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<div class="moz-cite-prefix">You likely have to derive the Kohm–Sham
equations and solve them for the wavefunction solutions (and look
into the WIEN2k source code) for the detailed answers to your
questions. I haven't done it myself, so I cannot help you there.
I think the go to references for that were:<br>
<br>
Planewaves, Pseudopotentials and the LAPW Method by David J. Singh
and Lars Nordström [
<a class="moz-txt-link-freetext" href="http://link.springer.com/book/10.1007%2F978-0-387-29684-5">http://link.springer.com/book/10.1007%2F978-0-387-29684-5</a> ] <br>
<a class="moz-txt-link-freetext" href="http://www.wien2k.at/reg_user/textbooks/double_counting.pdf">http://www.wien2k.at/reg_user/textbooks/double_counting.pdf</a><br>
<a class="moz-txt-link-freetext" href="http://www.wien2k.at/reg_user/textbooks/DFT_and_LAPW_2nd.pdf">http://www.wien2k.at/reg_user/textbooks/DFT_and_LAPW_2nd.pdf</a><br>
<br>
My attempt at general answers:<br>
<br>
No parameters are monitored to make the 2 densities equal. As
seen on slide 21 of
<a class="moz-txt-link-freetext" href="http://www.wien2k.at/events/ws2015/rolask_rela.pdf">http://www.wien2k.at/events/ws2015/rolask_rela.pdf</a> , there are two
equations, one for Psi_up and one for Psi_down, but for the
non-spin polarized case both equations are the same such that
Psi_up = Psi_down = Psi. So only one equation for the
wavefunction Psi needs to be solved for. As seen on slide 66 in
<a class="moz-txt-link-freetext" href="http://www.wien2k.at/events/ws2015/WS22-KS-DFT-LAPW.pdf">http://www.wien2k.at/events/ws2015/WS22-KS-DFT-LAPW.pdf</a> , the
calculation is given an initial charge density (during init_lapw),
then the charge (and spin) density should be computed from the
self consistent field (scf) cycles (run_lapw).<br>
<br>
On the other hand, the spin-polarized calculation (runsp_lapw) has
to solve two separate equations instead of one as shown on slide
24 in rolask_rela.pdf. Which is why for example there is lapw1 -up
and lapw1 -dn for the spin-polarized calculation and only just
lapw1 for the non-spin polarized. The simplified equations it
uses for the spin-polarized case was made possible by choosing the
z-axis for the direction of the magnetic field [ Ab Initio Study
of NiO-Fe Interfaces: Electron States and Magnetic Configurations
by L. D. Giustino,
<a class="moz-txt-link-freetext" href="http://www.nano-phdschool.unimore.it/site/home/phd-students/documento102017667.html">http://www.nano-phdschool.unimore.it/site/home/phd-students/documento102017667.html</a>
(page 24) ].<br>
<br>
The Bef term is crossed out on slide 21, so there should be no
exchange magnetic potential Bxc, since Bef = Bext + Bxc (from
slide 19). However, whether Bef term is not there or how the Bef
term is set to 0, I don't know and someone else might; I didn't
look into the source code to try to determine that.<br>
<br>
On 10/19/2016 5:18 PM, Abderrahmane Reggad wrote:<br>
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<div>Thank you Dr Gavin for your reply and also for your
interesting for my questions.</div>
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<div>I have checked the 2 presentations but I didn't find what I
look for .</div>
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<div>It's mentionned that in non spin-polarized calculation the
spin-up density = the spin-down density . Which parameters are
they monitored to make these 2 densities equal. I have read
that in this case the exchange magnetic potential will be
equal to zero. I want to know if it's so or not .</div>
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<div>Best regards</div>
-- <br>
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<div><font size="2"><span
style="color:rgb(0,0,255)">Mr:
A.Reggad</span>
<br>
<span
style="background-color:rgb(238,238,238)"><span
style="color:blue">Laboratoire
de Génie Physique</span></span><br>
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<font size="2"><span
style="color:rgb(0,0,255)">Université
Ibn Khaldoun - Tiaret</span><br>
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<div><font size="2"><span
style="font-family:monospace"><span
style="color:rgb(0,0,255)">Algerie</span></span></font><br>
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