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

Theory Notes

2012-06-22T23:55:52Z

Alkhaz: /* Matrix Elements of H */

==Helium Calculations==
<math> [-\frac{\hbar^2}{2m}(\nabla^2_1 +\nabla^2_2) - \frac{Ze^2}{r_1} -
\frac{Ze^2}{r_2}+\frac{e^2}{r^2_{12}} ]\psi = E\psi\nonumber
</math>

Define <math>\rho = \frac{Zr}{a_0}</math> where <math>a_0 = \frac{\hbar^2}{me^2}</math> (Bohr
radius). Then

<math>[-\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</math>

But <math>\frac{\hbar^2}{m}(\frac{me^2}{\hbar^2})^2 = \frac{e^2}{a_0}</math> is in
atomic units (au) of energy. Therefore

<math>[-\frac{1}{2}(\nabla^2_{\rho_1}+\nabla^2_{\rho_2}) -
\frac{1}{\rho_1} - \frac{1}{\rho_2} + \frac{Z^{-1}}{\rho_{12}}]\psi =
\varepsilon\psi\nonumber</math>
where <math>\varepsilon = \frac{Ea_0}{Z^2e^2}</math>

The problem to be solved is thus
<math>[\frac{1}{2}(\nabla^2_1+\nabla^2_2) -
\frac{1}{r_1}-\frac{1}{r_2} + \frac{Z^{-1}}{r_{12}}]\psi =
\varepsilon\psi\nonumber</math>

[figure to be inserted]

==The Hartree Fock Method==

Assume that <math>\psi({\bf r}_1,{\bf r}_2)</math> can be written in the form

<math style="horizontal-align:middle;">\psi({\bf r}_1,{\bf r}_2) = \frac{1}{\sqrt{2}}[u_1(r_1)u_2(r_2) \pm u_2(r_1)u_1(r_2)]\nonumber</math>

for the <math>1S^21S</math> ground state

<math>[-\frac{1}{2}(\nabla^2_1+\nabla^2_2) - \frac{1}{r_1}- \frac{1}{r_2} + \frac{Z^{-1}}{r_{12}}]\psi(r_1,r_2) = E\psi(r_1,r_2)\nonumber</math>

Substitute into <math><\psi|H-E|\psi></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>. ie

<math>\frac{1}{2}<\delta u_1(r_1)u_2(r_2) \pm u_2(r_1)\delta
u_1(r_2)|H-E|u_1(r_1)u_2(r_2)\pm u_2(r_1)u_1(r_2)>\nonumber</math>

<math>=\int\delta u_1(r_1)d{\bf r}_1\{\int d{\bf r}_2u_2(r_2)(H-E)[u_1(r_1)u_2(r_2)\pm u_2(r_1)u_1(r_2)]\}\nonumber</math>

<math>= 0 \ \ \ for \ arbitrary \ \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

<math>\int d{\bf r}_1 u_1(r_1)(H-E)[u_1(r_1)u_2(r_2) \pm
u_2(r_1)u_1(r_2)] = 0\nonumber</math>

Define

<math>I_{12} = \int dru_1(r)u_2(r), \nonumber</math>

<math>I_{21} = \int dru_1(r)u_2(r), \nonumber</math>

<math>H_{ij} = \int d{\bf r}u_i(-\frac{1}{2}\nabla - \frac{1}{r})u_j(r), \nonumber</math>

<math>G_{ij}(r) = \int d{\bf r}^\prime u_i(r^\prime)\frac{1}{|{\bf r} - {\bf r}\prime|}u_j(r^\prime)\nonumber</math>

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

<math>[ 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</math>

<math>[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</math>

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 H.F. energy is <math>E \simeq -2.87\cdots a.u.</math> while the exact energy is <math>E = -2.903724\cdots a.u.</math>

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
H.F. equations only describe how one electron moves in the average field
provided by the other.

==Configuration Interaction==
Expand <math> \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)\Upsilon^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)\Upsilon^0_{2,2,0}(\hat{\bf r}_1, \hat{\bf r}_2)+... \pm</math> exchange where <math> \Upsilon^M_{l_1,l_2,L}(\hat{\bf r}_1, \hat{\bf r}_2)&=&\Sigma_{m_1,m_2}\Upsilon^{m_1}_{l_1}({\bf r}_1)\Upsilon^{m_2}_{l_2}({\bf r}_2)\times <l_1l_2m_1m_2\mid LM> </math>.

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

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

and writing the trial functions in the form

<math>
\Psi({\bf r}_1,{\bf r}_2) = \sum^{1+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 exchange\nonumber
</math>

Diagonalizing H in this non-orthogonal basis set is equivalent to solving
<math>
\frac{\partial E}{\partial c_{i,j,k}} = 0\nonumber
</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
<math>$r_{12}^2 = r_1^2 + r_2^2 - 2r_1r_2\cos(\Theta_{12})$</math> and
<math>$\cos(\Theta_{12})=\frac{4\pi}{3}\sum^1_{m=-1}Y^{m*}_l(\theta_1,\varphi_1)Y^m_l(\theta_2,\varphi_2)$</math>
consider first 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
<math>
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
</math>

For P-states, one would have similarly

<math>\begin{eqnarray}
&r_{12}^0&\ \ \ \ \ \ \ \ (sp)P\nonumber\\
&r_{12}^2&\ \ \ \ \ \ \ \ (pd)P\nonumber\\
&r_{12}^4&\ \ \ \ \ \ \ \ (df)P\nonumber\\
&\vdots& \ \ \ \ \ \ \ \ \ \ \vdots\nonumber
\end{eqnarray}</math>

For D-states

<math>\begin{eqnarray}
&r_{12}^0&\ \ \ \ \ \ \ \ (sp)D\ \ \ \ \ \ \ \ (pp^\prime)D\nonumber\\
&r_{12}^2&\ \ \ \ \ \ \ \ (pd)D\ \ \ \ \ \ \ \ (dd^\prime)D\nonumber\\
&r_{12}^4&\ \ \ \ \ \ \ \ (df)D\ \ \ \ \ \ \ \ (ff^\prime)D\nonumber\\
&\vdots& \ \ \ \ \ \ \ \ \ \ \vdots\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \vdots\nonumber\end{eqnarray}</math>

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

<math>\begin{eqnarray}
\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)\nonumber\\
&+&\sum d_{ijk}r_1^{i+1}r_2^{j+1}r_{12}e^{-\alpha^\prime r_1 - \beta^\prime
r_2}\mathcal{Y}^M_{112}(\hat{r}_1,\hat{r}_2)\nonumber
\end{eqnarray}</math>

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
Stern-Liouville problem

<math>
\left(-\frac{1}{2}\nabla^2-\frac{\lambda}{r_s}-E\right)\psi({\bf r}) = 0\nonumber
</math>
or
<math>
\left(-\frac{1}{2}\frac{1}{r^2}\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
</math>
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>

INSERT FIGURE HERE

<math>
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
</math>

with <math>$\alpha = (-2E)^{1/2}$</math> and <math>$n\geq l+1$</math>.

