spectral function normalization
NikaRybin opened this issue · comments
Hello, @stepan-tsirkin!
I have a small question related to the normalization of the spectral function. When we use "density" option for plotting, the spectral function has intensity, which is not from 0 to 1. In my understanding, this is not a scaling issue (cause in principle, we can scale intensities from [0,x] to [0,1]), but it is related to the normalization of the spectral function when the delta function is presented as gaussian in the code. Am I right that if the spectral function is properly normalized it should be between 0 and 1 or I am missing something? Could you, please, clarify this point?
Hi @NickRybin ,
No, the densitydoes not have to be between 0 and 1. When a delta-function is represented by a gaussian, the integral should be fixed, not the value at maximum.
@NickRybin
More detailed : If a band is represented as a gaussian (or any other smeared peak) g(E) it should be
\int g(E) dE=1
but when doing unfolding, that gaussian is multiplied by the unfolding weight W, which is between 0 and 1. But if 2 or more bands are degenerate (within the threshold), then the weights add up, so the weight may be actually greater than 1.
Does it answer your question?
Hi, Stepan!
Yeap, this I understand, but then I calculate degeneracy wrongly. I have Si supercell (at equilibrium, i.e. zero temperature, no distortion, so all the weights should be either 0 or 1) and for some reason the valence band maxima has an intensity ~6, whereas I though that the VBM at gamma point is three fold degenerate and shold have intensity ~3. The same I observe for the CBM at Gamma point, where the degeneracy, as I count should be 3, but somehow the intensity is also ~6.
UPDATE: spin ...
Hi Nikita,
Yes, if you do a spinor calculation, the deeneracy is 6. The deeneracies are easier to see in the weights, e.g. here :
banduppy/tutorial/reference/bandstructure_unfolded.txt
Lines 476 to 503 in f26af6a
I see only upto 3-fold degeneracies, but that is a scalar calculation.
Yes, you are right! I completely forgot about the spin. Thanks for clarifications.