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Re: [abinit-forum] nkpt problem


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  • From: Matthieu Verstraete <mjv500@york.ac.uk>
  • To: forum@abinit.org
  • Subject: Re: [abinit-forum] nkpt problem
  • Date: Thu, 23 Aug 2007 02:08:51 +0100 (BST)


Hello Hua,

1) you should not set chkprim 0; use an irreducible unit cell unless you are certain of what you are doing

2) The default kpt shift is 1 shift at 0.5 0.5 0.5. This could be what breaks the symmetries, as you are in a hexagonal, not a cubic system. Specify shiftk explicitly as 0.0 0.0 0.5 for hex systems, or even 0.0 0.0 0.0. If you include 0.5 0.5 0.5, a single kpt shift is probably not enough. See the tutorials and tests (in v1) on generating good kpt grids.
Depending on what you are doing the code sometimes imposes that the kpt grid has the full lattice symmetry. In any case this is always a good idea, and is broken if your shift can be rotated out of the kpt grid by a symop.

In your case the full point group symmetry appears to be P1, which I think is wrong for corundum (probably your xangst are not precise enough - should probably try natrd and associated variables). For the kpoints the bravais lattice symmetries are used, ie for the hexagonal cell.

Matthieu

--
================================================================
Dr. Matthieu Verstraete mailto:mjv500@york.ac.uk
Dept. of Physics, University of York, tel: +44 1904 43 22 08
Heslington, YO10 5DD York, United Kingdom fax: +44 1904 43 22 14



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