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"*": "<math>$\\displaystyle\\sum\\limits_{a < b} \\langle \\delta^3 ({\\bf r_{ab}}) \\rangle $</math>\n\n<math>\\varepsilon_{\\infty}^{(5)}(log)</math>\n\n<math>\\varepsilon_{\\infty}^{(5)}(nlog)</math>\n\n<math>\\frac{Z^3}{\\pi n^3}</math>\n\n<math>$\\displaystyle\\sum\\limits_{a} \\langle \\delta^3 ({\\bf r_{a}}) \\rangle $</math>\n\n<math>$\\displaystyle\\sum\\limits_{a < b} \\langle \\delta^3 ({\\bf \\tilde{Q}}_{ab}) \\rangle $</math>"
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"*": "<center>\n<H1><p>Notes on solving the Schrödinger equation in Hylleraas coordinates</p><p> for heliumlike atoms</p>\n<p style=\"font-size: 75%;\">Gordon W.F. Drake</p>\n<p style=\"font-size: 75%;\">Department of Physics, University of Windsor</p>\n<p style=\"font-size: 75%;\">Windsor, Ontario, Canada N9B 3P4</p>\n<p style=\"font-size: 50%; text-align:left\">(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.)</p>\n</H1>\n</center>\n\n==Introduction==\n<p>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\nG.W.F. Drake, Phys. Rev. A <math>{\\bf 18}</math>, 820 (1978).</p>\n\nThe starting point is the two-electron Schrödinger equation for infinite nuclear mass\n\\begin{equation} \\left[-\\frac{\\hbar^2}{2m}(\\nabla^2_1 +\\nabla^2_2) - \\frac{Ze^2}{r_1} -\n \\frac{Ze^2}{r_2}+\\frac{e^2}{r_{12}} \\right]\\psi = E\\psi\\nonumber\n\\end{equation}\nwhere <math>m</math> is the electron mass, and <math>r_{12} = |{\\bf r}_1 - {\\bf r}_2|</math> (see diagram below).\n\n[[File:Pic1.jpg|250px]]\n\nBegin 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\n\n\\begin{equation}[-\\frac{\\hbar^2}{2m}Z^2(\\frac{me^2}{\\hbar^2})^2(\\nabla^2_{\\rho_1}+\\nabla^2_{\\rho_2})\n - Z^2\\frac{e^2}{a_0}\\rho^{-1}_1 - Z^2\\frac{e^2}{a_0}\\rho^{-1}_2 +\n \\frac{e^2}{a_0}Z\\rho^{-1}_{12}]\\psi= E\\psi\\nonumber\\end{equation}\n\nBut <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\n\n\\begin{equation} \\left[-\\frac{1}{2}(\\nabla^2_{\\rho_1}+\\nabla^2_{\\rho_2}) -\n \\frac{1}{\\rho_1} - \\frac{1}{\\rho_2} + \\frac{Z^{-1}}{\\rho_{12}}\\right]\\psi =\n \\varepsilon\\psi\\nonumber \\end{equation}\n\nwhere <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 \n\\begin{equation}\n H\\psi = \\varepsilon\\psi\\nonumber \\end{equation}\nwhere (using <math>r</math> in place of <math>\\rho</math> for the Z-scaled distance)\n\\begin{equation}\nH = -\\frac{1}{2}(\\nabla^2_1+\\nabla^2_2) - \\frac{1}{r_1}-\\frac{1}{r_2} + \\frac{Z^{-1}}{r_{12}}\n \\nonumber\\end{equation}\nis the atomic Hamiltonian for infinite nuclear mass.\n\n==The Hartree Fock Method==\n\nFor 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\n\\begin{equation}\n\\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\n\\end{equation}\nfor 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\n\\begin{equation}\nH\\psi({\\bf r}_1,{\\bf r}_2) = E\\psi({\\bf r}_1,{\\bf r}_2)\\nonumber\n\\end{equation}\nthat can nevertheless be expressed in this separable product form, where as before\n\\begin{equation}\nH = -\\frac{1}{2}(\\nabla^2_1+\\nabla^2_2) - \\frac{1}{r_1}- \\frac{1}{r_2} + \\frac{Z^{-1}}{r_{12}}\\nonumber\n\\end{equation} \nis 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\nwith respect to arbitrary infinitesimal variations <math>\\delta u_1</math> and <math>\\delta\nu_2</math> in <math>u_1</math> and <math>u_2</math> respectively; i.e.\n\n\\begin{equation}\n\\frac{1}{2} \\left\\langle \\delta u_1(r_1)u_2(r_2) \\pm u_2(r_1)\\delta\nu_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\n\\end{equation}\n\n\\begin{equation}\n=\\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\n\\end{equation}\n\n\\begin{equation}\n= 0 \\nonumber\n\\end{equation}\n\nfor arbitrary <math> \\delta u_1(r_1).\\nonumber </math> Therefore <math>\\{\\int d{\\bf r}_2 \\ldots \\} = 0</math>.\n\nSimilarly, the coefficient of <math>\\delta u_2</math> would give\n\n\\begin{equation}\n\\int d{\\bf r}_1 u_1(r_1) \\left( H-E \\right) \\left[ u_1(r_1) u_2(r_2) \\pm\n u_2(r_1) u_1(r_2) \\right] = 0\\nonumber\n\\end{equation}\n\nDefine \n\n\\begin{equation}\nI_{12} = I_{21} = \\int d{\\bf r}\\, u_1(r)u_2(r), \\nonumber\n\\end{equation}\n\n\\begin{equation}\nH_{ij} = \\int d{\\bf r}\\, u_i(r)(-\\frac{1}{2}\\nabla - \\frac{1}{r})u_j(r), \\nonumber\n\\end{equation}\n\n\\begin{equation}\nG_{ij}(r) = \\int d{\\bf r}^\\prime u_i(r^\\prime)\\frac{1}{|{\\bf r} - {\\bf r}\\prime|}u_j(r^\\prime)\\nonumber\n\\end{equation}\n\nThen the above equations become the pair of integro-differential equations\n\n\\begin{equation}\n[ 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\n\\end{equation}\n\n\\begin{equation}\n[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\n\\end{equation}\n\nThese must be solved self-consistently for the \"constants\" <math>I_{12}</math> and\n<math>H_{ij}</math> and the function <math>G_{ij}(r)</math>.\n\nThe 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.\n\nThe difference is called the \"correlation energy\" because it arises from the\nway in which the motion of one electron is correlated to the other. The\nHartree Fock equations only describe how one electron moves in the average field\nprovided by the other (mean-field theory).\n\n==Configuration Interaction==\nExpand\n\\begin{equation}\n\\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}\n\\end{equation}\nwhere \n\\begin{equation}\n{\\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>\n\\end{equation}\n\nThis works, but is slowly convergent, and very laborious. The best CI calculations are accurate to <math> ~10^{-7}</math> a.u.