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

-ie <math>\int^\infty_0 r^2dru_{n^\prime l}(r)\frac{1}{r}u_{nl}(r) = \delta_{n,n^\prime}\nonumber
</math>

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

[see B Klahn and W.A. Bingel Theo. Chim. Acta (Berlin) <math>{\bf 44}, 9</math> and <math>27
(1977)</math>].

==Solutions of the Eigenvalue Problem==

For convenience, write

<math>\Psi({\bf r}_1,{\bf r}_2) = \sum^N_{m=1}c_m\varphi_m\nonumber </math>

where <math>$m=$ m'th</math> combinations of <math>$i,j,k$</math>

<math>\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 exchange.\nonumber

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

Therefore
<math>
(\cos^2(\theta)-\sin^2(\theta))H_{12} =
\cos(\theta)\sin(\theta)(H_{11}-H_{22})\nonumber
</math>
and
<math>
\tan(2\theta) = \frac{2H_{12}}{H_{11}-H{22}}\nonumber
</math>

ie

<math>\begin{eqnarray}
\cos(\theta)=\left(\frac{r+\omega}{2r}\right)^{1/2}\nonumber,\\
\sin(\theta)=-sgn(H_{12})\left(\frac{r-\omega}{2r}\right)^{1/2}\nonumber
\end{eqnarray}\\</math>

where

<math>\begin{eqnarray}
\omega &=& H_{22}-H_{11}\nonumber\\
r&=&\left(\omega^2+4H_{12}^2\right)^{1/2}\nonumber\\
E_1&=&\frac{1}{2}\left(H_{11}+H_{22}-r\right)\nonumber\\
E_2&=&\frac{1}{2}\left(H_{11}+H_{22}+r\right)\nonumber
\end{eqnarray}</math>

===Brute Force Method===

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

-First orthonormalize the basis set - ie form linear combinations

<math>
\Phi_m = \sum_n\varphi_nR_{nm}\nonumber
</math>
such that <math>$<\Phi_m|\Phi_n> = \delta_{m,n}$\\</math>.
This can be done by finding an orthogonal tranformation, T, such that

<math>
T^TOT=I=\left(
\begin{array}{cccc}
I_1 & 0 & \ldots & 0\\
0 & I_2 & \ & 0 \\
\vdots & \ & I_3 & 0 \\
0 & 0 & 0 & \ddots
\end{array}\right);
\ \ O_{mn} = <\varphi_m|\varphi_n>\nonumber
</math>
and then applying a scale change matrix
<math>
S = \left(\begin{array}{cccc}
\frac{1}{I_1^{1/2}} & 0 & \ldots & 0\\
0 & \frac{1}{I_2^{1/2}} & \ & 0 \\
\vdots & \ & \frac{1}{I_3^{1/2}} & 0 \\
0 & 0 & 0 & \ddots
\end{array}\right)= S^T\nonumber
</math>

Then <math>$S^TT^TOTS = 1$\\</math>. ie <math>$R^TOR = 1$</math> with <math>$R=TS$</math>.

If H is the matrix with elements <math>$H_{mn}=<\varphi_m|\varphi_n>$</math>, then H
expressed in the <math>$\Phi_m$</math> basis set is
<math>
H^\prime = R^THR.\nonumber
</math>

We next diagonalize <math>$H^\prime$</math> by finding an orthogonal transformation W such
that
<math>
W^TH^\prime W = \lambda = \left(
\begin{array}{cccc}
\lambda_1 & 0 & \ldots & 0\\
0 & \lambda_2& \ & 0 \\
\vdots & \ & \ddots & 0 \\
0 & 0 & 0 & \lambda_N
\end{array}\right)\nonumber
</math>

The q'th eigenvector is

<math>\begin{eqnarray}
\Psi^{(q)} &=& \sum_n\Phi_n W_{n,q}\nonumber \\
&=& \sum_{n,n^\prime}\varphi_{n^\prime}R_{n^\prime ,n}W_{n,q}.\nonumber
\end{eqnarray}</math>
ie. <math>$c_{n^prime}^{(q)} = \sum_n R_{n^\prime n} W_{n,q}$</math>.

===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>$\chi = \left(\nonumber\\
\begin{array}{c}
a_1\\
a_2\\
\vdots\nonumber
\end{array}\right)\nonumber$</math> is an arbitrary starting vector, then <math>$\chi = \sum_q
x_q\Psi^{(q)}$\nonumber</math>.

<math>\begin{eqnarray}
(H)^n\chi &=& \sum_q x_q \lambda^n_q\Psi^{(q)}\nonumber\\
&\rightarrow& x_M\lambda_M^n\Psi^{(M)}\nonumber
\end{eqnarray}\\</math>
provided <math>$x_M\neq 0$\\</math>.

To pick out the eigenvector correspondng to any eigenvalue, with the original
problem in the form

<math>\begin{eqnarray}
H\Psi &=& \lambda O\Psi\nonumber\\
(H-\lambda)qO)\Psi\nonumber &=& (\lambda - \lambda_q)O\Psi\nonumber\\
\end{eqnarray}</math>

Therefore,
<math>
G\Psi = \frac{1}{\lambda-\lambda_q}\Psi\nonumber\\
</math>
where <math>$G=(H-\lambda_qO)^{-1}O$\nonumber</math> with eigenvalues
<math>$\frac{1}{\lambda_n-\lambda_q}\nonumber$\\</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>$n=n^\prime$\nonumber</math> than for any other value. The sequence is then

<math>\begin{eqnarray}
\chi_1&=&G\chi\nonumber\\
\chi_2&=&G\chi_1\nonumber\\
\chi_3&=&G\chi_2\nonumber\\
\vdots\nonumber
\end{eqnarray}\\ </math>

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

<math>
F\chi_n = (\lambda-\lambda_q)O\chi_{n-1}\nonumber\\
</math>

where <math>$F = H-\lambda_qO$</math>. The factor of <math>$(\lambda - \lambda_q)$</math> can be
dropped because this only affects the normalization of <math>$\chi_n$</math>. To find
<math>$\chi_n$\\</math>, solve
<math>
F\chi_n = O\chi_{n-1}\nonumber\\
</math>
(N equations in N unknowns). Then

