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
[email protected]. 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
_______________________________________________________________________________
The Quantum ESPRESSO Foundation stands in solidarity with all
civilians worldwide who are victims of terrorism, military
aggression, and indiscriminate warfare.
--------------------------------------------------------------------------------
Quantum ESPRESSO is supported by MaX (www.max-centre.eu)
users mailing list [email protected]
https://lists.quantum-espresso.org/mailman/listinfo/users
_______________________________________________________________________________
The Quantum ESPRESSO Foundation stands in solidarity with all civilians
worldwide who are victims of terrorism, military aggression, and indiscriminate
warfare.
--------------------------------------------------------------------------------
Quantum ESPRESSO is supported by MaX (www.max-centre.eu)
users mailing list [email protected]
https://lists.quantum-espresso.org/mailman/listinfo/users