\n\n==Hylleraas Coordinates==\n[E.A. Hylleraas, Z. Phys. <math>{\\bf 48}, 469(1928)</math> and <math>{\\bf 54}, 347(1929)</math>]\nsuggested using the co-ordinates <math>r_1</math>, <math>r_2</math> and <math>r_{12}</math> or equivalently \n\n\\begin{eqnarray}\ns &=& r_1 + r_2, \\nonumber\\\\\nt &=& r_1-r_2, \\nonumber\\\\\nu &=& r_{12}\\nonumber\n\\end{eqnarray}\n\nand writing the trial functions in the form\n\n\\begin{equation}\n\\Psi({\\bf r}_1,{\\bf r}_2) = \\sum^{i+j+k\\leq\n 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}\n \\mathcal{Y}^M_{l_1,l_2,L}(\\hat{r}_1,\\hat{r}_2)\\pm \\text{exchange}\\nonumber\n\\end{equation}\n\nDiagonalizing H in this non-orthogonal basis set is equivalent to solving\n<math> \\begin{eqnarray}\n\\frac{\\partial E}{\\partial c_{i,j,k}} = 0\\nonumber\n\\end{eqnarray} </math>\nfor fixed <math>\\alpha</math> and <math>\\beta</math>.\n\nThe diagonalization must be repeated for different values of <math>\\alpha</math> and\n<math>\\beta</math> in order to optimize the non-linear parameters.\n\n==Completeness==\nThe completeness of the above basis set can be shown by first writing\n\\begin{equation}\nr_{12}^2 = r_1^2 + r_2^2 - 2r_1r_2\\cos\\theta_{12}\n\\end{equation}\nand, from the spherical harmonic addition theorem,\n\\begin{equation}\n\\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)\n\\end{equation}\nConsider first the S-states. The <math>r_{12}^0</math> terms are like the ss terms in a CI calculation. The\n<math>r_{12}^2</math> terms bring in p-p type contributions, and the higher powers bring\nin d-d, f-f etc. type terms. In general\n\n\\begin{equation}\nP_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\n\\end{equation}\n\nFor P-states, one would have similarly\n\n\\begin{array}{lr}\nr_{12}^0 & (sp)P\\nonumber\\\\\nr_{12}^2 & (pd)P\\nonumber\\\\\nr_{12}^4 & (df)P\\nonumber\\\\\n\\vdots & \\vdots\\nonumber\n\\end{array}\n\nFor D-states\n\n\\begin{array}{lr}\nr_{12}^0 & (sd)D & (pp^\\prime)D\\nonumber\\\\\nr_{12}^2 & (pf)D & (dd^\\prime)D\\nonumber\\\\\nr_{12}^4 & (dg)D & (ff^\\prime)D\\nonumber\\\\\n\\vdots & \\vdots & \\vdots\\nonumber\n\\end{array}\n\nIn this case, since there are two \"lowest-order\" couplings to form a\nD-state, both must be present in the basis set, i.e.\n\n\\begin{equation}\n\\Psi(r_2,r_2) = \\sum c_{ijk}r_1^ir_2^{j+2}r_{12}^ke^{-\\alpha r_1-\\beta\n r_2}\\mathcal{Y}^M_{022}(\\hat{r}_1,\\hat{r}_2)\n+\\sum d_{ijk}r_1^{i+1}r_2^{j+1}r_{12^k}e^{-\\alpha^\\prime r_1 - \\beta^\\prime\n r_2}\\mathcal{Y}^M_{112}(\\hat{r}_1,\\hat{r}_2)\\nonumber\n\\end{equation}\n\nFor F-states, one would need <math>(sf)F</math> and <math>(pd)F</math> terms.\n\nFor G-states, one would need <math>(sg)G</math>, <math>(pf)G</math> and <math>(dd^\\prime)G</math> terms.\n\nCompleteness of the radial functions can be proven by considering the\nSturm-Liouville problem\n\n\\begin{equation}\n\\left(-\\frac{1}{2}\\nabla^2-\\frac{\\lambda}{r_s}-E\\right)\\psi({\\bf r}) = 0\\nonumber\n\\end{equation}\n\nor\n\n\\begin{equation}\n\\left(-\\frac{1}{2}\\frac{1}{r^2}\\frac{\\partial}{\\partial r}\\left({r^2}\\frac{\\partial}{\\partial r}\\right) -\n \\frac{l\\left(l+1\\right)}{2r^2} - \\frac{\\lambda}{r} - E\\right)u(r) = 0.\\nonumber\n\\end{equation}\n\nFor fixed E and variable <math>\\lambda</math> (nuclear charge).\n\nThe eigenvalues are <math>\\lambda_n = (E/E_n)^{1/2}</math>, where <math>E_n =- \\frac{1}{2n^2}</math>\n\n[[File:Pic2.jpg|500px]]\n\nand the eigenfunctions are\n\n\\begin{equation}\nu_{nl}(r) =\n\\frac{1}{(2l+1)!}\\left(\\frac{(n+l)!}{(n-l-1)2!}\\right)^{1/2}(2\\alpha)^{3/2}e^{-\\alpha\n r}\\nonumber\n\\times \\ \\left( 2 \\alpha r \\right)^l\\,_{1}F_{1}\\left(-n+l+1,2l+2; 2 \\alpha r \\right)\n\\end{equation}\n\n\nwith <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.\n\n\nUnlike the hydrogen spectrum, which has both a discrete part for <math>E<0</math> and a\ncontinuous part for <math>E>0</math>, this forms an entirely discrete set of finite\npolynomials, called Sturmian functions. They are orthogonal with respect to\nthe potential, i.e.\n \n\\begin{equation}\n\\int^\\infty_0 r^2dr \\left( u_{n^\\prime l}(r)\\frac{1}{r}u_{nl}(r) \\right) = \\delta_{n,n^\\prime}\\nonumber\n\\end{equation}\n\n\nSince they become complete in the limit <math>n\\rightarrow\\infty</math>, this assures the\ncompleteness of the variational basis set.\n\n[See also B Klahn and W.A. Bingel Theo. Chim. Acta (Berlin) <strong>44</strong>, 9 and 27\n(1977)].\n\n==Solutions of the Eigenvalue Problem==\n\nFor convenience, write\n\n\n\\[\n\\Psi({\\bf r}_1,{\\bf r}_2) = \\sum^N_{m=1}c_m\\varphi_m\\nonumber\n\\]\n\nwhere <math>m</math> represents the <math>m</math>'th combination of <math>i,j,k</math> and\n\n\\begin{equation}\n\\varphi_{ijk}=r_1^ir_2^jr_{12}^ke^{-\\alpha r_1 - \\beta\n r_2}\\mathcal{Y}^M_{l_1,l_2,L}(\\hat{r}_1,\\hat{r}_2) \\pm \\mbox{exchange}\\nonumber\n\\end{equation}\n\n\n\\begin{align}\n\n&\n\\left( \\begin{array}{lr}\n\\cos(\\theta) & \\sin(\\theta)\\\\\n-\\sin(\\theta) & \\cos(\\theta)\n\\end{array} \\right)\n\\left( \\begin{array}{lr}\nH_{11} & H_{12} \\\\\nH_{12} & H_{22}\n\\end{array}\\right)\n\\left( \\begin{array}{lr}\n\\cos(\\theta) & -\\sin(\\theta)\\\\\n\\sin(\\theta) & \\cos(\\theta)\n\\end{array} \\right)\n\\\\[5pt]\n= & \\left(\\begin{array}{lr}\ncH_{11}+sH_{12} & cH_{12} + sH_{22}\\\\\n-sH_{11} + cH_{12} & -sH_{12} + cH_{22}\n\\end{array}\\right)\n\\left( \\begin{array}{cc}\nc & -s \\\\\ns & c\n\\end{array}\\right)\n\\\\[5pt]\n= & \\left(\\begin{array}{lr}\nc^2H_{11}+s^2H_{22} + 2csH_{12} & (c^2-s^2)H_{12}+cs(H_{22}-H_{11})\\\\\n(c^2-s^2)H_{12}+cs(H_{22}-H_{11}) & s^2H_{11}+c^2H_{22}-2csH_{12}\n\\end{array}\\right)\n\\end{align}\n\n\n\nTherefore\n\n\\begin{equation}\n\\left[\\cos^2(\\theta)-\\sin^2(\\theta)\\right]H_{12} =\n\\cos(\\theta)\\sin(\\theta)(H_{11}-H_{22})\\nonumber\n\\end{equation}\n\nand\n\n\\begin{equation}\n\\tan(2\\theta) = \\frac{2H_{12}}{H_{11}-H_{22}}\\nonumber\n\\end{equation}\n\ni.e.