \right) = I_{l} \left( r_{1}^{a-1} r_{2}^{b-1} \ln r_{12} \right) \left( 2l + 1 \right) - I_{l-1} \left( r_{1}^{a} r_{2}^{b} r_{12}^{-2} \right)
\end{equation}

This allows all <math>I_{l}</math> integrals to be calculated from tables of <math>I_0</math> and
<math>I_1</math> integrals.

===The General Integral===

The above results for the angular and radial integrals can now be combined into a general formula for integrals of the type

\begin{equation}
I = \int \int d{\bf r}_{1} d{\bf r}_{2} R_{1} \mathcal{Y}^{M^{\prime}}_{l^{\prime}_{1} l^{\prime}_{2} L^{\prime}} \left( \hat{r}_{1}, \hat{r}_{2} \right) T_{k_{1} k_{2} K}^{Q} \left( {\bf r}_{1},{\bf r}_{2} \right) R_{2} \mathcal{Y}_{l_{1} l_{2} L}^{M} \left( \hat{r}_{1}, \hat{r}_{2} \right)
\end{equation}

where

\begin{equation}
\mathcal{Y}^{M}_{l_{1} l_{2} L} \left( \hat{r}_{1}, \hat{r}_{2} \right) = \sum_{m_1,m_2} \langle l_{1} l_{2} m_{1} m_{2} | LM \rangle Y^{m_{1}}_{l_{1}} \left( \hat{r}_{1} \right) Y^{m_{2}}_{l_{2}} \left( \hat{r}_{2} \right)
\end{equation}

and

\begin{equation}
T^Q_{k_{1} k_{2} K} \left( {\bf r}_{1}, {\bf r}_{2} \right) = \sum_{q_{1}, q_{2}} \langle k_{1} k_{2} q_{1} q_{2} | KQ \rangle Y_{k_{1}}^{q_{1}} \left( \hat{r}_{1} \right) Y^{q_{2}}_{k_{2}} \left( \hat{r}_{2} \right)
\end{equation}

The basic idea is to make repeated use of the formula

\begin{equation}
Y^{m_{1}}_{l_{1}} \left( \hat{r}_{1} \right) Y^{m_{2}}_{l_{2}} \left( \hat{r}_{1} \right) = \sum_{l m} \left( \frac{\left( 2l_{1}+1 \right) \left(2l_{2}+1 \right) \left(2l+1 \right)}{4 \pi} \right)^{1/2}
\times \left( \begin{array}{ccc}
l_1 & l_2 & l \\
m_1 & m_2 & m
\end{array} \right) \left(\begin{array}{ccc}
l_1 & l_2 & l \\
0 & 0 & 0
\end{array} \right)
Y^{m*}_{l} \left( \hat{r}_{1} \right)
\end{equation}

where

\begin{equation}
Y^{m*}(\hat{r}) = (-1)^mY_l^{-m}(\hat{r})
\end{equation}

and

\begin{equation}
\left(\begin{array}{ccc}
l_1 & l_2 & l \\
m_1 & m_2 & m
\end{array} \right) = \frac{\left( -1 \right)^{l_{1} - l_{2} - m}}{\left( 2l + 1 \right)^{1/2}} \left( l_{1} l_{2} m_{1} m_{2} | l, -m \right)
\end{equation}

is a 3-<math>j</math> symbol. In particular, write

\begin{array}{cc}
Y^{m_1^\prime}_{l_1^\prime}(\hat{r}_1)^{*} \underbrace{Y_{k_{1}}^{q_{1}} \left( \hat{r}_{1} \right) Y^{m_{1}}_{l_{1}} \left( \hat{r}_{1} \right)} & = & \sum_{\Lambda M}(\ldots)Y^{M}_\Lambda(\hat{r}_1)\\
\ \ \ \ \ \ \ \sum_{\lambda_1\mu_1}Y_{\lambda_1}^{\mu_1}(\hat{r}_1) & & \\
& & \\
Y^{m_{2}^{\prime}}_{l_{2}^{\prime}} \left( \hat{r}_{2} \right) \underbrace{Y^{q_{2}}_{k_{2}} \left( \hat{r}_{2} \right) Y_{l_{2}}^{m_{2}} \left( \hat{r}_{2} \right)} & = & \sum_{\Lambda^{\prime} M^{\prime}} \left( \ldots \right)Y_{\Lambda^{\prime}}^{M^{\prime}*} \left( \hat{r}_{2} \right) \\
\ \ \ \ \ \ \ \ \ \sum_{\lambda_{2} \mu_{2}}Y_{\lambda_{2}}^{\mu_{2}} \left( \hat{r}_{2} \right) & &
\end{array}

The angular integral then gives a factor of <math>2 \pi \delta_{\Lambda, \Lambda^{\prime}} \delta_{M, M^{\prime}} P_{\Lambda} \left( \cos \theta_{12} \right)</math>. The total integral therefore reduces to the form

\begin{equation}
I = \sum_{\Lambda} C_{\Lambda} I_{\Lambda} \left( R_{1} R_{2} \right)
\end{equation}

where <math>C_{\Lambda} = \sum_{\lambda_{1}, \lambda_{2}} C_{ \lambda_{1}, \lambda_{2}, \Lambda}</math>. For further details and derivations, including graphical representations, see G.W.F. Drake, Phys. Rev. A <math>{\bf 18}</math>, 820 (1978).

==Graphical Representation==

[[File:GraphRep1.jpg|500px]]

==Matrix Elements of H==
Recall that

\begin{equation}
H = -\frac{1}{2} \nabla^{2}_{1} - \frac{1}{2} \nabla^{2}_{2} - \frac{1}{r_1} - \frac{1}{r_2} + \frac{Z^{-1}}{r_{12}}
\end{equation}

Consider matrix elements of

\begin{equation}
\nabla^2_1 = \frac{1}{r_1^2} \frac{\partial}{\partial r_1}\left(r_1^2\frac{\partial}{\partial r_1} \right) + \frac{1}{r^2}\frac{\partial}{\partial r}\left(r^2\frac{\partial}{\partial r}\right) - \frac{(\vec{l}_1^Y)^2}{r_1^2} + \frac{2 \left( r_1 - r_2 \cos \theta \right)}{r} \frac{\partial^2}{\partial r_1 \partial r} - 2 \left( \nabla^Y_1 \cdot {\bf r}_2 \right) \frac{1}{r} \frac{\partial}{\partial r}
\end{equation}

where <math>r \equiv r_{12}</math> and <math>\cos\theta \equiv \cos\theta_{12}</math>. Also in what follows, define <math>\hat{\nabla}^Y_1 \equiv r_1\nabla^Y_1</math> where <math>\nabla^Y</math> operates only on the spherical harmonic part of the wave function, and similarly for <math>\vec{l}_1^Y</math>.

A general matrix element is

\begin{equation}
\langle r_1^{a^\prime} r_2^{b^\prime} r_{12}^{c^\prime} e^{-\alpha^\prime r_1 - \beta^\prime r_2} \mathcal{Y}^{M^\prime}_{l^\prime_1 l^\prime_2 L^\prime} \left( \hat{r}_1, \hat{r}_2 \right) \left| \nabla^2_1 \right| r_1^a r_2^b r_{12}^c e^{-\alpha r_1 - \beta r_2} \mathcal{Y}^M_{l_1 l_2 L} \left( \hat{r}_1, \hat{r}_2 \right) \rangle
\end{equation}

Since <math>\nabla^2_1</math> is rotationally invariant, this vanishes unless <math>L=L^\prime</math>, <math>M=M^\prime</math>. Also <math>\nabla_1^2</math> is Hermitian, so that the result must be the same whether it operates to the right or left, even though the results look very different. In fact, requiring the results to be the same yields some interesting and useful integral identities as follows:

Put

\begin{equation}
\mathcal{F} = F \mathcal{Y}^M_{l_1 l_2 L} \left( \hat{r}_1, \hat{r_2} \right)
\end{equation}

and

\begin{equation}
F = r_1^a r_2^b r_{12}^c e^{-\alpha r_1 - \beta r_2}
\end{equation}

Then

\begin{equation}
\nabla^2_1 \mathcal{F} = \left\{ \frac{1}{r_1^2} \left[ a \left( a + 1 \right) - l_1 \left( l_1 + 1 \right) \right] + \frac{c \left( c + 1 \right)}{r^2} + \alpha^2\frac{2\alpha \left( a + 1 \right)}{r_1} + \frac{2 \left(r_1 - r_2 \cos \theta \right) }{r_1r^2}c \left[ a - \alpha r_1 \right] - \frac{2c}{r^2} \left( \hat{\nabla}^Y_1 \cdot \hat{r}_2 \right) \frac{r_2}{r_1} \right\} \mathcal{F}
\end{equation}

and

\begin{align}
\langle \mathcal{F}^\prime \left| \nabla^2_1 \right| \mathcal{F} \rangle & = \sum_\Lambda \int d\tau_r F^\prime C_\Lambda \left( 1 \right) P_\Lambda \left( \cos \theta \right) \times \left\{ \frac{1}{r_1^2} \left[ a \left( a + 1 \right) - l_1 \left( l_1 + 1 \right) \right] -\frac{2 \alpha \left( a + 1 \right)}{r_1} + \frac{c \left( c + 1 \right)}{r^2} +\alpha^2 + \frac{2 \left( r_2 - r_2 \cos \theta \right)}{r_1r^2}c \left( a - \alpha r_1 \right) \right\} F \\
& + \sum_\Lambda \int d\tau_r F^\prime C_\Lambda \left( \hat{\nabla}^Y_1 \cdot \hat{r}_2 \right) P_\Lambda \left( \cos \theta \right) \left( \frac{-2c r_2}{r_1 r^2} \right) F
\end{align}

where

\begin{equation}
\int d\tau_r = \int^\infty_0 r_1 dr_1 \int^\infty_0 r_2 dr_2 \int^{r_1+r_2}_{\left| r_1 - r_2 \right|} r dr
\end{equation}

For brevity, let the sum over <math>\Lambda</math> and the radial integrations be understood, and let <math>\left( \nabla_1^2 \right)</math> stand for the terms which appear in the integrand. Then operating to the right gives

\begin{equation}
(\nabla_1^2)_R = \frac{1}{r_1^2} \left[ a \left( a + 1 \right) - l_1 \left( l_1 + 1 \right) \right] + \frac{c \left( c + 1 \right)}{r^2} + \alpha^2 - \frac{2 \alpha \left( a + 1 \right)}{r_1}
+ \frac{2 \left( r_1 - r_2 \cos \theta \right)}{r_1 r^2}c \left( a - \alpha r_1 \right) - \frac{2c r_2}{r_1 r^2} \left( \hat{\nabla}^Y_1 \cdot \hat{r}_2 \right)
\end{equation}

Operating to the left gives

\begin{equation}
(\nabla_1^2)_L = \frac{1}{r_1^2} \left[a^\prime \left( a^\prime + 1 \right) - l_1^\prime \left( l_1^\prime + 1 \right) \right] + \frac{c^\prime \left( c^\prime + 1 \right)}{r^2} + \alpha^{\prime 2} -\frac{2 \alpha^\prime \left( a^\prime + 1 \right)}{r_1}
+ \frac{2 \left( r_1 - r_2 \cos \theta \right)}{r_1 r^2} c^\prime \left( a^\prime - \alpha^\prime r_1 \right) - \frac{2 c^\prime r_2}{r_1 r^2} \left( \hat{\nabla}^{Y^\prime}_1 \cdot \hat{r}_2 \right)
\end{equation}

Now put

\begin{array}{cc}
a_+ = a+a^\prime, & \hat{\nabla}_1^+ = \hat{\nabla}_1^Y + \hat{\nabla}_1^{Y\prime} \\
a_{-} = a-a^\prime, & \hat{\nabla}_1^{-} = \hat{\nabla}_1^Y - \hat{\nabla}_1^{Y\prime}
\end{array}

etc., and substitute <math>a^\prime = a_+ - a</math>, <math>c^\prime = c_+ - c</math>, and <math>\alpha^\prime = \alpha_+ - \alpha</math> in <math>(\nabla^2_1)_L</math>. If <math>a_+</math>, <math>c_+</math> and <math>\alpha_+</math> are held fixed, then the equation

\begin{equation}
(\nabla^2_1)_R = (\nabla^2_1)_L
\end{equation}

must be true for arbitrary <math>a</math>, <math>c</math>, and <math>\alpha</math>. Their coefficients must thus vanish.

This yields the integral relations

\begin{equation}
\label{I}
\frac{\left( r_1 - r_2 \cos \theta \right)}{r_1 r^2} = \frac{1}{c_+} \left( \frac{- \left( a_+ + 1 \right)}{r_1^2} + \frac{\alpha_+}{r_1} \right) \tag{I}
\end{equation}

from the coefficient of <math>a</math> and

\begin{equation}
\label{II}
\frac{ \left( r_1 - r_2 \cos \theta \right) \left( a_+ - \alpha_+ r_1 \right)}{r_1 r^2} = \frac{r_2}{r_1 r^2}\left( \hat{r}_2 \cdot \hat{\nabla}_1^+ \right) - \frac{ \left( c_+ + 1 \right)}{r^2} \tag{II}
\end{equation}

from the coefficient of <math>c</math>. The coefficient of <math>\alpha</math> gives an equation equivalent to (I).

Furthermore, if can be show that (see problem)

\begin{align}
& \sum_\Lambda \int d\tau_r \frac{r^c}{r^2} C_\Lambda \left( \hat{r}_2 \cdot \hat{\nabla}_1^Y \right) P_\Lambda \left( \cos \theta \right) \\
= & \sum_\Lambda \int d\tau_r \frac{r^c}{c r_1 r_2}C_\Lambda \left( 1 \right)P_\Lambda \left( \cos \theta \right) \times \left( \frac{l_1^\prime \left( l_1^\prime + 1 \right) - l_1 \left( l_1 + 1 \right) -\Lambda \left( \Lambda + 1 \right)}{2} \right)
\end{align}

and similarly for <math>\left( \hat{r}_2 \cdot \hat{\nabla}^{Y\prime}_1 \right)</math> with <math>l_1</math> and <math>l_1^\prime</math> interchangeable, then it follows that

\begin{equation}
\frac{r^{c_+}}{r^2} \left( \hat{r}_2 \cdot \hat{\nabla}_1^+ \right) = - \frac{r^{c_+}}{c_+ r_1 r_2} \Lambda \left( \Lambda + 1 \right)
\end{equation}

and

\begin{equation}
\frac{r^{c_+}}{r^2} \left( \hat{r}_2 \cdot \hat{\nabla}_1^- \right) = \frac{r^{c_+}}{c_+ r_1 r_2} \left[ l_1^\prime \left( l_1^\prime + 1 \right) - l_1 \left( l_1 + 1 \right) \right]
\end{equation}

where equality applies after integration and summation over <math>\Lambda</math>.

Thus (II) becomes

\begin{equation}\label{IIa}
\frac{ \left( r_1 - r_2 \cos \theta \right) \left( a_+ - \alpha_+ r_1 \right)}{r_1 r^2} = - \frac{\Lambda \left( \Lambda + 1 \right)}{c_+ r^2_1} - \frac{ \left( c_+ + 1 \right)}{r^2} \tag{IIa}
\end{equation}

===Problem===

Prove the integral relation

\begin{align}
& \sum_\Lambda \int d\tau_r f \left( r_1, r_2 \right) \frac{1}{r} \left( \frac{d}{dr} q(r) \right) C_\Lambda \left( \hat{r}_2 \cdot \hat{\nabla}^Y_1 \right) P_\Lambda \left( \cos \theta \right) \\
= & \sum_\Lambda \int d\tau_r \frac{ f \left( r_1, r_2 \right)}{r_1, r_2} q(r) C_\Lambda \left( 1 \right) P_\Lambda \left( \cos \theta \right) \left(\frac{l_1^\prime \left( l_1^\prime + 1 \right) - l_1 \left( l_1 + 1 \right) - \Lambda \left( \Lambda + 1 \right)}{2}\right)
\end{align}

where the coefficients <math>C_\Lambda \left( 1 \right)</math> are the angular coefficients from the overlap integral

\begin{align}
\int & d\Omega\, \mathcal{Y}^{M*}_{l_1^\prime l_2^\prime L}(\hat{r}_1,\hat{r}_2)\, \mathcal{Y}^{M}_{l_1 l_2 L}(\hat{r}_1,\hat{r}_2)\\
= & \sum_\Lambda C_\Lambda(1)P_\Lambda(\cos\theta_{12}).
\end{align}

Hint: Use the fact that <math>l_1^2</math> is Hermitian so that
<math>
\int d\tau (l_1^2\mathcal{Y}^\prime)^*q(r)\mathcal{Y} = \int d\tau
\mathcal{Y}^{\prime *}l_1^2(q(r)\mathcal{Y})
</math>
with <math>\ \vec{l}_1= \frac{1}{i}\vec{r}_1\times\nabla_1</math>.

It is also useful to use
<math>\begin{eqnarray}\ \
\cos\theta\, P_L(\cos\theta) &=& \frac{1}{2L+1}[L\,P_{L-1}(\cos\theta) + (L+1)P_{L+1}(\cos\theta)]
\\
(\cos^2\theta-1)P_L(\cos\theta) &=&
\frac{L(L-1)}{(2L-1)(2L+1)}P_{L-2}(\cos\theta)\nonumber
-\frac{2(L^2+L-1)}{(2L-1)(2L+3)}P_L(\cos\theta) + \frac{(L+1)(L+2)}{(2L+1)(2L+3)}P_{L+2}(\cos\theta)\nonumber\\
&=& \frac{L(L-1)}{(2L-1)(2L+1)}[P_{L-2}(\cos\theta) - P_L(\cos\theta)]
+ \frac{(L+1)(L+2)}{(2L+1)(2L+3)}[P_{L+2}(\cos\theta) - P_L(\cos\theta)]
\end{eqnarray}</math>

together with a double application of the integral recursion relation
<math>
I_{L+1}\left(\frac{1}{r}\frac{d}{dr}q(r)\right) =
(2L+1)I_L\left(\frac{1}{r_1r_2}q(r)\right)+I_{L-1}\left(\frac{1}{r}\frac{d}{dr}q(r)\right)
</math>

Of course <math>l_1^2\mathcal{Y}^M_{l_1l_2L}(\hat{r}_1,\hat{r}_2)=l_1(l_1+1)\mathcal{Y}^M_{l_1l_2L}(\hat{r}_1,\hat{r}_2)</math>. Begin by expanding
<math>
l_1^2(g\mathcal{Y}) = \mathcal{Y}l_1^2g + g\,l_1^2\mathcal{Y} + 2(\vec{l}_1g)\cdot(\vec{l}_1\mathcal{Y})
</math>, and show that

<math>
\begin{eqnarray}\ \
(\vec{l}_1g)\cdot(\vec{l}_1\mathcal{Y}) & = & \frac{r_1^2}{r}\frac{dg}{dr}\vec{r}_2\cdot \nabla_1\mathcal{Y}\,\,\,\,\mbox{ and}\\
l_1^2g(r) &=& 2r_1r_2\cos\theta\frac{1}{r}\frac{dg}{dr} + r_1^2r_2^2(\cos^2\theta - 1)
\frac{1}{r}\frac{d}{dr}\left(\frac{1}{r}\frac{dg}{dr}\right).
\end{eqnarray}
</math>

The proof amounts to showing that the term <math>l_1^2g(r)</math> can be replaced by
<math>\Lambda(\Lambda+1)g(r)</math> after multiplying by <math>P_\Lambda(\cos\theta)</math>
and integrating by parts with respect to the radial integrations over <math>r_1</math>, <math>r_2</math> and
<math>r \equiv |\vec{r}_1 - \vec{r}_2|</math>. Remember that <math>\cos\theta</math> is just a short-hand notation for the radial function <math>(r_1^2 + r_2^2 - r^2)/(2r_1r_2)</math>.

==General Hermitian Property==

Each combination of terms of the form
\begin{equation}
<f>_R = a^2f_1+af_2+abf_3+b^2f_4+bf_5+af_6\nabla_1^Y+bf_7\nabla^Y_2+f_8(Y)
\end{equation}
can be rewritten
\begin{equation}
<f>_L= (a_+-a)^2f_1+(a_+-a)f_2 + (a_+-a)(b_+-b)f_3+(b_+-b)^2f_4
+(b_+-b)f_5+(a_+-a)f_6\nabla^{Y\prime}_1+(b_+-b)f_7\nabla^{Y\prime}_2+f_8)Y^\prime
\end{equation}
where as usual <math>\nabla_1^Y</math> acts only on the spherical harmonic part of the wave function denoted for short by <math>Y</math>.
Since these must be equal for arbitrary <math>a</math> and <math>b</math>,
\begin{equation}
a_+^2f_1+a_+f_2+_+b_+f_3+b_+f_4+b_+f_5+a_+f_6\nabla^{Y\prime}_1 +
b_+f_7\nabla^{Y\prime}_2+f_8(Y^\prime)-f_8(Y)=0
\end{equation}

Adding the corresponding expression with <math>Y</math> and <math>Y^\prime</math> interchanged
yields
<math>\label{III}
a_+^2f_1+a_+f_2+a_+b_+f_3+b_+^2f_4+b_+f_5+\frac{1}{2}a_+f_6\nabla^+_1+\frac{1}{2}b_+f_7\nabla^+_2=0\tag{III}
</math>

Subtracting gives
\[\label{IV}
f_8(Y)-f_8(Y^\prime)=-\frac{1}{2}[a_+f_6\nabla^-_1+b_+f_7\nabla_2^-]\tag{IV}
\]
\[\label{V}
a[-2a_+f_1-2f_2-b_+f_3-f_6\nabla^+_1]=0\tag{V}
\]
\[\label{VI}
b[-2b_+f_4-2f_5-a_+f_3-f_7\nabla^+_2]=0\tag{VI}
\]

Adding the two forms gives

\begin{eqnarray}
<f>_R+<f>_L=\frac{1}{2}(a_+^2+a_-^2)f_1+a_+f_2+\frac{1}{2}(a_+b_++a_-b_-)f_3\nonumber\\
\ +\frac{1}{2}(b_+^2+b_-^2)f_4+b_+f_5+\frac{1}{2}f_6(a_+\nabla_1^++a_-\nabla^-_1)\nonumber\\
\ +\frac{1}{2}(b_+\nabla_2^++b_-\nabla^-_2)+f_8(Y)+f_8(Y^\prime)\nonumber
\end{eqnarray}

Subtracting <math>\frac{x}{2}\times({\rm I})</math> gives

\begin{eqnarray}\ \
<f>_R+<f>_L = \frac{1}{2}[(1-x)a_+^2 + a_-^2]f_1 +
(1-\frac{x}{2})a_+f_2+\frac{1}{2}[(1-x)a_+b_++a_-b_-]f_3\nonumber\\
+\frac{1}{2}[(1-x)b_+^2+b_-^2]f_4+(1-\frac{x}{2})f_5b_++\frac{1}{2}f_6[(1-\frac{x}{2})a_+\nabla_1^++a_-\nabla^-_1]\nonumber\\
+\frac{1}{2}[(1-\frac{x}{2})b_+\nabla^+_2+b_-\nabla^-_2]+f_8(Y)+f_8(Y^\prime)\nonumber
\end{eqnarray}

If <math>x=1</math>,
\begin{eqnarray}
<f>_R+<f>_L &=&
\frac{1}{2}[a_-^2f_1+a_+f_2+a_-b_-f_3+b_-^2f_4+b_+f_5\nonumber\\
&+&\left(\frac{1}{2}a_+\nabla^+_1+a_-\nabla^-_1\right)f_6+\left(\frac{1}{2}b_+\nabla^+_2+b_-\nabla^-_2\right)f_7]+f_8(Y)+f_8(Y^\prime)\nonumber
\end{eqnarray}

The General Hermitian Property for arbitrary <math>x</math> gives
\begin{eqnarray}\ \
\nabla_1^2=
\frac{1}{4}\{\frac{1}{r_1^2}[(1-x)a_+^2+a_-^2+2(1-\frac{x}{2})a_+-2[l_1(l_1+1)+l_1^\prime(l_1^\prime+1)]\nonumber\\
-\frac{2}{r_1}[(1-x)\alpha_+a_++\alpha_-a_-+2(1-\frac{x}{2})\alpha_+]+(1-x)\alpha_+^2
+ \alpha^2_-]\nonumber\\
+\frac{2(r_1-r_2\cos\theta)}{r_1r^2}[(1-x)(a_+-\alpha_+r_1)c_++(a_--\alpha_-r_2)c_-]\nonumber\\
-\frac{2r_2}{r_1r^2}[(1-\frac{x}{2})c_+\hat{r}_2\cdot\hat{\nabla}_1^++c_-\hat{r}_2\hat{\nabla}_1^-]+\frac{(1-x)c_+^2+c_-^2+2(1-\frac{x}{2})c_+}{r^2}\}\nonumber
\end{eqnarray}.

Use
<math>\begin{eqnarray}\ \
\frac{-2r_2}{r_1r^2}[(1-\frac{x}{2})c_+\hat{r}_2\hat{\nabla}_1^++c_-\hat{r}_2\hat{\nabla}_1^-]=\frac{2}{r_1^2}\left[(1-\frac{x}{2})\Lambda(\Lambda+1)-\frac{c_-}{c_+}[l_1^\prime(l_1^\prime+1)+l_1(l_1+1)]\right]\end{eqnarray}</math>,

<math>\begin{eqnarray}
\frac{2(r_1-r_2\cos\theta)}{r_1r^2}(a_--\alpha_-r_1)c_- = 2\frac{c_-}{c_+}\left[\frac{-1}{r_1^2}[a_-(a_++1)]+\frac{1}{r_1}[a_-\alpha_++\alpha_-(a_++2)]-\alpha_-\alpha_+\right]\nonumber
\end{eqnarray}</math>,

and
<math>\begin{eqnarray}\ \
\frac{2(r_1-r_2\cos\theta)}{r_1r^2}(a_+-\alpha_+r_1)c_+ = 2\left[-\frac{1}{r_1^2}[a_+(a_++1)]+\frac{1}{r_1}[a_+\alpha_++\alpha_+(a_++2)]-\alpha_+^2\right]
\end{eqnarray}</math>.

Substituting into <math>\nabla_1^2</math> gives

\begin{eqnarray}
\nabla^2_1 = \frac{1}{4}\{\frac{1}{r_1^2}[-(1-x)a_+^2 + a_-^2 +xa_+ +
2(1-\frac{x}{2})\Lambda(\Lambda+1)\nonumber\\
-2l_1(l_1+1)(1-\frac{c_-}{c_+}) -
2l_1^\prime(l_1^\prime+1)(1+\frac{c_-}{c_+})-\frac{2c_-a_-}{c_+}(a_++1)]\nonumber\\
- \frac{2}{r_1}\left[-(1-x)\alpha_+(a_++2)+\alpha_-a_-+2(1-\frac{x}{2})\alpha_+
-\frac{c_-}{c_+}[a_-\alpha_++\alpha_-(a_++2)]\right]\nonumber\\
-(1-x)\alpha_+^2+\alpha_-^2-2\frac{c_-}{c_+}\alpha_-\alpha_+ +\frac{1}{r^2}\left[(1-x)c_+^2+c_-^2+2(1-\frac{x}{2})c_+\right]\}\nonumber
\end{eqnarray}

This has the form
<math>
\nabla_1^2 =
\frac{1}{4}\left[\frac{A_1}{r_1^2}+\frac{B_1}{r_1}+\frac{C_1}{r^2}+D_1\right]
</math>
with

<math>\begin{array}{ccc}
A_1=-(1-x)a_+^2+a_-^2
+xa_++2(1-\frac{x}{2})\Lambda(\Lambda+1)-2l_1(l_1+1)(1-\frac{c_-}{c_+})\
-2l_1^\prime(l_1^\prime+1)(1+\frac{c_-}{c_+})-\frac{2c_-a_-}{c_+}(a_++1) \ \
\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \
\end{array}
</math>

<math>
B_1 =
2\left[(1-x)\alpha_+(a_++2)-\alpha_-a_--2(1-\frac{x}{2})\alpha_++\frac{c_-}{c_+}[a_-\alpha_++\alpha_-(a_++2)]\right]
</math>

<math>
C_1 = (1-x)c_+^2+ c_-^2 +2(1-\frac{x}{2})c_+
</math>

<math>
D_1 = -1(1-x)\alpha_+^2 +\alpha_-^2 - 2\frac{c_-}{c_+}\alpha_-\alpha_+
</math>

The complete Hamiltonian is then (in Z-scaled a.u.)

<math>\begin{eqnarray}
H =-\frac{1}{2}\nabla^2_1 -\frac{1}{2}\nabla^2_2 -\frac{1}{r_1} -\frac{1}{r_2} +\frac{Z^{-1}}{r}\nonumber\\
= -\frac{1}{2}\nabla^2_1 -\frac{1}{2}\nabla^2_2 -\frac{1}{r_1} -\frac{Z-1}{Zr_2}+Z^{-1}\left(\frac{1}{r}-\frac{1}{r_2}\right)\nonumber\\
\ =-\frac{1}{8}\left[\frac{A_1}{r_1^2}+\frac{B_1+8}{r_1}+\frac{C_1}{r^2}+D_1+D_2 +\frac{A_2}{r_2^2} + \frac{B_2 + 8(Z-1)/Z}{r_2}+\frac{C_2}{r^2}\right]\nonumber
+ Z^{-1}\left(\frac{1}{r}-\frac{1}{r_2}\right)\nonumber
\end{eqnarray}</math>

The screened hydrogenic energy is
<math>\begin{eqnarray}
E_{SH}=-\frac{1}{2}\left[\frac{1}{n_1^2}+ \left(\frac{Z-1}{Z}\right)^2\frac{1}{n_2^2}\right]
\end{eqnarray}</math>
so that

<math>\begin{eqnarray}
H-E_{SH} &=& -\frac{1}{8}[
\frac{A_1}{r_1^2}+\frac{B_1+8}{r_1}+\frac{C_1}{r^2} + D_1 + D_2
-\frac{4}{n_1^2} -\left(\frac{Z-1}{Z}\right)^2\frac{4}{n_2^2}\nonumber
+\frac{A_2}{r_2}+\frac{B_2+8(Z-1)/Z}{r_2}+\frac{C_2}{r^2}] +
Z^{-1}\left(\frac{1}{r} -\frac{1}{r_2}\right)\nonumber
\end{eqnarray}</math>

==Optimization of Nonlinear Parameters==
- The traditional method of performing Hylleraas calculations is to write the
basis set in the form

<math>
\Psi = \sum_{i,j,k}c_{i,j,k}\varphi_{i,j,k}(\alpha,\beta)\pm exchange
</math>

with

<math>
\varphi_{i,j,k}(\alpha,\beta) = r_1^ir_2^jr_{12}^ke^{-\alpha r_1 - \beta
r_2}\mathcal{Y}^M_{l_1l_2L}.
</math>

The usual procedure is to set <math>\alpha = Z</math> so that it represents the inner
<math>1s</math> electron, and then to vary <math>\beta</math> so as to minimize the energy.

[[File:OptNon.jpg|500px]]

Since <math>\beta</math> appears in <math>\Psi</math> as a non-linear parameter, the entire
calculation must be repeated for each value of <math>\beta</math>. However, the minimum
becomes progressively smaller as the basis set is enlarged.

[[File:OptNon02.jpg|500px]]

*Difficulties

1. If the basis set is constructed so that <math> i +j +j \leq N</math>, the the
number of terms is
<math>
(N+1)(N+2)(N+3)/6
</math>
N=14 already gives 680 terms and an accuracy of about 1 part in <math>10^{10}</math> for
low-lying states. A substantial improvement in accuracy would require much
larger basis sets, together with multiple precision arithmetic to avoid loss
of significant figures when high powers are included.

2. The accuracy rapidly deteriorates as one goes to more highly excited
states - about 1 significant figure is lost each time the principle quantum
number is increased.

*Cure

We have found that writing basis sets in the form
\begin{eqnarray}
\Psi &=& \sum_{i,j,k}\left(c^{(1)}_{i,j,k}\varphi_{ijk}(\alpha_1\beta_1) +
c_{ijk}^{(2)}\varphi_{ijk}(\alpha_2\beta_2)\right) \pm exchange \nonumber\\
&=& \psi({\bf r}_1,{\bf r}_2) \pm \psi({\bf r}_2,{\bf r}_1)\nonumber
\end{eqnarray}
so that each combination of powers is included twice with different nonlinear
parameters gives a dramatic improvement in accuracy for basis sets of about
the same total size.

However, the optimization of the nonlinear parameters is now much more
difficult, and an automated procedure is needed.

If
<math>
E = \frac{<\Psi|H|\Psi>}{<\Psi|\Psi>}</math> and we assume <math><\Psi|\Psi> = 1</math>, then

<math>
\frac{\partial E}{\partial\alpha_t} = -2 <\Psi|H-E|r_1\psi({\bf r}_1,{\bf
r}_2;\alpha_t)\pm r_21\psi({\bf r}_2,{\bf
r}_1;\alpha_t)>
</math>
where
<math>
\psi({\bf r}_1,{\bf r}_2;\alpha_t) =
\sum_{i,j,k}c_{ijk}^{(t)}\varphi_{ijk}(\alpha_t\beta_t)
</math>

*Problem

1. Prove the above.

2. Prove that there is no contribution from the implicit dependence of
<math>c_{ijk}^{(t)}</math> on <math>\alpha_t</math> if the linear parameters have been optimized.

==The Screened Hydrogenic Term==

If the Hamiltonian is written in the form (in Z-scaled a.u.)
<math>\begin{eqnarray}\
H = H_0({\bf r}_1,Z) + H_0({\bf r}_2,Z-1) +
Z^{-1}\left(\frac{1}{r}-\frac{1}{r_2}\right)
\end{eqnarray}</math>
with
<math>\
H_0({\bf r}_1,Z) = -\frac{1}{2}\nabla_1^2 - \frac{1}{r_1}\nonumber
</math> and
<math>
H_0({\bf r}_2,Z-1) = -\frac{1}{2}\nabla_2^2 - \frac{Z-1}Z{r_2}\nonumber
</math>,
then the eigenvectors of <math>H_0({\bf r},Z) +H_0({\bf r}_2,Z-1)</math> are products of
hydrogenic orbitals
<math>
\Psi_0=\psi_0(1s,Z)\psi_0(nl,Z-1)
</math>
and the eigenvalue is
<math>\begin{eqnarray}
E_{SH} = \left[-\frac{1}{2} - \left(\frac{Z-1}{Z}\right)^2\frac{1}{2n^2}\right]Z^2 a.u.
\end{eqnarray}</math>
called the screened hydrogenic eigenvalue. From highly excited states,
<math>E_{SH}</math> and <math>\Psi_0</math>are already excellent approximations.

For example, for the 1s8d states, the energies are

<math>\begin{eqnarray}\\
E(1s8d^1D)&=& -2.007\,816\,512\,563\,81 {\rm\ a.u.}\nonumber\\
E(1s8d^3D)&=& - 2.007\,817\,934\,711\,71 {\rm\ a.u.}\nonumber\\
E_{SH} &=& -2.007\,812\,5 {\rm\ a.u.} \nonumber
\end{eqnarray}</math>

It is therefore advantageous to include the screened hydrogenic terms in the
basis set so that the complete trial function becomes
<math>
\Psi = c_0\Psi_0 +
\sum_{ijk}\left[c_{ijk}^{(1)}\varphi_{ijk}(\alpha_1,\beta_1)+c_{ijk}^{(2)}\varphi(\alpha_2,\beta_2)\right]\pm
{\rm exchange}
</math>

Also, the variational principal can be re-expressed in the form
<math>\begin{eqnarray}
E= E_{SH}+\frac{<\Psi|H-E_{SH}|\Psi>}{<\Psi|\Psi>}
\end{eqnarray}</math>
so that the <math>E_{SH}</math> term can be cancelled analytically from the matrix
elements.

For example
<math>
<\Psi_0|H-E_{SH}|\Psi_0> = 0
</math>

and
<math>
<\varphi_{ijk}|H-E_{SH}|\Psi_0> =
Z^{-1}<\varphi_{ijk}|\frac{1}{r}-\frac{1}{r_2}|\Psi_0>
</math>

Recall that
<math>\begin{eqnarray}\ \
I_0(a,b,c) = \frac{2}{c+2}\sum^{[c+1]/2}_{i=0}\left(\begin{array}{c}
c+2\\
2i+1
\end{array}\right) \left[f(p,q;\beta) + f(p^\prime,q^\prime;\alpha)\right]
\end{eqnarray}</math>
where

<math>\begin{eqnarray}
f(p,q;x) =
\frac{q!}{x^{q+1}(\alpha+\beta)^{p+1}}\sum^p_{j=0}\frac{(p+j)!}{j!}\left(\frac{x}{\alpha+\beta}\right)^j
\end{eqnarray}</math>

<math>
\begin{array}{cc}
p=a+2i+2 & p^\prime = b+2i+2\\
q= b+c-2i+2 & q^\prime =a+c-2i+2
\end{array}
</math>

If <math>\beta\ll\alpha</math>, the <math>f(p,q;\beta)\gg f(p^\prime,q^\prime;\alpha)</math> and the
<math>i=0</math> term is the dominant contribution to <math>f(p,q;\beta)</math>. However, since
this term depends only on the sum of powers <math>b+c</math> for <math>r_2</math>, it cancels exactly
from the matrix element of
<math>
\frac{1}{r_{12}} - \frac{1}{r_2}
</math>
and can therefore
be omitted in the calculations of integrands, thereby saving many significant
figures. This is especially valuable when <math>b+c</math> is large and <math>a</math> is small.

==Small Corrections==
*Mass Polarization
If the nuclear mass is not taken to be infinite, then the Hamiltonian is
<math>\begin{eqnarray}\ \
H = \frac{P_N^2}{2M} + \sum_{i=1}\frac{p_i^2}{2m} + V
\end{eqnarray}</math>

[[File:SmallCor.jpg|500px]]

where <math>M =M_A-nm</math> is the nuclear mass and <math>m</math> is the electron mass.

Change variables to the C of M
<math>\begin{eqnarray}
{\bf R} = \frac{1}{M+nm}\left[M{\bf r}_N + m({\bf r}_1 + {\bf r}_2 +
\ldots)\right]
\end{eqnarray}</math>

and relative variables

<math>
{\bf s}_i = {\bf r}_i - {\bf r}_N.
</math>

Then

<math>\begin{eqnarray}\ \ \ \ \
H = \frac{1}{2(M+nm)}{\bf p}_R^2+\frac{1}{2m}\sum_{i=1}^n{\bf p}_{s_i}^2
+\frac{1}{2M}\sum_{i,k}{\bf p}_{s_i}\cdot {\bf p}_{s_k} + V({\bf s}_1, {\bf
s}_2, \ldots)\nonumber
= \frac{1}{M+nm}{\bf p}_R^2+\frac{1}{2\mu}\sum_{i=1}^n{\bf
p}_{s_i}^2+\frac{1}{2M}\sum_{i\neq k}{\bf p}_{s_i}\cdot {\bf p}_{s_k}+V({\bf s}_1, {\bf
s}_2, \ldots)\nonumber
\end{eqnarray}</math>

where
<math>\begin{eqnarray}
\mu = \left(\frac{1}{M} + \frac{1}{m}\right)^{-1}
\end{eqnarray}</math>

If <math>n=2</math>, the Schrödinger equation is
<math>\begin{eqnarray}\ \ \
\left[ -\frac{\hbar^2}{2\mu}(\nabla^2_{s_1}+ \nabla^2_{s_2}) -
\frac{\hbar^2}{M}\nabla_{s_1}\nabla_{s_2}-\frac{Ze^2}{s_1}-\frac{Ze^2}{s_2}+\frac{e^2}{s_{12}}\right]\psi
= E\psi
\end{eqnarray}</math>.

Define the reduced mass Bohr radius
<math>\begin{eqnarray}
a_\mu = \frac{\hbar^2}{\mu e^2} = \frac{m}{\mu}a_0
\end{eqnarray}</math>
and

<math>\begin{eqnarray}
\rho_i = \frac{Zs_i}{a_\mu}\nonumber\\
\varepsilon = \frac{E}{Z^2(e^2/a_\mu)}\nonumber
\end{eqnarray}</math>

The Schrödinger equation then becomes
<math>\begin{eqnarray}\ \ \
\left[
-\frac{1}{2}(\nabla^2_{\rho_1}+\nabla^2_{\rho_2})-\frac{\mu}{M}\vec{\nabla}_{\rho_1}\cdot\vec{\nabla}_{\rho_2} - \frac{1}{\rho_1}-\frac{1}{\rho_2}+\frac{Z^{-1}}{\rho_{12}}\right]\psi = \varepsilon \psi
\end{eqnarray}</math>

The effects of finite mass are thus

1. A reduced mass correction to all energy levels of <math>(e^2/a_\mu)/(e^2/a_0)
= \mu/m</math>, i.e.
<math>\begin{eqnarray}
E_M=\frac{\mu}{m}E_\infty
\end{eqnarray}</math>

2. A specific mass shift given in first order perturbation theory by
<math>\begin{eqnarray}
\Delta\varepsilon =
-\frac{\mu}{M}<\psi|\nabla_{\rho_1}\cdot\nabla_{\rho_2}|\psi>
\end{eqnarray}</math>
or

<math>\begin{eqnarray}
\Delta\varepsilon =
-\frac{\mu}{M}<\psi|\nabla_{\rho_1}\cdot\nabla_{\rho_2}|\psi>\frac{e^2}{a_\mu}\nonumber\\
= -\frac{\mu^2}{mM}<\psi|\nabla_{\rho_1}\cdot\nabla_{\rho_2}|\psi>\frac{e^2}{a_0}\nonumber
\end{eqnarray}</math>

Since <math>\nabla_{\rho_1}\cdot\nabla_{\rho_2}</math> has the same angular properties as
<math>{\bf\rho}_1 \cdot {\bf\rho}_2</math>, the operator is like the product of two
dipole operators. For product type wave functions of the form
$
\psi = \psi(1s)\psi(nl) \pm \mbox{exchange},
$
the matrix element vanishes for all but P-states.

=Radiative Transitions=

==Transition Integrals and the Line Strength==

For an electric dipole transition <math>\gamma'\,L'\,S'\,J'\,M' \rightarrow \gamma\,L\,S\,J\,M</math>, the line strength is defined by
\begin{equation}
S = \sum_{M',\mu,M}|\langle\gamma'\,L'\,S'\,J'\,M'|r_\mu|\gamma\,L\,S\,J\,M\rangle|^2
\end{equation}
Using the Wigner-Eckart theorem, this is related to the reduced matrix element by
\begin{eqnarray}
S(J',J) = |\langle\gamma'\,L'\,S'\,J'||r||\gamma\,L\,S\,J\,M\rangle|^2
\sum_{M',\mu,M}\left|\left(\matrix{J' & 1 & J \cr
-M'& \mu &M \cr}\right)\right|^2
\end{eqnarray}
Using the sum rule for the 3-<math>j</math> symbols
\begin{equation}
\sum_{M',\mu,M}\left|\left(\matrix{J' & 1 & J \cr
-M'& \mu &M \cr}\right)\right|^2 = 1
\end{equation}
we see that the line strength is identical to the reduced matrix element
\begin{equation}
S(J',J) = |\langle\gamma'\,L'\,S'\,J'||r||\gamma\,L\,S\,J\,M\rangle|^2
\end{equation}
For a nonrelativistic calculation in <math>LS</math> coupling, <math>S = S'</math> and we can next strip out the dependence on <math>J</math> and <math>J'</math> by use of Eq.\ (7.1.7) of Edmonds { Angular Momentum in Quantum Mechanics} with <math>k = 1</math> to obtain
\begin{equation}
S(J',J) = (-1)^{L'+S+J+1}[(2J+1)(2J'+1)]^{1/2}\left\{
\matrix{L' & J'& S\cr
J & L & 1\cr}\right\}\langle\gamma' L'|| r || \gamma L\rangle
\end{equation}
where <math>\langle\gamma' L'|| r || \gamma L\rangle</math> is the reduced matrix element if spin is neglected. Note that if spin is neglected, then the line-strength is simply
\begin{equation}
S(L',L) = |\langle\gamma' L'|| r || \gamma L\rangle|^2
\end{equation}
To calculate the reduced matrix element, use a particular example along with the Wigner-Eckart theorem
such as
\begin{equation}
\langle\gamma'\, L'|| r || \gamma\, L\rangle = (-1)^{L'}\langle\gamma'\, L'\, 0| z | \gamma\, L\, 0\rangle{\Large /}\left(\matrix{L' & 1 & L \cr
0 & 0 & 0 \cr}\right)
\end{equation}

===Correlated Two-electron Integrals===
For the case of two-electron integrals, <math>z = z_1 + z_2</math> and <math>r = r_1 + r_2</math>. As discussed in G.W.F. Drake, Phys. Rev. A 18, 820 (1978), the general integral in Hylleraas coordinates for the integral over the <math>z_1</math> term can be written in the form
\begin{equation}
\label{int1}
\langle\gamma'\, L'\, 0| z_1 | \gamma\, L\, 0\rangle = \sum_\Lambda C_\Lambda I_\Lambda(\gamma',\gamma)
\end{equation}
where <math>I_\Lambda(\gamma',\gamma)</math> stands for the purely radial integral
\begin{equation}
\label{radial}
I_\Lambda(\gamma',\gamma) = \int_0^\infty r_1\,dr_1\int_0^\infty r_2\,dr_2\int_{|r_1-r_2|}^{r_1+r_2}r_{12}\, dr_{12}
R_{\gamma'}^*\,r_1\, R_\gamma P_\Lambda(\cos\theta_{12})
\end{equation}
and <math>\cos\theta_{12} = (r_1^2 + r_2^2 - r_{12}^2)/(2r_1r_2)</math> is a short-hand notation for a radial function of <math>r_1</math>, <math>r_2</math> and <math>r_{12}</math>. Also, <math>P_\Lambda(\cos\theta_{12})</math> is a Legendre polynomial,
and, <math>R_\gamma = R_\gamma(r_1,r_2,r_{12})</math> is the radial part of the Hylleraas wave function for state <math>\gamma</math> such
that the total wave function is <math>\psi({\bf r}_1,{\bf r}_2) = R_\gamma(r_1,r_2,r_{12}){\cal Y}_{l_1,l_2,L}^M(\hat{\bf r}_1,\hat{\bf r}_2)\pm {\rm exchange}</math>, normalized to unity, and <math>{\cal Y}_{l_1,l_2,L}^M(\hat{\bf r}_1,\hat{\bf r}_2)</math> is a vector coupled product of spherical harmonics. The coefficents <math>C_\Lambda</math> in Eq.\ (\ref{int1}) can be calculated from Eq.\ (32) of the above 1978 paper. They need to be calculated separately for he direct-direct and the direct-exchange contributions. For the reduced matrix element of <math>z_1</math>, the leading 3-<math>j</math> symbol can be dropped, leaving (for the case <math>k = 1</math>)
\begin{eqnarray}
C_\Lambda^{\rm reduced} &=& \frac{(-1)^{l_2'+L'+L+1}}{2} (l_1,l_1',l_2,l_2',L,L')^{1/2}
\sum_{\lambda={\rm max}(l_1-1,0)}^{l_1+1} (\lambda,\Lambda,1)
\left(\matrix{l_1 & 1 & \lambda \cr
0 & 0 & 0 \cr}\right)
\left(\matrix{l_1'& \lambda & \Lambda \cr
0 & 0 & 0 \cr}\right)\nonumber\\
&\times&\left(\matrix{l_2' & 1_2 & \Lambda \cr
0 & 0 & 0 \cr}\right)
\left\{\matrix{L' & \lambda & l_2 \cr
\Lambda& l_2' & l_1' \cr}\right\}
\left\{\matrix{L' & \lambda & l_2 \cr
l_1 & L & 1 \cr}\right\}
\end{eqnarray}
where the notation <math>(a,b,c,\cdots)</math> means <math>(2a+1)(2b+1)(2c+1)\cdots</math>.
A similar expression can be obtained for the angular coefficients for the <math>z_2</math> term by interchanging the subscripts 1 and 2 throughout and multiplying by <math>(-1)^{L+L'+1}</math>.

===Example===
As an example, for the <math>1s^2\;^1S - 1s2p\;^1P</math> transition of helium, the only nonvanishing <math>C_\Lambda</math> angular coefficients are <math>C_1 = 1/2</math> for the direct-direct term (i.e. <math>P_1(\cos\theta_{12}) = \cos\theta_{12}</math> is included in the radial integral (\ref{radial})) and <math>C_0 = 1/2</math> for the direct-exchange term (i.e. <math>P_0(\cos\theta_{12}) = 1</math> is included in the radial integral (\ref{radial})). The results for infinite nuclear mass are <math>S(1\;^1S-2\;^1P) = (0.781\,127\,158\,667)^2</math> and <math>f(1\;^1S-2\;^1P) = -0.276\,164\,703\,736</math> where
\begin{equation}
f(1\;^1S-2\;^1P) = \frac{2}{3}[E(1\;^1S) - E(2\;^1P)]S(1\;^1S-2\;^1P)
\end{equation}
is the absorption oscillator strength.

Theory Notes

2020-12-14T00:34:45Z

Drake: /* General Hermitian Property */

<center>
<H1>Notes on solving the Schrödinger equation in Hylleraas coordinates for heliumlike atoms
Gordon W.F. Drake
Department of Physics, University of Windsor
Windsor, Ontario, Canada N9B 3P4
(Transcribed from hand-written notes and edited by Lauren Moffatt. Last revised November 22, 2017 to correct a missing subscript in the expression for $B_1$ in the Hamiltonian in Sect. 11 General Hermitian Property.)
</H1>
</center>

==Introduction==
These notes describe some of the technical details involved in solving the Schrödinger equation for a heliumlike atom of nuclear charge Z in correlated Hylleraas coordinates. The standard for computational purposes is to express all quantities in a form that is valid for electronic states of arbitrary angular momentum, and to express matrix elements of the Hamiltonian in a manifestly Hermitian symmetrized form. Further information and derivations of the mathematical techniques for angular integrals and radial recursion relations can be found in the paper
G.W.F. Drake, Phys. Rev. A <math>{\bf 18}</math>, 820 (1978).

The starting point is the two-electron Schrödinger equation for infinite nuclear mass
\begin{equation} \left[-\frac{\hbar^2}{2m}(\nabla^2_1 +\nabla^2_2) - \frac{Ze^2}{r_1} -
\frac{Ze^2}{r_2}+\frac{e^2}{r_{12}} \right]\psi = E\psi\nonumber
\end{equation}
where <math>m</math> is the electron mass, and <math>r_{12} = |{\bf r}_1 - {\bf r}_2|</math> (see diagram below).

[[File:Pic1.jpg|250px]]

Begin by rescaling distances and energies so that the Schrödinger equation can be expressed in a dimensionless form. The dimensionless Z-scaled distance is defined by <math>\begin{eqnarray} \rho = \frac{Zr}{a_0} \end{eqnarray} </math> where <math> \begin{eqnarray} a_0 = \frac{\hbar^2}{me^2} \end{eqnarray} </math> is the atomic unit (a.u.) of distance equal to the Bohr radius <math>0.529\, 177\, 210\, 92(17) \times 10^{-10}</math> m. Then

\begin{equation}[-\frac{\hbar^2}{2m}Z^2(\frac{me^2}{\hbar^2})^2(\nabla^2_{\rho_1}+\nabla^2_{\rho_2})
- Z^2\frac{e^2}{a_0}\rho^{-1}_1 - Z^2\frac{e^2}{a_0}\rho^{-1}_2 +
\frac{e^2}{a_0}Z\rho^{-1}_{12}]\psi= E\psi\nonumber\end{equation}

But <math>\begin{eqnarray} \frac{\hbar^2}{m}(\frac{me^2}{\hbar^2})^2 = \frac{e^2}{a_0} \end{eqnarray} </math> is the hartree atomic unit of of energy <math>(E_h = 27.211\,385\,05(60)</math> eV, or equivalently <math>E_h/(hc) = 2194.746\,313\,708(11)</math> m<math>^{-1})</math>. Therefore, after multiplying through by <math>a_0/(Ze)^2</math>, the problem to be solved in Z-scaled dimensionless units becomes

\begin{equation} \left[-\frac{1}{2}(\nabla^2_{\rho_1}+\nabla^2_{\rho_2}) -
\frac{1}{\rho_1} - \frac{1}{\rho_2} + \frac{Z^{-1}}{\rho_{12}}\right]\psi =
\varepsilon\psi\nonumber \end{equation}

where <math>\begin{eqnarray}\varepsilon = \frac{Ea_0}{(Ze)^2} \end{eqnarray}</math> is the energy in Z-scaled atomic units. For convenience, rewrite this in the conventional form
\begin{equation}
H\psi = \varepsilon\psi\nonumber \end{equation}
where (using <math>r</math> in place of <math>\rho</math> for the Z-scaled distance)
\begin{equation}
H = -\frac{1}{2}(\nabla^2_1+\nabla^2_2) - \frac{1}{r_1}-\frac{1}{r_2} + \frac{Z^{-1}}{r_{12}}
\nonumber\end{equation}
is the atomic Hamiltonian for infinite nuclear mass.

==The Hartree Fock Method==

For purposes of comparison, and to define the correlation energy, assume that <math>\psi({\bf r}_1,{\bf r}_2)</math> can be written in the separable product form
\begin{equation}
\psi({\bf r}_1,{\bf r}_2) = \frac{1}{\sqrt{2}} \left[ u_1(r_1) u_2(r_2) \pm u_2(r_1) u_1(r_2) \right] \nonumber
\end{equation}
for the <math>1s^2\ ^1\!S</math> ground state. Because of the <math>\frac{1}{r_{12}}</math> term in the Schrödinger equation, the exact solution connot be expressed in this form as a separable product. However, the Hartree-Fock approximation corresponds to finding the best possible solution to the Schrödinger equation
\begin{equation}
H\psi({\bf r}_1,{\bf r}_2) = E\psi({\bf r}_1,{\bf r}_2)\nonumber
\end{equation}
that can nevertheless be expressed in this separable product form, where as before
\begin{equation}
H = -\frac{1}{2}(\nabla^2_1+\nabla^2_2) - \frac{1}{r_1}- \frac{1}{r_2} + \frac{Z^{-1}}{r_{12}}\nonumber
\end{equation}
is the full two-electron Hamiltonian. To find the best solution, substitute into <math>\left\langle\psi \lvert H-E \rvert \psi\right\rangle</math> and require this expression to be stationary
with respect to arbitrary infinitesimal variations <math>\delta u_1</math> and <math>\delta
u_2</math> in <math>u_1</math> and <math>u_2</math> respectively; i.e.

\begin{equation}
\frac{1}{2} \left\langle \delta u_1(r_1)u_2(r_2) \pm u_2(r_1)\delta
u_1(r_2) \lvert H-E \rvert u_1(r_1) u_2(r_2)\pm u_2(r_1) u_1(r_2)\right\rangle\nonumber
\end{equation}

\begin{equation}
=\int\delta u_1(r_1) d{\bf r}_1 \left\{ \int d{\bf r}_2 u_2(r_2) \left( H - E \right) \left[ u_1(r_1)u_2(r_2) \pm u_2(r_1) u_1(r_2) \right] \right\}\nonumber
\end{equation}

\begin{equation}
= 0 \nonumber
\end{equation}

for arbitrary <math> \delta u_1(r_1).\nonumber </math> Therefore <math>\{\int d{\bf r}_2 \ldots \} = 0</math>.

Similarly, the coefficient of <math>\delta u_2</math> would give

\begin{equation}
\int d{\bf r}_1 u_1(r_1) \left( H-E \right) \left[ u_1(r_1) u_2(r_2) \pm
u_2(r_1) u_1(r_2) \right] = 0\nonumber
\end{equation}

Define

\begin{equation}
I_{12} = I_{21} = \int d{\bf r}\, u_1(r)u_2(r), \nonumber
\end{equation}

\begin{equation}
H_{ij} = \int d{\bf r}\, u_i(r)(-\frac{1}{2}\nabla - \frac{1}{r})u_j(r), \nonumber
\end{equation}

\begin{equation}
G_{ij}(r) = \int d{\bf r}^\prime u_i(r^\prime)\frac{1}{|{\bf r} - {\bf r}\prime|}u_j(r^\prime)\nonumber
\end{equation}

Then the above equations become the pair of integro-differential equations

\begin{equation}
[ H_0 - E + H_{22}+G_{22}(r)]u_1(r) = \mp [ I_{12}(H_0-E) + H_{12}+G_{12}(r)]u_2(r)\nonumber
\end{equation}

\begin{equation}
[H_0-E+H_{11}+G_{11}(r)]u_2(r) = \mp [I_{12}(H_0-E) + H_{12}+G_{12}(r)]u_1(r)\nonumber
\end{equation}

These must be solved self-consistently for the "constants" <math>I_{12}</math> and
<math>H_{ij}</math> and the function <math>G_{ij}(r)</math>.

The Hartree Fock energy is <math>E \simeq -2.87\ldots </math>a.u. while the exact energy is <math>E = -2.903724\ldots </math>a.u.

The difference is called the "correlation energy" because it arises from the
way in which the motion of one electron is correlated to the other. The
Hartree Fock equations only describe how one electron moves in the average field
provided by the other (mean-field theory).

==Configuration Interaction==
Expand
\begin{equation}
\psi({\bf r}_1,{\bf r}_2)= C_0u^{(s)}_1(r_1)u^{(s)}_1(r_2) + C_1u^{(P)}_1({\bf r}_1)u^{(P)}_1({\bf r}_2){\cal Y}^0_{1,1,0}(\hat{\bf r}_1, \hat{\bf r}_2)+C_2u^{(d)}_1({\bf r}_1)u^{(d)}_2({\bf r}_2){\cal Y}^0_{2,2,0}(\hat{\bf r}_1, \hat{\bf r}_2)+... \pm {\rm\ exchange}
\end{equation}
where
\begin{equation}
{\cal Y}^M_{l_1,l_2,L}(\hat{\bf r}_1, \hat{\bf r}_2)=\sum_{m_1,m_2}Y^{m_1}_{l_1}({\bf r}_1)Y^{m_2}_{l_2}({\bf r}_2)\times <l_1l_2m_1m_2\mid LM>
\end{equation}

This works, but is slowly convergent, and very laborious. The best CI calculations are accurate to <math> ~10^{-7}</math> a.u.

==Hylleraas Coordinates==
[E.A. Hylleraas, Z. Phys. <math>{\bf 48}, 469(1928)</math> and <math>{\bf 54}, 347(1929)</math>]
suggested using the co-ordinates <math>r_1</math>, <math>r_2</math> and <math>r_{12}</math> or equivalently

\begin{eqnarray}
s &=& r_1 + r_2, \nonumber\\
t &=& r_1-r_2, \nonumber\\
u &=& r_{12}\nonumber
\end{eqnarray}

and writing the trial functions in the form

\begin{equation}
\Psi({\bf r}_1,{\bf r}_2) = \sum^{i+j+k\leq
N}_{i,j,k}c_{i,j,k}r_1^{i+l_1}r_2^{j+l_2}r_{12}^ke^{-\alpha r_1 - \beta r_2}
\mathcal{Y}^M_{l_1,l_2,L}(\hat{r}_1,\hat{r}_2)\pm \text{exchange}\nonumber
\end{equation}

Diagonalizing H in this non-orthogonal basis set is equivalent to solving
<math> \begin{eqnarray}
\frac{\partial E}{\partial c_{i,j,k}} = 0\nonumber
\end{eqnarray} </math>
for fixed <math>\alpha</math> and <math>\beta</math>.

The diagonalization must be repeated for different values of <math>\alpha</math> and
<math>\beta</math> in order to optimize the non-linear parameters.

==Completeness==
The completeness of the above basis set can be shown by first writing
\begin{equation}
r_{12}^2 = r_1^2 + r_2^2 - 2r_1r_2\cos\theta_{12}
\end{equation}
and, from the spherical harmonic addition theorem,
\begin{equation}
\cos(\theta_{12})=\frac{4\pi}{3}\sum^1_{m=-1}Y^{m*}_1(\theta_1,\varphi_1)Y^m_1(\theta_2,\varphi_2)
\end{equation}
Consider first the S-states. The <math>r_{12}^0</math> terms are like the ss terms in a CI calculation. The
<math>r_{12}^2</math> terms bring in p-p type contributions, and the higher powers bring
in d-d, f-f etc. type terms. In general

\begin{equation}
P_l(\cos\theta_{12}) = \frac{4\pi}{2l+1}\sum^l_{m=-l}{Y^{m}_l}^*(\theta_1,\varphi_1)Y^m_l(\theta_2,\varphi_2)\nonumber
\end{equation}

For P-states, one would have similarly

\begin{array}{lr}
r_{12}^0 & (sp)P\nonumber\\
r_{12}^2 & (pd)P\nonumber\\
r_{12}^4 & (df)P\nonumber\\
\vdots & \vdots\nonumber
\end{array}

For D-states

\begin{array}{lr}
r_{12}^0 & (sd)D & (pp^\prime)D\nonumber\\
r_{12}^2 & (pf)D & (dd^\prime)D\nonumber\\
r_{12}^4 & (dg)D & (ff^\prime)D\nonumber\\
\vdots & \vdots & \vdots\nonumber
\end{array}

In this case, since there are two "lowest-order" couplings to form a
D-state, both must be present in the basis set, i.e.

\begin{equation}
\Psi(r_2,r_2) = \sum c_{ijk}r_1^ir_2^{j+2}r_{12}^ke^{-\alpha r_1-\beta
r_2}\mathcal{Y}^M_{022}(\hat{r}_1,\hat{r}_2)
+\sum d_{ijk}r_1^{i+1}r_2^{j+1}r_{12^k}e^{-\alpha^\prime r_1 - \beta^\prime
r_2}\mathcal{Y}^M_{112}(\hat{r}_1,\hat{r}_2)\nonumber
\end{equation}

For F-states, one would need <math>(sf)F</math> and <math>(pd)F</math> terms.

For G-states, one would need <math>(sg)G</math>, <math>(pf)G</math> and <math>(dd^\prime)G</math> terms.

Completeness of the radial functions can be proven by considering the
Sturm-Liouville problem

\begin{equation}
\left(-\frac{1}{2}\nabla^2-\frac{\lambda}{r_s}-E\right)\psi({\bf r}) = 0\nonumber
\end{equation}

or

\begin{equation}
\left(-\frac{1}{2}\frac{1}{r^2}\frac{\partial}{\partial r}\left({r^2}\frac{\partial}{\partial r}\right) -
\frac{l\left(l+1\right)}{2r^2} - \frac{\lambda}{r} - E\right)u(r) = 0.\nonumber
\end{equation}

For fixed E and variable <math>\lambda</math> (nuclear charge).

The eigenvalues are <math>\lambda_n = (E/E_n)^{1/2}</math>, where <math>E_n =- \frac{1}{2n^2}</math>

[[File:Pic2.jpg|500px]]

and the eigenfunctions are

\begin{equation}
u_{nl}(r) =
\frac{1}{(2l+1)!}\left(\frac{(n+l)!}{(n-l-1)2!}\right)^{1/2}(2\alpha)^{3/2}e^{-\alpha
r}\nonumber
\times \ \left( 2 \alpha r \right)^l\,_{1}F_{1}\left(-n+l+1,2l+2; 2 \alpha r \right)
\end{equation}

with <math>\alpha = (-2E)^{1/2}</math> and <math>n\geq l+1</math>. The confluent hypergeometric function <math> _1F_1(a,b,;z)</math> then denotes a finite polynomial since <math>a=-n+l+1</math> is a negative integer or zero.

Unlike the hydrogen spectrum, which has both a discrete part for <math>E<0</math> and a
continuous part for <math>E>0</math>, this forms an entirely discrete set of finite
polynomials, called Sturmian functions. They are orthogonal with respect to
the potential, i.e.

\begin{equation}
\int^\infty_0 r^2dr \left( u_{n^\prime l}(r)\frac{1}{r}u_{nl}(r) \right) = \delta_{n,n^\prime}\nonumber
\end{equation}

Since they become complete in the limit <math>n\rightarrow\infty</math>, this assures the
completeness of the variational basis set.

[See also B Klahn and W.A. Bingel Theo. Chim. Acta (Berlin) 44, 9 and 27
(1977)].

==Solutions of the Eigenvalue Problem==

For convenience, write

\[
\Psi({\bf r}_1,{\bf r}_2) = \sum^N_{m=1}c_m\varphi_m\nonumber
\]

where <math>m</math> represents the <math>m</math>'th combination of <math>i,j,k</math> and

\begin{equation}
\varphi_{ijk}=r_1^ir_2^jr_{12}^ke^{-\alpha r_1 - \beta
r_2}\mathcal{Y}^M_{l_1,l_2,L}(\hat{r}_1,\hat{r}_2) \pm \mbox{exchange}\nonumber
\end{equation}

\begin{align}

&
\left( \begin{array}{lr}
\cos(\theta) & \sin(\theta)\\
-\sin(\theta) & \cos(\theta)
\end{array} \right)
\left( \begin{array}{lr}
H_{11} & H_{12} \\
H_{12} & H_{22}
\end{array}\right)
\left( \begin{array}{lr}
\cos(\theta) & -\sin(\theta)\\
\sin(\theta) & \cos(\theta)
\end{array} \right)
\\[5pt]
= & \left(\begin{array}{lr}
cH_{11}+sH_{12} & cH_{12} + sH_{22}\\
-sH_{11} + cH_{12} & -sH_{12} + cH_{22}
\end{array}\right)
\left( \begin{array}{cc}
c & -s \\
s & c
\end{array}\right)
\\[5pt]
= & \left(\begin{array}{lr}
c^2H_{11}+s^2H_{22} + 2csH_{12} & (c^2-s^2)H_{12}+cs(H_{22}-H_{11})\\
(c^2-s^2)H_{12}+cs(H_{22}-H_{11}) & s^2H_{11}+c^2H_{22}-2csH_{12}
\end{array}\right)
\end{align}

Therefore

\begin{equation}
\left[\cos^2(\theta)-\sin^2(\theta)\right]H_{12} =
\cos(\theta)\sin(\theta)(H_{11}-H_{22})\nonumber
\end{equation}

and

\begin{equation}
\tan(2\theta) = \frac{2H_{12}}{H_{11}-H_{22}}\nonumber
\end{equation}

i.e.

\begin{equation}
\cos(\theta)=\left(\frac{r+\omega}{2r}\right)^{1/2}\nonumber
\end{equation}

\begin{equation}
\sin(\theta)=-\mbox{sgn}(H_{12})\left(\frac{r-\omega}{2r}\right)^{1/2}\nonumber
\end{equation}

where

\begin{equation}
\omega = H_{22}-H_{11}\nonumber
\end{equation}

\begin{equation}
r=\left(\omega^2+4H_{12}^2\right)^{1/2}\nonumber
\end{equation}

\begin{equation}
E_1=\frac{1}{2}\left(H_{11}+H_{22}-r\right)\nonumber
\end{equation}

\begin{equation}
E_2=\frac{1}{2}\left(H_{11}+H_{22}+r\right)\nonumber
\end{equation}

===Brute Force Method===

- Gives all the eigenvalues and eigenvectors, but it is slow

- First orthonormalize the <math>N</math>-dimensional basis set; i.e. form linear combinations

\begin{equation}
\Phi_m = \sum_{n=1}^N\varphi_nR_{nm}\nonumber
\end{equation}

such that
\begin{equation}
\langle \Phi_m | \Phi_n \rangle = \delta_{m,n}
\end{equation}

This can be done by finding an orthogonal tranformation T such that

\begin{equation}
T^TOT = I = \left(
\begin{array}{ccccc}
I_1 & 0 & 0 & \ldots & 0 \\
0 & I_2 & 0 & \ldots & 0 \\
0 & 0 & I_3 & \ldots & 0 \\
\vdots & \vdots & \vdots & \ddots & \vdots \\
0 & 0 & 0 & \ldots & I_N
\end{array}\right);
\end{equation}

\begin{equation}
O_{mn} = \langle \varphi_m | \varphi_n \rangle \nonumber
\end{equation}

and then applying a scale change matrix

\begin{equation}
S = \left(\begin{array}{ccccc}
\frac{1}{I_1^{1/2}} & 0 & 0 & \ldots & 0\\
0 & \frac{1}{I_2^{1/2}} & 0 & \ldots & 0 \\
0 & 0 & \frac{1}{I_3^{1/2}} & \ldots & 0 \\
\vdots & \vdots & \vdots & \ddots & \vdots \\
0 & 0 & 0 & \ldots & I_N^{1/2}
\end{array}\right)= S^T\nonumber
\end{equation}

Then

\begin{equation}
S^TT^TOTS = 1
\end{equation}

i.e.

\begin{equation}
R^TOR = 1
\end{equation}

with <math>R=TS</math>.

If H is the matrix with elements <math>H_{mn} = \langle \varphi_m | \varphi_n \rangle </math>, then H
expressed in the <math>\Phi_m</math> basis set is

\begin{equation}
H^\prime = R^THR.\nonumber
\end{equation}

We next diagonalize <math>H^\prime</math> by finding an orthogonal transformation W such
that

\begin{equation}
W^T H^\prime W = \lambda = \left(
\begin{array}{cccc}
\lambda_1 & 0 & \ldots & 0 \\
0 & \lambda_2 & \ldots & 0 \\
\vdots & \vdots & \ddots & \vdots \\
0 & 0 & \ldots & \lambda_N
\end{array}\right)\nonumber
\end{equation}

The q'th eigenvector is

\begin{equation}
\Psi^{(q)} = \sum_{n=1}^N\Phi_n W_{n,q} = \sum_{n,n^\prime} \varphi_{n^\prime} R_{n^\prime ,n} W_{n,q} \nonumber
\end{equation}

i.e.

\begin{equation}
c_{n^\prime}^{(q)} = \sum_{n=1}^N R_{n^\prime n} W_{n,q}
\end{equation}

===The Power Method===

- Based on the observation that if H has one eigenvalue, <math>\lambda_M</math>, much bigger than all the rest, and <math>\chi = \left(
\begin{array}{c}a_1\\a_2\\\vdots\end{array}\right)</math>
is an arbitrary starting vector, then <math>\chi = \sum_{q=1}^N x_q\Psi^{(q)}</math>.

\begin{equation}
(H)^n\chi = \sum_{q=1}^N x_q \lambda^n_q\Psi^{(q)}\nonumber\rightarrow x_M\lambda_M^n\Psi^{(M)}\nonumber
\end{equation}

provided <math>x_M\neq 0</math>.

To pick out the eigenvector corresponding to any eigenvalue, write the original problem in the form

\begin{align}
H\Psi & = \lambda O \Psi \nonumber \\
\left( H - \lambda_q O \right) \Psi & = \left( \lambda - \lambda_q \right) O \Psi \nonumber
\end{align}

Therefore,

\begin{equation}
G \Psi = \frac{1}{\lambda - \lambda_q } \Psi \nonumber \\
\end{equation}

where <math>G=(H-\lambda_qO)^{-1}O</math> with eigenvalues <math>\frac{1}{\lambda_n - \lambda_q}</math>.

By picking <math>\lambda_q</math> close to any one of the <math>\lambda_n</math>, say
<math>\lambda_{n^\prime}</math>, then <math>\frac{1}{\lambda_n-\lambda_q}</math> is much larger for
<math>n=n^\prime</math> than for any other value. The sequence is then

\begin{array}{c}
\chi_1=G\chi\nonumber\\
\chi_2=G\chi_1\nonumber\\
\chi_3=G\chi_2\nonumber\\
\vdots\nonumber
\end{array}

until the ratios of components in <math>\chi_n</math> stop changing.

- To avoid matrix inversion and multiplication, note that the sequence is equivalent to
\begin{equation}
F \chi_n = \left( \lambda - \lambda_q \right) O \chi_{n-1} \nonumber \\
\end{equation}
where <math>F = H-\lambda_qO</math>. The factor of <math>\left( \lambda - \lambda_q \right)</math> can be
dropped because this only affects the normalization of <math>\chi_n</math>. To find <math>\chi_n</math>, solve
\begin{equation}
F \chi_n = O \chi_{n-1} \nonumber
\end{equation}
(N equations in N unknowns). Then
\begin{equation}
\lambda = \frac{ \langle \chi_n | H | \chi_n \rangle}{\langle \chi_n | \chi_n \rangle} \nonumber
\end{equation}

==Matrix Elements of H==

\begin{equation}
H=-\frac{1}{2}\nabla^2_1 -\frac{1}{2}\nabla^2_2 - \frac{1}{r_1} - \frac{1}{r_2} +\frac{Z^{-1}}{r_{12}}\nonumber
\end{equation}

Taking <math>r_1, r_2</math> and <math>r_{12}</math> as independent variables,

\begin{equation}
\nabla_1^2 = \frac{1}{r_1^2}\frac{\partial}{\partial r_1}\left(r_1^2\frac{\partial}{\partial r_1}\right)+ \frac{1}{r_{12}^2}\frac{\partial}{\partial r_{12}}
\left(r_{12}^2\frac{\partial}{\partial r_{12}}\right)\nonumber -\frac{l_1(l_1+1)}{r_1^2}+2(r_1 - r_2\cos(\theta_{12}))\frac{1}{r_{12}}\frac{\partial^2}{\partial r_1 \partial r_{12}} \nonumber - 2(\nabla_1^Y \cdot {\bf r}_2)\frac{1}{r_{12}}\frac{\partial}{\partial r_{12}}\nonumber
\end{equation}

where <math>\nabla_1^Y</math> acts only on the spherical harmonic part of the wave function and the diagram

[[File:Pic3.jpg|400px]]

defines the complete set of 6 independent variables is <math>r_1</math>, <math>r_2</math>, <math>r_{12}</math>, <math>\theta_1</math>, <math>\varphi_1</math>, <math>\chi</math>.

If <math>r_{12}</math> were not an independent variable, then one could take the volume element to be

\begin{equation}
d\tau = r_1^2dr_1\sin(\theta_1)d\theta_1d\varphi_1r_2^2dr_2\sin(\theta_2)d\theta_2d\varphi_2.\nonumber
\end{equation}

However, <math>\theta_2</math> and <math>\varphi_2</math> are no longer independent variables. To eliminate them, take the point <math>{\bf r}_1</math> as the
origin of a new polar co-ordinate system, and write

\begin{equation}
d\tau=-r_1^2dr_1\sin(\theta_1)d\theta_1d\varphi_1r_{12}^2dr_{12}\sin(\psi)d\psi d\chi\nonumber
\end{equation}

and use

\begin{equation}
r_2^2=r_1^2+r_{12}^2 +2r_1r_{12}\cos(\psi)
\end{equation}

Then for fixed <math>r_1</math> and <math>r_{12}</math>,

\begin{equation}
2r_2dr_2 = -2r_1r_{12}\sin(\psi)d\psi
\end{equation}

Thus

\begin{equation}
d\tau= r_1dr_1r_2dr_2r_{12}dr_{12}\sin(\theta_1)d\theta_1d\varphi_1d\chi
\end{equation}

The basic type of integral to be calculated is

\begin{align}
I \left( l_1, m_1, l_2, m_2; R \right) & = \int \sin \left( \theta_1 \right) d\theta_1 d\varphi_1 d\chi Y^{m_1}_{l_1} \left( \theta_1, \varphi_1 \right)^{*} Y^{m_2}_{l_2} \left( \theta_2, \varphi_2 \right) \\
& \times \int r_1dr_1r_2dr_2r_{12}dr_{12}R \left( r_1, r_2, r_{12} \right)
\end{align}

Consider first the angular integral. <math>Y^{m_2}_{l_2} \left( \theta_2, \varphi_2 \right)</math> can be expressed in terms of the independent variables <math>\theta_1, \varphi_1, \chi</math> by use of the rotation matrix relation
\begin{equation}
Y^{m_2}_{l_2}( \theta_2, \varphi_2) = \sum_m \mathcal{D}^{(l_2)}_{m_2,m}(\varphi_1,\theta_1,\chi)^*Y^m_{l_2}( \theta_{12}, \varphi)
\end{equation}
where <math>\theta, \varphi</math> are the polar angles of <math>{\bf r}_2</math> relative to <math>{\bf r}_1</math>. The angular integral is then

\begin{align}
I_{ang} & = \int^{2 \pi}_{0} d\chi \int^{2\pi}_{0} d\varphi_1 \int^{\pi}_{0} \sin \left( \theta_1 \right) d\theta_{1} Y^{m_1}_{l_1} \left( \theta_1, \varphi_1 \right)^{*} \\
& \times \sum_m \mathcal{D}^{(l_2)}_{m_2,m} \left( \varphi_1,\theta_1,\chi \right)^{*} Y^{m}_{l_2} \left( \theta_{12},\varphi \right) \\
\end{align}

Use

\begin{equation}
Y^{m_1}_{l_1} \left( \theta_1, \varphi_1 \right)^{*} = \sqrt{\frac{2l_1+1}{4\pi}}\mathcal{D}^{(l_1)}_{m_1,0} \left( \varphi_1, \theta_1, \chi \right)
\end{equation}

together with the orthogonality property of the rotation matrices (Brink and Satchler, p 147)

\begin{align}
& \int \mathcal{D}^{(j)*}_{m,m^\prime}\mathcal{D}^{(J)}_{M,M^\prime}\sin(\theta_1)d\theta_1d\varphi_1d\chi \\
& = \frac{8\pi^2}{2j+1}\delta_{jJ}\delta_{mM}\delta_{m^\prime M^\prime}\nonumber
\end{align}

to obtain

\begin{align}
I_{ang} & = \sqrt{ \frac{2l_1+1}{4\pi}} \frac{8\pi^2}{2l_1+1} \delta_{l_1,l_2} \delta_{m_1,m_2} Y^{0}_{l_2} \left( \theta_{12}, \varphi \right) \\
& = 2 \pi \delta_{l_1,l_2} \delta_{m_1,m_2} P_{l_2} \left( \cos \theta_{12} \right)
\end{align}

since

\begin{equation}
Y^0_{l_2} \left( \theta_{12}, \varphi \right) = \sqrt{\frac{2l_1+1}{4\pi}} P_{l_2} \left( \cos \left( \theta_{12} \right) \right)
\end{equation}

Note that <math>P_{l_2} \left( \cos \theta \right)</math> is just a short hand expression for a radial function because

\begin{equation}
\cos \theta_{12} = \frac{r_1^2 + r_2^2 - r_{12}^2}{2r_1r_2}
\end{equation}

The original integral is thus

\begin{equation}
I \left( l_1, m_1, l_2, m_2; R \right) = 2 \pi \delta_{l_1,l_2} \delta_{m_1,m_2} \int^{\infty}_{0} r_{1} dr_{1} \int^{\infty}_{0} r_{2} dr_{2} \int^{r_{1} + r_{2}}_{|r_{1} - r_{2}|} r_{12} dr_{12} R \left( r_{1}, r_{2}, r_{12} \right) P_{l_2} \left( \cos \theta_{12} \right)
\end{equation}

where again

\begin{equation}
\cos \theta_{12} = \frac{r_{1}^{2} +r^{2}_{2} - r_{12}^{2}}{2 r_{1} r_{2}}
\end{equation}

is a purely radial function.

The above would become quite complicated for large <math>l_2</math> because <math>P_{l_2}(\cos \theta_{12})</math> contains terms up to <math>\left( \cos \theta_{12} \right)^{l_2}</math>. However, recursion relations exist which allow any integral containing <math>P_l \left( \cos \theta_{12} \right)</math> in terms of those containing just <math>P_0 \left( \cos \theta_{12} \right) = 1</math> and <math>P_1 \left( \cos \theta_{12} \right) = \cos \theta_{12}</math>.

==Radial Integrals and Recursion Relations==

The basic radial integral is [see G.W.F. Drake, Phys. Rev. A 18, 820 (1978)]

\begin{align}
I_0 \left( a, b, c \right) & =
\int^{\infty}_{0} r_{1} dr_{1} \int^{\infty}_{r_{1}} r_{2} dr_{2} \int^{r_{1}+r_{2}}_{r_{2} - r_{1}} r_{12} dr_{12} r_{1}^{a} r_{2}^{b} r_{12}^{c} e^{-\alpha r_{1} - \beta r_{2}} + \int^{\infty}_{0} r_{2} dr_{2} \int^{\infty}_{r_{2}} r_{1} dr_{1} \int^{r_{1} + r_{2}}_{r_{1} - r_{2}} r_{12} dr_{12} r_{1}^{a} r_{2}^{b} r_{12}^{c} e^{- \alpha r_{1} - \beta r_{2}} \\
& = \frac{2}{c+2} \sum^{[(c+1)/2]}_{i=0}
\left(\begin{array}{c}
c+2 \\
2i+1
\end{array} \right)
\left\{ \frac{q!}{\beta^{q+1} \left( \alpha + \beta \right)^{p+1}} \sum_{j=0}^{q} \frac{\left( p + j \right)!}{j!} \left( \frac{\beta}{\alpha + \beta} \right)^j + \frac{q^{\prime}!}{\alpha^{q^{\prime}+1} \left( \alpha + \beta \right)^{p^{\prime}+1}} \sum^{q^{\prime}}_{j=0} \frac{\left( p^{\prime} + j \right)!}{j!} \left( \frac{\alpha}{\alpha + \beta} \right)^j \right\}
\end{align}

where

\begin{array}{ll}
p = a + 2i + 2 & p^{\prime} = b + 2i + 2 \\
q = b + c - 2i + 2 & q^{\prime} = a + c - 2i + 2
\end{array}

The above applies for <math> a, b \geq -2, c\geq -1</math>.
<math>[x]</math> means "greatest integer in" <math>x</math>.

