Talk:Hartree equation

There is apparently a connection between the Hartree equation and the Hartree-Fock equation, but I unfortunately am not aware of the details. Terry 13:32, 9 July 2007 (EDT)

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As a physicist I am somewhat unfamiliar with this form of Hartree equation. The reason for this is that this equation is about many-body problems (e.g. large atoms). This means that one has to consider many-body wavefunctions like

$u=\phi _{1}(x_{1})\phi _{2}(x_{2})...\phi _{n}(x_{n}).$

The modification due to Fock to the Hartree method is to use fully antisymmetric wavefunctions as happens to fermionic matter represented by a determinat (Slater determinant). This introduces a further term into the equation due to the particle exchange (exchange integral). --Jonlester 18:02, 9 July 2007 (EDT)