<math>
\lambda = \frac{<\chi_n|H|\chi_n>}{<\chi_n|\chi_n>}\nonumber
</math>

==Matrix Elements of H==

<math>
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\\
</math>

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

<math>\begin{eqnarray}
\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))\frac{1}{r_{12}}\frac{\partial^2}{\partial r_1 \partial r} \nonumber &-& 2(\nabla_1 \cdot {\bf r}_2)\frac{1}{r}\frac{\partial}{\partial r}\nonumber
\end{eqnarray}</math>

where
INSERT FIGURE HERE

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

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

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 <math> d\tau=-r_1^2dr_1\sin(\theta_1)d\theta_1d\varphi_1r_{12}^2dr_{12}\sin(\psi)d\psi d\chi\nonumber\\</math>

and use <math>r_2^2=r_1^2+r_{12}^2 +2r_1r_{12}\cos(\psi).\\ </math>

Then for fixed <math>$r_1$</math> and <math>$r_{12}$\\</math>, <math> 2r_2dr_2 = -2r_1r_{12}\sin(\psi)d\psi\\ </math>

Thus <math> d\tau= r_1dr_1r_2dr_2r_{12}dr_{12}\sin(\theta_1)d\theta_1d\varphi_1d\chi\\ </math>

The basic type of integral to be calculated is

<math>\begin{eqnarray}
I(l_1,m_1,l_2,m_2;R) &=&\int\sin(\theta_1)d\theta_1d\varphi_1d\chi Y^{m_1}_{l_1}(\theta_1,\varphi_1)^{*}Y^{m_2}_{l_2}(\theta_2,\varphi_2)\nonumber\\
&\times&\int r_1dr_1r_2dr_2r_{12}dr_{12}R(r_1,r_2,r_{12})\nonumber\\
\end{eqnarray}\\</math>

Consider first the angular integral. <math>$Y^{m_2}_{l_2}(\theta_2,\varphi_2)$\\</math> can be expressed in terms of the independent variables <math>$\theta_1, \varphi_1,\chi$\\</math> by use of the rotation matrix relation

<math> 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,\varphi)\nonumber\\</math>

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

<math>\begin{eqnarray}
I_{ang}&=&\int^{2\pi}_0d\chi\int^{2\pi}_0d\varphi_1\int^\pi_0\sin(\theta_1)d=theta_1Y^{m_1}_{l_1}(\theta_1,\varphi_1)^*\nonumber\\
&\times&\sum_m\mathcal{D}^{(l_2)}_{m_2,m}(\varphi_1,\theta_1,\chi)^*Y^m_{l_2}(\theta,\varphi)\nonumber\\
\end{eqnarray}\\</math>

Use

<math> Y^{m_1}_{l_1}(\theta_1,\varphi_1)^* = \sqrt{\frac{2l_1+1}{4\pi}}\mathcal{D}^{(l_1)}_{m_1,0}(\varphi_1,\theta_1,\chi)\nonumber\\ </math>

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

<math> \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\\ </math>

to obtain

<math>\begin{eqnarray}
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}(\theta,\varphi)\nonumber\\
&=&2\pi\delta_{l_1,l_2}\delta_{m_1,m_2}P_{l_2}(\cos\theta)\nonumber\\
\end{eqnarray}\\</math>

since

<math>$Y^0_{l_2}(\theta,\varphi)=\sqrt{\frac{2l_1+1}{4\pi}}P_{l_2}(\cos(\theta))$\nonumber\\</math>.

Note that <math>$P_{l_2}(\cos\theta)$</math> is just a short hand expression for a radial function because <math>$\cos\theta = \frac{r_1^2+r_2^2-r_{12}^2}
{2r_1r_2}$\nonumber\\</math>.

The original integral is thus

<math>\begin{eqnarray}
I(l_1,m_1,l_2,m_2;R)&=&2\pi\delta_{l_1l_2}\delta_{m_1m_2}\nonumber\\
&\times&\int^\infty_0r_1dr_1\int^\infty_0r_2dr_2\int^{r_1+r_2}_{|r_1-r_2|} r_{12}dr_{12}R(r_1,r_2,r_{12})P_{l_2}(\cos\theta)\nonumber\\
\end{eqnarray}\\</math>

where <math>$\cos\theta = (r_1^2 +r^2_2-r_{12}^2)/(2r_1r_2)$\nonumber</math> is a purely radial function.

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

==Radial Integrals and Recursion Relations==

===The Radial Recursion Relation===

===The General Integral===

==Graphical Representation==
[figure to be inserted]

==Matrix Elements of H==

===Problem===

==General Hermitean Property==

==Optimization of Non-linear Parameters==
*Difficulties
*Cure

==The Screened Hyrdogenic Term==

==Small Corrections==
*Mass Polarization

Theory Notes

2012-06-22T23:54:11Z

Alkhaz: /* Matrix Elements of H */

==Helium Calculations==
<math> [-\frac{\hbar^2}{2m}(\nabla^2_1 +\nabla^2_2) - \frac{Ze^2}{r_1} -
\frac{Ze^2}{r_2}+\frac{e^2}{r^2_{12}} ]\psi = E\psi\nonumber
</math>

Define <math>\rho = \frac{Zr}{a_0}</math> where <math>a_0 = \frac{\hbar^2}{me^2}</math> (Bohr
radius). Then

<math>[-\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</math>

But <math>\frac{\hbar^2}{m}(\frac{me^2}{\hbar^2})^2 = \frac{e^2}{a_0}</math> is in
atomic units (au) of energy. Therefore

<math>[-\frac{1}{2}(\nabla^2_{\rho_1}+\nabla^2_{\rho_2}) -
\frac{1}{\rho_1} - \frac{1}{\rho_2} + \frac{Z^{-1}}{\rho_{12}}]\psi =
\varepsilon\psi\nonumber</math>
where <math>\varepsilon = \frac{Ea_0}{Z^2e^2}</math>

The problem to be solved is thus
<math>[\frac{1}{2}(\nabla^2_1+\nabla^2_2) -
\frac{1}{r_1}-\frac{1}{r_2} + \frac{Z^{-1}}{r_{12}}]\psi =
\varepsilon\psi\nonumber</math>

[figure to be inserted]

