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

[Fecher, Gerhard]

I guess the answer is basic calculus
    B_eff = V_up - V_dn
is zero when V_up = V_dn and the equations in the spin polarized and the non spinpolarized case become the same, isn't it.
(Note: V_up=V_up(rho_up) and V_dn=V_dn(rho_dn) is used for short, the densities rho_up and rho_dn are calculated 
from the Kohn-Sham wave functions the V(rho) depend on the used exchange-correlation functional.)

Just take a pencil and write down the equations given in the trancparencies or textbooks and proof that I am right by
setting B_eff=0.

... or you finally did not understand what a selfconsistent field calculation means, 
    then you have to attend some basic courses on mathematics or theoretical physics.


Ciao
Gerhard

DEEP THOUGHT in D. Adams; Hitchhikers Guide to the Galaxy:
"I think the problem, to be quite honest with you,
is that you have never actually known what the question is."

====================================
Dr. Gerhard H. Fecher
Institut of Inorganic and Analytical Chemistry
Johannes Gutenberg - University
55099 Mainz
and
Max Planck Institute for Chemical Physics of Solids
01187 Dresden
________________________________________
Von: Wien [wien-bounces at zeus.theochem.tuwien.ac.at] im Auftrag von Gavin Abo [gsabo at crimson.ua.edu]
Gesendet: Donnerstag, 20. Oktober 2016 07:30
An: A Mailing list for WIEN2k users
Betreff: Re: [Wien] What's the difference between the spin-polarized and the non spin-polarized calculations

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