10
$\begingroup$

I'm new to DFT.

Does the DFT take into account the spin-orbit interaction? On the one hand, this is a relativistic effect, perhaps DFT does not take it into account. However, on the other hand, I read that DFT is an extremely accurate method and the error of this method lies only in the inaccuracy of the exchange-correlation potential.

Improve this question
$\endgroup$
5
  • 2
    $\begingroup$ +1. Standard DFT does not take into account spin-orbit interactions, but relativistic DFT does exist, and spin-orbit interactions can be taken into account. I do not know the best references on this, so I will wait for someone else (perhaps more qualified than me) to write an "answer". $\endgroup$
    – Nike Dattani
    May 30, 2020 at 17:59
  • $\begingroup$ Since DFT is inherently unable to describe orbital multiplet states, I am not sure that the description of the relativistic part is the only problem. $\endgroup$
    – Greg
    Jun 1, 2020 at 5:51
  • $\begingroup$ @Greg DFT is not inherently unable to describe orbital multiplet states, we just don't know what the appropriate density functional is for the appropriate observables. It shouldn't be bad in non-Kohn Sham DFT though. $\endgroup$
    – Phil Hasnip
    Jun 29, 2020 at 2:30
  • $\begingroup$ @PhilHasnip Thanks, I always though HK theorem is only for non-degenerate states, but you are correct $\endgroup$
    – Greg
    Jun 29, 2020 at 4:27
  • 1
    $\begingroup$ @Greg the original proof by Hohenberg and Kohn did assume a non-degenerate ground state, but this limitation is removed (along with a few others) in the rigorous proof by Mel Levy. $\endgroup$
    – Phil Hasnip
    Jul 5, 2020 at 22:27

1 Answer 1

Reset to default
15
$\begingroup$

There are options to include spin-orbit coupling in DFT. In general, there are two ways to do it:

  1. Solve Dirac's relativistic equation for the electrons
  2. Incorporate relativistic effects through the Pseudopotential

Most DFT codes employ (2) as it is easier. There are well tested and readily available 'fully-relativistic' pseudopotentials for LDA and GGA available now.

Regarding the second part of the question, it depends on what you want to calculate. KS-DFT for example, cannot predict optical properties faithfully since it is a strictly ground-state method. In fact, the accuracy of DFT is more nuanced than you think - It depends on a variety of factors including the exchange functional used. One functional is not unequivocally better than another - Check this post for a comparison of LDA vs GGA in the context of elastic constants. Another example I can think of is in the context of highly-correlated materials. Kohn-Sham DFT predicts some transition-metal oxides such as FeO and ZnO incorrectly as metals. The inclusion of a modest Hubbard 'U' faithfully treats the highly localized 'd' orbitals in these materials, opening up a band gap. The properties predicted in this 'LDA(or GGA)+U' scheme are found to agree well with experiments.

The most commonly used exchange functionals such as LDA and GGA are very well known to underestimate electronic band gap due to a spurious self-interaction in partially occupied orbitals that is not exactly cancelled out by the terms in the exchange correlation functional. This causes occupied bands to over-delocalize, and push them up in energy, hence reducing band gap. To get an accurate band gap, you would need to do the actual quasiparticle calculation - via the GW-BSE method. However, various 'hybrid' functionals have been developed in the past years to give a more reliable estimate of the band gap. However these methods are semi-empirical, and hence strictly speaking, you don't know what band gap it gives you.

Improve this answer
$\endgroup$
1
  • 2
    $\begingroup$ Just to add that including the effects through the pseudopotential is not just easier to implement, it is also much less computationally demanding and can be made to be just as accurate. $\endgroup$
    – Phil Hasnip
    Jun 29, 2020 at 2:26

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .