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I am a bit confused about the notions of exchange in the Hartree-Fock and Kohn-Sham Density Functional Theory schemes.

In Hartree-Fock, one just writes down the many-electron wavefunction as a hartree-product. This does not capture any correlation effects since each electron is treated to be independent of each other.

In Kohn-Sham DFT, what is generally mentioned in many texts (for example, this one [1]), is if you set correlation to zero, you recover the ground state energy and density corresponding to Hartree-Fock.

  1. My first question is, how different is 'exchange' in the HF scheme and the $E_x$ term (assume you separate $E_{xc}$ into $E_x$ and $E_c$) in the KS-DFT scheme?

  2. The term 'exact exchange' is often used in the context of Hybrid Functional calculations. Is this the same as the condition I initially mentioned (i.e. Set correlation=0 in the KS-DFT scheme).

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In Hartree-Fock, one just writes down the many-electron wavefunction as a hartree-product. This does not capture any correlation effects since each electron is treated to be independent of each other.

No, the Hartree-Fock many electron wave function is a single Slater determinant, which is completely antisymmetric to the exchange of two electrons and thereby satisfies the Pauli principle.

In Kohn-Sham DFT, what is generally mentioned in many texts (for example, this one [1]), is if you set correlation to zero, you recover the ground state energy and density corresponding to Hartree-Fock.

False. The concept of "exchange" is somewhat ill-defined in DFT, since the definition is inherently based on the Hartree-Fock wave function. Running an exchange-only Kohn-Sham calculation can give you a variety of results, depending on which density functional approximation you use for the exchange; remember, there is no correct choice for this!

My first question is, how different is 'exchange' in the HF scheme and the Ex term (assume you separate Exc into Ex and Ec) in the KS-DFT scheme?

This depends on the functional. If you use LDA exchange, you will get different results. Same with GGA and meta-GGA functionals. If and only if you pick the exchange part as Hartree-Fock exchange (which few functionals do as full Hartree-Fock is often too much), will you get behavior that matches Hartree-Fock in the lack of correlation.

Note: This is only in the case you are running a generalized Kohn-Sham calculation, where the exact exchange is evaluated the same way as in Hartree-Fock.

However, you could also run the calculation with an optimized effective potential (OEP), which is a local, scalar potential $V({\bf r})$ experienced by all electrons at variance to Hartree-Fock in which the occupied orbitals experience different potentials; again, you would see differences to Hartree-Fock.

The term 'exact exchange' is often used in the context of Hybrid Functional calculations. Is this the same as the condition I initially mentioned (i.e. Set correlation=0 in the KS-DFT scheme).

This is a term to watch out for, since it may mean either the generalized Kohn-Sham scheme in which the potential is orbital dependent, or the OEP scheme in which all orbitals experience the same potential.

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  • $\begingroup$ Thank you for the answer. I think I had mixed up 'Hartree' with 'Hartree-Fock' in the question. $\endgroup$ – livars98 Feb 8 at 0:19
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    $\begingroup$ @livars98 no, you mixed Hartree product and Slater determinant. $\endgroup$ – Susi Lehtola Feb 8 at 22:34
  • $\begingroup$ Ah, okay. Thanks for pointing it out. On that note, what's the distinction between Hartree and Hartree-Fock? $\endgroup$ – livars98 Feb 12 at 5:46
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    $\begingroup$ Hartree product / Hartree wave function is not antisymmetric to orbital permutations, and thereby does not obey Fermi statistics. Note that it's not a bosonic function either, since it's also not symmetric to particle interchange! $\endgroup$ – Susi Lehtola Feb 13 at 5:35

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