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Re: full GW+BSE calculation

Posted: Mon Apr 11, 2016 4:14 pm
by Daniele Varsano
Dear Khaled,

1) When you solve the BSE, the meaningful part of the exciton energy is the real part. If you are interested in exciton lifetime you need to do a model for it first.
2) If you want to extract the transition dipole matrix you can write them either looking into ./src/bse/K_diag_compute_epsilon.F, when the epsilon is built after the diagonalization of the dielectric matrix.
Otherwise with the postprocessing ypp, when excitons are sorted out (have a look in ./ypp/exciton_sort_and_report.F), the BS_R are read, normalized and printed out.

Best,

Daniele

Re: full GW+BSE calculation

Posted: Wed Apr 13, 2016 2:01 pm
by Khaled
Dear Dr. Daniele Varsano,

Thanks a lot Dr. Daniele for your great help. I was able to extract the transition dipole matrix, but I am not sure a bout the units of the transition dipole matrix $ | \sum_{vck} \langle vq-k | e^{-iqr} | ck \rangle A_{\lambda} |^{2} $
because I extracted this term alone (even I couldn't figure out if you used the electric dipole approimation for this term or not). Please, Can you help me for getting this information. Thank you in advance.

Best regards,

Khaled.

Re: full GW+BSE calculation

Posted: Fri Apr 15, 2016 3:36 pm
by Daniele Varsano
Dear Khaled,
the internal unit in the code is the atomic unit (bohr for length, Hartree for energy etc..).
The electric dipole approximation it is used for the q=0 term.

Best,

Daniele

Re: full GW+BSE calculation

Posted: Thu Aug 11, 2016 5:23 pm
by Flex
What is exactly the unit of the matrix elements ?

A transition dipole has SI units of Coulomb.meters, so here, should it be electron charge.bohr ?

Re: full GW+BSE calculation

Posted: Tue Aug 16, 2016 9:41 am
by Daniele Varsano
Dear Thierry,
yes, it should be as you wrote.

Daniele

Re: full GW+BSE calculation

Posted: Sat Mar 07, 2020 3:43 pm
by haseebphysics1
Dear Yambors,

I have a pretty simple question.

In the BSE input file, one of the important parameter is BSSmod:

BSSmod= "d" # [BSS] (h)aydock/(d)iagonalization/(i)nversion/(t)ddft`

I was curious about the last option (t), here t represetns TDDFT? since it is just the method to solve BSE equation, so what does TDDFT means here. I have done TDDFT using LRCs approx. already.


Thanks,

Re: full GW+BSE calculation

Posted: Sat Mar 07, 2020 6:52 pm
by Daniele Varsano
Dear Haseeb,
this variable was meant for a tddft kernel derived from the BSE equation: see e.g. Phys. Rev. Lett. 91, 256402 (2003). This feature is not present anymore in the code.
Best,
Daniele