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Question about the generation of q-vectors

Posted: Thu Jul 14, 2011 6:34 pm
by Boris
Hi all,

In the tutorial #2 for response function calculations, it is mentioned that the easiest way to determine the list of q-vectors is to "run a GS calculation with kptopt=1, nshiftk=1, shiftk=0 0 0 (to include gamma) and take the output kpt set file as this qpt set."

When dealing with a metal, the kpoint grid is rather fine, so that the list of q-vectors that are generated using the above method is huge. Is it possible to reduce the generated list of q-vectors by removing random points (for instance every other q-point)? If I remove half of the q-vectors, will it have a significant influence on the calculation results ?

Thank you for your help

Boris

Re: Question about the generation of q-vectors

Posted: Fri Jul 15, 2011 8:21 pm
by jzwanzig
I *think* the point of choosing q-points by the recommended method is to get q points that are expressible as the difference of two k points used in the electronic part of the problem. Then it would not wreck the calculation to throw away some of them. Whether this makes a difference in your results is probably system dependent. It would be very easy to try on something simple, like Al metal.

Re: Question about the generation of q-vectors

Posted: Wed Jul 27, 2011 9:40 pm
by Boris
Thank you for your answer,

I will then run some tests to check whether it has an influence on the calculations

Boris

Re: Question about the generation of q-vectors

Posted: Mon Apr 02, 2012 6:42 pm
by mverstra
Joe is correct, the prescription recommends that the q-grid be a sub grid of the k-grid. In general the phonons are much smoother and it's not necessary to go beyond 4x4x4 or 8x8x8 q-points. Sometimes it is necessary, however - you must converge each system.