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Does the Hartree approximation basically assume that electrons are independent and do not interact with each other?

Also, do Exchange integrals essentially measure the interaction between two different electrons?

Thanks.

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- Thread starter chemstudent09
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- #1

- 8

- 0

Does the Hartree approximation basically assume that electrons are independent and do not interact with each other?

Also, do Exchange integrals essentially measure the interaction between two different electrons?

Thanks.

- #2

alxm

Science Advisor

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Does the Hartree approximation basically assume that electrons are independent and do not interact with each other?

Also, do Exchange integrals essentially measure the interaction between two different electrons?

Well, the Hartree (or Hartree-Fock) method makes several approximations, but not that the electrons are non-interacting. If that were the case the resulting wave function of an atom would simply be a superposition of solutions to the hydrogen-like atom.

HF improves on that by including exchange (the effect of the Pauli principle), and the electrostatic interaction between the electrons (the Coulomb integral). So it includes two forms of electron-electron interaction. (although exchange is not, strictly speaking, an interaction, but a boundary condition placed on the solutions to the S.E.) The exchange integrals have no classical analog. They're simply a direct consequence of preserving the known boundary condition (antisymmetry).

However, HF does assume that the

The other approximation, implicit in HF, is that it's a single-determinant description of the system.

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