# Hartree-Fock exchange operator

1. Sep 13, 2005

### cire

I'm trying to understand the Hartree-Fock mathematical formulation I understand the Coulomb operator, but I dont understand the exchange operator:
$$\hat{K_{j}}[\Psi](\textbf{x})=\Phi_{j}(\textbf{x})\int d\textbf{x}'\frac{\Phi_{j}^{*}(\textbf{x}')\Psi(\textbf{x}')}{|\textbf{r}-\textbf{r}'|}$$
Can any one explain me why this operator is like this. I understand that it is the interaction of the j-th electron with the electrons' cloud but... how it come to be like that

thanks in advance

2. Sep 13, 2005

### Berislav

I don't know anything about that formulation, but this book on Google Print might help:

http://print.google.com/print?id=b8AzpUPopqQC&lpg=PA16&dq=Hartree-Fock+exchange+operator&prev=http://print.google.com/print%3Fq%3DHartree-Fock%2Bexchange%2Boperator%26btnG%3DSearch%2BPrint&pg=PA15&sig=Q_plYtBA58CUSQi5a6NQc5AExnY [Broken]

You can't see all the relevant information, but I think it might help you understand where the idea's headed.

Last edited by a moderator: May 2, 2017
3. Sep 26, 2005

### lalbatros

I guess the integral on the rhs is simply the matrix element <j|K|phi> of the electrostatic potential. Therefore K|phi> is indeed given by |j><j|K|ph> .

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