Assume that the wave function for the $1s^2$ ground configuration is a product of $Ne^{-\alpha r_1}, Ne^{-\alpha r_2}$, I would like to calculate the average energy $\langle e^2/4\pi\epsilon_0r_{12}\rangle$
Then:
$$\left\langle \frac{e^2}{4\pi\epsilon_0r_{12}} \right\rangle = \int|Ne^{-\alpha r_1}|^2\cdot|Ne^{-\alpha r_1}|^2\frac{1}{|\vec{r}_1-\vec{r}_2|}d^3r_1\,d^3r_2$$
Any good suggestions on how to solve this integral?
Best Answer
You can use the Laplace expansion of the inverse distance $$\frac{1}{|\vec{r}_1-\vec{r}_2|} =\sum_{\ell=0}^\infty\frac{4\pi}{2\ell+1}\frac{r_<^\ell}{r_>^{\ell+1}} \sum_{m=-\ell}^{+\ell}Y_{\ell m}^*(\theta_1,\phi_1)Y_{\ell m}(\theta_2,\phi_2)$$ where $r_<$ is the smaller one of $r_1$ and $r_2$, and $r_>$ is the bigger one of $r_1$ and $r_2$.
Insert this expansion into your integral, and you get an integral of a sum. You can reorder this to a sum of integrals. Then you can calculate each of these integrals with straight-forward methods. Luckily for you, in the sum the angular integrals over $(\theta_1,\phi_1)$ and $(\theta_2,\phi_2)$ are all zero, except the one with $\ell=0$ and $m=0$. So you arrive at $$\int_0^\infty\int_0^\infty |Ne^{-\alpha r_1}|^2\cdot |Ne^{-\alpha r_2}|^2\ \frac{1}{r_>} \ r_1\ dr_1\ r_2\ dr_2$$