Then

\begin{align}
I_1 \left( a, b, c \right) & = \int d\tau_{r} r_{1}^{a} r_{2}^{b} r_{12}^{c} e^{ -\alpha r_{1} - \beta r_{2}} P_{1}\left( \cos \theta \right) \\
& = \frac{1}{2} \left( I_{0} \left( a + 1, b - 1, c \right) + I_{0} \left( a - 1, b + 1, c \right) - I_{0} \left( a - 1, b - 1, c + 2 \right) \right)
\end{align}

===The Radial Recursion Relation===

Recall that the full integral is

\begin{equation}
I(l_1m_1,l_2m_2;R)=2\pi\delta_{l_1,l_2}\delta_{m_1,m_2}I_{l_2}(R)
\end{equation}

where

\begin{equation}
I_{l_2}(R) = \int d\tau_r R(r_1,r_2,r_{12})P_{l_2}(\cos \theta_{12})
\end{equation}

To obtain the recursion relation, use

\begin{equation}
P_l(x) = \frac{[P^{\prime}_{l+1}(x)-P^{\prime}_{l-1}(x)]}{2l+1}
\end{equation}

with

\begin{equation}
P_{l+1}^\prime (x) = \frac{d}{dx}P_{l+1}(x)
\end{equation}

Here <math>x=\cos\theta_{12}</math> and

\begin{align}
\frac{d}{d \cos \theta_{12}} & = \frac{dr_{12}}{d \cos \theta_{12}} \frac{d}{dr_{12}} \\
& = -\frac{r_{1} r_{2}}{r_{12}} \frac{d}{dr_{12}}
\end{align}

Then,

\begin{equation}
I_{l}\left( R \right) = - \int d\tau_r R \frac{r_{1} r_{2}}{r_{12}} \frac{d}{dr_{12}} \frac{ \left[ P_{l+1} \left( \cos \theta_{12} \right) - P_{l-1} \left( \cos \theta_{12} \right) \right]}{2l+1}
\end{equation}

The <math>r_{12}</math> part of the integral is

\begin{align}
& \int^{r_{1} + r_{2}}_{|r_{1} - r_{2}|} r_{12} dr_{12} R \frac{r_{1} r_{2}}{r_{12}} \frac{d}{dr_{12}} \left[P_{l+1}-P_{l-1} \right] \\
& = R r_{1} r_{2} \left[ P_{l+1} - P_{l-1} \right]^{r_{1} + r_{2}}_{|r_{1} - r_{2}|} - \int^{r_{1} + r_{2}}_{|r_{1} - r_{2}|} r_{12} dr_{12} \left( \frac{d}{dr_{12}} R \right) \frac{r_{1} r_{2}}{r_{12}} \frac{ \left[ P_{l+1} - P_{l-1} \right]}{2l+1}
\end{align}

The integrated term vanishes because

\begin{equation}
\cos \theta_{12} = \frac{r_{1}^{2} + r_{2}^{2} - r_{12}^{2}}{2 r_{1} r_{2}}
\begin{array}{lll}
& = -1 & {\rm when} & r_{12}^{2} = \left( r_{1} + r_{2} \right)^{2} \\
& = 1 & {\rm when} & r_{12}^{2} = \left( r_{1} - r_{2} \right)^{2}
\end{array}
\end{equation}

and <math>P_l(1)=1</math>, <math>P_{l}(-1)=(-1)^{l}</math>.

Thus,

\begin{equation}
I_{l+1}\left(\frac{r_1r_2}{r_{12}}R^\prime\right) = \left( 2l+1 \right) I_{l}(R) - I_{l-1} \left( \frac{r_{1} r_{2}}{r_{12}} R^{\prime} \right)
\end{equation}

If

\begin{equation}
R = r_{1}^{a-1} r_{2}^{b-1} r_{12}^{c+2} e^{-\alpha r_{1} - \beta r_{2}}
\end{equation}

then [G.W.F. Drake, Phys. Rev. A 18, 820 (1978)]

\begin{equation}
I_{l+1}(r_1^ar_2^br_{12}^c) =
\frac{2l+1}{c+2} I_{l} \left( r_{1}^{a-1} r_{2}^{b-2} r_{12}^{c+2} \right)+I_{l-1} \left( r_{1}^{a} r_{2}^{b} r_{12}^{c} \right)
\end{equation}

For the case <math>c=-2</math>, take

\begin{equation}
R = r_{1}^{a-1} r_{2}^{b-1} \ln r_{12} e^{ -\alpha r_{1} -\beta r_{2}}
\end{equation}

Then

\begin{equation}
I_{l+1} \left( r_{1}^{a} r_{2}^{b} r_{12}^{c} \right) = I_{l} \left( r_{1}^{a-1} r_{2}^{b-1} \ln r_{12} \right) \left( 2l + 1 \right) - I_{l-1} \left( r_{1}^{a} r_{2}^{b} r_{12}^{-2} \right)
\end{equation}

This allows all <math>I_{l}</math> integrals to be calculated from tables of <math>I_0</math> and
<math>I_1</math> integrals.

===The General Integral===

The above results for the angular and radial integrals can now be combined into a general formula for integrals of the type

\begin{equation}
I = \int \int d{\bf r}_{1} d{\bf r}_{2} R_{1} \mathcal{Y}^{M^{\prime}}_{l^{\prime}_{1} l^{\prime}_{2} L^{\prime}} \left( \hat{r}_{1}, \hat{r}_{2} \right) T_{k_{1} k_{2} K}^{Q} \left( {\bf r}_{1},{\bf r}_{2} \right) R_{2} \mathcal{Y}_{l_{1} l_{2} L}^{M} \left( \hat{r}_{1}, \hat{r}_{2} \right)
\end{equation}

where

\begin{equation}
\mathcal{Y}^{M}_{l_{1} l_{2} L} \left( \hat{r}_{1}, \hat{r}_{2} \right) = \sum_{m_1,m_2} \langle l_{1} l_{2} m_{1} m_{2} | LM \rangle Y^{m_{1}}_{l_{1}} \left( \hat{r}_{1} \right) Y^{m_{2}}_{l_{2}} \left( \hat{r}_{2} \right)
\end{equation}

and

\begin{equation}
T^Q_{k_{1} k_{2} K} \left( {\bf r}_{1}, {\bf r}_{2} \right) = \sum_{q_{1}, q_{2}} \langle k_{1} k_{2} q_{1} q_{2} | KQ \rangle Y_{k_{1}}^{q_{1}} \left( \hat{r}_{1} \right) Y^{q_{2}}_{k_{2}} \left( \hat{r}_{2} \right)
\end{equation}

The basic idea is to make repeated use of the formula

\begin{equation}
Y^{m_{1}}_{l_{1}} \left( \hat{r}_{1} \right) Y^{m_{2}}_{l_{2}} \left( \hat{r}_{1} \right) = \sum_{l m} \left( \frac{\left( 2l_{1}+1 \right) \left(2l_{2}+1 \right) \left(2l+1 \right)}{4 \pi} \right)^{1/2}
\times \left( \begin{array}{ccc}
l_1 & l_2 & l \\
m_1 & m_2 & m
\end{array} \right) \left(\begin{array}{ccc}
l_1 & l_2 & l \\
0 & 0 & 0
\end{array} \right)
Y^{m*}_{l} \left( \hat{r}_{1} \right)
\end{equation}

where

\begin{equation}
Y^{m*}(\hat{r}) = (-1)^mY_l^{-m}(\hat{r})
\end{equation}

and

\begin{equation}
\left(\begin{array}{ccc}
l_1 & l_2 & l \\
m_1 & m_2 & m
\end{array} \right) = \frac{\left( -1 \right)^{l_{1} - l_{2} - m}}{\left( 2l + 1 \right)^{1/2}} \left( l_{1} l_{2} m_{1} m_{2} | l, -m \right)
\end{equation}

is a 3-<math>j</math> symbol. In particular, write

\begin{array}{cc}
Y^{m_1^\prime}_{l_1^\prime}(\hat{r}_1)^{*} \underbrace{Y_{k_{1}}^{q_{1}} \left( \hat{r}_{1} \right) Y^{m_{1}}_{l_{1}} \left( \hat{r}_{1} \right)} & = & \sum_{\Lambda M}(\ldots)Y^{M}_\Lambda(\hat{r}_1)\\
\ \ \ \ \ \ \ \sum_{\lambda_1\mu_1}Y_{\lambda_1}^{\mu_1}(\hat{r}_1) & & \\
& & \\
Y^{m_{2}^{\prime}}_{l_{2}^{\prime}} \left( \hat{r}_{2} \right) \underbrace{Y^{q_{2}}_{k_{2}} \left( \hat{r}_{2} \right) Y_{l_{2}}^{m_{2}} \left( \hat{r}_{2} \right)} & = & \sum_{\Lambda^{\prime} M^{\prime}} \left( \ldots \right)Y_{\Lambda^{\prime}}^{M^{\prime}*} \left( \hat{r}_{2} \right) \\
\ \ \ \ \ \ \ \ \ \sum_{\lambda_{2} \mu_{2}}Y_{\lambda_{2}}^{\mu_{2}} \left( \hat{r}_{2} \right) & &
\end{array}

The angular integral then gives a factor of <math>2 \pi \delta_{\Lambda, \Lambda^{\prime}} \delta_{M, M^{\prime}} P_{\Lambda} \left( \cos \theta_{12} \right)</math>. The total integral therefore reduces to the form

\begin{equation}
I = \sum_{\Lambda} C_{\Lambda} I_{\Lambda} \left( R_{1} R_{2} \right)
\end{equation}

where <math>C_{\Lambda} = \sum_{\lambda_{1}, \lambda_{2}} C_{ \lambda_{1}, \lambda_{2}, \Lambda}</math>. For further details and derivations, including graphical representations, see G.W.F. Drake, Phys. Rev. A <math>{\bf 18}</math>, 820 (1978).

==Graphical Representation==

[[File:GraphRep1.jpg|500px]]

==Matrix Elements of H==
Recall that

\begin{equation}
H = -\frac{1}{2} \nabla^{2}_{1} - \frac{1}{2} \nabla^{2}_{2} - \frac{1}{r_1} - \frac{1}{r_2} + \frac{Z^{-1}}{r_{12}}
\end{equation}

Consider matrix elements of

\begin{equation}
\nabla^2_1 = \frac{1}{r_1^2} \frac{\partial}{\partial r_1}\left(r_1^2\frac{\partial}{\partial r_1} \right) + \frac{1}{r^2}\frac{\partial}{\partial r}\left(r^2\frac{\partial}{\partial r}\right) - \frac{(\vec{l}_1^Y)^2}{r_1^2} + \frac{2 \left( r_1 - r_2 \cos \theta \right)}{r} \frac{\partial^2}{\partial r_1 \partial r} - 2 \left( \nabla^Y_1 \cdot {\bf r}_2 \right) \frac{1}{r} \frac{\partial}{\partial r}
\end{equation}

where <math>r \equiv r_{12}</math> and <math>\cos\theta \equiv \cos\theta_{12}</math>. Also in what follows, define <math>\hat{\nabla}^Y_1 \equiv r_1\nabla^Y_1</math> where <math>\nabla^Y</math> operates only on the spherical harmonic part of the wave function, and similarly for <math>\vec{l}_1^Y</math>.

A general matrix element is

\begin{equation}
\langle r_1^{a^\prime} r_2^{b^\prime} r_{12}^{c^\prime} e^{-\alpha^\prime r_1 - \beta^\prime r_2} \mathcal{Y}^{M^\prime}_{l^\prime_1 l^\prime_2 L^\prime} \left( \hat{r}_1, \hat{r}_2 \right) \left| \nabla^2_1 \right| r_1^a r_2^b r_{12}^c e^{-\alpha r_1 - \beta r_2} \mathcal{Y}^M_{l_1 l_2 L} \left( \hat{r}_1, \hat{r}_2 \right) \rangle
\end{equation}

Since <math>\nabla^2_1</math> is rotationally invariant, this vanishes unless <math>L=L^\prime</math>, <math>M=M^\prime</math>. Also <math>\nabla_1^2</math> is Hermitian, so that the result must be the same whether it operates to the right or left, even though the results look very different. In fact, requiring the results to be the same yields some interesting and useful integral identities as follows:

Put

\begin{equation}
\mathcal{F} = F \mathcal{Y}^M_{l_1 l_2 L} \left( \hat{r}_1, \hat{r_2} \right)
\end{equation}

and

\begin{equation}
F = r_1^a r_2^b r_{12}^c e^{-\alpha r_1 - \beta r_2}
\end{equation}

Then

\begin{equation}
\nabla^2_1 \mathcal{F} = \left\{ \frac{1}{r_1^2} \left[ a \left( a + 1 \right) - l_1 \left( l_1 + 1 \right) \right] + \frac{c \left( c + 1 \right)}{r^2} + \alpha^2\frac{2\alpha \left( a + 1 \right)}{r_1} + \frac{2 \left(r_1 - r_2 \cos \theta \right) }{r_1r^2}c \left[ a - \alpha r_1 \right] - \frac{2c}{r^2} \left( \hat{\nabla}^Y_1 \cdot \hat{r}_2 \right) \frac{r_2}{r_1} \right\} \mathcal{F}
\end{equation}

and

\begin{align}
\langle \mathcal{F}^\prime \left| \nabla^2_1 \right| \mathcal{F} \rangle & = \sum_\Lambda \int d\tau_r F^\prime C_\Lambda \left( 1 \right) P_\Lambda \left( \cos \theta \right) \times \left\{ \frac{1}{r_1^2} \left[ a \left( a + 1 \right) - l_1 \left( l_1 + 1 \right) \right] -\frac{2 \alpha \left( a + 1 \right)}{r_1} + \frac{c \left( c + 1 \right)}{r^2} +\alpha^2 + \frac{2 \left( r_2 - r_2 \cos \theta \right)}{r_1r^2}c \left( a - \alpha r_1 \right) \right\} F \\
& + \sum_\Lambda \int d\tau_r F^\prime C_\Lambda \left( \hat{\nabla}^Y_1 \cdot \hat{r}_2 \right) P_\Lambda \left( \cos \theta \right) \left( \frac{-2c r_2}{r_1 r^2} \right) F
\end{align}

where

\begin{equation}
\int d\tau_r = \int^\infty_0 r_1 dr_1 \int^\infty_0 r_2 dr_2 \int^{r_1+r_2}_{\left| r_1 - r_2 \right|} r dr
\end{equation}

For brevity, let the sum over <math>\Lambda</math> and the radial integrations be understood, and let <math>\left( \nabla_1^2 \right)</math> stand for the terms which appear in the integrand. Then operating to the right gives

\begin{equation}
(\nabla_1^2)_R = \frac{1}{r_1^2} \left[ a \left( a + 1 \right) - l_1 \left( l_1 + 1 \right) \right] + \frac{c \left( c + 1 \right)}{r^2} + \alpha^2 - \frac{2 \alpha \left( a + 1 \right)}{r_1}
+ \frac{2 \left( r_1 - r_2 \cos \theta \right)}{r_1 r^2}c \left( a - \alpha r_1 \right) - \frac{2c r_2}{r_1 r^2} \left( \hat{\nabla}^Y_1 \cdot \hat{r}_2 \right)
\end{equation}

Operating to the left gives

\begin{equation}
(\nabla_1^2)_L = \frac{1}{r_1^2} \left[a^\prime \left( a^\prime + 1 \right) - l_1^\prime \left( l_1^\prime + 1 \right) \right] + \frac{c^\prime \left( c^\prime + 1 \right)}{r^2} + \alpha^{\prime 2} -\frac{2 \alpha^\prime \left( a^\prime + 1 \right)}{r_1}
+ \frac{2 \left( r_1 - r_2 \cos \theta \right)}{r_1 r^2} c^\prime \left( a^\prime - \alpha^\prime r_1 \right) - \frac{2 c^\prime r_2}{r_1 r^2} \left( \hat{\nabla}^{Y^\prime}_1 \cdot \hat{r}_2 \right)
\end{equation}

Now put

\begin{array}{cc}
a_+ = a+a^\prime, & \hat{\nabla}_1^+ = \hat{\nabla}_1^Y + \hat{\nabla}_1^{Y\prime} \\
a_{-} = a-a^\prime, & \hat{\nabla}_1^{-} = \hat{\nabla}_1^Y - \hat{\nabla}_1^{Y\prime}
\end{array}

etc., and substitute <math>a^\prime = a_+ - a</math>, <math>c^\prime = c_+ - c</math>, and <math>\alpha^\prime = \alpha_+ - \alpha</math> in <math>(\nabla^2_1)_L</math>. If <math>a_+</math>, <math>c_+</math> and <math>\alpha_+</math> are held fixed, then the equation

\begin{equation}
(\nabla^2_1)_R = (\nabla^2_1)_L
\end{equation}

must be true for arbitrary <math>a</math>, <math>c</math>, and <math>\alpha</math>. Their coefficients must thus vanish.

This yields the integral relations

\begin{equation}
\label{I}
\frac{\left( r_1 - r_2 \cos \theta \right)}{r_1 r^2} = \frac{1}{c_+} \left( \frac{- \left( a_+ + 1 \right)}{r_1^2} + \frac{\alpha_+}{r_1} \right) \tag{I}
\end{equation}

from the coefficient of <math>a</math> and

\begin{equation}
\label{II}
\frac{ \left( r_1 - r_2 \cos \theta \right) \left( a_+ - \alpha_+ r_1 \right)}{r_1 r^2} = \frac{r_2}{r_1 r^2}\left( \hat{r}_2 \cdot \hat{\nabla}_1^+ \right) - \frac{ \left( c_+ + 1 \right)}{r^2} \tag{II}
\end{equation}

from the coefficient of <math>c</math>. The coefficient of <math>\alpha</math> gives an equation equivalent to (I).

Furthermore, if can be show that (see problem)

\begin{align}
& \sum_\Lambda \int d\tau_r \frac{r^c}{r^2} C_\Lambda \left( \hat{r}_2 \cdot \hat{\nabla}_1^Y \right) P_\Lambda \left( \cos \theta \right) \\
= & \sum_\Lambda \int d\tau_r \frac{r^c}{c r_1 r_2}C_\Lambda \left( 1 \right)P_\Lambda \left( \cos \theta \right) \times \left( \frac{l_1^\prime \left( l_1^\prime + 1 \right) - l_1 \left( l_1 + 1 \right) -\Lambda \left( \Lambda + 1 \right)}{2} \right)
\end{align}

and similarly for <math>\left( \hat{r}_2 \cdot \hat{\nabla}^{Y\prime}_1 \right)</math> with <math>l_1</math> and <math>l_1^\prime</math> interchangeable, then it follows that

\begin{equation}
\frac{r^{c_+}}{r^2} \left( \hat{r}_2 \cdot \hat{\nabla}_1^+ \right) = - \frac{r^{c_+}}{c_+ r_1 r_2} \Lambda \left( \Lambda + 1 \right)
\end{equation}

and

\begin{equation}
\frac{r^{c_+}}{r^2} \left( \hat{r}_2 \cdot \hat{\nabla}_1^- \right) = \frac{r^{c_+}}{c_+ r_1 r_2} \left[ l_1^\prime \left( l_1^\prime + 1 \right) - l_1 \left( l_1 + 1 \right) \right]
\end{equation}

where equality applies after integration and summation over <math>\Lambda</math>.

Thus (II) becomes

\begin{equation}\label{IIa}
\frac{ \left( r_1 - r_2 \cos \theta \right) \left( a_+ - \alpha_+ r_1 \right)}{r_1 r^2} = - \frac{\Lambda \left( \Lambda + 1 \right)}{c_+ r^2_1} - \frac{ \left( c_+ + 1 \right)}{r^2} \tag{IIa}
\end{equation}

===Problem===

Prove the integral relation

\begin{align}
& \sum_\Lambda \int d\tau_r f \left( r_1, r_2 \right) \frac{1}{r} \left( \frac{d}{dr} q(r) \right) C_\Lambda \left( \hat{r}_2 \cdot \hat{\nabla}^Y_1 \right) P_\Lambda \left( \cos \theta \right) \\
= & \sum_\Lambda \int d\tau_r \frac{ f \left( r_1, r_2 \right)}{r_1, r_2} q(r) C_\Lambda \left( 1 \right) P_\Lambda \left( \cos \theta \right) \left(\frac{l_1^\prime \left( l_1^\prime + 1 \right) - l_1 \left( l_1 + 1 \right) - \Lambda \left( \Lambda + 1 \right)}{2}\right)
\end{align}

where the coefficients <math>C_\Lambda \left( 1 \right)</math> are the angular coefficients from the overlap integral

\begin{align}
\int & d\Omega\, \mathcal{Y}^{M*}_{l_1^\prime l_2^\prime L}(\hat{r}_1,\hat{r}_2)\, \mathcal{Y}^{M}_{l_1 l_2 L}(\hat{r}_1,\hat{r}_2)\\
= & \sum_\Lambda C_\Lambda(1)P_\Lambda(\cos\theta_{12}).
\end{align}

Hint: Use the fact that <math>l_1^2</math> is Hermitian so that
<math>
\int d\tau (l_1^2\mathcal{Y}^\prime)^*q(r)\mathcal{Y} = \int d\tau
\mathcal{Y}^{\prime *}l_1^2(q(r)\mathcal{Y})
</math>
with <math>\ \vec{l}_1= \frac{1}{i}\vec{r}_1\times\nabla_1</math>.

It is also useful to use
<math>\begin{eqnarray}\ \
\cos\theta\, P_L(\cos\theta) &=& \frac{1}{2L+1}[L\,P_{L-1}(\cos\theta) + (L+1)P_{L+1}(\cos\theta)]
\\
(\cos^2\theta-1)P_L(\cos\theta) &=&
\frac{L(L-1)}{(2L-1)(2L+1)}P_{L-2}(\cos\theta)\nonumber
-\frac{2(L^2+L-1)}{(2L-1)(2L+3)}P_L(\cos\theta) + \frac{(L+1)(L+2)}{(2L+1)(2L+3)}P_{L+2}(\cos\theta)\nonumber\\
&=& \frac{L(L-1)}{(2L-1)(2L+1)}[P_{L-2}(\cos\theta) - P_L(\cos\theta)]
+ \frac{(L+1)(L+2)}{(2L+1)(2L+3)}[P_{L+2}(\cos\theta) - P_L(\cos\theta)]
\end{eqnarray}</math>

together with a double application of the integral recursion relation
<math>
I_{L+1}\left(\frac{1}{r}\frac{d}{dr}q(r)\right) =
(2L+1)I_L\left(\frac{1}{r_1r_2}q(r)\right)+I_{L-1}\left(\frac{1}{r}\frac{d}{dr}q(r)\right)
</math>

Of course <math>l_1^2\mathcal{Y}^M_{l_1l_2L}(\hat{r}_1,\hat{r}_2)=l_1(l_1+1)\mathcal{Y}^M_{l_1l_2L}(\hat{r}_1,\hat{r}_2)</math>. Begin by expanding
<math>
l_1^2(g\mathcal{Y}) = \mathcal{Y}l_1^2g + g\,l_1^2\mathcal{Y} + 2(\vec{l}_1g)\cdot(\vec{l}_1\mathcal{Y})
</math>, and show that

<math>
\begin{eqnarray}\ \
(\vec{l}_1g)\cdot(\vec{l}_1\mathcal{Y}) & = & \frac{r_1^2}{r}\frac{dg}{dr}\vec{r}_2\cdot \nabla_1\mathcal{Y}\,\,\,\,\mbox{ and}\\
l_1^2g(r) &=& 2r_1r_2\cos\theta\frac{1}{r}\frac{dg}{dr} + r_1^2r_2^2(\cos^2\theta - 1)
\frac{1}{r}\frac{d}{dr}\left(\frac{1}{r}\frac{dg}{dr}\right).
\end{eqnarray}
</math>

The proof amounts to showing that the term <math>l_1^2g(r)</math> can be replaced by
<math>\Lambda(\Lambda+1)g(r)</math> after multiplying by <math>P_\Lambda(\cos\theta)</math>
and integrating by parts with respect to the radial integrations over <math>r_1</math>, <math>r_2</math> and
<math>r \equiv |\vec{r}_1 - \vec{r}_2|</math>. Remember that <math>\cos\theta</math> is just a short-hand notation for the radial function <math>(r_1^2 + r_2^2 - r^2)/(2r_1r_2)</math>.

==General Hermitian Property==

Each combination of terms of the form
\begin{equation}
<f>_R = a^2f_1+af_2+abf_3+b^2f_4+bf_5+af_6\nabla_1^Y+bf_7\nabla^Y_2+f_8(Y)
\end{equation}
can be rewritten
\begin{equation}
<f>_L= (a_+-a)^2f_1+(a_+-a)f_2 + (a_+-a)(b_+-b)f_3+(b_+-b)^2f_4
+(b_+-b)f_5+(a_+-a)f_6\nabla^{Y\prime}_1+(b_+-b)f_7\nabla^{Y\prime}_2+f_8)Y^\prime
\end{equation}
where as usual <math>\nabla_1^Y</math> acts only on the spherical harmonic part of the wave function denoted for short by <math>Y</math>.
Since these must be equal for arbitrary <math>a</math> and <math>b</math>,
\begin{equation}
a_+^2f_1+a_+f_2+_+b_+f_3+b_+f_4+b_+f_5+a_+f_6\nabla^{Y\prime}_1 +
b_+f_7\nabla^{Y\prime}_2+f_8(Y^\prime)-f_8(Y)=0
\end{equation}

Adding the corresponding expression with <math>Y</math> and <math>Y^\prime</math> interchanged
yields
<math>\label{III}
a_+^2f_1+a_+f_2+a_+b_+f_3+b_+^2f_4+b_+f_5+\frac{1}{2}a_+f_6\nabla^+_1+\frac{1}{2}b_+f_7\nabla^+_2=0\tag{III}
</math>

Subtracting gives
<math>\label{IV}
f_8(Y)-f_8(Y^\prime)=-\frac{1}{2}[a_+f_6\nabla^-_1+b_+f_7\nabla_2^-]\tag{IV}
</math>
<math>\label{V}
a[-2a_+f_1-2f_2-b_+f_3-f_6\nabla^+_1]=0\tag{V}
</math>
<math>\label{VI}
b[-2b_+f_4-2f_5-a_+f_3-f_7\nabla^+_2]=0\tag{VI}
</math>

Adding the two forms gives

\begin{eqnarray}
<f>_R+<f>_L=\frac{1}{2}(a_+^2+a_-^2)f_1+a_+f_2+\frac{1}{2}(a_+b_++a_-b_-)f_3\nonumber\\
\ +\frac{1}{2}(b_+^2+b_-^2)f_4+b_+f_5+\frac{1}{2}f_6(a_+\nabla_1^++a_-\nabla^-_1)\nonumber\\
\ +\frac{1}{2}(b_+\nabla_2^++b_-\nabla^-_2)+f_8(Y)+f_8(Y^\prime)\nonumber
\end{eqnarray}

Subtracting <math>\frac{x}{2}\times({\rm I})</math> gives

\begin{eqnarray}\ \
<f>_R+<f>_L = \frac{1}{2}[(1-x)a_+^2 + a_-^2]f_1 +
(1-\frac{x}{2})a_+f_2+\frac{1}{2}[(1-x)a_+b_++a_-b_-]f_3\nonumber\\
+\frac{1}{2}[(1-x)b_+^2+b_-^2]f_4+(1-\frac{x}{2})f_5b_++\frac{1}{2}f_6[(1-\frac{x}{2})a_+\nabla_1^++a_-\nabla^-_1]\nonumber\\
+\frac{1}{2}[(1-\frac{x}{2})b_+\nabla^+_2+b_-\nabla^-_2]+f_8(Y)+f_8(Y^\prime)\nonumber
\end{eqnarray}

If <math>x=1</math>,
\begin{eqnarray}
<f>_R+<f>_L &=&
\frac{1}{2}[a_-^2f_1+a_+f_2+a_-b_-f_3+b_-^2f_4+b_+f_5\nonumber\\
&+&\left(\frac{1}{2}a_+\nabla^+_1+a_-\nabla^-_1\right)f_6+\left(\frac{1}{2}b_+\nabla^+_2+b_-\nabla^-_2\right)f_7]+f_8(Y)+f_8(Y^\prime)\nonumber
\end{eqnarray}

The General Hermitian Property for arbitrary <math>x</math> gives
\begin{eqnarray}\ \
\nabla_1^2=
\frac{1}{4}\{\frac{1}{r_1^2}[(1-x)a_+^2+a_-^2+2(1-\frac{x}{2})a_+-2[l_1(l_1+1)+l_1^\prime(l_1^\prime+1)]\nonumber\\
-\frac{2}{r_1}[(1-x)\alpha_+a_++\alpha_-a_-+2(1-\frac{x}{2})\alpha_+]+(1-x)\alpha_+^2
+ \alpha^2_-]\nonumber\\
+\frac{2(r_1-r_2\cos\theta)}{r_1r^2}[(1-x)(a_+-\alpha_+r_1)c_++(a_--\alpha_-r_2)c_-]\nonumber\\
-\frac{2r_2}{r_1r^2}[(1-\frac{x}{2})c_+\hat{r}_2\cdot\hat{\nabla}_1^++c_-\hat{r}_2\hat{\nabla}_1^-]+\frac{(1-x)c_+^2+c_-^2+2(1-\frac{x}{2})c_+}{r^2}\}\nonumber
\end{eqnarray}.

Use
<math>\begin{eqnarray}\ \
\frac{-2r_2}{r_1r^2}[(1-\frac{x}{2})c_+\hat{r}_2\hat{\nabla}_1^++c_-\hat{r}_2\hat{\nabla}_1^-]=\frac{2}{r_1^2}\left[(1-\frac{x}{2})\Lambda(\Lambda+1)-\frac{c_-}{c_+}[l_1^\prime(l_1^\prime+1)+l_1(l_1+1)]\right]\end{eqnarray}</math>,

<math>\begin{eqnarray}
\frac{2(r_1-r_2\cos\theta)}{r_1r^2}(a_--\alpha_-r_1)c_- = 2\frac{c_-}{c_+}\left[\frac{-1}{r_1^2}[a_-(a_++1)]+\frac{1}{r_1}[a_-\alpha_++\alpha_-(a_++2)]-\alpha_-\alpha_+\right]\nonumber
\end{eqnarray}</math>,

and
<math>\begin{eqnarray}\ \
\frac{2(r_1-r_2\cos\theta)}{r_1r^2}(a_+-\alpha_+r_1)c_+ = 2\left[-\frac{1}{r_1^2}[a_+(a_++1)]+\frac{1}{r_1}[a_+\alpha_++\alpha_+(a_++2)]-\alpha_+^2\right]
\end{eqnarray}</math>.

Substituting into <math>\nabla_1^2</math> gives

\begin{eqnarray}
\nabla^2_1 = \frac{1}{4}\{\frac{1}{r_1^2}[-(1-x)a_+^2 + a_-^2 +xa_+ +
2(1-\frac{x}{2})\Lambda(\Lambda+1)\nonumber\\
-2l_1(l_1+1)(1-\frac{c_-}{c_+}) -
2l_1^\prime(l_1^\prime+1)(1+\frac{c_-}{c_+})-\frac{2c_-a_-}{c_+}(a_++1)]\nonumber\\
- \frac{2}{r_1}\left[-(1-x)\alpha_+(a_++2)+\alpha_-a_-+2(1-\frac{x}{2})\alpha_+
-\frac{c_-}{c_+}[a_-\alpha_++\alpha_-(a_++2)]\right]\nonumber\\
-(1-x)\alpha_+^2+\alpha_-^2-2\frac{c_-}{c_+}\alpha_-\alpha_+ +\frac{1}{r^2}\left[(1-x)c_+^2+c_-^2+2(1-\frac{x}{2})c_+\right]\}\nonumber
\end{eqnarray}

This has the form
<math>
\nabla_1^2 =
\frac{1}{4}\left[\frac{A_1}{r_1^2}+\frac{B_1}{r_1}+\frac{C_1}{r^2}+D_1\right]
</math>
with

<math>\begin{array}{ccc}
A_1=-(1-x)a_+^2+a_-^2
+xa_++2(1-\frac{x}{2})\Lambda(\Lambda+1)-2l_1(l_1+1)(1-\frac{c_-}{c_+})\
-2l_1^\prime(l_1^\prime+1)(1+\frac{c_-}{c_+})-\frac{2c_-a_-}{c_+}(a_++1) \ \
\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \
\end{array}
</math>

<math>
B_1 =
2\left[(1-x)\alpha_+(a_++2)-\alpha_-a_--2(1-\frac{x}{2})\alpha_++\frac{c_-}{c_+}[a_-\alpha_++\alpha_-(a_++2)]\right]
</math>

<math>
C_1 = (1-x)c_+^2+ c_-^2 +2(1-\frac{x}{2})c_+
</math>

<math>
D_1 = -1(1-x)\alpha_+^2 +\alpha_-^2 - 2\frac{c_-}{c_+}\alpha_-\alpha_+
</math>

The complete Hamiltonian is then (in Z-scaled a.u.)

<math>\begin{eqnarray}
H =-\frac{1}{2}\nabla^2_1 -\frac{1}{2}\nabla^2_2 -\frac{1}{r_1} -\frac{1}{r_2} +\frac{Z^{-1}}{r}\nonumber\\
= -\frac{1}{2}\nabla^2_1 -\frac{1}{2}\nabla^2_2 -\frac{1}{r_1} -\frac{Z-1}{Zr_2}+Z^{-1}\left(\frac{1}{r}-\frac{1}{r_2}\right)\nonumber\\
\ =-\frac{1}{8}\left[\frac{A_1}{r_1^2}+\frac{B_1+8}{r_1}+\frac{C_1}{r^2}+D_1+D_2 +\frac{A_2}{r_2^2} + \frac{B_2 + 8(Z-1)/Z}{r_2}+\frac{C_2}{r^2}\right]\nonumber
+ Z^{-1}\left(\frac{1}{r}-\frac{1}{r_2}\right)\nonumber
\end{eqnarray}</math>

The screened hydrogenic energy is
<math>\begin{eqnarray}
E_{SH}=-\frac{1}{2}\left[\frac{1}{n_1^2}+ \left(\frac{Z-1}{Z}\right)^2\frac{1}{n_2^2}\right]
\end{eqnarray}</math>
so that

<math>\begin{eqnarray}
H-E_{SH} &=& -\frac{1}{8}[
\frac{A_1}{r_1^2}+\frac{B_1+8}{r_1}+\frac{C_1}{r^2} + D_1 + D_2
-\frac{4}{n_1^2} -\left(\frac{Z-1}{Z}\right)^2\frac{4}{n_2^2}\nonumber
+\frac{A_2}{r_2}+\frac{B_2+8(Z-1)/Z}{r_2}+\frac{C_2}{r^2}] +
Z^{-1}\left(\frac{1}{r} -\frac{1}{r_2}\right)\nonumber
\end{eqnarray}</math>

==Optimization of Nonlinear Parameters==
- The traditional method of performing Hylleraas calculations is to write the
basis set in the form

<math>
\Psi = \sum_{i,j,k}c_{i,j,k}\varphi_{i,j,k}(\alpha,\beta)\pm exchange
</math>

with

<math>
\varphi_{i,j,k}(\alpha,\beta) = r_1^ir_2^jr_{12}^ke^{-\alpha r_1 - \beta
r_2}\mathcal{Y}^M_{l_1l_2L}.
</math>

The usual procedure is to set <math>\alpha = Z</math> so that it represents the inner
<math>1s</math> electron, and then to vary <math>\beta</math> so as to minimize the energy.

[[File:OptNon.jpg|500px]]

Since <math>\beta</math> appears in <math>\Psi</math> as a non-linear parameter, the entire
calculation must be repeated for each value of <math>\beta</math>. However, the minimum
becomes progressively smaller as the basis set is enlarged.

[[File:OptNon02.jpg|500px]]

*Difficulties

1. If the basis set is constructed so that <math> i +j +j \leq N</math>, the the
number of terms is
<math>
(N+1)(N+2)(N+3)/6
</math>
N=14 already gives 680 terms and an accuracy of about 1 part in <math>10^{10}</math> for
low-lying states. A substantial improvement in accuracy would require much
larger basis sets, together with multiple precision arithmetic to avoid loss
of significant figures when high powers are included.

2. The accuracy rapidly deteriorates as one goes to more highly excited
states - about 1 significant figure is lost each time the principle quantum
number is increased.

*Cure

We have found that writing basis sets in the form
\begin{eqnarray}
\Psi &=& \sum_{i,j,k}\left(c^{(1)}_{i,j,k}\varphi_{ijk}(\alpha_1\beta_1) +
c_{ijk}^{(2)}\varphi_{ijk}(\alpha_2\beta_2)\right) \pm exchange \nonumber\\
&=& \psi({\bf r}_1,{\bf r}_2) \pm \psi({\bf r}_2,{\bf r}_1)\nonumber
\end{eqnarray}
so that each combination of powers is included twice with different nonlinear
parameters gives a dramatic improvement in accuracy for basis sets of about
the same total size.

However, the optimization of the nonlinear parameters is now much more
difficult, and an automated procedure is needed.

If
<math>
E = \frac{<\Psi|H|\Psi>}{<\Psi|\Psi>}</math> and we assume <math><\Psi|\Psi> = 1</math>, then

<math>
\frac{\partial E}{\partial\alpha_t} = -2 <\Psi|H-E|r_1\psi({\bf r}_1,{\bf
r}_2;\alpha_t)\pm r_21\psi({\bf r}_2,{\bf
r}_1;\alpha_t)>
</math>
where
<math>
\psi({\bf r}_1,{\bf r}_2;\alpha_t) =
\sum_{i,j,k}c_{ijk}^{(t)}\varphi_{ijk}(\alpha_t\beta_t)
</math>

*Problem

1. Prove the above.

2. Prove that there is no contribution from the implicit dependence of
<math>c_{ijk}^{(t)}</math> on <math>\alpha_t</math> if the linear parameters have been optimized.

==The Screened Hydrogenic Term==

If the Hamiltonian is written in the form (in Z-scaled a.u.)
<math>\begin{eqnarray}\
H = H_0({\bf r}_1,Z) + H_0({\bf r}_2,Z-1) +
Z^{-1}\left(\frac{1}{r}-\frac{1}{r_2}\right)
\end{eqnarray}</math>
with
<math>\
H_0({\bf r}_1,Z) = -\frac{1}{2}\nabla_1^2 - \frac{1}{r_1}\nonumber
</math> and
<math>
H_0({\bf r}_2,Z-1) = -\frac{1}{2}\nabla_2^2 - \frac{Z-1}Z{r_2}\nonumber
</math>,
then the eigenvectors of <math>H_0({\bf r},Z) +H_0({\bf r}_2,Z-1)</math> are products of
hydrogenic orbitals
<math>
\Psi_0=\psi_0(1s,Z)\psi_0(nl,Z-1)
</math>
and the eigenvalue is
<math>\begin{eqnarray}
E_{SH} = \left[-\frac{1}{2} - \left(\frac{Z-1}{Z}\right)^2\frac{1}{2n^2}\right]Z^2 a.u.
\end{eqnarray}</math>
called the screened hydrogenic eigenvalue. From highly excited states,
<math>E_{SH}</math> and <math>\Psi_0</math>are already excellent approximations.

For example, for the 1s8d states, the energies are

<math>\begin{eqnarray}\\
E(1s8d^1D)&=& -2.007\,816\,512\,563\,81 {\rm\ a.u.}\nonumber\\
E(1s8d^3D)&=& - 2.007\,817\,934\,711\,71 {\rm\ a.u.}\nonumber\\
E_{SH} &=& -2.007\,812\,5 {\rm\ a.u.} \nonumber
\end{eqnarray}</math>

It is therefore advantageous to include the screened hydrogenic terms in the
basis set so that the complete trial function becomes
<math>
\Psi = c_0\Psi_0 +
\sum_{ijk}\left[c_{ijk}^{(1)}\varphi_{ijk}(\alpha_1,\beta_1)+c_{ijk}^{(2)}\varphi(\alpha_2,\beta_2)\right]\pm
{\rm exchange}
</math>

Also, the variational principal can be re-expressed in the form
<math>\begin{eqnarray}
E= E_{SH}+\frac{<\Psi|H-E_{SH}|\Psi>}{<\Psi|\Psi>}
\end{eqnarray}</math>
so that the <math>E_{SH}</math> term can be cancelled analytically from the matrix
elements.

For example
<math>
<\Psi_0|H-E_{SH}|\Psi_0> = 0
</math>

and
<math>
<\varphi_{ijk}|H-E_{SH}|\Psi_0> =
Z^{-1}<\varphi_{ijk}|\frac{1}{r}-\frac{1}{r_2}|\Psi_0>
</math>

Recall that
<math>\begin{eqnarray}\ \
I_0(a,b,c) = \frac{2}{c+2}\sum^{[c+1]/2}_{i=0}\left(\begin{array}{c}
c+2\\
2i+1
\end{array}\right) \left[f(p,q;\beta) + f(p^\prime,q^\prime;\alpha)\right]
\end{eqnarray}</math>
where

<math>\begin{eqnarray}
f(p,q;x) =
\frac{q!}{x^{q+1}(\alpha+\beta)^{p+1}}\sum^p_{j=0}\frac{(p+j)!}{j!}\left(\frac{x}{\alpha+\beta}\right)^j
\end{eqnarray}</math>

<math>
\begin{array}{cc}
p=a+2i+2 & p^\prime = b+2i+2\\
q= b+c-2i+2 & q^\prime =a+c-2i+2
\end{array}
</math>

If <math>\beta\ll\alpha</math>, the <math>f(p,q;\beta)\gg f(p^\prime,q^\prime;\alpha)</math> and the
<math>i=0</math> term is the dominant contribution to <math>f(p,q;\beta)</math>. However, since
this term depends only on the sum of powers <math>b+c</math> for <math>r_2</math>, it cancels exactly
from the matrix element of
<math>
\frac{1}{r_{12}} - \frac{1}{r_2}
</math>
and can therefore
be omitted in the calculations of integrands, thereby saving many significant
figures. This is especially valuable when <math>b+c</math> is large and <math>a</math> is small.

==Small Corrections==
*Mass Polarization
If the nuclear mass is not taken to be infinite, then the Hamiltonian is
<math>\begin{eqnarray}\ \
H = \frac{P_N^2}{2M} + \sum_{i=1}\frac{p_i^2}{2m} + V
\end{eqnarray}</math>

[[File:SmallCor.jpg|500px]]

where <math>M =M_A-nm</math> is the nuclear mass and <math>m</math> is the electron mass.

Change variables to the C of M
<math>\begin{eqnarray}
{\bf R} = \frac{1}{M+nm}\left[M{\bf r}_N + m({\bf r}_1 + {\bf r}_2 +
\ldots)\right]
\end{eqnarray}</math>

and relative variables

<math>
{\bf s}_i = {\bf r}_i - {\bf r}_N.
</math>

Then

<math>\begin{eqnarray}\ \ \ \ \
H = \frac{1}{2(M+nm)}{\bf p}_R^2+\frac{1}{2m}\sum_{i=1}^n{\bf p}_{s_i}^2
+\frac{1}{2M}\sum_{i,k}{\bf p}_{s_i}\cdot {\bf p}_{s_k} + V({\bf s}_1, {\bf
s}_2, \ldots)\nonumber
= \frac{1}{M+nm}{\bf p}_R^2+\frac{1}{2\mu}\sum_{i=1}^n{\bf
p}_{s_i}^2+\frac{1}{2M}\sum_{i\neq k}{\bf p}_{s_i}\cdot {\bf p}_{s_k}+V({\bf s}_1, {\bf
s}_2, \ldots)\nonumber
\end{eqnarray}</math>

where
<math>\begin{eqnarray}
\mu = \left(\frac{1}{M} + \frac{1}{m}\right)^{-1}
\end{eqnarray}</math>

If <math>n=2</math>, the Schrödinger equation is
<math>\begin{eqnarray}\ \ \
\left[ -\frac{\hbar^2}{2\mu}(\nabla^2_{s_1}+ \nabla^2_{s_2}) -
\frac{\hbar^2}{M}\nabla_{s_1}\nabla_{s_2}-\frac{Ze^2}{s_1}-\frac{Ze^2}{s_2}+\frac{e^2}{s_{12}}\right]\psi
= E\psi
\end{eqnarray}</math>.

Define the reduced mass Bohr radius
<math>\begin{eqnarray}
a_\mu = \frac{\hbar^2}{\mu e^2} = \frac{m}{\mu}a_0
\end{eqnarray}</math>
and

<math>\begin{eqnarray}
\rho_i = \frac{Zs_i}{a_\mu}\nonumber\\
\varepsilon = \frac{E}{Z^2(e^2/a_\mu)}\nonumber
\end{eqnarray}</math>

The Schrödinger equation then becomes
<math>\begin{eqnarray}\ \ \
\left[
-\frac{1}{2}(\nabla^2_{\rho_1}+\nabla^2_{\rho_2})-\frac{\mu}{M}\vec{\nabla}_{\rho_1}\cdot\vec{\nabla}_{\rho_2} - \frac{1}{\rho_1}-\frac{1}{\rho_2}+\frac{Z^{-1}}{\rho_{12}}\right]\psi = \varepsilon \psi
\end{eqnarray}</math>

The effects of finite mass are thus

1. A reduced mass correction to all energy levels of <math>(e^2/a_\mu)/(e^2/a_0)
= \mu/m</math>, i.e.
<math>\begin{eqnarray}
E_M=\frac{\mu}{m}E_\infty
\end{eqnarray}</math>

2. A specific mass shift given in first order perturbation theory by
<math>\begin{eqnarray}
\Delta\varepsilon =
-\frac{\mu}{M}<\psi|\nabla_{\rho_1}\cdot\nabla_{\rho_2}|\psi>
\end{eqnarray}</math>
or

<math>\begin{eqnarray}
\Delta\varepsilon =
-\frac{\mu}{M}<\psi|\nabla_{\rho_1}\cdot\nabla_{\rho_2}|\psi>\frac{e^2}{a_\mu}\nonumber\\
= -\frac{\mu^2}{mM}<\psi|\nabla_{\rho_1}\cdot\nabla_{\rho_2}|\psi>\frac{e^2}{a_0}\nonumber
\end{eqnarray}</math>

Since <math>\nabla_{\rho_1}\cdot\nabla_{\rho_2}</math> has the same angular properties as
<math>{\bf\rho}_1 \cdot {\bf\rho}_2</math>, the operator is like the product of two
dipole operators. For product type wave functions of the form
$
\psi = \psi(1s)\psi(nl) \pm \mbox{exchange},
$
the matrix element vanishes for all but P-states.

=Radiative Transitions=

==Transition Integrals and the Line Strength==

For an electric dipole transition <math>\gamma'\,L'\,S'\,J'\,M' \rightarrow \gamma\,L\,S\,J\,M</math>, the line strength is defined by
\begin{equation}
S = \sum_{M',\mu,M}|\langle\gamma'\,L'\,S'\,J'\,M'|r_\mu|\gamma\,L\,S\,J\,M\rangle|^2
\end{equation}
Using the Wigner-Eckart theorem, this is related to the reduced matrix element by
\begin{eqnarray}
S(J',J) = |\langle\gamma'\,L'\,S'\,J'||r||\gamma\,L\,S\,J\,M\rangle|^2
\sum_{M',\mu,M}\left|\left(\matrix{J' & 1 & J \cr
-M'& \mu &M \cr}\right)\right|^2
\end{eqnarray}
Using the sum rule for the 3-<math>j</math> symbols
\begin{equation}
\sum_{M',\mu,M}\left|\left(\matrix{J' & 1 & J \cr
-M'& \mu &M \cr}\right)\right|^2 = 1
\end{equation}
we see that the line strength is identical to the reduced matrix element
\begin{equation}
S(J',J) = |\langle\gamma'\,L'\,S'\,J'||r||\gamma\,L\,S\,J\,M\rangle|^2
\end{equation}
For a nonrelativistic calculation in <math>LS</math> coupling, <math>S = S'</math> and we can next strip out the dependence on <math>J</math> and <math>J'</math> by use of Eq.\ (7.1.7) of Edmonds { Angular Momentum in Quantum Mechanics} with <math>k = 1</math> to obtain
\begin{equation}
S(J',J) = (-1)^{L'+S+J+1}[(2J+1)(2J'+1)]^{1/2}\left\{
\matrix{L' & J'& S\cr
J & L & 1\cr}\right\}\langle\gamma' L'|| r || \gamma L\rangle
\end{equation}
where <math>\langle\gamma' L'|| r || \gamma L\rangle</math> is the reduced matrix element if spin is neglected. Note that if spin is neglected, then the line-strength is simply
\begin{equation}
S(L',L) = |\langle\gamma' L'|| r || \gamma L\rangle|^2
\end{equation}
To calculate the reduced matrix element, use a particular example along with the Wigner-Eckart theorem
such as
\begin{equation}
\langle\gamma'\, L'|| r || \gamma\, L\rangle = (-1)^{L'}\langle\gamma'\, L'\, 0| z | \gamma\, L\, 0\rangle{\Large /}\left(\matrix{L' & 1 & L \cr
0 & 0 & 0 \cr}\right)
\end{equation}

===Correlated Two-electron Integrals===
For the case of two-electron integrals, <math>z = z_1 + z_2</math> and <math>r = r_1 + r_2</math>. As discussed in G.W.F. Drake, Phys. Rev. A 18, 820 (1978), the general integral in Hylleraas coordinates for the integral over the <math>z_1</math> term can be written in the form
\begin{equation}
\label{int1}
\langle\gamma'\, L'\, 0| z_1 | \gamma\, L\, 0\rangle = \sum_\Lambda C_\Lambda I_\Lambda(\gamma',\gamma)
\end{equation}
where <math>I_\Lambda(\gamma',\gamma)</math> stands for the purely radial integral
\begin{equation}
\label{radial}
I_\Lambda(\gamma',\gamma) = \int_0^\infty r_1\,dr_1\int_0^\infty r_2\,dr_2\int_{|r_1-r_2|}^{r_1+r_2}r_{12}\, dr_{12}
R_{\gamma'}^*\,r_1\, R_\gamma P_\Lambda(\cos\theta_{12})
\end{equation}
and <math>\cos\theta_{12} = (r_1^2 + r_2^2 - r_{12}^2)/(2r_1r_2)</math> is a short-hand notation for a radial function of <math>r_1</math>, <math>r_2</math> and <math>r_{12}</math>. Also, <math>P_\Lambda(\cos\theta_{12})</math> is a Legendre polynomial,
and, <math>R_\gamma = R_\gamma(r_1,r_2,r_{12})</math> is the radial part of the Hylleraas wave function for state <math>\gamma</math> such
that the total wave function is <math>\psi({\bf r}_1,{\bf r}_2) = R_\gamma(r_1,r_2,r_{12}){\cal Y}_{l_1,l_2,L}^M(\hat{\bf r}_1,\hat{\bf r}_2)\pm {\rm exchange}</math>, normalized to unity, and <math>{\cal Y}_{l_1,l_2,L}^M(\hat{\bf r}_1,\hat{\bf r}_2)</math> is a vector coupled product of spherical harmonics. The coefficents <math>C_\Lambda</math> in Eq.\ (\ref{int1}) can be calculated from Eq.\ (32) of the above 1978 paper. They need to be calculated separately for he direct-direct and the direct-exchange contributions. For the reduced matrix element of <math>z_1</math>, the leading 3-<math>j</math> symbol can be dropped, leaving (for the case <math>k = 1</math>)
\begin{eqnarray}
C_\Lambda^{\rm reduced} &=& \frac{(-1)^{l_2'+L'+L+1}}{2} (l_1,l_1',l_2,l_2',L,L')^{1/2}
\sum_{\lambda={\rm max}(l_1-1,0)}^{l_1+1} (\lambda,\Lambda,1)
\left(\matrix{l_1 & 1 & \lambda \cr
0 & 0 & 0 \cr}\right)
\left(\matrix{l_1'& \lambda & \Lambda \cr
0 & 0 & 0 \cr}\right)\nonumber\\
&\times&\left(\matrix{l_2' & 1_2 & \Lambda \cr
0 & 0 & 0 \cr}\right)
\left\{\matrix{L' & \lambda & l_2 \cr
\Lambda& l_2' & l_1' \cr}\right\}
\left\{\matrix{L' & \lambda & l_2 \cr
l_1 & L & 1 \cr}\right\}
\end{eqnarray}
where the notation <math>(a,b,c,\cdots)</math> means <math>(2a+1)(2b+1)(2c+1)\cdots</math>.
A similar expression can be obtained for the angular coefficients for the <math>z_2</math> term by interchanging the subscripts 1 and 2 throughout and multiplying by <math>(-1)^{L+L'+1}</math>.

===Example===
As an example, for the <math>1s^2\;^1S - 1s2p\;^1P</math> transition of helium, the only nonvanishing <math>C_\Lambda</math> angular coefficients are <math>C_1 = 1/2</math> for the direct-direct term (i.e. <math>P_1(\cos\theta_{12}) = \cos\theta_{12}</math> is included in the radial integral (\ref{radial})) and <math>C_0 = 1/2</math> for the direct-exchange term (i.e. <math>P_0(\cos\theta_{12}) = 1</math> is included in the radial integral (\ref{radial})). The results for infinite nuclear mass are <math>S(1\;^1S-2\;^1P) = (0.781\,127\,158\,667)^2</math> and <math>f(1\;^1S-2\;^1P) = -0.276\,164\,703\,736</math> where
\begin{equation}
f(1\;^1S-2\;^1P) = \frac{2}{3}[E(1\;^1S) - E(2\;^1P)]S(1\;^1S-2\;^1P)
\end{equation}
is the absorption oscillator strength.

Theory Notes

2020-12-14T00:32:55Z

Drake: /* General Hermitian Property */

<center>
<H1>Notes on solving the Schrödinger equation in Hylleraas coordinates for heliumlike atoms
Gordon W.F. Drake
Department of Physics, University of Windsor
Windsor, Ontario, Canada N9B 3P4
(Transcribed from hand-written notes and edited by Lauren Moffatt. Last revised November 22, 2017 to correct a missing subscript in the expression for $B_1$ in the Hamiltonian in Sect. 11 General Hermitian Property.)
</H1>
</center>

==Introduction==
These notes describe some of the technical details involved in solving the Schrödinger equation for a heliumlike atom of nuclear charge Z in correlated Hylleraas coordinates. The standard for computational purposes is to express all quantities in a form that is valid for electronic states of arbitrary angular momentum, and to express matrix elements of the Hamiltonian in a manifestly Hermitian symmetrized form. Further information and derivations of the mathematical techniques for angular integrals and radial recursion relations can be found in the paper
G.W.F. Drake, Phys. Rev. A <math>{\bf 18}</math>, 820 (1978).

The starting point is the two-electron Schrödinger equation for infinite nuclear mass
\begin{equation} \left[-\frac{\hbar^2}{2m}(\nabla^2_1 +\nabla^2_2) - \frac{Ze^2}{r_1} -
\frac{Ze^2}{r_2}+\frac{e^2}{r_{12}} \right]\psi = E\psi\nonumber
\end{equation}
where <math>m</math> is the electron mass, and <math>r_{12} = |{\bf r}_1 - {\bf r}_2|</math> (see diagram below).

[[File:Pic1.jpg|250px]]

Begin by rescaling distances and energies so that the Schrödinger equation can be expressed in a dimensionless form. The dimensionless Z-scaled distance is defined by <math>\begin{eqnarray} \rho = \frac{Zr}{a_0} \end{eqnarray} </math> where <math> \begin{eqnarray} a_0 = \frac{\hbar^2}{me^2} \end{eqnarray} </math> is the atomic unit (a.u.) of distance equal to the Bohr radius <math>0.529\, 177\, 210\, 92(17) \times 10^{-10}</math> m. Then

\begin{equation}[-\frac{\hbar^2}{2m}Z^2(\frac{me^2}{\hbar^2})^2(\nabla^2_{\rho_1}+\nabla^2_{\rho_2})
- Z^2\frac{e^2}{a_0}\rho^{-1}_1 - Z^2\frac{e^2}{a_0}\rho^{-1}_2 +
\frac{e^2}{a_0}Z\rho^{-1}_{12}]\psi= E\psi\nonumber\end{equation}

But <math>\begin{eqnarray} \frac{\hbar^2}{m}(\frac{me^2}{\hbar^2})^2 = \frac{e^2}{a_0} \end{eqnarray} </math> is the hartree atomic unit of of energy <math>(E_h = 27.211\,385\,05(60)</math> eV, or equivalently <math>E_h/(hc) = 2194.746\,313\,708(11)</math> m<math>^{-1})</math>. Therefore, after multiplying through by <math>a_0/(Ze)^2</math>, the problem to be solved in Z-scaled dimensionless units becomes

\begin{equation} \left[-\frac{1}{2}(\nabla^2_{\rho_1}+\nabla^2_{\rho_2}) -
\frac{1}{\rho_1} - \frac{1}{\rho_2} + \frac{Z^{-1}}{\rho_{12}}\right]\psi =
\varepsilon\psi\nonumber \end{equation}

where <math>\begin{eqnarray}\varepsilon = \frac{Ea_0}{(Ze)^2} \end{eqnarray}</math> is the energy in Z-scaled atomic units. For convenience, rewrite this in the conventional form
\begin{equation}
H\psi = \varepsilon\psi\nonumber \end{equation}
where (using <math>r</math> in place of <math>\rho</math> for the Z-scaled distance)
\begin{equation}
H = -\frac{1}{2}(\nabla^2_1+\nabla^2_2) - \frac{1}{r_1}-\frac{1}{r_2} + \frac{Z^{-1}}{r_{12}}
\nonumber\end{equation}
is the atomic Hamiltonian for infinite nuclear mass.

==The Hartree Fock Method==

For purposes of comparison, and to define the correlation energy, assume that <math>\psi({\bf r}_1,{\bf r}_2)</math> can be written in the separable product form
\begin{equation}
\psi({\bf r}_1,{\bf r}_2) = \frac{1}{\sqrt{2}} \left[ u_1(r_1) u_2(r_2) \pm u_2(r_1) u_1(r_2) \right] \nonumber
\end{equation}
for the <math>1s^2\ ^1\!S</math> ground state. Because of the <math>\frac{1}{r_{12}}</math> term in the Schrödinger equation, the exact solution connot be expressed in this form as a separable product. However, the Hartree-Fock approximation corresponds to finding the best possible solution to the Schrödinger equation
\begin{equation}
H\psi({\bf r}_1,{\bf r}_2) = E\psi({\bf r}_1,{\bf r}_2)\nonumber
\end{equation}
that can nevertheless be expressed in this separable product form, where as before
\begin{equation}
H = -\frac{1}{2}(\nabla^2_1+\nabla^2_2) - \frac{1}{r_1}- \frac{1}{r_2} + \frac{Z^{-1}}{r_{12}}\nonumber
\end{equation}
is the full two-electron Hamiltonian. To find the best solution, substitute into <math>\left\langle\psi \lvert H-E \rvert \psi\right\rangle</math> and require this expression to be stationary
with respect to arbitrary infinitesimal variations <math>\delta u_1</math> and <math>\delta
u_2</math> in <math>u_1</math> and <math>u_2</math> respectively; i.e.

\begin{equation}
\frac{1}{2} \left\langle \delta u_1(r_1)u_2(r_2) \pm u_2(r_1)\delta
u_1(r_2) \lvert H-E \rvert u_1(r_1) u_2(r_2)\pm u_2(r_1) u_1(r_2)\right\rangle\nonumber
\end{equation}

\begin{equation}
=\int\delta u_1(r_1) d{\bf r}_1 \left\{ \int d{\bf r}_2 u_2(r_2) \left( H - E \right) \left[ u_1(r_1)u_2(r_2) \pm u_2(r_1) u_1(r_2) \right] \right\}\nonumber
\end{equation}

\begin{equation}
= 0 \nonumber
\end{equation}

for arbitrary <math> \delta u_1(r_1).\nonumber </math> Therefore <math>\{\int d{\bf r}_2 \ldots \} = 0</math>.

Similarly, the coefficient of <math>\delta u_2</math> would give

\begin{equation}
\int d{\bf r}_1 u_1(r_1) \left( H-E \right) \left[ u_1(r_1) u_2(r_2) \pm
u_2(r_1) u_1(r_2) \right] = 0\nonumber
\end{equation}

Define

\begin{equation}
I_{12} = I_{21} = \int d{\bf r}\, u_1(r)u_2(r), \nonumber
\end{equation}

\begin{equation}
H_{ij} = \int d{\bf r}\, u_i(r)(-\frac{1}{2}\nabla - \frac{1}{r})u_j(r), \nonumber
\end{equation}

\begin{equation}
G_{ij}(r) = \int d{\bf r}^\prime u_i(r^\prime)\frac{1}{|{\bf r} - {\bf r}\prime|}u_j(r^\prime)\nonumber
\end{equation}

Then the above equations become the pair of integro-differential equations

\begin{equation}
[ H_0 - E + H_{22}+G_{22}(r)]u_1(r) = \mp [ I_{12}(H_0-E) + H_{12}+G_{12}(r)]u_2(r)\nonumber
\end{equation}

\begin{equation}
[H_0-E+H_{11}+G_{11}(r)]u_2(r) = \mp [I_{12}(H_0-E) + H_{12}+G_{12}(r)]u_1(r)\nonumber
\end{equation}

These must be solved self-consistently for the "constants" <math>I_{12}</math> and
<math>H_{ij}</math> and the function <math>G_{ij}(r)</math>.

The Hartree Fock energy is <math>E \simeq -2.87\ldots </math>a.u. while the exact energy is <math>E = -2.903724\ldots </math>a.u.

The difference is called the "correlation energy" because it arises from the
way in which the motion of one electron is correlated to the other. The
Hartree Fock equations only describe how one electron moves in the average field
provided by the other (mean-field theory).

==Configuration Interaction==
Expand
\begin{equation}
\psi({\bf r}_1,{\bf r}_2)= C_0u^{(s)}_1(r_1)u^{(s)}_1(r_2) + C_1u^{(P)}_1({\bf r}_1)u^{(P)}_1({\bf r}_2){\cal Y}^0_{1,1,0}(\hat{\bf r}_1, \hat{\bf r}_2)+C_2u^{(d)}_1({\bf r}_1)u^{(d)}_2({\bf r}_2){\cal Y}^0_{2,2,0}(\hat{\bf r}_1, \hat{\bf r}_2)+... \pm {\rm\ exchange}
\end{equation}
where
\begin{equation}
{\cal Y}^M_{l_1,l_2,L}(\hat{\bf r}_1, \hat{\bf r}_2)=\sum_{m_1,m_2}Y^{m_1}_{l_1}({\bf r}_1)Y^{m_2}_{l_2}({\bf r}_2)\times <l_1l_2m_1m_2\mid LM>
\end{equation}

This works, but is slowly convergent, and very laborious. The best CI calculations are accurate to <math> ~10^{-7}</math> a.u.

==Hylleraas Coordinates==
[E.A. Hylleraas, Z. Phys. <math>{\bf 48}, 469(1928)</math> and <math>{\bf 54}, 347(1929)</math>]
suggested using the co-ordinates <math>r_1</math>, <math>r_2</math> and <math>r_{12}</math> or equivalently

\begin{eqnarray}
s &=& r_1 + r_2, \nonumber\\
t &=& r_1-r_2, \nonumber\\
u &=& r_{12}\nonumber
\end{eqnarray}

and writing the trial functions in the form

\begin{equation}
\Psi({\bf r}_1,{\bf r}_2) = \sum^{i+j+k\leq
N}_{i,j,k}c_{i,j,k}r_1^{i+l_1}r_2^{j+l_2}r_{12}^ke^{-\alpha r_1 - \beta r_2}
\mathcal{Y}^M_{l_1,l_2,L}(\hat{r}_1,\hat{r}_2)\pm \text{exchange}\nonumber
\end{equation}

Diagonalizing H in this non-orthogonal basis set is equivalent to solving
<math> \begin{eqnarray}
\frac{\partial E}{\partial c_{i,j,k}} = 0\nonumber
\end{eqnarray} </math>
for fixed <math>\alpha</math> and <math>\beta</math>.

The diagonalization must be repeated for different values of <math>\alpha</math> and
<math>\beta</math> in order to optimize the non-linear parameters.

==Completeness==
The completeness of the above basis set can be shown by first writing
\begin{equation}
r_{12}^2 = r_1^2 + r_2^2 - 2r_1r_2\cos\theta_{12}
\end{equation}
and, from the spherical harmonic addition theorem,
\begin{equation}
\cos(\theta_{12})=\frac{4\pi}{3}\sum^1_{m=-1}Y^{m*}_1(\theta_1,\varphi_1)Y^m_1(\theta_2,\varphi_2)
\end{equation}
Consider first the S-states. The <math>r_{12}^0</math> terms are like the ss terms in a CI calculation. The
<math>r_{12}^2</math> terms bring in p-p type contributions, and the higher powers bring
in d-d, f-f etc. type terms. In general

\begin{equation}
P_l(\cos\theta_{12}) = \frac{4\pi}{2l+1}\sum^l_{m=-l}{Y^{m}_l}^*(\theta_1,\varphi_1)Y^m_l(\theta_2,\varphi_2)\nonumber
\end{equation}

For P-states, one would have similarly

\begin{array}{lr}
r_{12}^0 & (sp)P\nonumber\\
r_{12}^2 & (pd)P\nonumber\\
r_{12}^4 & (df)P\nonumber\\
\vdots & \vdots\nonumber
\end{array}

For D-states

\begin{array}{lr}
r_{12}^0 & (sd)D & (pp^\prime)D\nonumber\\
r_{12}^2 & (pf)D & (dd^\prime)D\nonumber\\
r_{12}^4 & (dg)D & (ff^\prime)D\nonumber\\
\vdots & \vdots & \vdots\nonumber
\end{array}

In this case, since there are two "lowest-order" couplings to form a
D-state, both must be present in the basis set, i.e.

\begin{equation}
\Psi(r_2,r_2) = \sum c_{ijk}r_1^ir_2^{j+2}r_{12}^ke^{-\alpha r_1-\beta
r_2}\mathcal{Y}^M_{022}(\hat{r}_1,\hat{r}_2)
+\sum d_{ijk}r_1^{i+1}r_2^{j+1}r_{12^k}e^{-\alpha^\prime r_1 - \beta^\prime
r_2}\mathcal{Y}^M_{112}(\hat{r}_1,\hat{r}_2)\nonumber
\end{equation}

For F-states, one would need <math>(sf)F</math> and <math>(pd)F</math> terms.

For G-states, one would need <math>(sg)G</math>, <math>(pf)G</math> and <math>(dd^\prime)G</math> terms.

Completeness of the radial functions can be proven by considering the
Sturm-Liouville problem

\begin{equation}
\left(-\frac{1}{2}\nabla^2-\frac{\lambda}{r_s}-E\right)\psi({\bf r}) = 0\nonumber
\end{equation}

or

\begin{equation}
\left(-\frac{1}{2}\frac{1}{r^2}\frac{\partial}{\partial r}\left({r^2}\frac{\partial}{\partial r}\right) -
\frac{l\left(l+1\right)}{2r^2} - \frac{\lambda}{r} - E\right)u(r) = 0.\nonumber
\end{equation}

For fixed E and variable <math>\lambda</math> (nuclear charge).

The eigenvalues are <math>\lambda_n = (E/E_n)^{1/2}</math>, where <math>E_n =- \frac{1}{2n^2}</math>

[[File:Pic2.jpg|500px]]

and the eigenfunctions are

\begin{equation}
u_{nl}(r) =
\frac{1}{(2l+1)!}\left(\frac{(n+l)!}{(n-l-1)2!}\right)^{1/2}(2\alpha)^{3/2}e^{-\alpha
r}\nonumber
\times \ \left( 2 \alpha r \right)^l\,_{1}F_{1}\left(-n+l+1,2l+2; 2 \alpha r \right)
\end{equation}

with <math>\alpha = (-2E)^{1/2}</math> and <math>n\geq l+1</math>. The confluent hypergeometric function <math> _1F_1(a,b,;z)</math> then denotes a finite polynomial since <math>a=-n+l+1</math> is a negative integer or zero.

Unlike the hydrogen spectrum, which has both a discrete part for <math>E<0</math> and a
continuous part for <math>E>0</math>, this forms an entirely discrete set of finite
polynomials, called Sturmian functions. They are orthogonal with respect to
the potential, i.e.

\begin{equation}
\int^\infty_0 r^2dr \left( u_{n^\prime l}(r)\frac{1}{r}u_{nl}(r) \right) = \delta_{n,n^\prime}\nonumber
\end{equation}

Since they become complete in the limit <math>n\rightarrow\infty</math>, this assures the
completeness of the variational basis set.

[See also B Klahn and W.A. Bingel Theo. Chim. Acta (Berlin) 44, 9 and 27
(1977)].

==Solutions of the Eigenvalue Problem==

For convenience, write

\[
\Psi({\bf r}_1,{\bf r}_2) = \sum^N_{m=1}c_m\varphi_m\nonumber
\]

where <math>m</math> represents the <math>m</math>'th combination of <math>i,j,k</math> and

\begin{equation}
\varphi_{ijk}=r_1^ir_2^jr_{12}^ke^{-\alpha r_1 - \beta
r_2}\mathcal{Y}^M_{l_1,l_2,L}(\hat{r}_1,\hat{r}_2) \pm \mbox{exchange}\nonumber
\end{equation}

\begin{align}

&
\left( \begin{array}{lr}
\cos(\theta) & \sin(\theta)\\
-\sin(\theta) & \cos(\theta)
\end{array} \right)
\left( \begin{array}{lr}
H_{11} & H_{12} \\
H_{12} & H_{22}
\end{array}\right)
\left( \begin{array}{lr}
\cos(\theta) & -\sin(\theta)\\
\sin(\theta) & \cos(\theta)
\end{array} \right)
\\[5pt]
= & \left(\begin{array}{lr}
cH_{11}+sH_{12} & cH_{12} + sH_{22}\\
-sH_{11} + cH_{12} & -sH_{12} + cH_{22}
\end{array}\right)
\left( \begin{array}{cc}
c & -s \\
s & c
\end{array}\right)
\\[5pt]
= & \left(\begin{array}{lr}
c^2H_{11}+s^2H_{22} + 2csH_{12} & (c^2-s^2)H_{12}+cs(H_{22}-H_{11})\\
(c^2-s^2)H_{12}+cs(H_{22}-H_{11}) & s^2H_{11}+c^2H_{22}-2csH_{12}
\end{array}\right)
\end{align}

Therefore

\begin{equation}
\left[\cos^2(\theta)-\sin^2(\theta)\right]H_{12} =
\cos(\theta)\sin(\theta)(H_{11}-H_{22})\nonumber
\end{equation}

and

\begin{equation}
\tan(2\theta) = \frac{2H_{12}}{H_{11}-H_{22}}\nonumber
\end{equation}

i.e.

\begin{equation}
\cos(\theta)=\left(\frac{r+\omega}{2r}\right)^{1/2}\nonumber
\end{equation}

\begin{equation}
\sin(\theta)=-\mbox{sgn}(H_{12})\left(\frac{r-\omega}{2r}\right)^{1/2}\nonumber
\end{equation}

where

\begin{equation}
\omega = H_{22}-H_{11}\nonumber
\end{equation}

\begin{equation}
r=\left(\omega^2+4H_{12}^2\right)^{1/2}\nonumber
\end{equation}

\begin{equation}
E_1=\frac{1}{2}\left(H_{11}+H_{22}-r\right)\nonumber
\end{equation}

\begin{equation}
E_2=\frac{1}{2}\left(H_{11}+H_{22}+r\right)\nonumber
\end{equation}

===Brute Force Method===

- Gives all the eigenvalues and eigenvectors, but it is slow

- First orthonormalize the <math>N</math>-dimensional basis set; i.e. form linear combinations

\begin{equation}
\Phi_m = \sum_{n=1}^N\varphi_nR_{nm}\nonumber
\end{equation}

such that
\begin{equation}
\langle \Phi_m | \Phi_n \rangle = \delta_{m,n}
\end{equation}

This can be done by finding an orthogonal tranformation T such that

\begin{equation}
T^TOT = I = \left(
\begin{array}{ccccc}
I_1 & 0 & 0 & \ldots & 0 \\
0 & I_2 & 0 & \ldots & 0 \\
0 & 0 & I_3 & \ldots & 0 \\
\vdots & \vdots & \vdots & \ddots & \vdots \\
0 & 0 & 0 & \ldots & I_N
\end{array}\right);
\end{equation}

\begin{equation}
O_{mn} = \langle \varphi_m | \varphi_n \rangle \nonumber
\end{equation}

and then applying a scale change matrix

\begin{equation}
S = \left(\begin{array}{ccccc}
\frac{1}{I_1^{1/2}} & 0 & 0 & \ldots & 0\\
0 & \frac{1}{I_2^{1/2}} & 0 & \ldots & 0 \\
0 & 0 & \frac{1}{I_3^{1/2}} & \ldots & 0 \\
\vdots & \vdots & \vdots & \ddots & \vdots \\
0 & 0 & 0 & \ldots & I_N^{1/2}
\end{array}\right)= S^T\nonumber
\end{equation}

Then

\begin{equation}
S^TT^TOTS = 1
\end{equation}

i.e.

\begin{equation}
R^TOR = 1
\end{equation}

with <math>R=TS</math>.

If H is the matrix with elements <math>H_{mn} = \langle \varphi_m | \varphi_n \rangle </math>, then H
expressed in the <math>\Phi_m</math> basis set is

\begin{equation}
H^\prime = R^THR.\nonumber
\end{equation}

We next diagonalize <math>H^\prime</math> by finding an orthogonal transformation W such
that

\begin{equation}
W^T H^\prime W = \lambda = \left(
\begin{array}{cccc}
\lambda_1 & 0 & \ldots & 0 \\
0 & \lambda_2 & \ldots & 0 \\
\vdots & \vdots & \ddots & \vdots \\
0 & 0 & \ldots & \lambda_N
\end{array}\right)\nonumber
\end{equation}

The q'th eigenvector is

\begin{equation}
\Psi^{(q)} = \sum_{n=1}^N\Phi_n W_{n,q} = \sum_{n,n^\prime} \varphi_{n^\prime} R_{n^\prime ,n} W_{n,q} \nonumber
\end{equation}

i.e.

\begin{equation}
c_{n^\prime}^{(q)} = \sum_{n=1}^N R_{n^\prime n} W_{n,q}
\end{equation}

===The Power Method===

- Based on the observation that if H has one eigenvalue, <math>\lambda_M</math>, much bigger than all the rest, and <math>\chi = \left(
\begin{array}{c}a_1\\a_2\\\vdots\end{array}\right)</math>
is an arbitrary starting vector, then <math>\chi = \sum_{q=1}^N x_q\Psi^{(q)}</math>.

\begin{equation}
(H)^n\chi = \sum_{q=1}^N x_q \lambda^n_q\Psi^{(q)}\nonumber\rightarrow x_M\lambda_M^n\Psi^{(M)}\nonumber
\end{equation}

provided <math>x_M\neq 0</math>.

To pick out the eigenvector corresponding to any eigenvalue, write the original problem in the form

\begin{align}
H\Psi & = \lambda O \Psi \nonumber \\
\left( H - \lambda_q O \right) \Psi & = \left( \lambda - \lambda_q \right) O \Psi \nonumber
\end{align}

Therefore,

\begin{equation}
G \Psi = \frac{1}{\lambda - \lambda_q } \Psi \nonumber \\
\end{equation}

where <math>G=(H-\lambda_qO)^{-1}O</math> with eigenvalues <math>\frac{1}{\lambda_n - \lambda_q}</math>.

By picking <math>\lambda_q</math> close to any one of the <math>\lambda_n</math>, say
<math>\lambda_{n^\prime}</math>, then <math>\frac{1}{\lambda_n-\lambda_q}</math> is much larger for
<math>n=n^\prime</math> than for any other value. The sequence is then

\begin{array}{c}
\chi_1=G\chi\nonumber\\
\chi_2=G\chi_1\nonumber\\
\chi_3=G\chi_2\nonumber\\
\vdots\nonumber
\end{array}

until the ratios of components in <math>\chi_n</math> stop changing.

- To avoid matrix inversion and multiplication, note that the sequence is equivalent to
\begin{equation}
F \chi_n = \left( \lambda - \lambda_q \right) O \chi_{n-1} \nonumber \\
\end{equation}
where <math>F = H-\lambda_qO</math>. The factor of <math>\left( \lambda - \lambda_q \right)</math> can be
dropped because this only affects the normalization of <math>\chi_n</math>. To find <math>\chi_n</math>, solve
\begin{equation}
F \chi_n = O \chi_{n-1} \nonumber
\end{equation}
(N equations in N unknowns). Then
\begin{equation}
\lambda = \frac{ \langle \chi_n | H | \chi_n \rangle}{\langle \chi_n | \chi_n \rangle} \nonumber
\end{equation}

==Matrix Elements of H==

\begin{equation}
H=-\frac{1}{2}\nabla^2_1 -\frac{1}{2}\nabla^2_2 - \frac{1}{r_1} - \frac{1}{r_2} +\frac{Z^{-1}}{r_{12}}\nonumber
\end{equation}

Taking <math>r_1, r_2</math> and <math>r_{12}</math> as independent variables,

\begin{equation}
\nabla_1^2 = \frac{1}{r_1^2}\frac{\partial}{\partial r_1}\left(r_1^2\frac{\partial}{\partial r_1}\right)+ \frac{1}{r_{12}^2}\frac{\partial}{\partial r_{12}}
\left(r_{12}^2\frac{\partial}{\partial r_{12}}\right)\nonumber -\frac{l_1(l_1+1)}{r_1^2}+2(r_1 - r_2\cos(\theta_{12}))\frac{1}{r_{12}}\frac{\partial^2}{\partial r_1 \partial r_{12}} \nonumber - 2(\nabla_1^Y \cdot {\bf r}_2)\frac{1}{r_{12}}\frac{\partial}{\partial r_{12}}\nonumber
\end{equation}

where <math>\nabla_1^Y</math> acts only on the spherical harmonic part of the wave function and the diagram

[[File:Pic3.jpg|400px]]

defines the complete set of 6 independent variables is <math>r_1</math>, <math>r_2</math>, <math>r_{12}</math>, <math>\theta_1</math>, <math>\varphi_1</math>, <math>\chi</math>.

If <math>r_{12}</math> were not an independent variable, then one could take the volume element to be

\begin{equation}
d\tau = r_1^2dr_1\sin(\theta_1)d\theta_1d\varphi_1r_2^2dr_2\sin(\theta_2)d\theta_2d\varphi_2.\nonumber
\end{equation}

However, <math>\theta_2</math> and <math>\varphi_2</math> are no longer independent variables. To eliminate them, take the point <math>{\bf r}_1</math> as the
origin of a new polar co-ordinate system, and write

\begin{equation}
d\tau=-r_1^2dr_1\sin(\theta_1)d\theta_1d\varphi_1r_{12}^2dr_{12}\sin(\psi)d\psi d\chi\nonumber
\end{equation}

and use

\begin{equation}
r_2^2=r_1^2+r_{12}^2 +2r_1r_{12}\cos(\psi)
\end{equation}

Then for fixed <math>r_1</math> and <math>r_{12}</math>,

\begin{equation}
2r_2dr_2 = -2r_1r_{12}\sin(\psi)d\psi
\end{equation}

Thus

\begin{equation}
d\tau= r_1dr_1r_2dr_2r_{12}dr_{12}\sin(\theta_1)d\theta_1d\varphi_1d\chi
\end{equation}

The basic type of integral to be calculated is

\begin{align}
I \left( l_1, m_1, l_2, m_2; R \right) & = \int \sin \left( \theta_1 \right) d\theta_1 d\varphi_1 d\chi Y^{m_1}_{l_1} \left( \theta_1, \varphi_1 \right)^{*} Y^{m_2}_{l_2} \left( \theta_2, \varphi_2 \right) \\
& \times \int r_1dr_1r_2dr_2r_{12}dr_{12}R \left( r_1, r_2, r_{12} \right)
\end{align}

Consider first the angular integral. <math>Y^{m_2}_{l_2} \left( \theta_2, \varphi_2 \right)</math> can be expressed in terms of the independent variables <math>\theta_1, \varphi_1, \chi</math> by use of the rotation matrix relation
\begin{equation}
Y^{m_2}_{l_2}( \theta_2, \varphi_2) = \sum_m \mathcal{D}^{(l_2)}_{m_2,m}(\varphi_1,\theta_1,\chi)^*Y^m_{l_2}( \theta_{12}, \varphi)
\end{equation}
where <math>\theta, \varphi</math> are the polar angles of <math>{\bf r}_2</math> relative to <math>{\bf r}_1</math>. The angular integral is then

\begin{align}
I_{ang} & = \int^{2 \pi}_{0} d\chi \int^{2\pi}_{0} d\varphi_1 \int^{\pi}_{0} \sin \left( \theta_1 \right) d\theta_{1} Y^{m_1}_{l_1} \left( \theta_1, \varphi_1 \right)^{*} \\
& \times \sum_m \mathcal{D}^{(l_2)}_{m_2,m} \left( \varphi_1,\theta_1,\chi \right)^{*} Y^{m}_{l_2} \left( \theta_{12},\varphi \right) \\
\end{align}

Use

\begin{equation}
Y^{m_1}_{l_1} \left( \theta_1, \varphi_1 \right)^{*} = \sqrt{\frac{2l_1+1}{4\pi}}\mathcal{D}^{(l_1)}_{m_1,0} \left( \varphi_1, \theta_1, \chi \right)
\end{equation}

together with the orthogonality property of the rotation matrices (Brink and Satchler, p 147)

\begin{align}
& \int \mathcal{D}^{(j)*}_{m,m^\prime}\mathcal{D}^{(J)}_{M,M^\prime}\sin(\theta_1)d\theta_1d\varphi_1d\chi \\
& = \frac{8\pi^2}{2j+1}\delta_{jJ}\delta_{mM}\delta_{m^\prime M^\prime}\nonumber
\end{align}

to obtain

\begin{align}
I_{ang} & = \sqrt{ \frac{2l_1+1}{4\pi}} \frac{8\pi^2}{2l_1+1} \delta_{l_1,l_2} \delta_{m_1,m_2} Y^{0}_{l_2} \left( \theta_{12}, \varphi \right) \\
& = 2 \pi \delta_{l_1,l_2} \delta_{m_1,m_2} P_{l_2} \left( \cos \theta_{12} \right)
\end{align}

since

\begin{equation}
Y^0_{l_2} \left( \theta_{12}, \varphi \right) = \sqrt{\frac{2l_1+1}{4\pi}} P_{l_2} \left( \cos \left( \theta_{12} \right) \right)
\end{equation}

Note that <math>P_{l_2} \left( \cos \theta \right)</math> is just a short hand expression for a radial function because

\begin{equation}
\cos \theta_{12} = \frac{r_1^2 + r_2^2 - r_{12}^2}{2r_1r_2}
\end{equation}

The original integral is thus

\begin{equation}
I \left( l_1, m_1, l_2, m_2; R \right) = 2 \pi \delta_{l_1,l_2} \delta_{m_1,m_2} \int^{\infty}_{0} r_{1} dr_{1} \int^{\infty}_{0} r_{2} dr_{2} \int^{r_{1} + r_{2}}_{|r_{1} - r_{2}|} r_{12} dr_{12} R \left( r_{1}, r_{2}, r_{12} \right) P_{l_2} \left( \cos \theta_{12} \right)
\end{equation}

where again

\begin{equation}
\cos \theta_{12} = \frac{r_{1}^{2} +r^{2}_{2} - r_{12}^{2}}{2 r_{1} r_{2}}
\end{equation}

is a purely radial function.

