[Pw_forum] q-mesh breaks symmetry
Ari P Seitsonen
Ari.P.Seitsonen at iki.fi
Thu Apr 24 09:39:57 CEST 2014
Dear Nh Ben,
Sorry, I haven't found time to look at your system more carefully (well,
I tried, but it seemed to run, at least the first q vectors, with
PBE-PAWs).
If you are interested in my Thesis, the link
http://opus.kobv.de/tuberlin/volltexte/2002/152/ leads to it; discussion
about the symmetry and Monkhorst-Pack is mainly in the Appendices. That's
why I would recommend to use at least k1 = k2 in the input, ("4 4 3 0 0
0" or "4 4 3 0 0 1").
If you still need help please report your present input file(s) and the
version of QE you are trying to use, and I'll try to get back to it (or
some one else who is more of an expert on this).
Good Luck from Lille/France,
apsi
-=*=-=*=-=*=-=*=-=*=-=*=-=*=-=*=-=*=-=*=-=*=-=*=-=*=-=*=-=*=-=*=-=*=-=*=-=*=-
Ari Paavo Seitsonen / Ari.P.Seitsonen at iki.fi / http://www.iki.fi/~apsi/
Institut für Chemie der Universität Zürich, CH-8057 Zürich
Tel: +41 44 63 54 497 / Mobile: +41 79 71 90 935
On Tue, 22 Apr 2014, BENYAHIA NEZHA wrote:
> Dear Ari Paavo Seitsonen
>
> Firsly thank you for your response.
> I have tried again with K_POINTS (automatic) 4 4 3 1 1 1 on the scf calculation, and (nq1=4, nq2=4, nq3=4) on the phonon's input file,
> these are what i did, consequently the dynamicall matrix were calculated normally, however it gaves me a wrong dispersion curve. I dont
> really understand what do you mean with "try without shifting the k points, so without k1 = k2 = k3"??? why it displays me q-mesk break
> symmetry when i use (nq1=4, nq2=4, nq3=4) in phonon's input file and not for (nq1=4, nq2=4, nq3=3).
> I apreciate if i read your ancient PhD thesis but where can i find it?
>
>
>
> Regards
> Nh Ben
>
>
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