Ground state

The unperturbed ground state has $\vert nlm\rangle=\vert 100\rangle$ and $\vert n'l'm'\rangle=\vert 100\rangle$, i.e. $\alpha=\beta$ with a symmetrical orbital wave function $\phi_{\alpha\alpha}^{+}({\bf r}_1,{\bf r}_2)= \phi_{100}({\bf r}_1)\phi_{100}({\bf r}_2)$ and a singlet spinor $\vert S\rangle$. The energy to first order in $U$ therefore is

$$E^{(1)}_{\alpha\alpha} = E^{(0)}_{\alpha\alpha} + A_{\alpha\alpha},\alpha=(100)$$ (3.3)

$$A_{\alpha\alpha} = \int \int d{\bf r}_1 d{\bf r}_2 \vert\phi_{100}({\bf r}_1)\vert^2 U\left(\vert{\bf r}_1 -{\bf r}_2\vert\right) \vert\phi_{100}({\bf r}_2)\vert^2.$$ (3.4)

Calculation of $A$ yields

$$E^{(1)}_{100,100} = E^{(0)}_1 + E^{(0)}_1 + A_{100,100} = 2 \left( -\frac{1}{2}\frac{... ...4\pi\varepsilon_0 a_0} \right) + \frac{5}{8} \frac{Ze^2}{4\pi\varepsilon_0a_0}.$$ (3.5)

For $Z=2$, one has $2 E^{(0)}_1=-108.8$eV and $A_{100,100}=34$eV such that $E^{(1)}_{100,100}=-74.8$eV.

Exercise: Calculate the integral leading to the result Eq. (III.3.5). Solution hints are given in Gasiorowicz [3].