The above would become quite complicated for large <math>l_2</math> because <math>P_{l_2}(\cos \theta_{12})</math> contains terms up to <math>\left( \cos \theta_{12} \right)^{l_2}</math>. However, recursion relations exist which allow any integral containing <math>P_l \left( \cos \theta_{12} \right)</math> in terms of those containing just <math>P_0 \left( \cos \theta_{12} \right) = 1</math> and <math>P_1 \left( \cos \theta_{12} \right) = \cos \theta_{12}</math>.

==Radial Integrals and Recursion Relations==

The basic radial integral is [see G.W.F. Drake, Phys. Rev. A 18, 820 (1978)]

\begin{align}
I_0 \left( a, b, c \right) & =
\int^{\infty}_{0} r_{1} dr_{1} \int^{\infty}_{r_{1}} r_{2} dr_{2} \int^{r_{1}+r_{2}}_{r_{2} - r_{1}} r_{12} dr_{12} r_{1}^{a} r_{2}^{b} r_{12}^{c} e^{-\alpha r_{1} - \beta r_{2}} + \int^{\infty}_{0} r_{2} dr_{2} \int^{\infty}_{r_{2}} r_{1} dr_{1} \int^{r_{1} + r_{2}}_{r_{1} - r_{2}} r_{12} dr_{12} r_{1}^{a} r_{2}^{b} r_{12}^{c} e^{- \alpha r_{1} - \beta r_{2}} \\
& = \frac{2}{c+2} \sum^{[(c+1)/2]}_{i=0}
\left(\begin{array}{c}
c+2 \\
2i+1
\end{array} \right)
\left\{ \frac{q!}{\beta^{q+1} \left( \alpha + \beta \right)^{p+1}} \sum_{j=0}^{q} \frac{\left( p + j \right)!}{j!} \left( \frac{\beta}{\alpha + \beta} \right)^j + \frac{q^{\prime}!}{\alpha^{q^{\prime}+1} \left( \alpha + \beta \right)^{p^{\prime}+1}} \sum^{q^{\prime}}_{j=0} \frac{\left( p^{\prime} + j \right)!}{j!} \left( \frac{\alpha}{\alpha + \beta} \right)^j \right\}
\end{align}

where

\begin{array}{ll}
p = a + 2i + 2 & p^{\prime} = b + 2i + 2 \\
q = b + c - 2i + 2 & q^{\prime} = a + c - 2i + 2
\end{array}

The above applies for <math> a, b \geq -2, c\geq -1</math>.
<math>[x]</math> means "greatest integer in" <math>x</math>.

Then

\begin{align}
I_1 \left( a, b, c \right) & = \int d\tau_{r} r_{1}^{a} r_{2}^{b} r_{12}^{c} e^{ -\alpha r_{1} - \beta r_{2}} P_{1}\left( \cos \theta \right) \\
& = \frac{1}{2} \left( I_{0} \left( a + 1, b - 1, c \right) + I_{0} \left( a - 1, b + 1, c \right) - I_{0} \left( a - 1, b - 1, c + 2 \right) \right)
\end{align}

===The Radial Recursion Relation===

Recall that the full integral is

\begin{equation}
I(l_1m_1,l_2m_2;R)=2\pi\delta_{l_1,l_2}\delta_{m_1,m_2}I_{l_2}(R)
\end{equation}

where

\begin{equation}
I_{l_2}(R) = \int d\tau_r R(r_1,r_2,r_{12})P_{l_2}(\cos \theta_{12})
\end{equation}

To obtain the recursion relation, use

\begin{equation}
P_l(x) = \frac{[P^{\prime}_{l+1}(x)-P^{\prime}_{l-1}(x)]}{2l+1}
\end{equation}

with

\begin{equation}
P_{l+1}^\prime (x) = \frac{d}{dx}P_{l+1}(x)
\end{equation}

Here <math>x=\cos\theta_{12}</math> and

\begin{align}
\frac{d}{d \cos \theta_{12}} & = \frac{dr_{12}}{d \cos \theta_{12}} \frac{d}{dr_{12}} \\
& = -\frac{r_{1} r_{2}}{r_{12}} \frac{d}{dr_{12}}
\end{align}

Then,

\begin{equation}
I_{l}\left( R \right) = - \int d\tau_r R \frac{r_{1} r_{2}}{r_{12}} \frac{d}{dr_{12}} \frac{ \left[ P_{l+1} \left( \cos \theta_{12} \right) - P_{l-1} \left( \cos \theta_{12} \right) \right]}{2l+1}
\end{equation}

The <math>r_{12}</math> part of the integral is

\begin{align}
& \int^{r_{1} + r_{2}}_{|r_{1} - r_{2}|} r_{12} dr_{12} R \frac{r_{1} r_{2}}{r_{12}} \frac{d}{dr_{12}} \left[P_{l+1}-P_{l-1} \right] \\
& = R r_{1} r_{2} \left[ P_{l+1} - P_{l-1} \right]^{r_{1} + r_{2}}_{|r_{1} - r_{2}|} - \int^{r_{1} + r_{2}}_{|r_{1} - r_{2}|} r_{12} dr_{12} \left( \frac{d}{dr_{12}} R \right) \frac{r_{1} r_{2}}{r_{12}} \frac{ \left[ P_{l+1} - P_{l-1} \right]}{2l+1}
\end{align}

The integrated term vanishes because

\begin{equation}
\cos \theta_{12} = \frac{r_{1}^{2} + r_{2}^{2} - r_{12}^{2}}{2 r_{1} r_{2}}
\begin{array}{lll}
& = -1 & {\rm when} & r_{12}^{2} = \left( r_{1} + r_{2} \right)^{2} \\
& = 1 & {\rm when} & r_{12}^{2} = \left( r_{1} - r_{2} \right)^{2}
\end{array}
\end{equation}

and <math>P_l(1)=1</math>, <math>P_{l}(-1)=(-1)^{l}</math>.

Thus,

\begin{equation}
I_{l+1}\left(\frac{r_1r_2}{r_{12}}R^\prime\right) = \left( 2l+1 \right) I_{l}(R) - I_{l-1} \left( \frac{r_{1} r_{2}}{r_{12}} R^{\prime} \right)
\end{equation}

If

\begin{equation}
R = r_{1}^{a-1} r_{2}^{b-1} r_{12}^{c+2} e^{-\alpha r_{1} - \beta r_{2}}
\end{equation}

then [G.W.F. Drake, Phys. Rev. A 18, 820 (1978)]

\begin{equation}
I_{l+1}(r_1^ar_2^br_{12}^c) =
\frac{2l+1}{c+2} I_{l} \left( r_{1}^{a-1} r_{2}^{b-2} r_{12}^{c+2} \right)+I_{l-1} \left( r_{1}^{a} r_{2}^{b} r_{12}^{c} \right)
\end{equation}

For the case <math>c=-2</math>, take

\begin{equation}
R = r_{1}^{a-1} r_{2}^{b-1} \ln r_{12} e^{ -\alpha r_{1} -\beta r_{2}}
\end{equation}

Then

\begin{equation}
I_{l+1} \left( r_{1}^{a} r_{2}^{b} r_{12}^{c} \right) = I_{l} \left( r_{1}^{a-1} r_{2}^{b-1} \ln r_{12} \right) \left( 2l + 1 \right) - I_{l-1} \left( r_{1}^{a} r_{2}^{b} r_{12}^{-2} \right)
\end{equation}

This allows all <math>I_{l}</math> integrals to be calculated from tables of <math>I_0</math> and
<math>I_1</math> integrals.

===The General Integral===

The above results for the angular and radial integrals can now be combined into a general formula for integrals of the type

\begin{equation}
I = \int \int d{\bf r}_{1} d{\bf r}_{2} R_{1} \mathcal{Y}^{M^{\prime}}_{l^{\prime}_{1} l^{\prime}_{2} L^{\prime}} \left( \hat{r}_{1}, \hat{r}_{2} \right) T_{k_{1} k_{2} K}^{Q} \left( {\bf r}_{1},{\bf r}_{2} \right) R_{2} \mathcal{Y}_{l_{1} l_{2} L}^{M} \left( \hat{r}_{1}, \hat{r}_{2} \right)
\end{equation}

where

\begin{equation}
\mathcal{Y}^{M}_{l_{1} l_{2} L} \left( \hat{r}_{1}, \hat{r}_{2} \right) = \sum_{m_1,m_2} \langle l_{1} l_{2} m_{1} m_{2} | LM \rangle Y^{m_{1}}_{l_{1}} \left( \hat{r}_{1} \right) Y^{m_{2}}_{l_{2}} \left( \hat{r}_{2} \right)
\end{equation}

and

\begin{equation}
T^Q_{k_{1} k_{2} K} \left( {\bf r}_{1}, {\bf r}_{2} \right) = \sum_{q_{1}, q_{2}} \langle k_{1} k_{2} q_{1} q_{2} | KQ \rangle Y_{k_{1}}^{q_{1}} \left( \hat{r}_{1} \right) Y^{q_{2}}_{k_{2}} \left( \hat{r}_{2} \right)
\end{equation}

The basic idea is to make repeated use of the formula

\begin{equation}
Y^{m_{1}}_{l_{1}} \left( \hat{r}_{1} \right) Y^{m_{2}}_{l_{2}} \left( \hat{r}_{1} \right) = \sum_{l m} \left( \frac{\left( 2l_{1}+1 \right) \left(2l_{2}+1 \right) \left(2l+1 \right)}{4 \pi} \right)^{1/2}
\times \left( \begin{array}{ccc}
l_1 & l_2 & l \\
m_1 & m_2 & m
\end{array} \right) \left(\begin{array}{ccc}
l_1 & l_2 & l \\
0 & 0 & 0
\end{array} \right)
Y^{m*}_{l} \left( \hat{r}_{1} \right)
\end{equation}

where

\begin{equation}
Y^{m*}(\hat{r}) = (-1)^mY_l^{-m}(\hat{r})
\end{equation}

and

\begin{equation}
\left(\begin{array}{ccc}
l_1 & l_2 & l \\
m_1 & m_2 & m
\end{array} \right) = \frac{\left( -1 \right)^{l_{1} - l_{2} - m}}{\left( 2l + 1 \right)^{1/2}} \left( l_{1} l_{2} m_{1} m_{2} | l, -m \right)
\end{equation}

is a 3-<math>j</math> symbol. In particular, write

\begin{array}{cc}
Y^{m_1^\prime}_{l_1^\prime}(\hat{r}_1)^{*} \underbrace{Y_{k_{1}}^{q_{1}} \left( \hat{r}_{1} \right) Y^{m_{1}}_{l_{1}} \left( \hat{r}_{1} \right)} & = & \sum_{\Lambda M}(\ldots)Y^{M}_\Lambda(\hat{r}_1)\\
\ \ \ \ \ \ \ \sum_{\lambda_1\mu_1}Y_{\lambda_1}^{\mu_1}(\hat{r}_1) & & \\
& & \\
Y^{m_{2}^{\prime}}_{l_{2}^{\prime}} \left( \hat{r}_{2} \right) \underbrace{Y^{q_{2}}_{k_{2}} \left( \hat{r}_{2} \right) Y_{l_{2}}^{m_{2}} \left( \hat{r}_{2} \right)} & = & \sum_{\Lambda^{\prime} M^{\prime}} \left( \ldots \right)Y_{\Lambda^{\prime}}^{M^{\prime}*} \left( \hat{r}_{2} \right) \\
\ \ \ \ \ \ \ \ \ \sum_{\lambda_{2} \mu_{2}}Y_{\lambda_{2}}^{\mu_{2}} \left( \hat{r}_{2} \right) & &
\end{array}

The angular integral then gives a factor of <math>2 \pi \delta_{\Lambda, \Lambda^{\prime}} \delta_{M, M^{\prime}} P_{\Lambda} \left( \cos \theta_{12} \right)</math>. The total integral therefore reduces to the form

\begin{equation}
I = \sum_{\Lambda} C_{\Lambda} I_{\Lambda} \left( R_{1} R_{2} \right)
\end{equation}

where <math>C_{\Lambda} = \sum_{\lambda_{1}, \lambda_{2}} C_{ \lambda_{1}, \lambda_{2}, \Lambda}</math>. For further details and derivations, including graphical representations, see G.W.F. Drake, Phys. Rev. A <math>{\bf 18}</math>, 820 (1978).

==Graphical Representation==

[[File:GraphRep1.jpg|500px]]

==Matrix Elements of H==
Recall that

\begin{equation}
H = -\frac{1}{2} \nabla^{2}_{1} - \frac{1}{2} \nabla^{2}_{2} - \frac{1}{r_1} - \frac{1}{r_2} + \frac{Z^{-1}}{r_{12}}
\end{equation}

Consider matrix elements of

\begin{equation}
\nabla^2_1 = \frac{1}{r_1^2} \frac{\partial}{\partial r_1}\left(r_1^2\frac{\partial}{\partial r_1} \right) + \frac{1}{r^2}\frac{\partial}{\partial r}\left(r^2\frac{\partial}{\partial r}\right) - \frac{(\vec{l}_1^Y)^2}{r_1^2} + \frac{2 \left( r_1 - r_2 \cos \theta \right)}{r} \frac{\partial^2}{\partial r_1 \partial r} - 2 \left( \nabla^Y_1 \cdot {\bf r}_2 \right) \frac{1}{r} \frac{\partial}{\partial r}
\end{equation}

where <math>r \equiv r_{12}</math> and <math>\cos\theta \equiv \cos\theta_{12}</math>. Also in what follows, define <math>\hat{\nabla}^Y_1 \equiv r_1\nabla^Y_1</math> where <math>\nabla^Y</math> operates only on the spherical harmonic part of the wave function, and similarly for <math>\vec{l}_1^Y</math>.

A general matrix element is

\begin{equation}
\langle r_1^{a^\prime} r_2^{b^\prime} r_{12}^{c^\prime} e^{-\alpha^\prime r_1 - \beta^\prime r_2} \mathcal{Y}^{M^\prime}_{l^\prime_1 l^\prime_2 L^\prime} \left( \hat{r}_1, \hat{r}_2 \right) \left| \nabla^2_1 \right| r_1^a r_2^b r_{12}^c e^{-\alpha r_1 - \beta r_2} \mathcal{Y}^M_{l_1 l_2 L} \left( \hat{r}_1, \hat{r}_2 \right) \rangle
\end{equation}

Since <math>\nabla^2_1</math> is rotationally invariant, this vanishes unless <math>L=L^\prime</math>, <math>M=M^\prime</math>. Also <math>\nabla_1^2</math> is Hermitian, so that the result must be the same whether it operates to the right or left, even though the results look very different. In fact, requiring the results to be the same yields some interesting and useful integral identities as follows:

Put

\begin{equation}
\mathcal{F} = F \mathcal{Y}^M_{l_1 l_2 L} \left( \hat{r}_1, \hat{r_2} \right)
\end{equation}

and

\begin{equation}
F = r_1^a r_2^b r_{12}^c e^{-\alpha r_1 - \beta r_2}
\end{equation}

Then

\begin{equation}
\nabla^2_1 \mathcal{F} = \left\{ \frac{1}{r_1^2} \left[ a \left( a + 1 \right) - l_1 \left( l_1 + 1 \right) \right] + \frac{c \left( c + 1 \right)}{r^2} + \alpha^2\frac{2\alpha \left( a + 1 \right)}{r_1} + \frac{2 \left(r_1 - r_2 \cos \theta \right) }{r_1r^2}c \left[ a - \alpha r_1 \right] - \frac{2c}{r^2} \left( \hat{\nabla}^Y_1 \cdot \hat{r}_2 \right) \frac{r_2}{r_1} \right\} \mathcal{F}
\end{equation}

and

\begin{align}
\langle \mathcal{F}^\prime \left| \nabla^2_1 \right| \mathcal{F} \rangle & = \sum_\Lambda \int d\tau_r F^\prime C_\Lambda \left( 1 \right) P_\Lambda \left( \cos \theta \right) \times \left\{ \frac{1}{r_1^2} \left[ a \left( a + 1 \right) - l_1 \left( l_1 + 1 \right) \right] -\frac{2 \alpha \left( a + 1 \right)}{r_1} + \frac{c \left( c + 1 \right)}{r^2} +\alpha^2 + \frac{2 \left( r_2 - r_2 \cos \theta \right)}{r_1r^2}c \left( a - \alpha r_1 \right) \right\} F \\
& + \sum_\Lambda \int d\tau_r F^\prime C_\Lambda \left( \hat{\nabla}^Y_1 \cdot \hat{r}_2 \right) P_\Lambda \left( \cos \theta \right) \left( \frac{-2c r_2}{r_1 r^2} \right) F
\end{align}

where

\begin{equation}
\int d\tau_r = \int^\infty_0 r_1 dr_1 \int^\infty_0 r_2 dr_2 \int^{r_1+r_2}_{\left| r_1 - r_2 \right|} r dr
\end{equation}

For brevity, let the sum over <math>\Lambda</math> and the radial integrations be understood, and let <math>\left( \nabla_1^2 \right)</math> stand for the terms which appear in the integrand. Then operating to the right gives

\begin{equation}
(\nabla_1^2)_R = \frac{1}{r_1^2} \left[ a \left( a + 1 \right) - l_1 \left( l_1 + 1 \right) \right] + \frac{c \left( c + 1 \right)}{r^2} + \alpha^2 - \frac{2 \alpha \left( a + 1 \right)}{r_1}
+ \frac{2 \left( r_1 - r_2 \cos \theta \right)}{r_1 r^2}c \left( a - \alpha r_1 \right) - \frac{2c r_2}{r_1 r^2} \left( \hat{\nabla}^Y_1 \cdot \hat{r}_2 \right)
\end{equation}

Operating to the left gives

\begin{equation}
(\nabla_1^2)_L = \frac{1}{r_1^2} \left[a^\prime \left( a^\prime + 1 \right) - l_1^\prime \left( l_1^\prime + 1 \right) \right] + \frac{c^\prime \left( c^\prime + 1 \right)}{r^2} + \alpha^{\prime 2} -\frac{2 \alpha^\prime \left( a^\prime + 1 \right)}{r_1}
+ \frac{2 \left( r_1 - r_2 \cos \theta \right)}{r_1 r^2} c^\prime \left( a^\prime - \alpha^\prime r_1 \right) - \frac{2 c^\prime r_2}{r_1 r^2} \left( \hat{\nabla}^{Y^\prime}_1 \cdot \hat{r}_2 \right)
\end{equation}

Now put

\begin{array}{cc}
a_+ = a+a^\prime, & \hat{\nabla}_1^+ = \hat{\nabla}_1^Y + \hat{\nabla}_1^{Y\prime} \\
a_{-} = a-a^\prime, & \hat{\nabla}_1^{-} = \hat{\nabla}_1^Y - \hat{\nabla}_1^{Y\prime}
\end{array}

etc., and substitute <math>a^\prime = a_+ - a</math>, <math>c^\prime = c_+ - c</math>, and <math>\alpha^\prime = \alpha_+ - \alpha</math> in <math>(\nabla^2_1)_L</math>. If <math>a_+</math>, <math>c_+</math> and <math>\alpha_+</math> are held fixed, then the equation

\begin{equation}
(\nabla^2_1)_R = (\nabla^2_1)_L
\end{equation}

must be true for arbitrary <math>a</math>, <math>c</math>, and <math>\alpha</math>. Their coefficients must thus vanish.

This yields the integral relations

\begin{equation}
\label{I}
\frac{\left( r_1 - r_2 \cos \theta \right)}{r_1 r^2} = \frac{1}{c_+} \left( \frac{- \left( a_+ + 1 \right)}{r_1^2} + \frac{\alpha_+}{r_1} \right) \tag{I}
\end{equation}

from the coefficient of <math>a</math> and

\begin{equation}
\label{II}
\frac{ \left( r_1 - r_2 \cos \theta \right) \left( a_+ - \alpha_+ r_1 \right)}{r_1 r^2} = \frac{r_2}{r_1 r^2}\left( \hat{r}_2 \cdot \hat{\nabla}_1^+ \right) - \frac{ \left( c_+ + 1 \right)}{r^2} \tag{II}
\end{equation}

from the coefficient of <math>c</math>. The coefficient of <math>\alpha</math> gives an equation equivalent to (I).

Furthermore, if can be show that (see problem)

\begin{align}
& \sum_\Lambda \int d\tau_r \frac{r^c}{r^2} C_\Lambda \left( \hat{r}_2 \cdot \hat{\nabla}_1^Y \right) P_\Lambda \left( \cos \theta \right) \\
= & \sum_\Lambda \int d\tau_r \frac{r^c}{c r_1 r_2}C_\Lambda \left( 1 \right)P_\Lambda \left( \cos \theta \right) \times \left( \frac{l_1^\prime \left( l_1^\prime + 1 \right) - l_1 \left( l_1 + 1 \right) -\Lambda \left( \Lambda + 1 \right)}{2} \right)
\end{align}

and similarly for <math>\left( \hat{r}_2 \cdot \hat{\nabla}^{Y\prime}_1 \right)</math> with <math>l_1</math> and <math>l_1^\prime</math> interchangeable, then it follows that

\begin{equation}
\frac{r^{c_+}}{r^2} \left( \hat{r}_2 \cdot \hat{\nabla}_1^+ \right) = - \frac{r^{c_+}}{c_+ r_1 r_2} \Lambda \left( \Lambda + 1 \right)
\end{equation}

and

\begin{equation}
\frac{r^{c_+}}{r^2} \left( \hat{r}_2 \cdot \hat{\nabla}_1^- \right) = \frac{r^{c_+}}{c_+ r_1 r_2} \left[ l_1^\prime \left( l_1^\prime + 1 \right) - l_1 \left( l_1 + 1 \right) \right]
\end{equation}

where equality applies after integration and summation over <math>\Lambda</math>.

Thus (II) becomes

\begin{equation}\label{IIa}
\frac{ \left( r_1 - r_2 \cos \theta \right) \left( a_+ - \alpha_+ r_1 \right)}{r_1 r^2} = - \frac{\Lambda \left( \Lambda + 1 \right)}{c_+ r^2_1} - \frac{ \left( c_+ + 1 \right)}{r^2} \tag{IIa}
\end{equation}

===Problem===

Prove the integral relation

\begin{align}
& \sum_\Lambda \int d\tau_r f \left( r_1, r_2 \right) \frac{1}{r} \left( \frac{d}{dr} q(r) \right) C_\Lambda \left( \hat{r}_2 \cdot \hat{\nabla}^Y_1 \right) P_\Lambda \left( \cos \theta \right) \\
= & \sum_\Lambda \int d\tau_r \frac{ f \left( r_1, r_2 \right)}{r_1, r_2} q(r) C_\Lambda \left( 1 \right) P_\Lambda \left( \cos \theta \right) \left(\frac{l_1^\prime \left( l_1^\prime + 1 \right) - l_1 \left( l_1 + 1 \right) - \Lambda \left( \Lambda + 1 \right)}{2}\right)
\end{align}

where the coefficients <math>C_\Lambda \left( 1 \right)</math> are the angular coefficients from the overlap integral

\begin{align}
\int & d\Omega\, \mathcal{Y}^{M*}_{l_1^\prime l_2^\prime L}(\hat{r}_1,\hat{r}_2)\, \mathcal{Y}^{M}_{l_1 l_2 L}(\hat{r}_1,\hat{r}_2)\\
= & \sum_\Lambda C_\Lambda(1)P_\Lambda(\cos\theta_{12}).
\end{align}

Hint: Use the fact that <math>l_1^2</math> is Hermitian so that
<math>
\int d\tau (l_1^2\mathcal{Y}^\prime)^*q(r)\mathcal{Y} = \int d\tau
\mathcal{Y}^{\prime *}l_1^2(q(r)\mathcal{Y})
</math>
with <math>\ \vec{l}_1= \frac{1}{i}\vec{r}_1\times\nabla_1</math>.

It is also useful to use
<math>\begin{eqnarray}\ \
\cos\theta\, P_L(\cos\theta) &=& \frac{1}{2L+1}[L\,P_{L-1}(\cos\theta) + (L+1)P_{L+1}(\cos\theta)]
\\
(\cos^2\theta-1)P_L(\cos\theta) &=&
\frac{L(L-1)}{(2L-1)(2L+1)}P_{L-2}(\cos\theta)\nonumber
-\frac{2(L^2+L-1)}{(2L-1)(2L+3)}P_L(\cos\theta) + \frac{(L+1)(L+2)}{(2L+1)(2L+3)}P_{L+2}(\cos\theta)\nonumber\\
&=& \frac{L(L-1)}{(2L-1)(2L+1)}[P_{L-2}(\cos\theta) - P_L(\cos\theta)]
+ \frac{(L+1)(L+2)}{(2L+1)(2L+3)}[P_{L+2}(\cos\theta) - P_L(\cos\theta)]
\end{eqnarray}</math>

together with a double application of the integral recursion relation
<math>
I_{L+1}\left(\frac{1}{r}\frac{d}{dr}q(r)\right) =
(2L+1)I_L\left(\frac{1}{r_1r_2}q(r)\right)+I_{L-1}\left(\frac{1}{r}\frac{d}{dr}q(r)\right)
</math>

Of course <math>l_1^2\mathcal{Y}^M_{l_1l_2L}(\hat{r}_1,\hat{r}_2)=l_1(l_1+1)\mathcal{Y}^M_{l_1l_2L}(\hat{r}_1,\hat{r}_2)</math>. Begin by expanding
<math>
l_1^2(g\mathcal{Y}) = \mathcal{Y}l_1^2g + g\,l_1^2\mathcal{Y} + 2(\vec{l}_1g)\cdot(\vec{l}_1\mathcal{Y})
</math>, and show that

<math>
\begin{eqnarray}\ \
(\vec{l}_1g)\cdot(\vec{l}_1\mathcal{Y}) & = & \frac{r_1^2}{r}\frac{dg}{dr}\vec{r}_2\cdot \nabla_1\mathcal{Y}\,\,\,\,\mbox{ and}\\
l_1^2g(r) &=& 2r_1r_2\cos\theta\frac{1}{r}\frac{dg}{dr} + r_1^2r_2^2(\cos^2\theta - 1)
\frac{1}{r}\frac{d}{dr}\left(\frac{1}{r}\frac{dg}{dr}\right).
\end{eqnarray}
</math>

The proof amounts to showing that the term <math>l_1^2g(r)</math> can be replaced by
<math>\Lambda(\Lambda+1)g(r)</math> after multiplying by <math>P_\Lambda(\cos\theta)</math>
and integrating by parts with respect to the radial integrations over <math>r_1</math>, <math>r_2</math> and
<math>r \equiv |\vec{r}_1 - \vec{r}_2|</math>. Remember that <math>\cos\theta</math> is just a short-hand notation for the radial function <math>(r_1^2 + r_2^2 - r^2)/(2r_1r_2)</math>.

==General Hermitian Property==

Each combination of terms of the form
\begin{equation}
<f>_R = a^2f_1+af_2+abf_3+b^2f_4+bf_5+af_6\nabla_1^Y+bf_7\nabla^Y_2+f_8(Y)
\end{equation}
can be rewritten
<math>
<f>_L= (a_+-a)^2f_1+(a_+-a)f_2 + (a_+-a)(b_+-b)f_3+(b_+-b)^2f_4
+(b_+-b)f_5+(a_+-a)f_6\nabla^{Y\prime}_1+(b_+-b)f_7\nabla^{Y\prime}_2+f_8)Y^\prime
</math>
where as usual <math>\nabla_1^Y</math> acts only on the spherical harmonic part of the wave function denoted for short by <math>Y</math>.
Since these must be equal for arbitrary <math>a</math> and <math>b</math>,
<math>
a_+^2f_1+a_+f_2+_+b_+f_3+b_+f_4+b_+f_5+a_+f_6\nabla^{Y\prime}_1 +
b_+f_7\nabla^{Y\prime}_2+f_8(Y^\prime)-f_8(Y)=0
</math>

Adding the corresponding expression with <math>Y</math> and <math>Y^\prime</math> interchanged
yields
<math>\label{III}
a_+^2f_1+a_+f_2+a_+b_+f_3+b_+^2f_4+b_+f_5+\frac{1}{2}a_+f_6\nabla^+_1+\frac{1}{2}b_+f_7\nabla^+_2=0\tag{III}
</math>

Subtracting gives
<math>\label{IV}
f_8(Y)-f_8(Y^\prime)=-\frac{1}{2}[a_+f_6\nabla^-_1+b_+f_7\nabla_2^-]\tag{IV}
</math>
<math>\label{V}
a[-2a_+f_1-2f_2-b_+f_3-f_6\nabla^+_1]=0\tag{V}
</math>
<math>\label{VI}
b[-2b_+f_4-2f_5-a_+f_3-f_7\nabla^+_2]=0\tag{VI}
</math>

Adding the two forms gives

\begin{eqnarray}
<f>_R+<f>_L=\frac{1}{2}(a_+^2+a_-^2)f_1+a_+f_2+\frac{1}{2}(a_+b_++a_-b_-)f_3\nonumber\\
\ +\frac{1}{2}(b_+^2+b_-^2)f_4+b_+f_5+\frac{1}{2}f_6(a_+\nabla_1^++a_-\nabla^-_1)\nonumber\\
\ +\frac{1}{2}(b_+\nabla_2^++b_-\nabla^-_2)+f_8(Y)+f_8(Y^\prime)\nonumber
\end{eqnarray}

Subtracting <math>\frac{x}{2}\times({\rm I})</math> gives

\begin{eqnarray}\ \
<f>_R+<f>_L = \frac{1}{2}[(1-x)a_+^2 + a_-^2]f_1 +
(1-\frac{x}{2})a_+f_2+\frac{1}{2}[(1-x)a_+b_++a_-b_-]f_3\nonumber\\
+\frac{1}{2}[(1-x)b_+^2+b_-^2]f_4+(1-\frac{x}{2})f_5b_++\frac{1}{2}f_6[(1-\frac{x}{2})a_+\nabla_1^++a_-\nabla^-_1]\nonumber\\
+\frac{1}{2}[(1-\frac{x}{2})b_+\nabla^+_2+b_-\nabla^-_2]+f_8(Y)+f_8(Y^\prime)\nonumber
\end{eqnarray}

If <math>x=1</math>,
\begin{eqnarray}
<f>_R+<f>_L &=&
\frac{1}{2}[a_-^2f_1+a_+f_2+a_-b_-f_3+b_-^2f_4+b_+f_5\nonumber\\
&+&\left(\frac{1}{2}a_+\nabla^+_1+a_-\nabla^-_1\right)f_6+\left(\frac{1}{2}b_+\nabla^+_2+b_-\nabla^-_2\right)f_7]+f_8(Y)+f_8(Y^\prime)\nonumber
\end{eqnarray}

The General Hermitian Property for arbitrary <math>x</math> gives
\begin{eqnarray}\ \
\nabla_1^2=
\frac{1}{4}\{\frac{1}{r_1^2}[(1-x)a_+^2+a_-^2+2(1-\frac{x}{2})a_+-2[l_1(l_1+1)+l_1^\prime(l_1^\prime+1)]\nonumber\\
-\frac{2}{r_1}[(1-x)\alpha_+a_++\alpha_-a_-+2(1-\frac{x}{2})\alpha_+]+(1-x)\alpha_+^2
+ \alpha^2_-]\nonumber\\
+\frac{2(r_1-r_2\cos\theta)}{r_1r^2}[(1-x)(a_+-\alpha_+r_1)c_++(a_--\alpha_-r_2)c_-]\nonumber\\
-\frac{2r_2}{r_1r^2}[(1-\frac{x}{2})c_+\hat{r}_2\cdot\hat{\nabla}_1^++c_-\hat{r}_2\hat{\nabla}_1^-]+\frac{(1-x)c_+^2+c_-^2+2(1-\frac{x}{2})c_+}{r^2}\}\nonumber
\end{eqnarray}.

Use
<math>\begin{eqnarray}\ \
\frac{-2r_2}{r_1r^2}[(1-\frac{x}{2})c_+\hat{r}_2\hat{\nabla}_1^++c_-\hat{r}_2\hat{\nabla}_1^-]=\frac{2}{r_1^2}\left[(1-\frac{x}{2})\Lambda(\Lambda+1)-\frac{c_-}{c_+}[l_1^\prime(l_1^\prime+1)+l_1(l_1+1)]\right]\end{eqnarray}</math>,

<math>\begin{eqnarray}
\frac{2(r_1-r_2\cos\theta)}{r_1r^2}(a_--\alpha_-r_1)c_- = 2\frac{c_-}{c_+}\left[\frac{-1}{r_1^2}[a_-(a_++1)]+\frac{1}{r_1}[a_-\alpha_++\alpha_-(a_++2)]-\alpha_-\alpha_+\right]\nonumber
\end{eqnarray}</math>,

and
<math>\begin{eqnarray}\ \
\frac{2(r_1-r_2\cos\theta)}{r_1r^2}(a_+-\alpha_+r_1)c_+ = 2\left[-\frac{1}{r_1^2}[a_+(a_++1)]+\frac{1}{r_1}[a_+\alpha_++\alpha_+(a_++2)]-\alpha_+^2\right]
\end{eqnarray}</math>.

Substituting into <math>\nabla_1^2</math> gives

\begin{eqnarray}
\nabla^2_1 = \frac{1}{4}\{\frac{1}{r_1^2}[-(1-x)a_+^2 + a_-^2 +xa_+ +
2(1-\frac{x}{2})\Lambda(\Lambda+1)\nonumber\\
-2l_1(l_1+1)(1-\frac{c_-}{c_+}) -
2l_1^\prime(l_1^\prime+1)(1+\frac{c_-}{c_+})-\frac{2c_-a_-}{c_+}(a_++1)]\nonumber\\
- \frac{2}{r_1}\left[-(1-x)\alpha_+(a_++2)+\alpha_-a_-+2(1-\frac{x}{2})\alpha_+
-\frac{c_-}{c_+}[a_-\alpha_++\alpha_-(a_++2)]\right]\nonumber\\
-(1-x)\alpha_+^2+\alpha_-^2-2\frac{c_-}{c_+}\alpha_-\alpha_+ +\frac{1}{r^2}\left[(1-x)c_+^2+c_-^2+2(1-\frac{x}{2})c_+\right]\}\nonumber
\end{eqnarray}

This has the form
<math>
\nabla_1^2 =
\frac{1}{4}\left[\frac{A_1}{r_1^2}+\frac{B_1}{r_1}+\frac{C_1}{r^2}+D_1\right]
</math>
with

<math>\begin{array}{ccc}
A_1=-(1-x)a_+^2+a_-^2
+xa_++2(1-\frac{x}{2})\Lambda(\Lambda+1)-2l_1(l_1+1)(1-\frac{c_-}{c_+})\
-2l_1^\prime(l_1^\prime+1)(1+\frac{c_-}{c_+})-\frac{2c_-a_-}{c_+}(a_++1) \ \
\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \
\end{array}
</math>

<math>
B_1 =
2\left[(1-x)\alpha_+(a_++2)-\alpha_-a_--2(1-\frac{x}{2})\alpha_++\frac{c_-}{c_+}[a_-\alpha_++\alpha_-(a_++2)]\right]
</math>

<math>
C_1 = (1-x)c_+^2+ c_-^2 +2(1-\frac{x}{2})c_+
</math>

<math>
D_1 = -1(1-x)\alpha_+^2 +\alpha_-^2 - 2\frac{c_-}{c_+}\alpha_-\alpha_+
</math>

The complete Hamiltonian is then (in Z-scaled a.u.)

<math>\begin{eqnarray}
H =-\frac{1}{2}\nabla^2_1 -\frac{1}{2}\nabla^2_2 -\frac{1}{r_1} -\frac{1}{r_2} +\frac{Z^{-1}}{r}\nonumber\\
= -\frac{1}{2}\nabla^2_1 -\frac{1}{2}\nabla^2_2 -\frac{1}{r_1} -\frac{Z-1}{Zr_2}+Z^{-1}\left(\frac{1}{r}-\frac{1}{r_2}\right)\nonumber\\
\ =-\frac{1}{8}\left[\frac{A_1}{r_1^2}+\frac{B_1+8}{r_1}+\frac{C_1}{r^2}+D_1+D_2 +\frac{A_2}{r_2^2} + \frac{B_2 + 8(Z-1)/Z}{r_2}+\frac{C_2}{r^2}\right]\nonumber
+ Z^{-1}\left(\frac{1}{r}-\frac{1}{r_2}\right)\nonumber
\end{eqnarray}</math>

The screened hydrogenic energy is
<math>\begin{eqnarray}
E_{SH}=-\frac{1}{2}\left[\frac{1}{n_1^2}+ \left(\frac{Z-1}{Z}\right)^2\frac{1}{n_2^2}\right]
\end{eqnarray}</math>
so that

<math>\begin{eqnarray}
H-E_{SH} &=& -\frac{1}{8}[
\frac{A_1}{r_1^2}+\frac{B_1+8}{r_1}+\frac{C_1}{r^2} + D_1 + D_2
-\frac{4}{n_1^2} -\left(\frac{Z-1}{Z}\right)^2\frac{4}{n_2^2}\nonumber
+\frac{A_2}{r_2}+\frac{B_2+8(Z-1)/Z}{r_2}+\frac{C_2}{r^2}] +
Z^{-1}\left(\frac{1}{r} -\frac{1}{r_2}\right)\nonumber
\end{eqnarray}</math>

==Optimization of Nonlinear Parameters==
- The traditional method of performing Hylleraas calculations is to write the
basis set in the form

<math>
\Psi = \sum_{i,j,k}c_{i,j,k}\varphi_{i,j,k}(\alpha,\beta)\pm exchange
</math>

with

<math>
\varphi_{i,j,k}(\alpha,\beta) = r_1^ir_2^jr_{12}^ke^{-\alpha r_1 - \beta
r_2}\mathcal{Y}^M_{l_1l_2L}.
</math>

The usual procedure is to set <math>\alpha = Z</math> so that it represents the inner
<math>1s</math> electron, and then to vary <math>\beta</math> so as to minimize the energy.

[[File:OptNon.jpg|500px]]

Since <math>\beta</math> appears in <math>\Psi</math> as a non-linear parameter, the entire
calculation must be repeated for each value of <math>\beta</math>. However, the minimum
becomes progressively smaller as the basis set is enlarged.

[[File:OptNon02.jpg|500px]]

*Difficulties

1. If the basis set is constructed so that <math> i +j +j \leq N</math>, the the
number of terms is
<math>
(N+1)(N+2)(N+3)/6
</math>
N=14 already gives 680 terms and an accuracy of about 1 part in <math>10^{10}</math> for
low-lying states. A substantial improvement in accuracy would require much
larger basis sets, together with multiple precision arithmetic to avoid loss
of significant figures when high powers are included.

2. The accuracy rapidly deteriorates as one goes to more highly excited
states - about 1 significant figure is lost each time the principle quantum
number is increased.

*Cure

We have found that writing basis sets in the form
\begin{eqnarray}
\Psi &=& \sum_{i,j,k}\left(c^{(1)}_{i,j,k}\varphi_{ijk}(\alpha_1\beta_1) +
c_{ijk}^{(2)}\varphi_{ijk}(\alpha_2\beta_2)\right) \pm exchange \nonumber\\
&=& \psi({\bf r}_1,{\bf r}_2) \pm \psi({\bf r}_2,{\bf r}_1)\nonumber
\end{eqnarray}
so that each combination of powers is included twice with different nonlinear
parameters gives a dramatic improvement in accuracy for basis sets of about
the same total size.

However, the optimization of the nonlinear parameters is now much more
difficult, and an automated procedure is needed.

If
<math>
E = \frac{<\Psi|H|\Psi>}{<\Psi|\Psi>}</math> and we assume <math><\Psi|\Psi> = 1</math>, then

<math>
\frac{\partial E}{\partial\alpha_t} = -2 <\Psi|H-E|r_1\psi({\bf r}_1,{\bf
r}_2;\alpha_t)\pm r_21\psi({\bf r}_2,{\bf
r}_1;\alpha_t)>
</math>
where
<math>
\psi({\bf r}_1,{\bf r}_2;\alpha_t) =
\sum_{i,j,k}c_{ijk}^{(t)}\varphi_{ijk}(\alpha_t\beta_t)
</math>

*Problem

1. Prove the above.

2. Prove that there is no contribution from the implicit dependence of
<math>c_{ijk}^{(t)}</math> on <math>\alpha_t</math> if the linear parameters have been optimized.

==The Screened Hydrogenic Term==

If the Hamiltonian is written in the form (in Z-scaled a.u.)
<math>\begin{eqnarray}\
H = H_0({\bf r}_1,Z) + H_0({\bf r}_2,Z-1) +
Z^{-1}\left(\frac{1}{r}-\frac{1}{r_2}\right)
\end{eqnarray}</math>
with
<math>\
H_0({\bf r}_1,Z) = -\frac{1}{2}\nabla_1^2 - \frac{1}{r_1}\nonumber
</math> and
<math>
H_0({\bf r}_2,Z-1) = -\frac{1}{2}\nabla_2^2 - \frac{Z-1}Z{r_2}\nonumber
</math>,
then the eigenvectors of <math>H_0({\bf r},Z) +H_0({\bf r}_2,Z-1)</math> are products of
hydrogenic orbitals
<math>
\Psi_0=\psi_0(1s,Z)\psi_0(nl,Z-1)
</math>
and the eigenvalue is
<math>\begin{eqnarray}
E_{SH} = \left[-\frac{1}{2} - \left(\frac{Z-1}{Z}\right)^2\frac{1}{2n^2}\right]Z^2 a.u.
\end{eqnarray}</math>
called the screened hydrogenic eigenvalue. From highly excited states,
<math>E_{SH}</math> and <math>\Psi_0</math>are already excellent approximations.

For example, for the 1s8d states, the energies are

<math>\begin{eqnarray}\\
E(1s8d^1D)&=& -2.007\,816\,512\,563\,81 {\rm\ a.u.}\nonumber\\
E(1s8d^3D)&=& - 2.007\,817\,934\,711\,71 {\rm\ a.u.}\nonumber\\
E_{SH} &=& -2.007\,812\,5 {\rm\ a.u.} \nonumber
\end{eqnarray}</math>

It is therefore advantageous to include the screened hydrogenic terms in the
basis set so that the complete trial function becomes
<math>
\Psi = c_0\Psi_0 +
\sum_{ijk}\left[c_{ijk}^{(1)}\varphi_{ijk}(\alpha_1,\beta_1)+c_{ijk}^{(2)}\varphi(\alpha_2,\beta_2)\right]\pm
{\rm exchange}
</math>

Also, the variational principal can be re-expressed in the form
<math>\begin{eqnarray}
E= E_{SH}+\frac{<\Psi|H-E_{SH}|\Psi>}{<\Psi|\Psi>}
\end{eqnarray}</math>
so that the <math>E_{SH}</math> term can be cancelled analytically from the matrix
elements.

For example
<math>
<\Psi_0|H-E_{SH}|\Psi_0> = 0
</math>

and
<math>
<\varphi_{ijk}|H-E_{SH}|\Psi_0> =
Z^{-1}<\varphi_{ijk}|\frac{1}{r}-\frac{1}{r_2}|\Psi_0>
</math>

Recall that
<math>\begin{eqnarray}\ \
I_0(a,b,c) = \frac{2}{c+2}\sum^{[c+1]/2}_{i=0}\left(\begin{array}{c}
c+2\\
2i+1
\end{array}\right) \left[f(p,q;\beta) + f(p^\prime,q^\prime;\alpha)\right]
\end{eqnarray}</math>
where

<math>\begin{eqnarray}
f(p,q;x) =
\frac{q!}{x^{q+1}(\alpha+\beta)^{p+1}}\sum^p_{j=0}\frac{(p+j)!}{j!}\left(\frac{x}{\alpha+\beta}\right)^j
\end{eqnarray}</math>

<math>
\begin{array}{cc}
p=a+2i+2 & p^\prime = b+2i+2\\
q= b+c-2i+2 & q^\prime =a+c-2i+2
\end{array}
</math>

If <math>\beta\ll\alpha</math>, the <math>f(p,q;\beta)\gg f(p^\prime,q^\prime;\alpha)</math> and the
<math>i=0</math> term is the dominant contribution to <math>f(p,q;\beta)</math>. However, since
this term depends only on the sum of powers <math>b+c</math> for <math>r_2</math>, it cancels exactly
from the matrix element of
<math>
\frac{1}{r_{12}} - \frac{1}{r_2}
</math>
and can therefore
be omitted in the calculations of integrands, thereby saving many significant
figures. This is especially valuable when <math>b+c</math> is large and <math>a</math> is small.

==Small Corrections==
*Mass Polarization
If the nuclear mass is not taken to be infinite, then the Hamiltonian is
<math>\begin{eqnarray}\ \
H = \frac{P_N^2}{2M} + \sum_{i=1}\frac{p_i^2}{2m} + V
\end{eqnarray}</math>

[[File:SmallCor.jpg|500px]]

where <math>M =M_A-nm</math> is the nuclear mass and <math>m</math> is the electron mass.

Change variables to the C of M
<math>\begin{eqnarray}
{\bf R} = \frac{1}{M+nm}\left[M{\bf r}_N + m({\bf r}_1 + {\bf r}_2 +
\ldots)\right]
\end{eqnarray}</math>

and relative variables

<math>
{\bf s}_i = {\bf r}_i - {\bf r}_N.
</math>

Then

<math>\begin{eqnarray}\ \ \ \ \
H = \frac{1}{2(M+nm)}{\bf p}_R^2+\frac{1}{2m}\sum_{i=1}^n{\bf p}_{s_i}^2
+\frac{1}{2M}\sum_{i,k}{\bf p}_{s_i}\cdot {\bf p}_{s_k} + V({\bf s}_1, {\bf
s}_2, \ldots)\nonumber
= \frac{1}{M+nm}{\bf p}_R^2+\frac{1}{2\mu}\sum_{i=1}^n{\bf
p}_{s_i}^2+\frac{1}{2M}\sum_{i\neq k}{\bf p}_{s_i}\cdot {\bf p}_{s_k}+V({\bf s}_1, {\bf
s}_2, \ldots)\nonumber
\end{eqnarray}</math>

where
<math>\begin{eqnarray}
\mu = \left(\frac{1}{M} + \frac{1}{m}\right)^{-1}
\end{eqnarray}</math>

If <math>n=2</math>, the Schrödinger equation is
<math>\begin{eqnarray}\ \ \
\left[ -\frac{\hbar^2}{2\mu}(\nabla^2_{s_1}+ \nabla^2_{s_2}) -
\frac{\hbar^2}{M}\nabla_{s_1}\nabla_{s_2}-\frac{Ze^2}{s_1}-\frac{Ze^2}{s_2}+\frac{e^2}{s_{12}}\right]\psi
= E\psi
\end{eqnarray}</math>.

Define the reduced mass Bohr radius
<math>\begin{eqnarray}
a_\mu = \frac{\hbar^2}{\mu e^2} = \frac{m}{\mu}a_0
\end{eqnarray}</math>
and

<math>\begin{eqnarray}
\rho_i = \frac{Zs_i}{a_\mu}\nonumber\\
\varepsilon = \frac{E}{Z^2(e^2/a_\mu)}\nonumber
\end{eqnarray}</math>

The Schrödinger equation then becomes
<math>\begin{eqnarray}\ \ \
\left[
-\frac{1}{2}(\nabla^2_{\rho_1}+\nabla^2_{\rho_2})-\frac{\mu}{M}\vec{\nabla}_{\rho_1}\cdot\vec{\nabla}_{\rho_2} - \frac{1}{\rho_1}-\frac{1}{\rho_2}+\frac{Z^{-1}}{\rho_{12}}\right]\psi = \varepsilon \psi
\end{eqnarray}</math>

The effects of finite mass are thus

1. A reduced mass correction to all energy levels of <math>(e^2/a_\mu)/(e^2/a_0)
= \mu/m</math>, i.e.
<math>\begin{eqnarray}
E_M=\frac{\mu}{m}E_\infty
\end{eqnarray}</math>

2. A specific mass shift given in first order perturbation theory by
<math>\begin{eqnarray}
\Delta\varepsilon =
-\frac{\mu}{M}<\psi|\nabla_{\rho_1}\cdot\nabla_{\rho_2}|\psi>
\end{eqnarray}</math>
or

<math>\begin{eqnarray}
\Delta\varepsilon =
-\frac{\mu}{M}<\psi|\nabla_{\rho_1}\cdot\nabla_{\rho_2}|\psi>\frac{e^2}{a_\mu}\nonumber\\
= -\frac{\mu^2}{mM}<\psi|\nabla_{\rho_1}\cdot\nabla_{\rho_2}|\psi>\frac{e^2}{a_0}\nonumber
\end{eqnarray}</math>

Since <math>\nabla_{\rho_1}\cdot\nabla_{\rho_2}</math> has the same angular properties as
<math>{\bf\rho}_1 \cdot {\bf\rho}_2</math>, the operator is like the product of two
dipole operators. For product type wave functions of the form
$
\psi = \psi(1s)\psi(nl) \pm \mbox{exchange},
$
the matrix element vanishes for all but P-states.

=Radiative Transitions=

==Transition Integrals and the Line Strength==

For an electric dipole transition <math>\gamma'\,L'\,S'\,J'\,M' \rightarrow \gamma\,L\,S\,J\,M</math>, the line strength is defined by
\begin{equation}
S = \sum_{M',\mu,M}|\langle\gamma'\,L'\,S'\,J'\,M'|r_\mu|\gamma\,L\,S\,J\,M\rangle|^2
\end{equation}
Using the Wigner-Eckart theorem, this is related to the reduced matrix element by
\begin{eqnarray}
S(J',J) = |\langle\gamma'\,L'\,S'\,J'||r||\gamma\,L\,S\,J\,M\rangle|^2
\sum_{M',\mu,M}\left|\left(\matrix{J' & 1 & J \cr
-M'& \mu &M \cr}\right)\right|^2
\end{eqnarray}
Using the sum rule for the 3-<math>j</math> symbols
\begin{equation}
\sum_{M',\mu,M}\left|\left(\matrix{J' & 1 & J \cr
-M'& \mu &M \cr}\right)\right|^2 = 1
\end{equation}
we see that the line strength is identical to the reduced matrix element
\begin{equation}
S(J',J) = |\langle\gamma'\,L'\,S'\,J'||r||\gamma\,L\,S\,J\,M\rangle|^2
\end{equation}
For a nonrelativistic calculation in <math>LS</math> coupling, <math>S = S'</math> and we can next strip out the dependence on <math>J</math> and <math>J'</math> by use of Eq.\ (7.1.7) of Edmonds { Angular Momentum in Quantum Mechanics} with <math>k = 1</math> to obtain
\begin{equation}
S(J',J) = (-1)^{L'+S+J+1}[(2J+1)(2J'+1)]^{1/2}\left\{
\matrix{L' & J'& S\cr
J & L & 1\cr}\right\}\langle\gamma' L'|| r || \gamma L\rangle
\end{equation}
where <math>\langle\gamma' L'|| r || \gamma L\rangle</math> is the reduced matrix element if spin is neglected. Note that if spin is neglected, then the line-strength is simply
\begin{equation}
S(L',L) = |\langle\gamma' L'|| r || \gamma L\rangle|^2
\end{equation}
To calculate the reduced matrix element, use a particular example along with the Wigner-Eckart theorem
such as
\begin{equation}
\langle\gamma'\, L'|| r || \gamma\, L\rangle = (-1)^{L'}\langle\gamma'\, L'\, 0| z | \gamma\, L\, 0\rangle{\Large /}\left(\matrix{L' & 1 & L \cr
0 & 0 & 0 \cr}\right)
\end{equation}

===Correlated Two-electron Integrals===
For the case of two-electron integrals, <math>z = z_1 + z_2</math> and <math>r = r_1 + r_2</math>. As discussed in G.W.F. Drake, Phys. Rev. A 18, 820 (1978), the general integral in Hylleraas coordinates for the integral over the <math>z_1</math> term can be written in the form
\begin{equation}
\label{int1}
\langle\gamma'\, L'\, 0| z_1 | \gamma\, L\, 0\rangle = \sum_\Lambda C_\Lambda I_\Lambda(\gamma',\gamma)
\end{equation}
where <math>I_\Lambda(\gamma',\gamma)</math> stands for the purely radial integral
\begin{equation}
\label{radial}
I_\Lambda(\gamma',\gamma) = \int_0^\infty r_1\,dr_1\int_0^\infty r_2\,dr_2\int_{|r_1-r_2|}^{r_1+r_2}r_{12}\, dr_{12}
R_{\gamma'}^*\,r_1\, R_\gamma P_\Lambda(\cos\theta_{12})
\end{equation}
and <math>\cos\theta_{12} = (r_1^2 + r_2^2 - r_{12}^2)/(2r_1r_2)</math> is a short-hand notation for a radial function of <math>r_1</math>, <math>r_2</math> and <math>r_{12}</math>. Also, <math>P_\Lambda(\cos\theta_{12})</math> is a Legendre polynomial,
and, <math>R_\gamma = R_\gamma(r_1,r_2,r_{12})</math> is the radial part of the Hylleraas wave function for state <math>\gamma</math> such
that the total wave function is <math>\psi({\bf r}_1,{\bf r}_2) = R_\gamma(r_1,r_2,r_{12}){\cal Y}_{l_1,l_2,L}^M(\hat{\bf r}_1,\hat{\bf r}_2)\pm {\rm exchange}</math>, normalized to unity, and <math>{\cal Y}_{l_1,l_2,L}^M(\hat{\bf r}_1,\hat{\bf r}_2)</math> is a vector coupled product of spherical harmonics. The coefficents <math>C_\Lambda</math> in Eq.\ (\ref{int1}) can be calculated from Eq.\ (32) of the above 1978 paper. They need to be calculated separately for he direct-direct and the direct-exchange contributions. For the reduced matrix element of <math>z_1</math>, the leading 3-<math>j</math> symbol can be dropped, leaving (for the case <math>k = 1</math>)
\begin{eqnarray}
C_\Lambda^{\rm reduced} &=& \frac{(-1)^{l_2'+L'+L+1}}{2} (l_1,l_1',l_2,l_2',L,L')^{1/2}
\sum_{\lambda={\rm max}(l_1-1,0)}^{l_1+1} (\lambda,\Lambda,1)
\left(\matrix{l_1 & 1 & \lambda \cr
0 & 0 & 0 \cr}\right)
\left(\matrix{l_1'& \lambda & \Lambda \cr
0 & 0 & 0 \cr}\right)\nonumber\\
&\times&\left(\matrix{l_2' & 1_2 & \Lambda \cr
0 & 0 & 0 \cr}\right)
\left\{\matrix{L' & \lambda & l_2 \cr
\Lambda& l_2' & l_1' \cr}\right\}
\left\{\matrix{L' & \lambda & l_2 \cr
l_1 & L & 1 \cr}\right\}
\end{eqnarray}
where the notation <math>(a,b,c,\cdots)</math> means <math>(2a+1)(2b+1)(2c+1)\cdots</math>.
A similar expression can be obtained for the angular coefficients for the <math>z_2</math> term by interchanging the subscripts 1 and 2 throughout and multiplying by <math>(-1)^{L+L'+1}</math>.

===Example===
As an example, for the <math>1s^2\;^1S - 1s2p\;^1P</math> transition of helium, the only nonvanishing <math>C_\Lambda</math> angular coefficients are <math>C_1 = 1/2</math> for the direct-direct term (i.e. <math>P_1(\cos\theta_{12}) = \cos\theta_{12}</math> is included in the radial integral (\ref{radial})) and <math>C_0 = 1/2</math> for the direct-exchange term (i.e. <math>P_0(\cos\theta_{12}) = 1</math> is included in the radial integral (\ref{radial})). The results for infinite nuclear mass are <math>S(1\;^1S-2\;^1P) = (0.781\,127\,158\,667)^2</math> and <math>f(1\;^1S-2\;^1P) = -0.276\,164\,703\,736</math> where
\begin{equation}
f(1\;^1S-2\;^1P) = \frac{2}{3}[E(1\;^1S) - E(2\;^1P)]S(1\;^1S-2\;^1P)
\end{equation}
is the absorption oscillator strength.

Theory Notes

2020-12-14T00:30:07Z

Drake: /* Example */

<center>
<H1>Notes on solving the Schrödinger equation in Hylleraas coordinates for heliumlike atoms
Gordon W.F. Drake
Department of Physics, University of Windsor
Windsor, Ontario, Canada N9B 3P4
(Transcribed from hand-written notes and edited by Lauren Moffatt. Last revised November 22, 2017 to correct a missing subscript in the expression for $B_1$ in the Hamiltonian in Sect. 11 General Hermitian Property.)
</H1>
</center>

==Introduction==
These notes describe some of the technical details involved in solving the Schrödinger equation for a heliumlike atom of nuclear charge Z in correlated Hylleraas coordinates. The standard for computational purposes is to express all quantities in a form that is valid for electronic states of arbitrary angular momentum, and to express matrix elements of the Hamiltonian in a manifestly Hermitian symmetrized form. Further information and derivations of the mathematical techniques for angular integrals and radial recursion relations can be found in the paper
G.W.F. Drake, Phys. Rev. A <math>{\bf 18}</math>, 820 (1978).

The starting point is the two-electron Schrödinger equation for infinite nuclear mass
\begin{equation} \left[-\frac{\hbar^2}{2m}(\nabla^2_1 +\nabla^2_2) - \frac{Ze^2}{r_1} -
\frac{Ze^2}{r_2}+\frac{e^2}{r_{12}} \right]\psi = E\psi\nonumber
\end{equation}
where <math>m</math> is the electron mass, and <math>r_{12} = |{\bf r}_1 - {\bf r}_2|</math> (see diagram below).

[[File:Pic1.jpg|250px]]

Begin by rescaling distances and energies so that the Schrödinger equation can be expressed in a dimensionless form. The dimensionless Z-scaled distance is defined by <math>\begin{eqnarray} \rho = \frac{Zr}{a_0} \end{eqnarray} </math> where <math> \begin{eqnarray} a_0 = \frac{\hbar^2}{me^2} \end{eqnarray} </math> is the atomic unit (a.u.) of distance equal to the Bohr radius <math>0.529\, 177\, 210\, 92(17) \times 10^{-10}</math> m. Then

\begin{equation}[-\frac{\hbar^2}{2m}Z^2(\frac{me^2}{\hbar^2})^2(\nabla^2_{\rho_1}+\nabla^2_{\rho_2})
- Z^2\frac{e^2}{a_0}\rho^{-1}_1 - Z^2\frac{e^2}{a_0}\rho^{-1}_2 +
\frac{e^2}{a_0}Z\rho^{-1}_{12}]\psi= E\psi\nonumber\end{equation}

But <math>\begin{eqnarray} \frac{\hbar^2}{m}(\frac{me^2}{\hbar^2})^2 = \frac{e^2}{a_0} \end{eqnarray} </math> is the hartree atomic unit of of energy <math>(E_h = 27.211\,385\,05(60)</math> eV, or equivalently <math>E_h/(hc) = 2194.746\,313\,708(11)</math> m<math>^{-1})</math>. Therefore, after multiplying through by <math>a_0/(Ze)^2</math>, the problem to be solved in Z-scaled dimensionless units becomes

\begin{equation} \left[-\frac{1}{2}(\nabla^2_{\rho_1}+\nabla^2_{\rho_2}) -
\frac{1}{\rho_1} - \frac{1}{\rho_2} + \frac{Z^{-1}}{\rho_{12}}\right]\psi =
\varepsilon\psi\nonumber \end{equation}

where <math>\begin{eqnarray}\varepsilon = \frac{Ea_0}{(Ze)^2} \end{eqnarray}</math> is the energy in Z-scaled atomic units. For convenience, rewrite this in the conventional form
\begin{equation}
H\psi = \varepsilon\psi\nonumber \end{equation}
where (using <math>r</math> in place of <math>\rho</math> for the Z-scaled distance)
\begin{equation}
H = -\frac{1}{2}(\nabla^2_1+\nabla^2_2) - \frac{1}{r_1}-\frac{1}{r_2} + \frac{Z^{-1}}{r_{12}}
\nonumber\end{equation}
is the atomic Hamiltonian for infinite nuclear mass.

==The Hartree Fock Method==

For purposes of comparison, and to define the correlation energy, assume that <math>\psi({\bf r}_1,{\bf r}_2)</math> can be written in the separable product form
\begin{equation}
\psi({\bf r}_1,{\bf r}_2) = \frac{1}{\sqrt{2}} \left[ u_1(r_1) u_2(r_2) \pm u_2(r_1) u_1(r_2) \right] \nonumber
\end{equation}
for the <math>1s^2\ ^1\!S</math> ground state. Because of the <math>\frac{1}{r_{12}}</math> term in the Schrödinger equation, the exact solution connot be expressed in this form as a separable product. However, the Hartree-Fock approximation corresponds to finding the best possible solution to the Schrödinger equation
\begin{equation}
H\psi({\bf r}_1,{\bf r}_2) = E\psi({\bf r}_1,{\bf r}_2)\nonumber
\end{equation}
that can nevertheless be expressed in this separable product form, where as before
\begin{equation}
H = -\frac{1}{2}(\nabla^2_1+\nabla^2_2) - \frac{1}{r_1}- \frac{1}{r_2} + \frac{Z^{-1}}{r_{12}}\nonumber
\end{equation}
is the full two-electron Hamiltonian. To find the best solution, substitute into <math>\left\langle\psi \lvert H-E \rvert \psi\right\rangle</math> and require this expression to be stationary
with respect to arbitrary infinitesimal variations <math>\delta u_1</math> and <math>\delta
u_2</math> in <math>u_1</math> and <math>u_2</math> respectively; i.e.

\begin{equation}
\frac{1}{2} \left\langle \delta u_1(r_1)u_2(r_2) \pm u_2(r_1)\delta
u_1(r_2) \lvert H-E \rvert u_1(r_1) u_2(r_2)\pm u_2(r_1) u_1(r_2)\right\rangle\nonumber
\end{equation}

\begin{equation}
=\int\delta u_1(r_1) d{\bf r}_1 \left\{ \int d{\bf r}_2 u_2(r_2) \left( H - E \right) \left[ u_1(r_1)u_2(r_2) \pm u_2(r_1) u_1(r_2) \right] \right\}\nonumber
\end{equation}

\begin{equation}
= 0 \nonumber
\end{equation}

for arbitrary <math> \delta u_1(r_1).\nonumber </math> Therefore <math>\{\int d{\bf r}_2 \ldots \} = 0</math>.

Similarly, the coefficient of <math>\delta u_2</math> would give

\begin{equation}
\int d{\bf r}_1 u_1(r_1) \left( H-E \right) \left[ u_1(r_1) u_2(r_2) \pm
u_2(r_1) u_1(r_2) \right] = 0\nonumber
\end{equation}

Define

\begin{equation}
I_{12} = I_{21} = \int d{\bf r}\, u_1(r)u_2(r), \nonumber
\end{equation}

\begin{equation}
H_{ij} = \int d{\bf r}\, u_i(r)(-\frac{1}{2}\nabla - \frac{1}{r})u_j(r), \nonumber
\end{equation}

\begin{equation}
G_{ij}(r) = \int d{\bf r}^\prime u_i(r^\prime)\frac{1}{|{\bf r} - {\bf r}\prime|}u_j(r^\prime)\nonumber
\end{equation}

Then the above equations become the pair of integro-differential equations

\begin{equation}
[ H_0 - E + H_{22}+G_{22}(r)]u_1(r) = \mp [ I_{12}(H_0-E) + H_{12}+G_{12}(r)]u_2(r)\nonumber
\end{equation}

\begin{equation}
[H_0-E+H_{11}+G_{11}(r)]u_2(r) = \mp [I_{12}(H_0-E) + H_{12}+G_{12}(r)]u_1(r)\nonumber
\end{equation}

These must be solved self-consistently for the "constants" <math>I_{12}</math> and
<math>H_{ij}</math> and the function <math>G_{ij}(r)</math>.

The Hartree Fock energy is <math>E \simeq -2.87\ldots </math>a.u. while the exact energy is <math>E = -2.903724\ldots </math>a.u.

The difference is called the "correlation energy" because it arises from the
way in which the motion of one electron is correlated to the other. The
Hartree Fock equations only describe how one electron moves in the average field
provided by the other (mean-field theory).

==Configuration Interaction==
Expand
\begin{equation}
\psi({\bf r}_1,{\bf r}_2)= C_0u^{(s)}_1(r_1)u^{(s)}_1(r_2) + C_1u^{(P)}_1({\bf r}_1)u^{(P)}_1({\bf r}_2){\cal Y}^0_{1,1,0}(\hat{\bf r}_1, \hat{\bf r}_2)+C_2u^{(d)}_1({\bf r}_1)u^{(d)}_2({\bf r}_2){\cal Y}^0_{2,2,0}(\hat{\bf r}_1, \hat{\bf r}_2)+... \pm {\rm\ exchange}
\end{equation}
where
\begin{equation}
{\cal Y}^M_{l_1,l_2,L}(\hat{\bf r}_1, \hat{\bf r}_2)=\sum_{m_1,m_2}Y^{m_1}_{l_1}({\bf r}_1)Y^{m_2}_{l_2}({\bf r}_2)\times <l_1l_2m_1m_2\mid LM>
\end{equation}

This works, but is slowly convergent, and very laborious. The best CI calculations are accurate to <math> ~10^{-7}</math> a.u.

==Hylleraas Coordinates==
[E.A. Hylleraas, Z. Phys. <math>{\bf 48}, 469(1928)</math> and <math>{\bf 54}, 347(1929)</math>]
suggested using the co-ordinates <math>r_1</math>, <math>r_2</math> and <math>r_{12}</math> or equivalently

\begin{eqnarray}
s &=& r_1 + r_2, \nonumber\\
t &=& r_1-r_2, \nonumber\\
u &=& r_{12}\nonumber
\end{eqnarray}

and writing the trial functions in the form

\begin{equation}
\Psi({\bf r}_1,{\bf r}_2) = \sum^{i+j+k\leq
N}_{i,j,k}c_{i,j,k}r_1^{i+l_1}r_2^{j+l_2}r_{12}^ke^{-\alpha r_1 - \beta r_2}
\mathcal{Y}^M_{l_1,l_2,L}(\hat{r}_1,\hat{r}_2)\pm \text{exchange}\nonumber
\end{equation}

Diagonalizing H in this non-orthogonal basis set is equivalent to solving
<math> \begin{eqnarray}
\frac{\partial E}{\partial c_{i,j,k}} = 0\nonumber
\end{eqnarray} </math>
for fixed <math>\alpha</math> and <math>\beta</math>.

The diagonalization must be repeated for different values of <math>\alpha</math> and
<math>\beta</math> in order to optimize the non-linear parameters.

==Completeness==
The completeness of the above basis set can be shown by first writing
\begin{equation}
r_{12}^2 = r_1^2 + r_2^2 - 2r_1r_2\cos\theta_{12}
\end{equation}
and, from the spherical harmonic addition theorem,
\begin{equation}
\cos(\theta_{12})=\frac{4\pi}{3}\sum^1_{m=-1}Y^{m*}_1(\theta_1,\varphi_1)Y^m_1(\theta_2,\varphi_2)
\end{equation}
Consider first the S-states. The <math>r_{12}^0</math> terms are like the ss terms in a CI calculation. The
<math>r_{12}^2</math> terms bring in p-p type contributions, and the higher powers bring
in d-d, f-f etc. type terms. In general

\begin{equation}
P_l(\cos\theta_{12}) = \frac{4\pi}{2l+1}\sum^l_{m=-l}{Y^{m}_l}^*(\theta_1,\varphi_1)Y^m_l(\theta_2,\varphi_2)\nonumber
\end{equation}

For P-states, one would have similarly

\begin{array}{lr}
r_{12}^0 & (sp)P\nonumber\\
r_{12}^2 & (pd)P\nonumber\\
r_{12}^4 & (df)P\nonumber\\
\vdots & \vdots\nonumber
\end{array}

For D-states

\begin{array}{lr}
r_{12}^0 & (sd)D & (pp^\prime)D\nonumber\\
r_{12}^2 & (pf)D & (dd^\prime)D\nonumber\\
r_{12}^4 & (dg)D & (ff^\prime)D\nonumber\\
\vdots & \vdots & \vdots\nonumber
\end{array}

In this case, since there are two "lowest-order" couplings to form a
D-state, both must be present in the basis set, i.e.

\begin{equation}
\Psi(r_2,r_2) = \sum c_{ijk}r_1^ir_2^{j+2}r_{12}^ke^{-\alpha r_1-\beta
r_2}\mathcal{Y}^M_{022}(\hat{r}_1,\hat{r}_2)
+\sum d_{ijk}r_1^{i+1}r_2^{j+1}r_{12^k}e^{-\alpha^\prime r_1 - \beta^\prime
r_2}\mathcal{Y}^M_{112}(\hat{r}_1,\hat{r}_2)\nonumber
\end{equation}

For F-states, one would need <math>(sf)F</math> and <math>(pd)F</math> terms.

For G-states, one would need <math>(sg)G</math>, <math>(pf)G</math> and <math>(dd^\prime)G</math> terms.

Completeness of the radial functions can be proven by considering the
Sturm-Liouville problem

\begin{equation}
\left(-\frac{1}{2}\nabla^2-\frac{\lambda}{r_s}-E\right)\psi({\bf r}) = 0\nonumber
\end{equation}

or

\begin{equation}
\left(-\frac{1}{2}\frac{1}{r^2}\frac{\partial}{\partial r}\left({r^2}\frac{\partial}{\partial r}\right) -
\frac{l\left(l+1\right)}{2r^2} - \frac{\lambda}{r} - E\right)u(r) = 0.\nonumber
\end{equation}

For fixed E and variable <math>\lambda</math> (nuclear charge).

The eigenvalues are <math>\lambda_n = (E/E_n)^{1/2}</math>, where <math>E_n =- \frac{1}{2n^2}</math>

[[File:Pic2.jpg|500px]]

and the eigenfunctions are

\begin{equation}
u_{nl}(r) =
\frac{1}{(2l+1)!}\left(\frac{(n+l)!}{(n-l-1)2!}\right)^{1/2}(2\alpha)^{3/2}e^{-\alpha
r}\nonumber
\times \ \left( 2 \alpha r \right)^l\,_{1}F_{1}\left(-n+l+1,2l+2; 2 \alpha r \right)
\end{equation}

with <math>\alpha = (-2E)^{1/2}</math> and <math>n\geq l+1</math>. The confluent hypergeometric function <math> _1F_1(a,b,;z)</math> then denotes a finite polynomial since <math>a=-n+l+1</math> is a negative integer or zero.

Unlike the hydrogen spectrum, which has both a discrete part for <math>E<0</math> and a
continuous part for <math>E>0</math>, this forms an entirely discrete set of finite
polynomials, called Sturmian functions. They are orthogonal with respect to
the potential, i.e.

\begin{equation}
\int^\infty_0 r^2dr \left( u_{n^\prime l}(r)\frac{1}{r}u_{nl}(r) \right) = \delta_{n,n^\prime}\nonumber
\end{equation}

Since they become complete in the limit <math>n\rightarrow\infty</math>, this assures the
completeness of the variational basis set.

[See also B Klahn and W.A. Bingel Theo. Chim. Acta (Berlin) 44, 9 and 27
(1977)].

==Solutions of the Eigenvalue Problem==

For convenience, write

\[
\Psi({\bf r}_1,{\bf r}_2) = \sum^N_{m=1}c_m\varphi_m\nonumber
\]

where <math>m</math> represents the <math>m</math>'th combination of <math>i,j,k</math> and

\begin{equation}
\varphi_{ijk}=r_1^ir_2^jr_{12}^ke^{-\alpha r_1 - \beta
r_2}\mathcal{Y}^M_{l_1,l_2,L}(\hat{r}_1,\hat{r}_2) \pm \mbox{exchange}\nonumber
\end{equation}

\begin{align}

&
\left( \begin{array}{lr}
\cos(\theta) & \sin(\theta)\\
-\sin(\theta) & \cos(\theta)
\end{array} \right)
\left( \begin{array}{lr}
H_{11} & H_{12} \\
H_{12} & H_{22}
\end{array}\right)
\left( \begin{array}{lr}
\cos(\theta) & -\sin(\theta)\\
\sin(\theta) & \cos(\theta)
\end{array} \right)
\\[5pt]
= & \left(\begin{array}{lr}
cH_{11}+sH_{12} & cH_{12} + sH_{22}\\
-sH_{11} + cH_{12} & -sH_{12} + cH_{22}
\end{array}\right)
\left( \begin{array}{cc}
c & -s \\
s & c
\end{array}\right)
\\[5pt]
= & \left(\begin{array}{lr}
c^2H_{11}+s^2H_{22} + 2csH_{12} & (c^2-s^2)H_{12}+cs(H_{22}-H_{11})\\
(c^2-s^2)H_{12}+cs(H_{22}-H_{11}) & s^2H_{11}+c^2H_{22}-2csH_{12}
\end{array}\right)
\end{align}

Therefore

\begin{equation}
\left[\cos^2(\theta)-\sin^2(\theta)\right]H_{12} =
\cos(\theta)\sin(\theta)(H_{11}-H_{22})\nonumber
\end{equation}

and

\begin{equation}
\tan(2\theta) = \frac{2H_{12}}{H_{11}-H_{22}}\nonumber
\end{equation}

i.e.

\begin{equation}
\cos(\theta)=\left(\frac{r+\omega}{2r}\right)^{1/2}\nonumber
\end{equation}

\begin{equation}
\sin(\theta)=-\mbox{sgn}(H_{12})\left(\frac{r-\omega}{2r}\right)^{1/2}\nonumber
\end{equation}

where

\begin{equation}
\omega = H_{22}-H_{11}\nonumber
\end{equation}

\begin{equation}
r=\left(\omega^2+4H_{12}^2\right)^{1/2}\nonumber
\end{equation}

\begin{equation}
E_1=\frac{1}{2}\left(H_{11}+H_{22}-r\right)\nonumber
\end{equation}

\begin{equation}
E_2=\frac{1}{2}\left(H_{11}+H_{22}+r\right)\nonumber
\end{equation}

===Brute Force Method===

- Gives all the eigenvalues and eigenvectors, but it is slow

- First orthonormalize the <math>N</math>-dimensional basis set; i.e. form linear combinations

\begin{equation}
\Phi_m = \sum_{n=1}^N\varphi_nR_{nm}\nonumber
\end{equation}

such that
\begin{equation}
\langle \Phi_m | \Phi_n \rangle = \delta_{m,n}
\end{equation}

This can be done by finding an orthogonal tranformation T such that

\begin{equation}
T^TOT = I = \left(
\begin{array}{ccccc}
I_1 & 0 & 0 & \ldots & 0 \\
0 & I_2 & 0 & \ldots & 0 \\
0 & 0 & I_3 & \ldots & 0 \\
\vdots & \vdots & \vdots & \ddots & \vdots \\
0 & 0 & 0 & \ldots & I_N
\end{array}\right);
\end{equation}

\begin{equation}
O_{mn} = \langle \varphi_m | \varphi_n \rangle \nonumber
\end{equation}

and then applying a scale change matrix

\begin{equation}
S = \left(\begin{array}{ccccc}
\frac{1}{I_1^{1/2}} & 0 & 0 & \ldots & 0\\
0 & \frac{1}{I_2^{1/2}} & 0 & \ldots & 0 \\
0 & 0 & \frac{1}{I_3^{1/2}} & \ldots & 0 \\
\vdots & \vdots & \vdots & \ddots & \vdots \\
0 & 0 & 0 & \ldots & I_N^{1/2}
\end{array}\right)= S^T\nonumber
\end{equation}

Then

\begin{equation}
S^TT^TOTS = 1
\end{equation}

i.e.

\begin{equation}
R^TOR = 1
\end{equation}

with <math>R=TS</math>.

If H is the matrix with elements <math>H_{mn} = \langle \varphi_m | \varphi_n \rangle </math>, then H
expressed in the <math>\Phi_m</math> basis set is

\begin{equation}
H^\prime = R^THR.\nonumber
\end{equation}

We next diagonalize <math>H^\prime</math> by finding an orthogonal transformation W such
that

\begin{equation}
W^T H^\prime W = \lambda = \left(
\begin{array}{cccc}
\lambda_1 & 0 & \ldots & 0 \\
0 & \lambda_2 & \ldots & 0 \\
\vdots & \vdots & \ddots & \vdots \\
0 & 0 & \ldots & \lambda_N
\end{array}\right)\nonumber
\end{equation}

The q'th eigenvector is

\begin{equation}
\Psi^{(q)} = \sum_{n=1}^N\Phi_n W_{n,q} = \sum_{n,n^\prime} \varphi_{n^\prime} R_{n^\prime ,n} W_{n,q} \nonumber
\end{equation}

i.e.

\begin{equation}
c_{n^\prime}^{(q)} = \sum_{n=1}^N R_{n^\prime n} W_{n,q}
\end{equation}

===The Power Method===

- Based on the observation that if H has one eigenvalue, <math>\lambda_M</math>, much bigger than all the rest, and <math>\chi = \left(
\begin{array}{c}a_1\\a_2\\\vdots\end{array}\right)</math>
is an arbitrary starting vector, then <math>\chi = \sum_{q=1}^N x_q\Psi^{(q)}</math>.

\begin{equation}
(H)^n\chi = \sum_{q=1}^N x_q \lambda^n_q\Psi^{(q)}\nonumber\rightarrow x_M\lambda_M^n\Psi^{(M)}\nonumber
\end{equation}

provided <math>x_M\neq 0</math>.

To pick out the eigenvector corresponding to any eigenvalue, write the original problem in the form

\begin{align}
H\Psi & = \lambda O \Psi \nonumber \\
\left( H - \lambda_q O \right) \Psi & = \left( \lambda - \lambda_q \right) O \Psi \nonumber
\end{align}

Therefore,

\begin{equation}
G \Psi = \frac{1}{\lambda - \lambda_q } \Psi \nonumber \\
\end{equation}

where <math>G=(H-\lambda_qO)^{-1}O</math> with eigenvalues <math>\frac{1}{\lambda_n - \lambda_q}</math>.

By picking <math>\lambda_q</math> close to any one of the <math>\lambda_n</math>, say
<math>\lambda_{n^\prime}</math>, then <math>\frac{1}{\lambda_n-\lambda_q}</math> is much larger for
<math>n=n^\prime</math> than for any other value. The sequence is then

\begin{array}{c}
\chi_1=G\chi\nonumber\\
\chi_2=G\chi_1\nonumber\\
\chi_3=G\chi_2\nonumber\\
\vdots\nonumber
\end{array}

until the ratios of components in <math>\chi_n</math> stop changing.

