Outline the steps involved in the Hartree–Fock method for the calculation of molecular electronic structure.

I understand the basics of HF, in that it does not account for electron correlation (neither non-dynamical or dynamical) but I do not understand what the question means by steps.

Are the steps of the HF method just those of the self-consistent procedure or is this something difference?

In addition, does Hartree-Fock account for the antisymmetry requirement or not? I have notes that say it does, because it uses a Slater determinant, but I also have notes that say DFT accounts for the antisymmetry neglected by HF.

  • $\begingroup$ The SCF and HF are essentially the same, if you include specifying the nuclear coordinates/charges/#of e- and computing various integrals as part of the SCF. I have never heard of DFT including antisymmetry neglected by HF $\endgroup$
    – Tyberius
    Dec 5, 2019 at 16:53
  • $\begingroup$ Hope this would be helpful: vergil.chemistry.gatech.edu/notes/hf-intro/hf-intro.pdf $\endgroup$ Dec 5, 2019 at 17:07
  • $\begingroup$ HF accounts for exchange correlation exactly. DFT is a whole other story, see some more here: Are there any full worked examples of DFT calculations? $\endgroup$ Dec 5, 2019 at 17:58
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    $\begingroup$ Martin is correct here, though the terminology around this issue is easy to get confused. Exchange accounts for a certain type of correlation (same spin electrons avoid being in the same place). A lot of authors call the energy added by post-Hartree Fock methods the correlation energy, which is mostly true because it accounts for the rest of the correlation (electrons moving in response to each other electron rather than their average potential), but exchange is still a form of correlation. $\endgroup$
    – Tyberius
    Dec 5, 2019 at 20:08
  • $\begingroup$ For another phrasing of my previous comment, I would point you toward the second and third paragraphs of the "Atomic and Molecular Systems" section: en.m.wikipedia.org/wiki/Electronic_correlation $\endgroup$
    – Tyberius
    Dec 6, 2019 at 1:01

1 Answer 1


Yeah, I think it asks you to write the energy as expectation value, vary one orbital and get the Roothan equation. The Hartree Fock model is the extension of the Hartree model where the wavefunction was schematized by the following guess: \begin{equation} \psi=\phi_{a}(1)\phi_{b}(2)...\phi_{n}(n) \end{equation}
If you work out the equations, you'll see the equations look a lot like the HF ones, but you'll miss the exchange term. This is because the wavefunction is not in the correct fashion to describe a fermionic system not made by infinitely distanced parts. So you need to plug in the antisymmetric form of the wavefunctions (Slater determinants are a basis in the space of antisymmetric functions). Also, the HF approximation requires you to just approximate the wavefunction with just one determinant, just one piece of the expansion. This is clearly a simplification but it can be correct enough if there is a significative HOMO-LUMO gap and the system is closed shell. So the HF model gets some correlation (known as fermi hole) meaning that electrons having same spin do avoid each other (you enforced it using antisymmetric wavefunctions) but the electrons don't avoid themselves respecting the Coulomb repulsion (two electrons cannot be in the same place even if they have not the same spin) this is a first piece of correlation you lose, the other being the fact you simply are using one determinant. Unfortunately correlation is described as everything going beyond HF because of Quantum information theory, so you have to be careful when you talk of correlation of HF. As it regards DFT, you'll plug in it functionals, so you may include functionals in it that allow you to recover some more correlation, but remember that DFT functionals are, most of the time, semiempirical.


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