[QE-users] Irreducible set of k-points

[Stefano de Gironcoli]

a symmetry that is always present is k->-k due to time reversal which I 
think should account for the equivalence between [1/3, 0, 0] and [2/3, 
0, 0], up to a G vector as Paolo is mentioning.

stefano

On 10/02/25 15:11, Paolo Giannozzi wrote:
> Not sure I understand the problem: if k is a Bloch vector and G a 
> reciprocal-lattice vector, k+G is equivalent to k. This property can 
> be used to reduce the number of inequivalent k-points needed for the 
> sum over the (irreducible) Brillouin Zone
>
> Paolo
>
> On 10/02/2025 14:13, Lukas Cvitkovich wrote:
>>
>> Non ricevi spesso messaggi di posta elettronica da 
>> lukas.cvitkovich at physik.uni-regensburg.de. Scopri perché è importante 
>> <https://aka.ms/LearnAboutSenderIdentification>
>>
>>
>> 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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