- To avoid matrix inversion and multiplication, note that the sequence is equivalent to
\begin{equation}
F \chi_n = \left( \lambda - \lambda_q \right) O \chi_{n-1} \nonumber \\
\end{equation}
where <math>F = H-\lambda_qO</math>. The factor of <math>\left( \lambda - \lambda_q \right)</math> can be
dropped because this only affects the normalization of <math>\chi_n</math>. To find <math>\chi_n</math>, solve
\begin{equation}
F \chi_n = O \chi_{n-1} \nonumber
\end{equation}
(N equations in N unknowns). Then
\begin{equation}
\lambda = \frac{ \langle \chi_n | H | \chi_n \rangle}{\langle \chi_n | \chi_n \rangle} \nonumber
\end{equation}

==Matrix Elements of H==

\begin{equation}
H=-\frac{1}{2}\nabla^2_1 -\frac{1}{2}\nabla^2_2 - \frac{1}{r_1} - \frac{1}{r_2} +\frac{Z^{-1}}{r_{12}}\nonumber
\end{equation}

Taking <math>r_1, r_2</math> and <math>r_{12}</math> as independent variables,

\begin{equation}
\nabla_1^2 = \frac{1}{r_1^2}\frac{\partial}{\partial r_1}\left(r_1^2\frac{\partial}{\partial r_1}\right)+ \frac{1}{r_{12}^2}\frac{\partial}{\partial r_{12}}
\left(r_{12}^2\frac{\partial}{\partial r_{12}}\right)\nonumber -\frac{l_1(l_1+1)}{r_1^2}+2(r_1 - r_2\cos(\theta_{12}))\frac{1}{r_{12}}\frac{\partial^2}{\partial r_1 \partial r_{12}} \nonumber - 2(\nabla_1^Y \cdot {\bf r}_2)\frac{1}{r_{12}}\frac{\partial}{\partial r_{12}}\nonumber
\end{equation}

where <math>\nabla_1^Y</math> acts only on the spherical harmonic part of the wave function and the diagram

[[File:Pic3.jpg|400px]]

defines the complete set of 6 independent variables is <math>r_1</math>, <math>r_2</math>, <math>r_{12}</math>, <math>\theta_1</math>, <math>\varphi_1</math>, <math>\chi</math>.

If <math>r_{12}</math> were not an independent variable, then one could take the volume element to be

\begin{equation}
d\tau = r_1^2dr_1\sin(\theta_1)d\theta_1d\varphi_1r_2^2dr_2\sin(\theta_2)d\theta_2d\varphi_2.\nonumber
\end{equation}

However, <math>\theta_2</math> and <math>\varphi_2</math> are no longer independent variables. To eliminate them, take the point <math>{\bf r}_1</math> as the
origin of a new polar co-ordinate system, and write

\begin{equation}
d\tau=-r_1^2dr_1\sin(\theta_1)d\theta_1d\varphi_1r_{12}^2dr_{12}\sin(\psi)d\psi d\chi\nonumber
\end{equation}

and use

\begin{equation}
r_2^2=r_1^2+r_{12}^2 +2r_1r_{12}\cos(\psi)
\end{equation}

Then for fixed <math>r_1</math> and <math>r_{12}</math>,

\begin{equation}
2r_2dr_2 = -2r_1r_{12}\sin(\psi)d\psi
\end{equation}

Thus

\begin{equation}
d\tau= r_1dr_1r_2dr_2r_{12}dr_{12}\sin(\theta_1)d\theta_1d\varphi_1d\chi
\end{equation}

The basic type of integral to be calculated is

\begin{align}
I \left( l_1, m_1, l_2, m_2; R \right) & = \int \sin \left( \theta_1 \right) d\theta_1 d\varphi_1 d\chi Y^{m_1}_{l_1} \left( \theta_1, \varphi_1 \right)^{*} Y^{m_2}_{l_2} \left( \theta_2, \varphi_2 \right) \\
& \times \int r_1dr_1r_2dr_2r_{12}dr_{12}R \left( r_1, r_2, r_{12} \right)
\end{align}

Consider first the angular integral. <math>Y^{m_2}_{l_2} \left( \theta_2, \varphi_2 \right)</math> can be expressed in terms of the independent variables <math>\theta_1, \varphi_1, \chi</math> by use of the rotation matrix relation
\begin{equation}
Y^{m_2}_{l_2}( \theta_2, \varphi_2) = \sum_m \mathcal{D}^{(l_2)}_{m_2,m}(\varphi_1,\theta_1,\chi)^*Y^m_{l_2}( \theta_{12}, \varphi)
\end{equation}
where <math>\theta, \varphi</math> are the polar angles of <math>{\bf r}_2</math> relative to <math>{\bf r}_1</math>. The angular integral is then

\begin{align}
I_{ang} & = \int^{2 \pi}_{0} d\chi \int^{2\pi}_{0} d\varphi_1 \int^{\pi}_{0} \sin \left( \theta_1 \right) d\theta_{1} Y^{m_1}_{l_1} \left( \theta_1, \varphi_1 \right)^{*} \\
& \times \sum_m \mathcal{D}^{(l_2)}_{m_2,m} \left( \varphi_1,\theta_1,\chi \right)^{*} Y^{m}_{l_2} \left( \theta_{12},\varphi \right) \\
\end{align}

Use

\begin{equation}
Y^{m_1}_{l_1} \left( \theta_1, \varphi_1 \right)^{*} = \sqrt{\frac{2l_1+1}{4\pi}}\mathcal{D}^{(l_1)}_{m_1,0} \left( \varphi_1, \theta_1, \chi \right)
\end{equation}

together with the orthogonality property of the rotation matrices (Brink and Satchler, p 147)

\begin{align}
& \int \mathcal{D}^{(j)*}_{m,m^\prime}\mathcal{D}^{(J)}_{M,M^\prime}\sin(\theta_1)d\theta_1d\varphi_1d\chi \\
& = \frac{8\pi^2}{2j+1}\delta_{jJ}\delta_{mM}\delta_{m^\prime M^\prime}\nonumber
\end{align}

to obtain

\begin{align}
I_{ang} & = \sqrt{ \frac{2l_1+1}{4\pi}} \frac{8\pi^2}{2l_1+1} \delta_{l_1,l_2} \delta_{m_1,m_2} Y^{0}_{l_2} \left( \theta_{12}, \varphi \right) \\
& = 2 \pi \delta_{l_1,l_2} \delta_{m_1,m_2} P_{l_2} \left( \cos \theta_{12} \right)
\end{align}

since

\begin{equation}
Y^0_{l_2} \left( \theta_{12}, \varphi \right) = \sqrt{\frac{2l_1+1}{4\pi}} P_{l_2} \left( \cos \left( \theta_{12} \right) \right)
\end{equation}

Note that <math>P_{l_2} \left( \cos \theta \right)</math> is just a short hand expression for a radial function because

\begin{equation}
\cos \theta_{12} = \frac{r_1^2 + r_2^2 - r_{12}^2}{2r_1r_2}
\end{equation}

The original integral is thus

\begin{equation}
I \left( l_1, m_1, l_2, m_2; R \right) = 2 \pi \delta_{l_1,l_2} \delta_{m_1,m_2} \int^{\infty}_{0} r_{1} dr_{1} \int^{\infty}_{0} r_{2} dr_{2} \int^{r_{1} + r_{2}}_{|r_{1} - r_{2}|} r_{12} dr_{12} R \left( r_{1}, r_{2}, r_{12} \right) P_{l_2} \left( \cos \theta_{12} \right)
\end{equation}

where again

\begin{equation}
\cos \theta_{12} = \frac{r_{1}^{2} +r^{2}_{2} - r_{12}^{2}}{2 r_{1} r_{2}}
\end{equation}

is a purely radial function.

The above would become quite complicated for large <math>l_2</math> because <math>P_{l_2}(\cos \theta_{12})</math> contains terms up to <math>\left( \cos \theta_{12} \right)^{l_2}</math>. However, recursion relations exist which allow any integral containing <math>P_l \left( \cos \theta_{12} \right)</math> in terms of those containing just <math>P_0 \left( \cos \theta_{12} \right) = 1</math> and <math>P_1 \left( \cos \theta_{12} \right) = \cos \theta_{12}</math>.

==Radial Integrals and Recursion Relations==

The basic radial integral is [see G.W.F. Drake, Phys. Rev. A 18, 820 (1978)]

\begin{align}
I_0 \left( a, b, c \right) & =
\int^{\infty}_{0} r_{1} dr_{1} \int^{\infty}_{r_{1}} r_{2} dr_{2} \int^{r_{1}+r_{2}}_{r_{2} - r_{1}} r_{12} dr_{12} r_{1}^{a} r_{2}^{b} r_{12}^{c} e^{-\alpha r_{1} - \beta r_{2}} + \int^{\infty}_{0} r_{2} dr_{2} \int^{\infty}_{r_{2}} r_{1} dr_{1} \int^{r_{1} + r_{2}}_{r_{1} - r_{2}} r_{12} dr_{12} r_{1}^{a} r_{2}^{b} r_{12}^{c} e^{- \alpha r_{1} - \beta r_{2}} \\
& = \frac{2}{c+2} \sum^{[(c+1)/2]}_{i=0}
\left(\begin{array}{c}
c+2 \\
2i+1
\end{array} \right)
\left\{ \frac{q!}{\beta^{q+1} \left( \alpha + \beta \right)^{p+1}} \sum_{j=0}^{q} \frac{\left( p + j \right)!}{j!} \left( \frac{\beta}{\alpha + \beta} \right)^j + \frac{q^{\prime}!}{\alpha^{q^{\prime}+1} \left( \alpha + \beta \right)^{p^{\prime}+1}} \sum^{q^{\prime}}_{j=0} \frac{\left( p^{\prime} + j \right)!}{j!} \left( \frac{\alpha}{\alpha + \beta} \right)^j \right\}
\end{align}

where

\begin{array}{ll}
p = a + 2i + 2 & p^{\prime} = b + 2i + 2 \\
q = b + c - 2i + 2 & q^{\prime} = a + c - 2i + 2
\end{array}

The above applies for <math> a, b \geq -2, c\geq -1</math>.
<math>[x]</math> means "greatest integer in" <math>x</math>.

Then

\begin{align}
I_1 \left( a, b, c \right) & = \int d\tau_{r} r_{1}^{a} r_{2}^{b} r_{12}^{c} e^{ -\alpha r_{1} - \beta r_{2}} P_{1}\left( \cos \theta \right) \\
& = \frac{1}{2} \left( I_{0} \left( a + 1, b - 1, c \right) + I_{0} \left( a - 1, b + 1, c \right) - I_{0} \left( a - 1, b - 1, c + 2 \right) \right)
\end{align}

===The Radial Recursion Relation===

Recall that the full integral is

\begin{equation}
I(l_1m_1,l_2m_2;R)=2\pi\delta_{l_1,l_2}\delta_{m_1,m_2}I_{l_2}(R)
\end{equation}

where

\begin{equation}
I_{l_2}(R) = \int d\tau_r R(r_1,r_2,r_{12})P_{l_2}(\cos \theta_{12})
\end{equation}

To obtain the recursion relation, use

\begin{equation}
P_l(x) = \frac{[P^{\prime}_{l+1}(x)-P^{\prime}_{l-1}(x)]}{2l+1}
\end{equation}

with

\begin{equation}
P_{l+1}^\prime (x) = \frac{d}{dx}P_{l+1}(x)
\end{equation}

Here <math>x=\cos\theta_{12}</math> and

\begin{align}
\frac{d}{d \cos \theta_{12}} & = \frac{dr_{12}}{d \cos \theta_{12}} \frac{d}{dr_{12}} \\
& = -\frac{r_{1} r_{2}}{r_{12}} \frac{d}{dr_{12}}
\end{align}

Then,

\begin{equation}
I_{l}\left( R \right) = - \int d\tau_r R \frac{r_{1} r_{2}}{r_{12}} \frac{d}{dr_{12}} \frac{ \left[ P_{l+1} \left( \cos \theta_{12} \right) - P_{l-1} \left( \cos \theta_{12} \right) \right]}{2l+1}
\end{equation}

The <math>r_{12}</math> part of the integral is

\begin{align}
& \int^{r_{1} + r_{2}}_{|r_{1} - r_{2}|} r_{12} dr_{12} R \frac{r_{1} r_{2}}{r_{12}} \frac{d}{dr_{12}} \left[P_{l+1}-P_{l-1} \right] \\
& = R r_{1} r_{2} \left[ P_{l+1} - P_{l-1} \right]^{r_{1} + r_{2}}_{|r_{1} - r_{2}|} - \int^{r_{1} + r_{2}}_{|r_{1} - r_{2}|} r_{12} dr_{12} \left( \frac{d}{dr_{12}} R \right) \frac{r_{1} r_{2}}{r_{12}} \frac{ \left[ P_{l+1} - P_{l-1} \right]}{2l+1}
\end{align}

The integrated term vanishes because

\begin{equation}
\cos \theta_{12} = \frac{r_{1}^{2} + r_{2}^{2} - r_{12}^{2}}{2 r_{1} r_{2}}
\begin{array}{lll}
& = -1 & {\rm when} & r_{12}^{2} = \left( r_{1} + r_{2} \right)^{2} \\
& = 1 & {\rm when} & r_{12}^{2} = \left( r_{1} - r_{2} \right)^{2}
\end{array}
\end{equation}

and <math>P_l(1)=1</math>, <math>P_{l}(-1)=(-1)^{l}</math>.

Thus,

\begin{equation}
I_{l+1}\left(\frac{r_1r_2}{r_{12}}R^\prime\right) = \left( 2l+1 \right) I_{l}(R) - I_{l-1} \left( \frac{r_{1} r_{2}}{r_{12}} R^{\prime} \right)
\end{equation}

If

\begin{equation}
R = r_{1}^{a-1} r_{2}^{b-1} r_{12}^{c+2} e^{-\alpha r_{1} - \beta r_{2}}
\end{equation}

then [G.W.F. Drake, Phys. Rev. A 18, 820 (1978)]

\begin{equation}
I_{l+1}(r_1^ar_2^br_{12}^c) =
\frac{2l+1}{c+2} I_{l} \left( r_{1}^{a-1} r_{2}^{b-2} r_{12}^{c+2} \right)+I_{l-1} \left( r_{1}^{a} r_{2}^{b} r_{12}^{c} \right)
\end{equation}

For the case <math>c=-2</math>, take

\begin{equation}
R = r_{1}^{a-1} r_{2}^{b-1} \ln r_{12} e^{ -\alpha r_{1} -\beta r_{2}}
\end{equation}

Then

\begin{equation}
I_{l+1} \left( r_{1}^{a} r_{2}^{b} r_{12}^{c} \right) = I_{l} \left( r_{1}^{a-1} r_{2}^{b-1} \ln r_{12} \right) \left( 2l + 1 \right) - I_{l-1} \left( r_{1}^{a} r_{2}^{b} r_{12}^{-2} \right)
\end{equation}

This allows all <math>I_{l}</math> integrals to be calculated from tables of <math>I_0</math> and
<math>I_1</math> integrals.

===The General Integral===

The above results for the angular and radial integrals can now be combined into a general formula for integrals of the type

\begin{equation}
I = \int \int d{\bf r}_{1} d{\bf r}_{2} R_{1} \mathcal{Y}^{M^{\prime}}_{l^{\prime}_{1} l^{\prime}_{2} L^{\prime}} \left( \hat{r}_{1}, \hat{r}_{2} \right) T_{k_{1} k_{2} K}^{Q} \left( {\bf r}_{1},{\bf r}_{2} \right) R_{2} \mathcal{Y}_{l_{1} l_{2} L}^{M} \left( \hat{r}_{1}, \hat{r}_{2} \right)
\end{equation}

where

\begin{equation}
\mathcal{Y}^{M}_{l_{1} l_{2} L} \left( \hat{r}_{1}, \hat{r}_{2} \right) = \sum_{m_1,m_2} \langle l_{1} l_{2} m_{1} m_{2} | LM \rangle Y^{m_{1}}_{l_{1}} \left( \hat{r}_{1} \right) Y^{m_{2}}_{l_{2}} \left( \hat{r}_{2} \right)
\end{equation}

and

\begin{equation}
T^Q_{k_{1} k_{2} K} \left( {\bf r}_{1}, {\bf r}_{2} \right) = \sum_{q_{1}, q_{2}} \langle k_{1} k_{2} q_{1} q_{2} | KQ \rangle Y_{k_{1}}^{q_{1}} \left( \hat{r}_{1} \right) Y^{q_{2}}_{k_{2}} \left( \hat{r}_{2} \right)
\end{equation}

The basic idea is to make repeated use of the formula

\begin{equation}
Y^{m_{1}}_{l_{1}} \left( \hat{r}_{1} \right) Y^{m_{2}}_{l_{2}} \left( \hat{r}_{1} \right) = \sum_{l m} \left( \frac{\left( 2l_{1}+1 \right) \left(2l_{2}+1 \right) \left(2l+1 \right)}{4 \pi} \right)^{1/2}
\times \left( \begin{array}{ccc}
l_1 & l_2 & l \\
m_1 & m_2 & m
\end{array} \right) \left(\begin{array}{ccc}
l_1 & l_2 & l \\
0 & 0 & 0
\end{array} \right)
Y^{m*}_{l} \left( \hat{r}_{1} \right)
\end{equation}

where

\begin{equation}
Y^{m*}(\hat{r}) = (-1)^mY_l^{-m}(\hat{r})
\end{equation}

and

\begin{equation}
\left(\begin{array}{ccc}
l_1 & l_2 & l \\
m_1 & m_2 & m
\end{array} \right) = \frac{\left( -1 \right)^{l_{1} - l_{2} - m}}{\left( 2l + 1 \right)^{1/2}} \left( l_{1} l_{2} m_{1} m_{2} | l, -m \right)
\end{equation}

is a 3-<math>j</math> symbol. In particular, write

\begin{array}{cc}
Y^{m_1^\prime}_{l_1^\prime}(\hat{r}_1)^{*} \underbrace{Y_{k_{1}}^{q_{1}} \left( \hat{r}_{1} \right) Y^{m_{1}}_{l_{1}} \left( \hat{r}_{1} \right)} & = & \sum_{\Lambda M}(\ldots)Y^{M}_\Lambda(\hat{r}_1)\\
\ \ \ \ \ \ \ \sum_{\lambda_1\mu_1}Y_{\lambda_1}^{\mu_1}(\hat{r}_1) & & \\
& & \\
Y^{m_{2}^{\prime}}_{l_{2}^{\prime}} \left( \hat{r}_{2} \right) \underbrace{Y^{q_{2}}_{k_{2}} \left( \hat{r}_{2} \right) Y_{l_{2}}^{m_{2}} \left( \hat{r}_{2} \right)} & = & \sum_{\Lambda^{\prime} M^{\prime}} \left( \ldots \right)Y_{\Lambda^{\prime}}^{M^{\prime}*} \left( \hat{r}_{2} \right) \\
\ \ \ \ \ \ \ \ \ \sum_{\lambda_{2} \mu_{2}}Y_{\lambda_{2}}^{\mu_{2}} \left( \hat{r}_{2} \right) & &
\end{array}

The angular integral then gives a factor of <math>2 \pi \delta_{\Lambda, \Lambda^{\prime}} \delta_{M, M^{\prime}} P_{\Lambda} \left( \cos \theta_{12} \right)</math>. The total integral therefore reduces to the form

\begin{equation}
I = \sum_{\Lambda} C_{\Lambda} I_{\Lambda} \left( R_{1} R_{2} \right)
\end{equation}

where <math>C_{\Lambda} = \sum_{\lambda_{1}, \lambda_{2}} C_{ \lambda_{1}, \lambda_{2}, \Lambda}</math>. For further details and derivations, including graphical representations, see G.W.F. Drake, Phys. Rev. A <math>{\bf 18}</math>, 820 (1978).

==Graphical Representation==

[[File:GraphRep1.jpg|500px]]

==Matrix Elements of H==
Recall that

\begin{equation}
H = -\frac{1}{2} \nabla^{2}_{1} - \frac{1}{2} \nabla^{2}_{2} - \frac{1}{r_1} - \frac{1}{r_2} + \frac{Z^{-1}}{r_{12}}
\end{equation}

Consider matrix elements of

\begin{equation}
\nabla^2_1 = \frac{1}{r_1^2} \frac{\partial}{\partial r_1}\left(r_1^2\frac{\partial}{\partial r_1} \right) + \frac{1}{r^2}\frac{\partial}{\partial r}\left(r^2\frac{\partial}{\partial r}\right) - \frac{(\vec{l}_1^Y)^2}{r_1^2} + \frac{2 \left( r_1 - r_2 \cos \theta \right)}{r} \frac{\partial^2}{\partial r_1 \partial r} - 2 \left( \nabla^Y_1 \cdot {\bf r}_2 \right) \frac{1}{r} \frac{\partial}{\partial r}
\end{equation}

where <math>r \equiv r_{12}</math> and <math>\cos\theta \equiv \cos\theta_{12}</math>. Also in what follows, define <math>\hat{\nabla}^Y_1 \equiv r_1\nabla^Y_1</math> where <math>\nabla^Y</math> operates only on the spherical harmonic part of the wave function, and similarly for <math>\vec{l}_1^Y</math>.

A general matrix element is

\begin{equation}
\langle r_1^{a^\prime} r_2^{b^\prime} r_{12}^{c^\prime} e^{-\alpha^\prime r_1 - \beta^\prime r_2} \mathcal{Y}^{M^\prime}_{l^\prime_1 l^\prime_2 L^\prime} \left( \hat{r}_1, \hat{r}_2 \right) \left| \nabla^2_1 \right| r_1^a r_2^b r_{12}^c e^{-\alpha r_1 - \beta r_2} \mathcal{Y}^M_{l_1 l_2 L} \left( \hat{r}_1, \hat{r}_2 \right) \rangle
\end{equation}

Since <math>\nabla^2_1</math> is rotationally invariant, this vanishes unless <math>L=L^\prime</math>, <math>M=M^\prime</math>. Also <math>\nabla_1^2</math> is Hermitian, so that the result must be the same whether it operates to the right or left, even though the results look very different. In fact, requiring the results to be the same yields some interesting and useful integral identities as follows:

Put

\begin{equation}
\mathcal{F} = F \mathcal{Y}^M_{l_1 l_2 L} \left( \hat{r}_1, \hat{r_2} \right)
\end{equation}

and

\begin{equation}
F = r_1^a r_2^b r_{12}^c e^{-\alpha r_1 - \beta r_2}
\end{equation}

Then

\begin{equation}
\nabla^2_1 \mathcal{F} = \left\{ \frac{1}{r_1^2} \left[ a \left( a + 1 \right) - l_1 \left( l_1 + 1 \right) \right] + \frac{c \left( c + 1 \right)}{r^2} + \alpha^2\frac{2\alpha \left( a + 1 \right)}{r_1} + \frac{2 \left(r_1 - r_2 \cos \theta \right) }{r_1r^2}c \left[ a - \alpha r_1 \right] - \frac{2c}{r^2} \left( \hat{\nabla}^Y_1 \cdot \hat{r}_2 \right) \frac{r_2}{r_1} \right\} \mathcal{F}
\end{equation}

and

\begin{align}
\langle \mathcal{F}^\prime \left| \nabla^2_1 \right| \mathcal{F} \rangle & = \sum_\Lambda \int d\tau_r F^\prime C_\Lambda \left( 1 \right) P_\Lambda \left( \cos \theta \right) \times \left\{ \frac{1}{r_1^2} \left[ a \left( a + 1 \right) - l_1 \left( l_1 + 1 \right) \right] -\frac{2 \alpha \left( a + 1 \right)}{r_1} + \frac{c \left( c + 1 \right)}{r^2} +\alpha^2 + \frac{2 \left( r_2 - r_2 \cos \theta \right)}{r_1r^2}c \left( a - \alpha r_1 \right) \right\} F \\
& + \sum_\Lambda \int d\tau_r F^\prime C_\Lambda \left( \hat{\nabla}^Y_1 \cdot \hat{r}_2 \right) P_\Lambda \left( \cos \theta \right) \left( \frac{-2c r_2}{r_1 r^2} \right) F
\end{align}

where

\begin{equation}
\int d\tau_r = \int^\infty_0 r_1 dr_1 \int^\infty_0 r_2 dr_2 \int^{r_1+r_2}_{\left| r_1 - r_2 \right|} r dr
\end{equation}

For brevity, let the sum over <math>\Lambda</math> and the radial integrations be understood, and let <math>\left( \nabla_1^2 \right)</math> stand for the terms which appear in the integrand. Then operating to the right gives

\begin{equation}
(\nabla_1^2)_R = \frac{1}{r_1^2} \left[ a \left( a + 1 \right) - l_1 \left( l_1 + 1 \right) \right] + \frac{c \left( c + 1 \right)}{r^2} + \alpha^2 - \frac{2 \alpha \left( a + 1 \right)}{r_1}
+ \frac{2 \left( r_1 - r_2 \cos \theta \right)}{r_1 r^2}c \left( a - \alpha r_1 \right) - \frac{2c r_2}{r_1 r^2} \left( \hat{\nabla}^Y_1 \cdot \hat{r}_2 \right)
\end{equation}

Operating to the left gives

\begin{equation}
(\nabla_1^2)_L = \frac{1}{r_1^2} \left[a^\prime \left( a^\prime + 1 \right) - l_1^\prime \left( l_1^\prime + 1 \right) \right] + \frac{c^\prime \left( c^\prime + 1 \right)}{r^2} + \alpha^{\prime 2} -\frac{2 \alpha^\prime \left( a^\prime + 1 \right)}{r_1}
+ \frac{2 \left( r_1 - r_2 \cos \theta \right)}{r_1 r^2} c^\prime \left( a^\prime - \alpha^\prime r_1 \right) - \frac{2 c^\prime r_2}{r_1 r^2} \left( \hat{\nabla}^{Y^\prime}_1 \cdot \hat{r}_2 \right)
\end{equation}

Now put

\begin{array}{cc}
a_+ = a+a^\prime, & \hat{\nabla}_1^+ = \hat{\nabla}_1^Y + \hat{\nabla}_1^{Y\prime} \\
a_{-} = a-a^\prime, & \hat{\nabla}_1^{-} = \hat{\nabla}_1^Y - \hat{\nabla}_1^{Y\prime}
\end{array}

etc., and substitute <math>a^\prime = a_+ - a</math>, <math>c^\prime = c_+ - c</math>, and <math>\alpha^\prime = \alpha_+ - \alpha</math> in <math>(\nabla^2_1)_L</math>. If <math>a_+</math>, <math>c_+</math> and <math>\alpha_+</math> are held fixed, then the equation

\begin{equation}
(\nabla^2_1)_R = (\nabla^2_1)_L
\end{equation}

must be true for arbitrary <math>a</math>, <math>c</math>, and <math>\alpha</math>. Their coefficients must thus vanish.

This yields the integral relations

\begin{equation}
\label{I}
\frac{\left( r_1 - r_2 \cos \theta \right)}{r_1 r^2} = \frac{1}{c_+} \left( \frac{- \left( a_+ + 1 \right)}{r_1^2} + \frac{\alpha_+}{r_1} \right) \tag{I}
\end{equation}

from the coefficient of <math>a</math> and

\begin{equation}
\label{II}
\frac{ \left( r_1 - r_2 \cos \theta \right) \left( a_+ - \alpha_+ r_1 \right)}{r_1 r^2} = \frac{r_2}{r_1 r^2}\left( \hat{r}_2 \cdot \hat{\nabla}_1^+ \right) - \frac{ \left( c_+ + 1 \right)}{r^2} \tag{II}
\end{equation}

from the coefficient of <math>c</math>. The coefficient of <math>\alpha</math> gives an equation equivalent to (I).

Furthermore, if can be show that (see problem)

\begin{align}
& \sum_\Lambda \int d\tau_r \frac{r^c}{r^2} C_\Lambda \left( \hat{r}_2 \cdot \hat{\nabla}_1^Y \right) P_\Lambda \left( \cos \theta \right) \\
= & \sum_\Lambda \int d\tau_r \frac{r^c}{c r_1 r_2}C_\Lambda \left( 1 \right)P_\Lambda \left( \cos \theta \right) \times \left( \frac{l_1^\prime \left( l_1^\prime + 1 \right) - l_1 \left( l_1 + 1 \right) -\Lambda \left( \Lambda + 1 \right)}{2} \right)
\end{align}

and similarly for <math>\left( \hat{r}_2 \cdot \hat{\nabla}^{Y\prime}_1 \right)</math> with <math>l_1</math> and <math>l_1^\prime</math> interchangeable, then it follows that

\begin{equation}
\frac{r^{c_+}}{r^2} \left( \hat{r}_2 \cdot \hat{\nabla}_1^+ \right) = - \frac{r^{c_+}}{c_+ r_1 r_2} \Lambda \left( \Lambda + 1 \right)
\end{equation}

and

\begin{equation}
\frac{r^{c_+}}{r^2} \left( \hat{r}_2 \cdot \hat{\nabla}_1^- \right) = \frac{r^{c_+}}{c_+ r_1 r_2} \left[ l_1^\prime \left( l_1^\prime + 1 \right) - l_1 \left( l_1 + 1 \right) \right]
\end{equation}

where equality applies after integration and summation over <math>\Lambda</math>.

Thus (II) becomes

\begin{equation}\label{IIa}
\frac{ \left( r_1 - r_2 \cos \theta \right) \left( a_+ - \alpha_+ r_1 \right)}{r_1 r^2} = - \frac{\Lambda \left( \Lambda + 1 \right)}{c_+ r^2_1} - \frac{ \left( c_+ + 1 \right)}{r^2} \tag{IIa}
\end{equation}

===Problem===

Prove the integral relation

\begin{align}
& \sum_\Lambda \int d\tau_r f \left( r_1, r_2 \right) \frac{1}{r} \left( \frac{d}{dr} q(r) \right) C_\Lambda \left( \hat{r}_2 \cdot \hat{\nabla}^Y_1 \right) P_\Lambda \left( \cos \theta \right) \\
= & \sum_\Lambda \int d\tau_r \frac{ f \left( r_1, r_2 \right)}{r_1, r_2} q(r) C_\Lambda \left( 1 \right) P_\Lambda \left( \cos \theta \right) \left(\frac{l_1^\prime \left( l_1^\prime + 1 \right) - l_1 \left( l_1 + 1 \right) - \Lambda \left( \Lambda + 1 \right)}{2}\right)
\end{align}

where the coefficients <math>C_\Lambda \left( 1 \right)</math> are the angular coefficients from the overlap integral

\begin{align}
\int & d\Omega\, \mathcal{Y}^{M*}_{l_1^\prime l_2^\prime L}(\hat{r}_1,\hat{r}_2)\, \mathcal{Y}^{M}_{l_1 l_2 L}(\hat{r}_1,\hat{r}_2)\\
= & \sum_\Lambda C_\Lambda(1)P_\Lambda(\cos\theta_{12}).
\end{align}

Hint: Use the fact that <math>l_1^2</math> is Hermitian so that
<math>
\int d\tau (l_1^2\mathcal{Y}^\prime)^*q(r)\mathcal{Y} = \int d\tau
\mathcal{Y}^{\prime *}l_1^2(q(r)\mathcal{Y})
</math>
with <math>\ \vec{l}_1= \frac{1}{i}\vec{r}_1\times\nabla_1</math>.

It is also useful to use
<math>\begin{eqnarray}\ \
\cos\theta\, P_L(\cos\theta) &=& \frac{1}{2L+1}[L\,P_{L-1}(\cos\theta) + (L+1)P_{L+1}(\cos\theta)]
\\
(\cos^2\theta-1)P_L(\cos\theta) &=&
\frac{L(L-1)}{(2L-1)(2L+1)}P_{L-2}(\cos\theta)\nonumber
-\frac{2(L^2+L-1)}{(2L-1)(2L+3)}P_L(\cos\theta) + \frac{(L+1)(L+2)}{(2L+1)(2L+3)}P_{L+2}(\cos\theta)\nonumber\\
&=& \frac{L(L-1)}{(2L-1)(2L+1)}[P_{L-2}(\cos\theta) - P_L(\cos\theta)]
+ \frac{(L+1)(L+2)}{(2L+1)(2L+3)}[P_{L+2}(\cos\theta) - P_L(\cos\theta)]
\end{eqnarray}</math>

together with a double application of the integral recursion relation
<math>
I_{L+1}\left(\frac{1}{r}\frac{d}{dr}q(r)\right) =
(2L+1)I_L\left(\frac{1}{r_1r_2}q(r)\right)+I_{L-1}\left(\frac{1}{r}\frac{d}{dr}q(r)\right)
</math>

Of course <math>l_1^2\mathcal{Y}^M_{l_1l_2L}(\hat{r}_1,\hat{r}_2)=l_1(l_1+1)\mathcal{Y}^M_{l_1l_2L}(\hat{r}_1,\hat{r}_2)</math>. Begin by expanding
<math>
l_1^2(g\mathcal{Y}) = \mathcal{Y}l_1^2g + g\,l_1^2\mathcal{Y} + 2(\vec{l}_1g)\cdot(\vec{l}_1\mathcal{Y})
</math>, and show that

<math>
\begin{eqnarray}\ \
(\vec{l}_1g)\cdot(\vec{l}_1\mathcal{Y}) & = & \frac{r_1^2}{r}\frac{dg}{dr}\vec{r}_2\cdot \nabla_1\mathcal{Y}\,\,\,\,\mbox{ and}\\
l_1^2g(r) &=& 2r_1r_2\cos\theta\frac{1}{r}\frac{dg}{dr} + r_1^2r_2^2(\cos^2\theta - 1)
\frac{1}{r}\frac{d}{dr}\left(\frac{1}{r}\frac{dg}{dr}\right).
\end{eqnarray}
</math>

The proof amounts to showing that the term <math>l_1^2g(r)</math> can be replaced by
<math>\Lambda(\Lambda+1)g(r)</math> after multiplying by <math>P_\Lambda(\cos\theta)</math>
and integrating by parts with respect to the radial integrations over <math>r_1</math>, <math>r_2</math> and
<math>r \equiv |\vec{r}_1 - \vec{r}_2|</math>. Remember that <math>\cos\theta</math> is just a short-hand notation for the radial function <math>(r_1^2 + r_2^2 - r^2)/(2r_1r_2)</math>.

==General Hermitian Property==

Each combination of terms of the form
<math>
<f>_R = a^2f_1+af_2+abf_3+b^2f_4+bf_5+af_6\nabla_1^Y+bf_7\nabla^Y_2+f_8(Y)
</math>
can be rewritten
<math>
<f>_L= (a_+-a)^2f_1+(a_+-a)f_2 + (a_+-a)(b_+-b)f_3+(b_+-b)^2f_4
+(b_+-b)f_5+(a_+-a)f_6\nabla^{Y\prime}_1+(b_+-b)f_7\nabla^{Y\prime}_2+f_8)Y^\prime
</math>
where as usual <math>\nabla_1^Y</math> acts only on the spherical harmonic part of the wave function denoted for short by <math>Y</math>.
Since these must be equal for arbitrary <math>a</math> and <math>b</math>,
<math>
a_+^2f_1+a_+f_2+_+b_+f_3+b_+f_4+b_+f_5+a_+f_6\nabla^{Y\prime}_1 +
b_+f_7\nabla^{Y\prime}_2+f_8(Y^\prime)-f_8(Y)=0
</math>

Adding the corresponding expression with <math>Y</math> and <math>Y^\prime</math> interchanged
yields
<math>\label{III}
a_+^2f_1+a_+f_2+a_+b_+f_3+b_+^2f_4+b_+f_5+\frac{1}{2}a_+f_6\nabla^+_1+\frac{1}{2}b_+f_7\nabla^+_2=0\tag{III}
</math>

Subtracting gives
<math>\label{IV}
f_8(Y)-f_8(Y^\prime)=-\frac{1}{2}[a_+f_6\nabla^-_1+b_+f_7\nabla_2^-]\tag{IV}
</math>
<math>\label{V}
a[-2a_+f_1-2f_2-b_+f_3-f_6\nabla^+_1]=0\tag{V}
</math>
<math>\label{VI}
b[-2b_+f_4-2f_5-a_+f_3-f_7\nabla^+_2]=0\tag{VI}
</math>

Adding the two forms gives

\begin{eqnarray}
<f>_R+<f>_L=\frac{1}{2}(a_+^2+a_-^2)f_1+a_+f_2+\frac{1}{2}(a_+b_++a_-b_-)f_3\nonumber\\
\ +\frac{1}{2}(b_+^2+b_-^2)f_4+b_+f_5+\frac{1}{2}f_6(a_+\nabla_1^++a_-\nabla^-_1)\nonumber\\
\ +\frac{1}{2}(b_+\nabla_2^++b_-\nabla^-_2)+f_8(Y)+f_8(Y^\prime)\nonumber
\end{eqnarray}

Subtracting <math>\frac{x}{2}\times({\rm I})</math> gives

\begin{eqnarray}\ \
<f>_R+<f>_L = \frac{1}{2}[(1-x)a_+^2 + a_-^2]f_1 +
(1-\frac{x}{2})a_+f_2+\frac{1}{2}[(1-x)a_+b_++a_-b_-]f_3\nonumber\\
+\frac{1}{2}[(1-x)b_+^2+b_-^2]f_4+(1-\frac{x}{2})f_5b_++\frac{1}{2}f_6[(1-\frac{x}{2})a_+\nabla_1^++a_-\nabla^-_1]\nonumber\\
+\frac{1}{2}[(1-\frac{x}{2})b_+\nabla^+_2+b_-\nabla^-_2]+f_8(Y)+f_8(Y^\prime)\nonumber
\end{eqnarray}

If <math>x=1</math>,
\begin{eqnarray}
<f>_R+<f>_L &=&
\frac{1}{2}[a_-^2f_1+a_+f_2+a_-b_-f_3+b_-^2f_4+b_+f_5\nonumber\\
&+&\left(\frac{1}{2}a_+\nabla^+_1+a_-\nabla^-_1\right)f_6+\left(\frac{1}{2}b_+\nabla^+_2+b_-\nabla^-_2\right)f_7]+f_8(Y)+f_8(Y^\prime)\nonumber
\end{eqnarray}

The General Hermitian Property for arbitrary <math>x</math> gives
\begin{eqnarray}\ \
\nabla_1^2=
\frac{1}{4}\{\frac{1}{r_1^2}[(1-x)a_+^2+a_-^2+2(1-\frac{x}{2})a_+-2[l_1(l_1+1)+l_1^\prime(l_1^\prime+1)]\nonumber\\
-\frac{2}{r_1}[(1-x)\alpha_+a_++\alpha_-a_-+2(1-\frac{x}{2})\alpha_+]+(1-x)\alpha_+^2
+ \alpha^2_-]\nonumber\\
+\frac{2(r_1-r_2\cos\theta)}{r_1r^2}[(1-x)(a_+-\alpha_+r_1)c_++(a_--\alpha_-r_2)c_-]\nonumber\\
-\frac{2r_2}{r_1r^2}[(1-\frac{x}{2})c_+\hat{r}_2\cdot\hat{\nabla}_1^++c_-\hat{r}_2\hat{\nabla}_1^-]+\frac{(1-x)c_+^2+c_-^2+2(1-\frac{x}{2})c_+}{r^2}\}\nonumber
\end{eqnarray}.

Use
<math>\begin{eqnarray}\ \
\frac{-2r_2}{r_1r^2}[(1-\frac{x}{2})c_+\hat{r}_2\hat{\nabla}_1^++c_-\hat{r}_2\hat{\nabla}_1^-]=\frac{2}{r_1^2}\left[(1-\frac{x}{2})\Lambda(\Lambda+1)-\frac{c_-}{c_+}[l_1^\prime(l_1^\prime+1)+l_1(l_1+1)]\right]\end{eqnarray}</math>,

<math>\begin{eqnarray}
\frac{2(r_1-r_2\cos\theta)}{r_1r^2}(a_--\alpha_-r_1)c_- = 2\frac{c_-}{c_+}\left[\frac{-1}{r_1^2}[a_-(a_++1)]+\frac{1}{r_1}[a_-\alpha_++\alpha_-(a_++2)]-\alpha_-\alpha_+\right]\nonumber
\end{eqnarray}</math>,

and
<math>\begin{eqnarray}\ \
\frac{2(r_1-r_2\cos\theta)}{r_1r^2}(a_+-\alpha_+r_1)c_+ = 2\left[-\frac{1}{r_1^2}[a_+(a_++1)]+\frac{1}{r_1}[a_+\alpha_++\alpha_+(a_++2)]-\alpha_+^2\right]
\end{eqnarray}</math>.

Substituting into <math>\nabla_1^2</math> gives

\begin{eqnarray}
\nabla^2_1 = \frac{1}{4}\{\frac{1}{r_1^2}[-(1-x)a_+^2 + a_-^2 +xa_+ +
2(1-\frac{x}{2})\Lambda(\Lambda+1)\nonumber\\
-2l_1(l_1+1)(1-\frac{c_-}{c_+}) -
2l_1^\prime(l_1^\prime+1)(1+\frac{c_-}{c_+})-\frac{2c_-a_-}{c_+}(a_++1)]\nonumber\\
- \frac{2}{r_1}\left[-(1-x)\alpha_+(a_++2)+\alpha_-a_-+2(1-\frac{x}{2})\alpha_+
-\frac{c_-}{c_+}[a_-\alpha_++\alpha_-(a_++2)]\right]\nonumber\\
-(1-x)\alpha_+^2+\alpha_-^2-2\frac{c_-}{c_+}\alpha_-\alpha_+ +\frac{1}{r^2}\left[(1-x)c_+^2+c_-^2+2(1-\frac{x}{2})c_+\right]\}\nonumber
\end{eqnarray}

This has the form
<math>
\nabla_1^2 =
\frac{1}{4}\left[\frac{A_1}{r_1^2}+\frac{B_1}{r_1}+\frac{C_1}{r^2}+D_1\right]
</math>
with

<math>\begin{array}{ccc}
A_1=-(1-x)a_+^2+a_-^2
+xa_++2(1-\frac{x}{2})\Lambda(\Lambda+1)-2l_1(l_1+1)(1-\frac{c_-}{c_+})\
-2l_1^\prime(l_1^\prime+1)(1+\frac{c_-}{c_+})-\frac{2c_-a_-}{c_+}(a_++1) \ \
\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \
\end{array}
</math>

<math>
B_1 =
2\left[(1-x)\alpha_+(a_++2)-\alpha_-a_--2(1-\frac{x}{2})\alpha_++\frac{c_-}{c_+}[a_-\alpha_++\alpha_-(a_++2)]\right]
</math>

<math>
C_1 = (1-x)c_+^2+ c_-^2 +2(1-\frac{x}{2})c_+
</math>

<math>
D_1 = -1(1-x)\alpha_+^2 +\alpha_-^2 - 2\frac{c_-}{c_+}\alpha_-\alpha_+
</math>

The complete Hamiltonian is then (in Z-scaled a.u.)

<math>\begin{eqnarray}
H =-\frac{1}{2}\nabla^2_1 -\frac{1}{2}\nabla^2_2 -\frac{1}{r_1} -\frac{1}{r_2} +\frac{Z^{-1}}{r}\nonumber\\
= -\frac{1}{2}\nabla^2_1 -\frac{1}{2}\nabla^2_2 -\frac{1}{r_1} -\frac{Z-1}{Zr_2}+Z^{-1}\left(\frac{1}{r}-\frac{1}{r_2}\right)\nonumber\\
\ =-\frac{1}{8}\left[\frac{A_1}{r_1^2}+\frac{B_1+8}{r_1}+\frac{C_1}{r^2}+D_1+D_2 +\frac{A_2}{r_2^2} + \frac{B_2 + 8(Z-1)/Z}{r_2}+\frac{C_2}{r^2}\right]\nonumber
+ Z^{-1}\left(\frac{1}{r}-\frac{1}{r_2}\right)\nonumber
\end{eqnarray}</math>

The screened hydrogenic energy is
<math>\begin{eqnarray}
E_{SH}=-\frac{1}{2}\left[\frac{1}{n_1^2}+ \left(\frac{Z-1}{Z}\right)^2\frac{1}{n_2^2}\right]
\end{eqnarray}</math>
so that

<math>\begin{eqnarray}
H-E_{SH} &=& -\frac{1}{8}[
\frac{A_1}{r_1^2}+\frac{B_1+8}{r_1}+\frac{C_1}{r^2} + D_1 + D_2
-\frac{4}{n_1^2} -\left(\frac{Z-1}{Z}\right)^2\frac{4}{n_2^2}\nonumber
+\frac{A_2}{r_2}+\frac{B_2+8(Z-1)/Z}{r_2}+\frac{C_2}{r^2}] +
Z^{-1}\left(\frac{1}{r} -\frac{1}{r_2}\right)\nonumber
\end{eqnarray}</math>

==Optimization of Nonlinear Parameters==
- The traditional method of performing Hylleraas calculations is to write the
basis set in the form

<math>
\Psi = \sum_{i,j,k}c_{i,j,k}\varphi_{i,j,k}(\alpha,\beta)\pm exchange
</math>

with

<math>
\varphi_{i,j,k}(\alpha,\beta) = r_1^ir_2^jr_{12}^ke^{-\alpha r_1 - \beta
r_2}\mathcal{Y}^M_{l_1l_2L}.
</math>

The usual procedure is to set <math>\alpha = Z</math> so that it represents the inner
<math>1s</math> electron, and then to vary <math>\beta</math> so as to minimize the energy.

[[File:OptNon.jpg|500px]]

Since <math>\beta</math> appears in <math>\Psi</math> as a non-linear parameter, the entire
calculation must be repeated for each value of <math>\beta</math>. However, the minimum
becomes progressively smaller as the basis set is enlarged.

[[File:OptNon02.jpg|500px]]

*Difficulties

1. If the basis set is constructed so that <math> i +j +j \leq N</math>, the the
number of terms is
<math>
(N+1)(N+2)(N+3)/6
</math>
N=14 already gives 680 terms and an accuracy of about 1 part in <math>10^{10}</math> for
low-lying states. A substantial improvement in accuracy would require much
larger basis sets, together with multiple precision arithmetic to avoid loss
of significant figures when high powers are included.

2. The accuracy rapidly deteriorates as one goes to more highly excited
states - about 1 significant figure is lost each time the principle quantum
number is increased.

*Cure

We have found that writing basis sets in the form
\begin{eqnarray}
\Psi &=& \sum_{i,j,k}\left(c^{(1)}_{i,j,k}\varphi_{ijk}(\alpha_1\beta_1) +
c_{ijk}^{(2)}\varphi_{ijk}(\alpha_2\beta_2)\right) \pm exchange \nonumber\\
&=& \psi({\bf r}_1,{\bf r}_2) \pm \psi({\bf r}_2,{\bf r}_1)\nonumber
\end{eqnarray}
so that each combination of powers is included twice with different nonlinear
parameters gives a dramatic improvement in accuracy for basis sets of about
the same total size.

However, the optimization of the nonlinear parameters is now much more
difficult, and an automated procedure is needed.

If
<math>
E = \frac{<\Psi|H|\Psi>}{<\Psi|\Psi>}</math> and we assume <math><\Psi|\Psi> = 1</math>, then

<math>
\frac{\partial E}{\partial\alpha_t} = -2 <\Psi|H-E|r_1\psi({\bf r}_1,{\bf
r}_2;\alpha_t)\pm r_21\psi({\bf r}_2,{\bf
r}_1;\alpha_t)>
</math>
where
<math>
\psi({\bf r}_1,{\bf r}_2;\alpha_t) =
\sum_{i,j,k}c_{ijk}^{(t)}\varphi_{ijk}(\alpha_t\beta_t)
</math>

*Problem

1. Prove the above.

2. Prove that there is no contribution from the implicit dependence of
<math>c_{ijk}^{(t)}</math> on <math>\alpha_t</math> if the linear parameters have been optimized.

==The Screened Hydrogenic Term==

If the Hamiltonian is written in the form (in Z-scaled a.u.)
<math>\begin{eqnarray}\
H = H_0({\bf r}_1,Z) + H_0({\bf r}_2,Z-1) +
Z^{-1}\left(\frac{1}{r}-\frac{1}{r_2}\right)
\end{eqnarray}</math>
with
<math>\
H_0({\bf r}_1,Z) = -\frac{1}{2}\nabla_1^2 - \frac{1}{r_1}\nonumber
</math> and
<math>
H_0({\bf r}_2,Z-1) = -\frac{1}{2}\nabla_2^2 - \frac{Z-1}Z{r_2}\nonumber
</math>,
then the eigenvectors of <math>H_0({\bf r},Z) +H_0({\bf r}_2,Z-1)</math> are products of
hydrogenic orbitals
<math>
\Psi_0=\psi_0(1s,Z)\psi_0(nl,Z-1)
</math>
and the eigenvalue is
<math>\begin{eqnarray}
E_{SH} = \left[-\frac{1}{2} - \left(\frac{Z-1}{Z}\right)^2\frac{1}{2n^2}\right]Z^2 a.u.
\end{eqnarray}</math>
called the screened hydrogenic eigenvalue. From highly excited states,
<math>E_{SH}</math> and <math>\Psi_0</math>are already excellent approximations.

For example, for the 1s8d states, the energies are

<math>\begin{eqnarray}\\
E(1s8d^1D)&=& -2.007\,816\,512\,563\,81 {\rm\ a.u.}\nonumber\\
E(1s8d^3D)&=& - 2.007\,817\,934\,711\,71 {\rm\ a.u.}\nonumber\\
E_{SH} &=& -2.007\,812\,5 {\rm\ a.u.} \nonumber
\end{eqnarray}</math>

It is therefore advantageous to include the screened hydrogenic terms in the
basis set so that the complete trial function becomes
<math>
\Psi = c_0\Psi_0 +
\sum_{ijk}\left[c_{ijk}^{(1)}\varphi_{ijk}(\alpha_1,\beta_1)+c_{ijk}^{(2)}\varphi(\alpha_2,\beta_2)\right]\pm
{\rm exchange}
</math>

Also, the variational principal can be re-expressed in the form
<math>\begin{eqnarray}
E= E_{SH}+\frac{<\Psi|H-E_{SH}|\Psi>}{<\Psi|\Psi>}
\end{eqnarray}</math>
so that the <math>E_{SH}</math> term can be cancelled analytically from the matrix
elements.

For example
<math>
<\Psi_0|H-E_{SH}|\Psi_0> = 0
</math>

and
<math>
<\varphi_{ijk}|H-E_{SH}|\Psi_0> =
Z^{-1}<\varphi_{ijk}|\frac{1}{r}-\frac{1}{r_2}|\Psi_0>
</math>

Recall that
<math>\begin{eqnarray}\ \
I_0(a,b,c) = \frac{2}{c+2}\sum^{[c+1]/2}_{i=0}\left(\begin{array}{c}
c+2\\
2i+1
\end{array}\right) \left[f(p,q;\beta) + f(p^\prime,q^\prime;\alpha)\right]
\end{eqnarray}</math>
where

<math>\begin{eqnarray}
f(p,q;x) =
\frac{q!}{x^{q+1}(\alpha+\beta)^{p+1}}\sum^p_{j=0}\frac{(p+j)!}{j!}\left(\frac{x}{\alpha+\beta}\right)^j
\end{eqnarray}</math>

<math>
\begin{array}{cc}
p=a+2i+2 & p^\prime = b+2i+2\\
q= b+c-2i+2 & q^\prime =a+c-2i+2
\end{array}
</math>

If <math>\beta\ll\alpha</math>, the <math>f(p,q;\beta)\gg f(p^\prime,q^\prime;\alpha)</math> and the
<math>i=0</math> term is the dominant contribution to <math>f(p,q;\beta)</math>. However, since
this term depends only on the sum of powers <math>b+c</math> for <math>r_2</math>, it cancels exactly
from the matrix element of
<math>
\frac{1}{r_{12}} - \frac{1}{r_2}
</math>
and can therefore
be omitted in the calculations of integrands, thereby saving many significant
figures. This is especially valuable when <math>b+c</math> is large and <math>a</math> is small.

==Small Corrections==
*Mass Polarization
If the nuclear mass is not taken to be infinite, then the Hamiltonian is
<math>\begin{eqnarray}\ \
H = \frac{P_N^2}{2M} + \sum_{i=1}\frac{p_i^2}{2m} + V
\end{eqnarray}</math>

[[File:SmallCor.jpg|500px]]

where <math>M =M_A-nm</math> is the nuclear mass and <math>m</math> is the electron mass.

Change variables to the C of M
<math>\begin{eqnarray}
{\bf R} = \frac{1}{M+nm}\left[M{\bf r}_N + m({\bf r}_1 + {\bf r}_2 +
\ldots)\right]
\end{eqnarray}</math>

and relative variables

<math>
{\bf s}_i = {\bf r}_i - {\bf r}_N.
</math>

Then

<math>\begin{eqnarray}\ \ \ \ \
H = \frac{1}{2(M+nm)}{\bf p}_R^2+\frac{1}{2m}\sum_{i=1}^n{\bf p}_{s_i}^2
+\frac{1}{2M}\sum_{i,k}{\bf p}_{s_i}\cdot {\bf p}_{s_k} + V({\bf s}_1, {\bf
s}_2, \ldots)\nonumber
= \frac{1}{M+nm}{\bf p}_R^2+\frac{1}{2\mu}\sum_{i=1}^n{\bf
p}_{s_i}^2+\frac{1}{2M}\sum_{i\neq k}{\bf p}_{s_i}\cdot {\bf p}_{s_k}+V({\bf s}_1, {\bf
s}_2, \ldots)\nonumber
\end{eqnarray}</math>

where
<math>\begin{eqnarray}
\mu = \left(\frac{1}{M} + \frac{1}{m}\right)^{-1}
\end{eqnarray}</math>

If <math>n=2</math>, the Schrödinger equation is
<math>\begin{eqnarray}\ \ \
\left[ -\frac{\hbar^2}{2\mu}(\nabla^2_{s_1}+ \nabla^2_{s_2}) -
\frac{\hbar^2}{M}\nabla_{s_1}\nabla_{s_2}-\frac{Ze^2}{s_1}-\frac{Ze^2}{s_2}+\frac{e^2}{s_{12}}\right]\psi
= E\psi
\end{eqnarray}</math>.

Define the reduced mass Bohr radius
<math>\begin{eqnarray}
a_\mu = \frac{\hbar^2}{\mu e^2} = \frac{m}{\mu}a_0
\end{eqnarray}</math>
and

<math>\begin{eqnarray}
\rho_i = \frac{Zs_i}{a_\mu}\nonumber\\
\varepsilon = \frac{E}{Z^2(e^2/a_\mu)}\nonumber
\end{eqnarray}</math>

The Schrödinger equation then becomes
<math>\begin{eqnarray}\ \ \
\left[
-\frac{1}{2}(\nabla^2_{\rho_1}+\nabla^2_{\rho_2})-\frac{\mu}{M}\vec{\nabla}_{\rho_1}\cdot\vec{\nabla}_{\rho_2} - \frac{1}{\rho_1}-\frac{1}{\rho_2}+\frac{Z^{-1}}{\rho_{12}}\right]\psi = \varepsilon \psi
\end{eqnarray}</math>

The effects of finite mass are thus

1. A reduced mass correction to all energy levels of <math>(e^2/a_\mu)/(e^2/a_0)
= \mu/m</math>, i.e.
<math>\begin{eqnarray}
E_M=\frac{\mu}{m}E_\infty
\end{eqnarray}</math>

2. A specific mass shift given in first order perturbation theory by
<math>\begin{eqnarray}
\Delta\varepsilon =
-\frac{\mu}{M}<\psi|\nabla_{\rho_1}\cdot\nabla_{\rho_2}|\psi>
\end{eqnarray}</math>
or

<math>\begin{eqnarray}
\Delta\varepsilon =
-\frac{\mu}{M}<\psi|\nabla_{\rho_1}\cdot\nabla_{\rho_2}|\psi>\frac{e^2}{a_\mu}\nonumber\\
= -\frac{\mu^2}{mM}<\psi|\nabla_{\rho_1}\cdot\nabla_{\rho_2}|\psi>\frac{e^2}{a_0}\nonumber
\end{eqnarray}</math>

Since <math>\nabla_{\rho_1}\cdot\nabla_{\rho_2}</math> has the same angular properties as
<math>{\bf\rho}_1 \cdot {\bf\rho}_2</math>, the operator is like the product of two
dipole operators. For product type wave functions of the form
$
\psi = \psi(1s)\psi(nl) \pm \mbox{exchange},
$
the matrix element vanishes for all but P-states.

=Radiative Transitions=

==Transition Integrals and the Line Strength==

For an electric dipole transition <math>\gamma'\,L'\,S'\,J'\,M' \rightarrow \gamma\,L\,S\,J\,M</math>, the line strength is defined by
\begin{equation}
S = \sum_{M',\mu,M}|\langle\gamma'\,L'\,S'\,J'\,M'|r_\mu|\gamma\,L\,S\,J\,M\rangle|^2
\end{equation}
Using the Wigner-Eckart theorem, this is related to the reduced matrix element by
\begin{eqnarray}
S(J',J) = |\langle\gamma'\,L'\,S'\,J'||r||\gamma\,L\,S\,J\,M\rangle|^2
\sum_{M',\mu,M}\left|\left(\matrix{J' & 1 & J \cr
-M'& \mu &M \cr}\right)\right|^2
\end{eqnarray}
Using the sum rule for the 3-<math>j</math> symbols
\begin{equation}
\sum_{M',\mu,M}\left|\left(\matrix{J' & 1 & J \cr
-M'& \mu &M \cr}\right)\right|^2 = 1
\end{equation}
we see that the line strength is identical to the reduced matrix element
\begin{equation}
S(J',J) = |\langle\gamma'\,L'\,S'\,J'||r||\gamma\,L\,S\,J\,M\rangle|^2
\end{equation}
For a nonrelativistic calculation in <math>LS</math> coupling, <math>S = S'</math> and we can next strip out the dependence on <math>J</math> and <math>J'</math> by use of Eq.\ (7.1.7) of Edmonds { Angular Momentum in Quantum Mechanics} with <math>k = 1</math> to obtain
\begin{equation}
S(J',J) = (-1)^{L'+S+J+1}[(2J+1)(2J'+1)]^{1/2}\left\{
\matrix{L' & J'& S\cr
J & L & 1\cr}\right\}\langle\gamma' L'|| r || \gamma L\rangle
\end{equation}
where <math>\langle\gamma' L'|| r || \gamma L\rangle</math> is the reduced matrix element if spin is neglected. Note that if spin is neglected, then the line-strength is simply
\begin{equation}
S(L',L) = |\langle\gamma' L'|| r || \gamma L\rangle|^2
\end{equation}
To calculate the reduced matrix element, use a particular example along with the Wigner-Eckart theorem
such as
\begin{equation}
\langle\gamma'\, L'|| r || \gamma\, L\rangle = (-1)^{L'}\langle\gamma'\, L'\, 0| z | \gamma\, L\, 0\rangle{\Large /}\left(\matrix{L' & 1 & L \cr
0 & 0 & 0 \cr}\right)
\end{equation}

===Correlated Two-electron Integrals===
For the case of two-electron integrals, <math>z = z_1 + z_2</math> and <math>r = r_1 + r_2</math>. As discussed in G.W.F. Drake, Phys. Rev. A 18, 820 (1978), the general integral in Hylleraas coordinates for the integral over the <math>z_1</math> term can be written in the form
\begin{equation}
\label{int1}
\langle\gamma'\, L'\, 0| z_1 | \gamma\, L\, 0\rangle = \sum_\Lambda C_\Lambda I_\Lambda(\gamma',\gamma)
\end{equation}
where <math>I_\Lambda(\gamma',\gamma)</math> stands for the purely radial integral
\begin{equation}
\label{radial}
I_\Lambda(\gamma',\gamma) = \int_0^\infty r_1\,dr_1\int_0^\infty r_2\,dr_2\int_{|r_1-r_2|}^{r_1+r_2}r_{12}\, dr_{12}
R_{\gamma'}^*\,r_1\, R_\gamma P_\Lambda(\cos\theta_{12})
\end{equation}
and <math>\cos\theta_{12} = (r_1^2 + r_2^2 - r_{12}^2)/(2r_1r_2)</math> is a short-hand notation for a radial function of <math>r_1</math>, <math>r_2</math> and <math>r_{12}</math>. Also, <math>P_\Lambda(\cos\theta_{12})</math> is a Legendre polynomial,
and, <math>R_\gamma = R_\gamma(r_1,r_2,r_{12})</math> is the radial part of the Hylleraas wave function for state <math>\gamma</math> such
that the total wave function is <math>\psi({\bf r}_1,{\bf r}_2) = R_\gamma(r_1,r_2,r_{12}){\cal Y}_{l_1,l_2,L}^M(\hat{\bf r}_1,\hat{\bf r}_2)\pm {\rm exchange}</math>, normalized to unity, and <math>{\cal Y}_{l_1,l_2,L}^M(\hat{\bf r}_1,\hat{\bf r}_2)</math> is a vector coupled product of spherical harmonics. The coefficents <math>C_\Lambda</math> in Eq.\ (\ref{int1}) can be calculated from Eq.\ (32) of the above 1978 paper. They need to be calculated separately for he direct-direct and the direct-exchange contributions. For the reduced matrix element of <math>z_1</math>, the leading 3-<math>j</math> symbol can be dropped, leaving (for the case <math>k = 1</math>)
\begin{eqnarray}
C_\Lambda^{\rm reduced} &=& \frac{(-1)^{l_2'+L'+L+1}}{2} (l_1,l_1',l_2,l_2',L,L')^{1/2}
\sum_{\lambda={\rm max}(l_1-1,0)}^{l_1+1} (\lambda,\Lambda,1)
\left(\matrix{l_1 & 1 & \lambda \cr
0 & 0 & 0 \cr}\right)
\left(\matrix{l_1'& \lambda & \Lambda \cr
0 & 0 & 0 \cr}\right)\nonumber\\
&\times&\left(\matrix{l_2' & 1_2 & \Lambda \cr
0 & 0 & 0 \cr}\right)
\left\{\matrix{L' & \lambda & l_2 \cr
\Lambda& l_2' & l_1' \cr}\right\}
\left\{\matrix{L' & \lambda & l_2 \cr
l_1 & L & 1 \cr}\right\}
\end{eqnarray}
where the notation <math>(a,b,c,\cdots)</math> means <math>(2a+1)(2b+1)(2c+1)\cdots</math>.
A similar expression can be obtained for the angular coefficients for the <math>z_2</math> term by interchanging the subscripts 1 and 2 throughout and multiplying by <math>(-1)^{L+L'+1}</math>.

===Example===
As an example, for the <math>1s^2\;^1S - 1s2p\;^1P</math> transition of helium, the only nonvanishing <math>C_\Lambda</math> angular coefficients are <math>C_1 = 1/2</math> for the direct-direct term (i.e. <math>P_1(\cos\theta_{12}) = \cos\theta_{12}</math> is included in the radial integral (\ref{radial})) and <math>C_0 = 1/2</math> for the direct-exchange term (i.e. <math>P_0(\cos\theta_{12}) = 1</math> is included in the radial integral (\ref{radial})). The results for infinite nuclear mass are <math>S(1\;^1S-2\;^1P) = (0.781\,127\,158\,667)^2</math> and <math>f(1\;^1S-2\;^1P) = -0.276\,164\,703\,736</math> where
\begin{equation}
f(1\;^1S-2\;^1P) = \frac{2}{3}[E(1\;^1S) - E(2\;^1P)]S(1\;^1S-2\;^1P)
\end{equation}
is the absorption oscillator strength.

Theory Notes

2020-12-14T00:28:15Z

Drake: /* Introduction */

<center>
<H1>Notes on solving the Schrödinger equation in Hylleraas coordinates for heliumlike atoms
Gordon W.F. Drake
Department of Physics, University of Windsor
Windsor, Ontario, Canada N9B 3P4
(Transcribed from hand-written notes and edited by Lauren Moffatt. Last revised November 22, 2017 to correct a missing subscript in the expression for $B_1$ in the Hamiltonian in Sect. 11 General Hermitian Property.)
</H1>
</center>

==Introduction==
These notes describe some of the technical details involved in solving the Schrödinger equation for a heliumlike atom of nuclear charge Z in correlated Hylleraas coordinates. The standard for computational purposes is to express all quantities in a form that is valid for electronic states of arbitrary angular momentum, and to express matrix elements of the Hamiltonian in a manifestly Hermitian symmetrized form. Further information and derivations of the mathematical techniques for angular integrals and radial recursion relations can be found in the paper
G.W.F. Drake, Phys. Rev. A <math>{\bf 18}</math>, 820 (1978).

The starting point is the two-electron Schrödinger equation for infinite nuclear mass
\begin{equation} \left[-\frac{\hbar^2}{2m}(\nabla^2_1 +\nabla^2_2) - \frac{Ze^2}{r_1} -
\frac{Ze^2}{r_2}+\frac{e^2}{r_{12}} \right]\psi = E\psi\nonumber
\end{equation}
where <math>m</math> is the electron mass, and <math>r_{12} = |{\bf r}_1 - {\bf r}_2|</math> (see diagram below).

[[File:Pic1.jpg|250px]]

Begin by rescaling distances and energies so that the Schrödinger equation can be expressed in a dimensionless form. The dimensionless Z-scaled distance is defined by <math>\begin{eqnarray} \rho = \frac{Zr}{a_0} \end{eqnarray} </math> where <math> \begin{eqnarray} a_0 = \frac{\hbar^2}{me^2} \end{eqnarray} </math> is the atomic unit (a.u.) of distance equal to the Bohr radius <math>0.529\, 177\, 210\, 92(17) \times 10^{-10}</math> m. Then

\begin{equation}[-\frac{\hbar^2}{2m}Z^2(\frac{me^2}{\hbar^2})^2(\nabla^2_{\rho_1}+\nabla^2_{\rho_2})
- Z^2\frac{e^2}{a_0}\rho^{-1}_1 - Z^2\frac{e^2}{a_0}\rho^{-1}_2 +
\frac{e^2}{a_0}Z\rho^{-1}_{12}]\psi= E\psi\nonumber\end{equation}

But <math>\begin{eqnarray} \frac{\hbar^2}{m}(\frac{me^2}{\hbar^2})^2 = \frac{e^2}{a_0} \end{eqnarray} </math> is the hartree atomic unit of of energy <math>(E_h = 27.211\,385\,05(60)</math> eV, or equivalently <math>E_h/(hc) = 2194.746\,313\,708(11)</math> m<math>^{-1})</math>. Therefore, after multiplying through by <math>a_0/(Ze)^2</math>, the problem to be solved in Z-scaled dimensionless units becomes

\begin{equation} \left[-\frac{1}{2}(\nabla^2_{\rho_1}+\nabla^2_{\rho_2}) -
\frac{1}{\rho_1} - \frac{1}{\rho_2} + \frac{Z^{-1}}{\rho_{12}}\right]\psi =
\varepsilon\psi\nonumber \end{equation}

where <math>\begin{eqnarray}\varepsilon = \frac{Ea_0}{(Ze)^2} \end{eqnarray}</math> is the energy in Z-scaled atomic units. For convenience, rewrite this in the conventional form
\begin{equation}
H\psi = \varepsilon\psi\nonumber \end{equation}
where (using <math>r</math> in place of <math>\rho</math> for the Z-scaled distance)
\begin{equation}
H = -\frac{1}{2}(\nabla^2_1+\nabla^2_2) - \frac{1}{r_1}-\frac{1}{r_2} + \frac{Z^{-1}}{r_{12}}
\nonumber\end{equation}
is the atomic Hamiltonian for infinite nuclear mass.

==The Hartree Fock Method==

For purposes of comparison, and to define the correlation energy, assume that <math>\psi({\bf r}_1,{\bf r}_2)</math> can be written in the separable product form
\begin{equation}
\psi({\bf r}_1,{\bf r}_2) = \frac{1}{\sqrt{2}} \left[ u_1(r_1) u_2(r_2) \pm u_2(r_1) u_1(r_2) \right] \nonumber
\end{equation}
for the <math>1s^2\ ^1\!S</math> ground state. Because of the <math>\frac{1}{r_{12}}</math> term in the Schrödinger equation, the exact solution connot be expressed in this form as a separable product. However, the Hartree-Fock approximation corresponds to finding the best possible solution to the Schrödinger equation
\begin{equation}
H\psi({\bf r}_1,{\bf r}_2) = E\psi({\bf r}_1,{\bf r}_2)\nonumber
\end{equation}
that can nevertheless be expressed in this separable product form, where as before
\begin{equation}
H = -\frac{1}{2}(\nabla^2_1+\nabla^2_2) - \frac{1}{r_1}- \frac{1}{r_2} + \frac{Z^{-1}}{r_{12}}\nonumber
\end{equation}
is the full two-electron Hamiltonian. To find the best solution, substitute into <math>\left\langle\psi \lvert H-E \rvert \psi\right\rangle</math> and require this expression to be stationary
with respect to arbitrary infinitesimal variations <math>\delta u_1</math> and <math>\delta
u_2</math> in <math>u_1</math> and <math>u_2</math> respectively; i.e.

\begin{equation}
\frac{1}{2} \left\langle \delta u_1(r_1)u_2(r_2) \pm u_2(r_1)\delta
u_1(r_2) \lvert H-E \rvert u_1(r_1) u_2(r_2)\pm u_2(r_1) u_1(r_2)\right\rangle\nonumber
\end{equation}

\begin{equation}
=\int\delta u_1(r_1) d{\bf r}_1 \left\{ \int d{\bf r}_2 u_2(r_2) \left( H - E \right) \left[ u_1(r_1)u_2(r_2) \pm u_2(r_1) u_1(r_2) \right] \right\}\nonumber
\end{equation}

\begin{equation}
= 0 \nonumber
\end{equation}

for arbitrary <math> \delta u_1(r_1).\nonumber </math> Therefore <math>\{\int d{\bf r}_2 \ldots \} = 0</math>.

Similarly, the coefficient of <math>\delta u_2</math> would give

\begin{equation}
\int d{\bf r}_1 u_1(r_1) \left( H-E \right) \left[ u_1(r_1) u_2(r_2) \pm
u_2(r_1) u_1(r_2) \right] = 0\nonumber
\end{equation}

Define

\begin{equation}
I_{12} = I_{21} = \int d{\bf r}\, u_1(r)u_2(r), \nonumber
\end{equation}

\begin{equation}
H_{ij} = \int d{\bf r}\, u_i(r)(-\frac{1}{2}\nabla - \frac{1}{r})u_j(r), \nonumber
\end{equation}

\begin{equation}
G_{ij}(r) = \int d{\bf r}^\prime u_i(r^\prime)\frac{1}{|{\bf r} - {\bf r}\prime|}u_j(r^\prime)\nonumber
\end{equation}

Then the above equations become the pair of integro-differential equations

\begin{equation}
[ H_0 - E + H_{22}+G_{22}(r)]u_1(r) = \mp [ I_{12}(H_0-E) + H_{12}+G_{12}(r)]u_2(r)\nonumber
\end{equation}

\begin{equation}
[H_0-E+H_{11}+G_{11}(r)]u_2(r) = \mp [I_{12}(H_0-E) + H_{12}+G_{12}(r)]u_1(r)\nonumber
\end{equation}

These must be solved self-consistently for the "constants" <math>I_{12}</math> and
<math>H_{ij}</math> and the function <math>G_{ij}(r)</math>.

The Hartree Fock energy is <math>E \simeq -2.87\ldots </math>a.u. while the exact energy is <math>E = -2.903724\ldots </math>a.u.

The difference is called the "correlation energy" because it arises from the
way in which the motion of one electron is correlated to the other. The
Hartree Fock equations only describe how one electron moves in the average field
provided by the other (mean-field theory).

==Configuration Interaction==
Expand
\begin{equation}
\psi({\bf r}_1,{\bf r}_2)= C_0u^{(s)}_1(r_1)u^{(s)}_1(r_2) + C_1u^{(P)}_1({\bf r}_1)u^{(P)}_1({\bf r}_2){\cal Y}^0_{1,1,0}(\hat{\bf r}_1, \hat{\bf r}_2)+C_2u^{(d)}_1({\bf r}_1)u^{(d)}_2({\bf r}_2){\cal Y}^0_{2,2,0}(\hat{\bf r}_1, \hat{\bf r}_2)+... \pm {\rm\ exchange}
\end{equation}
where
\begin{equation}
{\cal Y}^M_{l_1,l_2,L}(\hat{\bf r}_1, \hat{\bf r}_2)=\sum_{m_1,m_2}Y^{m_1}_{l_1}({\bf r}_1)Y^{m_2}_{l_2}({\bf r}_2)\times <l_1l_2m_1m_2\mid LM>
\end{equation}

This works, but is slowly convergent, and very laborious. The best CI calculations are accurate to <math> ~10^{-7}</math> a.u.

==Hylleraas Coordinates==
[E.A. Hylleraas, Z. Phys. <math>{\bf 48}, 469(1928)</math> and <math>{\bf 54}, 347(1929)</math>]
suggested using the co-ordinates <math>r_1</math>, <math>r_2</math> and <math>r_{12}</math> or equivalently

\begin{eqnarray}
s &=& r_1 + r_2, \nonumber\\
t &=& r_1-r_2, \nonumber\\
u &=& r_{12}\nonumber
\end{eqnarray}

and writing the trial functions in the form

\begin{equation}
\Psi({\bf r}_1,{\bf r}_2) = \sum^{i+j+k\leq
N}_{i,j,k}c_{i,j,k}r_1^{i+l_1}r_2^{j+l_2}r_{12}^ke^{-\alpha r_1 - \beta r_2}
\mathcal{Y}^M_{l_1,l_2,L}(\hat{r}_1,\hat{r}_2)\pm \text{exchange}\nonumber
\end{equation}

Diagonalizing H in this non-orthogonal basis set is equivalent to solving
<math> \begin{eqnarray}
\frac{\partial E}{\partial c_{i,j,k}} = 0\nonumber
\end{eqnarray} </math>
for fixed <math>\alpha</math> and <math>\beta</math>.

The diagonalization must be repeated for different values of <math>\alpha</math> and
<math>\beta</math> in order to optimize the non-linear parameters.

==Completeness==
The completeness of the above basis set can be shown by first writing
\begin{equation}
r_{12}^2 = r_1^2 + r_2^2 - 2r_1r_2\cos\theta_{12}
\end{equation}
and, from the spherical harmonic addition theorem,
\begin{equation}
\cos(\theta_{12})=\frac{4\pi}{3}\sum^1_{m=-1}Y^{m*}_1(\theta_1,\varphi_1)Y^m_1(\theta_2,\varphi_2)
\end{equation}
Consider first the S-states. The <math>r_{12}^0</math> terms are like the ss terms in a CI calculation. The
<math>r_{12}^2</math> terms bring in p-p type contributions, and the higher powers bring
in d-d, f-f etc. type terms. In general

\begin{equation}
P_l(\cos\theta_{12}) = \frac{4\pi}{2l+1}\sum^l_{m=-l}{Y^{m}_l}^*(\theta_1,\varphi_1)Y^m_l(\theta_2,\varphi_2)\nonumber
\end{equation}

For P-states, one would have similarly

\begin{array}{lr}
r_{12}^0 & (sp)P\nonumber\\
r_{12}^2 & (pd)P\nonumber\\
r_{12}^4 & (df)P\nonumber\\
\vdots & \vdots\nonumber
\end{array}

For D-states

\begin{array}{lr}
r_{12}^0 & (sd)D & (pp^\prime)D\nonumber\\
r_{12}^2 & (pf)D & (dd^\prime)D\nonumber\\
r_{12}^4 & (dg)D & (ff^\prime)D\nonumber\\
\vdots & \vdots & \vdots\nonumber
\end{array}

In this case, since there are two "lowest-order" couplings to form a
D-state, both must be present in the basis set, i.e.

\begin{equation}
\Psi(r_2,r_2) = \sum c_{ijk}r_1^ir_2^{j+2}r_{12}^ke^{-\alpha r_1-\beta
r_2}\mathcal{Y}^M_{022}(\hat{r}_1,\hat{r}_2)
+\sum d_{ijk}r_1^{i+1}r_2^{j+1}r_{12^k}e^{-\alpha^\prime r_1 - \beta^\prime
r_2}\mathcal{Y}^M_{112}(\hat{r}_1,\hat{r}_2)\nonumber
\end{equation}

For F-states, one would need <math>(sf)F</math> and <math>(pd)F</math> terms.

For G-states, one would need <math>(sg)G</math>, <math>(pf)G</math> and <math>(dd^\prime)G</math> terms.

Completeness of the radial functions can be proven by considering the
Sturm-Liouville problem

\begin{equation}
\left(-\frac{1}{2}\nabla^2-\frac{\lambda}{r_s}-E\right)\psi({\bf r}) = 0\nonumber
\end{equation}

or

\begin{equation}
\left(-\frac{1}{2}\frac{1}{r^2}\frac{\partial}{\partial r}\left({r^2}\frac{\partial}{\partial r}\right) -
\frac{l\left(l+1\right)}{2r^2} - \frac{\lambda}{r} - E\right)u(r) = 0.\nonumber
\end{equation}

For fixed E and variable <math>\lambda</math> (nuclear charge).

The eigenvalues are <math>\lambda_n = (E/E_n)^{1/2}</math>, where <math>E_n =- \frac{1}{2n^2}</math>

[[File:Pic2.jpg|500px]]

and the eigenfunctions are

\begin{equation}
u_{nl}(r) =
\frac{1}{(2l+1)!}\left(\frac{(n+l)!}{(n-l-1)2!}\right)^{1/2}(2\alpha)^{3/2}e^{-\alpha
r}\nonumber
\times \ \left( 2 \alpha r \right)^l\,_{1}F_{1}\left(-n+l+1,2l+2; 2 \alpha r \right)
\end{equation}

with <math>\alpha = (-2E)^{1/2}</math> and <math>n\geq l+1</math>. The confluent hypergeometric function <math> _1F_1(a,b,;z)</math> then denotes a finite polynomial since <math>a=-n+l+1</math> is a negative integer or zero.

Unlike the hydrogen spectrum, which has both a discrete part for <math>E<0</math> and a
continuous part for <math>E>0</math>, this forms an entirely discrete set of finite
polynomials, called Sturmian functions. They are orthogonal with respect to
the potential, i.e.

\begin{equation}
\int^\infty_0 r^2dr \left( u_{n^\prime l}(r)\frac{1}{r}u_{nl}(r) \right) = \delta_{n,n^\prime}\nonumber
\end{equation}

Since they become complete in the limit <math>n\rightarrow\infty</math>, this assures the
completeness of the variational basis set.

[See also B Klahn and W.A. Bingel Theo. Chim. Acta (Berlin) 44, 9 and 27
(1977)].

==Solutions of the Eigenvalue Problem==

For convenience, write

\[
\Psi({\bf r}_1,{\bf r}_2) = \sum^N_{m=1}c_m\varphi_m\nonumber
\]

where <math>m</math> represents the <math>m</math>'th combination of <math>i,j,k</math> and

\begin{equation}
\varphi_{ijk}=r_1^ir_2^jr_{12}^ke^{-\alpha r_1 - \beta
r_2}\mathcal{Y}^M_{l_1,l_2,L}(\hat{r}_1,\hat{r}_2) \pm \mbox{exchange}\nonumber
\end{equation}

\begin{align}

&
\left( \begin{array}{lr}
\cos(\theta) & \sin(\theta)\\
-\sin(\theta) & \cos(\theta)
\end{array} \right)
\left( \begin{array}{lr}
H_{11} & H_{12} \\
H_{12} & H_{22}
\end{array}\right)
\left( \begin{array}{lr}
\cos(\theta) & -\sin(\theta)\\
\sin(\theta) & \cos(\theta)
\end{array} \right)
\\[5pt]
= & \left(\begin{array}{lr}
cH_{11}+sH_{12} & cH_{12} + sH_{22}\\
-sH_{11} + cH_{12} & -sH_{12} + cH_{22}
\end{array}\right)
\left( \begin{array}{cc}
c & -s \\
s & c
\end{array}\right)
\\[5pt]
= & \left(\begin{array}{lr}
c^2H_{11}+s^2H_{22} + 2csH_{12} & (c^2-s^2)H_{12}+cs(H_{22}-H_{11})\\
(c^2-s^2)H_{12}+cs(H_{22}-H_{11}) & s^2H_{11}+c^2H_{22}-2csH_{12}
\end{array}\right)
\end{align}

Therefore

\begin{equation}
\left[\cos^2(\theta)-\sin^2(\theta)\right]H_{12} =
\cos(\theta)\sin(\theta)(H_{11}-H_{22})\nonumber
\end{equation}

and

\begin{equation}
\tan(2\theta) = \frac{2H_{12}}{H_{11}-H_{22}}\nonumber
\end{equation}

i.e.

\begin{equation}
\cos(\theta)=\left(\frac{r+\omega}{2r}\right)^{1/2}\nonumber
\end{equation}

\begin{equation}
\sin(\theta)=-\mbox{sgn}(H_{12})\left(\frac{r-\omega}{2r}\right)^{1/2}\nonumber
\end{equation}

where

\begin{equation}
\omega = H_{22}-H_{11}\nonumber
\end{equation}

\begin{equation}
r=\left(\omega^2+4H_{12}^2\right)^{1/2}\nonumber
\end{equation}

\begin{equation}
E_1=\frac{1}{2}\left(H_{11}+H_{22}-r\right)\nonumber
\end{equation}

\begin{equation}
E_2=\frac{1}{2}\left(H_{11}+H_{22}+r\right)\nonumber
\end{equation}

===Brute Force Method===

- Gives all the eigenvalues and eigenvectors, but it is slow

- First orthonormalize the <math>N</math>-dimensional basis set; i.e. form linear combinations

\begin{equation}
\Phi_m = \sum_{n=1}^N\varphi_nR_{nm}\nonumber
\end{equation}

such that
\begin{equation}
\langle \Phi_m | \Phi_n \rangle = \delta_{m,n}
\end{equation}

This can be done by finding an orthogonal tranformation T such that

\begin{equation}
T^TOT = I = \left(
\begin{array}{ccccc}
I_1 & 0 & 0 & \ldots & 0 \\
0 & I_2 & 0 & \ldots & 0 \\
0 & 0 & I_3 & \ldots & 0 \\
\vdots & \vdots & \vdots & \ddots & \vdots \\
0 & 0 & 0 & \ldots & I_N
\end{array}\right);
\end{equation}

\begin{equation}
O_{mn} = \langle \varphi_m | \varphi_n \rangle \nonumber
\end{equation}

and then applying a scale change matrix

\begin{equation}
S = \left(\begin{array}{ccccc}
\frac{1}{I_1^{1/2}} & 0 & 0 & \ldots & 0\\
0 & \frac{1}{I_2^{1/2}} & 0 & \ldots & 0 \\
0 & 0 & \frac{1}{I_3^{1/2}} & \ldots & 0 \\
\vdots & \vdots & \vdots & \ddots & \vdots \\
0 & 0 & 0 & \ldots & I_N^{1/2}
\end{array}\right)= S^T\nonumber
\end{equation}

Then

\begin{equation}
S^TT^TOTS = 1
\end{equation}

i.e.

\begin{equation}
R^TOR = 1
\end{equation}

with <math>R=TS</math>.

