\documentclass[12pt]{article} %%%%%%%%%%%% \topmargin -.55in \oddsidemargin -.05in \textheight 9.6in \textwidth 6.0in \begin{document} \parindent 0pt \parskip 0.3cm \newcommand{\breef}[1]{eq.~(\ref{#1})} \newcommand{\be}{\begin{equation}} \newcommand{\en}{\end{equation}} \newcommand{\beq}{\begin{equation}} \newcommand{\eeq}{\end{equation}} \newcommand{\bref}[1]{(\ref{#1})} \newcommand{\bfs}[1]{\mbox{\scriptsize\bf #1}} \newcommand{\combover}[2]{\left(\!\! \begin{array}{c} #1 \\ #2 \end{array} \!\! \right)} \newcommand{\bea}{\begin{eqnarray}} \newcommand{\eea}{\end{eqnarray}} \newcommand{\dd}{{\rm d}} {\bf \large Evaluating matrix elements of Coulomb repulsion } We calculate matrix element of Coulomb interaction % % $$ % e^2 \frac { 1 } {\left| \vec{r}_1-\vec{r}_2 % {\bf r}_1 - {\bf r}_2 \right| } $$ % % when the electrons are in states $\phi_{a}(\vec{r}_1)$, $\phi_{b}(\vec{r}_2)$, % % \begin{equation} \langle \phi_{a} \phi_{b} \left|V_{12}\right| \phi_{a} \phi_{b} \rangle = \int\int \phi^{\star}_{a}(\vec{r}_1) \phi^{\star}_{b}(\vec{r}_2) \frac{1}{ {|\vec{r}_1-\vec{r}_2|}} \phi_{a}(\vec{r}_1)\phi_{b}(\vec{r}_2) d\vec{r}_1 d\vec{r}_2 \label{ee13} \end{equation} % % with % % \begin{equation} \phi_{a}(\vec{r}_1) \ = \ R_a(r_1) Y_{l_{a}m_{a}} (\hat{r}_1) \end{equation} and \begin{equation} \phi_{b}(\vec{r}_2) \ = \ R_b(r_2) Y_{l_{b}m_{b}} (\hat{r}_2) \end{equation} % % using the multipole expansion % % \begin{equation} \frac {1} { {|\vec{r}_1-\vec{r}_2 |}} = \sum_{LM}\frac{4\pi}{2L+1} \frac{r^{L}_{<}}{r^{L+1}_{>}} Y^{\star}_{LM} (\hat{r}_1) Y_{LM} (\hat{r}_2) \label{eq:v1} \end{equation} % % where $$r_< = r_1 ,\ \ \ r_> = r_2 \ \ \ {\rm for} \ \ \ | \vec{r}_1|\ <\ |\vec{r}_2|$$ $$r_< = r_2 ,\ \ \ r_> = r_1 \ \ \ {\rm for} \ \ \ | \vec{r}_2|\ <\ |\vec{r}_1|$$ % % Then this can be separated in several integrals % % \begin{eqnarray} & \ & \langle \phi_{a} \phi_{b} \left|V_{12}\right| \phi_{a} \phi_{b} \rangle = \sum_{LM}\frac{4\pi}{2L+1} \nonumber \\ & \ & \int r_1^{2} dr_1 \int r_2^{2} dr_2 R^*_a(r_1)R^*_b(r_2) \frac{r_<^{L}}{r_>^{L+1}}R_a(r_1)R_b(r_2) \nonumber \\ & \ & \int Y^{\star}_{l_{a}m_{a}}(\hat{r}_1) Y^{\star}_{LM}(\hat{r}_1) Y_{l_{a}m_{a}}(\hat{r}_1) d \hat{r}_1 \ \ \int Y^{\star}_{l_{b}m_{b}}(\hat{r}_2) Y_{LM}(\hat{r}_2) Y_{l_{b}m_{b}} (\hat{r}_2) d \hat{r}_2 \end{eqnarray} % % with the notation % % \begin{equation} \hat{r}_1 \ \longrightarrow \ (\ \theta_1 \ \ \phi_1 \ ) \end{equation} % % and the integrations % % \begin{equation} \int d \hat{r}_1 \ \longrightarrow \ \int_0^\pi \sin \theta_1 \ d \theta_1 \int_0^{2 \pi} \ d \phi \end{equation} % % For s-states, when $l_{a}=0$, $m_{a}=0$ each of the angular integrals simply give $\sqrt{4 \pi}$ and cancel that factor in the first term. What remains to be evaluated is the double integral % % \begin{equation} \int r_1^{2} dr_1 \int