Hartree-Fock exchange operator

  1. Sep 13, 2005 #1
    I'm trying to understand the Hartree-Fock mathematical formulation I understand the Coulomb operator, but I dont understand the exchange operator:
    [tex]
    \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}'|}[/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 :confused:
     
  2. jcsd
  3. Sep 13, 2005 #2
    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.
     
    Last edited: Sep 13, 2005
  4. Sep 26, 2005 #3
    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> .
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook

Have something to add?
Similar Discussions: Hartree-Fock exchange operator
  1. Hartree & Hartree-Fock (Replies: 1)

  2. Hartree Fock (Replies: 2)

  3. Hartree Fock iteration (Replies: 2)

Loading...