If H is the matrix with elements <math>H_{mn} = \langle \varphi_m | \varphi_n \rangle </math>, then H
expressed in the <math>\Phi_m</math> basis set is

\begin{equation}
H^\prime = R^THR.\nonumber
\end{equation}

We next diagonalize <math>H^\prime</math> by finding an orthogonal transformation W such
that

\begin{equation}
W^T H^\prime W = \lambda = \left(
\begin{array}{cccc}
\lambda_1 & 0 & \ldots & 0 \\
0 & \lambda_2 & \ldots & 0 \\
\vdots & \vdots & \ddots & \vdots \\
0 & 0 & \ldots & \lambda_N
\end{array}\right)\nonumber
\end{equation}

The q'th eigenvector is

\begin{equation}
\Psi^{(q)} = \sum_{n=1}^N\Phi_n W_{n,q} = \sum_{n,n^\prime} \varphi_{n^\prime} R_{n^\prime ,n} W_{n,q} \nonumber
\end{equation}

i.e.

\begin{equation}
c_{n^\prime}^{(q)} = \sum_{n=1}^N R_{n^\prime n} W_{n,q}
\end{equation}

===The Power Method===

- Based on the observation that if H has one eigenvalue, <math>\lambda_M</math>, much bigger than all the rest, and <math>\chi = \left(
\begin{array}{c}a_1\\a_2\\\vdots\end{array}\right)</math>
is an arbitrary starting vector, then <math>\chi = \sum_{q=1}^N x_q\Psi^{(q)}</math>.

\begin{equation}
(H)^n\chi = \sum_{q=1}^N x_q \lambda^n_q\Psi^{(q)}\nonumber\rightarrow x_M\lambda_M^n\Psi^{(M)}\nonumber
\end{equation}

provided <math>x_M\neq 0</math>.

To pick out the eigenvector corresponding to any eigenvalue, write the original problem in the form

\begin{align}
H\Psi & = \lambda O \Psi \nonumber \\
\left( H - \lambda_q O \right) \Psi & = \left( \lambda - \lambda_q \right) O \Psi \nonumber
\end{align}

Therefore,

\begin{equation}
G \Psi = \frac{1}{\lambda - \lambda_q } \Psi \nonumber \\
\end{equation}

where <math>G=(H-\lambda_qO)^{-1}O</math> with eigenvalues <math>\frac{1}{\lambda_n - \lambda_q}</math>.

By picking <math>\lambda_q</math> close to any one of the <math>\lambda_n</math>, say
<math>\lambda_{n^\prime}</math>, then <math>\frac{1}{\lambda_n-\lambda_q}</math> is much larger for
<math>n=n^\prime</math> than for any other value. The sequence is then

\begin{array}{c}
\chi_1=G\chi\nonumber\\
\chi_2=G\chi_1\nonumber\\
\chi_3=G\chi_2\nonumber\\
\vdots\nonumber
\end{array}

until the ratios of components in <math>\chi_n</math> stop changing.

- To avoid matrix inversion and multiplication, note that the sequence is equivalent to
\begin{equation}
F \chi_n = \left( \lambda - \lambda_q \right) O \chi_{n-1} \nonumber \\
\end{equation}
where <math>F = H-\lambda_qO</math>. The factor of <math>\left( \lambda - \lambda_q \right)</math> can be
dropped because this only affects the normalization of <math>\chi_n</math>. To find <math>\chi_n</math>, solve
\begin{equation}
F \chi_n = O \chi_{n-1} \nonumber
\end{equation}
(N equations in N unknowns). Then
\begin{equation}
\lambda = \frac{ \langle \chi_n | H | \chi_n \rangle}{\langle \chi_n | \chi_n \rangle} \nonumber
\end{equation}

==Matrix Elements of H==

\begin{equation}
H=-\frac{1}{2}\nabla^2_1 -\frac{1}{2}\nabla^2_2 - \frac{1}{r_1} - \frac{1}{r_2} +\frac{Z^{-1}}{r_{12}}\nonumber
\end{equation}

Taking <math>r_1, r_2</math> and <math>r_{12}</math> as independent variables,

\begin{equation}
\nabla_1^2 = \frac{1}{r_1^2}\frac{\partial}{\partial r_1}\left(r_1^2\frac{\partial}{\partial r_1}\right)+ \frac{1}{r_{12}^2}\frac{\partial}{\partial r_{12}}
\left(r_{12}^2\frac{\partial}{\partial r_{12}}\right)\nonumber -\frac{l_1(l_1+1)}{r_1^2}+2(r_1 - r_2\cos(\theta_{12}))\frac{1}{r_{12}}\frac{\partial^2}{\partial r_1 \partial r_{12}} \nonumber - 2(\nabla_1^Y \cdot {\bf r}_2)\frac{1}{r_{12}}\frac{\partial}{\partial r_{12}}\nonumber
\end{equation}

where <math>\nabla_1^Y</math> acts only on the spherical harmonic part of the wave function and the diagram

[[File:Pic3.jpg|400px]]

defines the complete set of 6 independent variables is <math>r_1</math>, <math>r_2</math>, <math>r_{12}</math>, <math>\theta_1</math>, <math>\varphi_1</math>, <math>\chi</math>.

If <math>r_{12}</math> were not an independent variable, then one could take the volume element to be

\begin{equation}
d\tau = r_1^2dr_1\sin(\theta_1)d\theta_1d\varphi_1r_2^2dr_2\sin(\theta_2)d\theta_2d\varphi_2.\nonumber
\end{equation}

However, <math>\theta_2</math> and <math>\varphi_2</math> are no longer independent variables. To eliminate them, take the point <math>{\bf r}_1</math> as the
origin of a new polar co-ordinate system, and write

\begin{equation}
d\tau=-r_1^2dr_1\sin(\theta_1)d\theta_1d\varphi_1r_{12}^2dr_{12}\sin(\psi)d\psi d\chi\nonumber
\end{equation}

and use

\begin{equation}
r_2^2=r_1^2+r_{12}^2 +2r_1r_{12}\cos(\psi)
\end{equation}

Then for fixed <math>r_1</math> and <math>r_{12}</math>,

\begin{equation}
2r_2dr_2 = -2r_1r_{12}\sin(\psi)d\psi
\end{equation}

Thus

\begin{equation}
d\tau= r_1dr_1r_2dr_2r_{12}dr_{12}\sin(\theta_1)d\theta_1d\varphi_1d\chi
\end{equation}

The basic type of integral to be calculated is

\begin{align}
I \left( l_1, m_1, l_2, m_2; R \right) & = \int \sin \left( \theta_1 \right) d\theta_1 d\varphi_1 d\chi Y^{m_1}_{l_1} \left( \theta_1, \varphi_1 \right)^{*} Y^{m_2}_{l_2} \left( \theta_2, \varphi_2 \right) \\
& \times \int r_1dr_1r_2dr_2r_{12}dr_{12}R \left( r_1, r_2, r_{12} \right)
\end{align}

Consider first the angular integral. <math>Y^{m_2}_{l_2} \left( \theta_2, \varphi_2 \right)</math> can be expressed in terms of the independent variables <math>\theta_1, \varphi_1, \chi</math> by use of the rotation matrix relation
\begin{equation}
Y^{m_2}_{l_2}( \theta_2, \varphi_2) = \sum_m \mathcal{D}^{(l_2)}_{m_2,m}(\varphi_1,\theta_1,\chi)^*Y^m_{l_2}( \theta_{12}, \varphi)
\end{equation}
where <math>\theta, \varphi</math> are the polar angles of <math>{\bf r}_2</math> relative to <math>{\bf r}_1</math>. The angular integral is then

\begin{align}
I_{ang} & = \int^{2 \pi}_{0} d\chi \int^{2\pi}_{0} d\varphi_1 \int^{\pi}_{0} \sin \left( \theta_1 \right) d\theta_{1} Y^{m_1}_{l_1} \left( \theta_1, \varphi_1 \right)^{*} \\
& \times \sum_m \mathcal{D}^{(l_2)}_{m_2,m} \left( \varphi_1,\theta_1,\chi \right)^{*} Y^{m}_{l_2} \left( \theta_{12},\varphi \right) \\
\end{align}

Use

\begin{equation}
Y^{m_1}_{l_1} \left( \theta_1, \varphi_1 \right)^{*} = \sqrt{\frac{2l_1+1}{4\pi}}\mathcal{D}^{(l_1)}_{m_1,0} \left( \varphi_1, \theta_1, \chi \right)
\end{equation}

together with the orthogonality property of the rotation matrices (Brink and Satchler, p 147)

\begin{align}
& \int \mathcal{D}^{(j)*}_{m,m^\prime}\mathcal{D}^{(J)}_{M,M^\prime}\sin(\theta_1)d\theta_1d\varphi_1d\chi \\
& = \frac{8\pi^2}{2j+1}\delta_{jJ}\delta_{mM}\delta_{m^\prime M^\prime}\nonumber
\end{align}

to obtain

\begin{align}
I_{ang} & = \sqrt{ \frac{2l_1+1}{4\pi}} \frac{8\pi^2}{2l_1+1} \delta_{l_1,l_2} \delta_{m_1,m_2} Y^{0}_{l_2} \left( \theta_{12}, \varphi \right) \\
& = 2 \pi \delta_{l_1,l_2} \delta_{m_1,m_2} P_{l_2} \left( \cos \theta_{12} \right)
\end{align}

since

\begin{equation}
Y^0_{l_2} \left( \theta_{12}, \varphi \right) = \sqrt{\frac{2l_1+1}{4\pi}} P_{l_2} \left( \cos \left( \theta_{12} \right) \right)
\end{equation}

Note that <math>P_{l_2} \left( \cos \theta \right)</math> is just a short hand expression for a radial function because

\begin{equation}
\cos \theta_{12} = \frac{r_1^2 + r_2^2 - r_{12}^2}{2r_1r_2}
\end{equation}

The original integral is thus

\begin{equation}
I \left( l_1, m_1, l_2, m_2; R \right) = 2 \pi \delta_{l_1,l_2} \delta_{m_1,m_2} \int^{\infty}_{0} r_{1} dr_{1} \int^{\infty}_{0} r_{2} dr_{2} \int^{r_{1} + r_{2}}_{|r_{1} - r_{2}|} r_{12} dr_{12} R \left( r_{1}, r_{2}, r_{12} \right) P_{l_2} \left( \cos \theta_{12} \right)
\end{equation}

where again

\begin{equation}
\cos \theta_{12} = \frac{r_{1}^{2} +r^{2}_{2} - r_{12}^{2}}{2 r_{1} r_{2}}
\end{equation}

is a purely radial function.

The above would become quite complicated for large <math>l_2</math> because <math>P_{l_2}(\cos \theta_{12})</math> contains terms up to <math>\left( \cos \theta_{12} \right)^{l_2}</math>. However, recursion relations exist which allow any integral containing <math>P_l \left( \cos \theta_{12} \right)</math> in terms of those containing just <math>P_0 \left( \cos \theta_{12} \right) = 1</math> and <math>P_1 \left( \cos \theta_{12} \right) = \cos \theta_{12}</math>.

==Radial Integrals and Recursion Relations==

The basic radial integral is [see G.W.F. Drake, Phys. Rev. A 18, 820 (1978)]

\begin{align}
I_0 \left( a, b, c \right) & =
\int^{\infty}_{0} r_{1} dr_{1} \int^{\infty}_{r_{1}} r_{2} dr_{2} \int^{r_{1}+r_{2}}_{r_{2} - r_{1}} r_{12} dr_{12} r_{1}^{a} r_{2}^{b} r_{12}^{c} e^{-\alpha r_{1} - \beta r_{2}} + \int^{\infty}_{0} r_{2} dr_{2} \int^{\infty}_{r_{2}} r_{1} dr_{1} \int^{r_{1} + r_{2}}_{r_{1} - r_{2}} r_{12} dr_{12} r_{1}^{a} r_{2}^{b} r_{12}^{c} e^{- \alpha r_{1} - \beta r_{2}} \\
& = \frac{2}{c+2} \sum^{[(c+1)/2]}_{i=0}
\left(\begin{array}{c}
c+2 \\
2i+1
\end{array} \right)
\left\{ \frac{q!}{\beta^{q+1} \left( \alpha + \beta \right)^{p+1}} \sum_{j=0}^{q} \frac{\left( p + j \right)!}{j!} \left( \frac{\beta}{\alpha + \beta} \right)^j + \frac{q^{\prime}!}{\alpha^{q^{\prime}+1} \left( \alpha + \beta \right)^{p^{\prime}+1}} \sum^{q^{\prime}}_{j=0} \frac{\left( p^{\prime} + j \right)!}{j!} \left( \frac{\alpha}{\alpha + \beta} \right)^j \right\}
\end{align}

where

\begin{array}{ll}
p = a + 2i + 2 & p^{\prime} = b + 2i + 2 \\
q = b + c - 2i + 2 & q^{\prime} = a + c - 2i + 2
\end{array}

The above applies for <math> a, b \geq -2, c\geq -1</math>.
<math>[x]</math> means "greatest integer in" <math>x</math>.

Then

\begin{align}
I_1 \left( a, b, c \right) & = \int d\tau_{r} r_{1}^{a} r_{2}^{b} r_{12}^{c} e^{ -\alpha r_{1} - \beta r_{2}} P_{1}\left( \cos \theta \right) \\
& = \frac{1}{2} \left( I_{0} \left( a + 1, b - 1, c \right) + I_{0} \left( a - 1, b + 1, c \right) - I_{0} \left( a - 1, b - 1, c + 2 \right) \right)
\end{align}

===The Radial Recursion Relation===

Recall that the full integral is

\begin{equation}
I(l_1m_1,l_2m_2;R)=2\pi\delta_{l_1,l_2}\delta_{m_1,m_2}I_{l_2}(R)
\end{equation}

where

\begin{equation}
I_{l_2}(R) = \int d\tau_r R(r_1,r_2,r_{12})P_{l_2}(\cos \theta_{12})
\end{equation}

To obtain the recursion relation, use

\begin{equation}
P_l(x) = \frac{[P^{\prime}_{l+1}(x)-P^{\prime}_{l-1}(x)]}{2l+1}
\end{equation}

with

\begin{equation}
P_{l+1}^\prime (x) = \frac{d}{dx}P_{l+1}(x)
\end{equation}

Here <math>x=\cos\theta_{12}</math> and

\begin{align}
\frac{d}{d \cos \theta_{12}} & = \frac{dr_{12}}{d \cos \theta_{12}} \frac{d}{dr_{12}} \\
& = -\frac{r_{1} r_{2}}{r_{12}} \frac{d}{dr_{12}}
\end{align}

Then,

\begin{equation}
I_{l}\left( R \right) = - \int d\tau_r R \frac{r_{1} r_{2}}{r_{12}} \frac{d}{dr_{12}} \frac{ \left[ P_{l+1} \left( \cos \theta_{12} \right) - P_{l-1} \left( \cos \theta_{12} \right) \right]}{2l+1}
\end{equation}

The <math>r_{12}</math> part of the integral is

\begin{align}
& \int^{r_{1} + r_{2}}_{|r_{1} - r_{2}|} r_{12} dr_{12} R \frac{r_{1} r_{2}}{r_{12}} \frac{d}{dr_{12}} \left[P_{l+1}-P_{l-1} \right] \\
& = R r_{1} r_{2} \left[ P_{l+1} - P_{l-1} \right]^{r_{1} + r_{2}}_{|r_{1} - r_{2}|} - \int^{r_{1} + r_{2}}_{|r_{1} - r_{2}|} r_{12} dr_{12} \left( \frac{d}{dr_{12}} R \right) \frac{r_{1} r_{2}}{r_{12}} \frac{ \left[ P_{l+1} - P_{l-1} \right]}{2l+1}
\end{align}

The integrated term vanishes because

\begin{equation}
\cos \theta_{12} = \frac{r_{1}^{2} + r_{2}^{2} - r_{12}^{2}}{2 r_{1} r_{2}}
\begin{array}{lll}
& = -1 & {\rm when} & r_{12}^{2} = \left( r_{1} + r_{2} \right)^{2} \\
& = 1 & {\rm when} & r_{12}^{2} = \left( r_{1} - r_{2} \right)^{2}
\end{array}
\end{equation}

and <math>P_l(1)=1</math>, <math>P_{l}(-1)=(-1)^{l}</math>.

Thus,

\begin{equation}
I_{l+1}\left(\frac{r_1r_2}{r_{12}}R^\prime\right) = \left( 2l+1 \right) I_{l}(R) - I_{l-1} \left( \frac{r_{1} r_{2}}{r_{12}} R^{\prime} \right)
\end{equation}

If

\begin{equation}
R = r_{1}^{a-1} r_{2}^{b-1} r_{12}^{c+2} e^{-\alpha r_{1} - \beta r_{2}}
\end{equation}

then [G.W.F. Drake, Phys. Rev. A 18, 820 (1978)]

\begin{equation}
I_{l+1}(r_1^ar_2^br_{12}^c) =
\frac{2l+1}{c+2} I_{l} \left( r_{1}^{a-1} r_{2}^{b-2} r_{12}^{c+2} \right)+I_{l-1} \left( r_{1}^{a} r_{2}^{b} r_{12}^{c} \right)
\end{equation}

For the case <math>c=-2</math>, take

\begin{equation}
R = r_{1}^{a-1} r_{2}^{b-1} \ln r_{12} e^{ -\alpha r_{1} -\beta r_{2}}
\end{equation}

Then

\begin{equation}
I_{l+1} \left( r_{1}^{a} r_{2}^{b} r_{12}^{c} \right) = I_{l} \left( r_{1}^{a-1} r_{2}^{b-1} \ln r_{12} \right) \left( 2l + 1 \right) - I_{l-1} \left( r_{1}^{a} r_{2}^{b} r_{12}^{-2} \right)
\end{equation}

This allows all <math>I_{l}</math> integrals to be calculated from tables of <math>I_0</math> and
<math>I_1</math> integrals.

===The General Integral===

The above results for the angular and radial integrals can now be combined into a general formula for integrals of the type

\begin{equation}
I = \int \int d{\bf r}_{1} d{\bf r}_{2} R_{1} \mathcal{Y}^{M^{\prime}}_{l^{\prime}_{1} l^{\prime}_{2} L^{\prime}} \left( \hat{r}_{1}, \hat{r}_{2} \right) T_{k_{1} k_{2} K}^{Q} \left( {\bf r}_{1},{\bf r}_{2} \right) R_{2} \mathcal{Y}_{l_{1} l_{2} L}^{M} \left( \hat{r}_{1}, \hat{r}_{2} \right)
\end{equation}

where

\begin{equation}
\mathcal{Y}^{M}_{l_{1} l_{2} L} \left( \hat{r}_{1}, \hat{r}_{2} \right) = \sum_{m_1,m_2} \langle l_{1} l_{2} m_{1} m_{2} | LM \rangle Y^{m_{1}}_{l_{1}} \left( \hat{r}_{1} \right) Y^{m_{2}}_{l_{2}} \left( \hat{r}_{2} \right)
\end{equation}

and

\begin{equation}
T^Q_{k_{1} k_{2} K} \left( {\bf r}_{1}, {\bf r}_{2} \right) = \sum_{q_{1}, q_{2}} \langle k_{1} k_{2} q_{1} q_{2} | KQ \rangle Y_{k_{1}}^{q_{1}} \left( \hat{r}_{1} \right) Y^{q_{2}}_{k_{2}} \left( \hat{r}_{2} \right)
\end{equation}

The basic idea is to make repeated use of the formula

\begin{equation}
Y^{m_{1}}_{l_{1}} \left( \hat{r}_{1} \right) Y^{m_{2}}_{l_{2}} \left( \hat{r}_{1} \right) = \sum_{l m} \left( \frac{\left( 2l_{1}+1 \right) \left(2l_{2}+1 \right) \left(2l+1 \right)}{4 \pi} \right)^{1/2}
\times \left( \begin{array}{ccc}
l_1 & l_2 & l \\
m_1 & m_2 & m
\end{array} \right) \left(\begin{array}{ccc}
l_1 & l_2 & l \\
0 & 0 & 0
\end{array} \right)
Y^{m*}_{l} \left( \hat{r}_{1} \right)
\end{equation}

where

\begin{equation}
Y^{m*}(\hat{r}) = (-1)^mY_l^{-m}(\hat{r})
\end{equation}

and

\begin{equation}
\left(\begin{array}{ccc}
l_1 & l_2 & l \\
m_1 & m_2 & m
\end{array} \right) = \frac{\left( -1 \right)^{l_{1} - l_{2} - m}}{\left( 2l + 1 \right)^{1/2}} \left( l_{1} l_{2} m_{1} m_{2} | l, -m \right)
\end{equation}

is a 3-<math>j</math> symbol. In particular, write

\begin{array}{cc}
Y^{m_1^\prime}_{l_1^\prime}(\hat{r}_1)^{*} \underbrace{Y_{k_{1}}^{q_{1}} \left( \hat{r}_{1} \right) Y^{m_{1}}_{l_{1}} \left( \hat{r}_{1} \right)} & = & \sum_{\Lambda M}(\ldots)Y^{M}_\Lambda(\hat{r}_1)\\
\ \ \ \ \ \ \ \sum_{\lambda_1\mu_1}Y_{\lambda_1}^{\mu_1}(\hat{r}_1) & & \\
& & \\
Y^{m_{2}^{\prime}}_{l_{2}^{\prime}} \left( \hat{r}_{2} \right) \underbrace{Y^{q_{2}}_{k_{2}} \left( \hat{r}_{2} \right) Y_{l_{2}}^{m_{2}} \left( \hat{r}_{2} \right)} & = & \sum_{\Lambda^{\prime} M^{\prime}} \left( \ldots \right)Y_{\Lambda^{\prime}}^{M^{\prime}*} \left( \hat{r}_{2} \right) \\
\ \ \ \ \ \ \ \ \ \sum_{\lambda_{2} \mu_{2}}Y_{\lambda_{2}}^{\mu_{2}} \left( \hat{r}_{2} \right) & &
\end{array}

The angular integral then gives a factor of <math>2 \pi \delta_{\Lambda, \Lambda^{\prime}} \delta_{M, M^{\prime}} P_{\Lambda} \left( \cos \theta_{12} \right)</math>. The total integral therefore reduces to the form

\begin{equation}
I = \sum_{\Lambda} C_{\Lambda} I_{\Lambda} \left( R_{1} R_{2} \right)
\end{equation}

where <math>C_{\Lambda} = \sum_{\lambda_{1}, \lambda_{2}} C_{ \lambda_{1}, \lambda_{2}, \Lambda}</math>. For further details and derivations, including graphical representations, see G.W.F. Drake, Phys. Rev. A <math>{\bf 18}</math>, 820 (1978).

==Graphical Representation==

[[File:GraphRep1.jpg|500px]]

==Matrix Elements of H==
Recall that

\begin{equation}
H = -\frac{1}{2} \nabla^{2}_{1} - \frac{1}{2} \nabla^{2}_{2} - \frac{1}{r_1} - \frac{1}{r_2} + \frac{Z^{-1}}{r_{12}}
\end{equation}

Consider matrix elements of

\begin{equation}
\nabla^2_1 = \frac{1}{r_1^2} \frac{\partial}{\partial r_1}\left(r_1^2\frac{\partial}{\partial r_1} \right) + \frac{1}{r^2}\frac{\partial}{\partial r}\left(r^2\frac{\partial}{\partial r}\right) - \frac{(\vec{l}_1^Y)^2}{r_1^2} + \frac{2 \left( r_1 - r_2 \cos \theta \right)}{r} \frac{\partial^2}{\partial r_1 \partial r} - 2 \left( \nabla^Y_1 \cdot {\bf r}_2 \right) \frac{1}{r} \frac{\partial}{\partial r}
\end{equation}

where <math>r \equiv r_{12}</math> and <math>\cos\theta \equiv \cos\theta_{12}</math>. Also in what follows, define <math>\hat{\nabla}^Y_1 \equiv r_1\nabla^Y_1</math> where <math>\nabla^Y</math> operates only on the spherical harmonic part of the wave function, and similarly for <math>\vec{l}_1^Y</math>.

A general matrix element is

\begin{equation}
\langle r_1^{a^\prime} r_2^{b^\prime} r_{12}^{c^\prime} e^{-\alpha^\prime r_1 - \beta^\prime r_2} \mathcal{Y}^{M^\prime}_{l^\prime_1 l^\prime_2 L^\prime} \left( \hat{r}_1, \hat{r}_2 \right) \left| \nabla^2_1 \right| r_1^a r_2^b r_{12}^c e^{-\alpha r_1 - \beta r_2} \mathcal{Y}^M_{l_1 l_2 L} \left( \hat{r}_1, \hat{r}_2 \right) \rangle
\end{equation}

Since <math>\nabla^2_1</math> is rotationally invariant, this vanishes unless <math>L=L^\prime</math>, <math>M=M^\prime</math>. Also <math>\nabla_1^2</math> is Hermitian, so that the result must be the same whether it operates to the right or left, even though the results look very different. In fact, requiring the results to be the same yields some interesting and useful integral identities as follows:

Put

\begin{equation}
\mathcal{F} = F \mathcal{Y}^M_{l_1 l_2 L} \left( \hat{r}_1, \hat{r_2} \right)
\end{equation}

and

\begin{equation}
F = r_1^a r_2^b r_{12}^c e^{-\alpha r_1 - \beta r_2}
\end{equation}

Then

\begin{equation}
\nabla^2_1 \mathcal{F} = \left\{ \frac{1}{r_1^2} \left[ a \left( a + 1 \right) - l_1 \left( l_1 + 1 \right) \right] + \frac{c \left( c + 1 \right)}{r^2} + \alpha^2\frac{2\alpha \left( a + 1 \right)}{r_1} + \frac{2 \left(r_1 - r_2 \cos \theta \right) }{r_1r^2}c \left[ a - \alpha r_1 \right] - \frac{2c}{r^2} \left( \hat{\nabla}^Y_1 \cdot \hat{r}_2 \right) \frac{r_2}{r_1} \right\} \mathcal{F}
\end{equation}

and

\begin{align}
\langle \mathcal{F}^\prime \left| \nabla^2_1 \right| \mathcal{F} \rangle & = \sum_\Lambda \int d\tau_r F^\prime C_\Lambda \left( 1 \right) P_\Lambda \left( \cos \theta \right) \times \left\{ \frac{1}{r_1^2} \left[ a \left( a + 1 \right) - l_1 \left( l_1 + 1 \right) \right] -\frac{2 \alpha \left( a + 1 \right)}{r_1} + \frac{c \left( c + 1 \right)}{r^2} +\alpha^2 + \frac{2 \left( r_2 - r_2 \cos \theta \right)}{r_1r^2}c \left( a - \alpha r_1 \right) \right\} F \\
& + \sum_\Lambda \int d\tau_r F^\prime C_\Lambda \left( \hat{\nabla}^Y_1 \cdot \hat{r}_2 \right) P_\Lambda \left( \cos \theta \right) \left( \frac{-2c r_2}{r_1 r^2} \right) F
\end{align}

where

\begin{equation}
\int d\tau_r = \int^\infty_0 r_1 dr_1 \int^\infty_0 r_2 dr_2 \int^{r_1+r_2}_{\left| r_1 - r_2 \right|} r dr
\end{equation}

For brevity, let the sum over <math>\Lambda</math> and the radial integrations be understood, and let <math>\left( \nabla_1^2 \right)</math> stand for the terms which appear in the integrand. Then operating to the right gives

\begin{equation}
(\nabla_1^2)_R = \frac{1}{r_1^2} \left[ a \left( a + 1 \right) - l_1 \left( l_1 + 1 \right) \right] + \frac{c \left( c + 1 \right)}{r^2} + \alpha^2 - \frac{2 \alpha \left( a + 1 \right)}{r_1}
+ \frac{2 \left( r_1 - r_2 \cos \theta \right)}{r_1 r^2}c \left( a - \alpha r_1 \right) - \frac{2c r_2}{r_1 r^2} \left( \hat{\nabla}^Y_1 \cdot \hat{r}_2 \right)
\end{equation}

Operating to the left gives

\begin{equation}
(\nabla_1^2)_L = \frac{1}{r_1^2} \left[a^\prime \left( a^\prime + 1 \right) - l_1^\prime \left( l_1^\prime + 1 \right) \right] + \frac{c^\prime \left( c^\prime + 1 \right)}{r^2} + \alpha^{\prime 2} -\frac{2 \alpha^\prime \left( a^\prime + 1 \right)}{r_1}
+ \frac{2 \left( r_1 - r_2 \cos \theta \right)}{r_1 r^2} c^\prime \left( a^\prime - \alpha^\prime r_1 \right) - \frac{2 c^\prime r_2}{r_1 r^2} \left( \hat{\nabla}^{Y^\prime}_1 \cdot \hat{r}_2 \right)
\end{equation}

Now put

\begin{array}{cc}
a_+ = a+a^\prime, & \hat{\nabla}_1^+ = \hat{\nabla}_1^Y + \hat{\nabla}_1^{Y\prime} \\
a_{-} = a-a^\prime, & \hat{\nabla}_1^{-} = \hat{\nabla}_1^Y - \hat{\nabla}_1^{Y\prime}
\end{array}

etc., and substitute <math>a^\prime = a_+ - a</math>, <math>c^\prime = c_+ - c</math>, and <math>\alpha^\prime = \alpha_+ - \alpha</math> in <math>(\nabla^2_1)_L</math>. If <math>a_+</math>, <math>c_+</math> and <math>\alpha_+</math> are held fixed, then the equation

\begin{equation}
(\nabla^2_1)_R = (\nabla^2_1)_L
\end{equation}

must be true for arbitrary <math>a</math>, <math>c</math>, and <math>\alpha</math>. Their coefficients must thus vanish.

This yields the integral relations

\begin{equation}
\label{I}
\frac{\left( r_1 - r_2 \cos \theta \right)}{r_1 r^2} = \frac{1}{c_+} \left( \frac{- \left( a_+ + 1 \right)}{r_1^2} + \frac{\alpha_+}{r_1} \right) \tag{I}
\end{equation}

from the coefficient of <math>a</math> and

\begin{equation}
\label{II}
\frac{ \left( r_1 - r_2 \cos \theta \right) \left( a_+ - \alpha_+ r_1 \right)}{r_1 r^2} = \frac{r_2}{r_1 r^2}\left( \hat{r}_2 \cdot \hat{\nabla}_1^+ \right) - \frac{ \left( c_+ + 1 \right)}{r^2} \tag{II}
\end{equation}

from the coefficient of <math>c</math>. The coefficient of <math>\alpha</math> gives an equation equivalent to (I).

Furthermore, if can be show that (see problem)

\begin{align}
& \sum_\Lambda \int d\tau_r \frac{r^c}{r^2} C_\Lambda \left( \hat{r}_2 \cdot \hat{\nabla}_1^Y \right) P_\Lambda \left( \cos \theta \right) \\
= & \sum_\Lambda \int d\tau_r \frac{r^c}{c r_1 r_2}C_\Lambda \left( 1 \right)P_\Lambda \left( \cos \theta \right) \times \left( \frac{l_1^\prime \left( l_1^\prime + 1 \right) - l_1 \left( l_1 + 1 \right) -\Lambda \left( \Lambda + 1 \right)}{2} \right)
\end{align}

and similarly for <math>\left( \hat{r}_2 \cdot \hat{\nabla}^{Y\prime}_1 \right)</math> with <math>l_1</math> and <math>l_1^\prime</math> interchangeable, then it follows that

\begin{equation}
\frac{r^{c_+}}{r^2} \left( \hat{r}_2 \cdot \hat{\nabla}_1^+ \right) = - \frac{r^{c_+}}{c_+ r_1 r_2} \Lambda \left( \Lambda + 1 \right)
\end{equation}

and

\begin{equation}
\frac{r^{c_+}}{r^2} \left( \hat{r}_2 \cdot \hat{\nabla}_1^- \right) = \frac{r^{c_+}}{c_+ r_1 r_2} \left[ l_1^\prime \left( l_1^\prime + 1 \right) - l_1 \left( l_1 + 1 \right) \right]
\end{equation}

where equality applies after integration and summation over <math>\Lambda</math>.

Thus (II) becomes

\begin{equation}\label{IIa}
\frac{ \left( r_1 - r_2 \cos \theta \right) \left( a_+ - \alpha_+ r_1 \right)}{r_1 r^2} = - \frac{\Lambda \left( \Lambda + 1 \right)}{c_+ r^2_1} - \frac{ \left( c_+ + 1 \right)}{r^2} \tag{IIa}
\end{equation}

===Problem===

Prove the integral relation

\begin{align}
& \sum_\Lambda \int d\tau_r f \left( r_1, r_2 \right) \frac{1}{r} \left( \frac{d}{dr} q(r) \right) C_\Lambda \left( \hat{r}_2 \cdot \hat{\nabla}^Y_1 \right) P_\Lambda \left( \cos \theta \right) \\
= & \sum_\Lambda \int d\tau_r \frac{ f \left( r_1, r_2 \right)}{r_1, r_2} q(r) C_\Lambda \left( 1 \right) P_\Lambda \left( \cos \theta \right) \left(\frac{l_1^\prime \left( l_1^\prime + 1 \right) - l_1 \left( l_1 + 1 \right) - \Lambda \left( \Lambda + 1 \right)}{2}\right)
\end{align}

where the coefficients <math>C_\Lambda \left( 1 \right)</math> are the angular coefficients from the overlap integral

\begin{align}
\int & d\Omega\, \mathcal{Y}^{M*}_{l_1^\prime l_2^\prime L}(\hat{r}_1,\hat{r}_2)\, \mathcal{Y}^{M}_{l_1 l_2 L}(\hat{r}_1,\hat{r}_2)\\
= & \sum_\Lambda C_\Lambda(1)P_\Lambda(\cos\theta_{12}).
\end{align}

Hint: Use the fact that <math>l_1^2</math> is Hermitian so that
<math>
\int d\tau (l_1^2\mathcal{Y}^\prime)^*q(r)\mathcal{Y} = \int d\tau
\mathcal{Y}^{\prime *}l_1^2(q(r)\mathcal{Y})
</math>
with <math>\ \vec{l}_1= \frac{1}{i}\vec{r}_1\times\nabla_1</math>.

It is also useful to use
<math>\begin{eqnarray}\ \
\cos\theta\, P_L(\cos\theta) &=& \frac{1}{2L+1}[L\,P_{L-1}(\cos\theta) + (L+1)P_{L+1}(\cos\theta)]
\\
(\cos^2\theta-1)P_L(\cos\theta) &=&
\frac{L(L-1)}{(2L-1)(2L+1)}P_{L-2}(\cos\theta)\nonumber
-\frac{2(L^2+L-1)}{(2L-1)(2L+3)}P_L(\cos\theta) + \frac{(L+1)(L+2)}{(2L+1)(2L+3)}P_{L+2}(\cos\theta)\nonumber\\
&=& \frac{L(L-1)}{(2L-1)(2L+1)}[P_{L-2}(\cos\theta) - P_L(\cos\theta)]
+ \frac{(L+1)(L+2)}{(2L+1)(2L+3)}[P_{L+2}(\cos\theta) - P_L(\cos\theta)]
\end{eqnarray}</math>

together with a double application of the integral recursion relation
<math>
I_{L+1}\left(\frac{1}{r}\frac{d}{dr}q(r)\right) =
(2L+1)I_L\left(\frac{1}{r_1r_2}q(r)\right)+I_{L-1}\left(\frac{1}{r}\frac{d}{dr}q(r)\right)
</math>

Of course <math>l_1^2\mathcal{Y}^M_{l_1l_2L}(\hat{r}_1,\hat{r}_2)=l_1(l_1+1)\mathcal{Y}^M_{l_1l_2L}(\hat{r}_1,\hat{r}_2)</math>. Begin by expanding
<math>
l_1^2(g\mathcal{Y}) = \mathcal{Y}l_1^2g + g\,l_1^2\mathcal{Y} + 2(\vec{l}_1g)\cdot(\vec{l}_1\mathcal{Y})
</math>, and show that

<math>
\begin{eqnarray}\ \
(\vec{l}_1g)\cdot(\vec{l}_1\mathcal{Y}) & = & \frac{r_1^2}{r}\frac{dg}{dr}\vec{r}_2\cdot \nabla_1\mathcal{Y}\,\,\,\,\mbox{ and}\\
l_1^2g(r) &=& 2r_1r_2\cos\theta\frac{1}{r}\frac{dg}{dr} + r_1^2r_2^2(\cos^2\theta - 1)
\frac{1}{r}\frac{d}{dr}\left(\frac{1}{r}\frac{dg}{dr}\right).
\end{eqnarray}
</math>

The proof amounts to showing that the term <math>l_1^2g(r)</math> can be replaced by
<math>\Lambda(\Lambda+1)g(r)</math> after multiplying by <math>P_\Lambda(\cos\theta)</math>
and integrating by parts with respect to the radial integrations over <math>r_1</math>, <math>r_2</math> and
<math>r \equiv |\vec{r}_1 - \vec{r}_2|</math>. Remember that <math>\cos\theta</math> is just a short-hand notation for the radial function <math>(r_1^2 + r_2^2 - r^2)/(2r_1r_2)</math>.

==General Hermitian Property==

Each combination of terms of the form
<math>
<f>_R = a^2f_1+af_2+abf_3+b^2f_4+bf_5+af_6\nabla_1^Y+bf_7\nabla^Y_2+f_8(Y)
</math>
can be rewritten
<math>
<f>_L= (a_+-a)^2f_1+(a_+-a)f_2 + (a_+-a)(b_+-b)f_3+(b_+-b)^2f_4
+(b_+-b)f_5+(a_+-a)f_6\nabla^{Y\prime}_1+(b_+-b)f_7\nabla^{Y\prime}_2+f_8)Y^\prime
</math>
where as usual <math>\nabla_1^Y</math> acts only on the spherical harmonic part of the wave function denoted for short by <math>Y</math>.
Since these must be equal for arbitrary <math>a</math> and <math>b</math>,
<math>
a_+^2f_1+a_+f_2+_+b_+f_3+b_+f_4+b_+f_5+a_+f_6\nabla^{Y\prime}_1 +
b_+f_7\nabla^{Y\prime}_2+f_8(Y^\prime)-f_8(Y)=0
</math>

Adding the corresponding expression with <math>Y</math> and <math>Y^\prime</math> interchanged
yields
<math>\label{III}
a_+^2f_1+a_+f_2+a_+b_+f_3+b_+^2f_4+b_+f_5+\frac{1}{2}a_+f_6\nabla^+_1+\frac{1}{2}b_+f_7\nabla^+_2=0\tag{III}
</math>

Subtracting gives
<math>\label{IV}
f_8(Y)-f_8(Y^\prime)=-\frac{1}{2}[a_+f_6\nabla^-_1+b_+f_7\nabla_2^-]\tag{IV}
</math>
<math>\label{V}
a[-2a_+f_1-2f_2-b_+f_3-f_6\nabla^+_1]=0\tag{V}
</math>
<math>\label{VI}
b[-2b_+f_4-2f_5-a_+f_3-f_7\nabla^+_2]=0\tag{VI}
</math>

Adding the two forms gives

\begin{eqnarray}
<f>_R+<f>_L=\frac{1}{2}(a_+^2+a_-^2)f_1+a_+f_2+\frac{1}{2}(a_+b_++a_-b_-)f_3\nonumber\\
\ +\frac{1}{2}(b_+^2+b_-^2)f_4+b_+f_5+\frac{1}{2}f_6(a_+\nabla_1^++a_-\nabla^-_1)\nonumber\\
\ +\frac{1}{2}(b_+\nabla_2^++b_-\nabla^-_2)+f_8(Y)+f_8(Y^\prime)\nonumber
\end{eqnarray}

Subtracting <math>\frac{x}{2}\times({\rm I})</math> gives

\begin{eqnarray}\ \
<f>_R+<f>_L = \frac{1}{2}[(1-x)a_+^2 + a_-^2]f_1 +
(1-\frac{x}{2})a_+f_2+\frac{1}{2}[(1-x)a_+b_++a_-b_-]f_3\nonumber\\
+\frac{1}{2}[(1-x)b_+^2+b_-^2]f_4+(1-\frac{x}{2})f_5b_++\frac{1}{2}f_6[(1-\frac{x}{2})a_+\nabla_1^++a_-\nabla^-_1]\nonumber\\
+\frac{1}{2}[(1-\frac{x}{2})b_+\nabla^+_2+b_-\nabla^-_2]+f_8(Y)+f_8(Y^\prime)\nonumber
\end{eqnarray}

If <math>x=1</math>,
\begin{eqnarray}
<f>_R+<f>_L &=&
\frac{1}{2}[a_-^2f_1+a_+f_2+a_-b_-f_3+b_-^2f_4+b_+f_5\nonumber\\
&+&\left(\frac{1}{2}a_+\nabla^+_1+a_-\nabla^-_1\right)f_6+\left(\frac{1}{2}b_+\nabla^+_2+b_-\nabla^-_2\right)f_7]+f_8(Y)+f_8(Y^\prime)\nonumber
\end{eqnarray}

The General Hermitian Property for arbitrary <math>x</math> gives
\begin{eqnarray}\ \
\nabla_1^2=
\frac{1}{4}\{\frac{1}{r_1^2}[(1-x)a_+^2+a_-^2+2(1-\frac{x}{2})a_+-2[l_1(l_1+1)+l_1^\prime(l_1^\prime+1)]\nonumber\\
-\frac{2}{r_1}[(1-x)\alpha_+a_++\alpha_-a_-+2(1-\frac{x}{2})\alpha_+]+(1-x)\alpha_+^2
+ \alpha^2_-]\nonumber\\
+\frac{2(r_1-r_2\cos\theta)}{r_1r^2}[(1-x)(a_+-\alpha_+r_1)c_++(a_--\alpha_-r_2)c_-]\nonumber\\
-\frac{2r_2}{r_1r^2}[(1-\frac{x}{2})c_+\hat{r}_2\cdot\hat{\nabla}_1^++c_-\hat{r}_2\hat{\nabla}_1^-]+\frac{(1-x)c_+^2+c_-^2+2(1-\frac{x}{2})c_+}{r^2}\}\nonumber
\end{eqnarray}.

Use
<math>\begin{eqnarray}\ \
\frac{-2r_2}{r_1r^2}[(1-\frac{x}{2})c_+\hat{r}_2\hat{\nabla}_1^++c_-\hat{r}_2\hat{\nabla}_1^-]=\frac{2}{r_1^2}\left[(1-\frac{x}{2})\Lambda(\Lambda+1)-\frac{c_-}{c_+}[l_1^\prime(l_1^\prime+1)+l_1(l_1+1)]\right]\end{eqnarray}</math>,

<math>\begin{eqnarray}
\frac{2(r_1-r_2\cos\theta)}{r_1r^2}(a_--\alpha_-r_1)c_- = 2\frac{c_-}{c_+}\left[\frac{-1}{r_1^2}[a_-(a_++1)]+\frac{1}{r_1}[a_-\alpha_++\alpha_-(a_++2)]-\alpha_-\alpha_+\right]\nonumber
\end{eqnarray}</math>,

and
<math>\begin{eqnarray}\ \
\frac{2(r_1-r_2\cos\theta)}{r_1r^2}(a_+-\alpha_+r_1)c_+ = 2\left[-\frac{1}{r_1^2}[a_+(a_++1)]+\frac{1}{r_1}[a_+\alpha_++\alpha_+(a_++2)]-\alpha_+^2\right]
\end{eqnarray}</math>.

Substituting into <math>\nabla_1^2</math> gives

\begin{eqnarray}
\nabla^2_1 = \frac{1}{4}\{\frac{1}{r_1^2}[-(1-x)a_+^2 + a_-^2 +xa_+ +
2(1-\frac{x}{2})\Lambda(\Lambda+1)\nonumber\\
-2l_1(l_1+1)(1-\frac{c_-}{c_+}) -
2l_1^\prime(l_1^\prime+1)(1+\frac{c_-}{c_+})-\frac{2c_-a_-}{c_+}(a_++1)]\nonumber\\
- \frac{2}{r_1}\left[-(1-x)\alpha_+(a_++2)+\alpha_-a_-+2(1-\frac{x}{2})\alpha_+
-\frac{c_-}{c_+}[a_-\alpha_++\alpha_-(a_++2)]\right]\nonumber\\
-(1-x)\alpha_+^2+\alpha_-^2-2\frac{c_-}{c_+}\alpha_-\alpha_+ +\frac{1}{r^2}\left[(1-x)c_+^2+c_-^2+2(1-\frac{x}{2})c_+\right]\}\nonumber
\end{eqnarray}

This has the form
<math>
\nabla_1^2 =
\frac{1}{4}\left[\frac{A_1}{r_1^2}+\frac{B_1}{r_1}+\frac{C_1}{r^2}+D_1\right]
</math>
with

<math>\begin{array}{ccc}
A_1=-(1-x)a_+^2+a_-^2
+xa_++2(1-\frac{x}{2})\Lambda(\Lambda+1)-2l_1(l_1+1)(1-\frac{c_-}{c_+})\
-2l_1^\prime(l_1^\prime+1)(1+\frac{c_-}{c_+})-\frac{2c_-a_-}{c_+}(a_++1) \ \
\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \
\end{array}
</math>

<math>
B_1 =
2\left[(1-x)\alpha_+(a_++2)-\alpha_-a_--2(1-\frac{x}{2})\alpha_++\frac{c_-}{c_+}[a_-\alpha_++\alpha_-(a_++2)]\right]
</math>

<math>
C_1 = (1-x)c_+^2+ c_-^2 +2(1-\frac{x}{2})c_+
</math>

<math>
D_1 = -1(1-x)\alpha_+^2 +\alpha_-^2 - 2\frac{c_-}{c_+}\alpha_-\alpha_+
</math>

The complete Hamiltonian is then (in Z-scaled a.u.)

<math>\begin{eqnarray}
H =-\frac{1}{2}\nabla^2_1 -\frac{1}{2}\nabla^2_2 -\frac{1}{r_1} -\frac{1}{r_2} +\frac{Z^{-1}}{r}\nonumber\\
= -\frac{1}{2}\nabla^2_1 -\frac{1}{2}\nabla^2_2 -\frac{1}{r_1} -\frac{Z-1}{Zr_2}+Z^{-1}\left(\frac{1}{r}-\frac{1}{r_2}\right)\nonumber\\
\ =-\frac{1}{8}\left[\frac{A_1}{r_1^2}+\frac{B_1+8}{r_1}+\frac{C_1}{r^2}+D_1+D_2 +\frac{A_2}{r_2^2} + \frac{B_2 + 8(Z-1)/Z}{r_2}+\frac{C_2}{r^2}\right]\nonumber
+ Z^{-1}\left(\frac{1}{r}-\frac{1}{r_2}\right)\nonumber
\end{eqnarray}</math>

The screened hydrogenic energy is
<math>\begin{eqnarray}
E_{SH}=-\frac{1}{2}\left[\frac{1}{n_1^2}+ \left(\frac{Z-1}{Z}\right)^2\frac{1}{n_2^2}\right]
\end{eqnarray}</math>
so that

<math>\begin{eqnarray}
H-E_{SH} &=& -\frac{1}{8}[
\frac{A_1}{r_1^2}+\frac{B_1+8}{r_1}+\frac{C_1}{r^2} + D_1 + D_2
-\frac{4}{n_1^2} -\left(\frac{Z-1}{Z}\right)^2\frac{4}{n_2^2}\nonumber
+\frac{A_2}{r_2}+\frac{B_2+8(Z-1)/Z}{r_2}+\frac{C_2}{r^2}] +
Z^{-1}\left(\frac{1}{r} -\frac{1}{r_2}\right)\nonumber
\end{eqnarray}</math>

==Optimization of Nonlinear Parameters==
- The traditional method of performing Hylleraas calculations is to write the
basis set in the form

<math>
\Psi = \sum_{i,j,k}c_{i,j,k}\varphi_{i,j,k}(\alpha,\beta)\pm exchange
</math>

with

<math>
\varphi_{i,j,k}(\alpha,\beta) = r_1^ir_2^jr_{12}^ke^{-\alpha r_1 - \beta
r_2}\mathcal{Y}^M_{l_1l_2L}.
</math>

The usual procedure is to set <math>\alpha = Z</math> so that it represents the inner
<math>1s</math> electron, and then to vary <math>\beta</math> so as to minimize the energy.

[[File:OptNon.jpg|500px]]

Since <math>\beta</math> appears in <math>\Psi</math> as a non-linear parameter, the entire
calculation must be repeated for each value of <math>\beta</math>. However, the minimum
becomes progressively smaller as the basis set is enlarged.

[[File:OptNon02.jpg|500px]]

*Difficulties

1. If the basis set is constructed so that <math> i +j +j \leq N</math>, the the
number of terms is
<math>
(N+1)(N+2)(N+3)/6
</math>
N=14 already gives 680 terms and an accuracy of about 1 part in <math>10^{10}</math> for
low-lying states. A substantial improvement in accuracy would require much
larger basis sets, together with multiple precision arithmetic to avoid loss
of significant figures when high powers are included.

2. The accuracy rapidly deteriorates as one goes to more highly excited
states - about 1 significant figure is lost each time the principle quantum
number is increased.

*Cure

We have found that writing basis sets in the form
\begin{eqnarray}
\Psi &=& \sum_{i,j,k}\left(c^{(1)}_{i,j,k}\varphi_{ijk}(\alpha_1\beta_1) +
c_{ijk}^{(2)}\varphi_{ijk}(\alpha_2\beta_2)\right) \pm exchange \nonumber\\
&=& \psi({\bf r}_1,{\bf r}_2) \pm \psi({\bf r}_2,{\bf r}_1)\nonumber
\end{eqnarray}
so that each combination of powers is included twice with different nonlinear
parameters gives a dramatic improvement in accuracy for basis sets of about
the same total size.

However, the optimization of the nonlinear parameters is now much more
difficult, and an automated procedure is needed.

If
<math>
E = \frac{<\Psi|H|\Psi>}{<\Psi|\Psi>}</math> and we assume <math><\Psi|\Psi> = 1</math>, then

<math>
\frac{\partial E}{\partial\alpha_t} = -2 <\Psi|H-E|r_1\psi({\bf r}_1,{\bf
r}_2;\alpha_t)\pm r_21\psi({\bf r}_2,{\bf
r}_1;\alpha_t)>
</math>
where
<math>
\psi({\bf r}_1,{\bf r}_2;\alpha_t) =
\sum_{i,j,k}c_{ijk}^{(t)}\varphi_{ijk}(\alpha_t\beta_t)
</math>

*Problem

1. Prove the above.

2. Prove that there is no contribution from the implicit dependence of
<math>c_{ijk}^{(t)}</math> on <math>\alpha_t</math> if the linear parameters have been optimized.

==The Screened Hydrogenic Term==

If the Hamiltonian is written in the form (in Z-scaled a.u.)
<math>\begin{eqnarray}\
H = H_0({\bf r}_1,Z) + H_0({\bf r}_2,Z-1) +
Z^{-1}\left(\frac{1}{r}-\frac{1}{r_2}\right)
\end{eqnarray}</math>
with
<math>\
H_0({\bf r}_1,Z) = -\frac{1}{2}\nabla_1^2 - \frac{1}{r_1}\nonumber
</math> and
<math>
H_0({\bf r}_2,Z-1) = -\frac{1}{2}\nabla_2^2 - \frac{Z-1}Z{r_2}\nonumber
</math>,
then the eigenvectors of <math>H_0({\bf r},Z) +H_0({\bf r}_2,Z-1)</math> are products of
hydrogenic orbitals
<math>
\Psi_0=\psi_0(1s,Z)\psi_0(nl,Z-1)
</math>
and the eigenvalue is
<math>\begin{eqnarray}
E_{SH} = \left[-\frac{1}{2} - \left(\frac{Z-1}{Z}\right)^2\frac{1}{2n^2}\right]Z^2 a.u.
\end{eqnarray}</math>
called the screened hydrogenic eigenvalue. From highly excited states,
<math>E_{SH}</math> and <math>\Psi_0</math>are already excellent approximations.

For example, for the 1s8d states, the energies are

<math>\begin{eqnarray}\\
E(1s8d^1D)&=& -2.007\,816\,512\,563\,81 {\rm\ a.u.}\nonumber\\
E(1s8d^3D)&=& - 2.007\,817\,934\,711\,71 {\rm\ a.u.}\nonumber\\
E_{SH} &=& -2.007\,812\,5 {\rm\ a.u.} \nonumber
\end{eqnarray}</math>

It is therefore advantageous to include the screened hydrogenic terms in the
basis set so that the complete trial function becomes
<math>
\Psi = c_0\Psi_0 +
\sum_{ijk}\left[c_{ijk}^{(1)}\varphi_{ijk}(\alpha_1,\beta_1)+c_{ijk}^{(2)}\varphi(\alpha_2,\beta_2)\right]\pm
{\rm exchange}
</math>

Also, the variational principal can be re-expressed in the form
<math>\begin{eqnarray}
E= E_{SH}+\frac{<\Psi|H-E_{SH}|\Psi>}{<\Psi|\Psi>}
\end{eqnarray}</math>
so that the <math>E_{SH}</math> term can be cancelled analytically from the matrix
elements.

For example
<math>
<\Psi_0|H-E_{SH}|\Psi_0> = 0
</math>

and
<math>
<\varphi_{ijk}|H-E_{SH}|\Psi_0> =
Z^{-1}<\varphi_{ijk}|\frac{1}{r}-\frac{1}{r_2}|\Psi_0>
</math>

Recall that
<math>\begin{eqnarray}\ \
I_0(a,b,c) = \frac{2}{c+2}\sum^{[c+1]/2}_{i=0}\left(\begin{array}{c}
c+2\\
2i+1
\end{array}\right) \left[f(p,q;\beta) + f(p^\prime,q^\prime;\alpha)\right]
\end{eqnarray}</math>
where

<math>\begin{eqnarray}
f(p,q;x) =
\frac{q!}{x^{q+1}(\alpha+\beta)^{p+1}}\sum^p_{j=0}\frac{(p+j)!}{j!}\left(\frac{x}{\alpha+\beta}\right)^j
\end{eqnarray}</math>

<math>
\begin{array}{cc}
p=a+2i+2 & p^\prime = b+2i+2\\
q= b+c-2i+2 & q^\prime =a+c-2i+2
\end{array}
</math>

If <math>\beta\ll\alpha</math>, the <math>f(p,q;\beta)\gg f(p^\prime,q^\prime;\alpha)</math> and the
<math>i=0</math> term is the dominant contribution to <math>f(p,q;\beta)</math>. However, since
this term depends only on the sum of powers <math>b+c</math> for <math>r_2</math>, it cancels exactly
from the matrix element of
<math>
\frac{1}{r_{12}} - \frac{1}{r_2}
</math>
and can therefore
be omitted in the calculations of integrands, thereby saving many significant
figures. This is especially valuable when <math>b+c</math> is large and <math>a</math> is small.

==Small Corrections==
*Mass Polarization
If the nuclear mass is not taken to be infinite, then the Hamiltonian is
<math>\begin{eqnarray}\ \
H = \frac{P_N^2}{2M} + \sum_{i=1}\frac{p_i^2}{2m} + V
\end{eqnarray}</math>

[[File:SmallCor.jpg|500px]]

where <math>M =M_A-nm</math> is the nuclear mass and <math>m</math> is the electron mass.

Change variables to the C of M
<math>\begin{eqnarray}
{\bf R} = \frac{1}{M+nm}\left[M{\bf r}_N + m({\bf r}_1 + {\bf r}_2 +
\ldots)\right]
\end{eqnarray}</math>

and relative variables

<math>
{\bf s}_i = {\bf r}_i - {\bf r}_N.
</math>

Then

<math>\begin{eqnarray}\ \ \ \ \
H = \frac{1}{2(M+nm)}{\bf p}_R^2+\frac{1}{2m}\sum_{i=1}^n{\bf p}_{s_i}^2
+\frac{1}{2M}\sum_{i,k}{\bf p}_{s_i}\cdot {\bf p}_{s_k} + V({\bf s}_1, {\bf
s}_2, \ldots)\nonumber
= \frac{1}{M+nm}{\bf p}_R^2+\frac{1}{2\mu}\sum_{i=1}^n{\bf
p}_{s_i}^2+\frac{1}{2M}\sum_{i\neq k}{\bf p}_{s_i}\cdot {\bf p}_{s_k}+V({\bf s}_1, {\bf
s}_2, \ldots)\nonumber
\end{eqnarray}</math>

where
<math>\begin{eqnarray}
\mu = \left(\frac{1}{M} + \frac{1}{m}\right)^{-1}
\end{eqnarray}</math>

If <math>n=2</math>, the Schrödinger equation is
<math>\begin{eqnarray}\ \ \
\left[ -\frac{\hbar^2}{2\mu}(\nabla^2_{s_1}+ \nabla^2_{s_2}) -
\frac{\hbar^2}{M}\nabla_{s_1}\nabla_{s_2}-\frac{Ze^2}{s_1}-\frac{Ze^2}{s_2}+\frac{e^2}{s_{12}}\right]\psi
= E\psi
\end{eqnarray}</math>.

Define the reduced mass Bohr radius
<math>\begin{eqnarray}
a_\mu = \frac{\hbar^2}{\mu e^2} = \frac{m}{\mu}a_0
\end{eqnarray}</math>
and

<math>\begin{eqnarray}
\rho_i = \frac{Zs_i}{a_\mu}\nonumber\\
\varepsilon = \frac{E}{Z^2(e^2/a_\mu)}\nonumber
\end{eqnarray}</math>

The Schrödinger equation then becomes
<math>\begin{eqnarray}\ \ \
\left[
-\frac{1}{2}(\nabla^2_{\rho_1}+\nabla^2_{\rho_2})-\frac{\mu}{M}\vec{\nabla}_{\rho_1}\cdot\vec{\nabla}_{\rho_2} - \frac{1}{\rho_1}-\frac{1}{\rho_2}+\frac{Z^{-1}}{\rho_{12}}\right]\psi = \varepsilon \psi
\end{eqnarray}</math>

The effects of finite mass are thus

1. A reduced mass correction to all energy levels of <math>(e^2/a_\mu)/(e^2/a_0)
= \mu/m</math>, i.e.
<math>\begin{eqnarray}
E_M=\frac{\mu}{m}E_\infty
\end{eqnarray}</math>

2. A specific mass shift given in first order perturbation theory by
<math>\begin{eqnarray}
\Delta\varepsilon =
-\frac{\mu}{M}<\psi|\nabla_{\rho_1}\cdot\nabla_{\rho_2}|\psi>
\end{eqnarray}</math>
or

<math>\begin{eqnarray}
\Delta\varepsilon =
-\frac{\mu}{M}<\psi|\nabla_{\rho_1}\cdot\nabla_{\rho_2}|\psi>\frac{e^2}{a_\mu}\nonumber\\
= -\frac{\mu^2}{mM}<\psi|\nabla_{\rho_1}\cdot\nabla_{\rho_2}|\psi>\frac{e^2}{a_0}\nonumber
\end{eqnarray}</math>

Since <math>\nabla_{\rho_1}\cdot\nabla_{\rho_2}</math> has the same angular properties as
<math>{\bf\rho}_1 \cdot {\bf\rho}_2</math>, the operator is like the product of two
dipole operators. For product type wave functions of the form
$
\psi = \psi(1s)\psi(nl) \pm \mbox{exchange},
$
the matrix element vanishes for all but P-states.

=Radiative Transitions=

==Transition Integrals and the Line Strength==

For an electric dipole transition <math>\gamma'\,L'\,S'\,J'\,M' \rightarrow \gamma\,L\,S\,J\,M</math>, the line strength is defined by
\begin{equation}
S = \sum_{M',\mu,M}|\langle\gamma'\,L'\,S'\,J'\,M'|r_\mu|\gamma\,L\,S\,J\,M\rangle|^2
\end{equation}
Using the Wigner-Eckart theorem, this is related to the reduced matrix element by
\begin{eqnarray}
S(J',J) = |\langle\gamma'\,L'\,S'\,J'||r||\gamma\,L\,S\,J\,M\rangle|^2
\sum_{M',\mu,M}\left|\left(\matrix{J' & 1 & J \cr
-M'& \mu &M \cr}\right)\right|^2
\end{eqnarray}
Using the sum rule for the 3-<math>j</math> symbols
\begin{equation}
\sum_{M',\mu,M}\left|\left(\matrix{J' & 1 & J \cr
-M'& \mu &M \cr}\right)\right|^2 = 1
\end{equation}
we see that the line strength is identical to the reduced matrix element
\begin{equation}
S(J',J) = |\langle\gamma'\,L'\,S'\,J'||r||\gamma\,L\,S\,J\,M\rangle|^2
\end{equation}
For a nonrelativistic calculation in <math>LS</math> coupling, <math>S = S'</math> and we can next strip out the dependence on <math>J</math> and <math>J'</math> by use of Eq.\ (7.1.7) of Edmonds { Angular Momentum in Quantum Mechanics} with <math>k = 1</math> to obtain
\begin{equation}
S(J',J) = (-1)^{L'+S+J+1}[(2J+1)(2J'+1)]^{1/2}\left\{
\matrix{L' & J'& S\cr
J & L & 1\cr}\right\}\langle\gamma' L'|| r || \gamma L\rangle
\end{equation}
where <math>\langle\gamma' L'|| r || \gamma L\rangle</math> is the reduced matrix element if spin is neglected. Note that if spin is neglected, then the line-strength is simply
\begin{equation}
S(L',L) = |\langle\gamma' L'|| r || \gamma L\rangle|^2
\end{equation}
To calculate the reduced matrix element, use a particular example along with the Wigner-Eckart theorem
such as
\begin{equation}
\langle\gamma'\, L'|| r || \gamma\, L\rangle = (-1)^{L'}\langle\gamma'\, L'\, 0| z | \gamma\, L\, 0\rangle{\Large /}\left(\matrix{L' & 1 & L \cr
0 & 0 & 0 \cr}\right)
\end{equation}

===Correlated Two-electron Integrals===
For the case of two-electron integrals, <math>z = z_1 + z_2</math> and <math>r = r_1 + r_2</math>. As discussed in G.W.F. Drake, Phys. Rev. A 18, 820 (1978), the general integral in Hylleraas coordinates for the integral over the <math>z_1</math> term can be written in the form
\begin{equation}
\label{int1}
\langle\gamma'\, L'\, 0| z_1 | \gamma\, L\, 0\rangle = \sum_\Lambda C_\Lambda I_\Lambda(\gamma',\gamma)
\end{equation}
where <math>I_\Lambda(\gamma',\gamma)</math> stands for the purely radial integral
\begin{equation}
\label{radial}
I_\Lambda(\gamma',\gamma) = \int_0^\infty r_1\,dr_1\int_0^\infty r_2\,dr_2\int_{|r_1-r_2|}^{r_1+r_2}r_{12}\, dr_{12}
R_{\gamma'}^*\,r_1\, R_\gamma P_\Lambda(\cos\theta_{12})
\end{equation}
and <math>\cos\theta_{12} = (r_1^2 + r_2^2 - r_{12}^2)/(2r_1r_2)</math> is a short-hand notation for a radial function of <math>r_1</math>, <math>r_2</math> and <math>r_{12}</math>. Also, <math>P_\Lambda(\cos\theta_{12})</math> is a Legendre polynomial,
and, <math>R_\gamma = R_\gamma(r_1,r_2,r_{12})</math> is the radial part of the Hylleraas wave function for state <math>\gamma</math> such
that the total wave function is <math>\psi({\bf r}_1,{\bf r}_2) = R_\gamma(r_1,r_2,r_{12}){\cal Y}_{l_1,l_2,L}^M(\hat{\bf r}_1,\hat{\bf r}_2)\pm {\rm exchange}</math>, normalized to unity, and <math>{\cal Y}_{l_1,l_2,L}^M(\hat{\bf r}_1,\hat{\bf r}_2)</math> is a vector coupled product of spherical harmonics. The coefficents <math>C_\Lambda</math> in Eq.\ (\ref{int1}) can be calculated from Eq.\ (32) of the above 1978 paper. They need to be calculated separately for he direct-direct and the direct-exchange contributions. For the reduced matrix element of <math>z_1</math>, the leading 3-<math>j</math> symbol can be dropped, leaving (for the case <math>k = 1</math>)
\begin{eqnarray}
C_\Lambda^{\rm reduced} &=& \frac{(-1)^{l_2'+L'+L+1}}{2} (l_1,l_1',l_2,l_2',L,L')^{1/2}
\sum_{\lambda={\rm max}(l_1-1,0)}^{l_1+1} (\lambda,\Lambda,1)
\left(\matrix{l_1 & 1 & \lambda \cr
0 & 0 & 0 \cr}\right)
\left(\matrix{l_1'& \lambda & \Lambda \cr
0 & 0 & 0 \cr}\right)\nonumber\\
&\times&\left(\matrix{l_2' & 1_2 & \Lambda \cr
0 & 0 & 0 \cr}\right)
\left\{\matrix{L' & \lambda & l_2 \cr
\Lambda& l_2' & l_1' \cr}\right\}
\left\{\matrix{L' & \lambda & l_2 \cr
l_1 & L & 1 \cr}\right\}
\end{eqnarray}
where the notation <math>(a,b,c,\cdots)</math> means <math>(2a+1)(2b+1)(2c+1)\cdots</math>.
A similar expression can be obtained for the angular coefficients for the <math>z_2</math> term by interchanging the subscripts 1 and 2 throughout and multiplying by <math>(-1)^{L+L'+1}</math>.

===Example===
As an example, for the <math>1s^2\;^1S - 1s2p\;^1P</math> transition of helium, the only nonvanishing <math>C_\Lambda</math> angular coefficients are <math>C_1 = 1/2</math> for the direct-direct term (i.e. <math>P_1(\cos\theta_{12}) = \cos\theta_{12}</math> is included in the radial integral (\ref{radial})) and <math>C_0 = 1/2</math> for the direct-exchange term (i.e. <math>P_0(\cos\theta_{12}) = 1</math> is included in the radial integral (\ref{radial})). The results for infinite nuclear mass are <math>S(1\;^1S-2\;^1P) = (0.781\,127\,158\,667)^2</math> and <math>f(1\;^1S-2\;^1P) = -0.276\,164\,703\,736</math> where <math>f(1\;^1S-2\;^1P) = \frac{2}{3}[E(1\;^1S) - E(2\;^1P)]S(1\;^1S-2\;^1P)</math> is the absorption oscillator strength.

Theory Notes

2020-12-14T00:27:00Z

Drake: /* Introduction */

<center>
<H1>Notes on solving the Schrödinger equation in Hylleraas coordinates for heliumlike atoms
Gordon W.F. Drake
Department of Physics, University of Windsor
Windsor, Ontario, Canada N9B 3P4
(Transcribed from hand-written notes and edited by Lauren Moffatt. Last revised November 22, 2017 to correct a missing subscript in the expression for $B_1$ in the Hamiltonian in Sect. 11 General Hermitian Property.)
</H1>
</center>

==Introduction==
These notes describe some of the technical details involved in solving the Schrödinger equation for a heliumlike atom of nuclear charge Z in correlated Hylleraas coordinates. The standard for computational purposes is to express all quantities in a form that is valid for electronic states of arbitrary angular momentum, and to express matrix elements of the Hamiltonian in a manifestly Hermitian symmetrized form. Further information and derivations of the mathematical techniques for angular integrals and radial recursion relations can be found in the paper
G.W.F. Drake, Phys. Rev. A <math>{\bf 18}</math>, 820 (1978).

These notes describe some of the technical details involved in solving the Schrödinger equation for a heliumlike atom of nuclear charge Z in correlated Hylleraas coordinates. The standard for computational purposes is to express all quantities in a form that is valid for electronic states of arbitrary angular momentum, and to express matrix elements of the Hamiltonian in a manifestly Hermitian symmetrized form. Further information and derivations of the mathematical techniques for angular integrals and radial recursion relations can be found in the paper
G.W.F. Drake, Phys. Rev. A <math>{\bf 18}</math>, 820 (1978).

The starting point is the two-electron Schrödinger equation for infinite nuclear mass
\begin{equation} \left[-\frac{\hbar^2}{2m}(\nabla^2_1 +\nabla^2_2) - \frac{Ze^2}{r_1} -
\frac{Ze^2}{r_2}+\frac{e^2}{r_{12}} \right]\psi = E\psi\nonumber
\end{equation}
where <math>m</math> is the electron mass, and <math>r_{12} = |{\bf r}_1 - {\bf r}_2|</math> (see diagram below).

[[File:Pic1.jpg|250px]]

Begin by rescaling distances and energies so that the Schrödinger equation can be expressed in a dimensionless form. The dimensionless Z-scaled distance is defined by <math>\begin{eqnarray} \rho = \frac{Zr}{a_0} \end{eqnarray} </math> where <math> \begin{eqnarray} a_0 = \frac{\hbar^2}{me^2} \end{eqnarray} </math> is the atomic unit (a.u.) of distance equal to the Bohr radius <math>0.529\, 177\, 210\, 92(17) \times 10^{-10}</math> m. Then

\begin{equation}[-\frac{\hbar^2}{2m}Z^2(\frac{me^2}{\hbar^2})^2(\nabla^2_{\rho_1}+\nabla^2_{\rho_2})
- Z^2\frac{e^2}{a_0}\rho^{-1}_1 - Z^2\frac{e^2}{a_0}\rho^{-1}_2 +
\frac{e^2}{a_0}Z\rho^{-1}_{12}]\psi= E\psi\nonumber\end{equation}

But <math>\begin{eqnarray} \frac{\hbar^2}{m}(\frac{me^2}{\hbar^2})^2 = \frac{e^2}{a_0} \end{eqnarray} </math> is the hartree atomic unit of of energy <math>(E_h = 27.211\,385\,05(60)</math> eV, or equivalently <math>E_h/(hc) = 2194.746\,313\,708(11)</math> m<math>^{-1})</math>. Therefore, after multiplying through by <math>a_0/(Ze)^2</math>, the problem to be solved in Z-scaled dimensionless units becomes

\begin{equation} \left[-\frac{1}{2}(\nabla^2_{\rho_1}+\nabla^2_{\rho_2}) -
\frac{1}{\rho_1} - \frac{1}{\rho_2} + \frac{Z^{-1}}{\rho_{12}}\right]\psi =
\varepsilon\psi\nonumber \end{equation}

where <math>\begin{eqnarray}\varepsilon = \frac{Ea_0}{(Ze)^2} \end{eqnarray}</math> is the energy in Z-scaled atomic units. For convenience, rewrite this in the conventional form
\begin{equation}
H\psi = \varepsilon\psi\nonumber \end{equation}
where (using <math>r</math> in place of <math>\rho</math> for the Z-scaled distance)
\begin{equation}
H = -\frac{1}{2}(\nabla^2_1+\nabla^2_2) - \frac{1}{r_1}-\frac{1}{r_2} + \frac{Z^{-1}}{r_{12}}
\nonumber\end{equation}
is the atomic Hamiltonian for infinite nuclear mass.

==The Hartree Fock Method==

For purposes of comparison, and to define the correlation energy, assume that <math>\psi({\bf r}_1,{\bf r}_2)</math> can be written in the separable product form
\begin{equation}
\psi({\bf r}_1,{\bf r}_2) = \frac{1}{\sqrt{2}} \left[ u_1(r_1) u_2(r_2) \pm u_2(r_1) u_1(r_2) \right] \nonumber
\end{equation}
for the <math>1s^2\ ^1\!S</math> ground state. Because of the <math>\frac{1}{r_{12}}</math> term in the Schrödinger equation, the exact solution connot be expressed in this form as a separable product. However, the Hartree-Fock approximation corresponds to finding the best possible solution to the Schrödinger equation
\begin{equation}
H\psi({\bf r}_1,{\bf r}_2) = E\psi({\bf r}_1,{\bf r}_2)\nonumber
\end{equation}
that can nevertheless be expressed in this separable product form, where as before
\begin{equation}
H = -\frac{1}{2}(\nabla^2_1+\nabla^2_2) - \frac{1}{r_1}- \frac{1}{r_2} + \frac{Z^{-1}}{r_{12}}\nonumber
\end{equation}
is the full two-electron Hamiltonian. To find the best solution, substitute into <math>\left\langle\psi \lvert H-E \rvert \psi\right\rangle</math> and require this expression to be stationary
with respect to arbitrary infinitesimal variations <math>\delta u_1</math> and <math>\delta
u_2</math> in <math>u_1</math> and <math>u_2</math> respectively; i.e.

\begin{equation}
\frac{1}{2} \left\langle \delta u_1(r_1)u_2(r_2) \pm u_2(r_1)\delta
u_1(r_2) \lvert H-E \rvert u_1(r_1) u_2(r_2)\pm u_2(r_1) u_1(r_2)\right\rangle\nonumber
\end{equation}

\begin{equation}
=\int\delta u_1(r_1) d{\bf r}_1 \left\{ \int d{\bf r}_2 u_2(r_2) \left( H - E \right) \left[ u_1(r_1)u_2(r_2) \pm u_2(r_1) u_1(r_2) \right] \right\}\nonumber
\end{equation}

\begin{equation}
= 0 \nonumber
\end{equation}

for arbitrary <math> \delta u_1(r_1).\nonumber </math> Therefore <math>\{\int d{\bf r}_2 \ldots \} = 0</math>.

Similarly, the coefficient of <math>\delta u_2</math> would give

\begin{equation}
\int d{\bf r}_1 u_1(r_1) \left( H-E \right) \left[ u_1(r_1) u_2(r_2) \pm
u_2(r_1) u_1(r_2) \right] = 0\nonumber
\end{equation}

Define

\begin{equation}
I_{12} = I_{21} = \int d{\bf r}\, u_1(r)u_2(r), \nonumber
\end{equation}

\begin{equation}
H_{ij} = \int d{\bf r}\, u_i(r)(-\frac{1}{2}\nabla - \frac{1}{r})u_j(r), \nonumber
\end{equation}

\begin{equation}
G_{ij}(r) = \int d{\bf r}^\prime u_i(r^\prime)\frac{1}{|{\bf r} - {\bf r}\prime|}u_j(r^\prime)\nonumber
\end{equation}

Then the above equations become the pair of integro-differential equations

\begin{equation}
[ H_0 - E + H_{22}+G_{22}(r)]u_1(r) = \mp [ I_{12}(H_0-E) + H_{12}+G_{12}(r)]u_2(r)\nonumber
\end{equation}

\begin{equation}
[H_0-E+H_{11}+G_{11}(r)]u_2(r) = \mp [I_{12}(H_0-E) + H_{12}+G_{12}(r)]u_1(r)\nonumber
\end{equation}

These must be solved self-consistently for the "constants" <math>I_{12}</math> and
<math>H_{ij}</math> and the function <math>G_{ij}(r)</math>.

The Hartree Fock energy is <math>E \simeq -2.87\ldots </math>a.u. while the exact energy is <math>E = -2.903724\ldots </math>a.u.

The difference is called the "correlation energy" because it arises from the
way in which the motion of one electron is correlated to the other. The
Hartree Fock equations only describe how one electron moves in the average field
provided by the other (mean-field theory).

==Configuration Interaction==
Expand
\begin{equation}
\psi({\bf r}_1,{\bf r}_2)= C_0u^{(s)}_1(r_1)u^{(s)}_1(r_2) + C_1u^{(P)}_1({\bf r}_1)u^{(P)}_1({\bf r}_2){\cal Y}^0_{1,1,0}(\hat{\bf r}_1, \hat{\bf r}_2)+C_2u^{(d)}_1({\bf r}_1)u^{(d)}_2({\bf r}_2){\cal Y}^0_{2,2,0}(\hat{\bf r}_1, \hat{\bf r}_2)+... \pm {\rm\ exchange}
\end{equation}
where
\begin{equation}
{\cal Y}^M_{l_1,l_2,L}(\hat{\bf r}_1, \hat{\bf r}_2)=\sum_{m_1,m_2}Y^{m_1}_{l_1}({\bf r}_1)Y^{m_2}_{l_2}({\bf r}_2)\times <l_1l_2m_1m_2\mid LM>
\end{equation}

This works, but is slowly convergent, and very laborious. The best CI calculations are accurate to <math> ~10^{-7}</math> a.u.

==Hylleraas Coordinates==
[E.A. Hylleraas, Z. Phys. <math>{\bf 48}, 469(1928)</math> and <math>{\bf 54}, 347(1929)</math>]
suggested using the co-ordinates <math>r_1</math>, <math>r_2</math> and <math>r_{12}</math> or equivalently

\begin{eqnarray}
s &=& r_1 + r_2, \nonumber\\
t &=& r_1-r_2, \nonumber\\
u &=& r_{12}\nonumber
\end{eqnarray}

and writing the trial functions in the form

\begin{equation}
\Psi({\bf r}_1,{\bf r}_2) = \sum^{i+j+k\leq
N}_{i,j,k}c_{i,j,k}r_1^{i+l_1}r_2^{j+l_2}r_{12}^ke^{-\alpha r_1 - \beta r_2}
\mathcal{Y}^M_{l_1,l_2,L}(\hat{r}_1,\hat{r}_2)\pm \text{exchange}\nonumber
\end{equation}

Diagonalizing H in this non-orthogonal basis set is equivalent to solving
<math> \begin{eqnarray}
\frac{\partial E}{\partial c_{i,j,k}} = 0\nonumber
\end{eqnarray} </math>
for fixed <math>\alpha</math> and <math>\beta</math>.

The diagonalization must be repeated for different values of <math>\alpha</math> and
<math>\beta</math> in order to optimize the non-linear parameters.

==Completeness==
The completeness of the above basis set can be shown by first writing
\begin{equation}
r_{12}^2 = r_1^2 + r_2^2 - 2r_1r_2\cos\theta_{12}
\end{equation}
and, from the spherical harmonic addition theorem,
\begin{equation}
\cos(\theta_{12})=\frac{4\pi}{3}\sum^1_{m=-1}Y^{m*}_1(\theta_1,\varphi_1)Y^m_1(\theta_2,\varphi_2)
\end{equation}
Consider first the S-states. The <math>r_{12}^0</math> terms are like the ss terms in a CI calculation. The
<math>r_{12}^2</math> terms bring in p-p type contributions, and the higher powers bring
in d-d, f-f etc. type terms. In general

\begin{equation}
P_l(\cos\theta_{12}) = \frac{4\pi}{2l+1}\sum^l_{m=-l}{Y^{m}_l}^*(\theta_1,\varphi_1)Y^m_l(\theta_2,\varphi_2)\nonumber
\end{equation}

For P-states, one would have similarly

\begin{array}{lr}
r_{12}^0 & (sp)P\nonumber\\
r_{12}^2 & (pd)P\nonumber\\
r_{12}^4 & (df)P\nonumber\\
\vdots & \vdots\nonumber
\end{array}

For D-states

\begin{array}{lr}
r_{12}^0 & (sd)D & (pp^\prime)D\nonumber\\
r_{12}^2 & (pf)D & (dd^\prime)D\nonumber\\
r_{12}^4 & (dg)D & (ff^\prime)D\nonumber\\
\vdots & \vdots & \vdots\nonumber
\end{array}

In this case, since there are two "lowest-order" couplings to form a
D-state, both must be present in the basis set, i.e.

\begin{equation}
\Psi(r_2,r_2) = \sum c_{ijk}r_1^ir_2^{j+2}r_{12}^ke^{-\alpha r_1-\beta
r_2}\mathcal{Y}^M_{022}(\hat{r}_1,\hat{r}_2)
+\sum d_{ijk}r_1^{i+1}r_2^{j+1}r_{12^k}e^{-\alpha^\prime r_1 - \beta^\prime
r_2}\mathcal{Y}^M_{112}(\hat{r}_1,\hat{r}_2)\nonumber
\end{equation}

For F-states, one would need <math>(sf)F</math> and <math>(pd)F</math> terms.

For G-states, one would need <math>(sg)G</math>, <math>(pf)G</math> and <math>(dd^\prime)G</math> terms.

Completeness of the radial functions can be proven by considering the
Sturm-Liouville problem

\begin{equation}
\left(-\frac{1}{2}\nabla^2-\frac{\lambda}{r_s}-E\right)\psi({\bf r}) = 0\nonumber
\end{equation}

or

\begin{equation}
\left(-\frac{1}{2}\frac{1}{r^2}\frac{\partial}{\partial r}\left({r^2}\frac{\partial}{\partial r}\right) -
\frac{l\left(l+1\right)}{2r^2} - \frac{\lambda}{r} - E\right)u(r) = 0.\nonumber
\end{equation}

For fixed E and variable <math>\lambda</math> (nuclear charge).

