On Fri, 4 May 2007, Vasilii Artyukhov wrote: | I was wondering how to make k-space sampling in SIESTA efficient.
Dear Vasilii: it is efficient enough... | Since the | code doesn't use symmetry to generate the k-points in the IBZ only, using | origin-centered Monkhorst-Pack grids seems a major waste of effort. It is "a major waste" if you have few k-points; with more k-points it becomes a moderate waste - but yes, there are (always?) less irreducible k-points when they are shifted, than when they are unshifted, for a given mesh density. But take care that in a shifted mesh, one loses the information about band features at symmetric points (e.g., band edges), which sometimes might be important to take into account properly. | Now, as | I understand, the grid could be shifted to make all k-points inequivalent, | giving a better sampling (up to 2x2x2 = 8 times better) at the same | computational effort. The idea is exactly the opposite - by shifting the points you make them equivalent with more other points then while keeping them high-symmetric. Think about a cube from -1 to 1 (in units of pi/2a) along each axis, and having cubic symmetry around zero. You define 2 divisions, and (unshifted) you get four non-equivalent points (0,0,0), (1,0,0), (1,1,0), (1,1,1), out of 8 in total. In a shifted mesh, you'll have just one point (0.5, 0.5, 0.5) which is equivalent to 7 others. If you have just inversion, out of symmetry operations (as in Siesta), you'll have (seems to... I might be wrong) all 8 unequivalent points unshifted and 4 points shifted. | Is it correct? Should then my kgrid block | look like this? | | %block kgrid_Monkhorst_Pack | 4 0 0 .125 | 0 4 0 .125 | 0 0 4 .125 | %endblock No. The displacement (in the last column) is given in units of k-mesh step (that resulting from the division of reciprocal lattice vector by 4, in your example). Defining a strange number of 0.125 will ensure every your k-point to become non-equivalent with any other. This is completely legal but will leave you with 4*4*4=64 points, and no gain from the symmetry (inversion) at all. In real life, you'd never want to use any other number for the displaceent than 0. or 0.5. Conclusing this issue, I think it is long due (for someone... :-) to programm tetrahedron integration in Siesta, it should be not so difficult..? Best regards, Andrei +-- Dr. Andrei Postnikov ---- Tel. +33-387315873 ----- mobile +33-666784053 ---+ | Paul Verlaine University - Institute de Physique Electronique et Chimie, | | Laboratoire de Physique des Milieux Denses, 1 Bd Arago, F-57078 Metz, France | +-- [EMAIL PROTECTED] ------ http://www.home.uni-osnabrueck.de/apostnik/ --+
