Calculation of the ground state energy shift

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

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

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

\begin{displaymath}\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\end{displaymath}

\begin{displaymath}\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\end{displaymath}

Do the \bgroup\color{black}$d\Omega_1$\egroup integral and prepare the other.

\begin{displaymath}\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\end{displaymath}

The angular integrals are not hard to do.

\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) \\

We can do the integral for \bgroup\color{black}$r_2<r_1$\egroup and simplify the expression. Because of the symmetry between \bgroup\color{black}$r_1$\egroup and \bgroup\color{black}$r_2$\egroup the rest of the integral just doubles the result.

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

Jim Branson 2013-04-22