this might be a stupid question but I am really confused at the moment. I'm currently preparing for an exam in computational chemistry. In the lecture I noted always down that the exchange term (Pauli-Principle), which arises as part of the two-electron functions is not determined in Hartree-Fock and added to the cusp-problem, when two electrons come close together. Those two then form the correlation energy, which is determined in post-Hartree-Fock methods.

Then in DFT we have several attempts to solve the exchange part, and I took a note back then, that basically in DFT or in the Kohn-Sham approach, as we do not know the DFT-functional, you create something, that looks like a Fock-operator, with the difference, that it also contains the exchange part.

Now I looked at some tasks and one question was why you do not simply take the HF expression for the exchange part and insert it into the DFT equations. I was a bit shocked what this expression might be and it turns out, when I look into other scripts I cannot find any note that in modern HF the exchange part would be neglected in any way.

So for simple HF (not post-HF) calculations, is the exchange term calculated as well and the only thing you miss would the the correlation?

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    $\begingroup$ If I'm understanding your question correctly, yes, HF does include exchange (sometimes referred to as exchange correlation) interactions between electrons with the same spin. It does not include the overall correlation of electron movement, as HF uses the mean field for the potential. $\endgroup$
    – Tyberius
    Commented Aug 5, 2017 at 19:35
  • $\begingroup$ Ah thank you, I found the problem, I thought xc was the abbreveation for exchange, not exchange correlation. So my Professor probably meant, that you seperate it into x + e and use post-HF methods from this point and I thought you also take the exchange part. $\endgroup$ Commented Aug 6, 2017 at 6:24

1 Answer 1


Most of the original question has been solved by discussion in the comments, but a pertinent question remains:

[W]hy you do not simply take the HF expression for the exchange part and insert it into the DFT equations [?]

The answer is that it has been tried and does not work well. In a sense, the DFT quantities "exchange energy" ($x$) and "correlation energy" ($c$) do not quite match the HF quantities "exchange energy" and "correlation energy".$^1$ In fact, even within DFT, the quantities do not match between functionals, as can be seen in some $x$- and $c$-functionals working together well and some not so well. In particular, those that have been fitted in combination with each other often only work in that combination. This suggests to me an illusion of a clear-cut way to differentiate the two in the DFT framework. It is also true that in some functionals, the $x$-term appears to account for some static correlation (which, broadly speaking, often becomes important when a single Lewis structure does not describe the molecule well: transition states, resonance structures). Entirely replacing it with Fock exchange forgoes this advantage.

Some programs, such as NWChem, and to some extent Gaussian, allow you to combine functionals and Fock exchange (often with arbitrary weights) to conduct your own experiment. A small benchmark set for main-group thermochemistry would probably give you an indication.

I wish I could you references for this, but this has been related to me verbally by my advisor back in the day and I no longer have access to most of the literature. If anybody has such references, I would be happy to see them included.

$^1$ The notation is a bit sloppy, but we're not striving for rigor in this answer.


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