BSE calculations in spin-polarization systems

[Jing Liu]

Dear all,
Considering the spin-polarization system, I have a question that how to obtain the imaginary part of the BSE dielectric function with contributions from individual spins in practice

BSE for individual spins.png? (example shows in the attachment) https://journals.aps.org/prl/abstract/10.1103/PhysRevLett.122.236402 [attachment=0]BSE for individual spins.png

I have done an attempt in a FM semiconductor but I am not sure whether I am right.
For the spin-up channel, the filled bands are 1-64, while the empty bands are 65-240. For the spin-down channel, the filled bands are 1-60. Therefore, when I want to obtain the absorption spectrum of the spin-up channel, I set the tag "% BSEBands" as 63 | 68 in which two valence bands and four conduction bands are considered to build the BSE Hamiltonian. In this way, no transitions from the spin-down channel are included, right?
In the same way, if I set the "% BSEBands" as 59 | 64, only the transitions from spin-down channel are considered, right?
Thanks so much for any kind suggestions~

Best,
Jing

[Daniele Varsano]

Dear Jing,

given your electronic configuration with the setting you are using you are indeed looking at transition coming from a selected spin channel. The problem I see anyway is that it is not guaranteed that the spectra are converged.

Best,
Daniele

[Jing Liu]

Dear Daniele,
Thanks for your reply.
But in which way I can use to obtain the BSE from individual spins correctly?

Best,
Jing

[Davide Sangalli]

This feature is not coded.
But you can open the file src/bse/K_Transitions_setup.F and change the loop

Code: Select all

do i_sp_c=1,n_sp_pol

to

Code: Select all

do i_sp_c=1,1 !(for spin up)

and to

Code: Select all

do i_sp_c=2,2 !(for spin down)

So you would have two different executables.

Best,
D.

[Jing Liu]

Dear Davide,
Thank you so much~ I'll try!

Best,
Jing

[Davide Sangalli]

Just a warning.
The BSE mixes spin-up and spin-down transition.
With the change I suggested you avoid such mixing.
So the resulting spectrum will not be just the full one projected on the spin channel, but it could be substantially different.