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I have a small thought I would like to discuss:

To my knowledge the density of states (DOS) is calculated from the Kohn-Sham eigenvalues. Which would mean (at 0 K) that integrating: (E - EF)*DOS(E - EF) between -inf to the Fermi level (EF) should (at least in theory) yield the the sum of the eigenvalues as written to the OUTCAR and any deviations should be attributed to discretization and integration method. At least I get a value that is quite close to the sum of the eigenvalues so I assume this statement is correct.

So based on this I wonder: Can I use the density of states at all to explain total energy trends if the corresponding trends in eigenvalues don't have the same behavior?

For example if I look at work of separation (interface system - slabs) trends between Fe and transition metal nitrides along a row in periodic system I get that the differences in sum of eigenvalues (interface system - slabs) has the trend:

Fe/ScN: 552.20 eV/area
Fe/TiN: 650.35 eV/area
Fe/VN: 538.08 eV/area

while after adding xc,Hartree contributions etc. gives the work of separation trend:

Fe/ScN: -2.96 eV/area
Fe/TiN: -3.76 eV/area
Fe/VN: -3.95 eV/area

which doesn't have the same trend as the eigenvalues. Can I expect to find the reason behind the monotomic behavior in work of separation by looking on the DOS if the differences in eigenvalues behaves in a different way? So my point is that if the DOS just reflect the eigenvalues contribution I can't be sure to use DOS to explain total energy trends. Is this an incorrect point of view?

Best regards,
/Dan Fors




<span class='smallblacktext'>[ Edited ]</span>

I suppose that as the absolute values of the KS eigenvalues of a bulk system are only given to within a constant offset (the E=0 level is not fixed in 3D periodic systems), you rather should rely on the second argument. I suppose you have obtained the energies by taking the differences of total energies of the respective sub-systems, haven't you?

Yes, I did indeed obtain the energies in that way. I see your point. Thanks for your answer.

Regards