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,

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

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

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

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