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.

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

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

$$\bgroup\color{black}\Delta E_{gs} = {4\pi\over{\pi^2}}e^2\lef... ...r_1/a_0} \int\limits^\infty_0 r^2_2dr_2 e^{-2Zr_2/a_0} \egroup$$

$$\bgroup\color{black}\Delta E_{gs} = {4\pi\over{\pi^2}}e^2\lef... ...er{r_1r_2}}\left[ - \vert r_1-r_2\vert+(r_1+r_2)\right] \egroup$$

$$\bgroup\color{black}\Delta E_{gs} = 8 e^2\left({Z\over{a_0}}\... ...ty_0 r_2dr_2 e^{-2Zr_2/a_0} (r_1+r_2-\vert r_1-r_2\vert)\egroup$$

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.

$$\bgroup\color{black}\Delta E_{gs} = 16 e^2\left({Z\over{a_0}}... .../a_0} \int\limits^{r_1}_0 r_2dr_2 e^{-2Zr_2/a_0} (2r_2)\egroup$$

$$\bgroup\color{black}\Delta E_{gs} = e^2{Z\over{a_0}} \int\li... ..._1dx_1 e^{-x_1} \int\limits^{x_1}_0 x_2^2dx_2 e^{-x_2} \egroup$$

$$\bgroup\color{black} = {Ze^2\over{a_0}} \int\limits^\infty_0 ... ...-x_1} + \int\limits^{x_1}_0 2x_2 dx_2 e^{-x_2} \right\} \egroup$$

$$\bgroup\color{black} = {Ze^2\over{a_0}} \int\limits^\infty_0 ... ...1e^{-x_1} + 2 \int\limits^{x_1}_0 e^{-x_2}dx_2 \right\} \egroup$$

$$\bgroup\color{black} = {Ze^2\over{a_0}} \int\limits^\infty_0 ... ...x_1} -2x_1e^{-x_1} -2\left( e^{-x_1} -1\right) \right\} \egroup$$

$$\bgroup\color{black} = -{Ze^2\over{a_0}} \int\limits^\infty_0... ...x^2_1+2x_1\right) e^{-2x_1} -2x_1 e^{-x_1} \right] dx_1 \egroup$$

$$\bgroup\color{black} = -{Ze^2\over{a_0}} \left[ {3\over 2}{3... ...+ 2{1\over 2}{1\over 2} - 2{1\over 1}{1\over 1} \right] \egroup$$

$$\bgroup\color{black} = -{Ze^2\over{a_0}} \left[ {3\over 8}+{... ...er 8}-{16\over 8}\right] = +{5\over 8}{Ze^2\over{a_0}} \egroup$$

$$\bgroup\color{black}= {5\over 4}Z(13.6\ {\mathrm eV})\qquad\rightarrow\mbox{ 34 eV for Z=2}\egroup$$