Calculation of the ground state energy shift

To calculate the first order correction to the He ground state energy, we gotta do this integral.

$$\bgroup\color{black} \Delta E_{gs} = \left<u_0 \vert V \vert ... ...rt^2 { e^2\over{ \vert\vec{r}_1 - \vec{r}_2\vert }} } \egroup$$

First, plug in the Hydrogen ground state wave function (twice).

$$\bgroup\color{black}\Delta E_{gs} =\left[ {1\over {4\pi}}4\le... ...d\Omega_2 { 1\over{ \vert\vec{r}_1 - \vec{r}_2\vert }} }\egroup$$

$$\bgroup\color{black} { 1\over{ \vert\vec{r}_1 - \vec{r}_2\ver... ... = {1\over\sqrt{ r^2_1 + r^2_2 - 2r_1r_2 \cos\theta }} \egroup$$

Do the $d\Omega_1$ integral and prepare the other.

$$\bgroup\color{black}\Delta E_{gs} = {4\pi\over{\pi^2}}e^2\lef... ...2 {1\over\sqrt{ r^2_1 + r^2_2 - 2r_1r_2 \cos\theta_2 }} \egroup$$

The angular integrals are not hard to do.

$$\begin{eqnarray*} \Delta E_{gs} &=& {4\pi\over{\pi^2}}e^2\left({Z\over{a_0}}\rig... ...\infty_0 r_2dr_2 e^{-2Zr_2/a_0} (r_1+r_2-\vert r_1-r_2\vert) \\ \end{eqnarray*}$$

We can do the integral for $r_2<r_1$ and simplify the expression. Because of the symmetry between $r_1$ and $r_2$ the rest of the integral just doubles the result.

$$\begin{eqnarray*} \Delta E_{gs} &=& 16 e^2\left({Z\over{a_0}}\right)^6 \int\li... ...}Z(13.6 {\mathrm eV})\qquad\rightarrow\mbox{ 34 eV for Z=2} \\ \end{eqnarray*}$$