On 30/06/2020 13:07, Hongyi Zhao wrote:
As you can see, it uses 0.0156250 as the weight of all k-points. I
can't figure out how to obtain this value.
How much did you try? You have 64 k-points, the weights are all equal,
what could possibly be this value? Anyhow, here is the magic:
1/64=0.015625
It said that for a non-scf calculation, weights do not affect the
results. If so, why they still set weights for the k-points in the
example I mentioned above?
Because one could go into a mode, in the future, where the wannier code
uses only the points in the irreducible wedge, or at the very least it
uses time reversal.
nicola
Any hints will be highly appreciated.
--
----------------------------------------------------------------------
Prof Nicola Marzari, Chair of Theory and Simulation of Materials, EPFL
Director, National Centre for Competence in Research NCCR MARVEL, EPFL
http://theossrv1.epfl.ch/Main/Contact http://nccr-marvel.ch/en/project
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