The eigenvalues are <math>\lambda_n = (E/E_n)^{1/2}</math>, where <math>E_n =- \frac{1}{2n^2}</math>

[[File:Pic2.jpg|500px]]

and the eigenfunctions are

\begin{equation}
u_{nl}(r) =
\frac{1}{(2l+1)!}\left(\frac{(n+l)!}{(n-l-1)2!}\right)^{1/2}(2\alpha)^{3/2}e^{-\alpha
r}\nonumber
\times \ \left( 2 \alpha r \right)^l\,_{1}F_{1}\left(-n+l+1,2l+2; 2 \alpha r \right)
\end{equation}

with <math>\alpha = (-2E)^{1/2}</math> and <math>n\geq l+1</math>. The confluent hypergeometric function <math> _1F_1(a,b,;z)</math> then denotes a finite polynomial since <math>a=-n+l+1</math> is a negative integer or zero.

Unlike the hydrogen spectrum, which has both a discrete part for <math>E<0</math> and a
continuous part for <math>E>0</math>, this forms an entirely discrete set of finite
polynomials, called Sturmian functions. They are orthogonal with respect to
the potential, i.e.

\begin{equation}
\int^\infty_0 r^2dr \left( u_{n^\prime l}(r)\frac{1}{r}u_{nl}(r) \right) = \delta_{n,n^\prime}\nonumber
\end{equation}

Since they become complete in the limit <math>n\rightarrow\infty</math>, this assures the
completeness of the variational basis set.

[See also B Klahn and W.A. Bingel Theo. Chim. Acta (Berlin) 44, 9 and 27
(1977)].

==Solutions of the Eigenvalue Problem==

For convenience, write

\[
\Psi({\bf r}_1,{\bf r}_2) = \sum^N_{m=1}c_m\varphi_m\nonumber
\]

where <math>m</math> represents the <math>m</math>'th combination of <math>i,j,k</math> and

\begin{equation}
\varphi_{ijk}=r_1^ir_2^jr_{12}^ke^{-\alpha r_1 - \beta
r_2}\mathcal{Y}^M_{l_1,l_2,L}(\hat{r}_1,\hat{r}_2) \pm \mbox{exchange}\nonumber
\end{equation}

\begin{align}

&
\left( \begin{array}{lr}
\cos(\theta) & \sin(\theta)\\
-\sin(\theta) & \cos(\theta)
\end{array} \right)
\left( \begin{array}{lr}
H_{11} & H_{12} \\
H_{12} & H_{22}
\end{array}\right)
\left( \begin{array}{lr}
\cos(\theta) & -\sin(\theta)\\
\sin(\theta) & \cos(\theta)
\end{array} \right)
\\[5pt]
= & \left(\begin{array}{lr}
cH_{11}+sH_{12} & cH_{12} + sH_{22}\\
-sH_{11} + cH_{12} & -sH_{12} + cH_{22}
\end{array}\right)
\left( \begin{array}{cc}
c & -s \\
s & c
\end{array}\right)
\\[5pt]
= & \left(\begin{array}{lr}
c^2H_{11}+s^2H_{22} + 2csH_{12} & (c^2-s^2)H_{12}+cs(H_{22}-H_{11})\\
(c^2-s^2)H_{12}+cs(H_{22}-H_{11}) & s^2H_{11}+c^2H_{22}-2csH_{12}
\end{array}\right)
\end{align}

Therefore

\begin{equation}
\left[\cos^2(\theta)-\sin^2(\theta)\right]H_{12} =
\cos(\theta)\sin(\theta)(H_{11}-H_{22})\nonumber
\end{equation}

and

\begin{equation}
\tan(2\theta) = \frac{2H_{12}}{H_{11}-H_{22}}\nonumber
\end{equation}

i.e.

\begin{equation}
\cos(\theta)=\left(\frac{r+\omega}{2r}\right)^{1/2}\nonumber
\end{equation}

\begin{equation}
\sin(\theta)=-\mbox{sgn}(H_{12})\left(\frac{r-\omega}{2r}\right)^{1/2}\nonumber
\end{equation}

where

\begin{equation}
\omega = H_{22}-H_{11}\nonumber
\end{equation}

\begin{equation}
r=\left(\omega^2+4H_{12}^2\right)^{1/2}\nonumber
\end{equation}

\begin{equation}
E_1=\frac{1}{2}\left(H_{11}+H_{22}-r\right)\nonumber
\end{equation}

\begin{equation}
E_2=\frac{1}{2}\left(H_{11}+H_{22}+r\right)\nonumber
\end{equation}

===Brute Force Method===

- Gives all the eigenvalues and eigenvectors, but it is slow

- First orthonormalize the <math>N</math>-dimensional basis set; i.e. form linear combinations

\begin{equation}
\Phi_m = \sum_{n=1}^N\varphi_nR_{nm}\nonumber
\end{equation}

such that
\begin{equation}
\langle \Phi_m | \Phi_n \rangle = \delta_{m,n}
\end{equation}

This can be done by finding an orthogonal tranformation T such that

\begin{equation}
T^TOT = I = \left(
\begin{array}{ccccc}
I_1 & 0 & 0 & \ldots & 0 \\
0 & I_2 & 0 & \ldots & 0 \\
0 & 0 & I_3 & \ldots & 0 \\
\vdots & \vdots & \vdots & \ddots & \vdots \\
0 & 0 & 0 & \ldots & I_N
\end{array}\right);
\end{equation}

\begin{equation}
O_{mn} = \langle \varphi_m | \varphi_n \rangle \nonumber
\end{equation}

and then applying a scale change matrix

\begin{equation}
S = \left(\begin{array}{ccccc}
\frac{1}{I_1^{1/2}} & 0 & 0 & \ldots & 0\\
0 & \frac{1}{I_2^{1/2}} & 0 & \ldots & 0 \\
0 & 0 & \frac{1}{I_3^{1/2}} & \ldots & 0 \\
\vdots & \vdots & \vdots & \ddots & \vdots \\
0 & 0 & 0 & \ldots & I_N^{1/2}
\end{array}\right)= S^T\nonumber
\end{equation}

Then

\begin{equation}
S^TT^TOTS = 1
\end{equation}

i.e.

\begin{equation}
R^TOR = 1
\end{equation}

with <math>R=TS</math>.

If H is the matrix with elements <math>H_{mn} = \langle \varphi_m | \varphi_n \rangle </math>, then H
expressed in the <math>\Phi_m</math> basis set is

\begin{equation}
H^\prime = R^THR.\nonumber
\end{equation}

We next diagonalize <math>H^\prime</math> by finding an orthogonal transformation W such
that

\begin{equation}
W^T H^\prime W = \lambda = \left(
\begin{array}{cccc}
\lambda_1 & 0 & \ldots & 0 \\
0 & \lambda_2 & \ldots & 0 \\
\vdots & \vdots & \ddots & \vdots \\
0 & 0 & \ldots & \lambda_N
\end{array}\right)\nonumber
\end{equation}

The q'th eigenvector is

\begin{equation}
\Psi^{(q)} = \sum_{n=1}^N\Phi_n W_{n,q} = \sum_{n,n^\prime} \varphi_{n^\prime} R_{n^\prime ,n} W_{n,q} \nonumber
\end{equation}

i.e.

\begin{equation}
c_{n^\prime}^{(q)} = \sum_{n=1}^N R_{n^\prime n} W_{n,q}
\end{equation}

===The Power Method===

- Based on the observation that if H has one eigenvalue, <math>\lambda_M</math>, much bigger than all the rest, and <math>\chi = \left(
\begin{array}{c}a_1\\a_2\\\vdots\end{array}\right)</math>
is an arbitrary starting vector, then <math>\chi = \sum_{q=1}^N x_q\Psi^{(q)}</math>.

\begin{equation}
(H)^n\chi = \sum_{q=1}^N x_q \lambda^n_q\Psi^{(q)}\nonumber\rightarrow x_M\lambda_M^n\Psi^{(M)}\nonumber
\end{equation}

provided <math>x_M\neq 0</math>.

To pick out the eigenvector corresponding to any eigenvalue, write the original problem in the form

\begin{align}
H\Psi & = \lambda O \Psi \nonumber \\
\left( H - \lambda_q O \right) \Psi & = \left( \lambda - \lambda_q \right) O \Psi \nonumber
\end{align}

Therefore,

\begin{equation}
G \Psi = \frac{1}{\lambda - \lambda_q } \Psi \nonumber \\
\end{equation}

where <math>G=(H-\lambda_qO)^{-1}O</math> with eigenvalues <math>\frac{1}{\lambda_n - \lambda_q}</math>.

By picking <math>\lambda_q</math> close to any one of the <math>\lambda_n</math>, say
<math>\lambda_{n^\prime}</math>, then <math>\frac{1}{\lambda_n-\lambda_q}</math> is much larger for
<math>n=n^\prime</math> than for any other value. The sequence is then

\begin{array}{c}
\chi_1=G\chi\nonumber\\
\chi_2=G\chi_1\nonumber\\
\chi_3=G\chi_2\nonumber\\
\vdots\nonumber
\end{array}

until the ratios of components in <math>\chi_n</math> stop changing.

- To avoid matrix inversion and multiplication, note that the sequence is equivalent to
\begin{equation}
F \chi_n = \left( \lambda - \lambda_q \right) O \chi_{n-1} \nonumber \\
\end{equation}
where <math>F = H-\lambda_qO</math>. The factor of <math>\left( \lambda - \lambda_q \right)</math> can be
dropped because this only affects the normalization of <math>\chi_n</math>. To find <math>\chi_n</math>, solve
\begin{equation}
F \chi_n = O \chi_{n-1} \nonumber
\end{equation}
(N equations in N unknowns). Then
\begin{equation}
\lambda = \frac{ \langle \chi_n | H | \chi_n \rangle}{\langle \chi_n | \chi_n \rangle} \nonumber
\end{equation}

==Matrix Elements of H==

\begin{equation}
H=-\frac{1}{2}\nabla^2_1 -\frac{1}{2}\nabla^2_2 - \frac{1}{r_1} - \frac{1}{r_2} +\frac{Z^{-1}}{r_{12}}\nonumber
\end{equation}

Taking <math>r_1, r_2</math> and <math>r_{12}</math> as independent variables,

\begin{equation}
\nabla_1^2 = \frac{1}{r_1^2}\frac{\partial}{\partial r_1}\left(r_1^2\frac{\partial}{\partial r_1}\right)+ \frac{1}{r_{12}^2}\frac{\partial}{\partial r_{12}}
\left(r_{12}^2\frac{\partial}{\partial r_{12}}\right)\nonumber -\frac{l_1(l_1+1)}{r_1^2}+2(r_1 - r_2\cos(\theta_{12}))\frac{1}{r_{12}}\frac{\partial^2}{\partial r_1 \partial r_{12}} \nonumber - 2(\nabla_1^Y \cdot {\bf r}_2)\frac{1}{r_{12}}\frac{\partial}{\partial r_{12}}\nonumber
\end{equation}

where <math>\nabla_1^Y</math> acts only on the spherical harmonic part of the wave function and the diagram

[[File:Pic3.jpg|400px]]

defines the complete set of 6 independent variables is <math>r_1</math>, <math>r_2</math>, <math>r_{12}</math>, <math>\theta_1</math>, <math>\varphi_1</math>, <math>\chi</math>.

If <math>r_{12}</math> were not an independent variable, then one could take the volume element to be

\begin{equation}
d\tau = r_1^2dr_1\sin(\theta_1)d\theta_1d\varphi_1r_2^2dr_2\sin(\theta_2)d\theta_2d\varphi_2.\nonumber
\end{equation}

However, <math>\theta_2</math> and <math>\varphi_2</math> are no longer independent variables. To eliminate them, take the point <math>{\bf r}_1</math> as the
origin of a new polar co-ordinate system, and write

\begin{equation}
d\tau=-r_1^2dr_1\sin(\theta_1)d\theta_1d\varphi_1r_{12}^2dr_{12}\sin(\psi)d\psi d\chi\nonumber
\end{equation}

and use

\begin{equation}
r_2^2=r_1^2+r_{12}^2 +2r_1r_{12}\cos(\psi)
\end{equation}

Then for fixed <math>r_1</math> and <math>r_{12}</math>,

\begin{equation}
2r_2dr_2 = -2r_1r_{12}\sin(\psi)d\psi
\end{equation}

Thus

\begin{equation}
d\tau= r_1dr_1r_2dr_2r_{12}dr_{12}\sin(\theta_1)d\theta_1d\varphi_1d\chi
\end{equation}

The basic type of integral to be calculated is

\begin{align}
I \left( l_1, m_1, l_2, m_2; R \right) & = \int \sin \left( \theta_1 \right) d\theta_1 d\varphi_1 d\chi Y^{m_1}_{l_1} \left( \theta_1, \varphi_1 \right)^{*} Y^{m_2}_{l_2} \left( \theta_2, \varphi_2 \right) \\
& \times \int r_1dr_1r_2dr_2r_{12}dr_{12}R \left( r_1, r_2, r_{12} \right)
\end{align}

Consider first the angular integral. <math>Y^{m_2}_{l_2} \left( \theta_2, \varphi_2 \right)</math> can be expressed in terms of the independent variables <math>\theta_1, \varphi_1, \chi</math> by use of the rotation matrix relation
\begin{equation}
Y^{m_2}_{l_2}( \theta_2, \varphi_2) = \sum_m \mathcal{D}^{(l_2)}_{m_2,m}(\varphi_1,\theta_1,\chi)^*Y^m_{l_2}( \theta_{12}, \varphi)
\end{equation}
where <math>\theta, \varphi</math> are the polar angles of <math>{\bf r}_2</math> relative to <math>{\bf r}_1</math>. The angular integral is then

\begin{align}
I_{ang} & = \int^{2 \pi}_{0} d\chi \int^{2\pi}_{0} d\varphi_1 \int^{\pi}_{0} \sin \left( \theta_1 \right) d\theta_{1} Y^{m_1}_{l_1} \left( \theta_1, \varphi_1 \right)^{*} \\
& \times \sum_m \mathcal{D}^{(l_2)}_{m_2,m} \left( \varphi_1,\theta_1,\chi \right)^{*} Y^{m}_{l_2} \left( \theta_{12},\varphi \right) \\
\end{align}

Use

\begin{equation}
Y^{m_1}_{l_1} \left( \theta_1, \varphi_1 \right)^{*} = \sqrt{\frac{2l_1+1}{4\pi}}\mathcal{D}^{(l_1)}_{m_1,0} \left( \varphi_1, \theta_1, \chi \right)
\end{equation}

together with the orthogonality property of the rotation matrices (Brink and Satchler, p 147)

\begin{align}
& \int \mathcal{D}^{(j)*}_{m,m^\prime}\mathcal{D}^{(J)}_{M,M^\prime}\sin(\theta_1)d\theta_1d\varphi_1d\chi \\
& = \frac{8\pi^2}{2j+1}\delta_{jJ}\delta_{mM}\delta_{m^\prime M^\prime}\nonumber
\end{align}

to obtain

\begin{align}
I_{ang} & = \sqrt{ \frac{2l_1+1}{4\pi}} \frac{8\pi^2}{2l_1+1} \delta_{l_1,l_2} \delta_{m_1,m_2} Y^{0}_{l_2} \left( \theta_{12}, \varphi \right) \\
& = 2 \pi \delta_{l_1,l_2} \delta_{m_1,m_2} P_{l_2} \left( \cos \theta_{12} \right)
\end{align}

since

\begin{equation}
Y^0_{l_2} \left( \theta_{12}, \varphi \right) = \sqrt{\frac{2l_1+1}{4\pi}} P_{l_2} \left( \cos \left( \theta_{12} \right) \right)
\end{equation}

Note that <math>P_{l_2} \left( \cos \theta \right)</math> is just a short hand expression for a radial function because

\begin{equation}
\cos \theta_{12} = \frac{r_1^2 + r_2^2 - r_{12}^2}{2r_1r_2}
\end{equation}

The original integral is thus

\begin{equation}
I \left( l_1, m_1, l_2, m_2; R \right) = 2 \pi \delta_{l_1,l_2} \delta_{m_1,m_2} \int^{\infty}_{0} r_{1} dr_{1} \int^{\infty}_{0} r_{2} dr_{2} \int^{r_{1} + r_{2}}_{|r_{1} - r_{2}|} r_{12} dr_{12} R \left( r_{1}, r_{2}, r_{12} \right) P_{l_2} \left( \cos \theta_{12} \right)
\end{equation}

where again

\begin{equation}
\cos \theta_{12} = \frac{r_{1}^{2} +r^{2}_{2} - r_{12}^{2}}{2 r_{1} r_{2}}
\end{equation}

is a purely radial function.

The above would become quite complicated for large <math>l_2</math> because <math>P_{l_2}(\cos \theta_{12})</math> contains terms up to <math>\left( \cos \theta_{12} \right)^{l_2}</math>. However, recursion relations exist which allow any integral containing <math>P_l \left( \cos \theta_{12} \right)</math> in terms of those containing just <math>P_0 \left( \cos \theta_{12} \right) = 1</math> and <math>P_1 \left( \cos \theta_{12} \right) = \cos \theta_{12}</math>.

==Radial Integrals and Recursion Relations==

The basic radial integral is [see G.W.F. Drake, Phys. Rev. A 18, 820 (1978)]

\begin{align}
I_0 \left( a, b, c \right) & =
\int^{\infty}_{0} r_{1} dr_{1} \int^{\infty}_{r_{1}} r_{2} dr_{2} \int^{r_{1}+r_{2}}_{r_{2} - r_{1}} r_{12} dr_{12} r_{1}^{a} r_{2}^{b} r_{12}^{c} e^{-\alpha r_{1} - \beta r_{2}} + \int^{\infty}_{0} r_{2} dr_{2} \int^{\infty}_{r_{2}} r_{1} dr_{1} \int^{r_{1} + r_{2}}_{r_{1} - r_{2}} r_{12} dr_{12} r_{1}^{a} r_{2}^{b} r_{12}^{c} e^{- \alpha r_{1} - \beta r_{2}} \\
& = \frac{2}{c+2} \sum^{[(c+1)/2]}_{i=0}
\left(\begin{array}{c}
c+2 \\
2i+1
\end{array} \right)
\left\{ \frac{q!}{\beta^{q+1} \left( \alpha + \beta \right)^{p+1}} \sum_{j=0}^{q} \frac{\left( p + j \right)!}{j!} \left( \frac{\beta}{\alpha + \beta} \right)^j + \frac{q^{\prime}!}{\alpha^{q^{\prime}+1} \left( \alpha + \beta \right)^{p^{\prime}+1}} \sum^{q^{\prime}}_{j=0} \frac{\left( p^{\prime} + j \right)!}{j!} \left( \frac{\alpha}{\alpha + \beta} \right)^j \right\}
\end{align}

where

\begin{array}{ll}
p = a + 2i + 2 & p^{\prime} = b + 2i + 2 \\
q = b + c - 2i + 2 & q^{\prime} = a + c - 2i + 2
\end{array}

The above applies for <math> a, b \geq -2, c\geq -1</math>.
<math>[x]</math> means "greatest integer in" <math>x</math>.

Then

\begin{align}
I_1 \left( a, b, c \right) & = \int d\tau_{r} r_{1}^{a} r_{2}^{b} r_{12}^{c} e^{ -\alpha r_{1} - \beta r_{2}} P_{1}\left( \cos \theta \right) \\
& = \frac{1}{2} \left( I_{0} \left( a + 1, b - 1, c \right) + I_{0} \left( a - 1, b + 1, c \right) - I_{0} \left( a - 1, b - 1, c + 2 \right) \right)
\end{align}

===The Radial Recursion Relation===

Recall that the full integral is

\begin{equation}
I(l_1m_1,l_2m_2;R)=2\pi\delta_{l_1,l_2}\delta_{m_1,m_2}I_{l_2}(R)
\end{equation}

where

\begin{equation}
I_{l_2}(R) = \int d\tau_r R(r_1,r_2,r_{12})P_{l_2}(\cos \theta_{12})
\end{equation}

To obtain the recursion relation, use

\begin{equation}
P_l(x) = \frac{[P^{\prime}_{l+1}(x)-P^{\prime}_{l-1}(x)]}{2l+1}
\end{equation}

with

\begin{equation}
P_{l+1}^\prime (x) = \frac{d}{dx}P_{l+1}(x)
\end{equation}

Here <math>x=\cos\theta_{12}</math> and

\begin{align}
\frac{d}{d \cos \theta_{12}} & = \frac{dr_{12}}{d \cos \theta_{12}} \frac{d}{dr_{12}} \\
& = -\frac{r_{1} r_{2}}{r_{12}} \frac{d}{dr_{12}}
\end{align}

Then,

\begin{equation}
I_{l}\left( R \right) = - \int d\tau_r R \frac{r_{1} r_{2}}{r_{12}} \frac{d}{dr_{12}} \frac{ \left[ P_{l+1} \left( \cos \theta_{12} \right) - P_{l-1} \left( \cos \theta_{12} \right) \right]}{2l+1}
\end{equation}

The <math>r_{12}</math> part of the integral is

\begin{align}
& \int^{r_{1} + r_{2}}_{|r_{1} - r_{2}|} r_{12} dr_{12} R \frac{r_{1} r_{2}}{r_{12}} \frac{d}{dr_{12}} \left[P_{l+1}-P_{l-1} \right] \\
& = R r_{1} r_{2} \left[ P_{l+1} - P_{l-1} \right]^{r_{1} + r_{2}}_{|r_{1} - r_{2}|} - \int^{r_{1} + r_{2}}_{|r_{1} - r_{2}|} r_{12} dr_{12} \left( \frac{d}{dr_{12}} R \right) \frac{r_{1} r_{2}}{r_{12}} \frac{ \left[ P_{l+1} - P_{l-1} \right]}{2l+1}
\end{align}

The integrated term vanishes because

\begin{equation}
\cos \theta_{12} = \frac{r_{1}^{2} + r_{2}^{2} - r_{12}^{2}}{2 r_{1} r_{2}}
\begin{array}{lll}
& = -1 & {\rm when} & r_{12}^{2} = \left( r_{1} + r_{2} \right)^{2} \\
& = 1 & {\rm when} & r_{12}^{2} = \left( r_{1} - r_{2} \right)^{2}
\end{array}
\end{equation}

and <math>P_l(1)=1</math>, <math>P_{l}(-1)=(-1)^{l}</math>.

Thus,

\begin{equation}
I_{l+1}\left(\frac{r_1r_2}{r_{12}}R^\prime\right) = \left( 2l+1 \right) I_{l}(R) - I_{l-1} \left( \frac{r_{1} r_{2}}{r_{12}} R^{\prime} \right)
\end{equation}

If

\begin{equation}
R = r_{1}^{a-1} r_{2}^{b-1} r_{12}^{c+2} e^{-\alpha r_{1} - \beta r_{2}}
\end{equation}

then [G.W.F. Drake, Phys. Rev. A 18, 820 (1978)]

\begin{equation}
I_{l+1}(r_1^ar_2^br_{12}^c) =
\frac{2l+1}{c+2} I_{l} \left( r_{1}^{a-1} r_{2}^{b-2} r_{12}^{c+2} \right)+I_{l-1} \left( r_{1}^{a} r_{2}^{b} r_{12}^{c} \right)
\end{equation}

For the case <math>c=-2</math>, take

\begin{equation}
R = r_{1}^{a-1} r_{2}^{b-1} \ln r_{12} e^{ -\alpha r_{1} -\beta r_{2}}
\end{equation}

Then

\begin{equation}
I_{l+1} \left( r_{1}^{a} r_{2}^{b} r_{12}^{c} \right) = I_{l} \left( r_{1}^{a-1} r_{2}^{b-1} \ln r_{12} \right) \left( 2l + 1 \right) - I_{l-1} \left( r_{1}^{a} r_{2}^{b} r_{12}^{-2} \right)
\end{equation}

This allows all <math>I_{l}</math> integrals to be calculated from tables of <math>I_0</math> and
<math>I_1</math> integrals.

===The General Integral===

The above results for the angular and radial integrals can now be combined into a general formula for integrals of the type

\begin{equation}
I = \int \int d{\bf r}_{1} d{\bf r}_{2} R_{1} \mathcal{Y}^{M^{\prime}}_{l^{\prime}_{1} l^{\prime}_{2} L^{\prime}} \left( \hat{r}_{1}, \hat{r}_{2} \right) T_{k_{1} k_{2} K}^{Q} \left( {\bf r}_{1},{\bf r}_{2} \right) R_{2} \mathcal{Y}_{l_{1} l_{2} L}^{M} \left( \hat{r}_{1}, \hat{r}_{2} \right)
\end{equation}

where

\begin{equation}
\mathcal{Y}^{M}_{l_{1} l_{2} L} \left( \hat{r}_{1}, \hat{r}_{2} \right) = \sum_{m_1,m_2} \langle l_{1} l_{2} m_{1} m_{2} | LM \rangle Y^{m_{1}}_{l_{1}} \left( \hat{r}_{1} \right) Y^{m_{2}}_{l_{2}} \left( \hat{r}_{2} \right)
\end{equation}

and

\begin{equation}
T^Q_{k_{1} k_{2} K} \left( {\bf r}_{1}, {\bf r}_{2} \right) = \sum_{q_{1}, q_{2}} \langle k_{1} k_{2} q_{1} q_{2} | KQ \rangle Y_{k_{1}}^{q_{1}} \left( \hat{r}_{1} \right) Y^{q_{2}}_{k_{2}} \left( \hat{r}_{2} \right)
\end{equation}

The basic idea is to make repeated use of the formula

\begin{equation}
Y^{m_{1}}_{l_{1}} \left( \hat{r}_{1} \right) Y^{m_{2}}_{l_{2}} \left( \hat{r}_{1} \right) = \sum_{l m} \left( \frac{\left( 2l_{1}+1 \right) \left(2l_{2}+1 \right) \left(2l+1 \right)}{4 \pi} \right)^{1/2}
\times \left( \begin{array}{ccc}
l_1 & l_2 & l \\
m_1 & m_2 & m
\end{array} \right) \left(\begin{array}{ccc}
l_1 & l_2 & l \\
0 & 0 & 0
\end{array} \right)
Y^{m*}_{l} \left( \hat{r}_{1} \right)
\end{equation}

where

\begin{equation}
Y^{m*}(\hat{r}) = (-1)^mY_l^{-m}(\hat{r})
\end{equation}

and

\begin{equation}
\left(\begin{array}{ccc}
l_1 & l_2 & l \\
m_1 & m_2 & m
\end{array} \right) = \frac{\left( -1 \right)^{l_{1} - l_{2} - m}}{\left( 2l + 1 \right)^{1/2}} \left( l_{1} l_{2} m_{1} m_{2} | l, -m \right)
\end{equation}

is a 3-<math>j</math> symbol. In particular, write

\begin{array}{cc}
Y^{m_1^\prime}_{l_1^\prime}(\hat{r}_1)^{*} \underbrace{Y_{k_{1}}^{q_{1}} \left( \hat{r}_{1} \right) Y^{m_{1}}_{l_{1}} \left( \hat{r}_{1} \right)} & = & \sum_{\Lambda M}(\ldots)Y^{M}_\Lambda(\hat{r}_1)\\
\ \ \ \ \ \ \ \sum_{\lambda_1\mu_1}Y_{\lambda_1}^{\mu_1}(\hat{r}_1) & & \\
& & \\
Y^{m_{2}^{\prime}}_{l_{2}^{\prime}} \left( \hat{r}_{2} \right) \underbrace{Y^{q_{2}}_{k_{2}} \left( \hat{r}_{2} \right) Y_{l_{2}}^{m_{2}} \left( \hat{r}_{2} \right)} & = & \sum_{\Lambda^{\prime} M^{\prime}} \left( \ldots \right)Y_{\Lambda^{\prime}}^{M^{\prime}*} \left( \hat{r}_{2} \right) \\
\ \ \ \ \ \ \ \ \ \sum_{\lambda_{2} \mu_{2}}Y_{\lambda_{2}}^{\mu_{2}} \left( \hat{r}_{2} \right) & &
\end{array}

The angular integral then gives a factor of <math>2 \pi \delta_{\Lambda, \Lambda^{\prime}} \delta_{M, M^{\prime}} P_{\Lambda} \left( \cos \theta_{12} \right)</math>. The total integral therefore reduces to the form

\begin{equation}
I = \sum_{\Lambda} C_{\Lambda} I_{\Lambda} \left( R_{1} R_{2} \right)
\end{equation}

where <math>C_{\Lambda} = \sum_{\lambda_{1}, \lambda_{2}} C_{ \lambda_{1}, \lambda_{2}, \Lambda}</math>. For further details and derivations, including graphical representations, see G.W.F. Drake, Phys. Rev. A <math>{\bf 18}</math>, 820 (1978).

==Graphical Representation==

[[File:GraphRep1.jpg|500px]]

==Matrix Elements of H==
Recall that

\begin{equation}
H = -\frac{1}{2} \nabla^{2}_{1} - \frac{1}{2} \nabla^{2}_{2} - \frac{1}{r_1} - \frac{1}{r_2} + \frac{Z^{-1}}{r_{12}}
\end{equation}

Consider matrix elements of

\begin{equation}
\nabla^2_1 = \frac{1}{r_1^2} \frac{\partial}{\partial r_1}\left(r_1^2\frac{\partial}{\partial r_1} \right) + \frac{1}{r^2}\frac{\partial}{\partial r}\left(r^2\frac{\partial}{\partial r}\right) - \frac{(\vec{l}_1^Y)^2}{r_1^2} + \frac{2 \left( r_1 - r_2 \cos \theta \right)}{r} \frac{\partial^2}{\partial r_1 \partial r} - 2 \left( \nabla^Y_1 \cdot {\bf r}_2 \right) \frac{1}{r} \frac{\partial}{\partial r}
\end{equation}

where <math>r \equiv r_{12}</math> and <math>\cos\theta \equiv \cos\theta_{12}</math>. Also in what follows, define <math>\hat{\nabla}^Y_1 \equiv r_1\nabla^Y_1</math> where <math>\nabla^Y</math> operates only on the spherical harmonic part of the wave function, and similarly for <math>\vec{l}_1^Y</math>.

A general matrix element is

\begin{equation}
\langle r_1^{a^\prime} r_2^{b^\prime} r_{12}^{c^\prime} e^{-\alpha^\prime r_1 - \beta^\prime r_2} \mathcal{Y}^{M^\prime}_{l^\prime_1 l^\prime_2 L^\prime} \left( \hat{r}_1, \hat{r}_2 \right) \left| \nabla^2_1 \right| r_1^a r_2^b r_{12}^c e^{-\alpha r_1 - \beta r_2} \mathcal{Y}^M_{l_1 l_2 L} \left( \hat{r}_1, \hat{r}_2 \right) \rangle
\end{equation}

Since <math>\nabla^2_1</math> is rotationally invariant, this vanishes unless <math>L=L^\prime</math>, <math>M=M^\prime</math>. Also <math>\nabla_1^2</math> is Hermitian, so that the result must be the same whether it operates to the right or left, even though the results look very different. In fact, requiring the results to be the same yields some interesting and useful integral identities as follows:

Put

\begin{equation}
\mathcal{F} = F \mathcal{Y}^M_{l_1 l_2 L} \left( \hat{r}_1, \hat{r_2} \right)
\end{equation}

and

\begin{equation}
F = r_1^a r_2^b r_{12}^c e^{-\alpha r_1 - \beta r_2}
\end{equation}

Then

\begin{equation}
\nabla^2_1 \mathcal{F} = \left\{ \frac{1}{r_1^2} \left[ a \left( a + 1 \right) - l_1 \left( l_1 + 1 \right) \right] + \frac{c \left( c + 1 \right)}{r^2} + \alpha^2\frac{2\alpha \left( a + 1 \right)}{r_1} + \frac{2 \left(r_1 - r_2 \cos \theta \right) }{r_1r^2}c \left[ a - \alpha r_1 \right] - \frac{2c}{r^2} \left( \hat{\nabla}^Y_1 \cdot \hat{r}_2 \right) \frac{r_2}{r_1} \right\} \mathcal{F}
\end{equation}

and

\begin{align}
\langle \mathcal{F}^\prime \left| \nabla^2_1 \right| \mathcal{F} \rangle & = \sum_\Lambda \int d\tau_r F^\prime C_\Lambda \left( 1 \right) P_\Lambda \left( \cos \theta \right) \times \left\{ \frac{1}{r_1^2} \left[ a \left( a + 1 \right) - l_1 \left( l_1 + 1 \right) \right] -\frac{2 \alpha \left( a + 1 \right)}{r_1} + \frac{c \left( c + 1 \right)}{r^2} +\alpha^2 + \frac{2 \left( r_2 - r_2 \cos \theta \right)}{r_1r^2}c \left( a - \alpha r_1 \right) \right\} F \\
& + \sum_\Lambda \int d\tau_r F^\prime C_\Lambda \left( \hat{\nabla}^Y_1 \cdot \hat{r}_2 \right) P_\Lambda \left( \cos \theta \right) \left( \frac{-2c r_2}{r_1 r^2} \right) F
\end{align}

where

\begin{equation}
\int d\tau_r = \int^\infty_0 r_1 dr_1 \int^\infty_0 r_2 dr_2 \int^{r_1+r_2}_{\left| r_1 - r_2 \right|} r dr
\end{equation}

For brevity, let the sum over <math>\Lambda</math> and the radial integrations be understood, and let <math>\left( \nabla_1^2 \right)</math> stand for the terms which appear in the integrand. Then operating to the right gives

\begin{equation}
(\nabla_1^2)_R = \frac{1}{r_1^2} \left[ a \left( a + 1 \right) - l_1 \left( l_1 + 1 \right) \right] + \frac{c \left( c + 1 \right)}{r^2} + \alpha^2 - \frac{2 \alpha \left( a + 1 \right)}{r_1}
+ \frac{2 \left( r_1 - r_2 \cos \theta \right)}{r_1 r^2}c \left( a - \alpha r_1 \right) - \frac{2c r_2}{r_1 r^2} \left( \hat{\nabla}^Y_1 \cdot \hat{r}_2 \right)
\end{equation}

Operating to the left gives

\begin{equation}
(\nabla_1^2)_L = \frac{1}{r_1^2} \left[a^\prime \left( a^\prime + 1 \right) - l_1^\prime \left( l_1^\prime + 1 \right) \right] + \frac{c^\prime \left( c^\prime + 1 \right)}{r^2} + \alpha^{\prime 2} -\frac{2 \alpha^\prime \left( a^\prime + 1 \right)}{r_1}
+ \frac{2 \left( r_1 - r_2 \cos \theta \right)}{r_1 r^2} c^\prime \left( a^\prime - \alpha^\prime r_1 \right) - \frac{2 c^\prime r_2}{r_1 r^2} \left( \hat{\nabla}^{Y^\prime}_1 \cdot \hat{r}_2 \right)
\end{equation}

Now put

\begin{array}{cc}
a_+ = a+a^\prime, & \hat{\nabla}_1^+ = \hat{\nabla}_1^Y + \hat{\nabla}_1^{Y\prime} \\
a_{-} = a-a^\prime, & \hat{\nabla}_1^{-} = \hat{\nabla}_1^Y - \hat{\nabla}_1^{Y\prime}
\end{array}

etc., and substitute <math>a^\prime = a_+ - a</math>, <math>c^\prime = c_+ - c</math>, and <math>\alpha^\prime = \alpha_+ - \alpha</math> in <math>(\nabla^2_1)_L</math>. If <math>a_+</math>, <math>c_+</math> and <math>\alpha_+</math> are held fixed, then the equation

\begin{equation}
(\nabla^2_1)_R = (\nabla^2_1)_L
\end{equation}

must be true for arbitrary <math>a</math>, <math>c</math>, and <math>\alpha</math>. Their coefficients must thus vanish.

This yields the integral relations

\begin{equation}
\label{I}
\frac{\left( r_1 - r_2 \cos \theta \right)}{r_1 r^2} = \frac{1}{c_+} \left( \frac{- \left( a_+ + 1 \right)}{r_1^2} + \frac{\alpha_+}{r_1} \right) \tag{I}
\end{equation}

from the coefficient of <math>a</math> and

\begin{equation}
\label{II}
\frac{ \left( r_1 - r_2 \cos \theta \right) \left( a_+ - \alpha_+ r_1 \right)}{r_1 r^2} = \frac{r_2}{r_1 r^2}\left( \hat{r}_2 \cdot \hat{\nabla}_1^+ \right) - \frac{ \left( c_+ + 1 \right)}{r^2} \tag{II}
\end{equation}

from the coefficient of <math>c</math>. The coefficient of <math>\alpha</math> gives an equation equivalent to (I).

Furthermore, if can be show that (see problem)

\begin{align}
& \sum_\Lambda \int d\tau_r \frac{r^c}{r^2} C_\Lambda \left( \hat{r}_2 \cdot \hat{\nabla}_1^Y \right) P_\Lambda \left( \cos \theta \right) \\
= & \sum_\Lambda \int d\tau_r \frac{r^c}{c r_1 r_2}C_\Lambda \left( 1 \right)P_\Lambda \left( \cos \theta \right) \times \left( \frac{l_1^\prime \left( l_1^\prime + 1 \right) - l_1 \left( l_1 + 1 \right) -\Lambda \left( \Lambda + 1 \right)}{2} \right)
\end{align}

and similarly for <math>\left( \hat{r}_2 \cdot \hat{\nabla}^{Y\prime}_1 \right)</math> with <math>l_1</math> and <math>l_1^\prime</math> interchangeable, then it follows that

\begin{equation}
\frac{r^{c_+}}{r^2} \left( \hat{r}_2 \cdot \hat{\nabla}_1^+ \right) = - \frac{r^{c_+}}{c_+ r_1 r_2} \Lambda \left( \Lambda + 1 \right)
\end{equation}

and

\begin{equation}
\frac{r^{c_+}}{r^2} \left( \hat{r}_2 \cdot \hat{\nabla}_1^- \right) = \frac{r^{c_+}}{c_+ r_1 r_2} \left[ l_1^\prime \left( l_1^\prime + 1 \right) - l_1 \left( l_1 + 1 \right) \right]
\end{equation}

where equality applies after integration and summation over <math>\Lambda</math>.

Thus (II) becomes

\begin{equation}\label{IIa}
\frac{ \left( r_1 - r_2 \cos \theta \right) \left( a_+ - \alpha_+ r_1 \right)}{r_1 r^2} = - \frac{\Lambda \left( \Lambda + 1 \right)}{c_+ r^2_1} - \frac{ \left( c_+ + 1 \right)}{r^2} \tag{IIa}
\end{equation}

===Problem===

Prove the integral relation

\begin{align}
& \sum_\Lambda \int d\tau_r f \left( r_1, r_2 \right) \frac{1}{r} \left( \frac{d}{dr} q(r) \right) C_\Lambda \left( \hat{r}_2 \cdot \hat{\nabla}^Y_1 \right) P_\Lambda \left( \cos \theta \right) \\
= & \sum_\Lambda \int d\tau_r \frac{ f \left( r_1, r_2 \right)}{r_1, r_2} q(r) C_\Lambda \left( 1 \right) P_\Lambda \left( \cos \theta \right) \left(\frac{l_1^\prime \left( l_1^\prime + 1 \right) - l_1 \left( l_1 + 1 \right) - \Lambda \left( \Lambda + 1 \right)}{2}\right)
\end{align}

where the coefficients <math>C_\Lambda \left( 1 \right)</math> are the angular coefficients from the overlap integral

\begin{align}
\int & d\Omega\, \mathcal{Y}^{M*}_{l_1^\prime l_2^\prime L}(\hat{r}_1,\hat{r}_2)\, \mathcal{Y}^{M}_{l_1 l_2 L}(\hat{r}_1,\hat{r}_2)\\
= & \sum_\Lambda C_\Lambda(1)P_\Lambda(\cos\theta_{12}).
\end{align}

Hint: Use the fact that <math>l_1^2</math> is Hermitian so that
<math>
\int d\tau (l_1^2\mathcal{Y}^\prime)^*q(r)\mathcal{Y} = \int d\tau
\mathcal{Y}^{\prime *}l_1^2(q(r)\mathcal{Y})
</math>
with <math>\ \vec{l}_1= \frac{1}{i}\vec{r}_1\times\nabla_1</math>.

It is also useful to use
<math>\begin{eqnarray}\ \
\cos\theta\, P_L(\cos\theta) &=& \frac{1}{2L+1}[L\,P_{L-1}(\cos\theta) + (L+1)P_{L+1}(\cos\theta)]
\\
(\cos^2\theta-1)P_L(\cos\theta) &=&
\frac{L(L-1)}{(2L-1)(2L+1)}P_{L-2}(\cos\theta)\nonumber
-\frac{2(L^2+L-1)}{(2L-1)(2L+3)}P_L(\cos\theta) + \frac{(L+1)(L+2)}{(2L+1)(2L+3)}P_{L+2}(\cos\theta)\nonumber\\
&=& \frac{L(L-1)}{(2L-1)(2L+1)}[P_{L-2}(\cos\theta) - P_L(\cos\theta)]
+ \frac{(L+1)(L+2)}{(2L+1)(2L+3)}[P_{L+2}(\cos\theta) - P_L(\cos\theta)]
\end{eqnarray}</math>

together with a double application of the integral recursion relation
<math>
I_{L+1}\left(\frac{1}{r}\frac{d}{dr}q(r)\right) =
(2L+1)I_L\left(\frac{1}{r_1r_2}q(r)\right)+I_{L-1}\left(\frac{1}{r}\frac{d}{dr}q(r)\right)
</math>

Of course <math>l_1^2\mathcal{Y}^M_{l_1l_2L}(\hat{r}_1,\hat{r}_2)=l_1(l_1+1)\mathcal{Y}^M_{l_1l_2L}(\hat{r}_1,\hat{r}_2)</math>. Begin by expanding
<math>
l_1^2(g\mathcal{Y}) = \mathcal{Y}l_1^2g + g\,l_1^2\mathcal{Y} + 2(\vec{l}_1g)\cdot(\vec{l}_1\mathcal{Y})
</math>, and show that

<math>
\begin{eqnarray}\ \
(\vec{l}_1g)\cdot(\vec{l}_1\mathcal{Y}) & = & \frac{r_1^2}{r}\frac{dg}{dr}\vec{r}_2\cdot \nabla_1\mathcal{Y}\,\,\,\,\mbox{ and}\\
l_1^2g(r) &=& 2r_1r_2\cos\theta\frac{1}{r}\frac{dg}{dr} + r_1^2r_2^2(\cos^2\theta - 1)
\frac{1}{r}\frac{d}{dr}\left(\frac{1}{r}\frac{dg}{dr}\right).
\end{eqnarray}
</math>

The proof amounts to showing that the term <math>l_1^2g(r)</math> can be replaced by
<math>\Lambda(\Lambda+1)g(r)</math> after multiplying by <math>P_\Lambda(\cos\theta)</math>
and integrating by parts with respect to the radial integrations over <math>r_1</math>, <math>r_2</math> and
<math>r \equiv |\vec{r}_1 - \vec{r}_2|</math>. Remember that <math>\cos\theta</math> is just a short-hand notation for the radial function <math>(r_1^2 + r_2^2 - r^2)/(2r_1r_2)</math>.

==General Hermitian Property==

Each combination of terms of the form
<math>
<f>_R = a^2f_1+af_2+abf_3+b^2f_4+bf_5+af_6\nabla_1^Y+bf_7\nabla^Y_2+f_8(Y)
</math>
can be rewritten
<math>
<f>_L= (a_+-a)^2f_1+(a_+-a)f_2 + (a_+-a)(b_+-b)f_3+(b_+-b)^2f_4
+(b_+-b)f_5+(a_+-a)f_6\nabla^{Y\prime}_1+(b_+-b)f_7\nabla^{Y\prime}_2+f_8)Y^\prime
</math>
where as usual <math>\nabla_1^Y</math> acts only on the spherical harmonic part of the wave function denoted for short by <math>Y</math>.
Since these must be equal for arbitrary <math>a</math> and <math>b</math>,
<math>
a_+^2f_1+a_+f_2+_+b_+f_3+b_+f_4+b_+f_5+a_+f_6\nabla^{Y\prime}_1 +
b_+f_7\nabla^{Y\prime}_2+f_8(Y^\prime)-f_8(Y)=0
</math>

Adding the corresponding expression with <math>Y</math> and <math>Y^\prime</math> interchanged
yields
<math>\label{III}
a_+^2f_1+a_+f_2+a_+b_+f_3+b_+^2f_4+b_+f_5+\frac{1}{2}a_+f_6\nabla^+_1+\frac{1}{2}b_+f_7\nabla^+_2=0\tag{III}
</math>

Subtracting gives
<math>\label{IV}
f_8(Y)-f_8(Y^\prime)=-\frac{1}{2}[a_+f_6\nabla^-_1+b_+f_7\nabla_2^-]\tag{IV}
</math>
<math>\label{V}
a[-2a_+f_1-2f_2-b_+f_3-f_6\nabla^+_1]=0\tag{V}
</math>
<math>\label{VI}
b[-2b_+f_4-2f_5-a_+f_3-f_7\nabla^+_2]=0\tag{VI}
</math>

Adding the two forms gives

\begin{eqnarray}
<f>_R+<f>_L=\frac{1}{2}(a_+^2+a_-^2)f_1+a_+f_2+\frac{1}{2}(a_+b_++a_-b_-)f_3\nonumber\\
\ +\frac{1}{2}(b_+^2+b_-^2)f_4+b_+f_5+\frac{1}{2}f_6(a_+\nabla_1^++a_-\nabla^-_1)\nonumber\\
\ +\frac{1}{2}(b_+\nabla_2^++b_-\nabla^-_2)+f_8(Y)+f_8(Y^\prime)\nonumber
\end{eqnarray}

Subtracting <math>\frac{x}{2}\times({\rm I})</math> gives

\begin{eqnarray}\ \
<f>_R+<f>_L = \frac{1}{2}[(1-x)a_+^2 + a_-^2]f_1 +
(1-\frac{x}{2})a_+f_2+\frac{1}{2}[(1-x)a_+b_++a_-b_-]f_3\nonumber\\
+\frac{1}{2}[(1-x)b_+^2+b_-^2]f_4+(1-\frac{x}{2})f_5b_++\frac{1}{2}f_6[(1-\frac{x}{2})a_+\nabla_1^++a_-\nabla^-_1]\nonumber\\
+\frac{1}{2}[(1-\frac{x}{2})b_+\nabla^+_2+b_-\nabla^-_2]+f_8(Y)+f_8(Y^\prime)\nonumber
\end{eqnarray}

If <math>x=1</math>,
\begin{eqnarray}
<f>_R+<f>_L &=&
\frac{1}{2}[a_-^2f_1+a_+f_2+a_-b_-f_3+b_-^2f_4+b_+f_5\nonumber\\
&+&\left(\frac{1}{2}a_+\nabla^+_1+a_-\nabla^-_1\right)f_6+\left(\frac{1}{2}b_+\nabla^+_2+b_-\nabla^-_2\right)f_7]+f_8(Y)+f_8(Y^\prime)\nonumber
\end{eqnarray}

The General Hermitian Property for arbitrary <math>x</math> gives
\begin{eqnarray}\ \
\nabla_1^2=
\frac{1}{4}\{\frac{1}{r_1^2}[(1-x)a_+^2+a_-^2+2(1-\frac{x}{2})a_+-2[l_1(l_1+1)+l_1^\prime(l_1^\prime+1)]\nonumber\\
-\frac{2}{r_1}[(1-x)\alpha_+a_++\alpha_-a_-+2(1-\frac{x}{2})\alpha_+]+(1-x)\alpha_+^2
+ \alpha^2_-]\nonumber\\
+\frac{2(r_1-r_2\cos\theta)}{r_1r^2}[(1-x)(a_+-\alpha_+r_1)c_++(a_--\alpha_-r_2)c_-]\nonumber\\
-\frac{2r_2}{r_1r^2}[(1-\frac{x}{2})c_+\hat{r}_2\cdot\hat{\nabla}_1^++c_-\hat{r}_2\hat{\nabla}_1^-]+\frac{(1-x)c_+^2+c_-^2+2(1-\frac{x}{2})c_+}{r^2}\}\nonumber
\end{eqnarray}.

Use
<math>\begin{eqnarray}\ \
\frac{-2r_2}{r_1r^2}[(1-\frac{x}{2})c_+\hat{r}_2\hat{\nabla}_1^++c_-\hat{r}_2\hat{\nabla}_1^-]=\frac{2}{r_1^2}\left[(1-\frac{x}{2})\Lambda(\Lambda+1)-\frac{c_-}{c_+}[l_1^\prime(l_1^\prime+1)+l_1(l_1+1)]\right]\end{eqnarray}</math>,

<math>\begin{eqnarray}
\frac{2(r_1-r_2\cos\theta)}{r_1r^2}(a_--\alpha_-r_1)c_- = 2\frac{c_-}{c_+}\left[\frac{-1}{r_1^2}[a_-(a_++1)]+\frac{1}{r_1}[a_-\alpha_++\alpha_-(a_++2)]-\alpha_-\alpha_+\right]\nonumber
\end{eqnarray}</math>,

and
<math>\begin{eqnarray}\ \
\frac{2(r_1-r_2\cos\theta)}{r_1r^2}(a_+-\alpha_+r_1)c_+ = 2\left[-\frac{1}{r_1^2}[a_+(a_++1)]+\frac{1}{r_1}[a_+\alpha_++\alpha_+(a_++2)]-\alpha_+^2\right]
\end{eqnarray}</math>.

Substituting into <math>\nabla_1^2</math> gives

\begin{eqnarray}
\nabla^2_1 = \frac{1}{4}\{\frac{1}{r_1^2}[-(1-x)a_+^2 + a_-^2 +xa_+ +
2(1-\frac{x}{2})\Lambda(\Lambda+1)\nonumber\\
-2l_1(l_1+1)(1-\frac{c_-}{c_+}) -
2l_1^\prime(l_1^\prime+1)(1+\frac{c_-}{c_+})-\frac{2c_-a_-}{c_+}(a_++1)]\nonumber\\
- \frac{2}{r_1}\left[-(1-x)\alpha_+(a_++2)+\alpha_-a_-+2(1-\frac{x}{2})\alpha_+
-\frac{c_-}{c_+}[a_-\alpha_++\alpha_-(a_++2)]\right]\nonumber\\
-(1-x)\alpha_+^2+\alpha_-^2-2\frac{c_-}{c_+}\alpha_-\alpha_+ +\frac{1}{r^2}\left[(1-x)c_+^2+c_-^2+2(1-\frac{x}{2})c_+\right]\}\nonumber
\end{eqnarray}

This has the form
<math>
\nabla_1^2 =
\frac{1}{4}\left[\frac{A_1}{r_1^2}+\frac{B_1}{r_1}+\frac{C_1}{r^2}+D_1\right]
</math>
with

<math>\begin{array}{ccc}
A_1=-(1-x)a_+^2+a_-^2
+xa_++2(1-\frac{x}{2})\Lambda(\Lambda+1)-2l_1(l_1+1)(1-\frac{c_-}{c_+})\
-2l_1^\prime(l_1^\prime+1)(1+\frac{c_-}{c_+})-\frac{2c_-a_-}{c_+}(a_++1) \ \
\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \
\end{array}
</math>

<math>
B_1 =
2\left[(1-x)\alpha_+(a_++2)-\alpha_-a_--2(1-\frac{x}{2})\alpha_++\frac{c_-}{c_+}[a_-\alpha_++\alpha_-(a_++2)]\right]
</math>

<math>
C_1 = (1-x)c_+^2+ c_-^2 +2(1-\frac{x}{2})c_+
</math>

<math>
D_1 = -1(1-x)\alpha_+^2 +\alpha_-^2 - 2\frac{c_-}{c_+}\alpha_-\alpha_+
</math>

The complete Hamiltonian is then (in Z-scaled a.u.)

<math>\begin{eqnarray}
H =-\frac{1}{2}\nabla^2_1 -\frac{1}{2}\nabla^2_2 -\frac{1}{r_1} -\frac{1}{r_2} +\frac{Z^{-1}}{r}\nonumber\\
= -\frac{1}{2}\nabla^2_1 -\frac{1}{2}\nabla^2_2 -\frac{1}{r_1} -\frac{Z-1}{Zr_2}+Z^{-1}\left(\frac{1}{r}-\frac{1}{r_2}\right)\nonumber\\
\ =-\frac{1}{8}\left[\frac{A_1}{r_1^2}+\frac{B_1+8}{r_1}+\frac{C_1}{r^2}+D_1+D_2 +\frac{A_2}{r_2^2} + \frac{B_2 + 8(Z-1)/Z}{r_2}+\frac{C_2}{r^2}\right]\nonumber
+ Z^{-1}\left(\frac{1}{r}-\frac{1}{r_2}\right)\nonumber
\end{eqnarray}</math>

The screened hydrogenic energy is
<math>\begin{eqnarray}
E_{SH}=-\frac{1}{2}\left[\frac{1}{n_1^2}+ \left(\frac{Z-1}{Z}\right)^2\frac{1}{n_2^2}\right]
\end{eqnarray}</math>
so that

<math>\begin{eqnarray}
H-E_{SH} &=& -\frac{1}{8}[
\frac{A_1}{r_1^2}+\frac{B_1+8}{r_1}+\frac{C_1}{r^2} + D_1 + D_2
-\frac{4}{n_1^2} -\left(\frac{Z-1}{Z}\right)^2\frac{4}{n_2^2}\nonumber
+\frac{A_2}{r_2}+\frac{B_2+8(Z-1)/Z}{r_2}+\frac{C_2}{r^2}] +
Z^{-1}\left(\frac{1}{r} -\frac{1}{r_2}\right)\nonumber
\end{eqnarray}</math>

==Optimization of Nonlinear Parameters==
- The traditional method of performing Hylleraas calculations is to write the
basis set in the form

<math>
\Psi = \sum_{i,j,k}c_{i,j,k}\varphi_{i,j,k}(\alpha,\beta)\pm exchange
</math>

with

<math>
\varphi_{i,j,k}(\alpha,\beta) = r_1^ir_2^jr_{12}^ke^{-\alpha r_1 - \beta
r_2}\mathcal{Y}^M_{l_1l_2L}.
</math>

The usual procedure is to set <math>\alpha = Z</math> so that it represents the inner
<math>1s</math> electron, and then to vary <math>\beta</math> so as to minimize the energy.

[[File:OptNon.jpg|500px]]

Since <math>\beta</math> appears in <math>\Psi</math> as a non-linear parameter, the entire
calculation must be repeated for each value of <math>\beta</math>. However, the minimum
becomes progressively smaller as the basis set is enlarged.

[[File:OptNon02.jpg|500px]]

*Difficulties

1. If the basis set is constructed so that <math> i +j +j \leq N</math>, the the
number of terms is
<math>
(N+1)(N+2)(N+3)/6
</math>
N=14 already gives 680 terms and an accuracy of about 1 part in <math>10^{10}</math> for
low-lying states. A substantial improvement in accuracy would require much
larger basis sets, together with multiple precision arithmetic to avoid loss
of significant figures when high powers are included.

2. The accuracy rapidly deteriorates as one goes to more highly excited
states - about 1 significant figure is lost each time the principle quantum
number is increased.

*Cure

We have found that writing basis sets in the form
\begin{eqnarray}
\Psi &=& \sum_{i,j,k}\left(c^{(1)}_{i,j,k}\varphi_{ijk}(\alpha_1\beta_1) +
c_{ijk}^{(2)}\varphi_{ijk}(\alpha_2\beta_2)\right) \pm exchange \nonumber\\
&=& \psi({\bf r}_1,{\bf r}_2) \pm \psi({\bf r}_2,{\bf r}_1)\nonumber
\end{eqnarray}
so that each combination of powers is included twice with different nonlinear
parameters gives a dramatic improvement in accuracy for basis sets of about
the same total size.

However, the optimization of the nonlinear parameters is now much more
difficult, and an automated procedure is needed.

If
<math>
E = \frac{<\Psi|H|\Psi>}{<\Psi|\Psi>}</math> and we assume <math><\Psi|\Psi> = 1</math>, then

<math>
\frac{\partial E}{\partial\alpha_t} = -2 <\Psi|H-E|r_1\psi({\bf r}_1,{\bf
r}_2;\alpha_t)\pm r_21\psi({\bf r}_2,{\bf
r}_1;\alpha_t)>
</math>
where
<math>
\psi({\bf r}_1,{\bf r}_2;\alpha_t) =
\sum_{i,j,k}c_{ijk}^{(t)}\varphi_{ijk}(\alpha_t\beta_t)
</math>

*Problem

1. Prove the above.

2. Prove that there is no contribution from the implicit dependence of
<math>c_{ijk}^{(t)}</math> on <math>\alpha_t</math> if the linear parameters have been optimized.

==The Screened Hydrogenic Term==

If the Hamiltonian is written in the form (in Z-scaled a.u.)
<math>\begin{eqnarray}\
H = H_0({\bf r}_1,Z) + H_0({\bf r}_2,Z-1) +
Z^{-1}\left(\frac{1}{r}-\frac{1}{r_2}\right)
\end{eqnarray}</math>
with
<math>\
H_0({\bf r}_1,Z) = -\frac{1}{2}\nabla_1^2 - \frac{1}{r_1}\nonumber
</math> and
<math>
H_0({\bf r}_2,Z-1) = -\frac{1}{2}\nabla_2^2 - \frac{Z-1}Z{r_2}\nonumber
</math>,
then the eigenvectors of <math>H_0({\bf r},Z) +H_0({\bf r}_2,Z-1)</math> are products of
hydrogenic orbitals
<math>
\Psi_0=\psi_0(1s,Z)\psi_0(nl,Z-1)
</math>
and the eigenvalue is
<math>\begin{eqnarray}
E_{SH} = \left[-\frac{1}{2} - \left(\frac{Z-1}{Z}\right)^2\frac{1}{2n^2}\right]Z^2 a.u.
\end{eqnarray}</math>
called the screened hydrogenic eigenvalue. From highly excited states,
<math>E_{SH}</math> and <math>\Psi_0</math>are already excellent approximations.

For example, for the 1s8d states, the energies are

<math>\begin{eqnarray}\\
E(1s8d^1D)&=& -2.007\,816\,512\,563\,81 {\rm\ a.u.}\nonumber\\
E(1s8d^3D)&=& - 2.007\,817\,934\,711\,71 {\rm\ a.u.}\nonumber\\
E_{SH} &=& -2.007\,812\,5 {\rm\ a.u.} \nonumber
\end{eqnarray}</math>

It is therefore advantageous to include the screened hydrogenic terms in the
basis set so that the complete trial function becomes
<math>
\Psi = c_0\Psi_0 +
\sum_{ijk}\left[c_{ijk}^{(1)}\varphi_{ijk}(\alpha_1,\beta_1)+c_{ijk}^{(2)}\varphi(\alpha_2,\beta_2)\right]\pm
{\rm exchange}
</math>

Also, the variational principal can be re-expressed in the form
<math>\begin{eqnarray}
E= E_{SH}+\frac{<\Psi|H-E_{SH}|\Psi>}{<\Psi|\Psi>}
\end{eqnarray}</math>
so that the <math>E_{SH}</math> term can be cancelled analytically from the matrix
elements.

For example
<math>
<\Psi_0|H-E_{SH}|\Psi_0> = 0
</math>

and
<math>
<\varphi_{ijk}|H-E_{SH}|\Psi_0> =
Z^{-1}<\varphi_{ijk}|\frac{1}{r}-\frac{1}{r_2}|\Psi_0>
</math>

Recall that
<math>\begin{eqnarray}\ \
I_0(a,b,c) = \frac{2}{c+2}\sum^{[c+1]/2}_{i=0}\left(\begin{array}{c}
c+2\\
2i+1
\end{array}\right) \left[f(p,q;\beta) + f(p^\prime,q^\prime;\alpha)\right]
\end{eqnarray}</math>
where

<math>\begin{eqnarray}
f(p,q;x) =
\frac{q!}{x^{q+1}(\alpha+\beta)^{p+1}}\sum^p_{j=0}\frac{(p+j)!}{j!}\left(\frac{x}{\alpha+\beta}\right)^j
\end{eqnarray}</math>

<math>
\begin{array}{cc}
p=a+2i+2 & p^\prime = b+2i+2\\
q= b+c-2i+2 & q^\prime =a+c-2i+2
\end{array}
</math>

If <math>\beta\ll\alpha</math>, the <math>f(p,q;\beta)\gg f(p^\prime,q^\prime;\alpha)</math> and the
<math>i=0</math> term is the dominant contribution to <math>f(p,q;\beta)</math>. However, since
this term depends only on the sum of powers <math>b+c</math> for <math>r_2</math>, it cancels exactly
from the matrix element of
<math>
\frac{1}{r_{12}} - \frac{1}{r_2}
</math>
and can therefore
be omitted in the calculations of integrands, thereby saving many significant
figures. This is especially valuable when <math>b+c</math> is large and <math>a</math> is small.

==Small Corrections==
*Mass Polarization
If the nuclear mass is not taken to be infinite, then the Hamiltonian is
<math>\begin{eqnarray}\ \
H = \frac{P_N^2}{2M} + \sum_{i=1}\frac{p_i^2}{2m} + V
\end{eqnarray}</math>

[[File:SmallCor.jpg|500px]]

where <math>M =M_A-nm</math> is the nuclear mass and <math>m</math> is the electron mass.

Change variables to the C of M
<math>\begin{eqnarray}
{\bf R} = \frac{1}{M+nm}\left[M{\bf r}_N + m({\bf r}_1 + {\bf r}_2 +
\ldots)\right]
\end{eqnarray}</math>

and relative variables

<math>
{\bf s}_i = {\bf r}_i - {\bf r}_N.
</math>

Then

<math>\begin{eqnarray}\ \ \ \ \
H = \frac{1}{2(M+nm)}{\bf p}_R^2+\frac{1}{2m}\sum_{i=1}^n{\bf p}_{s_i}^2
+\frac{1}{2M}\sum_{i,k}{\bf p}_{s_i}\cdot {\bf p}_{s_k} + V({\bf s}_1, {\bf
s}_2, \ldots)\nonumber
= \frac{1}{M+nm}{\bf p}_R^2+\frac{1}{2\mu}\sum_{i=1}^n{\bf
p}_{s_i}^2+\frac{1}{2M}\sum_{i\neq k}{\bf p}_{s_i}\cdot {\bf p}_{s_k}+V({\bf s}_1, {\bf
s}_2, \ldots)\nonumber
\end{eqnarray}</math>

where
<math>\begin{eqnarray}
\mu = \left(\frac{1}{M} + \frac{1}{m}\right)^{-1}
\end{eqnarray}</math>

If <math>n=2</math>, the Schrödinger equation is
<math>\begin{eqnarray}\ \ \
\left[ -\frac{\hbar^2}{2\mu}(\nabla^2_{s_1}+ \nabla^2_{s_2}) -
\frac{\hbar^2}{M}\nabla_{s_1}\nabla_{s_2}-\frac{Ze^2}{s_1}-\frac{Ze^2}{s_2}+\frac{e^2}{s_{12}}\right]\psi
= E\psi
\end{eqnarray}</math>.

Define the reduced mass Bohr radius
<math>\begin{eqnarray}
a_\mu = \frac{\hbar^2}{\mu e^2} = \frac{m}{\mu}a_0
\end{eqnarray}</math>
and

<math>\begin{eqnarray}
\rho_i = \frac{Zs_i}{a_\mu}\nonumber\\
\varepsilon = \frac{E}{Z^2(e^2/a_\mu)}\nonumber
\end{eqnarray}</math>

The Schrödinger equation then becomes
<math>\begin{eqnarray}\ \ \
\left[
-\frac{1}{2}(\nabla^2_{\rho_1}+\nabla^2_{\rho_2})-\frac{\mu}{M}\vec{\nabla}_{\rho_1}\cdot\vec{\nabla}_{\rho_2} - \frac{1}{\rho_1}-\frac{1}{\rho_2}+\frac{Z^{-1}}{\rho_{12}}\right]\psi = \varepsilon \psi
\end{eqnarray}</math>

The effects of finite mass are thus

1. A reduced mass correction to all energy levels of <math>(e^2/a_\mu)/(e^2/a_0)
= \mu/m</math>, i.e.
<math>\begin{eqnarray}
E_M=\frac{\mu}{m}E_\infty
\end{eqnarray}</math>

2. A specific mass shift given in first order perturbation theory by
<math>\begin{eqnarray}
\Delta\varepsilon =
-\frac{\mu}{M}<\psi|\nabla_{\rho_1}\cdot\nabla_{\rho_2}|\psi>
\end{eqnarray}</math>
or

<math>\begin{eqnarray}
\Delta\varepsilon =
-\frac{\mu}{M}<\psi|\nabla_{\rho_1}\cdot\nabla_{\rho_2}|\psi>\frac{e^2}{a_\mu}\nonumber\\
= -\frac{\mu^2}{mM}<\psi|\nabla_{\rho_1}\cdot\nabla_{\rho_2}|\psi>\frac{e^2}{a_0}\nonumber
\end{eqnarray}</math>

Since <math>\nabla_{\rho_1}\cdot\nabla_{\rho_2}</math> has the same angular properties as
<math>{\bf\rho}_1 \cdot {\bf\rho}_2</math>, the operator is like the product of two
dipole operators. For product type wave functions of the form
$
\psi = \psi(1s)\psi(nl) \pm \mbox{exchange},
$
the matrix element vanishes for all but P-states.

=Radiative Transitions=

==Transition Integrals and the Line Strength==

For an electric dipole transition <math>\gamma'\,L'\,S'\,J'\,M' \rightarrow \gamma\,L\,S\,J\,M</math>, the line strength is defined by
\begin{equation}
S = \sum_{M',\mu,M}|\langle\gamma'\,L'\,S'\,J'\,M'|r_\mu|\gamma\,L\,S\,J\,M\rangle|^2
\end{equation}
Using the Wigner-Eckart theorem, this is related to the reduced matrix element by
\begin{eqnarray}
S(J',J) = |\langle\gamma'\,L'\,S'\,J'||r||\gamma\,L\,S\,J\,M\rangle|^2
\sum_{M',\mu,M}\left|\left(\matrix{J' & 1 & J \cr
-M'& \mu &M \cr}\right)\right|^2
\end{eqnarray}
Using the sum rule for the 3-<math>j</math> symbols
\begin{equation}
\sum_{M',\mu,M}\left|\left(\matrix{J' & 1 & J \cr
-M'& \mu &M \cr}\right)\right|^2 = 1
\end{equation}
we see that the line strength is identical to the reduced matrix element
\begin{equation}
S(J',J) = |\langle\gamma'\,L'\,S'\,J'||r||\gamma\,L\,S\,J\,M\rangle|^2
\end{equation}
For a nonrelativistic calculation in <math>LS</math> coupling, <math>S = S'</math> and we can next strip out the dependence on <math>J</math> and <math>J'</math> by use of Eq.\ (7.1.7) of Edmonds { Angular Momentum in Quantum Mechanics} with <math>k = 1</math> to obtain
\begin{equation}
S(J',J) = (-1)^{L'+S+J+1}[(2J+1)(2J'+1)]^{1/2}\left\{
\matrix{L' & J'& S\cr
J & L & 1\cr}\right\}\langle\gamma' L'|| r || \gamma L\rangle
\end{equation}
where <math>\langle\gamma' L'|| r || \gamma L\rangle</math> is the reduced matrix element if spin is neglected. Note that if spin is neglected, then the line-strength is simply
\begin{equation}
S(L',L) = |\langle\gamma' L'|| r || \gamma L\rangle|^2
\end{equation}
To calculate the reduced matrix element, use a particular example along with the Wigner-Eckart theorem
such as
\begin{equation}
\langle\gamma'\, L'|| r || \gamma\, L\rangle = (-1)^{L'}\langle\gamma'\, L'\, 0| z | \gamma\, L\, 0\rangle{\Large /}\left(\matrix{L' & 1 & L \cr
0 & 0 & 0 \cr}\right)
\end{equation}

===Correlated Two-electron Integrals===
For the case of two-electron integrals, <math>z = z_1 + z_2</math> and <math>r = r_1 + r_2</math>. As discussed in G.W.F. Drake, Phys. Rev. A 18, 820 (1978), the general integral in Hylleraas coordinates for the integral over the <math>z_1</math> term can be written in the form
\begin{equation}
\label{int1}
\langle\gamma'\, L'\, 0| z_1 | \gamma\, L\, 0\rangle = \sum_\Lambda C_\Lambda I_\Lambda(\gamma',\gamma)
\end{equation}
where <math>I_\Lambda(\gamma',\gamma)</math> stands for the purely radial integral
\begin{equation}
\label{radial}
I_\Lambda(\gamma',\gamma) = \int_0^\infty r_1\,dr_1\int_0^\infty r_2\,dr_2\int_{|r_1-r_2|}^{r_1+r_2}r_{12}\, dr_{12}
R_{\gamma'}^*\,r_1\, R_\gamma P_\Lambda(\cos\theta_{12})
\end{equation}
and <math>\cos\theta_{12} = (r_1^2 + r_2^2 - r_{12}^2)/(2r_1r_2)</math> is a short-hand notation for a radial function of <math>r_1</math>, <math>r_2</math> and <math>r_{12}</math>. Also, <math>P_\Lambda(\cos\theta_{12})</math> is a Legendre polynomial,
and, <math>R_\gamma = R_\gamma(r_1,r_2,r_{12})</math> is the radial part of the Hylleraas wave function for state <math>\gamma</math> such
that the total wave function is <math>\psi({\bf r}_1,{\bf r}_2) = R_\gamma(r_1,r_2,r_{12}){\cal Y}_{l_1,l_2,L}^M(\hat{\bf r}_1,\hat{\bf r}_2)\pm {\rm exchange}</math>, normalized to unity, and <math>{\cal Y}_{l_1,l_2,L}^M(\hat{\bf r}_1,\hat{\bf r}_2)</math> is a vector coupled product of spherical harmonics. The coefficents <math>C_\Lambda</math> in Eq.\ (\ref{int1}) can be calculated from Eq.\ (32) of the above 1978 paper. They need to be calculated separately for he direct-direct and the direct-exchange contributions. For the reduced matrix element of <math>z_1</math>, the leading 3-<math>j</math> symbol can be dropped, leaving (for the case <math>k = 1</math>)
\begin{eqnarray}
C_\Lambda^{\rm reduced} &=& \frac{(-1)^{l_2'+L'+L+1}}{2} (l_1,l_1',l_2,l_2',L,L')^{1/2}
\sum_{\lambda={\rm max}(l_1-1,0)}^{l_1+1} (\lambda,\Lambda,1)
\left(\matrix{l_1 & 1 & \lambda \cr
0 & 0 & 0 \cr}\right)
\left(\matrix{l_1'& \lambda & \Lambda \cr
0 & 0 & 0 \cr}\right)\nonumber\\
&\times&\left(\matrix{l_2' & 1_2 & \Lambda \cr
0 & 0 & 0 \cr}\right)
\left\{\matrix{L' & \lambda & l_2 \cr
\Lambda& l_2' & l_1' \cr}\right\}
\left\{\matrix{L' & \lambda & l_2 \cr
l_1 & L & 1 \cr}\right\}
\end{eqnarray}
where the notation <math>(a,b,c,\cdots)</math> means <math>(2a+1)(2b+1)(2c+1)\cdots</math>.
A similar expression can be obtained for the angular coefficients for the <math>z_2</math> term by interchanging the subscripts 1 and 2 throughout and multiplying by <math>(-1)^{L+L'+1}</math>.

===Example===
As an example, for the <math>1s^2\;^1S - 1s2p\;^1P</math> transition of helium, the only nonvanishing <math>C_\Lambda</math> angular coefficients are <math>C_1 = 1/2</math> for the direct-direct term (i.e. <math>P_1(\cos\theta_{12}) = \cos\theta_{12}</math> is included in the radial integral (\ref{radial})) and <math>C_0 = 1/2</math> for the direct-exchange term (i.e. <math>P_0(\cos\theta_{12}) = 1</math> is included in the radial integral (\ref{radial})). The results for infinite nuclear mass are <math>S(1\;^1S-2\;^1P) = (0.781\,127\,158\,667)^2</math> and <math>f(1\;^1S-2\;^1P) = -0.276\,164\,703\,736</math> where <math>f(1\;^1S-2\;^1P) = \frac{2}{3}[E(1\;^1S) - E(2\;^1P)]S(1\;^1S-2\;^1P)</math> is the absorption oscillator strength.

Theory Notes

2020-12-14T00:24:57Z

Drake: /* Introduction */

<center>
<H1>Notes on solving the Schrödinger equation in Hylleraas coordinates for heliumlike atoms
Gordon W.F. Drake
Department of Physics, University of Windsor
Windsor, Ontario, Canada N9B 3P4
(Transcribed from hand-written notes and edited by Lauren Moffatt. Last revised November 22, 2017 to correct a missing subscript in the expression for $B_1$ in the Hamiltonian in Sect. 11 General Hermitian Property.)
</H1>
</center>

==Introduction==
These notes describe some of the technical details involved in solving the Schrödinger equation for a heliumlike atom of nuclear charge Z in correlated Hylleraas coordinates. The standard for computational purposes is to express all quantities in a form that is valid for electronic states of arbitrary angular momentum, and to express matrix elements of the Hamiltonian in a manifestly Hermitian symmetrized form. Further information and derivations of the mathematical techniques for angular integrals and radial recursion relations can be found in the paper
G.W.F. Drake, Phys. Rev. A <math>{\bf 18}</math>, 820 (1978).

The starting point is the two-electron Schrödinger equation for infinite nuclear mass
\begin{equation} \left[-\frac{\hbar^2}{2m}(\nabla^2_1 +\nabla^2_2) - \frac{Ze^2}{r_1} -
\frac{Ze^2}{r_2}+\frac{e^2}{r_{12}} \right]\psi = E\psi\nonumber
\end{equation}
where <math>m</math> is the electron mass, and <math>r_{12} = |{\bf r}_1 - {\bf r}_2|</math> (see diagram below).

[[File:Pic1.jpg|250px]]

Begin by rescaling distances and energies so that the Schrödinger equation can be expressed in a dimensionless form. The dimensionless Z-scaled distance is defined by <math>\begin{eqnarray} \rho = \frac{Zr}{a_0} \end{eqnarray} </math> where <math> \begin{eqnarray} a_0 = \frac{\hbar^2}{me^2} \end{eqnarray} </math> is the atomic unit (a.u.) of distance equal to the Bohr radius <math>0.529\, 177\, 210\, 92(17) \times 10^{-10}</math> m. Then

\begin{equation}[-\frac{\hbar^2}{2m}Z^2(\frac{me^2}{\hbar^2})^2(\nabla^2_{\rho_1}+\nabla^2_{\rho_2})
- Z^2\frac{e^2}{a_0}\rho^{-1}_1 - Z^2\frac{e^2}{a_0}\rho^{-1}_2 +
\frac{e^2}{a_0}Z\rho^{-1}_{12}]\psi= E\psi\nonumber\end{equation}

But <math>\begin{eqnarray} \frac{\hbar^2}{m}(\frac{me^2}{\hbar^2})^2 = \frac{e^2}{a_0} \end{eqnarray} </math> is the hartree atomic unit of of energy <math>(E_h = 27.211\,385\,05(60)</math> eV, or equivalently <math>E_h/(hc) = 2194.746\,313\,708(11)</math> m<math>^{-1})</math>. Therefore, after multiplying through by <math>a_0/(Ze)^2</math>, the problem to be solved in Z-scaled dimensionless units becomes

\begin{equation} \left[-\frac{1}{2}(\nabla^2_{\rho_1}+\nabla^2_{\rho_2}) -
\frac{1}{\rho_1} - \frac{1}{\rho_2} + \frac{Z^{-1}}{\rho_{12}}\right]\psi =
\varepsilon\psi\nonumber \end{equation}

where <math>\begin{eqnarray}\varepsilon = \frac{Ea_0}{(Ze)^2} \end{eqnarray}</math> is the energy in Z-scaled atomic units. For convenience, rewrite this in the conventional form
\begin{equation}
H\psi = \varepsilon\psi\nonumber \end{equation}
where (using <math>r</math> in place of <math>\rho</math> for the Z-scaled distance)
\begin{equation}
H = -\frac{1}{2}(\nabla^2_1+\nabla^2_2) - \frac{1}{r_1}-\frac{1}{r_2} + \frac{Z^{-1}}{r_{12}}
\nonumber\end{equation}
is the atomic Hamiltonian for infinite nuclear mass.

==The Hartree Fock Method==

For purposes of comparison, and to define the correlation energy, assume that <math>\psi({\bf r}_1,{\bf r}_2)</math> can be written in the separable product form
\begin{equation}
\psi({\bf r}_1,{\bf r}_2) = \frac{1}{\sqrt{2}} \left[ u_1(r_1) u_2(r_2) \pm u_2(r_1) u_1(r_2) \right] \nonumber
\end{equation}
for the <math>1s^2\ ^1\!S</math> ground state. Because of the <math>\frac{1}{r_{12}}</math> term in the Schrödinger equation, the exact solution connot be expressed in this form as a separable product. However, the Hartree-Fock approximation corresponds to finding the best possible solution to the Schrödinger equation
\begin{equation}
H\psi({\bf r}_1,{\bf r}_2) = E\psi({\bf r}_1,{\bf r}_2)\nonumber
\end{equation}
that can nevertheless be expressed in this separable product form, where as before
\begin{equation}
H = -\frac{1}{2}(\nabla^2_1+\nabla^2_2) - \frac{1}{r_1}- \frac{1}{r_2} + \frac{Z^{-1}}{r_{12}}\nonumber
\end{equation}
is the full two-electron Hamiltonian. To find the best solution, substitute into <math>\left\langle\psi \lvert H-E \rvert \psi\right\rangle</math> and require this expression to be stationary
with respect to arbitrary infinitesimal variations <math>\delta u_1</math> and <math>\delta
u_2</math> in <math>u_1</math> and <math>u_2</math> respectively; i.e.

\begin{equation}
\frac{1}{2} \left\langle \delta u_1(r_1)u_2(r_2) \pm u_2(r_1)\delta
u_1(r_2) \lvert H-E \rvert u_1(r_1) u_2(r_2)\pm u_2(r_1) u_1(r_2)\right\rangle\nonumber
\end{equation}

\begin{equation}
=\int\delta u_1(r_1) d{\bf r}_1 \left\{ \int d{\bf r}_2 u_2(r_2) \left( H - E \right) \left[ u_1(r_1)u_2(r_2) \pm u_2(r_1) u_1(r_2) \right] \right\}\nonumber
\end{equation}

\begin{equation}
= 0 \nonumber
\end{equation}

for arbitrary <math> \delta u_1(r_1).\nonumber </math> Therefore <math>\{\int d{\bf r}_2 \ldots \} = 0</math>.

Similarly, the coefficient of <math>\delta u_2</math> would give

\begin{equation}
\int d{\bf r}_1 u_1(r_1) \left( H-E \right) \left[ u_1(r_1) u_2(r_2) \pm
u_2(r_1) u_1(r_2) \right] = 0\nonumber
\end{equation}

Define

\begin{equation}
I_{12} = I_{21} = \int d{\bf r}\, u_1(r)u_2(r), \nonumber
\end{equation}

\begin{equation}
H_{ij} = \int d{\bf r}\, u_i(r)(-\frac{1}{2}\nabla - \frac{1}{r})u_j(r), \nonumber
\end{equation}

\begin{equation}
G_{ij}(r) = \int d{\bf r}^\prime u_i(r^\prime)\frac{1}{|{\bf r} - {\bf r}\prime|}u_j(r^\prime)\nonumber
\end{equation}

Then the above equations become the pair of integro-differential equations

\begin{equation}
[ H_0 - E + H_{22}+G_{22}(r)]u_1(r) = \mp [ I_{12}(H_0-E) + H_{12}+G_{12}(r)]u_2(r)\nonumber
\end{equation}

\begin{equation}
[H_0-E+H_{11}+G_{11}(r)]u_2(r) = \mp [I_{12}(H_0-E) + H_{12}+G_{12}(r)]u_1(r)\nonumber
\end{equation}

These must be solved self-consistently for the "constants" <math>I_{12}</math> and
<math>H_{ij}</math> and the function <math>G_{ij}(r)</math>.

The Hartree Fock energy is <math>E \simeq -2.87\ldots </math>a.u. while the exact energy is <math>E = -2.903724\ldots </math>a.u.

The difference is called the "correlation energy" because it arises from the
way in which the motion of one electron is correlated to the other. The
Hartree Fock equations only describe how one electron moves in the average field
provided by the other (mean-field theory).

==Configuration Interaction==
Expand
\begin{equation}
\psi({\bf r}_1,{\bf r}_2)= C_0u^{(s)}_1(r_1)u^{(s)}_1(r_2) + C_1u^{(P)}_1({\bf r}_1)u^{(P)}_1({\bf r}_2){\cal Y}^0_{1,1,0}(\hat{\bf r}_1, \hat{\bf r}_2)+C_2u^{(d)}_1({\bf r}_1)u^{(d)}_2({\bf r}_2){\cal Y}^0_{2,2,0}(\hat{\bf r}_1, \hat{\bf r}_2)+... \pm {\rm\ exchange}
\end{equation}
where
\begin{equation}
{\cal Y}^M_{l_1,l_2,L}(\hat{\bf r}_1, \hat{\bf r}_2)=\sum_{m_1,m_2}Y^{m_1}_{l_1}({\bf r}_1)Y^{m_2}_{l_2}({\bf r}_2)\times <l_1l_2m_1m_2\mid LM>
\end{equation}

This works, but is slowly convergent, and very laborious. The best CI calculations are accurate to <math> ~10^{-7}</math> a.u.

