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Question about Hartree Approx

  1. Apr 19, 2009 #1
    I'm not even sure if I am asking this the right way, but..

    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?

  2. jcsd
  3. Apr 20, 2009 #2


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    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 kinetic energy of the electrons are independent. Or in other words, that the motion of the electrons are independent. But they're not, since electrons 'avoid' each other due to their charges. So you have a coupling (termed correlation) between the kinetic and potential energy, which is neglected. Each electron moves in a 'mean field' of the charge of every other electron, but the non-linear dynamical effects of their interactions is not included.

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