D.56 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,

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

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

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

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