==Hylleraas Coordinates==
[E.A. Hylleraas, Z. Phys. <math>{\bf 48}, 469(1928)</math> and <math>{\bf 54}, 347(1929)</math>]
suggested using the co-ordinates <math>r_1</math>, <math>r_2</math> and <math>r_{12}</math> or equivalently

\begin{eqnarray}
s &=& r_1 + r_2, \nonumber\\
t &=& r_1-r_2, \nonumber\\
u &=& r_{12}\nonumber
\end{eqnarray}

and writing the trial functions in the form

\begin{equation}
\Psi({\bf r}_1,{\bf r}_2) = \sum^{i+j+k\leq
N}_{i,j,k}c_{i,j,k}r_1^{i+l_1}r_2^{j+l_2}r_{12}^ke^{-\alpha r_1 - \beta r_2}
\mathcal{Y}^M_{l_1,l_2,L}(\hat{r}_1,\hat{r}_2)\pm \text{exchange}\nonumber
\end{equation}

Diagonalizing H in this non-orthogonal basis set is equivalent to solving
<math> \begin{eqnarray}
\frac{\partial E}{\partial c_{i,j,k}} = 0\nonumber
\end{eqnarray} </math>
for fixed <math>\alpha</math> and <math>\beta</math>.

The diagonalization must be repeated for different values of <math>\alpha</math> and
<math>\beta</math> in order to optimize the non-linear parameters.

==Completeness==
The completeness of the above basis set can be shown by first writing
\begin{equation}
r_{12}^2 = r_1^2 + r_2^2 - 2r_1r_2\cos\theta_{12}
\end{equation}
and, from the spherical harmonic addition theorem,
\begin{equation}
\cos(\theta_{12})=\frac{4\pi}{3}\sum^1_{m=-1}Y^{m*}_1(\theta_1,\varphi_1)Y^m_1(\theta_2,\varphi_2)
\end{equation}
Consider first the S-states. The <math>r_{12}^0</math> terms are like the ss terms in a CI calculation. The
<math>r_{12}^2</math> terms bring in p-p type contributions, and the higher powers bring
in d-d, f-f etc. type terms. In general

\begin{equation}
P_l(\cos\theta_{12}) = \frac{4\pi}{2l+1}\sum^l_{m=-l}{Y^{m}_l}^*(\theta_1,\varphi_1)Y^m_l(\theta_2,\varphi_2)\nonumber
\end{equation}

For P-states, one would have similarly

\begin{array}{lr}
r_{12}^0 & (sp)P\nonumber\\
r_{12}^2 & (pd)P\nonumber\\
r_{12}^4 & (df)P\nonumber\\
\vdots & \vdots\nonumber
\end{array}

For D-states

\begin{array}{lr}
r_{12}^0 & (sd)D & (pp^\prime)D\nonumber\\
r_{12}^2 & (pf)D & (dd^\prime)D\nonumber\\
r_{12}^4 & (dg)D & (ff^\prime)D\nonumber\\
\vdots & \vdots & \vdots\nonumber
\end{array}

In this case, since there are two "lowest-order" couplings to form a
D-state, both must be present in the basis set, i.e.

\begin{equation}
\Psi(r_2,r_2) = \sum c_{ijk}r_1^ir_2^{j+2}r_{12}^ke^{-\alpha r_1-\beta
r_2}\mathcal{Y}^M_{022}(\hat{r}_1,\hat{r}_2)
+\sum d_{ijk}r_1^{i+1}r_2^{j+1}r_{12^k}e^{-\alpha^\prime r_1 - \beta^\prime
r_2}\mathcal{Y}^M_{112}(\hat{r}_1,\hat{r}_2)\nonumber
\end{equation}

For F-states, one would need <math>(sf)F</math> and <math>(pd)F</math> terms.

For G-states, one would need <math>(sg)G</math>, <math>(pf)G</math> and <math>(dd^\prime)G</math> terms.

Completeness of the radial functions can be proven by considering the
Sturm-Liouville problem

\begin{equation}
\left(-\frac{1}{2}\nabla^2-\frac{\lambda}{r_s}-E\right)\psi({\bf r}) = 0\nonumber
\end{equation}

or

\begin{equation}
\left(-\frac{1}{2}\frac{1}{r^2}\frac{\partial}{\partial r}\left({r^2}\frac{\partial}{\partial r}\right) -
\frac{l\left(l+1\right)}{2r^2} - \frac{\lambda}{r} - E\right)u(r) = 0.\nonumber
\end{equation}

For fixed E and variable <math>\lambda</math> (nuclear charge).

The eigenvalues are <math>\lambda_n = (E/E_n)^{1/2}</math>, where <math>E_n =- \frac{1}{2n^2}</math>

[[File:Pic2.jpg|500px]]

and the eigenfunctions are

\begin{equation}
u_{nl}(r) =
\frac{1}{(2l+1)!}\left(\frac{(n+l)!}{(n-l-1)2!}\right)^{1/2}(2\alpha)^{3/2}e^{-\alpha
r}\nonumber
\times \ \left( 2 \alpha r \right)^l\,_{1}F_{1}\left(-n+l+1,2l+2; 2 \alpha r \right)
\end{equation}

with <math>\alpha = (-2E)^{1/2}</math> and <math>n\geq l+1</math>. The confluent hypergeometric function <math> _1F_1(a,b,;z)</math> then denotes a finite polynomial since <math>a=-n+l+1</math> is a negative integer or zero.

Unlike the hydrogen spectrum, which has both a discrete part for <math>E<0</math> and a
continuous part for <math>E>0</math>, this forms an entirely discrete set of finite
polynomials, called Sturmian functions. They are orthogonal with respect to
the potential, i.e.

\begin{equation}
\int^\infty_0 r^2dr \left( u_{n^\prime l}(r)\frac{1}{r}u_{nl}(r) \right) = \delta_{n,n^\prime}\nonumber
\end{equation}

Since they become complete in the limit <math>n\rightarrow\infty</math>, this assures the
completeness of the variational basis set.

[See also B Klahn and W.A. Bingel Theo. Chim. Acta (Berlin) 44, 9 and 27
(1977)].

==Solutions of the Eigenvalue Problem==

For convenience, write

\[
\Psi({\bf r}_1,{\bf r}_2) = \sum^N_{m=1}c_m\varphi_m\nonumber
\]

where <math>m</math> represents the <math>m</math>'th combination of <math>i,j,k</math> and

\begin{equation}
\varphi_{ijk}=r_1^ir_2^jr_{12}^ke^{-\alpha r_1 - \beta
r_2}\mathcal{Y}^M_{l_1,l_2,L}(\hat{r}_1,\hat{r}_2) \pm \mbox{exchange}\nonumber
\end{equation}

\begin{align}

&
\left( \begin{array}{lr}
\cos(\theta) & \sin(\theta)\\
-\sin(\theta) & \cos(\theta)
\end{array} \right)
\left( \begin{array}{lr}
H_{11} & H_{12} \\
H_{12} & H_{22}
\end{array}\right)
\left( \begin{array}{lr}
\cos(\theta) & -\sin(\theta)\\
\sin(\theta) & \cos(\theta)
\end{array} \right)
\\[5pt]
= & \left(\begin{array}{lr}
cH_{11}+sH_{12} & cH_{12} + sH_{22}\\
-sH_{11} + cH_{12} & -sH_{12} + cH_{22}
\end{array}\right)
\left( \begin{array}{cc}
c & -s \\
s & c
\end{array}\right)
\\[5pt]
= & \left(\begin{array}{lr}
c^2H_{11}+s^2H_{22} + 2csH_{12} & (c^2-s^2)H_{12}+cs(H_{22}-H_{11})\\
(c^2-s^2)H_{12}+cs(H_{22}-H_{11}) & s^2H_{11}+c^2H_{22}-2csH_{12}
\end{array}\right)
\end{align}

Therefore

\begin{equation}
\left[\cos^2(\theta)-\sin^2(\theta)\right]H_{12} =
\cos(\theta)\sin(\theta)(H_{11}-H_{22})\nonumber
\end{equation}

and

\begin{equation}
\tan(2\theta) = \frac{2H_{12}}{H_{11}-H_{22}}\nonumber
\end{equation}

i.e.

\begin{equation}
\cos(\theta)=\left(\frac{r+\omega}{2r}\right)^{1/2}\nonumber
\end{equation}

\begin{equation}
\sin(\theta)=-\mbox{sgn}(H_{12})\left(\frac{r-\omega}{2r}\right)^{1/2}\nonumber
\end{equation}

where

\begin{equation}
\omega = H_{22}-H_{11}\nonumber
\end{equation}

\begin{equation}
r=\left(\omega^2+4H_{12}^2\right)^{1/2}\nonumber
\end{equation}

\begin{equation}
E_1=\frac{1}{2}\left(H_{11}+H_{22}-r\right)\nonumber
\end{equation}

\begin{equation}
E_2=\frac{1}{2}\left(H_{11}+H_{22}+r\right)\nonumber
\end{equation}

===Brute Force Method===

- Gives all the eigenvalues and eigenvectors, but it is slow

- First orthonormalize the <math>N</math>-dimensional basis set; i.e. form linear combinations

\begin{equation}
\Phi_m = \sum_{n=1}^N\varphi_nR_{nm}\nonumber
\end{equation}

such that
\begin{equation}
\langle \Phi_m | \Phi_n \rangle = \delta_{m,n}
\end{equation}

This can be done by finding an orthogonal tranformation T such that

\begin{equation}
T^TOT = I = \left(
\begin{array}{ccccc}
I_1 & 0 & 0 & \ldots & 0 \\
0 & I_2 & 0 & \ldots & 0 \\
0 & 0 & I_3 & \ldots & 0 \\
\vdots & \vdots & \vdots & \ddots & \vdots \\
0 & 0 & 0 & \ldots & I_N
\end{array}\right);
\end{equation}

\begin{equation}
O_{mn} = \langle \varphi_m | \varphi_n \rangle \nonumber
\end{equation}

and then applying a scale change matrix

\begin{equation}
S = \left(\begin{array}{ccccc}
\frac{1}{I_1^{1/2}} & 0 & 0 & \ldots & 0\\
0 & \frac{1}{I_2^{1/2}} & 0 & \ldots & 0 \\
0 & 0 & \frac{1}{I_3^{1/2}} & \ldots & 0 \\
\vdots & \vdots & \vdots & \ddots & \vdots \\
0 & 0 & 0 & \ldots & I_N^{1/2}
\end{array}\right)= S^T\nonumber
\end{equation}

Then

\begin{equation}
S^TT^TOTS = 1
\end{equation}

i.e.

\begin{equation}
R^TOR = 1
\end{equation}

with <math>R=TS</math>.

If H is the matrix with elements <math>H_{mn} = \langle \varphi_m | \varphi_n \rangle </math>, then H
expressed in the <math>\Phi_m</math> basis set is

\begin{equation}
H^\prime = R^THR.\nonumber
\end{equation}

We next diagonalize <math>H^\prime</math> by finding an orthogonal transformation W such
that

\begin{equation}
W^T H^\prime W = \lambda = \left(
\begin{array}{cccc}
\lambda_1 & 0 & \ldots & 0 \\
0 & \lambda_2 & \ldots & 0 \\
\vdots & \vdots & \ddots & \vdots \\
0 & 0 & \ldots & \lambda_N
\end{array}\right)\nonumber
\end{equation}

The q'th eigenvector is

\begin{equation}
\Psi^{(q)} = \sum_{n=1}^N\Phi_n W_{n,q} = \sum_{n,n^\prime} \varphi_{n^\prime} R_{n^\prime ,n} W_{n,q} \nonumber
\end{equation}

i.e.

\begin{equation}
c_{n^\prime}^{(q)} = \sum_{n=1}^N R_{n^\prime n} W_{n,q}
\end{equation}

===The Power Method===

- Based on the observation that if H has one eigenvalue, <math>\lambda_M</math>, much bigger than all the rest, and <math>\chi = \left(
\begin{array}{c}a_1\\a_2\\\vdots\end{array}\right)</math>
is an arbitrary starting vector, then <math>\chi = \sum_{q=1}^N x_q\Psi^{(q)}</math>.

\begin{equation}
(H)^n\chi = \sum_{q=1}^N x_q \lambda^n_q\Psi^{(q)}\nonumber\rightarrow x_M\lambda_M^n\Psi^{(M)}\nonumber
\end{equation}

provided <math>x_M\neq 0</math>.

To pick out the eigenvector corresponding to any eigenvalue, write the original problem in the form

\begin{align}
H\Psi & = \lambda O \Psi \nonumber \\
\left( H - \lambda_q O \right) \Psi & = \left( \lambda - \lambda_q \right) O \Psi \nonumber
\end{align}

Therefore,

\begin{equation}
G \Psi = \frac{1}{\lambda - \lambda_q } \Psi \nonumber \\
\end{equation}

where <math>G=(H-\lambda_qO)^{-1}O</math> with eigenvalues <math>\frac{1}{\lambda_n - \lambda_q}</math>.

By picking <math>\lambda_q</math> close to any one of the <math>\lambda_n</math>, say
<math>\lambda_{n^\prime}</math>, then <math>\frac{1}{\lambda_n-\lambda_q}</math> is much larger for
<math>n=n^\prime</math> than for any other value. The sequence is then

\begin{array}{c}
\chi_1=G\chi\nonumber\\
\chi_2=G\chi_1\nonumber\\
\chi_3=G\chi_2\nonumber\\
\vdots\nonumber
\end{array}

until the ratios of components in <math>\chi_n</math> stop changing.

- To avoid matrix inversion and multiplication, note that the sequence is equivalent to
\begin{equation}
F \chi_n = \left( \lambda - \lambda_q \right) O \chi_{n-1} \nonumber \\
\end{equation}
where <math>F = H-\lambda_qO</math>. The factor of <math>\left( \lambda - \lambda_q \right)</math> can be
dropped because this only affects the normalization of <math>\chi_n</math>. To find <math>\chi_n</math>, solve
\begin{equation}
F \chi_n = O \chi_{n-1} \nonumber
\end{equation}
(N equations in N unknowns). Then
\begin{equation}
\lambda = \frac{ \langle \chi_n | H | \chi_n \rangle}{\langle \chi_n | \chi_n \rangle} \nonumber
\end{equation}

==Matrix Elements of H==

\begin{equation}
H=-\frac{1}{2}\nabla^2_1 -\frac{1}{2}\nabla^2_2 - \frac{1}{r_1} - \frac{1}{r_2} +\frac{Z^{-1}}{r_{12}}\nonumber
\end{equation}

Taking <math>r_1, r_2</math> and <math>r_{12}</math> as independent variables,

\begin{equation}
\nabla_1^2 = \frac{1}{r_1^2}\frac{\partial}{\partial r_1}\left(r_1^2\frac{\partial}{\partial r_1}\right)+ \frac{1}{r_{12}^2}\frac{\partial}{\partial r_{12}}
\left(r_{12}^2\frac{\partial}{\partial r_{12}}\right)\nonumber -\frac{l_1(l_1+1)}{r_1^2}+2(r_1 - r_2\cos(\theta_{12}))\frac{1}{r_{12}}\frac{\partial^2}{\partial r_1 \partial r_{12}} \nonumber - 2(\nabla_1^Y \cdot {\bf r}_2)\frac{1}{r_{12}}\frac{\partial}{\partial r_{12}}\nonumber
\end{equation}

where <math>\nabla_1^Y</math> acts only on the spherical harmonic part of the wave function and the diagram

[[File:Pic3.jpg|400px]]

defines the complete set of 6 independent variables is <math>r_1</math>, <math>r_2</math>, <math>r_{12}</math>, <math>\theta_1</math>, <math>\varphi_1</math>, <math>\chi</math>.

If <math>r_{12}</math> were not an independent variable, then one could take the volume element to be

\begin{equation}
d\tau = r_1^2dr_1\sin(\theta_1)d\theta_1d\varphi_1r_2^2dr_2\sin(\theta_2)d\theta_2d\varphi_2.\nonumber
\end{equation}

However, <math>\theta_2</math> and <math>\varphi_2</math> are no longer independent variables. To eliminate them, take the point <math>{\bf r}_1</math> as the
origin of a new polar co-ordinate system, and write

\begin{equation}
d\tau=-r_1^2dr_1\sin(\theta_1)d\theta_1d\varphi_1r_{12}^2dr_{12}\sin(\psi)d\psi d\chi\nonumber
\end{equation}

and use

\begin{equation}
r_2^2=r_1^2+r_{12}^2 +2r_1r_{12}\cos(\psi)
\end{equation}

Then for fixed <math>r_1</math> and <math>r_{12}</math>,

\begin{equation}
2r_2dr_2 = -2r_1r_{12}\sin(\psi)d\psi
\end{equation}

Thus

\begin{equation}
d\tau= r_1dr_1r_2dr_2r_{12}dr_{12}\sin(\theta_1)d\theta_1d\varphi_1d\chi
\end{equation}

The basic type of integral to be calculated is

\begin{align}
I \left( l_1, m_1, l_2, m_2; R \right) & = \int \sin \left( \theta_1 \right) d\theta_1 d\varphi_1 d\chi Y^{m_1}_{l_1} \left( \theta_1, \varphi_1 \right)^{*} Y^{m_2}_{l_2} \left( \theta_2, \varphi_2 \right) \\
& \times \int r_1dr_1r_2dr_2r_{12}dr_{12}R \left( r_1, r_2, r_{12} \right)
\end{align}

Consider first the angular integral. <math>Y^{m_2}_{l_2} \left( \theta_2, \varphi_2 \right)</math> can be expressed in terms of the independent variables <math>\theta_1, \varphi_1, \chi</math> by use of the rotation matrix relation
\begin{equation}
Y^{m_2}_{l_2}( \theta_2, \varphi_2) = \sum_m \mathcal{D}^{(l_2)}_{m_2,m}(\varphi_1,\theta_1,\chi)^*Y^m_{l_2}( \theta_{12}, \varphi)
\end{equation}
where <math>\theta, \varphi</math> are the polar angles of <math>{\bf r}_2</math> relative to <math>{\bf r}_1</math>. The angular integral is then

\begin{align}
I_{ang} & = \int^{2 \pi}_{0} d\chi \int^{2\pi}_{0} d\varphi_1 \int^{\pi}_{0} \sin \left( \theta_1 \right) d\theta_{1} Y^{m_1}_{l_1} \left( \theta_1, \varphi_1 \right)^{*} \\
& \times \sum_m \mathcal{D}^{(l_2)}_{m_2,m} \left( \varphi_1,\theta_1,\chi \right)^{*} Y^{m}_{l_2} \left( \theta_{12},\varphi \right) \\
\end{align}

Use

\begin{equation}
Y^{m_1}_{l_1} \left( \theta_1, \varphi_1 \right)^{*} = \sqrt{\frac{2l_1+1}{4\pi}}\mathcal{D}^{(l_1)}_{m_1,0} \left( \varphi_1, \theta_1, \chi \right)
\end{equation}

together with the orthogonality property of the rotation matrices (Brink and Satchler, p 147)

\begin{align}
& \int \mathcal{D}^{(j)*}_{m,m^\prime}\mathcal{D}^{(J)}_{M,M^\prime}\sin(\theta_1)d\theta_1d\varphi_1d\chi \\
& = \frac{8\pi^2}{2j+1}\delta_{jJ}\delta_{mM}\delta_{m^\prime M^\prime}\nonumber
\end{align}

to obtain

\begin{align}
I_{ang} & = \sqrt{ \frac{2l_1+1}{4\pi}} \frac{8\pi^2}{2l_1+1} \delta_{l_1,l_2} \delta_{m_1,m_2} Y^{0}_{l_2} \left( \theta_{12}, \varphi \right) \\
& = 2 \pi \delta_{l_1,l_2} \delta_{m_1,m_2} P_{l_2} \left( \cos \theta_{12} \right)
\end{align}

since

\begin{equation}
Y^0_{l_2} \left( \theta_{12}, \varphi \right) = \sqrt{\frac{2l_1+1}{4\pi}} P_{l_2} \left( \cos \left( \theta_{12} \right) \right)
\end{equation}

Note that <math>P_{l_2} \left( \cos \theta \right)</math> is just a short hand expression for a radial function because

\begin{equation}
\cos \theta_{12} = \frac{r_1^2 + r_2^2 - r_{12}^2}{2r_1r_2}
\end{equation}

The original integral is thus

\begin{equation}
I \left( l_1, m_1, l_2, m_2; R \right) = 2 \pi \delta_{l_1,l_2} \delta_{m_1,m_2} \int^{\infty}_{0} r_{1} dr_{1} \int^{\infty}_{0} r_{2} dr_{2} \int^{r_{1} + r_{2}}_{|r_{1} - r_{2}|} r_{12} dr_{12} R \left( r_{1}, r_{2}, r_{12} \right) P_{l_2} \left( \cos \theta_{12} \right)
\end{equation}

where again

\begin{equation}
\cos \theta_{12} = \frac{r_{1}^{2} +r^{2}_{2} - r_{12}^{2}}{2 r_{1} r_{2}}
\end{equation}

is a purely radial function.

The above would become quite complicated for large <math>l_2</math> because <math>P_{l_2}(\cos \theta_{12})</math> contains terms up to <math>\left( \cos \theta_{12} \right)^{l_2}</math>. However, recursion relations exist which allow any integral containing <math>P_l \left( \cos \theta_{12} \right)</math> in terms of those containing just <math>P_0 \left( \cos \theta_{12} \right) = 1</math> and <math>P_1 \left( \cos \theta_{12} \right) = \cos \theta_{12}</math>.

==Radial Integrals and Recursion Relations==

The basic radial integral is [see G.W.F. Drake, Phys. Rev. A 18, 820 (1978)]

\begin{align}
I_0 \left( a, b, c \right) & =
\int^{\infty}_{0} r_{1} dr_{1} \int^{\infty}_{r_{1}} r_{2} dr_{2} \int^{r_{1}+r_{2}}_{r_{2} - r_{1}} r_{12} dr_{12} r_{1}^{a} r_{2}^{b} r_{12}^{c} e^{-\alpha r_{1} - \beta r_{2}} + \int^{\infty}_{0} r_{2} dr_{2} \int^{\infty}_{r_{2}} r_{1} dr_{1} \int^{r_{1} + r_{2}}_{r_{1} - r_{2}} r_{12} dr_{12} r_{1}^{a} r_{2}^{b} r_{12}^{c} e^{- \alpha r_{1} - \beta r_{2}} \\
& = \frac{2}{c+2} \sum^{[(c+1)/2]}_{i=0}
\left(\begin{array}{c}
c+2 \\
2i+1
\end{array} \right)
\left\{ \frac{q!}{\beta^{q+1} \left( \alpha + \beta \right)^{p+1}} \sum_{j=0}^{q} \frac{\left( p + j \right)!}{j!} \left( \frac{\beta}{\alpha + \beta} \right)^j + \frac{q^{\prime}!}{\alpha^{q^{\prime}+1} \left( \alpha + \beta \right)^{p^{\prime}+1}} \sum^{q^{\prime}}_{j=0} \frac{\left( p^{\prime} + j \right)!}{j!} \left( \frac{\alpha}{\alpha + \beta} \right)^j \right\}
\end{align}

where

\begin{array}{ll}
p = a + 2i + 2 & p^{\prime} = b + 2i + 2 \\
q = b + c - 2i + 2 & q^{\prime} = a + c - 2i + 2
\end{array}

The above applies for <math> a, b \geq -2, c\geq -1</math>.
<math>[x]</math> means "greatest integer in" <math>x</math>.

Then

\begin{align}
I_1 \left( a, b, c \right) & = \int d\tau_{r} r_{1}^{a} r_{2}^{b} r_{12}^{c} e^{ -\alpha r_{1} - \beta r_{2}} P_{1}\left( \cos \theta \right) \\
& = \frac{1}{2} \left( I_{0} \left( a + 1, b - 1, c \right) + I_{0} \left( a - 1, b + 1, c \right) - I_{0} \left( a - 1, b - 1, c + 2 \right) \right)
\end{align}

===The Radial Recursion Relation===

Recall that the full integral is

\begin{equation}
I(l_1m_1,l_2m_2;R)=2\pi\delta_{l_1,l_2}\delta_{m_1,m_2}I_{l_2}(R)
\end{equation}

where

\begin{equation}
I_{l_2}(R) = \int d\tau_r R(r_1,r_2,r_{12})P_{l_2}(\cos \theta_{12})
\end{equation}

To obtain the recursion relation, use

\begin{equation}
P_l(x) = \frac{[P^{\prime}_{l+1}(x)-P^{\prime}_{l-1}(x)]}{2l+1}
\end{equation}

with

\begin{equation}
P_{l+1}^\prime (x) = \frac{d}{dx}P_{l+1}(x)
\end{equation}

Here <math>x=\cos\theta_{12}</math> and

\begin{align}
\frac{d}{d \cos \theta_{12}} & = \frac{dr_{12}}{d \cos \theta_{12}} \frac{d}{dr_{12}} \\
& = -\frac{r_{1} r_{2}}{r_{12}} \frac{d}{dr_{12}}
\end{align}

Then,

\begin{equation}
I_{l}\left( R \right) = - \int d\tau_r R \frac{r_{1} r_{2}}{r_{12}} \frac{d}{dr_{12}} \frac{ \left[ P_{l+1} \left( \cos \theta_{12} \right) - P_{l-1} \left( \cos \theta_{12} \right) \right]}{2l+1}
\end{equation}

The <math>r_{12}</math> part of the integral is

\begin{align}
& \int^{r_{1} + r_{2}}_{|r_{1} - r_{2}|} r_{12} dr_{12} R \frac{r_{1} r_{2}}{r_{12}} \frac{d}{dr_{12}} \left[P_{l+1}-P_{l-1} \right] \\
& = R r_{1} r_{2} \left[ P_{l+1} - P_{l-1} \right]^{r_{1} + r_{2}}_{|r_{1} - r_{2}|} - \int^{r_{1} + r_{2}}_{|r_{1} - r_{2}|} r_{12} dr_{12} \left( \frac{d}{dr_{12}} R \right) \frac{r_{1} r_{2}}{r_{12}} \frac{ \left[ P_{l+1} - P_{l-1} \right]}{2l+1}
\end{align}

The integrated term vanishes because

\begin{equation}
\cos \theta_{12} = \frac{r_{1}^{2} + r_{2}^{2} - r_{12}^{2}}{2 r_{1} r_{2}}
\begin{array}{lll}
& = -1 & {\rm when} & r_{12}^{2} = \left( r_{1} + r_{2} \right)^{2} \\
& = 1 & {\rm when} & r_{12}^{2} = \left( r_{1} - r_{2} \right)^{2}
\end{array}
\end{equation}

and <math>P_l(1)=1</math>, <math>P_{l}(-1)=(-1)^{l}</math>.

Thus,

\begin{equation}
I_{l+1}\left(\frac{r_1r_2}{r_{12}}R^\prime\right) = \left( 2l+1 \right) I_{l}(R) - I_{l-1} \left( \frac{r_{1} r_{2}}{r_{12}} R^{\prime} \right)
\end{equation}

If

\begin{equation}
R = r_{1}^{a-1} r_{2}^{b-1} r_{12}^{c+2} e^{-\alpha r_{1} - \beta r_{2}}
\end{equation}

then [G.W.F. Drake, Phys. Rev. A 18, 820 (1978)]

\begin{equation}
I_{l+1}(r_1^ar_2^br_{12}^c) =
\frac{2l+1}{c+2} I_{l} \left( r_{1}^{a-1} r_{2}^{b-2} r_{12}^{c+2} \right)+I_{l-1} \left( r_{1}^{a} r_{2}^{b} r_{12}^{c} \right)
\end{equation}

For the case <math>c=-2</math>, take

\begin{equation}
R = r_{1}^{a-1} r_{2}^{b-1} \ln r_{12} e^{ -\alpha r_{1} -\beta r_{2}}
\end{equation}

Then

\begin{equation}
I_{l+1} \left( r_{1}^{a} r_{2}^{b} r_{12}^{c} \right) = I_{l} \left( r_{1}^{a-1} r_{2}^{b-1} \ln r_{12} \right) \left( 2l + 1 \right) - I_{l-1} \left( r_{1}^{a} r_{2}^{b} r_{12}^{-2} \right)
\end{equation}

This allows all <math>I_{l}</math> integrals to be calculated from tables of <math>I_0</math> and
<math>I_1</math> integrals.

===The General Integral===

The above results for the angular and radial integrals can now be combined into a general formula for integrals of the type

\begin{equation}
I = \int \int d{\bf r}_{1} d{\bf r}_{2} R_{1} \mathcal{Y}^{M^{\prime}}_{l^{\prime}_{1} l^{\prime}_{2} L^{\prime}} \left( \hat{r}_{1}, \hat{r}_{2} \right) T_{k_{1} k_{2} K}^{Q} \left( {\bf r}_{1},{\bf r}_{2} \right) R_{2} \mathcal{Y}_{l_{1} l_{2} L}^{M} \left( \hat{r}_{1}, \hat{r}_{2} \right)
\end{equation}

where

\begin{equation}
\mathcal{Y}^{M}_{l_{1} l_{2} L} \left( \hat{r}_{1}, \hat{r}_{2} \right) = \sum_{m_1,m_2} \langle l_{1} l_{2} m_{1} m_{2} | LM \rangle Y^{m_{1}}_{l_{1}} \left( \hat{r}_{1} \right) Y^{m_{2}}_{l_{2}} \left( \hat{r}_{2} \right)
\end{equation}

and

\begin{equation}
T^Q_{k_{1} k_{2} K} \left( {\bf r}_{1}, {\bf r}_{2} \right) = \sum_{q_{1}, q_{2}} \langle k_{1} k_{2} q_{1} q_{2} | KQ \rangle Y_{k_{1}}^{q_{1}} \left( \hat{r}_{1} \right) Y^{q_{2}}_{k_{2}} \left( \hat{r}_{2} \right)
\end{equation}

The basic idea is to make repeated use of the formula

\begin{equation}
Y^{m_{1}}_{l_{1}} \left( \hat{r}_{1} \right) Y^{m_{2}}_{l_{2}} \left( \hat{r}_{1} \right) = \sum_{l m} \left( \frac{\left( 2l_{1}+1 \right) \left(2l_{2}+1 \right) \left(2l+1 \right)}{4 \pi} \right)^{1/2}
\times \left( \begin{array}{ccc}
l_1 & l_2 & l \\
m_1 & m_2 & m
\end{array} \right) \left(\begin{array}{ccc}
l_1 & l_2 & l \\
0 & 0 & 0
\end{array} \right)
Y^{m*}_{l} \left( \hat{r}_{1} \right)
\end{equation}

where

\begin{equation}
Y^{m*}(\hat{r}) = (-1)^mY_l^{-m}(\hat{r})
\end{equation}

and

\begin{equation}
\left(\begin{array}{ccc}
l_1 & l_2 & l \\
m_1 & m_2 & m
\end{array} \right) = \frac{\left( -1 \right)^{l_{1} - l_{2} - m}}{\left( 2l + 1 \right)^{1/2}} \left( l_{1} l_{2} m_{1} m_{2} | l, -m \right)
\end{equation}

is a 3-<math>j</math> symbol. In particular, write

\begin{array}{cc}
Y^{m_1^\prime}_{l_1^\prime}(\hat{r}_1)^{*} \underbrace{Y_{k_{1}}^{q_{1}} \left( \hat{r}_{1} \right) Y^{m_{1}}_{l_{1}} \left( \hat{r}_{1} \right)} & = & \sum_{\Lambda M}(\ldots)Y^{M}_\Lambda(\hat{r}_1)\\
\ \ \ \ \ \ \ \sum_{\lambda_1\mu_1}Y_{\lambda_1}^{\mu_1}(\hat{r}_1) & & \\
& & \\
Y^{m_{2}^{\prime}}_{l_{2}^{\prime}} \left( \hat{r}_{2} \right) \underbrace{Y^{q_{2}}_{k_{2}} \left( \hat{r}_{2} \right) Y_{l_{2}}^{m_{2}} \left( \hat{r}_{2} \right)} & = & \sum_{\Lambda^{\prime} M^{\prime}} \left( \ldots \right)Y_{\Lambda^{\prime}}^{M^{\prime}*} \left( \hat{r}_{2} \right) \\
\ \ \ \ \ \ \ \ \ \sum_{\lambda_{2} \mu_{2}}Y_{\lambda_{2}}^{\mu_{2}} \left( \hat{r}_{2} \right) & &
\end{array}

The angular integral then gives a factor of <math>2 \pi \delta_{\Lambda, \Lambda^{\prime}} \delta_{M, M^{\prime}} P_{\Lambda} \left( \cos \theta_{12} \right)</math>. The total integral therefore reduces to the form

\begin{equation}
I = \sum_{\Lambda} C_{\Lambda} I_{\Lambda} \left( R_{1} R_{2} \right)
\end{equation}

where <math>C_{\Lambda} = \sum_{\lambda_{1}, \lambda_{2}} C_{ \lambda_{1}, \lambda_{2}, \Lambda}</math>. For further details and derivations, including graphical representations, see G.W.F. Drake, Phys. Rev. A <math>{\bf 18}</math>, 820 (1978).

==Graphical Representation==

[[File:GraphRep1.jpg|500px]]

==Matrix Elements of H==
Recall that

\begin{equation}
H = -\frac{1}{2} \nabla^{2}_{1} - \frac{1}{2} \nabla^{2}_{2} - \frac{1}{r_1} - \frac{1}{r_2} + \frac{Z^{-1}}{r_{12}}
\end{equation}

Consider matrix elements of

\begin{equation}
\nabla^2_1 = \frac{1}{r_1^2} \frac{\partial}{\partial r_1}\left(r_1^2\frac{\partial}{\partial r_1} \right) + \frac{1}{r^2}\frac{\partial}{\partial r}\left(r^2\frac{\partial}{\partial r}\right) - \frac{(\vec{l}_1^Y)^2}{r_1^2} + \frac{2 \left( r_1 - r_2 \cos \theta \right)}{r} \frac{\partial^2}{\partial r_1 \partial r} - 2 \left( \nabla^Y_1 \cdot {\bf r}_2 \right) \frac{1}{r} \frac{\partial}{\partial r}
\end{equation}

where <math>r \equiv r_{12}</math> and <math>\cos\theta \equiv \cos\theta_{12}</math>. Also in what follows, define <math>\hat{\nabla}^Y_1 \equiv r_1\nabla^Y_1</math> where <math>\nabla^Y</math> operates only on the spherical harmonic part of the wave function, and similarly for <math>\vec{l}_1^Y</math>.

A general matrix element is

\begin{equation}
\langle r_1^{a^\prime} r_2^{b^\prime} r_{12}^{c^\prime} e^{-\alpha^\prime r_1 - \beta^\prime r_2} \mathcal{Y}^{M^\prime}_{l^\prime_1 l^\prime_2 L^\prime} \left( \hat{r}_1, \hat{r}_2 \right) \left| \nabla^2_1 \right| r_1^a r_2^b r_{12}^c e^{-\alpha r_1 - \beta r_2} \mathcal{Y}^M_{l_1 l_2 L} \left( \hat{r}_1, \hat{r}_2 \right) \rangle
\end{equation}

Since <math>\nabla^2_1</math> is rotationally invariant, this vanishes unless <math>L=L^\prime</math>, <math>M=M^\prime</math>. Also <math>\nabla_1^2</math> is Hermitian, so that the result must be the same whether it operates to the right or left, even though the results look very different. In fact, requiring the results to be the same yields some interesting and useful integral identities as follows:

Put

\begin{equation}
\mathcal{F} = F \mathcal{Y}^M_{l_1 l_2 L} \left( \hat{r}_1, \hat{r_2} \right)
\end{equation}

and

\begin{equation}
F = r_1^a r_2^b r_{12}^c e^{-\alpha r_1 - \beta r_2}
\end{equation}

Then

\begin{equation}
\nabla^2_1 \mathcal{F} = \left\{ \frac{1}{r_1^2} \left[ a \left( a + 1 \right) - l_1 \left( l_1 + 1 \right) \right] + \frac{c \left( c + 1 \right)}{r^2} + \alpha^2\frac{2\alpha \left( a + 1 \right)}{r_1} + \frac{2 \left(r_1 - r_2 \cos \theta \right) }{r_1r^2}c \left[ a - \alpha r_1 \right] - \frac{2c}{r^2} \left( \hat{\nabla}^Y_1 \cdot \hat{r}_2 \right) \frac{r_2}{r_1} \right\} \mathcal{F}
\end{equation}

and

\begin{align}
\langle \mathcal{F}^\prime \left| \nabla^2_1 \right| \mathcal{F} \rangle & = \sum_\Lambda \int d\tau_r F^\prime C_\Lambda \left( 1 \right) P_\Lambda \left( \cos \theta \right) \times \left\{ \frac{1}{r_1^2} \left[ a \left( a + 1 \right) - l_1 \left( l_1 + 1 \right) \right] -\frac{2 \alpha \left( a + 1 \right)}{r_1} + \frac{c \left( c + 1 \right)}{r^2} +\alpha^2 + \frac{2 \left( r_2 - r_2 \cos \theta \right)}{r_1r^2}c \left( a - \alpha r_1 \right) \right\} F \\
& + \sum_\Lambda \int d\tau_r F^\prime C_\Lambda \left( \hat{\nabla}^Y_1 \cdot \hat{r}_2 \right) P_\Lambda \left( \cos \theta \right) \left( \frac{-2c r_2}{r_1 r^2} \right) F
\end{align}

where

\begin{equation}
\int d\tau_r = \int^\infty_0 r_1 dr_1 \int^\infty_0 r_2 dr_2 \int^{r_1+r_2}_{\left| r_1 - r_2 \right|} r dr
\end{equation}

For brevity, let the sum over <math>\Lambda</math> and the radial integrations be understood, and let <math>\left( \nabla_1^2 \right)</math> stand for the terms which appear in the integrand. Then operating to the right gives

\begin{equation}
(\nabla_1^2)_R = \frac{1}{r_1^2} \left[ a \left( a + 1 \right) - l_1 \left( l_1 + 1 \right) \right] + \frac{c \left( c + 1 \right)}{r^2} + \alpha^2 - \frac{2 \alpha \left( a + 1 \right)}{r_1}
+ \frac{2 \left( r_1 - r_2 \cos \theta \right)}{r_1 r^2}c \left( a - \alpha r_1 \right) - \frac{2c r_2}{r_1 r^2} \left( \hat{\nabla}^Y_1 \cdot \hat{r}_2 \right)
\end{equation}

Operating to the left gives

\begin{equation}
(\nabla_1^2)_L = \frac{1}{r_1^2} \left[a^\prime \left( a^\prime + 1 \right) - l_1^\prime \left( l_1^\prime + 1 \right) \right] + \frac{c^\prime \left( c^\prime + 1 \right)}{r^2} + \alpha^{\prime 2} -\frac{2 \alpha^\prime \left( a^\prime + 1 \right)}{r_1}
+ \frac{2 \left( r_1 - r_2 \cos \theta \right)}{r_1 r^2} c^\prime \left( a^\prime - \alpha^\prime r_1 \right) - \frac{2 c^\prime r_2}{r_1 r^2} \left( \hat{\nabla}^{Y^\prime}_1 \cdot \hat{r}_2 \right)
\end{equation}

Now put

\begin{array}{cc}
a_+ = a+a^\prime, & \hat{\nabla}_1^+ = \hat{\nabla}_1^Y + \hat{\nabla}_1^{Y\prime} \\
a_{-} = a-a^\prime, & \hat{\nabla}_1^{-} = \hat{\nabla}_1^Y - \hat{\nabla}_1^{Y\prime}
\end{array}

etc., and substitute <math>a^\prime = a_+ - a</math>, <math>c^\prime = c_+ - c</math>, and <math>\alpha^\prime = \alpha_+ - \alpha</math> in <math>(\nabla^2_1)_L</math>. If <math>a_+</math>, <math>c_+</math> and <math>\alpha_+</math> are held fixed, then the equation

\begin{equation}
(\nabla^2_1)_R = (\nabla^2_1)_L
\end{equation}

must be true for arbitrary <math>a</math>, <math>c</math>, and <math>\alpha</math>. Their coefficients must thus vanish.

This yields the integral relations

\begin{equation}
\label{I}
\frac{\left( r_1 - r_2 \cos \theta \right)}{r_1 r^2} = \frac{1}{c_+} \left( \frac{- \left( a_+ + 1 \right)}{r_1^2} + \frac{\alpha_+}{r_1} \right) \tag{I}
\end{equation}

from the coefficient of <math>a</math> and

\begin{equation}
\label{II}
\frac{ \left( r_1 - r_2 \cos \theta \right) \left( a_+ - \alpha_+ r_1 \right)}{r_1 r^2} = \frac{r_2}{r_1 r^2}\left( \hat{r}_2 \cdot \hat{\nabla}_1^+ \right) - \frac{ \left( c_+ + 1 \right)}{r^2} \tag{II}
\end{equation}

from the coefficient of <math>c</math>. The coefficient of <math>\alpha</math> gives an equation equivalent to (I).

Furthermore, if can be show that (see problem)

\begin{align}
& \sum_\Lambda \int d\tau_r \frac{r^c}{r^2} C_\Lambda \left( \hat{r}_2 \cdot \hat{\nabla}_1^Y \right) P_\Lambda \left( \cos \theta \right) \\
= & \sum_\Lambda \int d\tau_r \frac{r^c}{c r_1 r_2}C_\Lambda \left( 1 \right)P_\Lambda \left( \cos \theta \right) \times \left( \frac{l_1^\prime \left( l_1^\prime + 1 \right) - l_1 \left( l_1 + 1 \right) -\Lambda \left( \Lambda + 1 \right)}{2} \right)
\end{align}

and similarly for <math>\left( \hat{r}_2 \cdot \hat{\nabla}^{Y\prime}_1 \right)</math> with <math>l_1</math> and <math>l_1^\prime</math> interchangeable, then it follows that

\begin{equation}
\frac{r^{c_+}}{r^2} \left( \hat{r}_2 \cdot \hat{\nabla}_1^+ \right) = - \frac{r^{c_+}}{c_+ r_1 r_2} \Lambda \left( \Lambda + 1 \right)
\end{equation}

and

\begin{equation}
\frac{r^{c_+}}{r^2} \left( \hat{r}_2 \cdot \hat{\nabla}_1^- \right) = \frac{r^{c_+}}{c_+ r_1 r_2} \left[ l_1^\prime \left( l_1^\prime + 1 \right) - l_1 \left( l_1 + 1 \right) \right]
\end{equation}

where equality applies after integration and summation over <math>\Lambda</math>.

Thus (II) becomes

\begin{equation}\label{IIa}
\frac{ \left( r_1 - r_2 \cos \theta \right) \left( a_+ - \alpha_+ r_1 \right)}{r_1 r^2} = - \frac{\Lambda \left( \Lambda + 1 \right)}{c_+ r^2_1} - \frac{ \left( c_+ + 1 \right)}{r^2} \tag{IIa}
\end{equation}

===Problem===

Prove the integral relation

\begin{align}
& \sum_\Lambda \int d\tau_r f \left( r_1, r_2 \right) \frac{1}{r} \left( \frac{d}{dr} q(r) \right) C_\Lambda \left( \hat{r}_2 \cdot \hat{\nabla}^Y_1 \right) P_\Lambda \left( \cos \theta \right) \\
= & \sum_\Lambda \int d\tau_r \frac{ f \left( r_1, r_2 \right)}{r_1, r_2} q(r) C_\Lambda \left( 1 \right) P_\Lambda \left( \cos \theta \right) \left(\frac{l_1^\prime \left( l_1^\prime + 1 \right) - l_1 \left( l_1 + 1 \right) - \Lambda \left( \Lambda + 1 \right)}{2}\right)
\end{align}

where the coefficients <math>C_\Lambda \left( 1 \right)</math> are the angular coefficients from the overlap integral

\begin{align}
\int & d\Omega\, \mathcal{Y}^{M*}_{l_1^\prime l_2^\prime L}(\hat{r}_1,\hat{r}_2)\, \mathcal{Y}^{M}_{l_1 l_2 L}(\hat{r}_1,\hat{r}_2)\\
= & \sum_\Lambda C_\Lambda(1)P_\Lambda(\cos\theta_{12}).
\end{align}

Hint: Use the fact that <math>l_1^2</math> is Hermitian so that
<math>
\int d\tau (l_1^2\mathcal{Y}^\prime)^*q(r)\mathcal{Y} = \int d\tau
\mathcal{Y}^{\prime *}l_1^2(q(r)\mathcal{Y})
</math>
with <math>\ \vec{l}_1= \frac{1}{i}\vec{r}_1\times\nabla_1</math>.

It is also useful to use
<math>\begin{eqnarray}\ \
\cos\theta\, P_L(\cos\theta) &=& \frac{1}{2L+1}[L\,P_{L-1}(\cos\theta) + (L+1)P_{L+1}(\cos\theta)]
\\
(\cos^2\theta-1)P_L(\cos\theta) &=&
\frac{L(L-1)}{(2L-1)(2L+1)}P_{L-2}(\cos\theta)\nonumber
-\frac{2(L^2+L-1)}{(2L-1)(2L+3)}P_L(\cos\theta) + \frac{(L+1)(L+2)}{(2L+1)(2L+3)}P_{L+2}(\cos\theta)\nonumber\\
&=& \frac{L(L-1)}{(2L-1)(2L+1)}[P_{L-2}(\cos\theta) - P_L(\cos\theta)]
+ \frac{(L+1)(L+2)}{(2L+1)(2L+3)}[P_{L+2}(\cos\theta) - P_L(\cos\theta)]
\end{eqnarray}</math>

together with a double application of the integral recursion relation
<math>
I_{L+1}\left(\frac{1}{r}\frac{d}{dr}q(r)\right) =
(2L+1)I_L\left(\frac{1}{r_1r_2}q(r)\right)+I_{L-1}\left(\frac{1}{r}\frac{d}{dr}q(r)\right)
</math>

Of course <math>l_1^2\mathcal{Y}^M_{l_1l_2L}(\hat{r}_1,\hat{r}_2)=l_1(l_1+1)\mathcal{Y}^M_{l_1l_2L}(\hat{r}_1,\hat{r}_2)</math>. Begin by expanding
<math>
l_1^2(g\mathcal{Y}) = \mathcal{Y}l_1^2g + g\,l_1^2\mathcal{Y} + 2(\vec{l}_1g)\cdot(\vec{l}_1\mathcal{Y})
</math>, and show that

<math>
\begin{eqnarray}\ \
(\vec{l}_1g)\cdot(\vec{l}_1\mathcal{Y}) & = & \frac{r_1^2}{r}\frac{dg}{dr}\vec{r}_2\cdot \nabla_1\mathcal{Y}\,\,\,\,\mbox{ and}\\
l_1^2g(r) &=& 2r_1r_2\cos\theta\frac{1}{r}\frac{dg}{dr} + r_1^2r_2^2(\cos^2\theta - 1)
\frac{1}{r}\frac{d}{dr}\left(\frac{1}{r}\frac{dg}{dr}\right).
\end{eqnarray}
</math>

The proof amounts to showing that the term <math>l_1^2g(r)</math> can be replaced by
<math>\Lambda(\Lambda+1)g(r)</math> after multiplying by <math>P_\Lambda(\cos\theta)</math>
and integrating by parts with respect to the radial integrations over <math>r_1</math>, <math>r_2</math> and
<math>r \equiv |\vec{r}_1 - \vec{r}_2|</math>. Remember that <math>\cos\theta</math> is just a short-hand notation for the radial function <math>(r_1^2 + r_2^2 - r^2)/(2r_1r_2)</math>.

==General Hermitian Property==

Each combination of terms of the form
<math>
<f>_R = a^2f_1+af_2+abf_3+b^2f_4+bf_5+af_6\nabla_1^Y+bf_7\nabla^Y_2+f_8(Y)
</math>
can be rewritten
<math>
<f>_L= (a_+-a)^2f_1+(a_+-a)f_2 + (a_+-a)(b_+-b)f_3+(b_+-b)^2f_4
+(b_+-b)f_5+(a_+-a)f_6\nabla^{Y\prime}_1+(b_+-b)f_7\nabla^{Y\prime}_2+f_8)Y^\prime
</math>
where as usual <math>\nabla_1^Y</math> acts only on the spherical harmonic part of the wave function denoted for short by <math>Y</math>.
Since these must be equal for arbitrary <math>a</math> and <math>b</math>,
<math>
a_+^2f_1+a_+f_2+_+b_+f_3+b_+f_4+b_+f_5+a_+f_6\nabla^{Y\prime}_1 +
b_+f_7\nabla^{Y\prime}_2+f_8(Y^\prime)-f_8(Y)=0
</math>

Adding the corresponding expression with <math>Y</math> and <math>Y^\prime</math> interchanged
yields
<math>\label{III}
a_+^2f_1+a_+f_2+a_+b_+f_3+b_+^2f_4+b_+f_5+\frac{1}{2}a_+f_6\nabla^+_1+\frac{1}{2}b_+f_7\nabla^+_2=0\tag{III}
</math>

Subtracting gives
<math>\label{IV}
f_8(Y)-f_8(Y^\prime)=-\frac{1}{2}[a_+f_6\nabla^-_1+b_+f_7\nabla_2^-]\tag{IV}
</math>
<math>\label{V}
a[-2a_+f_1-2f_2-b_+f_3-f_6\nabla^+_1]=0\tag{V}
</math>
<math>\label{VI}
b[-2b_+f_4-2f_5-a_+f_3-f_7\nabla^+_2]=0\tag{VI}
</math>

Adding the two forms gives

\begin{eqnarray}
<f>_R+<f>_L=\frac{1}{2}(a_+^2+a_-^2)f_1+a_+f_2+\frac{1}{2}(a_+b_++a_-b_-)f_3\nonumber\\
\ +\frac{1}{2}(b_+^2+b_-^2)f_4+b_+f_5+\frac{1}{2}f_6(a_+\nabla_1^++a_-\nabla^-_1)\nonumber\\
\ +\frac{1}{2}(b_+\nabla_2^++b_-\nabla^-_2)+f_8(Y)+f_8(Y^\prime)\nonumber
\end{eqnarray}

Subtracting <math>\frac{x}{2}\times({\rm I})</math> gives

\begin{eqnarray}\ \
<f>_R+<f>_L = \frac{1}{2}[(1-x)a_+^2 + a_-^2]f_1 +
(1-\frac{x}{2})a_+f_2+\frac{1}{2}[(1-x)a_+b_++a_-b_-]f_3\nonumber\\
+\frac{1}{2}[(1-x)b_+^2+b_-^2]f_4+(1-\frac{x}{2})f_5b_++\frac{1}{2}f_6[(1-\frac{x}{2})a_+\nabla_1^++a_-\nabla^-_1]\nonumber\\
+\frac{1}{2}[(1-\frac{x}{2})b_+\nabla^+_2+b_-\nabla^-_2]+f_8(Y)+f_8(Y^\prime)\nonumber
\end{eqnarray}

If <math>x=1</math>,
\begin{eqnarray}
<f>_R+<f>_L &=&
\frac{1}{2}[a_-^2f_1+a_+f_2+a_-b_-f_3+b_-^2f_4+b_+f_5\nonumber\\
&+&\left(\frac{1}{2}a_+\nabla^+_1+a_-\nabla^-_1\right)f_6+\left(\frac{1}{2}b_+\nabla^+_2+b_-\nabla^-_2\right)f_7]+f_8(Y)+f_8(Y^\prime)\nonumber
\end{eqnarray}

The General Hermitian Property for arbitrary <math>x</math> gives
\begin{eqnarray}\ \
\nabla_1^2=
\frac{1}{4}\{\frac{1}{r_1^2}[(1-x)a_+^2+a_-^2+2(1-\frac{x}{2})a_+-2[l_1(l_1+1)+l_1^\prime(l_1^\prime+1)]\nonumber\\
-\frac{2}{r_1}[(1-x)\alpha_+a_++\alpha_-a_-+2(1-\frac{x}{2})\alpha_+]+(1-x)\alpha_+^2
+ \alpha^2_-]\nonumber\\
+\frac{2(r_1-r_2\cos\theta)}{r_1r^2}[(1-x)(a_+-\alpha_+r_1)c_++(a_--\alpha_-r_2)c_-]\nonumber\\
-\frac{2r_2}{r_1r^2}[(1-\frac{x}{2})c_+\hat{r}_2\cdot\hat{\nabla}_1^++c_-\hat{r}_2\hat{\nabla}_1^-]+\frac{(1-x)c_+^2+c_-^2+2(1-\frac{x}{2})c_+}{r^2}\}\nonumber
\end{eqnarray}.

Use
<math>\begin{eqnarray}\ \
\frac{-2r_2}{r_1r^2}[(1-\frac{x}{2})c_+\hat{r}_2\hat{\nabla}_1^++c_-\hat{r}_2\hat{\nabla}_1^-]=\frac{2}{r_1^2}\left[(1-\frac{x}{2})\Lambda(\Lambda+1)-\frac{c_-}{c_+}[l_1^\prime(l_1^\prime+1)+l_1(l_1+1)]\right]\end{eqnarray}</math>,

<math>\begin{eqnarray}
\frac{2(r_1-r_2\cos\theta)}{r_1r^2}(a_--\alpha_-r_1)c_- = 2\frac{c_-}{c_+}\left[\frac{-1}{r_1^2}[a_-(a_++1)]+\frac{1}{r_1}[a_-\alpha_++\alpha_-(a_++2)]-\alpha_-\alpha_+\right]\nonumber
\end{eqnarray}</math>,

and
<math>\begin{eqnarray}\ \
\frac{2(r_1-r_2\cos\theta)}{r_1r^2}(a_+-\alpha_+r_1)c_+ = 2\left[-\frac{1}{r_1^2}[a_+(a_++1)]+\frac{1}{r_1}[a_+\alpha_++\alpha_+(a_++2)]-\alpha_+^2\right]
\end{eqnarray}</math>.

Substituting into <math>\nabla_1^2</math> gives

\begin{eqnarray}
\nabla^2_1 = \frac{1}{4}\{\frac{1}{r_1^2}[-(1-x)a_+^2 + a_-^2 +xa_+ +
2(1-\frac{x}{2})\Lambda(\Lambda+1)\nonumber\\
-2l_1(l_1+1)(1-\frac{c_-}{c_+}) -
2l_1^\prime(l_1^\prime+1)(1+\frac{c_-}{c_+})-\frac{2c_-a_-}{c_+}(a_++1)]\nonumber\\
- \frac{2}{r_1}\left[-(1-x)\alpha_+(a_++2)+\alpha_-a_-+2(1-\frac{x}{2})\alpha_+
-\frac{c_-}{c_+}[a_-\alpha_++\alpha_-(a_++2)]\right]\nonumber\\
-(1-x)\alpha_+^2+\alpha_-^2-2\frac{c_-}{c_+}\alpha_-\alpha_+ +\frac{1}{r^2}\left[(1-x)c_+^2+c_-^2+2(1-\frac{x}{2})c_+\right]\}\nonumber
\end{eqnarray}

This has the form
<math>
\nabla_1^2 =
\frac{1}{4}\left[\frac{A_1}{r_1^2}+\frac{B_1}{r_1}+\frac{C_1}{r^2}+D_1\right]
</math>
with

<math>\begin{array}{ccc}
A_1=-(1-x)a_+^2+a_-^2
+xa_++2(1-\frac{x}{2})\Lambda(\Lambda+1)-2l_1(l_1+1)(1-\frac{c_-}{c_+})\
-2l_1^\prime(l_1^\prime+1)(1+\frac{c_-}{c_+})-\frac{2c_-a_-}{c_+}(a_++1) \ \
\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \
\end{array}
</math>

<math>
B_1 =
2\left[(1-x)\alpha_+(a_++2)-\alpha_-a_--2(1-\frac{x}{2})\alpha_++\frac{c_-}{c_+}[a_-\alpha_++\alpha_-(a_++2)]\right]
</math>

<math>
C_1 = (1-x)c_+^2+ c_-^2 +2(1-\frac{x}{2})c_+
</math>

<math>
D_1 = -1(1-x)\alpha_+^2 +\alpha_-^2 - 2\frac{c_-}{c_+}\alpha_-\alpha_+
</math>

The complete Hamiltonian is then (in Z-scaled a.u.)

<math>\begin{eqnarray}
H =-\frac{1}{2}\nabla^2_1 -\frac{1}{2}\nabla^2_2 -\frac{1}{r_1} -\frac{1}{r_2} +\frac{Z^{-1}}{r}\nonumber\\
= -\frac{1}{2}\nabla^2_1 -\frac{1}{2}\nabla^2_2 -\frac{1}{r_1} -\frac{Z-1}{Zr_2}+Z^{-1}\left(\frac{1}{r}-\frac{1}{r_2}\right)\nonumber\\
\ =-\frac{1}{8}\left[\frac{A_1}{r_1^2}+\frac{B_1+8}{r_1}+\frac{C_1}{r^2}+D_1+D_2 +\frac{A_2}{r_2^2} + \frac{B_2 + 8(Z-1)/Z}{r_2}+\frac{C_2}{r^2}\right]\nonumber
+ Z^{-1}\left(\frac{1}{r}-\frac{1}{r_2}\right)\nonumber
\end{eqnarray}</math>

The screened hydrogenic energy is
<math>\begin{eqnarray}
E_{SH}=-\frac{1}{2}\left[\frac{1}{n_1^2}+ \left(\frac{Z-1}{Z}\right)^2\frac{1}{n_2^2}\right]
\end{eqnarray}</math>
so that

<math>\begin{eqnarray}
H-E_{SH} &=& -\frac{1}{8}[
\frac{A_1}{r_1^2}+\frac{B_1+8}{r_1}+\frac{C_1}{r^2} + D_1 + D_2
-\frac{4}{n_1^2} -\left(\frac{Z-1}{Z}\right)^2\frac{4}{n_2^2}\nonumber
+\frac{A_2}{r_2}+\frac{B_2+8(Z-1)/Z}{r_2}+\frac{C_2}{r^2}] +
Z^{-1}\left(\frac{1}{r} -\frac{1}{r_2}\right)\nonumber
\end{eqnarray}</math>

==Optimization of Nonlinear Parameters==
- The traditional method of performing Hylleraas calculations is to write the
basis set in the form

<math>
\Psi = \sum_{i,j,k}c_{i,j,k}\varphi_{i,j,k}(\alpha,\beta)\pm exchange
</math>

with

<math>
\varphi_{i,j,k}(\alpha,\beta) = r_1^ir_2^jr_{12}^ke^{-\alpha r_1 - \beta
r_2}\mathcal{Y}^M_{l_1l_2L}.
</math>

The usual procedure is to set <math>\alpha = Z</math> so that it represents the inner
<math>1s</math> electron, and then to vary <math>\beta</math> so as to minimize the energy.

[[File:OptNon.jpg|500px]]

Since <math>\beta</math> appears in <math>\Psi</math> as a non-linear parameter, the entire
calculation must be repeated for each value of <math>\beta</math>. However, the minimum
becomes progressively smaller as the basis set is enlarged.

[[File:OptNon02.jpg|500px]]

*Difficulties

1. If the basis set is constructed so that <math> i +j +j \leq N</math>, the the
number of terms is
<math>
(N+1)(N+2)(N+3)/6
</math>
N=14 already gives 680 terms and an accuracy of about 1 part in <math>10^{10}</math> for
low-lying states. A substantial improvement in accuracy would require much
larger basis sets, together with multiple precision arithmetic to avoid loss
of significant figures when high powers are included.

2. The accuracy rapidly deteriorates as one goes to more highly excited
states - about 1 significant figure is lost each time the principle quantum
number is increased.

*Cure

We have found that writing basis sets in the form
\begin{eqnarray}
\Psi &=& \sum_{i,j,k}\left(c^{(1)}_{i,j,k}\varphi_{ijk}(\alpha_1\beta_1) +
c_{ijk}^{(2)}\varphi_{ijk}(\alpha_2\beta_2)\right) \pm exchange \nonumber\\
&=& \psi({\bf r}_1,{\bf r}_2) \pm \psi({\bf r}_2,{\bf r}_1)\nonumber
\end{eqnarray}
so that each combination of powers is included twice with different nonlinear
parameters gives a dramatic improvement in accuracy for basis sets of about
the same total size.

However, the optimization of the nonlinear parameters is now much more
difficult, and an automated procedure is needed.

If
<math>
E = \frac{<\Psi|H|\Psi>}{<\Psi|\Psi>}</math> and we assume <math><\Psi|\Psi> = 1</math>, then

<math>
\frac{\partial E}{\partial\alpha_t} = -2 <\Psi|H-E|r_1\psi({\bf r}_1,{\bf
r}_2;\alpha_t)\pm r_21\psi({\bf r}_2,{\bf
r}_1;\alpha_t)>
</math>
where
<math>
\psi({\bf r}_1,{\bf r}_2;\alpha_t) =
\sum_{i,j,k}c_{ijk}^{(t)}\varphi_{ijk}(\alpha_t\beta_t)
</math>

*Problem

1. Prove the above.

2. Prove that there is no contribution from the implicit dependence of
<math>c_{ijk}^{(t)}</math> on <math>\alpha_t</math> if the linear parameters have been optimized.

==The Screened Hydrogenic Term==

If the Hamiltonian is written in the form (in Z-scaled a.u.)
<math>\begin{eqnarray}\
H = H_0({\bf r}_1,Z) + H_0({\bf r}_2,Z-1) +
Z^{-1}\left(\frac{1}{r}-\frac{1}{r_2}\right)
\end{eqnarray}</math>
with
<math>\
H_0({\bf r}_1,Z) = -\frac{1}{2}\nabla_1^2 - \frac{1}{r_1}\nonumber
</math> and
<math>
H_0({\bf r}_2,Z-1) = -\frac{1}{2}\nabla_2^2 - \frac{Z-1}Z{r_2}\nonumber
</math>,
then the eigenvectors of <math>H_0({\bf r},Z) +H_0({\bf r}_2,Z-1)</math> are products of
hydrogenic orbitals
<math>
\Psi_0=\psi_0(1s,Z)\psi_0(nl,Z-1)
</math>
and the eigenvalue is
<math>\begin{eqnarray}
E_{SH} = \left[-\frac{1}{2} - \left(\frac{Z-1}{Z}\right)^2\frac{1}{2n^2}\right]Z^2 a.u.
\end{eqnarray}</math>
called the screened hydrogenic eigenvalue. From highly excited states,
<math>E_{SH}</math> and <math>\Psi_0</math>are already excellent approximations.

For example, for the 1s8d states, the energies are

<math>\begin{eqnarray}\\
E(1s8d^1D)&=& -2.007\,816\,512\,563\,81 {\rm\ a.u.}\nonumber\\
E(1s8d^3D)&=& - 2.007\,817\,934\,711\,71 {\rm\ a.u.}\nonumber\\
E_{SH} &=& -2.007\,812\,5 {\rm\ a.u.} \nonumber
\end{eqnarray}</math>

It is therefore advantageous to include the screened hydrogenic terms in the
basis set so that the complete trial function becomes
<math>
\Psi = c_0\Psi_0 +
\sum_{ijk}\left[c_{ijk}^{(1)}\varphi_{ijk}(\alpha_1,\beta_1)+c_{ijk}^{(2)}\varphi(\alpha_2,\beta_2)\right]\pm
{\rm exchange}
</math>

Also, the variational principal can be re-expressed in the form
<math>\begin{eqnarray}
E= E_{SH}+\frac{<\Psi|H-E_{SH}|\Psi>}{<\Psi|\Psi>}
\end{eqnarray}</math>
so that the <math>E_{SH}</math> term can be cancelled analytically from the matrix
elements.

For example
<math>
<\Psi_0|H-E_{SH}|\Psi_0> = 0
</math>

and
<math>
<\varphi_{ijk}|H-E_{SH}|\Psi_0> =
Z^{-1}<\varphi_{ijk}|\frac{1}{r}-\frac{1}{r_2}|\Psi_0>
</math>

Recall that
<math>\begin{eqnarray}\ \
I_0(a,b,c) = \frac{2}{c+2}\sum^{[c+1]/2}_{i=0}\left(\begin{array}{c}
c+2\\
2i+1
\end{array}\right) \left[f(p,q;\beta) + f(p^\prime,q^\prime;\alpha)\right]
\end{eqnarray}</math>
where

<math>\begin{eqnarray}
f(p,q;x) =
\frac{q!}{x^{q+1}(\alpha+\beta)^{p+1}}\sum^p_{j=0}\frac{(p+j)!}{j!}\left(\frac{x}{\alpha+\beta}\right)^j
\end{eqnarray}</math>

<math>
\begin{array}{cc}
p=a+2i+2 & p^\prime = b+2i+2\\
q= b+c-2i+2 & q^\prime =a+c-2i+2
\end{array}
</math>

If <math>\beta\ll\alpha</math>, the <math>f(p,q;\beta)\gg f(p^\prime,q^\prime;\alpha)</math> and the
<math>i=0</math> term is the dominant contribution to <math>f(p,q;\beta)</math>. However, since
this term depends only on the sum of powers <math>b+c</math> for <math>r_2</math>, it cancels exactly
from the matrix element of
<math>
\frac{1}{r_{12}} - \frac{1}{r_2}
</math>
and can therefore
be omitted in the calculations of integrands, thereby saving many significant
figures. This is especially valuable when <math>b+c</math> is large and <math>a</math> is small.

==Small Corrections==
*Mass Polarization
If the nuclear mass is not taken to be infinite, then the Hamiltonian is
<math>\begin{eqnarray}\ \
H = \frac{P_N^2}{2M} + \sum_{i=1}\frac{p_i^2}{2m} + V
\end{eqnarray}</math>

[[File:SmallCor.jpg|500px]]

where <math>M =M_A-nm</math> is the nuclear mass and <math>m</math> is the electron mass.

Change variables to the C of M
<math>\begin{eqnarray}
{\bf R} = \frac{1}{M+nm}\left[M{\bf r}_N + m({\bf r}_1 + {\bf r}_2 +
\ldots)\right]
\end{eqnarray}</math>

and relative variables

<math>
{\bf s}_i = {\bf r}_i - {\bf r}_N.
</math>

Then

<math>\begin{eqnarray}\ \ \ \ \
H = \frac{1}{2(M+nm)}{\bf p}_R^2+\frac{1}{2m}\sum_{i=1}^n{\bf p}_{s_i}^2
+\frac{1}{2M}\sum_{i,k}{\bf p}_{s_i}\cdot {\bf p}_{s_k} + V({\bf s}_1, {\bf
s}_2, \ldots)\nonumber
= \frac{1}{M+nm}{\bf p}_R^2+\frac{1}{2\mu}\sum_{i=1}^n{\bf
p}_{s_i}^2+\frac{1}{2M}\sum_{i\neq k}{\bf p}_{s_i}\cdot {\bf p}_{s_k}+V({\bf s}_1, {\bf
s}_2, \ldots)\nonumber
\end{eqnarray}</math>

where
<math>\begin{eqnarray}
\mu = \left(\frac{1}{M} + \frac{1}{m}\right)^{-1}
\end{eqnarray}</math>

If <math>n=2</math>, the Schrödinger equation is
<math>\begin{eqnarray}\ \ \
\left[ -\frac{\hbar^2}{2\mu}(\nabla^2_{s_1}+ \nabla^2_{s_2}) -
\frac{\hbar^2}{M}\nabla_{s_1}\nabla_{s_2}-\frac{Ze^2}{s_1}-\frac{Ze^2}{s_2}+\frac{e^2}{s_{12}}\right]\psi
= E\psi
\end{eqnarray}</math>.

Define the reduced mass Bohr radius
<math>\begin{eqnarray}
a_\mu = \frac{\hbar^2}{\mu e^2} = \frac{m}{\mu}a_0
\end{eqnarray}</math>
and

<math>\begin{eqnarray}
\rho_i = \frac{Zs_i}{a_\mu}\nonumber\\
\varepsilon = \frac{E}{Z^2(e^2/a_\mu)}\nonumber
\end{eqnarray}</math>

The Schrödinger equation then becomes
<math>\begin{eqnarray}\ \ \
\left[
-\frac{1}{2}(\nabla^2_{\rho_1}+\nabla^2_{\rho_2})-\frac{\mu}{M}\vec{\nabla}_{\rho_1}\cdot\vec{\nabla}_{\rho_2} - \frac{1}{\rho_1}-\frac{1}{\rho_2}+\frac{Z^{-1}}{\rho_{12}}\right]\psi = \varepsilon \psi
\end{eqnarray}</math>

The effects of finite mass are thus

1. A reduced mass correction to all energy levels of <math>(e^2/a_\mu)/(e^2/a_0)
= \mu/m</math>, i.e.
<math>\begin{eqnarray}
E_M=\frac{\mu}{m}E_\infty
\end{eqnarray}</math>

2. A specific mass shift given in first order perturbation theory by
<math>\begin{eqnarray}
\Delta\varepsilon =
-\frac{\mu}{M}<\psi|\nabla_{\rho_1}\cdot\nabla_{\rho_2}|\psi>
\end{eqnarray}</math>
or

<math>\begin{eqnarray}
\Delta\varepsilon =
-\frac{\mu}{M}<\psi|\nabla_{\rho_1}\cdot\nabla_{\rho_2}|\psi>\frac{e^2}{a_\mu}\nonumber\\
= -\frac{\mu^2}{mM}<\psi|\nabla_{\rho_1}\cdot\nabla_{\rho_2}|\psi>\frac{e^2}{a_0}\nonumber
\end{eqnarray}</math>

Since <math>\nabla_{\rho_1}\cdot\nabla_{\rho_2}</math> has the same angular properties as
<math>{\bf\rho}_1 \cdot {\bf\rho}_2</math>, the operator is like the product of two
dipole operators. For product type wave functions of the form
$
\psi = \psi(1s)\psi(nl) \pm \mbox{exchange},
$
the matrix element vanishes for all but P-states.

=Radiative Transitions=

==Transition Integrals and the Line Strength==

For an electric dipole transition <math>\gamma'\,L'\,S'\,J'\,M' \rightarrow \gamma\,L\,S\,J\,M</math>, the line strength is defined by
\begin{equation}
S = \sum_{M',\mu,M}|\langle\gamma'\,L'\,S'\,J'\,M'|r_\mu|\gamma\,L\,S\,J\,M\rangle|^2
\end{equation}
Using the Wigner-Eckart theorem, this is related to the reduced matrix element by
\begin{eqnarray}
S(J',J) = |\langle\gamma'\,L'\,S'\,J'||r||\gamma\,L\,S\,J\,M\rangle|^2
\sum_{M',\mu,M}\left|\left(\matrix{J' & 1 & J \cr
-M'& \mu &M \cr}\right)\right|^2
\end{eqnarray}
Using the sum rule for the 3-<math>j</math> symbols
\begin{equation}
\sum_{M',\mu,M}\left|\left(\matrix{J' & 1 & J \cr
-M'& \mu &M \cr}\right)\right|^2 = 1
\end{equation}
we see that the line strength is identical to the reduced matrix element
\begin{equation}
S(J',J) = |\langle\gamma'\,L'\,S'\,J'||r||\gamma\,L\,S\,J\,M\rangle|^2
\end{equation}
For a nonrelativistic calculation in <math>LS</math> coupling, <math>S = S'</math> and we can next strip out the dependence on <math>J</math> and <math>J'</math> by use of Eq.\ (7.1.7) of Edmonds { Angular Momentum in Quantum Mechanics} with <math>k = 1</math> to obtain
\begin{equation}
S(J',J) = (-1)^{L'+S+J+1}[(2J+1)(2J'+1)]^{1/2}\left\{
\matrix{L' & J'& S\cr
J & L & 1\cr}\right\}\langle\gamma' L'|| r || \gamma L\rangle
\end{equation}
where <math>\langle\gamma' L'|| r || \gamma L\rangle</math> is the reduced matrix element if spin is neglected. Note that if spin is neglected, then the line-strength is simply
\begin{equation}
S(L',L) = |\langle\gamma' L'|| r || \gamma L\rangle|^2
\end{equation}
To calculate the reduced matrix element, use a particular example along with the Wigner-Eckart theorem
such as
\begin{equation}
\langle\gamma'\, L'|| r || \gamma\, L\rangle = (-1)^{L'}\langle\gamma'\, L'\, 0| z | \gamma\, L\, 0\rangle{\Large /}\left(\matrix{L' & 1 & L \cr
0 & 0 & 0 \cr}\right)
\end{equation}

===Correlated Two-electron Integrals===
For the case of two-electron integrals, <math>z = z_1 + z_2</math> and <math>r = r_1 + r_2</math>. As discussed in G.W.F. Drake, Phys. Rev. A 18, 820 (1978), the general integral in Hylleraas coordinates for the integral over the <math>z_1</math> term can be written in the form
\begin{equation}
\label{int1}
\langle\gamma'\, L'\, 0| z_1 | \gamma\, L\, 0\rangle = \sum_\Lambda C_\Lambda I_\Lambda(\gamma',\gamma)
\end{equation}
where <math>I_\Lambda(\gamma',\gamma)</math> stands for the purely radial integral
\begin{equation}
\label{radial}
I_\Lambda(\gamma',\gamma) = \int_0^\infty r_1\,dr_1\int_0^\infty r_2\,dr_2\int_{|r_1-r_2|}^{r_1+r_2}r_{12}\, dr_{12}
R_{\gamma'}^*\,r_1\, R_\gamma P_\Lambda(\cos\theta_{12})
\end{equation}
and <math>\cos\theta_{12} = (r_1^2 + r_2^2 - r_{12}^2)/(2r_1r_2)</math> is a short-hand notation for a radial function of <math>r_1</math>, <math>r_2</math> and <math>r_{12}</math>. Also, <math>P_\Lambda(\cos\theta_{12})</math> is a Legendre polynomial,
and, <math>R_\gamma = R_\gamma(r_1,r_2,r_{12})</math> is the radial part of the Hylleraas wave function for state <math>\gamma</math> such
that the total wave function is <math>\psi({\bf r}_1,{\bf r}_2) = R_\gamma(r_1,r_2,r_{12}){\cal Y}_{l_1,l_2,L}^M(\hat{\bf r}_1,\hat{\bf r}_2)\pm {\rm exchange}</math>, normalized to unity, and <math>{\cal Y}_{l_1,l_2,L}^M(\hat{\bf r}_1,\hat{\bf r}_2)</math> is a vector coupled product of spherical harmonics. The coefficents <math>C_\Lambda</math> in Eq.\ (\ref{int1}) can be calculated from Eq.\ (32) of the above 1978 paper. They need to be calculated separately for he direct-direct and the direct-exchange contributions. For the reduced matrix element of <math>z_1</math>, the leading 3-<math>j</math> symbol can be dropped, leaving (for the case <math>k = 1</math>)
\begin{eqnarray}
C_\Lambda^{\rm reduced} &=& \frac{(-1)^{l_2'+L'+L+1}}{2} (l_1,l_1',l_2,l_2',L,L')^{1/2}
\sum_{\lambda={\rm max}(l_1-1,0)}^{l_1+1} (\lambda,\Lambda,1)
\left(\matrix{l_1 & 1 & \lambda \cr
0 & 0 & 0 \cr}\right)
\left(\matrix{l_1'& \lambda & \Lambda \cr
0 & 0 & 0 \cr}\right)\nonumber\\
&\times&\left(\matrix{l_2' & 1_2 & \Lambda \cr
0 & 0 & 0 \cr}\right)
\left\{\matrix{L' & \lambda & l_2 \cr
\Lambda& l_2' & l_1' \cr}\right\}
\left\{\matrix{L' & \lambda & l_2 \cr
l_1 & L & 1 \cr}\right\}
\end{eqnarray}
where the notation <math>(a,b,c,\cdots)</math> means <math>(2a+1)(2b+1)(2c+1)\cdots</math>.
A similar expression can be obtained for the angular coefficients for the <math>z_2</math> term by interchanging the subscripts 1 and 2 throughout and multiplying by <math>(-1)^{L+L'+1}</math>.

===Example===
As an example, for the <math>1s^2\;^1S - 1s2p\;^1P</math> transition of helium, the only nonvanishing <math>C_\Lambda</math> angular coefficients are <math>C_1 = 1/2</math> for the direct-direct term (i.e. <math>P_1(\cos\theta_{12}) = \cos\theta_{12}</math> is included in the radial integral (\ref{radial})) and <math>C_0 = 1/2</math> for the direct-exchange term (i.e. <math>P_0(\cos\theta_{12}) = 1</math> is included in the radial integral (\ref{radial})). The results for infinite nuclear mass are <math>S(1\;^1S-2\;^1P) = (0.781\,127\,158\,667)^2</math> and <math>f(1\;^1S-2\;^1P) = -0.276\,164\,703\,736</math> where <math>f(1\;^1S-2\;^1P) = \frac{2}{3}[E(1\;^1S) - E(2\;^1P)]S(1\;^1S-2\;^1P)</math> is the absorption oscillator strength.

Theory Notes

2020-12-14T00:24:21Z

Drake: /* Introduction */

<center>
<H1>Notes on solving the Schrödinger equation in Hylleraas coordinates for heliumlike atoms
Gordon W.F. Drake
Department of Physics, University of Windsor
Windsor, Ontario, Canada N9B 3P4
(Transcribed from hand-written notes and edited by Lauren Moffatt. Last revised November 22, 2017 to correct a missing subscript in the expression for $B_1$ in the Hamiltonian in Sect. 11 General Hermitian Property.)
</H1>
</center>

==Introduction==
These notes describe some of the technical details involved in solving the Schrödinger equation for a heliumlike atom of nuclear charge Z in correlated Hylleraas coordinates. The standard for computational purposes is to express all quantities in a form that is valid for electronic states of arbitrary angular momentum, and to express matrix elements of the Hamiltonian in a manifestly Hermitian symmetrized form. Further information and derivations of the mathematical techniques for angular integrals and radial recursion relations can be found in the paper
G.W.F. Drake, Phys. Rev. A <math>{\bf 18}</math>, 820 (1978).

The starting point is the two-electron Schrödinger equation for infinite nuclear mass
\begin{equation} \left[-\frac{\hbar^2}{2m}(\nabla^2_1 +\nabla^2_2) - \frac{Ze^2}{r_1} -
\frac{Ze^2}{r_2}+\frac{e^2}{r_{12}} \right]\psi = E\psi\nonumber
\end{equation}
where <math>m</math> is the electron mass, and <math>r_{12} = |{\bf r}_1 - {\bf r}_2|<\math> (see diagram below).

[[File:Pic1.jpg|250px]]

Begin by rescaling distances and energies so that the Schrödinger equation can be expressed in a dimensionless form. The dimensionless Z-scaled distance is defined by <math>\begin{eqnarray} \rho = \frac{Zr}{a_0} \end{eqnarray} </math> where <math> \begin{eqnarray} a_0 = \frac{\hbar^2}{me^2} \end{eqnarray} </math> is the atomic unit (a.u.) of distance equal to the Bohr radius <math>0.529\, 177\, 210\, 92(17) \times 10^{-10}</math> m. Then

\begin{equation}[-\frac{\hbar^2}{2m}Z^2(\frac{me^2}{\hbar^2})^2(\nabla^2_{\rho_1}+\nabla^2_{\rho_2})
- Z^2\frac{e^2}{a_0}\rho^{-1}_1 - Z^2\frac{e^2}{a_0}\rho^{-1}_2 +
\frac{e^2}{a_0}Z\rho^{-1}_{12}]\psi= E\psi\nonumber\end{equation}

But <math>\begin{eqnarray} \frac{\hbar^2}{m}(\frac{me^2}{\hbar^2})^2 = \frac{e^2}{a_0} \end{eqnarray} </math> is the hartree atomic unit of of energy <math>(E_h = 27.211\,385\,05(60)</math> eV, or equivalently <math>E_h/(hc) = 2194.746\,313\,708(11)</math> m<math>^{-1})</math>. Therefore, after multiplying through by <math>a_0/(Ze)^2</math>, the problem to be solved in Z-scaled dimensionless units becomes

\begin{equation} \left[-\frac{1}{2}(\nabla^2_{\rho_1}+\nabla^2_{\rho_2}) -
\frac{1}{\rho_1} - \frac{1}{\rho_2} + \frac{Z^{-1}}{\rho_{12}}\right]\psi =
\varepsilon\psi\nonumber \end{equation}

where <math>\begin{eqnarray}\varepsilon = \frac{Ea_0}{(Ze)^2} \end{eqnarray}</math> is the energy in Z-scaled atomic units. For convenience, rewrite this in the conventional form
\begin{equation}
H\psi = \varepsilon\psi\nonumber \end{equation}
where (using <math>r</math> in place of <math>\rho</math> for the Z-scaled distance)
\begin{equation}
H = -\frac{1}{2}(\nabla^2_1+\nabla^2_2) - \frac{1}{r_1}-\frac{1}{r_2} + \frac{Z^{-1}}{r_{12}}
\nonumber\end{equation}
is the atomic Hamiltonian for infinite nuclear mass.

