On Tue, Jun 30, 2020 at 7:12 PM Nicola Marzari <[email protected]> wrote: > > 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
OMG. I really did not try this. And if so, do you mean this is an initial conjecture so that it can help wannier90 to do the job step by step? > > > 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. Do you mean this is just a future possibly compatible usage but still not implemented yet? -- Hongyi Zhao <[email protected]> _______________________________________________ Quantum ESPRESSO is supported by MaX (www.max-centre.eu/quantum-espresso) users mailing list [email protected] https://lists.quantum-espresso.org/mailman/listinfo/users