\n \n\\begin{equation}\n\\cos(\\theta)=\\left(\\frac{r+\\omega}{2r}\\right)^{1/2}\\nonumber\n\\end{equation}\n\n\\begin{equation}\n\\sin(\\theta)=-\\mbox{sgn}(H_{12})\\left(\\frac{r-\\omega}{2r}\\right)^{1/2}\\nonumber\n\\end{equation}\n\nwhere\n\n\\begin{equation}\n\\omega = H_{22}-H_{11}\\nonumber\n\\end{equation}\n\n\\begin{equation}\nr=\\left(\\omega^2+4H_{12}^2\\right)^{1/2}\\nonumber\n\\end{equation}\n\n\\begin{equation}\nE_1=\\frac{1}{2}\\left(H_{11}+H_{22}-r\\right)\\nonumber\n\\end{equation}\n\n\\begin{equation}\nE_2=\\frac{1}{2}\\left(H_{11}+H_{22}+r\\right)\\nonumber\n\\end{equation}\n\n===Brute Force Method===\n\n- Gives all the eigenvalues and eigenvectors, but it is slow\n\n- First orthonormalize the <math>N</math>-dimensional basis set; i.e. form linear combinations\n\n\\begin{equation}\n\\Phi_m = \\sum_{n=1}^N\\varphi_nR_{nm}\\nonumber\n\\end{equation}\n\nsuch that\n\\begin{equation}\n\\langle \\Phi_m | \\Phi_n \\rangle = \\delta_{m,n}\n\\end{equation}\n\nThis can be done by finding an orthogonal tranformation T such that\n\n\\begin{equation}\nT^TOT = I = \\left(\n\\begin{array}{ccccc}\nI_1 & 0 & 0 & \\ldots & 0 \\\\\n0 & I_2 & 0 & \\ldots & 0 \\\\\n0 & 0 & I_3 & \\ldots & 0 \\\\\n\\vdots & \\vdots & \\vdots & \\ddots & \\vdots \\\\\n0 & 0 & 0 & \\ldots & I_N\n\\end{array}\\right);\n\\end{equation}\n\n\\begin{equation}\nO_{mn} = \\langle \\varphi_m | \\varphi_n \\rangle \\nonumber\n\\end{equation}\n\nand then applying a scale change matrix\n\n\\begin{equation}\nS = \\left(\\begin{array}{ccccc}\n\\frac{1}{I_1^{1/2}} & 0 & 0 & \\ldots & 0\\\\\n0 & \\frac{1}{I_2^{1/2}} & 0 & \\ldots & 0 \\\\\n0 & 0 & \\frac{1}{I_3^{1/2}} & \\ldots & 0 \\\\\n\\vdots & \\vdots & \\vdots & \\ddots & \\vdots \\\\\n0 & 0 & 0 & \\ldots & I_N^{1/2}\n\\end{array}\\right)= S^T\\nonumber\n\\end{equation}\n\nThen\n\n\\begin{equation}\nS^TT^TOTS = 1\n\\end{equation}\n\ni.e.\n\n\\begin{equation}\nR^TOR = 1\n\\end{equation}\n\nwith <math>R=TS</math>.\n\nIf H is the matrix with elements <math>H_{mn} = \\langle \\varphi_m | \\varphi_n \\rangle </math>, then H\nexpressed in the <math>\\Phi_m</math> basis set is\n\n\\begin{equation}\nH^\\prime = R^THR.\\nonumber\n\\end{equation}\n\nWe next diagonalize <math>H^\\prime</math> by finding an orthogonal transformation W such\nthat\n\n\\begin{equation}\nW^T H^\\prime W = \\lambda = \\left(\n\\begin{array}{cccc}\n\\lambda_1 & 0 & \\ldots & 0 \\\\\n0 & \\lambda_2 & \\ldots & 0 \\\\\n\\vdots & \\vdots & \\ddots & \\vdots \\\\\n0 & 0 & \\ldots & \\lambda_N\n\\end{array}\\right)\\nonumber\n\\end{equation}\n\nThe q'th eigenvector is\n\n\\begin{equation}\n\\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\n\\end{equation}\n\ni.e.\n\n\\begin{equation}\nc_{n^\\prime}^{(q)} = \\sum_{n=1}^N R_{n^\\prime n} W_{n,q}\n\\end{equation}\n\n===The Power Method===\n\n- Based on the observation that if H has one eigenvalue, <math>\\lambda_M</math>, much bigger than all the rest, and <math>\\chi = \\left(\n\\begin{array}{c}a_1\\\\a_2\\\\\\vdots\\end{array}\\right)</math>\nis an arbitrary starting vector, then <math>\\chi = \\sum_{q=1}^N x_q\\Psi^{(q)}</math>.\n\n\\begin{equation}\n(H)^n\\chi = \\sum_{q=1}^N x_q \\lambda^n_q\\Psi^{(q)}\\nonumber\\rightarrow x_M\\lambda_M^n\\Psi^{(M)}\\nonumber\n\\end{equation}\n\nprovided <math>x_M\\neq 0</math>.\n\nTo pick out the eigenvector corresponding to any eigenvalue, write the original problem in the form\n\n\\begin{align}\nH\\Psi & = \\lambda O \\Psi \\nonumber \\\\\n\\left( H - \\lambda_q O \\right) \\Psi & = \\left( \\lambda - \\lambda_q \\right) O \\Psi \\nonumber\n\\end{align}\n\nTherefore,\n\n\\begin{equation}\nG \\Psi = \\frac{1}{\\lambda - \\lambda_q } \\Psi \\nonumber \\\\\n\\end{equation}\n\nwhere <math>G=(H-\\lambda_qO)^{-1}O</math> with eigenvalues <math>\\frac{1}{\\lambda_n - \\lambda_q}</math>.\n\nBy picking <math>\\lambda_q</math> close to any one of the <math>\\lambda_n</math>, say\n<math>\\lambda_{n^\\prime}</math>, then <math>\\frac{1}{\\lambda_n-\\lambda_q}</math> is much larger for\n<math>n=n^\\prime</math> than for any other value. The sequence is then\n\n\\begin{array}{c}\n\\chi_1=G\\chi\\nonumber\\\\\n\\chi_2=G\\chi_1\\nonumber\\\\\n\\chi_3=G\\chi_2\\nonumber\\\\\n\\vdots\\nonumber\n\\end{array}\n\nuntil the ratios of components in <math>\\chi_n</math> stop changing.\n\n- To avoid matrix inversion and multiplication, note that the sequence is equivalent to\n\\begin{equation}\nF \\chi_n = \\left( \\lambda - \\lambda_q \\right) O \\chi_{n-1} \\nonumber \\\\\n\\end{equation}\nwhere <math>F = H-\\lambda_qO</math>. The factor of <math>\\left( \\lambda - \\lambda_q \\right)</math> can be\ndropped because this only affects the normalization of <math>\\chi_n</math>. To find <math>\\chi_n</math>, solve\n\\begin{equation}\nF \\chi_n = O \\chi_{n-1} \\nonumber\n\\end{equation}\n(N equations in N unknowns). Then\n\\begin{equation}\n\\lambda = \\frac{ \\langle \\chi_n | H | \\chi_n \\rangle}{\\langle \\chi_n | \\chi_n \\rangle} \\nonumber\n\\end{equation}\n\n==Matrix Elements of H==\n\n\\begin{equation}\nH=-\\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\n\\end{equation}\n\nTaking <math>r_1, r_2</math> and <math>r_{12}</math> as independent variables,\n\n\\begin{equation}\n\\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}}\n\\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\n\\end{equation}\n\nwhere <math>\\nabla_1^Y</math> acts only on the spherical harmonic part of the wave function and the diagram\n\n[[File:Pic3.jpg|400px]]\n\ndefines 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>.\n\nIf <math>r_{12}</math> were not an independent variable, then one could take the volume element to be\n\n\\begin{equation}\nd\\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\n\\end{equation}\n\nHowever, <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 \norigin of a new polar co-ordinate system, and write\n\n\\begin{equation}\nd\\tau=-r_1^2dr_1\\sin(\\theta_1)d\\theta_1d\\varphi_1r_{12}^2dr_{12}\\sin(\\psi)d\\psi d\\chi\\nonumber\n\\end{equation}\n\nand use\n\n\\begin{equation}\nr_2^2=r_1^2+r_{12}^2 +2r_1r_{12}\\cos(\\psi)\n\\end{equation}\n\nThen for fixed <math>r_1</math> and <math>r_{12}</math>,\n\n\\begin{equation}\n2r_2dr_2 = -2r_1r_{12}\\sin(\\psi)d\\psi\n\\end{equation}\n\nThus\n\n\\begin{equation}\nd\\tau= r_1dr_1r_2dr_2r_{12}dr_{12}\\sin(\\theta_1)d\\theta_1d\\varphi_1d\\chi\n\\end{equation}\n\nThe basic type of integral to be calculated is\n\n\\begin{align}\nI \\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) \\\\ \n& \\times \\int r_1dr_1r_2dr_2r_{12}dr_{12}R \\left( r_1, r_2, r_{12} \\right)\n\\end{align}\n\nConsider 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\n\\begin{equation}\nY^{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)\n\\end{equation}\nwhere <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\n\n\\begin{align}\nI_{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)^{*} \\\\\n& \\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) \\\\\n\\end{align}\n\nUse\n\n\\begin{equation}\nY^{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)\n\\end{equation}\n\ntogether with the orthogonality property of the rotation matrices (Brink and Satchler, p 147)\n\n\\begin{align}\n& \\int \\mathcal{D}^{(j)*}_{m,m^\\prime}\\mathcal{D}^{(J)}_{M,M^\\prime}\\sin(\\theta_1)d\\theta_1d\\varphi_1d\\chi \\\\\n& = \\frac{8\\pi^2}{2j+1}\\delta_{jJ}\\delta_{mM}\\delta_{m^\\prime M^\\prime}\\nonumber\n\\end{align}\n\nto obtain\n\n\\begin{align}\nI_{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) \\\\\n& = 2 \\pi \\delta_{l_1,l_2} \\delta_{m_1,m_2} P_{l_2} \\left( \\cos \\theta_{12} \\right)\n\\end{align}\n\nsince\n\n\\begin{equation}\nY^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)\n\\end{equation}\n\nNote that <math>P_{l_2} \\left( \\cos \\theta \\right)</math> is just a short hand expression for a radial function because\n\n\\begin{equation}\n\\cos \\theta_{12} = \\frac{r_1^2 + r_2^2 - r_{12}^2}{2r_1r_2}\n\\end{equation}\n\nThe original integral is thus\n\n\\begin{equation}\nI \\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)\n\\end{equation}\n\nwhere again\n\n\\begin{equation}\n\\cos \\theta_{12} = \\frac{r_{1}^{2} +r^{2}_{2} - r_{12}^{2}}{2 r_{1} r_{2}}\n\\end{equation}\n\nis a purely radial function.\n\nThe 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>.\n\n==Radial Integrals and Recursion Relations==\n\nThe basic radial integral is [see G.W.F. Drake, Phys. Rev. A 18, 820 (1978)]\n\n\\begin{align}\nI_0 \\left( a, b, c \\right) & =\n\\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}} \\\\\n& = \\frac{2}{c+2} \\sum^{[(c+1)/2]}_{i=0}\n\\left(\\begin{array}{c}\nc+2 \\\\\n2i+1\n\\end{array} \\right)\n\\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\\}\n\\end{align}\n\nwhere\n\n\\begin{array}{ll}\np = a + 2i + 2 & p^{\\prime} = b + 2i + 2 \\\\\nq = b + c - 2i + 2 & q^{\\prime} = a + c - 2i + 2\n\\end{array}\n\nThe above applies for <math> a, b \\geq -2, c\\geq -1</math>.\n<math>[x]</math> means \"greatest integer in\" <math>x</math>.\n\nThen \n\n\\begin{align}\nI_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) \\\\\n& = \\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)\n\\end{align}\n\n===The Radial Recursion Relation===\n\nRecall that the full integral is\n\n\\begin{equation}\nI(l_1m_1,l_2m_2;R)=2\\pi\\delta_{l_1,l_2}\\delta_{m_1,m_2}I_{l_2}(R)\n\\end{equation}\n\nwhere\n\n\\begin{equation}\nI_{l_2}(R) = \\int d\\tau_r R(r_1,r_2,r_{12})P_{l_2}(\\cos \\theta_{12})\n\\end{equation}\n\nTo obtain the recursion relation, use\n\n\\begin{equation}\nP_l(x) = \\frac{[P^{\\prime}_{l+1}(x)-P^{\\prime}_{l-1}(x)]}{2l+1}\n\\end{equation}\n\nwith\n\n\\begin{equation}\nP_{l+1}^\\prime (x) = \\frac{d}{dx}P_{l+1}(x)\n\\end{equation}\n\nHere <math>x=\\cos\\theta_{12}</math> and\n\n\\begin{align}\n\\frac{d}{d \\cos \\theta_{12}} & = \\frac{dr_{12}}{d \\cos \\theta_{12}} \\frac{d}{dr_{12}} \\\\\n& = -\\frac{r_{1} r_{2}}{r_{12}} \\frac{d}{dr_{12}}\n\\end{align}\n\nThen,\n\n\\begin{equation}\nI_{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}\n\\end{equation}\n\nThe <math>r_{12}</math> part of the integral is\n\n\\begin{align}\n& \\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] \\\\\n& = 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}\n\\end{align}\n\nThe integrated term vanishes because\n\n\\begin{equation} \n\\cos \\theta_{12} = \\frac{r_{1}^{2} + r_{2}^{2} - r_{12}^{2}}{2 r_{1} r_{2}}\n\\begin{array}{lll}\n& = -1 & {\\rm when} & r_{12}^{2} = \\left( r_{1} + r_{2} \\right)^{2} \\\\\n& = 1 & {\\rm when} & r_{12}^{2} = \\left( r_{1} - r_{2} \\right)^{2}\n\\end{array}\n\\end{equation}\n\nand <math>P_l(1)=1</math>, <math>P_{l}(-1)=(-1)^{l}</math>.\n\nThus,\n\n\\begin{equation}\nI_{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)\n\\end{equation}\n\nIf\n\n\\begin{equation}\nR = r_{1}^{a-1} r_{2}^{b-1} r_{12}^{c+2} e^{-\\alpha r_{1} - \\beta r_{2}}\n\\end{equation}\n\nthen [G.W.F. Drake, Phys. Rev. A 18, 820 (1978)]\n\n\\begin{equation}\nI_{l+1}(r_1^ar_2^br_{12}^c) =\n\\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)\n\\end{equation}\n\nFor the case <math>c=-2</math>, take\n\n\\begin{equation}\nR = r_{1}^{a-1} r_{2}^{b-1} \\ln r_{12} e^{ -\\alpha r_{1} -\\beta r_{2}}\n\\end{equation}\n\nThen\n\n\\begin{equation}\nI_{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)\n\\end{equation}\n\nThis allows all <math>I_{l}</math> integrals to be calculated from tables of <math>I_0</math> and\n<math>I_1</math> integrals.\n\n===The General Integral===\n\nThe above results for the angular and radial integrals can now be combined into a general formula for integrals of the type\n\n\\begin{equation}\nI = \\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)\n\\end{equation}\n\nwhere \n\n\\begin{equation}\n\\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)\n\\end{equation}\n\nand\n\n\\begin{equation}\nT^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)\n\\end{equation}\n\nThe basic idea is to make repeated use of the formula\n\n\\begin{equation}\nY^{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}\n\\times \\left( \\begin{array}{ccc}\nl_1 & l_2 & l \\\\\nm_1 & m_2 & m\n\\end{array} \\right) \\left(\\begin{array}{ccc}\nl_1 & l_2 & l \\\\\n0 & 0 & 0\n\\end{array} \\right)\nY^{m*}_{l} \\left( \\hat{r}_{1} \\right)\n\\end{equation}\n\nwhere\n\n\\begin{equation}\nY^{m*}(\\hat{r}) = (-1)^mY_l^{-m}(\\hat{r})\n\\end{equation}\n\nand \n\n\\begin{equation}\n\\left(\\begin{array}{ccc}\nl_1 & l_2 & l \\\\\nm_1 & m_2 & m\n\\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)\n\\end{equation}\n\nis a 3-<math>j</math> symbol. In particular, write\n\n\\begin{array}{cc}\nY^{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)\\\\\n\\ \\ \\ \\ \\ \\ \\ \\sum_{\\lambda_1\\mu_1}Y_{\\lambda_1}^{\\mu_1}(\\hat{r}_1) & & \\\\\n& & \\\\\nY^{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) \\\\\n\\ \\ \\ \\ \\ \\ \\ \\ \\ \\sum_{\\lambda_{2} \\mu_{2}}Y_{\\lambda_{2}}^{\\mu_{2}} \\left( \\hat{r}_{2} \\right) & &\n\\end{array}\n\nThe 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\n\n\\begin{equation}\nI = \\sum_{\\Lambda} C_{\\Lambda} I_{\\Lambda} \\left( R_{1} R_{2} \\right)\n\\end{equation}\n\nwhere <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).