==The Hartree Fock Method==

Assume that <math>\psi({\bf r}_1,{\bf r}_2)</math> can be written in the form

<math style="horizontal-align:middle;">\psi({\bf r}_1,{\bf r}_2) = \frac{1}{\sqrt{2}}[u_1(r_1)u_2(r_2) \pm u_2(r_1)u_1(r_2)]\nonumber</math>

for the <math>1S^21S</math> ground state

<math>[-\frac{1}{2}(\nabla^2_1+\nabla^2_2) - \frac{1}{r_1}- \frac{1}{r_2} + \frac{Z^{-1}}{r_{12}}]\psi(r_1,r_2) = E\psi(r_1,r_2)\nonumber</math>

Substitute into <math><\psi|H-E|\psi></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>. ie

<math>\frac{1}{2}<\delta u_1(r_1)u_2(r_2) \pm u_2(r_1)\delta
u_1(r_2)|H-E|u_1(r_1)u_2(r_2)\pm u_2(r_1)u_1(r_2)>\nonumber</math>

<math>=\int\delta u_1(r_1)d{\bf r}_1\{\int d{\bf r}_2u_2(r_2)(H-E)[u_1(r_1)u_2(r_2)\pm u_2(r_1)u_1(r_2)]\}\nonumber</math>

<math>= 0 \ \ \ for \ arbitrary \ \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

<math>\int d{\bf r}_1 u_1(r_1)(H-E)[u_1(r_1)u_2(r_2) \pm
u_2(r_1)u_1(r_2)] = 0\nonumber</math>

Define

<math>I_{12} = \int dru_1(r)u_2(r), \nonumber</math>

<math>I_{21} = \int dru_1(r)u_2(r), \nonumber</math>

<math>H_{ij} = \int d{\bf r}u_i(-\frac{1}{2}\nabla - \frac{1}{r})u_j(r), \nonumber</math>

<math>G_{ij}(r) = \int d{\bf r}^\prime u_i(r^\prime)\frac{1}{|{\bf r} - {\bf r}\prime|}u_j(r^\prime)\nonumber</math>

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

<math>[ 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</math>

<math>[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</math>

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 H.F. energy is <math>E \simeq -2.87\cdots a.u.</math> while the exact energy is <math>E = -2.903724\cdots a.u.</math>

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
H.F. equations only describe how one electron moves in the average field
provided by the other.

==Configuration Interaction==
Expand <math> \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)\Upsilon^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)\Upsilon^0_{2,2,0}(\hat{\bf r}_1, \hat{\bf r}_2)+... \pm</math> exchange where <math> \Upsilon^M_{l_1,l_2,L}(\hat{\bf r}_1, \hat{\bf r}_2)&=&\Sigma_{m_1,m_2}\Upsilon^{m_1}_{l_1}({\bf r}_1)\Upsilon^{m_2}_{l_2}({\bf r}_2)\times <l_1l_2m_1m_2\mid LM> </math>.

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

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

and writing the trial functions in the form

<math>
\Psi({\bf r}_1,{\bf r}_2) = \sum^{1+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 exchange\nonumber
</math>

Diagonalizing H in this non-orthogonal basis set is equivalent to solving
<math>
\frac{\partial E}{\partial c_{i,j,k}} = 0\nonumber
</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
<math>$r_{12}^2 = r_1^2 + r_2^2 - 2r_1r_2\cos(\Theta_{12})$</math> and
<math>$\cos(\Theta_{12})=\frac{4\pi}{3}\sum^1_{m=-1}Y^{m*}_l(\theta_1,\varphi_1)Y^m_l(\theta_2,\varphi_2)$</math>
consider first 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
<math>
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
</math>

For P-states, one would have similarly

<math>\begin{eqnarray}
&r_{12}^0&\ \ \ \ \ \ \ \ (sp)P\nonumber\\
&r_{12}^2&\ \ \ \ \ \ \ \ (pd)P\nonumber\\
&r_{12}^4&\ \ \ \ \ \ \ \ (df)P\nonumber\\
&\vdots& \ \ \ \ \ \ \ \ \ \ \vdots\nonumber
\end{eqnarray}</math>

For D-states

<math>\begin{eqnarray}
&r_{12}^0&\ \ \ \ \ \ \ \ (sp)D\ \ \ \ \ \ \ \ (pp^\prime)D\nonumber\\
&r_{12}^2&\ \ \ \ \ \ \ \ (pd)D\ \ \ \ \ \ \ \ (dd^\prime)D\nonumber\\
&r_{12}^4&\ \ \ \ \ \ \ \ (df)D\ \ \ \ \ \ \ \ (ff^\prime)D\nonumber\\
&\vdots& \ \ \ \ \ \ \ \ \ \ \vdots\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \vdots\nonumber\end{eqnarray}</math>

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

<math>\begin{eqnarray}
\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)\nonumber\\
&+&\sum d_{ijk}r_1^{i+1}r_2^{j+1}r_{12}e^{-\alpha^\prime r_1 - \beta^\prime
r_2}\mathcal{Y}^M_{112}(\hat{r}_1,\hat{r}_2)\nonumber
\end{eqnarray}</math>

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
Stern-Liouville problem

<math>
\left(-\frac{1}{2}\nabla^2-\frac{\lambda}{r_s}-E\right)\psi({\bf r}) = 0\nonumber
</math>
or
<math>
\left(-\frac{1}{2}\frac{1}{r^2}\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
</math>
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>

INSERT FIGURE HERE

<math>
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
</math>

with <math>$\alpha = (-2E)^{1/2}$</math> and <math>$n\geq l+1$</math>.

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

-ie <math>\int^\infty_0 r^2dru_{n^\prime l}(r)\frac{1}{r}u_{nl}(r) = \delta_{n,n^\prime}\nonumber
</math>

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

[see B Klahn and W.A. Bingel Theo. Chim. Acta (Berlin) <math>{\bf 44}, 9</math> and <math>27
(1977)</math>].

