[Wannier] Energy mismatch among symmetry equivalent states in silicon
Vahid Askarpour
vh261281 at dal.ca
Fri Sep 2 00:55:22 CEST 2016
If I understand you correctly, the only way to force the symmetry equivalent states to have identical energies is to use more k points.
I am actually trying to apply symmetry in EPW to calculate the scattering rates for kpoints in the irreducible wedge and use symmetry to generate the scattering rates for the k points outside the wedge. However, I noticed that for silicon, the wannier-interpolated energies at symmetrically equivalent points are not identical.
Increasing the k points, as you suggested, would increase the size of the electron-phonon matrix elements for the coarse grids (terabyte range) beyond our capability to store.
It seems that I cannot use symmetry to simplify EPW calculations.
Thank you so much for your insightful comments.
Vahid
> On Sep 1, 2016, at 7:11 PM, Nicola Marzari <nicola.marzari at epfl.ch> wrote:
>
> On 01/09/2016 21:27, Vahid Askarpour wrote:
>> So with 8 bands, I get smaller spreads but the energy mismatch appears. It seems that I get better WFs with 8 bands than 4 bands. Or perhaps I am
>> missing something here.
>>
>
> Well, what does better mean? More localized allows for smoother
> interpolation, but symmetry is also important.
>
> The part of the manifold that comes from conduction is disentangled
> from a larger group of empty bands, and the resulting Wannier functions
> do not satisfy the symmetries of the crystal.
>
> Of course, the more kpoints you use, the more you will force the
> interpolation to go through the real calculated eigenvalues.
>
>
> nicola
>
>
> ----------------------------------------------------------------------
> Prof Nicola Marzari, Chair of Theory and Simulation of Materials, EPFL
> Director, National Centre for Competence in Research NCCR MARVEL, EPFL
> http://theossrv1.epfl.ch/Main/Contact http://nccr-marvel.ch/en/project
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