[Pw_forum] inhomogeneous K-point sampling

[Stefano Baroni]

On May 28, 2009, at 6:00 AM, Aritz Leonardo Liceranzu wrote:

> Dear prof. Baroni,

Aritz:

> For this purpose, I need to be able to construct nonuniform  
> reciprocal grids in the IBZ with their corresponding wafunctions and  
> weights. I am trying to figure out how to do this so I guess that  
> first I will need to understand how Monkhorst-Pack grids are  
> generated and more precisely the corresponding the weights of each  
> kpoint.

MP grids are uniform grids in reciprocal space. Doing Simpson  
integration on a regular grid would just require equal weigths. The  
reason why weights are not equal is because one does not use all the  
points of the grid, but just those that are inequivelent by symmetry  
(i.e. that belong to the "irreducible wedge" of the BZ). The weight in  
this case is just equal to the total number of points that are  
equivalent to a given point ("the star" of a given point). An example  
will explain this. Suppose you are in cubic symmetry and that you have  
the points (xyz), (xy0), (x00). What are the corresponding weights?  
Well, there are 48 points equivalent to (xyz), 24 points equivalent to  
(xy0) and 6 equivalent to (x00). The relative wights will thus be 8,  
4, 1. A special care must be paid to points that lie on the border of  
the BZ. In this case, a symmetry operation may send a point into  
itself (modulo a reciprocal-lattice vector). For instance, the star of  
(1/2,0,0) has six elements for an FCC lattice, whereas it has only 3  
for a SC lattice [because (1/2,0,0) and (-1/2,0,0) differ by (100)  
which is a reciprocal-lattice vector for SC but not for FCC].

SO far, so good, but not easily generalizable to non-uniform grids.  
The generalization requires a slight shift of point of view. The  
weight can be seen as the measure of a small volume associated to each  
integration points. Points in the interior of the irreducible wedge  
all have the same weight, because the grid is uniform. Points on the  
surface of the IW will have different weights according to the way the  
surface cuts the elementary volumes of the uniform grids. If you are  
good at geometry (I am not) you can recover the same weights that you  
get from symmetry considerations. This way of proceeding has the  
advantage of being easily generalizable (at least in principle) to non- 
uniform grids. You have to devise a way to tile the entire space with  
small volumes, each of them associated with each point of your grid.  
The "weight" of the point will the be just the volume of the 3D tile  
associated to it.

> ANY REFERENCE AND/OR TEXTBOOK IS REALLY APPRECIATED. (I just printed  
> the original M-P paper so I can read it... )

Non-uniform grids are treated in texts on the numerical solution of  
partial differential equations (search for "finite differences" or  
"finite elements"). I think that the most straightorward way of tiling  
a 3D volume is by using tetrahedra, wich are 3D simplexes. A simplex  
is the nD polyhedron delimited by the smallest number of (n-1)D  
surfaces. 2D simplexes are triangles, 3D simplexes are tetrahedra. I  
think that there automatic ways of generating tetrahedron tiles, but I  
do not know exactly. If I were you, I would try to figure out by  
myself. If you fail after 1-2 days of hard trying, then a library (or,  
may be, google) could come to the rescue.

Hope this helps.
Coraggio!
Stefano



---
Stefano Baroni - SISSA  &  DEMOCRITOS National Simulation Center -  
Trieste
http://stefano.baroni.me [+39] 040 3787 406 (tel) -528 (fax) /  
stefanobaroni (skype)

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