\n\n==Graphical Representation==\n\n[[File:GraphRep1.jpg|500px]]\n\n==Matrix Elements of H==\nRecall that\n\n\\begin{equation}\nH = -\\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}}\n\\end{equation}\n\nConsider matrix elements of \n\n\\begin{equation}\n\\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}\n\\end{equation}\n\nwhere <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>.\n\nA general matrix element is\n\n\\begin{equation}\n\\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\n\\end{equation}\n\nSince <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:\n\nPut\n\n\\begin{equation}\n\\mathcal{F} = F \\mathcal{Y}^M_{l_1 l_2 L} \\left( \\hat{r}_1, \\hat{r_2} \\right)\n\\end{equation}\n\nand\n\n\\begin{equation}\nF = r_1^a r_2^b r_{12}^c e^{-\\alpha r_1 - \\beta r_2}\n\\end{equation}\n\nThen\n\n\\begin{equation}\n\\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}\n\\end{equation}\n\nand\n\n\\begin{align}\n\\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 \\\\\n& + \\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\n\\end{align}\n\nwhere\n\n\\begin{equation}\n\\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\n\\end{equation}\n\nFor 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\n\n\\begin{equation}\n(\\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}\n+ \\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)\n\\end{equation}\n\nOperating to the left gives\n\n\\begin{equation}\n(\\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}\n+ \\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)\n\\end{equation}\n\nNow put\n\n\\begin{array}{cc}\na_+ = a+a^\\prime, & \\hat{\\nabla}_1^+ = \\hat{\\nabla}_1^Y + \\hat{\\nabla}_1^{Y\\prime} \\\\\na_{-} = a-a^\\prime, & \\hat{\\nabla}_1^{-} = \\hat{\\nabla}_1^Y - \\hat{\\nabla}_1^{Y\\prime}\n\\end{array}\n\netc., 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\n\n\\begin{equation}\n(\\nabla^2_1)_R = (\\nabla^2_1)_L\n\\end{equation}\n\nmust be true for arbitrary <math>a</math>, <math>c</math>, and <math>\\alpha</math>. Their coefficients must thus vanish.\n\nThis yields the integral relations\n\n\\begin{equation}\n\\label{I}\n\\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}\n\\end{equation}\n\nfrom the coefficient of <math>a</math> and \n\n\\begin{equation}\n\\label{II}\n\\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}\n\\end{equation}\n\nfrom the coefficient of <math>c</math>. The coefficient of <math>\\alpha</math> gives an equation equivalent to (I).\n\nFurthermore, if can be show that (see problem)\n\n\\begin{align}\n& \\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) \\\\\n= & \\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)\n\\end{align}\n\nand 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\n\n\\begin{equation}\n\\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)\n\\end{equation}\n\nand\n\n\\begin{equation}\n\\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]\n\\end{equation}\n\nwhere equality applies after integration and summation over <math>\\Lambda</math>.\n\nThus (II) becomes \n\n\\begin{equation}\\label{IIa}\n\\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}\n\\end{equation}\n\n===Problem===\n\nProve the integral relation\n\n\\begin{align}\n& \\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) \\\\\n= & \\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)\n\\end{align}\n\nwhere the coefficients <math>C_\\Lambda \\left( 1 \\right)</math> are the angular coefficients from the overlap integral \n\n\\begin{align}\n\\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)\\\\\n= & \\sum_\\Lambda C_\\Lambda(1)P_\\Lambda(\\cos\\theta_{12}).\n\\end{align}\n\nHint: Use the fact that <math>l_1^2</math> is Hermitian so that \n<math>\n\\int d\\tau (l_1^2\\mathcal{Y}^\\prime)^*q(r)\\mathcal{Y} = \\int d\\tau\n\\mathcal{Y}^{\\prime *}l_1^2(q(r)\\mathcal{Y})\n</math>\nwith <math>\\ \\vec{l}_1= \\frac{1}{i}\\vec{r}_1\\times\\nabla_1</math>.\n\nIt is also useful to use\n<math>\\begin{eqnarray}\\ \\ \n\\cos\\theta\\, P_L(\\cos\\theta) &=& \\frac{1}{2L+1}[L\\,P_{L-1}(\\cos\\theta) + (L+1)P_{L+1}(\\cos\\theta)]\n\\\\\n(\\cos^2\\theta-1)P_L(\\cos\\theta) &=&\n\\frac{L(L-1)}{(2L-1)(2L+1)}P_{L-2}(\\cos\\theta)\\nonumber\n-\\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\\\\\n&=& \\frac{L(L-1)}{(2L-1)(2L+1)}[P_{L-2}(\\cos\\theta) - P_L(\\cos\\theta)]\n + \\frac{(L+1)(L+2)}{(2L+1)(2L+3)}[P_{L+2}(\\cos\\theta) - P_L(\\cos\\theta)]\n\\end{eqnarray}</math>\n\ntogether with a double application of the integral recursion relation\n<math>\nI_{L+1}\\left(\\frac{1}{r}\\frac{d}{dr}q(r)\\right) =\n(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)\n</math>\n\nOf 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\n<math>\nl_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})\n</math>, and show that\n\n<math>\n\\begin{eqnarray}\\ \\\n(\\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}\\\\\nl_1^2g(r) &=& 2r_1r_2\\cos\\theta\\frac{1}{r}\\frac{dg}{dr} + r_1^2r_2^2(\\cos^2\\theta - 1)\n\\frac{1}{r}\\frac{d}{dr}\\left(\\frac{1}{r}\\frac{dg}{dr}\\right).