==Solutions of the Eigenvalue Problem==

For convenience, write

<math>\Psi({\bf r}_1,{\bf r}_2) = \sum^N_{m=1}c_m\varphi_m\nonumber </math>

where <math>$m=$ m'th</math> combinations of <math>$i,j,k$</math>

<math>\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 exchange.\nonumber

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

Therefore
<math>
(\cos^2(\theta)-\sin^2(\theta))H_{12} =
\cos(\theta)\sin(\theta)(H_{11}-H_{22})\nonumber
</math>
and
<math>
\tan(2\theta) = \frac{2H_{12}}{H_{11}-H{22}}\nonumber
</math>

ie

<math>\begin{eqnarray}
\cos(\theta)=\left(\frac{r+\omega}{2r}\right)^{1/2}\nonumber,\\
\sin(\theta)=-sgn(H_{12})\left(\frac{r-\omega}{2r}\right)^{1/2}\nonumber
\end{eqnarray}\\</math>

where

<math>\begin{eqnarray}
\omega &=& H_{22}-H_{11}\nonumber\\
r&=&\left(\omega^2+4H_{12}^2\right)^{1/2}\nonumber\\
E_1&=&\frac{1}{2}\left(H_{11}+H_{22}-r\right)\nonumber\\
E_2&=&\frac{1}{2}\left(H_{11}+H_{22}+r\right)\nonumber
\end{eqnarray}</math>

===Brute Force Method===

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

-First orthonormalize the basis set - ie form linear combinations

<math>
\Phi_m = \sum_n\varphi_nR_{nm}\nonumber
</math>
such that <math>$<\Phi_m|\Phi_n> = \delta_{m,n}$\\</math>.
This can be done by finding an orthogonal tranformation, T, such that

<math>
T^TOT=I=\left(
\begin{array}{cccc}
I_1 & 0 & \ldots & 0\\
0 & I_2 & \ & 0 \\
\vdots & \ & I_3 & 0 \\
0 & 0 & 0 & \ddots
\end{array}\right);
\ \ O_{mn} = <\varphi_m|\varphi_n>\nonumber
</math>
and then applying a scale change matrix
<math>
S = \left(\begin{array}{cccc}
\frac{1}{I_1^{1/2}} & 0 & \ldots & 0\\
0 & \frac{1}{I_2^{1/2}} & \ & 0 \\
\vdots & \ & \frac{1}{I_3^{1/2}} & 0 \\
0 & 0 & 0 & \ddots
\end{array}\right)= S^T\nonumber
</math>

Then <math>$S^TT^TOTS = 1$\\</math>. ie <math>$R^TOR = 1$</math> with <math>$R=TS$</math>.

If H is the matrix with elements <math>$H_{mn}=<\varphi_m|\varphi_n>$</math>, then H
expressed in the <math>$\Phi_m$</math> basis set is
<math>
H^\prime = R^THR.\nonumber
</math>

We next diagonalize <math>$H^\prime$</math> by finding an orthogonal transformation W such
that
<math>
W^TH^\prime W = \lambda = \left(
\begin{array}{cccc}
\lambda_1 & 0 & \ldots & 0\\
0 & \lambda_2& \ & 0 \\
\vdots & \ & \ddots & 0 \\
0 & 0 & 0 & \lambda_N
\end{array}\right)\nonumber
</math>

The q'th eigenvector is

<math>\begin{eqnarray}
\Psi^{(q)} &=& \sum_n\Phi_n W_{n,q}\nonumber \\
&=& \sum_{n,n^\prime}\varphi_{n^\prime}R_{n^\prime ,n}W_{n,q}.\nonumber
\end{eqnarray}</math>
ie. <math>$c_{n^prime}^{(q)} = \sum_n R_{n^\prime n} W_{n,q}$</math>.

===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>$\chi = \left(\nonumber\\
\begin{array}{c}
a_1\\
a_2\\
\vdots\nonumber
\end{array}\right)\nonumber$</math> is an arbitrary starting vector, then <math>$\chi = \sum_q
x_q\Psi^{(q)}$\nonumber</math>.

<math>\begin{eqnarray}
(H)^n\chi &=& \sum_q x_q \lambda^n_q\Psi^{(q)}\nonumber\\
&\rightarrow& x_M\lambda_M^n\Psi^{(M)}\nonumber
\end{eqnarray}\\</math>
provided <math>$x_M\neq 0$\\</math>.

To pick out the eigenvector correspondng to any eigenvalue, with the original
problem in the form

<math>\begin{eqnarray}
H\Psi &=& \lambda O\Psi\nonumber\\
(H-\lambda)qO)\Psi\nonumber &=& (\lambda - \lambda_q)O\Psi\nonumber\\
\end{eqnarray}</math>

Therefore,
<math>
G\Psi = \frac{1}{\lambda-\lambda_q}\Psi\nonumber\\
</math>
where <math>$G=(H-\lambda_qO)^{-1}O$\nonumber</math> with eigenvalues
<math>$\frac{1}{\lambda_n-\lambda_q}\nonumber$\\</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>$n=n^\prime$\nonumber</math> than for any other value. The sequence is then

<math>\begin{eqnarray}
\chi_1&=&G\chi\nonumber\\
\chi_2&=&G\chi_1\nonumber\\
\chi_3&=&G\chi_2\nonumber\\
\vdots\nonumber
\end{eqnarray}\\ </math>

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

<math>
F\chi_n = (\lambda-\lambda_q)O\chi_{n-1}\nonumber\\
</math>

where <math>$F = H-\lambda_qO$</math>. The factor of <math>$(\lambda - \lambda_q)$</math> can be
dropped because this only affects the normalization of <math>$\chi_n$</math>. To find
<math>$\chi_n$\\</math>, solve
<math>
F\chi_n = O\chi_{n-1}\nonumber\\
</math>
(N equations in N unknowns). Then

<math>
\lambda = \frac{<\chi_n|H|\chi_n>}{<\chi_n|\chi_n>}\nonumber
</math>

==Matrix Elements of H==

<math>
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\\
</math>

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

<math>\begin{eqnarray}
\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))\frac{1}{r_{12}}\frac{\partial^2}{\partial r_1 \partial r} \nonumber &-& 2(\nabla_1 \cdot {\bf r}_2)\frac{1}{r}\frac{\partial}{\partial r}\nonumber
\end{eqnarray}</math>

where
INSERT FIGURE HERE

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

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

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 <math> d\tau=-r_1^2dr_1\sin(\theta_1)d\theta_1d\varphi_1r_{12}^2dr_{12}\sin(\psi)d\psi d\chi\nonumber\\</math>

and use <math>r_2^2=r_1^2+r_{12}^2 +2r_1r_{12}\cos(\psi).\\ </math>

Then for fixed <math>$r_1$</math> and <math>$r_{12}$\\</math>, <math> 2r_2dr_2 = -2r_1r_{12}\sin(\psi)d\psi\\ </math>

Thus <math> d\tau= r_1dr_1r_2dr_2r_{12}dr_{12}\sin(\theta_1)d\theta_1d\varphi_1d\chi\\ </math>

The basic type of integral to be calculated is

<math>\begin{eqnarray}
I(l_1,m_1,l_2,m_2;R) &=&\int\sin(\theta_1)d\theta_1d\varphi_1d\chi Y^{m_1}_{l_1}(\theta_1,\varphi_1)^{*}Y^{m_2}_{l_2}(\theta_2,\varphi_2)\nonumber\\
&\times&\int r_1dr_1r_2dr_2r_{12}dr_{12}R(r_1,r_2,r_{12})\nonumber\\
\end{eqnarray}\\</math>

