[QE-users] Magnetic anisotropy energy in QE 6.3

[BARRETEAU Cyrille]

Of course to get reliable magnetic anisotropy you should drastically increase the number of Kpoints with respect to the example..

Cyrille

========================
Cyrille Barreteau
CEA Saclay, IRAMIS, SPEC Bat. 771
91191 Gif sur Yvette Cedex, FRANCE
~~~~~~~~~~~~~~~~~~~~~~~~~~~~
+33 1 69 08 38 56 /+33  6 47 53 66 52  (mobile)
email:     cyrille.barreteau at cea.fr
Web:     http://iramis.cea.fr/Pisp/cyrille.barreteau/
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________________________________
De : users [users-bounces at lists.quantum-espresso.org] de la part de Marcos Veríssimo Alves [marcos.verissimo.alves at gmail.com]
Envoyé : jeudi 12 juillet 2018 12:10
À : Quantum Espresso users Forum
Objet : Re: [QE-users] Magnetic anisotropy energy in QE 6.3

Hello Cyrille and Paolo,

Thanks for the quick response. I was interested really about the inner workings, which should be described in the paper that Cyrille mentioned. I will take a look at it. The actual execution of the calculation should be quite straightforward, from what I saw yesterday in the examples folder.

Once again, thanks to both of you for the response. If I run into any problems I'll ask, but hopefully all will be fine.

Best,

Marcos

Em qui, 12 de jul de 2018 10:37, BARRETEAU Cyrille <cyrille.barreteau at cea.fr<mailto:cyrille.barreteau at cea.fr>> escreveu:
Hi Marcos

The implementation of the Force Theorem has been described in the following paper:
https://journals.aps.org/prb/abstract/10.1103/PhysRevB.90.205409

The procedure is the following:
first perform a scf calculation with scalar relativistic pseudo
then perform  nscf calculation with fully relativistic pseudo (option lforcet=.true., nosym=.true') starting from previous scf charge  (startingpot='file') with various spin orientations (theta=0,90 for example)
Finally perform a projwfc calculation with  lforcet=.true. and the value of the Fermi level from the nscf calculation (same ef_0 for all calculations).

Then you get a file with the energy decomposed over the various atoms and orbitals of the system..
The anisotropy is obtained by difference between the different spin orientations.

hope it helps..

Cyrille


========================
Cyrille Barreteau
CEA Saclay, IRAMIS, SPEC Bat. 771
91191 Gif sur Yvette Cedex, FRANCE
~~~~~~~~~~~~~~~~~~~~~~~~~~~~
+33 1 69 08 38 56 /+33  6 47 53 66 52  (mobile)
email:     cyrille.barreteau at cea.fr<mailto:cyrille.barreteau at cea.fr>
Web:     http://iramis.cea.fr/Pisp/cyrille.barreteau/
========================
________________________________
De : users [users-bounces at lists.quantum-espresso.org<mailto:users-bounces at lists.quantum-espresso.org>] de la part de Paolo Giannozzi [p.giannozzi at gmail.com<mailto:p.giannozzi at gmail.com>]
Envoyé : jeudi 12 juillet 2018 10:04
À : Quantum Espresso users Forum
Objet : Re: [QE-users] Magnetic anisotropy energy in QE 6.3

here? PP/examples/ForceTheorem_example/

P.

On Thu, Jul 12, 2018 at 10:01 AM, Marcos Veríssimo Alves <marcos.verissimo.alves at gmail.com<mailto:marcos.verissimo.alves at gmail.com>> wrote:
Hi all,

Browsing QE 6.3's documentation, I saw that MAE can be calculated as a post-processing step to a pw.x scf calculation. What is the exact procedure followed? I.e., in the post-processing calculation is the spin density rotated, and then SOC is included? Is there any reference that details the procedure used when the MFT is applied in Espresso?

Best regards,

Marcos

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Paolo Giannozzi, Dip. Scienze Matematiche Informatiche e Fisiche,
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