D.55 Why the Fock operator is Hermitian

To verify that the Fock operator is Hermitian, first note that $h^{\rm e}$ is Hermitian since it is an Hamiltonian. Next if you form the inner product $\langle\overline{\psi^{\rm e}{\updownarrow}}\vert v^{\rm {HF}}\pe//b//\rangle$, the first term in $v^{\rm {HF}}$, the Coulomb term, can be taken to the other side since it is just a real function. The second term, the exchange one, produces the inner product,

\begin{displaymath}
- \sum_{{\underline n}=1}^I
\bigg\langle\overline{\psi^{...
...rangle
\pe{\underline n}/{\skew0\vec r}/b/z/
\bigg\rangle
\end{displaymath}

and if you take the operator to the other side, you get

\begin{displaymath}
- \sum_{{\underline n}=1}^I
\bigg\langle
\langle\pe{\u...
...gg\vert
\pe/{\underline{\skew0\vec r}}/b/z1/
\bigg\rangle
\end{displaymath}

and writing out these inner products as six-di­men­sion­al spatial integrals and sums over spin, you see that they are the same.