Hartree-Fock exchange operator

by cire
Tags: exchange, hartreefock, operator
Sep13-05, 08:31 PM
P: n/a
I'm trying to understand the Hartree-Fock mathematical formulation I understand the Coulomb operator, but I dont understand the exchange operator:
d\textbf{x}'\frac{\Phi_{j}^{*}(\textbf{x}')\Psi(\textbf{x}')}{|\textbf{ r}-\textbf{r}'|}[/tex]
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
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Berislav is offline
Sep13-05, 09:46 PM
P: 243
I don't know anything about that formulation, but this book on Google Print might help:

Google Print

You can't see all the relevant information, but I think it might help you understand where the idea's headed.
lalbatros is offline
Sep26-05, 08:47 AM
P: 1,235
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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