Chapter 9: Problem 8
Consider two Slater determinants \(\Phi_{1}\) and \(\Phi_{2}\) that differ by only one spinorbital; that is, $$\begin{array}{l} \Phi_{1}=\left\|\cdots \varphi_{m} \varphi_{i} \cdots\right\| \\ \Phi_{2}=\left\|\cdots \varphi_{p} \varphi_{i} \cdots\right\| \end{array}$$ Derive the following Condon-Slater rule: $$\begin{aligned} \left\langle\Phi_{1}|H| \Phi_{2}\right\rangle=&\left\langle\varphi_{m}(1)\left|h_{1}\right| \varphi_{p}(1)\right\rangle+\sum_{i}\left\\{\left[\varphi_{m} \varphi_{i} | \varphi_{p} \varphi_{i}\right]\right.\\\ &\left.-\left[\varphi_{m} \varphi_{i} | \varphi, \varphi_{p}\right]\right\\} \end{aligned}$$ where we have used the notation $$\left[\varphi_{4} \varphi_{b} | \varphi_{c} \varphi_{d}\right]=j_{0} \int \varphi_{a}^{*}(1) \varphi_{b}^{*}(2)\left(\frac{1}{r_{12}}\right) \varphi_{c}(1) \varphi_{d}(2) \mathrm{d} r_{1} \mathrm{d} r_{2}$$