[Pw_forum] K-Points and dipole correction
giannozz at nest.sns.it
Thu Mar 15 14:54:00 CET 2007
On Mar 13, 2007, at 13:04 , Daniele Passerone wrote:
> In the first case, I get 39 K points, in the second case I get 13 K
> There is no symmetry found in the slab in both cases.
You have a uniform grid that is not centered on the origin
but it is translated half step in the 1 and 2 directions. In
absence of an electric field, the code calculates the irreducible
grid by applying all lattice symmetries to all k-points and keeping
only those that are not found to be equivalent by a symmetry
operation of the actual crystal symmetry. This yields 39 points
in your case (no symmetry, apart the k <=> -k symmetry)
In presence of an electric field, the code should remove all
symmetries that do not leave the electric field unchanged.
A quick way to achieve this is to set internally the "nosym"
option to .true. ; for a uniform grid, however, nosym=.true.
means "take the entire uniform grid with no further processing".
This yields 13 points in your case : the 5 5 1 grid contains 25
points, 1 is gamma and of the remaining 24 only 12 are
independent (due to the k <=> -k symmetry), so 1+12=13.
If you use a centered grid, for instance 5 5 1 0 0 0, you
get 13 points in both cases.
So there is a logic, although I am not sure it is the correct one.
Paolo Giannozzi, Democritos and University of Udine, Italy
More information about the users