Talk:Hartree equation: Difference between revisions

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(Explanation about Hartree-Fock method)
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There is apparently a connection between the Hartree equation and the Hartree-Fock equation, but I unfortunately am not aware of the details.  [[User:Tao|Terry]] 13:32, 9 July 2007 (EDT)
There is apparently a connection between the Hartree equation and the Hartree-Fock equation, but I unfortunately am not aware of the details.  [[User:Tao|Terry]] 13:32, 9 July 2007 (EDT)
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
<math>
u=\phi_1(x_1)\phi_2(x_2)...\phi_n(x_n)
</math>
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).

Revision as of 22:01, 9 July 2007

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)

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

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).