[Q-e-developers] Symmetry in phonon and number of k-points

[Lorenzo Paulatto]

Dear all,
we have a problem here which I cannot solve, it may be bug in the phonon  
code or something weird.

Computing the phonon at q-point 2/3 -1/3 1 the number of kpoints used by  
ph.x explodes unreasonably. The k-point grid in pw.x was 4x4x4  
with/without shift, with the symmetry operatiosn they reduce to 10/8  
k-pointys respectively. Hence I would expect that at the very maximum ph.x  
would need 64x2=128 kpoints, yet this is not the case! See the following  
table with the number of k-points:

		pw	ph
444 111
	sym	10	256 (!!)
	nosym	64	128
444 000
	sym	8	40
	nosym	64	128

The situation is grossly the same for other q-points (part of the 6x6x6  
regular grid) and is generally worst for shifted grids. The paradox is  
that a phonon calculation in these specific case would be twice as fast  
WITHOUT symmetry than with.

Furthermore, tampering a bit with the system it is possible to get even  
more point (i.e. if fractional traslations are suppressed due to the FFT  
grid you can get 512 of them)

I've had a look at the relative subroutines, but without finding anything  
useful. Input/outpus attached.

cheers


-- 
Lorenzo Paulatto (IdR)
IMPMC - CNRS UMR 7590 & Université P&M Curie
T23-C13/23-5e27 - 4 place Jussieu - 75252 Paris Cedex5
phone: +33 (0)144 27 5211
www:   http://www-int.impmc.upmc.fr/~paulatto/
-------------- next part --------------
A non-text attachment was scrubbed...
Name: phxym.tgz
Type: application/x-gzip
Size: 42133 bytes
Desc: not available
URL: <http://lists.quantum-espresso.org/pipermail/developers/attachments/20110304/fb89ec1d/attachment.bin>