Talk:Hartree equation: Difference between revisions
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Marco Frasca (talk | contribs) (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).
