Hello,
in addition to what Gavin wrote, I'd remark that the implementation of
finite q AC conductivity (which is simply related to the dielectric
function) should not be difficult. Instead of taking (2.17) in
https://arxiv.org/pdf/cond-mat/0402523v1.pdf
you'd use Eq. 2.15 before the limit in q is taken, that's all. This is a
standard Kubo formula result (for conductivity or epsilon). However, you
should first sit back and think how large the q of interest will be. If it
is much smaller than any reasonable k-mesh finesse (I mean \Delta k
between two adjacent k-points) then you are out of luck.
Karel
--- x ---
dr. Karel Vyborny
Fyzikalni ustav AV CR, v.v.i.
Cukrovarnicka 10
Praha 6, CZ-16253
tel: +420220318459
Date: Mon, 23 May 2016 06:25:56 -0600
From: Gavin Abo <gs...@crimson.ua.edu>
To: ben amara imen <imen.benama...@gmail.com>
Cc: Karel Vyborny <vybor...@fzu.cz>
Subject: Re: [Wien] imaginary dielectric function
In the article by Ambrosch-Draxl et al. [
https://arxiv.org/pdf/cond-mat/0402523v1.pdf [arxiv.org] ], you can see on
page 26:
Furthermore, for the high energy photons the q = 0 limit is not justified
any more and thus a finite momentum transfer should be considered in
future
work.
I think that future work for q not equal to zero was never implemented for
WIEN2k. I believe that is because they started working on their own DFT
program called EXCITING [ https://en.wikipedia.org/wiki/EXCITING
[en.wikipedia.org] ].
If you look on the webpage for the elk branch [
http://elk.sourceforge.net/
[elk.sourceforge.net] ], you should see:
added dielectric function calculation for arbitrary q-vector, see
'LiF-Yambo' example (experimental)
On Sat, 21 May 2016, ben amara imen wrote:
Hi,
Using Wien2k , I have calculate the imaginary part of dielectric function
(epsilon2) as function of photon energy. I have two questions: :
1) How I can know at which high point symmetry , the epsilon2 is calculated
2 ) I need to calculate , with wien2k, the imaginary part of dielectric
function versus q ( epsilon 2=f(q)). How can I do this ?
Thanks for your help
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