Let us carry out a perturbation treatment of the ground state of the lithium atom. Defining $e^{\prime}$ as $e^{\prime} \equiv e / 4 \pi \varepsilon{0}$, we take
$\hat{H}^{0}=-\frac{\hbar^{2}}{2 m{e}} \nabla{1}^{2}-\frac{\hbar^{2}}{2 m{e}} \nabla{2}^{2}-\frac{\hbar^{2}}{2 m{e}} \nabla{3}^{2}-\frac{Z e^{\prime 2}}{r{1}}-\frac{Z e^{\prime 2}}{r{2}}-\frac{Z e^{\prime 2}}{r{3}}, \quad \hat{H}^{\prime}=\frac{e^{\prime 2}}{r{12}}+\frac{e^{\prime 2}}{r{23}}+\frac{e^{\prime 2}}{r_{13}}$
We found in Section 10.5 that to satisfy the antisymmetry requirement, the ground-state configuration must be $1 s^{2} 2 s$. The correct zeroth-order wave function is (10.40):
\(
\begin{align}
\psi^{(0)}= & 6^{-1 / 2}[1 s(1) 1 s(2) 2 s(3) \alpha(1) \beta(2) \alpha(3)-1 s(1) 2 s(2) 1 s(3) \alpha(1) \alpha(2) \beta(3) \
& -1 s(1) 1 s(2) 2 s(3) \beta(1) \alpha(2) \alpha(3)+1 s(1) 2 s(2) 1 s(3) \beta(1) \alpha(2) \alpha(3) \tag{10.48}\
& +2 s(1) 1 s(2) 1 s(3) \alpha(1) \alpha(2) \beta(3)-2 s(1) 1 s(2) 1 s(3) \alpha(1) \beta(2) \alpha(3)]
\end{align}
\)
What is $E^{(0)}$ ? Each term in $\psi^{(0)}$ contains the product of two $1 s$ hydrogenlike functions and one $2 s$ hydrogenlike function, multiplied by a spin factor. $\hat{H}^{0}$ is the sum of three hydrogenlike Hamiltonians, one for each electron, and does not involve spin. Thus $\psi^{(0)}$ is a linear combination of terms, each of which is an eigenfunction of $\hat{H}^{0}$ with eigenvalue $E{1 s}^{(0)}+E{1 s}^{(0)}+E{2 s}^{(0)}$, where these are hydrogenlike energies. Hence $\psi^{(0)}$ is an eigenfunction of $\hat{H}^{0}$ with eigenvalue $E{1 s}^{(0)}+E{1 s}^{(0)}+E{2 s}^{(0)}$. Therefore [Eq. (6.94)],
\(
\begin{equation}
E^{(0)}=-\left(\frac{1}{1^{2}}+\frac{1}{1^{2}}+\frac{1}{2^{2}}\right)\left(\frac{Z^{2} e^{\prime 2}}{2 a_{0}}\right)=-\frac{81}{4}(13.606 \mathrm{eV})=-275.5 \mathrm{eV} \tag{10.49}
\end{equation}
\)
The evaluation of $E^{(1)}=\left\langle\psi^{(0)}\right| \hat{H}^{\prime}\left|\psi^{(0)}\right\rangle$ is outlined in Prob. 10.15. One finds
\(
\begin{align}
E^{(1)}=2 \iint 1 s^{2}(1) 2 s^{2}(2) \frac{e^{\prime 2}}{r{12}} d v{1} d v{2} & +\iint 1 s^{2}(1) 1 s^{2}(2) \frac{e^{\prime 2}}{r{12}} d v{1} d v{2} \
& -\iint 1 s(1) 2 s(2) 1 s(2) 2 s(1) \frac{e^{\prime 2}}{r{12}} d v{1} d v_{2} \tag{10.50}
\end{align}
\)
These integrals are Coulomb and exchange integrals:
\(
\begin{equation}
E^{(1)}=2 J{1 s 2 s}+J{1 s 1 s}-K_{1 s 2 s} \tag{10.51}
\end{equation}
\)
We have [Eqs. (9.52), (9.53), and (9.111)]
\(
\begin{gathered}
J{1 s 1 s}=\frac{5}{8} \frac{Z e^{\prime 2}}{a{0}}, \quad J{1 s 2 s}=\frac{17}{81} \frac{Z e^{\prime 2}}{a{0}}, \quad K{1 s 2 s}=\frac{16}{729} \frac{Z e^{\prime 2}}{a{0}} \
E^{(1)}=\frac{5965}{972}\left(\frac{e^{\prime 2}}{2 a_{0}}\right)=83.5 \mathrm{eV}
\end{gathered}
\)
The energy through first order is -192.0 eV , as compared with the true ground-state energy of lithium, -203.5 eV . To improve on this result, we could calculate higher-order wave-function and energy corrections. This will mix into the wave function contributions from Slater determinants involving configurations besides $1 s^{2} 2 s$ (configuration interaction).