8
$\begingroup$

The Hartree-Fock approximation (HFA) works by assuming each electron sees the effective electric field from all the other electrons as some self-consistent field.

The HFA is known to give pretty decent results, but often gets the last quantitative bit of accuracy wrong. This last bit is attributed to mysterious many-body or "strongly-correlated" effects.

There are many examples of HFA quantitatively giving wrong answers (energies are wrong, incorrect numerical results, etc.). But, are there general phenomena that HFA gets totally wrong? I'm thinking of something where the chemistry/physics is fundamentally different from what HFA would tell you. Said differently, when do "strong correlations" actually matter and fundamentally change things in a chemical context?

I've tried reading up on this topic in the physics literature, and it is incomprehensible despite being an ex-physicist myself. Most of the discussion is about high temperature superconductors and complicated transition metal oxides. I can't understand the basic phenomena without all the specific details.

$\endgroup$
8
  • $\begingroup$ I suppose there is meant strong correlation of wave functions of some electrons, while about independence may be expected..But I am a dumb chemist with settled dust on lectures of quantum chemistry. $\endgroup$
    – Poutnik
    Commented Apr 25, 2019 at 6:47
  • 1
    $\begingroup$ Dispersion is another aspect $\endgroup$ Commented Apr 25, 2019 at 8:53
  • 1
    $\begingroup$ Funny you should ask that the same time my old question got some activity (Why do post-Hartree-Fock methods fail to predict the direction of the dipole moment of carbon monoxide?) Takeaway from this is the dipole moment of CO is completely wrong. In general I would not consider HF giving decent results except for molecular structures (for molecules). The neglect of most of the correlation energy in HF produces errors way beyond chemical accuracy (5 kJ/mol). (Quantitative accuracy is very hard to achieve in any case.) $\endgroup$ Commented Apr 25, 2019 at 9:31
  • 2
    $\begingroup$ @feodoran HF accounts for exchange correlation exactly. Admittedly the correlation energy is defined differently, but saying that it doesn't account for correlation at all is not quite correct. I agree that strong correlation is a different thing. $\endgroup$ Commented Apr 25, 2019 at 17:46
  • 1
    $\begingroup$ @Martin-マーチン If you consider exchange to be correlation ... yes. But then the terminology of having separate exchange and correlation functionals in DFT does not seem very consistent. $\endgroup$
    – Feodoran
    Commented Apr 25, 2019 at 18:02

4 Answers 4

4
$\begingroup$

Restricted HF (RHF) fails anytime you have partial occupancy due to (near) degeneracy of the HOMO. For example, $\ce{H2}$ dissociates to a mix of $\ce{2H}$ and $\ce{H- + H+}$ and the dissociation energy is way off. Similarly, singlet $\ce{O2}$ is not well described by RHF if you can get it to converge at all.

The solution is to use more than one MO wavefunction (determinant) and this is sometimes referred to as "non-dynamical" (rather than "strong") correlation.

Unrestricted HF (UHF) can usually treat these systems qualitatively correct.

$\endgroup$
5
  • $\begingroup$ If I understand correctly, are you saying that RHF fails because the basis it uses is too limited? What about gross qualitative failures of unrestricted HF? $\endgroup$
    – user157879
    Commented Apr 26, 2019 at 11:40
  • 1
    $\begingroup$ If by basis you mean basis set, then no. RHF will fail no matter how big the basis set is. The problem is that RHF requires double occupancy of all MOs. That's no good when the correct answer is single or partial occupancy. You need more than one determinant. I don't know of any gross qualitative failures for UHF $\endgroup$
    – Jan Jensen
    Commented Apr 26, 2019 at 12:04
  • $\begingroup$ So does UHF treat correlations exactly? Or are there still approximations being made there? $\endgroup$
    – user157879
    Commented Apr 26, 2019 at 22:35
  • $\begingroup$ The correlation energy is defined as the difference between the exact solution and the HF solution in the complete basis set limit. So there is no correlation in HF methods. They are mean field methods. Some refer to the exchange energy as "spin correlation" but then you are using the term "correlation" differently from the original definition. $\endgroup$
    – Jan Jensen
    Commented Apr 27, 2019 at 8:07
  • 1
    $\begingroup$ OK, so it sounds like my question has yet to be answered then. $\endgroup$
    – user157879
    Commented Apr 27, 2019 at 22:03
2
$\begingroup$

There are also examples of qualitative failures in unrestricted HF (UHF). For example, UHF produces an unbound potential for the dissociation of $\mathrm{F}_2$ and predicts a square geometry for cyclobutadiene.

$\endgroup$
2
$\begingroup$

Anions (negatively charged atoms or molecules) are great simple examples where the HFA fails. Very often, the HFA does not even predict the anion to be stable (that the excess electron remains bound to the neutral core). The excess electron is bound to the neutral core by various interactions. Taking a classical view, these can roughly be either (static-)multipole-charge (e.g. dipole-charge) interactions or multipole-induced-charged interactions (the interplay between the charge of the excess electron and the electron cloud of the neutral core). If the primary interaction term that allows the excess electron to remain bound is of the second nature, you can expect the HFA to fail predicting its stability. The reason is that, in order to describe the complicated interplay of the electron cloud, you need to account for electron correlation.

Here are two examples:

  • CO-: closed shell, core with dipole moment → is predicted to be bound at the HF level
  • C2-:open shell, no permanent multipole moment → not predicted to be bound at the HF level
$\endgroup$
1
$\begingroup$

HFA really struggles when the correlation energy is too great. I dont know every example where this is true but I do know when you have a heavy metal it is. The increased mass of the metal imposes a relativistic effect as the inner electrons have much more energy (because they move much faster) and thus the correlation energy is also increased dramatically. This in turn makes HFA fundamentally wrong for a distincy physical reason.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.