==The Hartree Fock Method==

For purposes of comparison, and to define the correlation energy, assume that <math>\psi({\bf r}_1,{\bf r}_2)</math> can be written in the separable product form
\begin{equation}
\psi({\bf r}_1,{\bf r}_2) = \frac{1}{\sqrt{2}} \left[ u_1(r_1) u_2(r_2) \pm u_2(r_1) u_1(r_2) \right] \nonumber
\end{equation}
for the <math>1s^2\ ^1\!S</math> ground state. Because of the <math>\frac{1}{r_{12}}</math> term in the Schrödinger equation, the exact solution connot be expressed in this form as a separable product. However, the Hartree-Fock approximation corresponds to finding the best possible solution to the Schrödinger equation
\begin{equation}
H\psi({\bf r}_1,{\bf r}_2) = E\psi({\bf r}_1,{\bf r}_2)\nonumber
\end{equation}
that can nevertheless be expressed in this separable product form, where as before
\begin{equation}
H = -\frac{1}{2}(\nabla^2_1+\nabla^2_2) - \frac{1}{r_1}- \frac{1}{r_2} + \frac{Z^{-1}}{r_{12}}\nonumber
\end{equation}
is the full two-electron Hamiltonian. To find the best solution, substitute into <math>\left\langle\psi \lvert H-E \rvert \psi\right\rangle</math> and require this expression to be stationary
with respect to arbitrary infinitesimal variations <math>\delta u_1</math> and <math>\delta
u_2</math> in <math>u_1</math> and <math>u_2</math> respectively; i.e.

\begin{equation}
\frac{1}{2} \left\langle \delta u_1(r_1)u_2(r_2) \pm u_2(r_1)\delta
u_1(r_2) \lvert H-E \rvert u_1(r_1) u_2(r_2)\pm u_2(r_1) u_1(r_2)\right\rangle\nonumber
\end{equation}

\begin{equation}
=\int\delta u_1(r_1) d{\bf r}_1 \left\{ \int d{\bf r}_2 u_2(r_2) \left( H - E \right) \left[ u_1(r_1)u_2(r_2) \pm u_2(r_1) u_1(r_2) \right] \right\}\nonumber
\end{equation}

\begin{equation}
= 0 \nonumber
\end{equation}

for arbitrary <math> \delta u_1(r_1).\nonumber </math> Therefore <math>\{\int d{\bf r}_2 \ldots \} = 0</math>.

Similarly, the coefficient of <math>\delta u_2</math> would give

\begin{equation}
\int d{\bf r}_1 u_1(r_1) \left( H-E \right) \left[ u_1(r_1) u_2(r_2) \pm
u_2(r_1) u_1(r_2) \right] = 0\nonumber
\end{equation}

Define

\begin{equation}
I_{12} = I_{21} = \int d{\bf r}\, u_1(r)u_2(r), \nonumber
\end{equation}

\begin{equation}
H_{ij} = \int d{\bf r}\, u_i(r)(-\frac{1}{2}\nabla - \frac{1}{r})u_j(r), \nonumber
\end{equation}

\begin{equation}
G_{ij}(r) = \int d{\bf r}^\prime u_i(r^\prime)\frac{1}{|{\bf r} - {\bf r}\prime|}u_j(r^\prime)\nonumber
\end{equation}

Then the above equations become the pair of integro-differential equations

\begin{equation}
[ H_0 - E + H_{22}+G_{22}(r)]u_1(r) = \mp [ I_{12}(H_0-E) + H_{12}+G_{12}(r)]u_2(r)\nonumber
\end{equation}

\begin{equation}
[H_0-E+H_{11}+G_{11}(r)]u_2(r) = \mp [I_{12}(H_0-E) + H_{12}+G_{12}(r)]u_1(r)\nonumber
\end{equation}

These must be solved self-consistently for the "constants" <math>I_{12}</math> and
<math>H_{ij}</math> and the function <math>G_{ij}(r)</math>.

The Hartree Fock energy is <math>E \simeq -2.87\ldots </math>a.u. while the exact energy is <math>E = -2.903724\ldots </math>a.u.

The difference is called the "correlation energy" because it arises from the
way in which the motion of one electron is correlated to the other. The
Hartree Fock equations only describe how one electron moves in the average field
provided by the other (mean-field theory).

==Configuration Interaction==
Expand
\begin{equation}
\psi({\bf r}_1,{\bf r}_2)= C_0u^{(s)}_1(r_1)u^{(s)}_1(r_2) + C_1u^{(P)}_1({\bf r}_1)u^{(P)}_1({\bf r}_2){\cal Y}^0_{1,1,0}(\hat{\bf r}_1, \hat{\bf r}_2)+C_2u^{(d)}_1({\bf r}_1)u^{(d)}_2({\bf r}_2){\cal Y}^0_{2,2,0}(\hat{\bf r}_1, \hat{\bf r}_2)+... \pm {\rm\ exchange}
\end{equation}
where
\begin{equation}
{\cal Y}^M_{l_1,l_2,L}(\hat{\bf r}_1, \hat{\bf r}_2)=\sum_{m_1,m_2}Y^{m_1}_{l_1}({\bf r}_1)Y^{m_2}_{l_2}({\bf r}_2)\times <l_1l_2m_1m_2\mid LM>
\end{equation}

This works, but is slowly convergent, and very laborious. The best CI calculations are accurate to <math> ~10^{-7}</math> a.u.

==Hylleraas Coordinates==
[E.A. Hylleraas, Z. Phys. <math>{\bf 48}, 469(1928)</math> and <math>{\bf 54}, 347(1929)</math>]
suggested using the co-ordinates <math>r_1</math>, <math>r_2</math> and <math>r_{12}</math> or equivalently

\begin{eqnarray}
s &=& r_1 + r_2, \nonumber\\
t &=& r_1-r_2, \nonumber\\
u &=& r_{12}\nonumber
\end{eqnarray}

and writing the trial functions in the form

\begin{equation}
\Psi({\bf r}_1,{\bf r}_2) = \sum^{i+j+k\leq
N}_{i,j,k}c_{i,j,k}r_1^{i+l_1}r_2^{j+l_2}r_{12}^ke^{-\alpha r_1 - \beta r_2}
\mathcal{Y}^M_{l_1,l_2,L}(\hat{r}_1,\hat{r}_2)\pm \text{exchange}\nonumber
\end{equation}

Diagonalizing H in this non-orthogonal basis set is equivalent to solving
<math> \begin{eqnarray}
\frac{\partial E}{\partial c_{i,j,k}} = 0\nonumber
\end{eqnarray} </math>
for fixed <math>\alpha</math> and <math>\beta</math>.

The diagonalization must be repeated for different values of <math>\alpha</math> and
<math>\beta</math> in order to optimize the non-linear parameters.

==Completeness==
The completeness of the above basis set can be shown by first writing
\begin{equation}
r_{12}^2 = r_1^2 + r_2^2 - 2r_1r_2\cos\theta_{12}
\end{equation}
and, from the spherical harmonic addition theorem,
\begin{equation}
\cos(\theta_{12})=\frac{4\pi}{3}\sum^1_{m=-1}Y^{m*}_1(\theta_1,\varphi_1)Y^m_1(\theta_2,\varphi_2)
\end{equation}
Consider first the S-states. The <math>r_{12}^0</math> terms are like the ss terms in a CI calculation. The
<math>r_{12}^2</math> terms bring in p-p type contributions, and the higher powers bring
in d-d, f-f etc. type terms. In general

\begin{equation}
P_l(\cos\theta_{12}) = \frac{4\pi}{2l+1}\sum^l_{m=-l}{Y^{m}_l}^*(\theta_1,\varphi_1)Y^m_l(\theta_2,\varphi_2)\nonumber
\end{equation}

For P-states, one would have similarly

\begin{array}{lr}
r_{12}^0 & (sp)P\nonumber\\
r_{12}^2 & (pd)P\nonumber\\
r_{12}^4 & (df)P\nonumber\\
\vdots & \vdots\nonumber
\end{array}

For D-states

\begin{array}{lr}
r_{12}^0 & (sd)D & (pp^\prime)D\nonumber\\
r_{12}^2 & (pf)D & (dd^\prime)D\nonumber\\
r_{12}^4 & (dg)D & (ff^\prime)D\nonumber\\
\vdots & \vdots & \vdots\nonumber
\end{array}

In this case, since there are two "lowest-order" couplings to form a
D-state, both must be present in the basis set, i.e.

\begin{equation}
\Psi(r_2,r_2) = \sum c_{ijk}r_1^ir_2^{j+2}r_{12}^ke^{-\alpha r_1-\beta
r_2}\mathcal{Y}^M_{022}(\hat{r}_1,\hat{r}_2)
+\sum d_{ijk}r_1^{i+1}r_2^{j+1}r_{12^k}e^{-\alpha^\prime r_1 - \beta^\prime
r_2}\mathcal{Y}^M_{112}(\hat{r}_1,\hat{r}_2)\nonumber
\end{equation}

For F-states, one would need <math>(sf)F</math> and <math>(pd)F</math> terms.

For G-states, one would need <math>(sg)G</math>, <math>(pf)G</math> and <math>(dd^\prime)G</math> terms.

Completeness of the radial functions can be proven by considering the
Sturm-Liouville problem

\begin{equation}
\left(-\frac{1}{2}\nabla^2-\frac{\lambda}{r_s}-E\right)\psi({\bf r}) = 0\nonumber
\end{equation}

or

\begin{equation}
\left(-\frac{1}{2}\frac{1}{r^2}\frac{\partial}{\partial r}\left({r^2}\frac{\partial}{\partial r}\right) -
\frac{l\left(l+1\right)}{2r^2} - \frac{\lambda}{r} - E\right)u(r) = 0.\nonumber
\end{equation}

For fixed E and variable <math>\lambda</math> (nuclear charge).

The eigenvalues are <math>\lambda_n = (E/E_n)^{1/2}</math>, where <math>E_n =- \frac{1}{2n^2}</math>

[[File:Pic2.jpg|500px]]

and the eigenfunctions are

\begin{equation}
u_{nl}(r) =
\frac{1}{(2l+1)!}\left(\frac{(n+l)!}{(n-l-1)2!}\right)^{1/2}(2\alpha)^{3/2}e^{-\alpha
r}\nonumber
\times \ \left( 2 \alpha r \right)^l\,_{1}F_{1}\left(-n+l+1,2l+2; 2 \alpha r \right)
\end{equation}

with <math>\alpha = (-2E)^{1/2}</math> and <math>n\geq l+1</math>. The confluent hypergeometric function <math> _1F_1(a,b,;z)</math> then denotes a finite polynomial since <math>a=-n+l+1</math> is a negative integer or zero.

Unlike the hydrogen spectrum, which has both a discrete part for <math>E<0</math> and a
continuous part for <math>E>0</math>, this forms an entirely discrete set of finite
polynomials, called Sturmian functions. They are orthogonal with respect to
the potential, i.e.

\begin{equation}
\int^\infty_0 r^2dr \left( u_{n^\prime l}(r)\frac{1}{r}u_{nl}(r) \right) = \delta_{n,n^\prime}\nonumber
\end{equation}

Since they become complete in the limit <math>n\rightarrow\infty</math>, this assures the
completeness of the variational basis set.

[See also B Klahn and W.A. Bingel Theo. Chim. Acta (Berlin) 44, 9 and 27
(1977)].

==Solutions of the Eigenvalue Problem==

For convenience, write

\[
\Psi({\bf r}_1,{\bf r}_2) = \sum^N_{m=1}c_m\varphi_m\nonumber
\]

where <math>m</math> represents the <math>m</math>'th combination of <math>i,j,k</math> and

\begin{equation}
\varphi_{ijk}=r_1^ir_2^jr_{12}^ke^{-\alpha r_1 - \beta
r_2}\mathcal{Y}^M_{l_1,l_2,L}(\hat{r}_1,\hat{r}_2) \pm \mbox{exchange}\nonumber
\end{equation}

\begin{align}

&
\left( \begin{array}{lr}
\cos(\theta) & \sin(\theta)\\
-\sin(\theta) & \cos(\theta)
\end{array} \right)
\left( \begin{array}{lr}
H_{11} & H_{12} \\
H_{12} & H_{22}
\end{array}\right)
\left( \begin{array}{lr}
\cos(\theta) & -\sin(\theta)\\
\sin(\theta) & \cos(\theta)
\end{array} \right)
\\[5pt]
= & \left(\begin{array}{lr}
cH_{11}+sH_{12} & cH_{12} + sH_{22}\\
-sH_{11} + cH_{12} & -sH_{12} + cH_{22}
\end{array}\right)
\left( \begin{array}{cc}
c & -s \\
s & c
\end{array}\right)
\\[5pt]
= & \left(\begin{array}{lr}
c^2H_{11}+s^2H_{22} + 2csH_{12} & (c^2-s^2)H_{12}+cs(H_{22}-H_{11})\\
(c^2-s^2)H_{12}+cs(H_{22}-H_{11}) & s^2H_{11}+c^2H_{22}-2csH_{12}
\end{array}\right)
\end{align}

Therefore

\begin{equation}
\left[\cos^2(\theta)-\sin^2(\theta)\right]H_{12} =
\cos(\theta)\sin(\theta)(H_{11}-H_{22})\nonumber
\end{equation}

and

\begin{equation}
\tan(2\theta) = \frac{2H_{12}}{H_{11}-H_{22}}\nonumber
\end{equation}

i.e.

\begin{equation}
\cos(\theta)=\left(\frac{r+\omega}{2r}\right)^{1/2}\nonumber
\end{equation}

\begin{equation}
\sin(\theta)=-\mbox{sgn}(H_{12})\left(\frac{r-\omega}{2r}\right)^{1/2}\nonumber
\end{equation}

where

\begin{equation}
\omega = H_{22}-H_{11}\nonumber
\end{equation}

\begin{equation}
r=\left(\omega^2+4H_{12}^2\right)^{1/2}\nonumber
\end{equation}

\begin{equation}
E_1=\frac{1}{2}\left(H_{11}+H_{22}-r\right)\nonumber
\end{equation}

\begin{equation}
E_2=\frac{1}{2}\left(H_{11}+H_{22}+r\right)\nonumber
\end{equation}

===Brute Force Method===

- Gives all the eigenvalues and eigenvectors, but it is slow

- First orthonormalize the <math>N</math>-dimensional basis set; i.e. form linear combinations

\begin{equation}
\Phi_m = \sum_{n=1}^N\varphi_nR_{nm}\nonumber
\end{equation}

such that
\begin{equation}
\langle \Phi_m | \Phi_n \rangle = \delta_{m,n}
\end{equation}

This can be done by finding an orthogonal tranformation T such that

\begin{equation}
T^TOT = I = \left(
\begin{array}{ccccc}
I_1 & 0 & 0 & \ldots & 0 \\
0 & I_2 & 0 & \ldots & 0 \\
0 & 0 & I_3 & \ldots & 0 \\
\vdots & \vdots & \vdots & \ddots & \vdots \\
0 & 0 & 0 & \ldots & I_N
\end{array}\right);
\end{equation}

\begin{equation}
O_{mn} = \langle \varphi_m | \varphi_n \rangle \nonumber
\end{equation}

and then applying a scale change matrix

\begin{equation}
S = \left(\begin{array}{ccccc}
\frac{1}{I_1^{1/2}} & 0 & 0 & \ldots & 0\\
0 & \frac{1}{I_2^{1/2}} & 0 & \ldots & 0 \\
0 & 0 & \frac{1}{I_3^{1/2}} & \ldots & 0 \\
\vdots & \vdots & \vdots & \ddots & \vdots \\
0 & 0 & 0 & \ldots & I_N^{1/2}
\end{array}\right)= S^T\nonumber
\end{equation}

Then

\begin{equation}
S^TT^TOTS = 1
\end{equation}

i.e.

\begin{equation}
R^TOR = 1
\end{equation}

with <math>R=TS</math>.

If H is the matrix with elements <math>H_{mn} = \langle \varphi_m | \varphi_n \rangle </math>, then H
expressed in the <math>\Phi_m</math> basis set is

\begin{equation}
H^\prime = R^THR.\nonumber
\end{equation}

We next diagonalize <math>H^\prime</math> by finding an orthogonal transformation W such
that

\begin{equation}
W^T H^\prime W = \lambda = \left(
\begin{array}{cccc}
\lambda_1 & 0 & \ldots & 0 \\
0 & \lambda_2 & \ldots & 0 \\
\vdots & \vdots & \ddots & \vdots \\
0 & 0 & \ldots & \lambda_N
\end{array}\right)\nonumber
\end{equation}

The q'th eigenvector is

\begin{equation}
\Psi^{(q)} = \sum_{n=1}^N\Phi_n W_{n,q} = \sum_{n,n^\prime} \varphi_{n^\prime} R_{n^\prime ,n} W_{n,q} \nonumber
\end{equation}

i.e.

\begin{equation}
c_{n^\prime}^{(q)} = \sum_{n=1}^N R_{n^\prime n} W_{n,q}
\end{equation}

===The Power Method===

- Based on the observation that if H has one eigenvalue, <math>\lambda_M</math>, much bigger than all the rest, and <math>\chi = \left(
\begin{array}{c}a_1\\a_2\\\vdots\end{array}\right)</math>
is an arbitrary starting vector, then <math>\chi = \sum_{q=1}^N x_q\Psi^{(q)}</math>.

\begin{equation}
(H)^n\chi = \sum_{q=1}^N x_q \lambda^n_q\Psi^{(q)}\nonumber\rightarrow x_M\lambda_M^n\Psi^{(M)}\nonumber
\end{equation}

provided <math>x_M\neq 0</math>.

To pick out the eigenvector corresponding to any eigenvalue, write the original problem in the form

\begin{align}
H\Psi & = \lambda O \Psi \nonumber \\
\left( H - \lambda_q O \right) \Psi & = \left( \lambda - \lambda_q \right) O \Psi \nonumber
\end{align}

Therefore,

\begin{equation}
G \Psi = \frac{1}{\lambda - \lambda_q } \Psi \nonumber \\
\end{equation}

where <math>G=(H-\lambda_qO)^{-1}O</math> with eigenvalues <math>\frac{1}{\lambda_n - \lambda_q}</math>.

By picking <math>\lambda_q</math> close to any one of the <math>\lambda_n</math>, say
<math>\lambda_{n^\prime}</math>, then <math>\frac{1}{\lambda_n-\lambda_q}</math> is much larger for
<math>n=n^\prime</math> than for any other value. The sequence is then

\begin{array}{c}
\chi_1=G\chi\nonumber\\
\chi_2=G\chi_1\nonumber\\
\chi_3=G\chi_2\nonumber\\
\vdots\nonumber
\end{array}

until the ratios of components in <math>\chi_n</math> stop changing.

- To avoid matrix inversion and multiplication, note that the sequence is equivalent to
\begin{equation}
F \chi_n = \left( \lambda - \lambda_q \right) O \chi_{n-1} \nonumber \\
\end{equation}
where <math>F = H-\lambda_qO</math>. The factor of <math>\left( \lambda - \lambda_q \right)</math> can be
dropped because this only affects the normalization of <math>\chi_n</math>. To find <math>\chi_n</math>, solve
\begin{equation}
F \chi_n = O \chi_{n-1} \nonumber
\end{equation}
(N equations in N unknowns). Then
\begin{equation}
\lambda = \frac{ \langle \chi_n | H | \chi_n \rangle}{\langle \chi_n | \chi_n \rangle} \nonumber
\end{equation}

==Matrix Elements of H==

\begin{equation}
H=-\frac{1}{2}\nabla^2_1 -\frac{1}{2}\nabla^2_2 - \frac{1}{r_1} - \frac{1}{r_2} +\frac{Z^{-1}}{r_{12}}\nonumber
\end{equation}

Taking <math>r_1, r_2</math> and <math>r_{12}</math> as independent variables,

\begin{equation}
\nabla_1^2 = \frac{1}{r_1^2}\frac{\partial}{\partial r_1}\left(r_1^2\frac{\partial}{\partial r_1}\right)+ \frac{1}{r_{12}^2}\frac{\partial}{\partial r_{12}}
\left(r_{12}^2\frac{\partial}{\partial r_{12}}\right)\nonumber -\frac{l_1(l_1+1)}{r_1^2}+2(r_1 - r_2\cos(\theta_{12}))\frac{1}{r_{12}}\frac{\partial^2}{\partial r_1 \partial r_{12}} \nonumber - 2(\nabla_1^Y \cdot {\bf r}_2)\frac{1}{r_{12}}\frac{\partial}{\partial r_{12}}\nonumber
\end{equation}

where <math>\nabla_1^Y</math> acts only on the spherical harmonic part of the wave function and the diagram

[[File:Pic3.jpg|400px]]

defines the complete set of 6 independent variables is <math>r_1</math>, <math>r_2</math>, <math>r_{12}</math>, <math>\theta_1</math>, <math>\varphi_1</math>, <math>\chi</math>.

If <math>r_{12}</math> were not an independent variable, then one could take the volume element to be

\begin{equation}
d\tau = r_1^2dr_1\sin(\theta_1)d\theta_1d\varphi_1r_2^2dr_2\sin(\theta_2)d\theta_2d\varphi_2.\nonumber
\end{equation}

However, <math>\theta_2</math> and <math>\varphi_2</math> are no longer independent variables. To eliminate them, take the point <math>{\bf r}_1</math> as the
origin of a new polar co-ordinate system, and write

\begin{equation}
d\tau=-r_1^2dr_1\sin(\theta_1)d\theta_1d\varphi_1r_{12}^2dr_{12}\sin(\psi)d\psi d\chi\nonumber
\end{equation}

and use

\begin{equation}
r_2^2=r_1^2+r_{12}^2 +2r_1r_{12}\cos(\psi)
\end{equation}

Then for fixed <math>r_1</math> and <math>r_{12}</math>,

\begin{equation}
2r_2dr_2 = -2r_1r_{12}\sin(\psi)d\psi
\end{equation}

Thus

\begin{equation}
d\tau= r_1dr_1r_2dr_2r_{12}dr_{12}\sin(\theta_1)d\theta_1d\varphi_1d\chi
\end{equation}

The basic type of integral to be calculated is

\begin{align}
I \left( l_1, m_1, l_2, m_2; R \right) & = \int \sin \left( \theta_1 \right) d\theta_1 d\varphi_1 d\chi Y^{m_1}_{l_1} \left( \theta_1, \varphi_1 \right)^{*} Y^{m_2}_{l_2} \left( \theta_2, \varphi_2 \right) \\
& \times \int r_1dr_1r_2dr_2r_{12}dr_{12}R \left( r_1, r_2, r_{12} \right)
\end{align}

Consider first the angular integral. <math>Y^{m_2}_{l_2} \left( \theta_2, \varphi_2 \right)</math> can be expressed in terms of the independent variables <math>\theta_1, \varphi_1, \chi</math> by use of the rotation matrix relation
\begin{equation}
Y^{m_2}_{l_2}( \theta_2, \varphi_2) = \sum_m \mathcal{D}^{(l_2)}_{m_2,m}(\varphi_1,\theta_1,\chi)^*Y^m_{l_2}( \theta_{12}, \varphi)
\end{equation}
where <math>\theta, \varphi</math> are the polar angles of <math>{\bf r}_2</math> relative to <math>{\bf r}_1</math>. The angular integral is then

\begin{align}
I_{ang} & = \int^{2 \pi}_{0} d\chi \int^{2\pi}_{0} d\varphi_1 \int^{\pi}_{0} \sin \left( \theta_1 \right) d\theta_{1} Y^{m_1}_{l_1} \left( \theta_1, \varphi_1 \right)^{*} \\
& \times \sum_m \mathcal{D}^{(l_2)}_{m_2,m} \left( \varphi_1,\theta_1,\chi \right)^{*} Y^{m}_{l_2} \left( \theta_{12},\varphi \right) \\
\end{align}

Use

\begin{equation}
Y^{m_1}_{l_1} \left( \theta_1, \varphi_1 \right)^{*} = \sqrt{\frac{2l_1+1}{4\pi}}\mathcal{D}^{(l_1)}_{m_1,0} \left( \varphi_1, \theta_1, \chi \right)
\end{equation}

together with the orthogonality property of the rotation matrices (Brink and Satchler, p 147)

\begin{align}
& \int \mathcal{D}^{(j)*}_{m,m^\prime}\mathcal{D}^{(J)}_{M,M^\prime}\sin(\theta_1)d\theta_1d\varphi_1d\chi \\
& = \frac{8\pi^2}{2j+1}\delta_{jJ}\delta_{mM}\delta_{m^\prime M^\prime}\nonumber
\end{align}

to obtain

\begin{align}
I_{ang} & = \sqrt{ \frac{2l_1+1}{4\pi}} \frac{8\pi^2}{2l_1+1} \delta_{l_1,l_2} \delta_{m_1,m_2} Y^{0}_{l_2} \left( \theta_{12}, \varphi \right) \\
& = 2 \pi \delta_{l_1,l_2} \delta_{m_1,m_2} P_{l_2} \left( \cos \theta_{12} \right)
\end{align}

since

\begin{equation}
Y^0_{l_2} \left( \theta_{12}, \varphi \right) = \sqrt{\frac{2l_1+1}{4\pi}} P_{l_2} \left( \cos \left( \theta_{12} \right) \right)
\end{equation}

Note that <math>P_{l_2} \left( \cos \theta \right)</math> is just a short hand expression for a radial function because

\begin{equation}
\cos \theta_{12} = \frac{r_1^2 + r_2^2 - r_{12}^2}{2r_1r_2}
\end{equation}

The original integral is thus

\begin{equation}
I \left( l_1, m_1, l_2, m_2; R \right) = 2 \pi \delta_{l_1,l_2} \delta_{m_1,m_2} \int^{\infty}_{0} r_{1} dr_{1} \int^{\infty}_{0} r_{2} dr_{2} \int^{r_{1} + r_{2}}_{|r_{1} - r_{2}|} r_{12} dr_{12} R \left( r_{1}, r_{2}, r_{12} \right) P_{l_2} \left( \cos \theta_{12} \right)
\end{equation}

where again

\begin{equation}
\cos \theta_{12} = \frac{r_{1}^{2} +r^{2}_{2} - r_{12}^{2}}{2 r_{1} r_{2}}
\end{equation}

is a purely radial function.

The above would become quite complicated for large <math>l_2</math> because <math>P_{l_2}(\cos \theta_{12})</math> contains terms up to <math>\left( \cos \theta_{12} \right)^{l_2}</math>. However, recursion relations exist which allow any integral containing <math>P_l \left( \cos \theta_{12} \right)</math> in terms of those containing just <math>P_0 \left( \cos \theta_{12} \right) = 1</math> and <math>P_1 \left( \cos \theta_{12} \right) = \cos \theta_{12}</math>.

==Radial Integrals and Recursion Relations==

The basic radial integral is [see G.W.F. Drake, Phys. Rev. A 18, 820 (1978)]

\begin{align}
I_0 \left( a, b, c \right) & =
\int^{\infty}_{0} r_{1} dr_{1} \int^{\infty}_{r_{1}} r_{2} dr_{2} \int^{r_{1}+r_{2}}_{r_{2} - r_{1}} r_{12} dr_{12} r_{1}^{a} r_{2}^{b} r_{12}^{c} e^{-\alpha r_{1} - \beta r_{2}} + \int^{\infty}_{0} r_{2} dr_{2} \int^{\infty}_{r_{2}} r_{1} dr_{1} \int^{r_{1} + r_{2}}_{r_{1} - r_{2}} r_{12} dr_{12} r_{1}^{a} r_{2}^{b} r_{12}^{c} e^{- \alpha r_{1} - \beta r_{2}} \\
& = \frac{2}{c+2} \sum^{[(c+1)/2]}_{i=0}
\left(\begin{array}{c}
c+2 \\
2i+1
\end{array} \right)
\left\{ \frac{q!}{\beta^{q+1} \left( \alpha + \beta \right)^{p+1}} \sum_{j=0}^{q} \frac{\left( p + j \right)!}{j!} \left( \frac{\beta}{\alpha + \beta} \right)^j + \frac{q^{\prime}!}{\alpha^{q^{\prime}+1} \left( \alpha + \beta \right)^{p^{\prime}+1}} \sum^{q^{\prime}}_{j=0} \frac{\left( p^{\prime} + j \right)!}{j!} \left( \frac{\alpha}{\alpha + \beta} \right)^j \right\}
\end{align}

where

\begin{array}{ll}
p = a + 2i + 2 & p^{\prime} = b + 2i + 2 \\
q = b + c - 2i + 2 & q^{\prime} = a + c - 2i + 2
\end{array}

The above applies for <math> a, b \geq -2, c\geq -1</math>.
<math>[x]</math> means "greatest integer in" <math>x</math>.

Then

\begin{align}
I_1 \left( a, b, c \right) & = \int d\tau_{r} r_{1}^{a} r_{2}^{b} r_{12}^{c} e^{ -\alpha r_{1} - \beta r_{2}} P_{1}\left( \cos \theta \right) \\
& = \frac{1}{2} \left( I_{0} \left( a + 1, b - 1, c \right) + I_{0} \left( a - 1, b + 1, c \right) - I_{0} \left( a - 1, b - 1, c + 2 \right) \right)
\end{align}

===The Radial Recursion Relation===

Recall that the full integral is

\begin{equation}
I(l_1m_1,l_2m_2;R)=2\pi\delta_{l_1,l_2}\delta_{m_1,m_2}I_{l_2}(R)
\end{equation}

where

\begin{equation}
I_{l_2}(R) = \int d\tau_r R(r_1,r_2,r_{12})P_{l_2}(\cos \theta_{12})
\end{equation}

To obtain the recursion relation, use

\begin{equation}
P_l(x) = \frac{[P^{\prime}_{l+1}(x)-P^{\prime}_{l-1}(x)]}{2l+1}
\end{equation}

with

\begin{equation}
P_{l+1}^\prime (x) = \frac{d}{dx}P_{l+1}(x)
\end{equation}

Here <math>x=\cos\theta_{12}</math> and

\begin{align}
\frac{d}{d \cos \theta_{12}} & = \frac{dr_{12}}{d \cos \theta_{12}} \frac{d}{dr_{12}} \\
& = -\frac{r_{1} r_{2}}{r_{12}} \frac{d}{dr_{12}}
\end{align}

Then,

\begin{equation}
I_{l}\left( R \right) = - \int d\tau_r R \frac{r_{1} r_{2}}{r_{12}} \frac{d}{dr_{12}} \frac{ \left[ P_{l+1} \left( \cos \theta_{12} \right) - P_{l-1} \left( \cos \theta_{12} \right) \right]}{2l+1}
\end{equation}

The <math>r_{12}</math> part of the integral is

\begin{align}
& \int^{r_{1} + r_{2}}_{|r_{1} - r_{2}|} r_{12} dr_{12} R \frac{r_{1} r_{2}}{r_{12}} \frac{d}{dr_{12}} \left[P_{l+1}-P_{l-1} \right] \\
& = R r_{1} r_{2} \left[ P_{l+1} - P_{l-1} \right]^{r_{1} + r_{2}}_{|r_{1} - r_{2}|} - \int^{r_{1} + r_{2}}_{|r_{1} - r_{2}|} r_{12} dr_{12} \left( \frac{d}{dr_{12}} R \right) \frac{r_{1} r_{2}}{r_{12}} \frac{ \left[ P_{l+1} - P_{l-1} \right]}{2l+1}
\end{align}

The integrated term vanishes because

\begin{equation}
\cos \theta_{12} = \frac{r_{1}^{2} + r_{2}^{2} - r_{12}^{2}}{2 r_{1} r_{2}}
\begin{array}{lll}
& = -1 & {\rm when} & r_{12}^{2} = \left( r_{1} + r_{2} \right)^{2} \\
& = 1 & {\rm when} & r_{12}^{2} = \left( r_{1} - r_{2} \right)^{2}
\end{array}
\end{equation}

and <math>P_l(1)=1</math>, <math>P_{l}(-1)=(-1)^{l}</math>.

Thus,

\begin{equation}
I_{l+1}\left(\frac{r_1r_2}{r_{12}}R^\prime\right) = \left( 2l+1 \right) I_{l}(R) - I_{l-1} \left( \frac{r_{1} r_{2}}{r_{12}} R^{\prime} \right)
\end{equation}

If

\begin{equation}
R = r_{1}^{a-1} r_{2}^{b-1} r_{12}^{c+2} e^{-\alpha r_{1} - \beta r_{2}}
\end{equation}

then [G.W.F. Drake, Phys. Rev. A 18, 820 (1978)]

\begin{equation}
I_{l+1}(r_1^ar_2^br_{12}^c) =
\frac{2l+1}{c+2} I_{l} \left( r_{1}^{a-1} r_{2}^{b-2} r_{12}^{c+2} \right)+I_{l-1} \left( r_{1}^{a} r_{2}^{b} r_{12}^{c} \right)
\end{equation}

For the case <math>c=-2</math>, take

\begin{equation}
R = r_{1}^{a-1} r_{2}^{b-1} \ln r_{12} e^{ -\alpha r_{1} -\beta r_{2}}
\end{equation}

Then

\begin{equation}
I_{l+1} \left( r_{1}^{a} r_{2}^{b} r_{12}^{c} \right) = I_{l} \left( r_{1}^{a-1} r_{2}^{b-1} \ln r_{12} \right) \left( 2l + 1 \right) - I_{l-1} \left( r_{1}^{a} r_{2}^{b} r_{12}^{-2} \right)
\end{equation}

This allows all <math>I_{l}</math> integrals to be calculated from tables of <math>I_0</math> and
<math>I_1</math> integrals.

===The General Integral===

The above results for the angular and radial integrals can now be combined into a general formula for integrals of the type

\begin{equation}
I = \int \int d{\bf r}_{1} d{\bf r}_{2} R_{1} \mathcal{Y}^{M^{\prime}}_{l^{\prime}_{1} l^{\prime}_{2} L^{\prime}} \left( \hat{r}_{1}, \hat{r}_{2} \right) T_{k_{1} k_{2} K}^{Q} \left( {\bf r}_{1},{\bf r}_{2} \right) R_{2} \mathcal{Y}_{l_{1} l_{2} L}^{M} \left( \hat{r}_{1}, \hat{r}_{2} \right)
\end{equation}

where

\begin{equation}
\mathcal{Y}^{M}_{l_{1} l_{2} L} \left( \hat{r}_{1}, \hat{r}_{2} \right) = \sum_{m_1,m_2} \langle l_{1} l_{2} m_{1} m_{2} | LM \rangle Y^{m_{1}}_{l_{1}} \left( \hat{r}_{1} \right) Y^{m_{2}}_{l_{2}} \left( \hat{r}_{2} \right)
\end{equation}

and

\begin{equation}
T^Q_{k_{1} k_{2} K} \left( {\bf r}_{1}, {\bf r}_{2} \right) = \sum_{q_{1}, q_{2}} \langle k_{1} k_{2} q_{1} q_{2} | KQ \rangle Y_{k_{1}}^{q_{1}} \left( \hat{r}_{1} \right) Y^{q_{2}}_{k_{2}} \left( \hat{r}_{2} \right)
\end{equation}

The basic idea is to make repeated use of the formula

\begin{equation}
Y^{m_{1}}_{l_{1}} \left( \hat{r}_{1} \right) Y^{m_{2}}_{l_{2}} \left( \hat{r}_{1} \right) = \sum_{l m} \left( \frac{\left( 2l_{1}+1 \right) \left(2l_{2}+1 \right) \left(2l+1 \right)}{4 \pi} \right)^{1/2}
\times \left( \begin{array}{ccc}
l_1 & l_2 & l \\
m_1 & m_2 & m
\end{array} \right) \left(\begin{array}{ccc}
l_1 & l_2 & l \\
0 & 0 & 0
\end{array} \right)
Y^{m*}_{l} \left( \hat{r}_{1} \right)
\end{equation}

where

\begin{equation}
Y^{m*}(\hat{r}) = (-1)^mY_l^{-m}(\hat{r})
\end{equation}

and

\begin{equation}
\left(\begin{array}{ccc}
l_1 & l_2 & l \\
m_1 & m_2 & m
\end{array} \right) = \frac{\left( -1 \right)^{l_{1} - l_{2} - m}}{\left( 2l + 1 \right)^{1/2}} \left( l_{1} l_{2} m_{1} m_{2} | l, -m \right)
\end{equation}

is a 3-<math>j</math> symbol. In particular, write

\begin{array}{cc}
Y^{m_1^\prime}_{l_1^\prime}(\hat{r}_1)^{*} \underbrace{Y_{k_{1}}^{q_{1}} \left( \hat{r}_{1} \right) Y^{m_{1}}_{l_{1}} \left( \hat{r}_{1} \right)} & = & \sum_{\Lambda M}(\ldots)Y^{M}_\Lambda(\hat{r}_1)\\
\ \ \ \ \ \ \ \sum_{\lambda_1\mu_1}Y_{\lambda_1}^{\mu_1}(\hat{r}_1) & & \\
& & \\
Y^{m_{2}^{\prime}}_{l_{2}^{\prime}} \left( \hat{r}_{2} \right) \underbrace{Y^{q_{2}}_{k_{2}} \left( \hat{r}_{2} \right) Y_{l_{2}}^{m_{2}} \left( \hat{r}_{2} \right)} & = & \sum_{\Lambda^{\prime} M^{\prime}} \left( \ldots \right)Y_{\Lambda^{\prime}}^{M^{\prime}*} \left( \hat{r}_{2} \right) \\
\ \ \ \ \ \ \ \ \ \sum_{\lambda_{2} \mu_{2}}Y_{\lambda_{2}}^{\mu_{2}} \left( \hat{r}_{2} \right) & &
\end{array}

The angular integral then gives a factor of <math>2 \pi \delta_{\Lambda, \Lambda^{\prime}} \delta_{M, M^{\prime}} P_{\Lambda} \left( \cos \theta_{12} \right)</math>. The total integral therefore reduces to the form

\begin{equation}
I = \sum_{\Lambda} C_{\Lambda} I_{\Lambda} \left( R_{1} R_{2} \right)
\end{equation}

where <math>C_{\Lambda} = \sum_{\lambda_{1}, \lambda_{2}} C_{ \lambda_{1}, \lambda_{2}, \Lambda}</math>. For further details and derivations, including graphical representations, see G.W.F. Drake, Phys. Rev. A <math>{\bf 18}</math>, 820 (1978).

==Graphical Representation==

[[File:GraphRep1.jpg|500px]]

==Matrix Elements of H==
Recall that

\begin{equation}
H = -\frac{1}{2} \nabla^{2}_{1} - \frac{1}{2} \nabla^{2}_{2} - \frac{1}{r_1} - \frac{1}{r_2} + \frac{Z^{-1}}{r_{12}}
\end{equation}

Consider matrix elements of

\begin{equation}
\nabla^2_1 = \frac{1}{r_1^2} \frac{\partial}{\partial r_1}\left(r_1^2\frac{\partial}{\partial r_1} \right) + \frac{1}{r^2}\frac{\partial}{\partial r}\left(r^2\frac{\partial}{\partial r}\right) - \frac{(\vec{l}_1^Y)^2}{r_1^2} + \frac{2 \left( r_1 - r_2 \cos \theta \right)}{r} \frac{\partial^2}{\partial r_1 \partial r} - 2 \left( \nabla^Y_1 \cdot {\bf r}_2 \right) \frac{1}{r} \frac{\partial}{\partial r}
\end{equation}

where <math>r \equiv r_{12}</math> and <math>\cos\theta \equiv \cos\theta_{12}</math>. Also in what follows, define <math>\hat{\nabla}^Y_1 \equiv r_1\nabla^Y_1</math> where <math>\nabla^Y</math> operates only on the spherical harmonic part of the wave function, and similarly for <math>\vec{l}_1^Y</math>.

A general matrix element is

\begin{equation}
\langle r_1^{a^\prime} r_2^{b^\prime} r_{12}^{c^\prime} e^{-\alpha^\prime r_1 - \beta^\prime r_2} \mathcal{Y}^{M^\prime}_{l^\prime_1 l^\prime_2 L^\prime} \left( \hat{r}_1, \hat{r}_2 \right) \left| \nabla^2_1 \right| r_1^a r_2^b r_{12}^c e^{-\alpha r_1 - \beta r_2} \mathcal{Y}^M_{l_1 l_2 L} \left( \hat{r}_1, \hat{r}_2 \right) \rangle
\end{equation}

Since <math>\nabla^2_1</math> is rotationally invariant, this vanishes unless <math>L=L^\prime</math>, <math>M=M^\prime</math>. Also <math>\nabla_1^2</math> is Hermitian, so that the result must be the same whether it operates to the right or left, even though the results look very different. In fact, requiring the results to be the same yields some interesting and useful integral identities as follows:

Put

\begin{equation}
\mathcal{F} = F \mathcal{Y}^M_{l_1 l_2 L} \left( \hat{r}_1, \hat{r_2} \right)
\end{equation}

and

\begin{equation}
F = r_1^a r_2^b r_{12}^c e^{-\alpha r_1 - \beta r_2}
\end{equation}

Then

\begin{equation}
\nabla^2_1 \mathcal{F} = \left\{ \frac{1}{r_1^2} \left[ a \left( a + 1 \right) - l_1 \left( l_1 + 1 \right) \right] + \frac{c \left( c + 1 \right)}{r^2} + \alpha^2\frac{2\alpha \left( a + 1 \right)}{r_1} + \frac{2 \left(r_1 - r_2 \cos \theta \right) }{r_1r^2}c \left[ a - \alpha r_1 \right] - \frac{2c}{r^2} \left( \hat{\nabla}^Y_1 \cdot \hat{r}_2 \right) \frac{r_2}{r_1} \right\} \mathcal{F}
\end{equation}

and

\begin{align}
\langle \mathcal{F}^\prime \left| \nabla^2_1 \right| \mathcal{F} \rangle & = \sum_\Lambda \int d\tau_r F^\prime C_\Lambda \left( 1 \right) P_\Lambda \left( \cos \theta \right) \times \left\{ \frac{1}{r_1^2} \left[ a \left( a + 1 \right) - l_1 \left( l_1 + 1 \right) \right] -\frac{2 \alpha \left( a + 1 \right)}{r_1} + \frac{c \left( c + 1 \right)}{r^2} +\alpha^2 + \frac{2 \left( r_2 - r_2 \cos \theta \right)}{r_1r^2}c \left( a - \alpha r_1 \right) \right\} F \\
& + \sum_\Lambda \int d\tau_r F^\prime C_\Lambda \left( \hat{\nabla}^Y_1 \cdot \hat{r}_2 \right) P_\Lambda \left( \cos \theta \right) \left( \frac{-2c r_2}{r_1 r^2} \right) F
\end{align}

where

\begin{equation}
\int d\tau_r = \int^\infty_0 r_1 dr_1 \int^\infty_0 r_2 dr_2 \int^{r_1+r_2}_{\left| r_1 - r_2 \right|} r dr
\end{equation}

For brevity, let the sum over <math>\Lambda</math> and the radial integrations be understood, and let <math>\left( \nabla_1^2 \right)</math> stand for the terms which appear in the integrand. Then operating to the right gives

\begin{equation}
(\nabla_1^2)_R = \frac{1}{r_1^2} \left[ a \left( a + 1 \right) - l_1 \left( l_1 + 1 \right) \right] + \frac{c \left( c + 1 \right)}{r^2} + \alpha^2 - \frac{2 \alpha \left( a + 1 \right)}{r_1}
+ \frac{2 \left( r_1 - r_2 \cos \theta \right)}{r_1 r^2}c \left( a - \alpha r_1 \right) - \frac{2c r_2}{r_1 r^2} \left( \hat{\nabla}^Y_1 \cdot \hat{r}_2 \right)
\end{equation}

Operating to the left gives

\begin{equation}
(\nabla_1^2)_L = \frac{1}{r_1^2} \left[a^\prime \left( a^\prime + 1 \right) - l_1^\prime \left( l_1^\prime + 1 \right) \right] + \frac{c^\prime \left( c^\prime + 1 \right)}{r^2} + \alpha^{\prime 2} -\frac{2 \alpha^\prime \left( a^\prime + 1 \right)}{r_1}
+ \frac{2 \left( r_1 - r_2 \cos \theta \right)}{r_1 r^2} c^\prime \left( a^\prime - \alpha^\prime r_1 \right) - \frac{2 c^\prime r_2}{r_1 r^2} \left( \hat{\nabla}^{Y^\prime}_1 \cdot \hat{r}_2 \right)
\end{equation}

Now put

\begin{array}{cc}
a_+ = a+a^\prime, & \hat{\nabla}_1^+ = \hat{\nabla}_1^Y + \hat{\nabla}_1^{Y\prime} \\
a_{-} = a-a^\prime, & \hat{\nabla}_1^{-} = \hat{\nabla}_1^Y - \hat{\nabla}_1^{Y\prime}
\end{array}

etc., and substitute <math>a^\prime = a_+ - a</math>, <math>c^\prime = c_+ - c</math>, and <math>\alpha^\prime = \alpha_+ - \alpha</math> in <math>(\nabla^2_1)_L</math>. If <math>a_+</math>, <math>c_+</math> and <math>\alpha_+</math> are held fixed, then the equation

\begin{equation}
(\nabla^2_1)_R = (\nabla^2_1)_L
\end{equation}

must be true for arbitrary <math>a</math>, <math>c</math>, and <math>\alpha</math>. Their coefficients must thus vanish.

This yields the integral relations

\begin{equation}
\label{I}
\frac{\left( r_1 - r_2 \cos \theta \right)}{r_1 r^2} = \frac{1}{c_+} \left( \frac{- \left( a_+ + 1 \right)}{r_1^2} + \frac{\alpha_+}{r_1} \right) \tag{I}
\end{equation}

from the coefficient of <math>a</math> and

\begin{equation}
\label{II}
\frac{ \left( r_1 - r_2 \cos \theta \right) \left( a_+ - \alpha_+ r_1 \right)}{r_1 r^2} = \frac{r_2}{r_1 r^2}\left( \hat{r}_2 \cdot \hat{\nabla}_1^+ \right) - \frac{ \left( c_+ + 1 \right)}{r^2} \tag{II}
\end{equation}

from the coefficient of <math>c</math>. The coefficient of <math>\alpha</math> gives an equation equivalent to (I).

Furthermore, if can be show that (see problem)

\begin{align}
& \sum_\Lambda \int d\tau_r \frac{r^c}{r^2} C_\Lambda \left( \hat{r}_2 \cdot \hat{\nabla}_1^Y \right) P_\Lambda \left( \cos \theta \right) \\
= & \sum_\Lambda \int d\tau_r \frac{r^c}{c r_1 r_2}C_\Lambda \left( 1 \right)P_\Lambda \left( \cos \theta \right) \times \left( \frac{l_1^\prime \left( l_1^\prime + 1 \right) - l_1 \left( l_1 + 1 \right) -\Lambda \left( \Lambda + 1 \right)}{2} \right)
\end{align}

and similarly for <math>\left( \hat{r}_2 \cdot \hat{\nabla}^{Y\prime}_1 \right)</math> with <math>l_1</math> and <math>l_1^\prime</math> interchangeable, then it follows that

\begin{equation}
\frac{r^{c_+}}{r^2} \left( \hat{r}_2 \cdot \hat{\nabla}_1^+ \right) = - \frac{r^{c_+}}{c_+ r_1 r_2} \Lambda \left( \Lambda + 1 \right)
\end{equation}

and

\begin{equation}
\frac{r^{c_+}}{r^2} \left( \hat{r}_2 \cdot \hat{\nabla}_1^- \right) = \frac{r^{c_+}}{c_+ r_1 r_2} \left[ l_1^\prime \left( l_1^\prime + 1 \right) - l_1 \left( l_1 + 1 \right) \right]
\end{equation}

where equality applies after integration and summation over <math>\Lambda</math>.

Thus (II) becomes

\begin{equation}\label{IIa}
\frac{ \left( r_1 - r_2 \cos \theta \right) \left( a_+ - \alpha_+ r_1 \right)}{r_1 r^2} = - \frac{\Lambda \left( \Lambda + 1 \right)}{c_+ r^2_1} - \frac{ \left( c_+ + 1 \right)}{r^2} \tag{IIa}
\end{equation}

===Problem===

Prove the integral relation

\begin{align}
& \sum_\Lambda \int d\tau_r f \left( r_1, r_2 \right) \frac{1}{r} \left( \frac{d}{dr} q(r) \right) C_\Lambda \left( \hat{r}_2 \cdot \hat{\nabla}^Y_1 \right) P_\Lambda \left( \cos \theta \right) \\
= & \sum_\Lambda \int d\tau_r \frac{ f \left( r_1, r_2 \right)}{r_1, r_2} q(r) C_\Lambda \left( 1 \right) P_\Lambda \left( \cos \theta \right) \left(\frac{l_1^\prime \left( l_1^\prime + 1 \right) - l_1 \left( l_1 + 1 \right) - \Lambda \left( \Lambda + 1 \right)}{2}\right)
\end{align}

where the coefficients <math>C_\Lambda \left( 1 \right)</math> are the angular coefficients from the overlap integral

\begin{align}
\int & d\Omega\, \mathcal{Y}^{M*}_{l_1^\prime l_2^\prime L}(\hat{r}_1,\hat{r}_2)\, \mathcal{Y}^{M}_{l_1 l_2 L}(\hat{r}_1,\hat{r}_2)\\
= & \sum_\Lambda C_\Lambda(1)P_\Lambda(\cos\theta_{12}).
\end{align}

Hint: Use the fact that <math>l_1^2</math> is Hermitian so that
<math>
\int d\tau (l_1^2\mathcal{Y}^\prime)^*q(r)\mathcal{Y} = \int d\tau
\mathcal{Y}^{\prime *}l_1^2(q(r)\mathcal{Y})
</math>
with <math>\ \vec{l}_1= \frac{1}{i}\vec{r}_1\times\nabla_1</math>.

It is also useful to use
<math>\begin{eqnarray}\ \
\cos\theta\, P_L(\cos\theta) &=& \frac{1}{2L+1}[L\,P_{L-1}(\cos\theta) + (L+1)P_{L+1}(\cos\theta)]
\\
(\cos^2\theta-1)P_L(\cos\theta) &=&
\frac{L(L-1)}{(2L-1)(2L+1)}P_{L-2}(\cos\theta)\nonumber
-\frac{2(L^2+L-1)}{(2L-1)(2L+3)}P_L(\cos\theta) + \frac{(L+1)(L+2)}{(2L+1)(2L+3)}P_{L+2}(\cos\theta)\nonumber\\
&=& \frac{L(L-1)}{(2L-1)(2L+1)}[P_{L-2}(\cos\theta) - P_L(\cos\theta)]
+ \frac{(L+1)(L+2)}{(2L+1)(2L+3)}[P_{L+2}(\cos\theta) - P_L(\cos\theta)]
\end{eqnarray}</math>

together with a double application of the integral recursion relation
<math>
I_{L+1}\left(\frac{1}{r}\frac{d}{dr}q(r)\right) =
(2L+1)I_L\left(\frac{1}{r_1r_2}q(r)\right)+I_{L-1}\left(\frac{1}{r}\frac{d}{dr}q(r)\right)
</math>

Of course <math>l_1^2\mathcal{Y}^M_{l_1l_2L}(\hat{r}_1,\hat{r}_2)=l_1(l_1+1)\mathcal{Y}^M_{l_1l_2L}(\hat{r}_1,\hat{r}_2)</math>. Begin by expanding
<math>
l_1^2(g\mathcal{Y}) = \mathcal{Y}l_1^2g + g\,l_1^2\mathcal{Y} + 2(\vec{l}_1g)\cdot(\vec{l}_1\mathcal{Y})
</math>, and show that

<math>
\begin{eqnarray}\ \
(\vec{l}_1g)\cdot(\vec{l}_1\mathcal{Y}) & = & \frac{r_1^2}{r}\frac{dg}{dr}\vec{r}_2\cdot \nabla_1\mathcal{Y}\,\,\,\,\mbox{ and}\\
l_1^2g(r) &=& 2r_1r_2\cos\theta\frac{1}{r}\frac{dg}{dr} + r_1^2r_2^2(\cos^2\theta - 1)
\frac{1}{r}\frac{d}{dr}\left(\frac{1}{r}\frac{dg}{dr}\right).
\end{eqnarray}
</math>

The proof amounts to showing that the term <math>l_1^2g(r)</math> can be replaced by
<math>\Lambda(\Lambda+1)g(r)</math> after multiplying by <math>P_\Lambda(\cos\theta)</math>
and integrating by parts with respect to the radial integrations over <math>r_1</math>, <math>r_2</math> and
<math>r \equiv |\vec{r}_1 - \vec{r}_2|</math>. Remember that <math>\cos\theta</math> is just a short-hand notation for the radial function <math>(r_1^2 + r_2^2 - r^2)/(2r_1r_2)</math>.

==General Hermitian Property==

Each combination of terms of the form
<math>
<f>_R = a^2f_1+af_2+abf_3+b^2f_4+bf_5+af_6\nabla_1^Y+bf_7\nabla^Y_2+f_8(Y)
</math>
can be rewritten
<math>
<f>_L= (a_+-a)^2f_1+(a_+-a)f_2 + (a_+-a)(b_+-b)f_3+(b_+-b)^2f_4
+(b_+-b)f_5+(a_+-a)f_6\nabla^{Y\prime}_1+(b_+-b)f_7\nabla^{Y\prime}_2+f_8)Y^\prime
</math>
where as usual <math>\nabla_1^Y</math> acts only on the spherical harmonic part of the wave function denoted for short by <math>Y</math>.
Since these must be equal for arbitrary <math>a</math> and <math>b</math>,
<math>
a_+^2f_1+a_+f_2+_+b_+f_3+b_+f_4+b_+f_5+a_+f_6\nabla^{Y\prime}_1 +
b_+f_7\nabla^{Y\prime}_2+f_8(Y^\prime)-f_8(Y)=0
</math>

Adding the corresponding expression with <math>Y</math> and <math>Y^\prime</math> interchanged
yields
<math>\label{III}
a_+^2f_1+a_+f_2+a_+b_+f_3+b_+^2f_4+b_+f_5+\frac{1}{2}a_+f_6\nabla^+_1+\frac{1}{2}b_+f_7\nabla^+_2=0\tag{III}
</math>

Subtracting gives
<math>\label{IV}
f_8(Y)-f_8(Y^\prime)=-\frac{1}{2}[a_+f_6\nabla^-_1+b_+f_7\nabla_2^-]\tag{IV}
</math>
<math>\label{V}
a[-2a_+f_1-2f_2-b_+f_3-f_6\nabla^+_1]=0\tag{V}
</math>
<math>\label{VI}
b[-2b_+f_4-2f_5-a_+f_3-f_7\nabla^+_2]=0\tag{VI}
</math>

Adding the two forms gives

\begin{eqnarray}
<f>_R+<f>_L=\frac{1}{2}(a_+^2+a_-^2)f_1+a_+f_2+\frac{1}{2}(a_+b_++a_-b_-)f_3\nonumber\\
\ +\frac{1}{2}(b_+^2+b_-^2)f_4+b_+f_5+\frac{1}{2}f_6(a_+\nabla_1^++a_-\nabla^-_1)\nonumber\\
\ +\frac{1}{2}(b_+\nabla_2^++b_-\nabla^-_2)+f_8(Y)+f_8(Y^\prime)\nonumber
\end{eqnarray}

Subtracting <math>\frac{x}{2}\times({\rm I})</math> gives

\begin{eqnarray}\ \
<f>_R+<f>_L = \frac{1}{2}[(1-x)a_+^2 + a_-^2]f_1 +
(1-\frac{x}{2})a_+f_2+\frac{1}{2}[(1-x)a_+b_++a_-b_-]f_3\nonumber\\
+\frac{1}{2}[(1-x)b_+^2+b_-^2]f_4+(1-\frac{x}{2})f_5b_++\frac{1}{2}f_6[(1-\frac{x}{2})a_+\nabla_1^++a_-\nabla^-_1]\nonumber\\
+\frac{1}{2}[(1-\frac{x}{2})b_+\nabla^+_2+b_-\nabla^-_2]+f_8(Y)+f_8(Y^\prime)\nonumber
\end{eqnarray}

If <math>x=1</math>,
\begin{eqnarray}
<f>_R+<f>_L &=&
\frac{1}{2}[a_-^2f_1+a_+f_2+a_-b_-f_3+b_-^2f_4+b_+f_5\nonumber\\
&+&\left(\frac{1}{2}a_+\nabla^+_1+a_-\nabla^-_1\right)f_6+\left(\frac{1}{2}b_+\nabla^+_2+b_-\nabla^-_2\right)f_7]+f_8(Y)+f_8(Y^\prime)\nonumber
\end{eqnarray}

The General Hermitian Property for arbitrary <math>x</math> gives
\begin{eqnarray}\ \
\nabla_1^2=
\frac{1}{4}\{\frac{1}{r_1^2}[(1-x)a_+^2+a_-^2+2(1-\frac{x}{2})a_+-2[l_1(l_1+1)+l_1^\prime(l_1^\prime+1)]\nonumber\\
-\frac{2}{r_1}[(1-x)\alpha_+a_++\alpha_-a_-+2(1-\frac{x}{2})\alpha_+]+(1-x)\alpha_+^2
+ \alpha^2_-]\nonumber\\
+\frac{2(r_1-r_2\cos\theta)}{r_1r^2}[(1-x)(a_+-\alpha_+r_1)c_++(a_--\alpha_-r_2)c_-]\nonumber\\
-\frac{2r_2}{r_1r^2}[(1-\frac{x}{2})c_+\hat{r}_2\cdot\hat{\nabla}_1^++c_-\hat{r}_2\hat{\nabla}_1^-]+\frac{(1-x)c_+^2+c_-^2+2(1-\frac{x}{2})c_+}{r^2}\}\nonumber
\end{eqnarray}.

Use
<math>\begin{eqnarray}\ \
\frac{-2r_2}{r_1r^2}[(1-\frac{x}{2})c_+\hat{r}_2\hat{\nabla}_1^++c_-\hat{r}_2\hat{\nabla}_1^-]=\frac{2}{r_1^2}\left[(1-\frac{x}{2})\Lambda(\Lambda+1)-\frac{c_-}{c_+}[l_1^\prime(l_1^\prime+1)+l_1(l_1+1)]\right]\end{eqnarray}</math>,

<math>\begin{eqnarray}
\frac{2(r_1-r_2\cos\theta)}{r_1r^2}(a_--\alpha_-r_1)c_- = 2\frac{c_-}{c_+}\left[\frac{-1}{r_1^2}[a_-(a_++1)]+\frac{1}{r_1}[a_-\alpha_++\alpha_-(a_++2)]-\alpha_-\alpha_+\right]\nonumber
\end{eqnarray}</math>,

and
<math>\begin{eqnarray}\ \
\frac{2(r_1-r_2\cos\theta)}{r_1r^2}(a_+-\alpha_+r_1)c_+ = 2\left[-\frac{1}{r_1^2}[a_+(a_++1)]+\frac{1}{r_1}[a_+\alpha_++\alpha_+(a_++2)]-\alpha_+^2\right]
\end{eqnarray}</math>.

Substituting into <math>\nabla_1^2</math> gives

\begin{eqnarray}
\nabla^2_1 = \frac{1}{4}\{\frac{1}{r_1^2}[-(1-x)a_+^2 + a_-^2 +xa_+ +
2(1-\frac{x}{2})\Lambda(\Lambda+1)\nonumber\\
-2l_1(l_1+1)(1-\frac{c_-}{c_+}) -
2l_1^\prime(l_1^\prime+1)(1+\frac{c_-}{c_+})-\frac{2c_-a_-}{c_+}(a_++1)]\nonumber\\
- \frac{2}{r_1}\left[-(1-x)\alpha_+(a_++2)+\alpha_-a_-+2(1-\frac{x}{2})\alpha_+
-\frac{c_-}{c_+}[a_-\alpha_++\alpha_-(a_++2)]\right]\nonumber\\
-(1-x)\alpha_+^2+\alpha_-^2-2\frac{c_-}{c_+}\alpha_-\alpha_+ +\frac{1}{r^2}\left[(1-x)c_+^2+c_-^2+2(1-\frac{x}{2})c_+\right]\}\nonumber
\end{eqnarray}

This has the form
<math>
\nabla_1^2 =
\frac{1}{4}\left[\frac{A_1}{r_1^2}+\frac{B_1}{r_1}+\frac{C_1}{r^2}+D_1\right]
</math>
with

<math>\begin{array}{ccc}
A_1=-(1-x)a_+^2+a_-^2
+xa_++2(1-\frac{x}{2})\Lambda(\Lambda+1)-2l_1(l_1+1)(1-\frac{c_-}{c_+})\
-2l_1^\prime(l_1^\prime+1)(1+\frac{c_-}{c_+})-\frac{2c_-a_-}{c_+}(a_++1) \ \
\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \
\end{array}
</math>

<math>
B_1 =
2\left[(1-x)\alpha_+(a_++2)-\alpha_-a_--2(1-\frac{x}{2})\alpha_++\frac{c_-}{c_+}[a_-\alpha_++\alpha_-(a_++2)]\right]
</math>

<math>
C_1 = (1-x)c_+^2+ c_-^2 +2(1-\frac{x}{2})c_+
</math>

<math>
D_1 = -1(1-x)\alpha_+^2 +\alpha_-^2 - 2\frac{c_-}{c_+}\alpha_-\alpha_+
</math>

The complete Hamiltonian is then (in Z-scaled a.u.)

<math>\begin{eqnarray}
H =-\frac{1}{2}\nabla^2_1 -\frac{1}{2}\nabla^2_2 -\frac{1}{r_1} -\frac{1}{r_2} +\frac{Z^{-1}}{r}\nonumber\\
= -\frac{1}{2}\nabla^2_1 -\frac{1}{2}\nabla^2_2 -\frac{1}{r_1} -\frac{Z-1}{Zr_2}+Z^{-1}\left(\frac{1}{r}-\frac{1}{r_2}\right)\nonumber\\
\ =-\frac{1}{8}\left[\frac{A_1}{r_1^2}+\frac{B_1+8}{r_1}+\frac{C_1}{r^2}+D_1+D_2 +\frac{A_2}{r_2^2} + \frac{B_2 + 8(Z-1)/Z}{r_2}+\frac{C_2}{r^2}\right]\nonumber
+ Z^{-1}\left(\frac{1}{r}-\frac{1}{r_2}\right)\nonumber
\end{eqnarray}</math>

The screened hydrogenic energy is
<math>\begin{eqnarray}
E_{SH}=-\frac{1}{2}\left[\frac{1}{n_1^2}+ \left(\frac{Z-1}{Z}\right)^2\frac{1}{n_2^2}\right]
\end{eqnarray}</math>
so that

<math>\begin{eqnarray}
H-E_{SH} &=& -\frac{1}{8}[
\frac{A_1}{r_1^2}+\frac{B_1+8}{r_1}+\frac{C_1}{r^2} + D_1 + D_2
-\frac{4}{n_1^2} -\left(\frac{Z-1}{Z}\right)^2\frac{4}{n_2^2}\nonumber
+\frac{A_2}{r_2}+\frac{B_2+8(Z-1)/Z}{r_2}+\frac{C_2}{r^2}] +
Z^{-1}\left(\frac{1}{r} -\frac{1}{r_2}\right)\nonumber
\end{eqnarray}</math>

==Optimization of Nonlinear Parameters==
- The traditional method of performing Hylleraas calculations is to write the
basis set in the form

<math>
\Psi = \sum_{i,j,k}c_{i,j,k}\varphi_{i,j,k}(\alpha,\beta)\pm exchange
</math>

with

<math>
\varphi_{i,j,k}(\alpha,\beta) = r_1^ir_2^jr_{12}^ke^{-\alpha r_1 - \beta
r_2}\mathcal{Y}^M_{l_1l_2L}.
</math>

The usual procedure is to set <math>\alpha = Z</math> so that it represents the inner
<math>1s</math> electron, and then to vary <math>\beta</math> so as to minimize the energy.

[[File:OptNon.jpg|500px]]

Since <math>\beta</math> appears in <math>\Psi</math> as a non-linear parameter, the entire
calculation must be repeated for each value of <math>\beta</math>. However, the minimum
becomes progressively smaller as the basis set is enlarged.

[[File:OptNon02.jpg|500px]]

*Difficulties

1. If the basis set is constructed so that <math> i +j +j \leq N</math>, the the
number of terms is
<math>
(N+1)(N+2)(N+3)/6
</math>
N=14 already gives 680 terms and an accuracy of about 1 part in <math>10^{10}</math> for
low-lying states. A substantial improvement in accuracy would require much
larger basis sets, together with multiple precision arithmetic to avoid loss
of significant figures when high powers are included.

2. The accuracy rapidly deteriorates as one goes to more highly excited
states - about 1 significant figure is lost each time the principle quantum
number is increased.

*Cure

We have found that writing basis sets in the form
\begin{eqnarray}
\Psi &=& \sum_{i,j,k}\left(c^{(1)}_{i,j,k}\varphi_{ijk}(\alpha_1\beta_1) +
c_{ijk}^{(2)}\varphi_{ijk}(\alpha_2\beta_2)\right) \pm exchange \nonumber\\
&=& \psi({\bf r}_1,{\bf r}_2) \pm \psi({\bf r}_2,{\bf r}_1)\nonumber
\end{eqnarray}
so that each combination of powers is included twice with different nonlinear
parameters gives a dramatic improvement in accuracy for basis sets of about
the same total size.

However, the optimization of the nonlinear parameters is now much more
difficult, and an automated procedure is needed.

If
<math>
E = \frac{<\Psi|H|\Psi>}{<\Psi|\Psi>}</math> and we assume <math><\Psi|\Psi> = 1</math>, then

<math>
\frac{\partial E}{\partial\alpha_t} = -2 <\Psi|H-E|r_1\psi({\bf r}_1,{\bf
r}_2;\alpha_t)\pm r_21\psi({\bf r}_2,{\bf
r}_1;\alpha_t)>
</math>
where
<math>
\psi({\bf r}_1,{\bf r}_2;\alpha_t) =
\sum_{i,j,k}c_{ijk}^{(t)}\varphi_{ijk}(\alpha_t\beta_t)
</math>

*Problem

1. Prove the above.

2. Prove that there is no contribution from the implicit dependence of
<math>c_{ijk}^{(t)}</math> on <math>\alpha_t</math> if the linear parameters have been optimized.

==The Screened Hydrogenic Term==

If the Hamiltonian is written in the form (in Z-scaled a.u.)
<math>\begin{eqnarray}\
H = H_0({\bf r}_1,Z) + H_0({\bf r}_2,Z-1) +
Z^{-1}\left(\frac{1}{r}-\frac{1}{r_2}\right)
\end{eqnarray}</math>
with
<math>\
H_0({\bf r}_1,Z) = -\frac{1}{2}\nabla_1^2 - \frac{1}{r_1}\nonumber
</math> and
<math>
H_0({\bf r}_2,Z-1) = -\frac{1}{2}\nabla_2^2 - \frac{Z-1}Z{r_2}\nonumber
</math>,
then the eigenvectors of <math>H_0({\bf r},Z) +H_0({\bf r}_2,Z-1)</math> are products of
hydrogenic orbitals
<math>
\Psi_0=\psi_0(1s,Z)\psi_0(nl,Z-1)
</math>
and the eigenvalue is
<math>\begin{eqnarray}
E_{SH} = \left[-\frac{1}{2} - \left(\frac{Z-1}{Z}\right)^2\frac{1}{2n^2}\right]Z^2 a.u.
\end{eqnarray}</math>
called the screened hydrogenic eigenvalue. From highly excited states,
<math>E_{SH}</math> and <math>\Psi_0</math>are already excellent approximations.

For example, for the 1s8d states, the energies are

<math>\begin{eqnarray}\\
E(1s8d^1D)&=& -2.007\,816\,512\,563\,81 {\rm\ a.u.}\nonumber\\
E(1s8d^3D)&=& - 2.007\,817\,934\,711\,71 {\rm\ a.u.}\nonumber\\
E_{SH} &=& -2.007\,812\,5 {\rm\ a.u.} \nonumber
\end{eqnarray}</math>

It is therefore advantageous to include the screened hydrogenic terms in the
basis set so that the complete trial function becomes
<math>
\Psi = c_0\Psi_0 +
\sum_{ijk}\left[c_{ijk}^{(1)}\varphi_{ijk}(\alpha_1,\beta_1)+c_{ijk}^{(2)}\varphi(\alpha_2,\beta_2)\right]\pm
{\rm exchange}
</math>

Also, the variational principal can be re-expressed in the form
<math>\begin{eqnarray}
E= E_{SH}+\frac{<\Psi|H-E_{SH}|\Psi>}{<\Psi|\Psi>}
\end{eqnarray}</math>
so that the <math>E_{SH}</math> term can be cancelled analytically from the matrix
elements.

For example
<math>
<\Psi_0|H-E_{SH}|\Psi_0> = 0
</math>

and
<math>
<\varphi_{ijk}|H-E_{SH}|\Psi_0> =
Z^{-1}<\varphi_{ijk}|\frac{1}{r}-\frac{1}{r_2}|\Psi_0>
</math>

Recall that
<math>\begin{eqnarray}\ \
I_0(a,b,c) = \frac{2}{c+2}\sum^{[c+1]/2}_{i=0}\left(\begin{array}{c}
c+2\\
2i+1
\end{array}\right) \left[f(p,q;\beta) + f(p^\prime,q^\prime;\alpha)\right]
\end{eqnarray}</math>
where

<math>\begin{eqnarray}
f(p,q;x) =
\frac{q!}{x^{q+1}(\alpha+\beta)^{p+1}}\sum^p_{j=0}\frac{(p+j)!}{j!}\left(\frac{x}{\alpha+\beta}\right)^j
\end{eqnarray}</math>

<math>
\begin{array}{cc}
p=a+2i+2 & p^\prime = b+2i+2\\
q= b+c-2i+2 & q^\prime =a+c-2i+2
\end{array}
</math>

If <math>\beta\ll\alpha</math>, the <math>f(p,q;\beta)\gg f(p^\prime,q^\prime;\alpha)</math> and the
<math>i=0</math> term is the dominant contribution to <math>f(p,q;\beta)</math>. However, since
this term depends only on the sum of powers <math>b+c</math> for <math>r_2</math>, it cancels exactly
from the matrix element of
<math>
\frac{1}{r_{12}} - \frac{1}{r_2}
</math>
and can therefore
be omitted in the calculations of integrands, thereby saving many significant
figures. This is especially valuable when <math>b+c</math> is large and <math>a</math> is small.

==Small Corrections==
*Mass Polarization
If the nuclear mass is not taken to be infinite, then the Hamiltonian is
<math>\begin{eqnarray}\ \
H = \frac{P_N^2}{2M} + \sum_{i=1}\frac{p_i^2}{2m} + V
\end{eqnarray}</math>

[[File:SmallCor.jpg|500px]]

where <math>M =M_A-nm</math> is the nuclear mass and <math>m</math> is the electron mass.

Change variables to the C of M
<math>\begin{eqnarray}
{\bf R} = \frac{1}{M+nm}\left[M{\bf r}_N + m({\bf r}_1 + {\bf r}_2 +
\ldots)\right]
\end{eqnarray}</math>

and relative variables

<math>
{\bf s}_i = {\bf r}_i - {\bf r}_N.
</math>

Then

<math>\begin{eqnarray}\ \ \ \ \
H = \frac{1}{2(M+nm)}{\bf p}_R^2+\frac{1}{2m}\sum_{i=1}^n{\bf p}_{s_i}^2
+\frac{1}{2M}\sum_{i,k}{\bf p}_{s_i}\cdot {\bf p}_{s_k} + V({\bf s}_1, {\bf
s}_2, \ldots)\nonumber
= \frac{1}{M+nm}{\bf p}_R^2+\frac{1}{2\mu}\sum_{i=1}^n{\bf
p}_{s_i}^2+\frac{1}{2M}\sum_{i\neq k}{\bf p}_{s_i}\cdot {\bf p}_{s_k}+V({\bf s}_1, {\bf
s}_2, \ldots)\nonumber
\end{eqnarray}</math>

where
<math>\begin{eqnarray}
\mu = \left(\frac{1}{M} + \frac{1}{m}\right)^{-1}
\end{eqnarray}</math>

If <math>n=2</math>, the Schrödinger equation is
<math>\begin{eqnarray}\ \ \
\left[ -\frac{\hbar^2}{2\mu}(\nabla^2_{s_1}+ \nabla^2_{s_2}) -
\frac{\hbar^2}{M}\nabla_{s_1}\nabla_{s_2}-\frac{Ze^2}{s_1}-\frac{Ze^2}{s_2}+\frac{e^2}{s_{12}}\right]\psi
= E\psi
\end{eqnarray}</math>.

Define the reduced mass Bohr radius
<math>\begin{eqnarray}
a_\mu = \frac{\hbar^2}{\mu e^2} = \frac{m}{\mu}a_0
\end{eqnarray}</math>
and

<math>\begin{eqnarray}
\rho_i = \frac{Zs_i}{a_\mu}\nonumber\\
\varepsilon = \frac{E}{Z^2(e^2/a_\mu)}\nonumber
\end{eqnarray}</math>

The Schrödinger equation then becomes
<math>\begin{eqnarray}\ \ \
\left[
-\frac{1}{2}(\nabla^2_{\rho_1}+\nabla^2_{\rho_2})-\frac{\mu}{M}\vec{\nabla}_{\rho_1}\cdot\vec{\nabla}_{\rho_2} - \frac{1}{\rho_1}-\frac{1}{\rho_2}+\frac{Z^{-1}}{\rho_{12}}\right]\psi = \varepsilon \psi
\end{eqnarray}</math>

The effects of finite mass are thus

1. A reduced mass correction to all energy levels of <math>(e^2/a_\mu)/(e^2/a_0)
= \mu/m</math>, i.e.
<math>\begin{eqnarray}
E_M=\frac{\mu}{m}E_\infty
\end{eqnarray}</math>

2. A specific mass shift given in first order perturbation theory by
<math>\begin{eqnarray}
\Delta\varepsilon =
-\frac{\mu}{M}<\psi|\nabla_{\rho_1}\cdot\nabla_{\rho_2}|\psi>
\end{eqnarray}</math>
or

<math>\begin{eqnarray}
\Delta\varepsilon =
-\frac{\mu}{M}<\psi|\nabla_{\rho_1}\cdot\nabla_{\rho_2}|\psi>\frac{e^2}{a_\mu}\nonumber\\
= -\frac{\mu^2}{mM}<\psi|\nabla_{\rho_1}\cdot\nabla_{\rho_2}|\psi>\frac{e^2}{a_0}\nonumber
\end{eqnarray}</math>

Since <math>\nabla_{\rho_1}\cdot\nabla_{\rho_2}</math> has the same angular properties as
<math>{\bf\rho}_1 \cdot {\bf\rho}_2</math>, the operator is like the product of two
dipole operators. For product type wave functions of the form
$
\psi = \psi(1s)\psi(nl) \pm \mbox{exchange},
$
the matrix element vanishes for all but P-states.

=Radiative Transitions=

==Transition Integrals and the Line Strength==

For an electric dipole transition <math>\gamma'\,L'\,S'\,J'\,M' \rightarrow \gamma\,L\,S\,J\,M</math>, the line strength is defined by
\begin{equation}
S = \sum_{M',\mu,M}|\langle\gamma'\,L'\,S'\,J'\,M'|r_\mu|\gamma\,L\,S\,J\,M\rangle|^2
\end{equation}
Using the Wigner-Eckart theorem, this is related to the reduced matrix element by
\begin{eqnarray}
S(J',J) = |\langle\gamma'\,L'\,S'\,J'||r||\gamma\,L\,S\,J\,M\rangle|^2
\sum_{M',\mu,M}\left|\left(\matrix{J' & 1 & J \cr
-M'& \mu &M \cr}\right)\right|^2
\end{eqnarray}
Using the sum rule for the 3-<math>j</math> symbols
\begin{equation}
\sum_{M',\mu,M}\left|\left(\matrix{J' & 1 & J \cr
-M'& \mu &M \cr}\right)\right|^2 = 1
\end{equation}
we see that the line strength is identical to the reduced matrix element
\begin{equation}
S(J',J) = |\langle\gamma'\,L'\,S'\,J'||r||\gamma\,L\,S\,J\,M\rangle|^2
\end{equation}
For a nonrelativistic calculation in <math>LS</math> coupling, <math>S = S'</math> and we can next strip out the dependence on <math>J</math> and <math>J'</math> by use of Eq.\ (7.1.7) of Edmonds { Angular Momentum in Quantum Mechanics} with <math>k = 1</math> to obtain
\begin{equation}
S(J',J) = (-1)^{L'+S+J+1}[(2J+1)(2J'+1)]^{1/2}\left\{
\matrix{L' & J'& S\cr
J & L & 1\cr}\right\}\langle\gamma' L'|| r || \gamma L\rangle
\end{equation}
where <math>\langle\gamma' L'|| r || \gamma L\rangle</math> is the reduced matrix element if spin is neglected. Note that if spin is neglected, then the line-strength is simply
\begin{equation}
S(L',L) = |\langle\gamma' L'|| r || \gamma L\rangle|^2
\end{equation}
To calculate the reduced matrix element, use a particular example along with the Wigner-Eckart theorem
such as
\begin{equation}
\langle\gamma'\, L'|| r || \gamma\, L\rangle = (-1)^{L'}\langle\gamma'\, L'\, 0| z | \gamma\, L\, 0\rangle{\Large /}\left(\matrix{L' & 1 & L \cr
0 & 0 & 0 \cr}\right)
\end{equation}

===Correlated Two-electron Integrals===
For the case of two-electron integrals, <math>z = z_1 + z_2</math> and <math>r = r_1 + r_2</math>. As discussed in G.W.F. Drake, Phys. Rev. A 18, 820 (1978), the general integral in Hylleraas coordinates for the integral over the <math>z_1</math> term can be written in the form
\begin{equation}
\label{int1}
\langle\gamma'\, L'\, 0| z_1 | \gamma\, L\, 0\rangle = \sum_\Lambda C_\Lambda I_\Lambda(\gamma',\gamma)
\end{equation}
where <math>I_\Lambda(\gamma',\gamma)</math> stands for the purely radial integral
\begin{equation}
\label{radial}
I_\Lambda(\gamma',\gamma) = \int_0^\infty r_1\,dr_1\int_0^\infty r_2\,dr_2\int_{|r_1-r_2|}^{r_1+r_2}r_{12}\, dr_{12}
R_{\gamma'}^*\,r_1\, R_\gamma P_\Lambda(\cos\theta_{12})
\end{equation}
and <math>\cos\theta_{12} = (r_1^2 + r_2^2 - r_{12}^2)/(2r_1r_2)</math> is a short-hand notation for a radial function of <math>r_1</math>, <math>r_2</math> and <math>r_{12}</math>. Also, <math>P_\Lambda(\cos\theta_{12})</math> is a Legendre polynomial,
and, <math>R_\gamma = R_\gamma(r_1,r_2,r_{12})</math> is the radial part of the Hylleraas wave function for state <math>\gamma</math> such
that the total wave function is <math>\psi({\bf r}_1,{\bf r}_2) = R_\gamma(r_1,r_2,r_{12}){\cal Y}_{l_1,l_2,L}^M(\hat{\bf r}_1,\hat{\bf r}_2)\pm {\rm exchange}</math>, normalized to unity, and <math>{\cal Y}_{l_1,l_2,L}^M(\hat{\bf r}_1,\hat{\bf r}_2)</math> is a vector coupled product of spherical harmonics. The coefficents <math>C_\Lambda</math> in Eq.\ (\ref{int1}) can be calculated from Eq.\ (32) of the above 1978 paper. They need to be calculated separately for he direct-direct and the direct-exchange contributions. For the reduced matrix element of <math>z_1</math>, the leading 3-<math>j</math> symbol can be dropped, leaving (for the case <math>k = 1</math>)
\begin{eqnarray}
C_\Lambda^{\rm reduced} &=& \frac{(-1)^{l_2'+L'+L+1}}{2} (l_1,l_1',l_2,l_2',L,L')^{1/2}
\sum_{\lambda={\rm max}(l_1-1,0)}^{l_1+1} (\lambda,\Lambda,1)
\left(\matrix{l_1 & 1 & \lambda \cr
0 & 0 & 0 \cr}\right)
\left(\matrix{l_1'& \lambda & \Lambda \cr
0 & 0 & 0 \cr}\right)\nonumber\\
&\times&\left(\matrix{l_2' & 1_2 & \Lambda \cr
0 & 0 & 0 \cr}\right)
\left\{\matrix{L' & \lambda & l_2 \cr
\Lambda& l_2' & l_1' \cr}\right\}
\left\{\matrix{L' & \lambda & l_2 \cr
l_1 & L & 1 \cr}\right\}
\end{eqnarray}
where the notation <math>(a,b,c,\cdots)</math> means <math>(2a+1)(2b+1)(2c+1)\cdots</math>.
A similar expression can be obtained for the angular coefficients for the <math>z_2</math> term by interchanging the subscripts 1 and 2 throughout and multiplying by <math>(-1)^{L+L'+1}</math>.

===Example===
As an example, for the <math>1s^2\;^1S - 1s2p\;^1P</math> transition of helium, the only nonvanishing <math>C_\Lambda</math> angular coefficients are <math>C_1 = 1/2</math> for the direct-direct term (i.e. <math>P_1(\cos\theta_{12}) = \cos\theta_{12}</math> is included in the radial integral (\ref{radial})) and <math>C_0 = 1/2</math> for the direct-exchange term (i.e. <math>P_0(\cos\theta_{12}) = 1</math> is included in the radial integral (\ref{radial})). The results for infinite nuclear mass are <math>S(1\;^1S-2\;^1P) = (0.781\,127\,158\,667)^2</math> and <math>f(1\;^1S-2\;^1P) = -0.276\,164\,703\,736</math> where <math>f(1\;^1S-2\;^1P) = \frac{2}{3}[E(1\;^1S) - E(2\;^1P)]S(1\;^1S-2\;^1P)</math> is the absorption oscillator strength.