We are using a k-mesh consisting of points only in the irreducible part
of the BZ (IBZ). Therefore this set of k-points does not contain the
full cubic symmetry, but a symmetrization is necessary.
This is done automatically in lapw2, and thus you should use x lapw2
-qtl to produce the properly symmetrized partial charges + PDOS.
You will see, that lapw2 produces only a eg and t2g PDOS, since they
correspond to the 2 irreducible representations for d-states in cubic
symmetry. If you want eg. the d-z2 PDOS (which does not really make
sense, since this is not a irreducible representation), it is simply 1/2
of the eg-PDOS.
If you want to use the qtl program, you have to do this symmetrization
yourself. Add up the d-z2 and d-x2-y2PDOS and you get the eg-PDOS. Only
the sum is meaningful, not the individual contributions.
If you don't believe me, just create a k-mesh in the full-BZ (x kgen
-fbz). after x lapw1; x qtl can now produce qtls (and PDOS) which
fulfills your expectations. But of course, the computer time is ~20
times bigger and the results are identical to the symmetrized ones.
Am 03.01.2022 um 14:03 schrieb Asif Ali:
Dear all
I am doing GGA calculation for ABO3 perovskite with cubic structure
(SrVO3). After using the qtl program, I am able to calculate PDOS for
dxy,dyz,dxz, dz2 and dx2-y2 separately. However, the calculated dxy,
dyz and dxz PDOS are non-generate (do not overlap). I understand that
due to crystal symmetry the PDOS should be degenerate for these three.
I read here
https://www.mail-archive.com/wien@zeus.theochem.tuwien.ac.at/msg08943.html
<https://www.mail-archive.com/wien@zeus.theochem.tuwien.ac.at/msg08943.html>,
but i am not able to understand why dx, dyz and dxz should not be
degenerate. Please help me with this.
Regards
Asif
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