[ADF-LIST] A question about the "KSpace" in the DFTB

Pier Philipsen philipse at scm.com
Mon Jun 6 14:39:57 CEST 2016

Dear DFTB user,

DFTB uses the old symmetric tetrahedron method of band. It splits up the irreducible wedge of the Brillouin zone (IBZ) into tetrahedra. For each of these tetrahedrons the k-space parameter is applied. Along all the edges of the tetrahedrons you will get 5 points (with kspace=5). The resulting grid depends strongly on the shape of the IBZ and cannot be simply compared to a regular grid such as 5*5*5. 

The kspace parameter has no unit, just like with a normal grid n1*n2*n3 the numbers have no dimension.

We plan to bring the regular grid also to DFTB, because with this scheme it is more easy to set a sensible defaults for a particular system.

Best regards, 
Pier Philipsen



> Op 6 Jun 2016, om 13:22 heeft ydinghz <ydinghz at hznu.edu.cn> het volgende geschreven:
> 
> Dear ADF:
>         
>   In the DFTB Manual, it shows the KSpace parameter controls the used k-points in the DFTB calculations. However, what is the unit for this parameter. 
>   For example, I want to set a 5*5*5 k-points. In the BAND Manual, the KSpace can be set as  Grid n1 {n2} {n3} by hand. 
>   However, for the DFTB calculation, such Grid setting can not be applied.
>   If I set KSpace 5, does it correspond to 5*5*5 k-points or not?
>   Thank you for your help.
>  
> Yi Ding
> Hangzhou Normal Univeristy, China
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