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)
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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
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
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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).
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). --[[User:Jonlester|Jonlester]] 18:02, 9 July 2007 (EDT)

Revision as of 22:02, 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). --Jonlester 18:02, 9 July 2007 (EDT)