I know in DFT the real system is mapped onto a fictitious non-interacting system under an effective KS potential.

My question is: When the Schrodinger equation is being solved (in each iteration of the self-consistent cycle) for each "free" electron under the effective KS potential, if the wave functions obtained have energy bands? Is that how they approximate the band-gap in DFT? ... or do they only find the gs wave function for each electron?

Thanks for any information you might have!

Edit: Under periodic systems, for which having bands makes sense.


I think you are mixing up things here a bit. First of all, without periodic boundary conditions there are no bands, but still electronic states which as you correctly say are effective single particle states. In general, these states have to be taken with care, since they have been derived from a single particle electron density representation, i.e.

$\rho = \sum\limits_l f_l|\psi_l|^2$

where $f_l$ is the occupation of the states and the $\psi_l$'s are the single electron wave functions. The physical meaning of the resulting states is mostly unclear. It has been however shown that the HOMO and LUMU correspond to vertical ionization potential and vertical electron affinity. From this fact, one can derive the HOMO-LUMU gap.

Under the influence of periodic boundary conditions, each state is broadened into bands. The HOMO-LUMU gap becomes the band gap.

The HOMO-LUMU gas is indeed the actual band gap, under the consideration of electronic entropy which smears out the bands . There are of course systems where certain functional are bad, but this is not a fault of DFT, but it's parametrization. For example semiconductors are intrinsically difficult to describe on the GGA level due to electron correlation having a huge influence on the band gap and GGA's artificially smearing out the electron density.

Concerning excited states, DFT can be modified by approximating excitation energies (e.g. Delta SCF) or extended to TDDFT for a more exact answer.

  • $\begingroup$ I forgot to add to the question, that yes, I was referring to periodic systems, crystals specifically. So under periodic conditions, bands are calculated for each wave function, and the HOMO-LUMU gap becomes the actual band gap? Everywhere I read it says it can never be the actual band gap and that it is a fault of DFT. $\endgroup$ – M.O. Mar 8 '19 at 1:10
  • $\begingroup$ Also, if it calculated bands for each state, meaning more than one solution for the energy for each k point considered, wouldn't it mean it is accessing "excited states" of this fictitious system? Wouldnt that violate the premise that DFT calculates the gs? $\endgroup$ – M.O. Mar 8 '19 at 1:11
  • $\begingroup$ I extended my answer in light of your questions $\endgroup$ – Guiste Mar 8 '19 at 6:22
  • $\begingroup$ Hey. I have another question that I hope you won't mind answering. I think it is implicit in your answer but I need to make sure. Once I have the effective KS periodic potential, does DFT solve the one electron Schrodinger equation, gets it's solutions and fills them up with electrons of the system? Is that how it gets the HOMO and LUMU? $\endgroup$ – M.O. Mar 21 '19 at 20:12
  • $\begingroup$ Yes that is right. The Fermi occupation function is needed when updating the electron density from the single electron states. The electron density then enters the Poisson equation which you need to update the potential. $\endgroup$ – Guiste Mar 24 '19 at 10:33

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

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