Consider first the angular integral. <math>$Y^{m_2}_{l_2}(\theta_2,\varphi_2)$\\</math> can be expressed in terms of the independent variables <math>$\theta_1, \varphi_1,\chi$\\</math> by use of the rotation matrix relation

<math> 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,\varphi)\nonumber\\</math>

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

<math>\begin{eqnarray}
I_{ang}&=&\int^{2\pi}_0d\chi\int^{2\pi}_0d\varphi_1\int^\pi_0\sin(\theta_1)d=theta_1Y^{m_1}_{l_1}(\theta_1,\varphi_1)^*\nonumber\\
&\times&\sum_m\mathcal{D}^{(l_2)}_{m_2,m}(\varphi_1,\theta_1,\chi)^*Y^m_{l_2}(\theta,\varphi)\nonumber\\
\end{eqnarray}\\</math>

Use

<math> Y^{m_1}_{l_1}(\theta_1,\varphi_1)^* = \sqrt{\frac{2l_1+1}{4\pi}}\mathcal{D}^{(l_1)}_{m_1,0}(\varphi_1,\theta_1,\chi)\nonumber\\ </math>

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

<math> \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\\ </math>

to obtain

<math>\begin{eqnarray}
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}(\theta,\varphi)\nonumber\\
&=&2\pi\delta_{l_1,l_2}\delta_{m_1,m_2}P_{l_2}(\cos\theta)\nonumber\\
\end{eqnarray}\\</math>

since

<math>$Y^0_{l_2}(\theta,\varphi)=\sqrt{\frac{2l_1+1}{4\pi}}P_{l_2}(\cos(\theta))$\nonumber\\</math>.

Note that <math>$P_{l_2}(\cos\theta)$</math> is just a short hand expression for a radial function because <math>$\cos\theta = \frac{r_1^2+r_2^2-r_{12}^2}
{2r_1r_2}$\nonumber\\</math>.

The original integral is thus

<math>\begin{eqnarray}
I(l_1,m_1,l_2,m_2;R)&=&2\pi\delta_{l_1l_2}\delta_{m_1m_2}\nonumber\\
&\times&\int^\infty_0r_1dr_1\int^\infty_0r_2dr_2\int^{r_1+r_2}_{|r_1-r_2|} r_{12}dr_{12}R(r_1,r_2,r_{12})P_{l_2}(\cos\theta)\nonumber\\
\end{eqnarray}\\</math>

where <math>$\cos\theta = (r_1^2 +r^2_2-r_{12}^2)/(2r_1r_2)$\nonumber</math> is a purely radial function.

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

<math>$P_1(\cos\theta)=\cos\theta$\nonumber\\</math>.

==Radial Integrals and Recursion Relations==

===The Radial Recursion Relation===

===The General Integral===

==Graphical Representation==
[figure to be inserted]

==Matrix Elements of H==

===Problem===

==General Hermitean Property==

==Optimization of Non-linear Parameters==
*Difficulties
*Cure

==The Screened Hyrdogenic Term==

==Small Corrections==
*Mass Polarization

Theory Notes

2012-06-22T22:40:43Z

Alkhaz: /* The Power Method */

==Helium Calculations==
<math> [-\frac{\hbar^2}{2m}(\nabla^2_1 +\nabla^2_2) - \frac{Ze^2}{r_1} -
\frac{Ze^2}{r_2}+\frac{e^2}{r^2_{12}} ]\psi = E\psi\nonumber
</math>

Define <math>\rho = \frac{Zr}{a_0}</math> where <math>a_0 = \frac{\hbar^2}{me^2}</math> (Bohr
radius). Then

<math>[-\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</math>

But <math>\frac{\hbar^2}{m}(\frac{me^2}{\hbar^2})^2 = \frac{e^2}{a_0}</math> is in
atomic units (au) of energy. Therefore

<math>[-\frac{1}{2}(\nabla^2_{\rho_1}+\nabla^2_{\rho_2}) -
\frac{1}{\rho_1} - \frac{1}{\rho_2} + \frac{Z^{-1}}{\rho_{12}}]\psi =
\varepsilon\psi\nonumber</math>
where <math>\varepsilon = \frac{Ea_0}{Z^2e^2}</math>

The problem to be solved is thus
<math>[\frac{1}{2}(\nabla^2_1+\nabla^2_2) -
\frac{1}{r_1}-\frac{1}{r_2} + \frac{Z^{-1}}{r_{12}}]\psi =
\varepsilon\psi\nonumber</math>

[figure to be inserted]

==The Hartree Fock Method==

Assume that <math>\psi({\bf r}_1,{\bf r}_2)</math> can be written in the form

<math style="horizontal-align:middle;">\psi({\bf r}_1,{\bf r}_2) = \frac{1}{\sqrt{2}}[u_1(r_1)u_2(r_2) \pm u_2(r_1)u_1(r_2)]\nonumber</math>

for the <math>1S^21S</math> ground state

<math>[-\frac{1}{2}(\nabla^2_1+\nabla^2_2) - \frac{1}{r_1}- \frac{1}{r_2} + \frac{Z^{-1}}{r_{12}}]\psi(r_1,r_2) = E\psi(r_1,r_2)\nonumber</math>

Substitute into <math><\psi|H-E|\psi></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>. ie

<math>\frac{1}{2}<\delta u_1(r_1)u_2(r_2) \pm u_2(r_1)\delta
u_1(r_2)|H-E|u_1(r_1)u_2(r_2)\pm u_2(r_1)u_1(r_2)>\nonumber</math>

<math>=\int\delta u_1(r_1)d{\bf r}_1\{\int d{\bf r}_2u_2(r_2)(H-E)[u_1(r_1)u_2(r_2)\pm u_2(r_1)u_1(r_2)]\}\nonumber</math>

<math>= 0 \ \ \ for \ arbitrary \ \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

<math>\int d{\bf r}_1 u_1(r_1)(H-E)[u_1(r_1)u_2(r_2) \pm
u_2(r_1)u_1(r_2)] = 0\nonumber</math>

Define

<math>I_{12} = \int dru_1(r)u_2(r), \nonumber</math>

<math>I_{21} = \int dru_1(r)u_2(r), \nonumber</math>

<math>H_{ij} = \int d{\bf r}u_i(-\frac{1}{2}\nabla - \frac{1}{r})u_j(r), \nonumber</math>

<math>G_{ij}(r) = \int d{\bf r}^\prime u_i(r^\prime)\frac{1}{|{\bf r} - {\bf r}\prime|}u_j(r^\prime)\nonumber</math>

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

<math>[ 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</math>

<math>[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</math>

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 H.F. energy is <math>E \simeq -2.87\cdots a.u.</math> while the exact energy is <math>E = -2.903724\cdots a.u.</math>

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
H.F. equations only describe how one electron moves in the average field
provided by the other.