\n\\end{eqnarray}\n</math>\n\nThe proof amounts to showing that the term <math>l_1^2g(r)</math> can be replaced by \n<math>\\Lambda(\\Lambda+1)g(r)</math> after multiplying by <math>P_\\Lambda(\\cos\\theta)</math>\nand integrating by parts with respect to the radial integrations over <math>r_1</math>, <math>r_2</math> and \n<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>.\n\n==General Hermitian Property==\n\nEach combination of terms of the form\n\\begin{equation}\n<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)\n\\end{equation}\ncan be rewritten\n\\begin{equation}\n<f>_L= (a_+-a)^2f_1+(a_+-a)f_2 + (a_+-a)(b_+-b)f_3+(b_+-b)^2f_4\n+(b_+-b)f_5+(a_+-a)f_6\\nabla^{Y\\prime}_1+(b_+-b)f_7\\nabla^{Y\\prime}_2+f_8)Y^\\prime\n\\end{equation}\nwhere 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>.\nSince these must be equal for arbitrary <math>a</math> and <math>b</math>,\n\\begin{equation}\na_+^2f_1+a_+f_2+_+b_+f_3+b_+f_4+b_+f_5+a_+f_6\\nabla^{Y\\prime}_1 +\nb_+f_7\\nabla^{Y\\prime}_2+f_8(Y^\\prime)-f_8(Y)=0\n\\end{equation}\n\nAdding the corresponding expression with <math>Y</math> and <math>Y^\\prime</math> interchanged\nyields\n<math>\\label{III}\na_+^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}\n</math>\n\nSubtracting gives\n\\begin{eqnarray*}\nf_8(Y)-f_8(Y^\\prime)&=&-\\frac{1}{2}[a_+f_6\\nabla^-_1+b_+f_7\\nabla_2^-]\\tag{IV}\\\\\na[-2a_+f_1-2f_2-b_+f_3-f_6\\nabla^+_1]&=&0\\tag{V}\\\\\nb[-2b_+f_4-2f_5-a_+f_3-f_7\\nabla^+_2]&=&0\\tag{VI}\\\\\n\\end{eqnarray*}\n\nAdding the two forms gives\n\n\\begin{eqnarray}\n<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\\\\\n\\ +\\frac{1}{2}(b_+^2+b_-^2)f_4+b_+f_5+\\frac{1}{2}f_6(a_+\\nabla_1^++a_-\\nabla^-_1)\\nonumber\\\\\n\\ +\\frac{1}{2}(b_+\\nabla_2^++b_-\\nabla^-_2)+f_8(Y)+f_8(Y^\\prime)\\nonumber\n\\end{eqnarray}\n\nSubtracting <math>\\frac{x}{2}\\times({\\rm I})</math> gives\n\n\\begin{eqnarray}\\ \\ \n<f>_R+<f>_L = \\frac{1}{2}[(1-x)a_+^2 + a_-^2]f_1 +\n(1-\\frac{x}{2})a_+f_2+\\frac{1}{2}[(1-x)a_+b_++a_-b_-]f_3\\nonumber\\\\\n+\\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\\\\\n+\\frac{1}{2}[(1-\\frac{x}{2})b_+\\nabla^+_2+b_-\\nabla^-_2]+f_8(Y)+f_8(Y^\\prime)\\nonumber\n\\end{eqnarray}\n\nIf <math>x=1</math>,\n\\begin{eqnarray}\n<f>_R+<f>_L &=&\n\\frac{1}{2}[a_-^2f_1+a_+f_2+a_-b_-f_3+b_-^2f_4+b_+f_5\\nonumber\\\\\n&+&\\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\n\\end{eqnarray}\n\nThe General Hermitian Property for arbitrary <math>x</math> gives\n\\begin{eqnarray}\\ \\ \n\\nabla_1^2=\n\\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\\\\\n-\\frac{2}{r_1}[(1-x)\\alpha_+a_++\\alpha_-a_-+2(1-\\frac{x}{2})\\alpha_+]+(1-x)\\alpha_+^2\n+ \\alpha^2_-]\\nonumber\\\\ \n+\\frac{2(r_1-r_2\\cos\\theta)}{r_1r^2}[(1-x)(a_+-\\alpha_+r_1)c_++(a_--\\alpha_-r_2)c_-]\\nonumber\\\\\n-\\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\n\\end{eqnarray}.\n\nUse\n<math>\\begin{eqnarray}\\ \\ \n\\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>,\n\n<math>\\begin{eqnarray}\n\\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\n\\end{eqnarray}</math>,\n\nand\n<math>\\begin{eqnarray}\\ \\ \n\\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]\n\\end{eqnarray}</math>.\n\nSubstituting into <math>\\nabla_1^2</math> gives\n\n\\begin{eqnarray}\n\\nabla^2_1 = \\frac{1}{4}\\{\\frac{1}{r_1^2}[-(1-x)a_+^2 + a_-^2 +xa_+ +\n2(1-\\frac{x}{2})\\Lambda(\\Lambda+1)\\nonumber\\\\\n-2l_1(l_1+1)(1-\\frac{c_-}{c_+}) -\n2l_1^\\prime(l_1^\\prime+1)(1+\\frac{c_-}{c_+})-\\frac{2c_-a_-}{c_+}(a_++1)]\\nonumber\\\\\n- \\frac{2}{r_1}\\left[-(1-x)\\alpha_+(a_++2)+\\alpha_-a_-+2(1-\\frac{x}{2})\\alpha_+\n-\\frac{c_-}{c_+}[a_-\\alpha_++\\alpha_-(a_++2)]\\right]\\nonumber\\\\\n-(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\n\\end{eqnarray}\n\nThis has the form\n<math>\n\\nabla_1^2 =\n\\frac{1}{4}\\left[\\frac{A_1}{r_1^2}+\\frac{B_1}{r_1}+\\frac{C_1}{r^2}+D_1\\right]\n</math>\nwith\n\n<math>\\begin{array}{ccc}\nA_1=-(1-x)a_+^2+a_-^2\n+xa_++2(1-\\frac{x}{2})\\Lambda(\\Lambda+1)-2l_1(l_1+1)(1-\\frac{c_-}{c_+})\\\n-2l_1^\\prime(l_1^\\prime+1)(1+\\frac{c_-}{c_+})-\\frac{2c_-a_-}{c_+}(a_++1) \\ \\\n\\ \\ \\ \\ \\ \\ \\ \\ \\ \\ \\ \\ \\ \\ \\ \\ \\ \\ \\ \\ \\ \\ \\ \\ \\ \\ \\ \\ \\ \\ \\ \\ \\ \\ \\ \\ \\ \\ \n\\end{array}\n</math>\n\n<math>\nB_1 =\n2\\left[(1-x)\\alpha_+(a_++2)-\\alpha_-a_--2(1-\\frac{x}{2})\\alpha_++\\frac{c_-}{c_+}[a_-\\alpha_++\\alpha_-(a_++2)]\\right]\n</math>\n\n<math>\nC_1 = (1-x)c_+^2+ c_-^2 +2(1-\\frac{x}{2})c_+\n</math>\n\n<math>\nD_1 = -(1-x)\\alpha_+^2 +\\alpha_-^2 - 2\\frac{c_-}{c_+}\\alpha_-\\alpha_+\n</math>\n\nThe complete Hamiltonian is then (in Z-scaled a.u.)\n\n<math>\\begin{eqnarray}\nH =-\\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\\\\\n= -\\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\\\\\n\\ =-\\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 \n+ Z^{-1}\\left(\\frac{1}{r}-\\frac{1}{r_2}\\right)\\nonumber\n\\end{eqnarray}</math>\n\nThe screened hydrogenic energy is \n<math>\\begin{eqnarray}\nE_{SH}=-\\frac{1}{2}\\left[\\frac{1}{n_1^2}+ \\left(\\frac{Z-1}{Z}\\right)^2\\frac{1}{n_2^2}\\right]\n\\end{eqnarray}</math>\nso that\n\n<math>\\begin{eqnarray}\nH-E_{SH} &=& -\\frac{1}{8}[\n \\frac{A_1}{r_1^2}+\\frac{B_1+8}{r_1}+\\frac{C_1}{r^2} + D_1 + D_2\n -\\frac{4}{n_1^2} -\\left(\\frac{Z-1}{Z}\\right)^2\\frac{4}{n_2^2}\\nonumber\n+\\frac{A_2}{r_2}+\\frac{B_2+8(Z-1)/Z}{r_2}+\\frac{C_2}{r^2}] +\n Z^{-1}\\left(\\frac{1}{r} -\\frac{1}{r_2}\\right)\\nonumber\n\\end{eqnarray}</math>\n\n==Optimization of Nonlinear Parameters==\n- The traditional method of performing Hylleraas calculations is to write the\nbasis set in the form\n\n<math>\n\\Psi = \\sum_{i,j,k}c_{i,j,k}\\varphi_{i,j,k}(\\alpha,\\beta)\\pm exchange\n</math>\n\nwith\n\n<math>\n\\varphi_{i,j,k}(\\alpha,\\beta) = r_1^ir_2^jr_{12}^ke^{-\\alpha r_1 - \\beta\n r_2}\\mathcal{Y}^M_{l_1l_2L}.\n</math>\n\nThe usual procedure is to set <math>\\alpha = Z</math> so that it represents the inner\n<math>1s</math> electron, and then to vary <math>\\beta</math> so as to minimize the energy.\n\n[[File:OptNon.jpg|500px]]\n\nSince <math>\\beta</math> appears in <math>\\Psi</math> as a non-linear parameter, the entire\ncalculation must be repeated for each value of <math>\\beta</math>. However, the minimum\nbecomes progressively smaller as the basis set is enlarged.\n\n[[File:OptNon02.jpg|500px]]\n\n*Difficulties\n\n1. If the basis set is constructed so that <math> i +j +j \\leq N</math>, the the\nnumber of terms is \n<math>\n(N+1)(N+2)(N+3)/6\n</math>\nN=14 already gives 680 terms and an accuracy of about 1 part in <math>10^{10}</math> for\nlow-lying states. A substantial improvement in accuracy would require much\nlarger basis sets, together with multiple precision arithmetic to avoid loss\nof significant figures when high powers are included.\n\n2. The accuracy rapidly deteriorates as one goes to more highly excited\nstates - about 1 significant figure is lost each time the principle quantum\nnumber is increased.\n\n*Cure\n\nWe have found that writing basis sets in the form\n\\begin{eqnarray}\n\\Psi &=& \\sum_{i,j,k}\\left(c^{(1)}_{i,j,k}\\varphi_{ijk}(\\alpha_1\\beta_1) +\n c_{ijk}^{(2)}\\varphi_{ijk}(\\alpha_2\\beta_2)\\right) \\pm exchange \\nonumber\\\\\n&=& \\psi({\\bf r}_1,{\\bf r}_2) \\pm \\psi({\\bf r}_2,{\\bf r}_1)\\nonumber\n\\end{eqnarray}\nso that each combination of powers is included twice with different nonlinear\nparameters gives a dramatic improvement in accuracy for basis sets of about\nthe same total size.\n\nHowever, the optimization of the nonlinear parameters is now much more\ndifficult, and an automated procedure is needed.\n\nIf\n<math>\nE = \\frac{<\\Psi|H|\\Psi>}{<\\Psi|\\Psi>}</math> and we assume <math><\\Psi|\\Psi> = 1</math>, then \n\n<math>\n\\frac{\\partial E}{\\partial\\alpha_t} = -2 <\\Psi|H-E|r_1\\psi({\\bf r}_1,{\\bf\n r}_2;\\alpha_t)\\pm r_21\\psi({\\bf r}_2,{\\bf\n r}_1;\\alpha_t)>\n</math>\nwhere \n<math>\n\\psi({\\bf r}_1,{\\bf r}_2;\\alpha_t) =\n\\sum_{i,j,k}c_{ijk}^{(t)}\\varphi_{ijk}(\\alpha_t\\beta_t)\n</math>\n\n*Problem\n\n1. Prove the above.\n\n2. Prove that there is no contribution from the implicit dependence of\n<math>c_{ijk}^{(t)}</math> on <math>\\alpha_t</math> if the linear parameters have been optimized.\n\n==The Screened Hydrogenic Term==\n\nIf the Hamiltonian is written in the form (in Z-scaled a.u.)\n<math>\\begin{eqnarray}\\ \nH = H_0({\\bf r}_1,Z) + H_0({\\bf r}_2,Z-1) +\nZ^{-1}\\left(\\frac{1}{r}-\\frac{1}{r_2}\\right)\n\\end{eqnarray}</math>\nwith\n<math>\\ \nH_0({\\bf r}_1,Z) = -\\frac{1}{2}\\nabla_1^2 - \\frac{1}{r_1}\\nonumber\n</math> and\n<math>\nH_0({\\bf r}_2,Z-1) = -\\frac{1}{2}\\nabla_2^2 - \\frac{Z-1}Z{r_2}\\nonumber\n</math>,\nthen the eigenvectors of <math>H_0({\\bf r},Z) +H_0({\\bf r}_2,Z-1)</math> are products of\nhydrogenic orbitals\n<math>\n\\Psi_0=\\psi_0(1s,Z)\\psi_0(nl,Z-1)\n</math>\nand the eigenvalue is \n<math>\\begin{eqnarray}\nE_{SH} = \\left[-\\frac{1}{2} - \\left(\\frac{Z-1}{Z}\\right)^2\\frac{1}{2n^2}\\right]Z^2 a.u.\n\\end{eqnarray}</math>\ncalled the screened hydrogenic eigenvalue. From highly excited states,\n<math>E_{SH}</math> and <math>\\Psi_0</math>are already excellent approximations.\n\nFor example, for the 1s8d states, the energies are\n\n<math>\\begin{eqnarray}\\\\\nE(1s8d^1D)&=& -2.007\\,816\\,512\\,563\\,81 {\\rm\\ a.u.}\\nonumber\\\\\nE(1s8d^3D)&=& - 2.007\\,817\\,934\\,711\\,71 {\\rm\\ a.u.}\\nonumber\\\\\nE_{SH} &=& -2.007\\,812\\,5 {\\rm\\ a.u.} \\nonumber\n\\end{eqnarray}</math>\n\nIt is therefore advantageous to include the screened hydrogenic terms in the\nbasis set so that the complete trial function becomes\n<math>\n\\Psi = c_0\\Psi_0 +\n\\sum_{ijk}\\left[c_{ijk}^{(1)}\\varphi_{ijk}(\\alpha_1,\\beta_1)+c_{ijk}^{(2)}\\varphi(\\alpha_2,\\beta_2)\\right]\\pm\n{\\rm exchange}\n</math>\n\nAlso, the variational principal can be re-expressed in the form\n<math>\\begin{eqnarray}\nE= E_{SH}+\\frac{<\\Psi|H-E_{SH}|\\Psi>}{<\\Psi|\\Psi>}\n\\end{eqnarray}</math>\nso that the <math>E_{SH}</math> term can be cancelled analytically from the matrix\nelements.\n\nFor example\n<math>\n<\\Psi_0|H-E_{SH}|\\Psi_0> = 0\n</math>\n\nand\n<math>\n<\\varphi_{ijk}|H-E_{SH}|\\Psi_0> =\nZ^{-1}<\\varphi_{ijk}|\\frac{1}{r}-\\frac{1}{r_2}|\\Psi_0>\n</math>\n\nRecall that\n<math>\\begin{eqnarray}\\ \\ \nI_0(a,b,c) = \\frac{2}{c+2}\\sum^{[c+1]/2}_{i=0}\\left(\\begin{array}{c}\nc+2\\\\\n2i+1\n\\end{array}\\right) \\left[f(p,q;\\beta) + f(p^\\prime,q^\\prime;\\alpha)\\right]\n\\end{eqnarray}</math>\nwhere\n\n<math>\\begin{eqnarray}\nf(p,q;x) =\n\\frac{q!}{x^{q+1}(\\alpha+\\beta)^{p+1}}\\sum^p_{j=0}\\frac{(p+j)!}{j!}\\left(\\frac{x}{\\alpha+\\beta}\\right)^j\n\\end{eqnarray}</math>\n\n<math>\n\\begin{array}{cc}\np=a+2i+2 & p^\\prime = b+2i+2\\\\\nq= b+c-2i+2 & q^\\prime =a+c-2i+2\n\\end{array}\n</math>\n\nIf <math>\\beta\\ll\\alpha</math>, the <math>f(p,q;\\beta)\\gg f(p^\\prime,q^\\prime;\\alpha)</math> and the\n<math>i=0</math> term is the dominant contribution to <math>f(p,q;\\beta)</math>. However, since\nthis term depends only on the sum of powers <math>b+c</math> for <math>r_2</math>, it cancels exactly\nfrom the matrix element of \n<math>\n\\frac{1}{r_{12}} - \\frac{1}{r_2}\n</math> \nand can therefore\nbe omitted in the calculations of integrands, thereby saving many significant\nfigures. This is especially valuable when <math>b+c</math> is large and <math>a</math> is small.\n\n==Small Corrections==\n*Mass Polarization\nIf the nuclear mass is not taken to be infinite, then the Hamiltonian is \n<math>\\begin{eqnarray}\\ \\ \nH = \\frac{P_N^2}{2M} + \\sum_{i=1}\\frac{p_i^2}{2m} + V\n\\end{eqnarray}</math>\n\n[[File:SmallCor.jpg|500px]]\n\nwhere <math>M =M_A-nm</math> is the nuclear mass and <math>m</math> is the electron mass.