[QE-users] Discrepancy in k-point sampling

[Giovanni Cantele]

The number of k-points is reduced or not reduced according to symmetries.
So, if for two systems that should be exactly equivalent,
symmetries are different, that means that either atomic positions are
different, or lattices are different or both.
Consider that symmetries are checked according to a threshold. I'm not very
aware of that, but for sure what the code does is to
apply symmetry matrices to atomic positions and then compare the obtained
positions with the initial ones (that should differ by at most lattice
translations).
When a comparison is carried out, a threshold is setup, in that quantities
(in our case atomic positions) are considered numerically the same if they
differ
by less than the threshold.

Now, in your case, the first think that comes to the eye is :
vs2_nscf.out:     12 Sym. Ops. (no inversion) found
vse2_nscf.out:      2 Sym. Ops. (no inversion) found

So, the two systems definitely have different symmetries, so they are
different!
After a closer inspection in the output files (similar conclusion could be
drawn
from input files) we see that:
vs2_nscf.out:
     crystal axes: (cart. coord. in units of alat)
               a(1) = (   1.000000   0.000000   0.000000 )
               a(2) = (  -0.500000   0.866025   0.000000 )

vse2_nscf.out:
     crystal axes: (cart. coord. in units of alat)
               a(1) = (   1.000000   0.000000   0.000000 )
               a(2) = (  -0.500097   0.866068   0.000000 )

As you might see, while in the first case a(2)_x = -1/2, a(2)_y =
sqrt(3)/2, in the second case there are little deviations
in the last digits that should be enough to prevent pw.x from finding
symmetries in the second case.

As a general rule, if I know that a systems is hexagonal, I wouldn't use
ibrav = 0 if there is a setting (ibrav = 4) that does the job for you!

As you said that you tried ibrav = 4, in that case what could be happened
is that you forgot to change the atomic positions as well. Indeed,
those positions are in crystal units, so they were referred to axes that
were very close to but not exactly (within the threshold) hexagonal ones.
So,
he symmetry check, that in the first case was failing because of the
crystal axes, in the second one is likely to fail because of the atomic
positions.

Use in the second case ibrav = 4 and exactly the same (in-plane) atomic
positions as in the first case, and for sure you'll get the same symmetries.

Giovanni
-- 

Giovanni Cantele, PhD
CNR-SPIN
c/o Dipartimento di Fisica
Universita' di Napoli "Federico II"
Complesso Universitario M. S. Angelo - Ed. 6
Via Cintia, I-80126, Napoli, Italy
e-mail: giovanni.cantele at spin.cnr.it <giovanni.cantele at spin.cnr.it>
Phone: +39 081 676910
Skype contact: giocan74

ResearcherID: http://www.researcherid.com/rid/A-1951-2009
Web page: https://sites.google.com/view/giovanni-cantele/home


Il giorno lun 3 mar 2025 alle ore 17:19 Zimmi Singh <
singh.60 at kgpian.iitkgp.ac.in> ha scritto:

> Dear QE Community,
>
> I am facing an unexpected discrepancy in k-point sampling during
> calculations for VS₂ and VSe₂, both of which have hexagonal symmetry. I
> used the same Monkhorst-Pack k-point grid (K_POINTS automatic 18x18x1)
> for both, yet the number of k-points in the irreducible Brillouin zone
> (IBZ) differs significantly:
>
>    - VS₂: 37 k-points
>    - VSe₂: 164 k-points
>
> To investigate, I also defined the lattice structure of VSe₂ using ibrav=4,
> but the IBZ k-point count remains 164. I have attached the nscf input and
> output files for both systems for reference.
> I would appreciate any insights on this matter, particularly on how to
> reduce the IBZ k-points for VSe₂, as it is critical for GW calculations. I
> also appreciate your time and suggestions.
>
> --
> *Best Regards*
> Zimmi Singh
>
> *Research_Scholar *
>
>
> *Department of Metallurgical and Materials Engineering Indian Institute of
> Technology, Kharagpur Kharagpur, India*
>
>
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