==Configuration Interaction==
Expand <math> \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)\Upsilon^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)\Upsilon^0_{2,2,0}(\hat{\bf r}_1, \hat{\bf r}_2)+... \pm</math> exchange where <math> \Upsilon^M_{l_1,l_2,L}(\hat{\bf r}_1, \hat{\bf r}_2)&=&\Sigma_{m_1,m_2}\Upsilon^{m_1}_{l_1}({\bf r}_1)\Upsilon^{m_2}_{l_2}({\bf r}_2)\times <l_1l_2m_1m_2\mid LM> </math>.

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

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

and writing the trial functions in the form

<math>
\Psi({\bf r}_1,{\bf r}_2) = \sum^{1+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 exchange\nonumber
</math>

Diagonalizing H in this non-orthogonal basis set is equivalent to solving
<math>
\frac{\partial E}{\partial c_{i,j,k}} = 0\nonumber
</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
<math>$r_{12}^2 = r_1^2 + r_2^2 - 2r_1r_2\cos(\Theta_{12})$</math> and
<math>$\cos(\Theta_{12})=\frac{4\pi}{3}\sum^1_{m=-1}Y^{m*}_l(\theta_1,\varphi_1)Y^m_l(\theta_2,\varphi_2)$</math>
consider first 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
<math>
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
</math>

For P-states, one would have similarly

<math>\begin{eqnarray}
&r_{12}^0&\ \ \ \ \ \ \ \ (sp)P\nonumber\\
&r_{12}^2&\ \ \ \ \ \ \ \ (pd)P\nonumber\\
&r_{12}^4&\ \ \ \ \ \ \ \ (df)P\nonumber\\
&\vdots& \ \ \ \ \ \ \ \ \ \ \vdots\nonumber
\end{eqnarray}</math>

For D-states

<math>\begin{eqnarray}
&r_{12}^0&\ \ \ \ \ \ \ \ (sp)D\ \ \ \ \ \ \ \ (pp^\prime)D\nonumber\\
&r_{12}^2&\ \ \ \ \ \ \ \ (pd)D\ \ \ \ \ \ \ \ (dd^\prime)D\nonumber\\
&r_{12}^4&\ \ \ \ \ \ \ \ (df)D\ \ \ \ \ \ \ \ (ff^\prime)D\nonumber\\
&\vdots& \ \ \ \ \ \ \ \ \ \ \vdots\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \vdots\nonumber\end{eqnarray}</math>

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

<math>\begin{eqnarray}
\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)\nonumber\\
&+&\sum d_{ijk}r_1^{i+1}r_2^{j+1}r_{12}e^{-\alpha^\prime r_1 - \beta^\prime
r_2}\mathcal{Y}^M_{112}(\hat{r}_1,\hat{r}_2)\nonumber
\end{eqnarray}</math>

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
Stern-Liouville problem

<math>
\left(-\frac{1}{2}\nabla^2-\frac{\lambda}{r_s}-E\right)\psi({\bf r}) = 0\nonumber
</math>
or
<math>
\left(-\frac{1}{2}\frac{1}{r^2}\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
</math>
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>

INSERT FIGURE HERE

<math>
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
</math>

with <math>$\alpha = (-2E)^{1/2}$</math> and <math>$n\geq l+1$</math>.

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

-ie <math>\int^\infty_0 r^2dru_{n^\prime l}(r)\frac{1}{r}u_{nl}(r) = \delta_{n,n^\prime}\nonumber
</math>

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

[see B Klahn and W.A. Bingel Theo. Chim. Acta (Berlin) <math>{\bf 44}, 9</math> and <math>27
(1977)</math>].

==Solutions of the Eigenvalue Problem==

For convenience, write

<math>\Psi({\bf r}_1,{\bf r}_2) = \sum^N_{m=1}c_m\varphi_m\nonumber </math>

where <math>$m=$ m'th</math> combinations of <math>$i,j,k$</math>

<math>\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 exchange.\nonumber

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

Therefore
<math>
(\cos^2(\theta)-\sin^2(\theta))H_{12} =
\cos(\theta)\sin(\theta)(H_{11}-H_{22})\nonumber
</math>
and
<math>
\tan(2\theta) = \frac{2H_{12}}{H_{11}-H{22}}\nonumber
</math>

ie

<math>\begin{eqnarray}
\cos(\theta)=\left(\frac{r+\omega}{2r}\right)^{1/2}\nonumber,\\
\sin(\theta)=-sgn(H_{12})\left(\frac{r-\omega}{2r}\right)^{1/2}\nonumber
\end{eqnarray}\\</math>

where

<math>\begin{eqnarray}
\omega &=& H_{22}-H_{11}\nonumber\\
r&=&\left(\omega^2+4H_{12}^2\right)^{1/2}\nonumber\\
E_1&=&\frac{1}{2}\left(H_{11}+H_{22}-r\right)\nonumber\\
E_2&=&\frac{1}{2}\left(H_{11}+H_{22}+r\right)\nonumber
\end{eqnarray}</math>

===Brute Force Method===

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

-First orthonormalize the basis set - ie form linear combinations

<math>
\Phi_m = \sum_n\varphi_nR_{nm}\nonumber
</math>
such that <math>$<\Phi_m|\Phi_n> = \delta_{m,n}$\\</math>.
This can be done by finding an orthogonal tranformation, T, such that

<math>
T^TOT=I=\left(
\begin{array}{cccc}
I_1 & 0 & \ldots & 0\\
0 & I_2 & \ & 0 \\
\vdots & \ & I_3 & 0 \\
0 & 0 & 0 & \ddots
\end{array}\right);
\ \ O_{mn} = <\varphi_m|\varphi_n>\nonumber
</math>
and then applying a scale change matrix
<math>
S = \left(\begin{array}{cccc}
\frac{1}{I_1^{1/2}} & 0 & \ldots & 0\\
0 & \frac{1}{I_2^{1/2}} & \ & 0 \\
\vdots & \ & \frac{1}{I_3^{1/2}} & 0 \\
0 & 0 & 0 & \ddots
\end{array}\right)= S^T\nonumber
</math>

Then <math>$S^TT^TOTS = 1$\\</math>. ie <math>$R^TOR = 1$</math> with <math>$R=TS$</math>.

