[Wien] What's the difference between the spin-polarized and the non spin-polarized calculations

[Gavin Abo]

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:

Planewaves, Pseudopotentials and the LAPW Method by David J. Singh and 
Lars Nordström [ 
http://link.springer.com/book/10.1007%2F978-0-387-29684-5 ]
http://www.wien2k.at/reg_user/textbooks/double_counting.pdf
http://www.wien2k.at/reg_user/textbooks/DFT_and_LAPW_2nd.pdf

My attempt at general answers:

No parameters are monitored to make the 2 densities equal.  As seen on 
slide 21 of http://www.wien2k.at/events/ws2015/rolask_rela.pdf , 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 
http://www.wien2k.at/events/ws2015/WS22-KS-DFT-LAPW.pdf , 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).

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, 
http://www.nano-phdschool.unimore.it/site/home/phd-students/documento102017667.html 
(page 24) ].

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.

On 10/19/2016 5:18 PM, Abderrahmane Reggad wrote:
>
> Thank you Dr Gavin for your reply and also for your interesting for my 
> questions.
>
> I have checked the 2 presentations but I didn't find what I look for .
>
> 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 .
>
> Best regards
> -- 
> Mr: A.Reggad
> Laboratoire de Génie Physique
> Université Ibn Khaldoun - Tiaret
> Algerie
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