[QE-users] Irreducible set of k-points

[Giovanni Cantele]

Hi,

I'm not very sure that this is correct, but I would say that what you're
missing is time reversal symmetry (that indeed is not
listed in the output file because it is not a lattice-related symmetry): k
<=> -k

[1/3,0,0] is equivalent to [1/3-1,0,0]=[-2/3,0,0] because they differ by a
lattice translation
[-2/3,0,0] is equivalent to [2/3,0,0] (k <=> -k)
So, [1/3,0,0] is equivalent to [2/3,0,0]

In the attached image:
- blue hexagon: 1st BZ
- blue dot: Gamma
- orange dot: [1/3,0]
- red dot: [2/3,0]
- green dot: [-1/3,0]

Red is outside the 1st BZ, because it is equivalent to green (by a lattice
translation).
Green in turn is equivalent to orange because k <=> -k

Giovanni

[image: image.png]
-- 

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 10 feb 2025 alle ore 14:13 Lukas Cvitkovich <
lukas.cvitkovich at physik.uni-regensburg.de> ha scritto:

> Dear QE users and developers,
>
> I am currently trying to reproduce the construction of an irreducible
> k-point set as done by QE.
> For this, I set "verbosity = high" to get the symmetry operations printed
> in the output file.
> I start from a uniform k-point mesh. Then, using the same symmetry
> operations as QE, I transform every k-point and fold it back in the first
> Brillouin zone.
> If the resulting k-point falls on another k-point of the uniform grid, it
> is NOT irreducible.
> In this manner, as also described by Blöchl et al (Phys. Rev. B *49*,
> 16223, 1994) I find the set of irreducible kpoints.
>
> My code agrees with QE for a simple structure (fcc crystal tested and
> verified) but I have problems with a more complicated case (the 2D magnet
> FGT).
> In this example, 6 symmetry operations are found (see attached QE-output
> file).
> Starting from a 3x3x3 uniform grid, the irreducible set of kpoints -
> according to QE - contains 7 points. However, I find 12 irreducible
> k-points.
>
> First, please note, that every point found by QE is also contained in my
> set. But I find additional points which (according to QE) should be related
> by some symmetry operation. By looking at the weights, I could figure out
> which kpoints should belong together.
> For instance: According to QE, the kpoints [1/3, 0, 0] and [2/3, 0, 0] are
> equivalent, as well as [0, 0, 1/3] and [0, 0, 2/3] should be equivalent
> too. I recognized that all the extra points could be transformed into each
> other by translating the lattice. However, applying all the symmetry
> operations from the QE output file (these are exclusively rotations and not
> translations), I cannot transform these points into each other. You might
> try for yourself.
>
> So the question that I would like to ask is: Are there any "hidden"
> symmetry operations which are not explicitly printed in the output file?
> Could fractional translations be the reason? Is it maybe related to
> differences between point group and space group? Any other hints to what I
> am missing?
> Thank you! Your help would be highly appreciated!
>
> Best,
> Lukas
>
>
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