If H is the matrix with elements <math>$H_{mn}=<\varphi_m|\varphi_n>$</math>, then H
expressed in the <math>$\Phi_m$</math> basis set is
<math>
H^\prime = R^THR.\nonumber
</math>

We next diagonalize <math>$H^\prime$</math> by finding an orthogonal transformation W such
that
<math>
W^TH^\prime W = \lambda = \left(
\begin{array}{cccc}
\lambda_1 & 0 & \ldots & 0\\
0 & \lambda_2& \ & 0 \\
\vdots & \ & \ddots & 0 \\
0 & 0 & 0 & \lambda_N
\end{array}\right)\nonumber
</math>

The q'th eigenvector is

<math>\begin{eqnarray}
\Psi^{(q)} &=& \sum_n\Phi_n W_{n,q}\nonumber \\
&=& \sum_{n,n^\prime}\varphi_{n^\prime}R_{n^\prime ,n}W_{n,q}.\nonumber
\end{eqnarray}</math>
ie. <math>$c_{n^prime}^{(q)} = \sum_n R_{n^\prime n} W_{n,q}$</math>.

===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>$\chi = \left(\nonumber\\
\begin{array}{c}
a_1\\
a_2\\
\vdots\nonumber
\end{array}\right)\nonumber$</math> is an arbitrary starting vector, then <math>$\chi = \sum_q
x_q\Psi^{(q)}$\nonumber</math>.

<math>\begin{eqnarray}
(H)^n\chi &=& \sum_q x_q \lambda^n_q\Psi^{(q)}\nonumber\\
&\rightarrow& x_M\lambda_M^n\Psi^{(M)}\nonumber
\end{eqnarray}\\</math>
provided <math>$x_M\neq 0$\\</math>.

To pick out the eigenvector correspondng to any eigenvalue, with the original
problem in the form

<math>\begin{eqnarray}
H\Psi &=& \lambda O\Psi\nonumber\\
(H-\lambda)qO)\Psi\nonumber &=& (\lambda - \lambda_q)O\Psi\nonumber\\
\end{eqnarray}</math>

Therefore,
<math>
G\Psi = \frac{1}{\lambda-\lambda_q}\Psi\nonumber\\
</math>
where <math>$G=(H-\lambda_qO)^{-1}O$\nonumber</math> with eigenvalues
<math>$\frac{1}{\lambda_n-\lambda_q}\nonumber$\\</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>$n=n^\prime$\nonumber</math> than for any other value. The sequence is then

<math>\begin{eqnarray}
\chi_1&=&G\chi\nonumber\\
\chi_2&=&G\chi_1\nonumber\\
\chi_3&=&G\chi_2\nonumber\\
\vdots\nonumber
\end{eqnarray}\\ </math>

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

<math>
F\chi_n = (\lambda-\lambda_q)O\chi_{n-1}\nonumber\\
</math>

where <math>$F = H-\lambda_qO$</math>. The factor of <math>$(\lambda - \lambda_q)$</math> can be
dropped because this only affects the normalization of <math>$\chi_n$</math>. To find
<math>$\chi_n$\\</math>, solve
<math>
F\chi_n = O\chi_{n-1}\nonumber\\
</math>
(N equations in N unknowns). Then

<math>
\lambda = \frac{<\chi_n|H|\chi_n>}{<\chi_n|\chi_n>}\nonumber
</math>

==Matrix Elements of H==

==Radial Integrals and Recursion Relations==

===The Radial Recursion Relation===

===The General Integral===

==Graphical Representation==
[figure to be inserted]

==Matrix Elements of H==

===Problem===

==General Hermitean Property==

==Optimization of Non-linear Parameters==
*Difficulties
*Cure

==The Screened Hyrdogenic Term==

==Small Corrections==
*Mass Polarization

Theory Notes

2012-06-22T22:09:33Z

Alkhaz:

Theory Notes

2012-06-22T21:58:48Z

Alkhaz:

Theory Notes

2012-06-22T21:33:42Z

Alkhaz:

Theory Notes

2012-06-20T15:13:51Z

Alkhaz:

Theory Notes

2012-06-20T15:06:23Z

Alkhaz: /* Hylleraas Coordinates */

Theory Notes

2012-06-20T15:05:20Z

Alkhaz:

Theory Notes

2012-06-20T14:57:53Z

Alkhaz:

Theory Notes

2012-06-20T14:54:02Z

Alkhaz:

Theory Notes

2012-06-20T14:48:25Z

Alkhaz: /* Hylleraas Coordinates */

Theory Notes

2012-06-20T14:47:04Z

Alkhaz: /* Completeness */

Theory Notes

2012-06-20T14:39:09Z

Alkhaz: /* Hylleras Co-ordinates */

Theory Notes

2012-06-20T14:30:25Z

Alkhaz: /* Hylleras Co-ordinates */

Theory Notes

2012-06-20T14:16:20Z

Alkhaz: /* Hylleras Co-ordinates */

Theory Notes

2012-06-20T14:15:42Z

Alkhaz: /* Hylleras Co-ordinates */

Theory Notes

2011-03-10T03:16:49Z

Alkhaz:

Theory Notes

2011-03-10T03:11:40Z

Alkhaz: /* Configuration Interaction */

Theory Notes

2011-03-10T03:08:22Z

Alkhaz:

Theory Notes

2011-03-10T03:00:31Z

Alkhaz:

Theory Notes

2011-03-10T02:50:27Z

Alkhaz: /* Configuration Interaction */

Theory Notes

2011-03-10T02:47:24Z

Alkhaz: /* Configuration Interaction */

Theory Notes

2011-03-10T02:46:45Z

Alkhaz: /* Configuration Interaction */

Theory Notes

2011-03-10T02:32:57Z

Alkhaz: /* Configuration Interaction */

Theory Notes

2011-03-10T02:31:30Z

Alkhaz: /* Configuration Interaction */

Theory Notes

2011-03-10T02:30:10Z

Alkhaz: /* Configuration Interaction */

Theory Notes

2011-03-10T02:26:14Z

Alkhaz:

Theory Notes

2011-03-10T02:13:16Z

Alkhaz:

Theory Notes

2011-03-10T02:05:29Z

Alkhaz:

Theory Notes

2011-03-01T01:37:54Z

Alkhaz:

Theory Notes

2011-03-01T01:32:44Z

Alkhaz: