Does nempty effect RPA NLF dielectric function?

Elk Users
Rob N
2014-01-28
2014-01-30
  • Rob N
    Rob N
    2014-01-28

    Does nempty effect the RPA NLF dielectric function (EPSILON_11) or only the tddft calculations that include local field effects (EPSILON_TDDFT)?

    If it affects both, is the value needed to converge the RPA-NLF not necessarily the same value required for TDDFT Epsilon?

    Best,

    RN

     
  • Markus
    Markus
    2014-01-29

    Hi Rob,

    it affects both. Just play with Si and see what happens. I would not bet that the same nempty will converge both. You might have a nicely converged head element of your response matrix (the RPA NLF), but the local-fields may still need some more. That's also true for BSE calculations, which require the full response function at w=0 for the screened Coulomb interaction.

    Regards,
    Markus

     
  • Rob N
    Rob N
    2014-01-29

    Markus,

    I played with it on my system and it didn't seem to affect my RPA-NLF calculations. I did notice a slight decrease in signal intensity in the TDDFT EELS (RPA-LFE) data past 25eV as I dropped below 25 down to 15 and 10. The relative shape of the plasmon remained the same, just the intensity was every so slightly decreased on the higher energy side.

    I will try dropping it even lower. I'm actually surprised nempty=10 didn't show larger differences for such a large system (28 atoms)

    Appreciatively,
    RN

     
    Last edit: Rob N 2014-01-29
  • Rob N
    Rob N
    2014-01-30

    nempty = 7 was as low as I could go. I attempted it at 5, but the signal past 25-30eV became unreasonable. Even with nempty=5 the lower portion (<22eV) was matched perfectly with nempty>25. So if a person is interested in this lower portion of the spectrum (say direct optical transitions) they may be able to keep this multiple of their computational expense very small.

    This may not be the case with all materials, and the energy at which this is true - if at all true, may in your system be quite different.

    Best,
    RN

     
    Last edit: Rob N 2014-01-30
  • Markus
    Markus
    2014-01-30

    Hi Rob,

    be aware that nempty is a scaling for the actual number of empty states:
    empty states = nempty * natoms * nspins. The default nempty=4 is already not bad. Even if you don't see a difference in the imaginary part of epsilon, you should notice a difference in the real part within the gap. This is quite sensitive to the number of transitions that you include. However, you'll have to go to quite low nempty to actually see the effect.

    Markus

     
  • Rob N
    Rob N
    2014-01-30

    Markus,

    That makes sense then. I thought nempty was the actual number of empty states being included, which is why I was surprised to see nempty=7 giving such good agreement.

     
  • Markus
    Markus
    2014-01-30

    Hi Rob,

    this was the case in elk's 1.x.x series. Since version 2.x.x it's defined as this type of scaling. This was meant to make working with the code a bit easier by automatic scaling of the number of empty states with the number of atoms. And since spin polarization is treated in second variation, the number of empty states is simply doubled then.

    Markus