\n\nChange variables to the C of M\n<math>\\begin{eqnarray}\n{\\bf R} = \\frac{1}{M+nm}\\left[M{\\bf r}_N + m({\\bf r}_1 + {\\bf r}_2 +\n \\ldots)\\right]\n\\end{eqnarray}</math>\n\nand relative variables\n\n<math>\n{\\bf s}_i = {\\bf r}_i - {\\bf r}_N. \n</math>\n\nThen\n\n<math>\\begin{eqnarray}\\ \\ \\ \\ \\ \nH = \\frac{1}{2(M+nm)}{\\bf p}_R^2+\\frac{1}{2m}\\sum_{i=1}^n{\\bf p}_{s_i}^2\n+\\frac{1}{2M}\\sum_{i,k}{\\bf p}_{s_i}\\cdot {\\bf p}_{s_k} + V({\\bf s}_1, {\\bf\n s}_2, \\ldots)\\nonumber\n= \\frac{1}{M+nm}{\\bf p}_R^2+\\frac{1}{2\\mu}\\sum_{i=1}^n{\\bf\n 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\n s}_2, \\ldots)\\nonumber\n\\end{eqnarray}</math>\n\nwhere\n<math>\\begin{eqnarray}\n\\mu = \\left(\\frac{1}{M} + \\frac{1}{m}\\right)^{-1}\n\\end{eqnarray}</math>\n\nIf <math>n=2</math>, the Schrödinger equation is\n<math>\\begin{eqnarray}\\ \\ \\ \n\\left[ -\\frac{\\hbar^2}{2\\mu}(\\nabla^2_{s_1}+ \\nabla^2_{s_2}) -\n\\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\n= E\\psi\n\\end{eqnarray}</math>.\n\nDefine the reduced mass Bohr radius\n<math>\\begin{eqnarray}\na_\\mu = \\frac{\\hbar^2}{\\mu e^2} = \\frac{m}{\\mu}a_0\n\\end{eqnarray}</math>\nand\n\n<math>\\begin{eqnarray}\n\\rho_i = \\frac{Zs_i}{a_\\mu}\\nonumber\\\\\n\\varepsilon = \\frac{E}{Z^2(e^2/a_\\mu)}\\nonumber\n\\end{eqnarray}</math>\n\nThe Schrödinger equation then becomes\n<math>\\begin{eqnarray}\\ \\ \\ \n\\left[\n -\\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\n\\end{eqnarray}</math>\n\nThe effects of finite mass are thus\n\n1. A reduced mass correction to all energy levels of <math>(e^2/a_\\mu)/(e^2/a_0)\n= \\mu/m</math>, i.e.\n<math>\\begin{eqnarray}\nE_M=\\frac{\\mu}{m}E_\\infty\n\\end{eqnarray}</math>\n\n2. A specific mass shift given in first order perturbation theory by\n<math>\\begin{eqnarray}\n\\Delta\\varepsilon =\n-\\frac{\\mu}{M}<\\psi|\\nabla_{\\rho_1}\\cdot\\nabla_{\\rho_2}|\\psi>\n\\end{eqnarray}</math>\nor\n\n<math>\\begin{eqnarray}\n\\Delta\\varepsilon =\n-\\frac{\\mu}{M}<\\psi|\\nabla_{\\rho_1}\\cdot\\nabla_{\\rho_2}|\\psi>\\frac{e^2}{a_\\mu}\\nonumber\\\\\n= -\\frac{\\mu^2}{mM}<\\psi|\\nabla_{\\rho_1}\\cdot\\nabla_{\\rho_2}|\\psi>\\frac{e^2}{a_0}\\nonumber\n\\end{eqnarray}</math>\n\nSince <math>\\nabla_{\\rho_1}\\cdot\\nabla_{\\rho_2}</math> has the same angular properties as\n<math>{\\bf\\rho}_1 \\cdot {\\bf\\rho}_2</math>, the operator is like the product of two\ndipole operators. For product type wave functions of the form\n$\n\\psi = \\psi(1s)\\psi(nl) \\pm \\mbox{exchange},\n$\nthe matrix element vanishes for all but P-states.\n\n=Radiative Transitions=\n\n==Transition Integrals and the Line Strength==\n\nFor an electric dipole transition <math>\\gamma'\\,L'\\,S'\\,J'\\,M' \\rightarrow \\gamma\\,L\\,S\\,J\\,M</math>, the line strength is defined by\n\\begin{equation}\nS = \\sum_{M',\\mu,M}|\\langle\\gamma'\\,L'\\,S'\\,J'\\,M'|r_\\mu|\\gamma\\,L\\,S\\,J\\,M\\rangle|^2\n\\end{equation}\nUsing the Wigner-Eckart theorem, this is related to the reduced matrix element by\n\\begin{eqnarray}\nS(J',J) = |\\langle\\gamma'\\,L'\\,S'\\,J'||r||\\gamma\\,L\\,S\\,J\\,M\\rangle|^2\n\\sum_{M',\\mu,M}\\left|\\left(\\matrix{J' & 1 & J \\cr\n -M'& \\mu &M \\cr}\\right)\\right|^2\n\\end{eqnarray}\nUsing the sum rule for the 3-<math>j</math> symbols\n\\begin{equation}\n\\sum_{M',\\mu,M}\\left|\\left(\\matrix{J' & 1 & J \\cr\n -M'& \\mu &M \\cr}\\right)\\right|^2 = 1\n\\end{equation}\nwe see that the line strength is identical to the reduced matrix element\n\\begin{equation}\nS(J',J) = |\\langle\\gamma'\\,L'\\,S'\\,J'||r||\\gamma\\,L\\,S\\,J\\,M\\rangle|^2\n\\end{equation}\nFor 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 {<i> Angular Momentum in Quantum Mechanics</i>} with <math>k = 1</math> to obtain\n\\begin{equation}\nS(J',J) = (-1)^{L'+S+J+1}[(2J+1)(2J'+1)]^{1/2}\\left\\{\n\\matrix{L' & J'& S\\cr\n J & L & 1\\cr}\\right\\}\\langle\\gamma' L'|| r || \\gamma L\\rangle\n\\end{equation}\nwhere <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\n\\begin{equation}\nS(L',L) = |\\langle\\gamma' L'|| r || \\gamma L\\rangle|^2\n\\end{equation}\nTo calculate the reduced matrix element, use a particular example along with the Wigner-Eckart theorem\nsuch as\n\\begin{equation}\n\\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\n 0 & 0 & 0 \\cr}\\right)\n\\end{equation}\n\n===Correlated Two-electron Integrals===\nFor 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 <strong>18</strong>, 820 (1978), the general integral in Hylleraas coordinates for the integral over the <math>z_1</math> term can be written in the form\n\\begin{equation}\n\\label{int1}\n\\langle\\gamma'\\, L'\\, 0| z_1 | \\gamma\\, L\\, 0\\rangle = \\sum_\\Lambda C_\\Lambda I_\\Lambda(\\gamma',\\gamma)\n\\end{equation}\nwhere <math>I_\\Lambda(\\gamma',\\gamma)</math> stands for the purely radial integral\n\\begin{equation}\n\\label{radial}\nI_\\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}\nR_{\\gamma'}^*\\,r_1\\, R_\\gamma P_\\Lambda(\\cos\\theta_{12})\n\\end{equation}\nand <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,\nand, <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\nthat 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>)\n\\begin{eqnarray}\nC_\\Lambda^{\\rm reduced} &=& \\frac{(-1)^{l_2'+L'+L+1}}{2} (l_1,l_1',l_2,l_2',L,L')^{1/2}\n \\sum_{\\lambda={\\rm max}(l_1-1,0)}^{l_1+1} (\\lambda,\\Lambda,1)\n\\left(\\matrix{l_1 & 1 & \\lambda \\cr\n 0 & 0 & 0 \\cr}\\right)\n\\left(\\matrix{l_1'& \\lambda & \\Lambda \\cr\n 0 & 0 & 0 \\cr}\\right)\\nonumber\\\\\n&\\times&\\left(\\matrix{l_2' & 1_2 & \\Lambda \\cr\n 0 & 0 & 0 \\cr}\\right)\n\\left\\{\\matrix{L' & \\lambda & l_2 \\cr\n \\Lambda& l_2' & l_1' \\cr}\\right\\}\n\\left\\{\\matrix{L' & \\lambda & l_2 \\cr\n l_1 & L & 1 \\cr}\\right\\}\n\\end{eqnarray}\nwhere the notation <math>(a,b,c,\\cdots)</math> means <math>(2a+1)(2b+1)(2c+1)\\cdots</math>.\nA 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>.\n\n===Example===\nAs 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 \n\\begin{equation}\nS(1\\;^1S-2\\;^1P) = (0.781\\,127\\,158\\,667)^2\n\\end{equation}\n\\begin{equation}f(1\\;^1S-2\\;^1P) = -0.276\\,164\\,703\\,736\n\\end{equation}\nwhere \n\\begin{equation}\nf(1\\;^1S-2\\;^1P) = \\frac{2}{3}[E(1\\;^1S) - E(2\\;^1P)]S(1\\;^1S-2\\;^1P)\n\\end{equation}\nis the absorption oscillator strength."
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