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In Density Functional Theory courses, one is often reminded that Kohn-Sham orbitals are often said to bear no any physical meaning. They only represent a noninteracting reference system which has the same electron density as the real interacting system.

That being said, there are plenty of studies in that field’s literature that given KS orbitals a physical interpretation, often after a disclaimer similar to what I said above. To give only two examples, KS orbitals of H2O[1] and CO2 closely resemble the well-known molecular orbitals.

Thus, I wonder: What good (by virtue of being intuitive, striking or famous) examples can one give as a warning of interpreting the KS orbitals resulting from a DFT calculation?


[1] “What Do the Kohn-Sham Orbitals and Eigenvalues Mean?”, R. Stowasser and R. Hoffmann, J. Am. Chem. Soc. 1999, 121, 3414–3420.

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    $\begingroup$ +1,000: The use of the KS orbitals drives me crazy and no one ever really tries to justify it. Conversely, I have not seen a good example of it failing, and everyone seems quite happy with it in practice. It's really frustrating: if it works it should have an explanation and if doesn't work people should stop doing it. $\endgroup$
    – BebopButUnsteady
    Commented Mar 16, 2012 at 21:38
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    $\begingroup$ Phil Anderson's latest book has a chapter on the issue of Slater vs. Mott, which is really DFT vs. reality. The KS orbitals would be exact in the absence of interactions. In cases where the excitations can be adiabatically linked to non-interacting electrons the error can only be quantitative and may be renormalised away; I think this is what the LDA+U work done by the Cambridge group is doing --- you use bond lengths to establish the "right" U for the current problem. One would expect that on the insulating side of a Mott transition the correspondence would be qualitatively poor. $\endgroup$
    – genneth
    Commented Mar 18, 2012 at 0:55
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    $\begingroup$ Cont.: I'm not a DFT expert (just a general theorist) but I would be surprised if accurate results are available for metal oxides, especially ones with d-orbital or f-orbitals. $\endgroup$
    – genneth
    Commented Mar 18, 2012 at 0:56
  • $\begingroup$ @genneth: Isn't that the answer? The mott insulator has localized states, but don't the Kohn Sham orbitals in a periodic potential end up delocalized? $\endgroup$
    – Ron Maimon
    Commented Mar 24, 2012 at 8:19
  • $\begingroup$ @RonMaimon: I would like to think so --- but I'm really not qualified to make a research-level statement. I'd like someone who really works on these things to say something intelligent. $\endgroup$
    – genneth
    Commented Mar 24, 2012 at 18:44

2 Answers 2

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When people say that Kohn-Sham orbitals bear no physical meaning, they mean it in the sense that nobody has proved mathematically that they mean anything. However, it has been empirically observed that many times, Kohn-Sham orbitals often do look very much like Hartree-Fock orbitals, which do have accepted physical interpretations in molecular orbital theory. In fact, the reference in the OP lends evidence to precisely this latter viewpoint.

To say that orbitals are "good" or "bad" is not really that meaningful in the first place. A basic fact that can be found in any electronic structure textbook is that in theories that use determinantal wavefunctions such as Hartree-Fock theory or Kohn-Sham DFT, the occupied orbitals form an invariant subspace in that any (unitary) rotation can be applied to the collection of occupied orbitals while leaving the overall density matrix unchanged. Since any observable you would care to construct is a functional of the density matrix in SCF theories, this means that individuals orbitals themselves aren't physical observables, and therefore interpretations of any orbitals should always be undertaken with caution.

Even the premise of this question is not quite true. The energies of Kohn-Sham orbitals are known to correspond to ionization energies and electron affinities of the true electronic system due to Janak's theorem, which is the DFT analogue of Koopmans' theorem. It would be exceedingly strange if the eigenvalues were meaningful while their corresponding eigenvectors were completely meaningless.

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  • $\begingroup$ So, in summary, there is no counter-example known where the KS orbitals clearly lack physical meaning, even though it's not theoretically guaranteed. Interesting… $\endgroup$
    – F'x
    Commented May 11, 2012 at 9:22
  • $\begingroup$ Well, not to my knowledge anyway. $\endgroup$ Commented May 12, 2012 at 2:46
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This is a complex issue, particularly because people often like to think in terms of an indepedent-particle picture (i.e. the aufbau filling up orbitals), even though the exact many-body wavefunction has strong electron-electron correlations. So let me rephrase your question:

What is the relationship between the KS eigenfunctions and the exact many-body wavefunction?

Mathematically, as you say, the KS eigenfunctions strictly speaking have no physical meaning (as far as we know). However, the KS eigenfunctions do give a useful qualitative (and sometimes quantitative) picture. The reason for this is that the KS eigenfunctions are a pretty good approximation to something in many-body perturbation theory called the quasiparticle wavefunction. The quasiparticle wavefunction is a well-defined physical property of a system that essentially tells you if you add (or remove) an electron with a certain amount of energy, where it will go. For example, see Phys. Rev. B 74, 045102 (2006).

Are there examples of when the KS eigenfunctions don't give a good desription of the quasiparticle wavefunctions? Well there are certainly many situations where the approximations we typically use in DFT (such as the local density approximation) lead to serious problems. However, I don't know of any examples where someone has shown that the exact KS eigenfunctions (i.e. those obtained with the true exchange-correlation functional) don't agree at least qualitatively with the quasiparticle wavefunctions.

As an aside, everything I have said above applies equally well to the Hartree-Fock wavefunctions. In fact, there is a solid mathematical basis for interpreting the HF wavefunctions as an approximation to the quasiparticle wavefunctions. See Chapter 4 of Fetter's Quantum Theory of Many-Particle Systems.

What about the KS eigenvalues? Strictly speaking, in general they do not correspond to ionization energies (or any other physically useful quantity). The one exception is the highest occupied eigenvalue, which is exactly equal to the ionization energy of the system. Janak's theorem tells us that the other eigenvalues are related to the derivative of the energy with respect to the occupancy of that eigenfunction:

$$\epsilon_i=\frac{dE}{dn_i}$$

See Phys. Rev. B 18, 7165 (1978) and Phys. Rev. B 56, 16021 (1997). It turns out that empirically these eigenvalues are nonetheless pretty good approximations to the true energy levels of the system with some caveats. In particular, the band gaps of solids are systematically underestimated.

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    $\begingroup$ Informative, thank you. Would Fetter's book be overkill for someone interested in DFT? $\endgroup$
    – CHM
    Commented May 21, 2013 at 14:20
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    $\begingroup$ Fetter's book is not a good resource to learn about DFT, because it is really about many-body perturbation theory. For someone interested in DFT, I would recommend Sholl and Steckel's introductory text or Richard Martin's book. Sholl is more practical, while Martin is more comprehensive and mathematical. $\endgroup$
    – Max Radin
    Commented May 26, 2013 at 17:35

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