r_2^{2} dr_2 R^*_a(r_1)R^*_b(r_2) \frac{r_<^{L}}{r_>^{L+1}}R_a(r_1)R_b(r_2) \end{equation} % % which for hydrogen-like states reduces to elementary integrations. EXERCISE: by expanding the inequalities, write out the two double integrals to be evaluated. \end{document} %%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%% %%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%% %%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%% %%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%% %%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%% %%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%% %%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%% %%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%% \begin{eqnarray} \langle \phi_{a} \phi_{b} \left|V_{12}\right| \phi_{a} \phi_{b} \rangle & = & \sum_{LM}\frac{4\pi}{2L+1} \\ \int r_1^{2} dr_1^{2} \int r_1^{2} dr_1^{2} \frac{r^{L}}{r^{L+1}} R^{\star}_{f}(\vec{r})[Y^{\star}_{l_{f}m_{f}}(\hat{r}) Y^{\star}_{LM}(\hat{R})Y_{LM}(\hat{r})] R_{i}(\vec{r})Y_{l_{i}m_{i}}(\hat{r})+$$ \int^{\infty}_{R}r^{2}dr \frac{R^{L}}{r^{L+1}} R^{\star}_{f}(\vec{r})[Y^{\star}_{l_{f}m_{f}}(\hat{r}) Y_{LM}(\hat{r})Y^{\star}_{LM}(\hat{R})] R_{i}(\vec{r})Y_{l_{i}m_{i}}(\hat{r}) \end{eqnarray} This can be described by a simple notation \begin{equation} V_{fi} = \sum_{LM}\frac{4\pi}{2L+1}[G^{L}_{fi}(R(t))] [Y^{\star}_{LM}(\hat{R})] [C^{L}] \label{e13} \end{equation} where $G^{L}_{fi}(R(t))$ is called The G-function and $C^{L}$ is composed of Clebsch-Gordan coefficients. The matrix element is different from zero only if: $$ M=m_{i}+m_{f}$$ and $$|l_{i}-l_{f}|\leq L \leq |l_{i}+l_{f}|$$ on $$l_{f}+L+l_{i} $$ is even. As can be seen above, we denote $$ G^{L}_{fi}[R(t)] = \int^{\infty}_{0}R^{\star}_{f}(\vec{r}) \frac{r^{L}<}{r^{L+1}>}R_{i}(\vec{r})r^{2}dr = $$ \begin{equation} \frac{1}{R^{L+1}} \int^{R}_{0}r^{L}r^{2}dr R^{\star}_{f}(\vec{r})R_{i}(\vec{r}) +R^{L}\int^{\infty}_{R} \frac{1}{r^{L+1}}r^{2}dr R^{\star}_{f}(\vec{r})R_{i}(\vec{r}) \end{equation} $R_{i}(\vec{r})$ and $R_{f}(\vec{r})$ are the radial wave functions for initial and final states. The integration over the angular parts, which is the integral over three spherical harmonics, gives \beq C^{L} =\int Y^{\star}_{l_{f}m_{f}}(\theta,\varphi) Y_{LM}(\theta,\varphi)Y_{l_{i}m_{i}}(\theta,\varphi)d\Omega \label{eq:CL} \eeq $$(-1)^{m_{f}}[\frac{(2l_{f}+1)(2L+1)(2l_{i}+1)}{4\pi}]^{\frac{1}{2}}$$ $$\left(\begin{array}{clcr} l_{f} & L & l_{i}\\ -m_{f}& M & m_{i} \end{array}\right)\left(\begin{array}{clcr} l_{f}& L & l_{i}\\ 0 & 0 & 0 \label{eq:spherical} \end{array}\right)$$ This is known as Gaunts formula, and the numerical values $\left(\begin{array}{clcr} a & b & c\\ d & e & f \end{array}\right)$ are Wigner-3j symbols, up to a factor equal to Clebsch-Gordan coefficients. \end{document}