Yambo Community Forum

Dear all,
When calculating the BSE containing momentum, for example，the Absorption @ Q(99) iku: 0.281250030 0.453125030 0.00000000
in the output file o.eps_q99_diago_bse corresponds to the transition from which point in the VB to which point in the CB, and what rules are used to arrange it ？
Thanks!
Pangrt

Dear pangrt,
please add your full name in your signature, it is a rule of the forum.

The excitation energies from which the spectrum is build are calculated by diagonalization of the BSE equations so different transitions do participate with a certain weight (i.e. it is not an independent particle spectrum). The BSE matrix is built considering all the transition in the BZ connecting valence bands at (k-q) to conduction bands at k (v(k-q)-->c(k)).

You can inspect the weight of each transition to a given excitation using the ypp post processing:

ypp -e a

In order to do that, you need to store the eigenvector of the BSE solution, so remember you need to add the flag WRbsWF in your yambo BSE diagonalization input file.

Best,
Daniele

Dear Daniele,
Thank you very much for your help. But when I use ypp -e a to generate the file o.exc_qpt12_weights_at_31, the content of the file is
# Band_V Band_C Kv-q ibz Symm_kv Kc ibz Symm_kc Weight Energy
#
72.0000000 73.0000000 96.0000000 1.00000000 39.0000000 3.00000000 0.135884881 1.41313267
72.0000000 73.0000000 97.0000000 1.00000000 25.0000000 3.00000000 0.105637647 1.42940605
72.0000000 73.0000000 91.0000000 1.00000000 51.0000000 3.00000000 0.990237594E-1 1.41552889.
while the smallest direct band gap is at the 102nd K point.I am confused if the transition from v(96)->c(39) is included, then why there are only 102 absorption spectra instead of 102×102.
Thanks!
Pangrt

Dear RongTianPang,

Here the output shows that the excitation number 31 at qpt =12 is composed by several transitions.
The most relevant (weight=0.13) is the transition from a valence (72) band at kpt=96 to conduction (73) at kpt=39 rotated by the symmetry operation number 3. Then, also other transitions contribute.
The kpt and symmetry operation are reported in the report file.
sorry, but I cannot understand your doubts, can you reformulate in a more clear way?

Best,
Daniele

Dear Daniele,
Thank you very much, with your help, I have understood.

Thanks!
Pangrt

Dear Daniele,
I cannot understand why the exciton oscillator is several orders of magnitude stronger when considering the transition of momentum. In the direct transition, my calculation result is displayed in the file o.exc_qpt1_E_sorted as After considering the momentum, the oscillator strength of the exciton is displayed in the file o.exc_qpt2_E_sorted as I don’t know if my understanding is wrong or the calculation result is a problem.
Thanks!
Pangrt

Dear RongTianPang,

thanks for reporting, there could be a prefactor problem, I will investigate it.
Here the reason is that generalised dipoles are different from the q=0 and finite q case, being q.<v|r|c> for q=0 (dipole approximation), and <v|e^iq.r|c> for a finite q case. I need to check the prefactors in the finite q case. In any case it is just a multiplication factor. Let me check.

Best,

Daniele

Dear RongTianPang,
indeed there is a problem we will fix soon.

The q=0 max strength is correct, while for finite q the pre-factor is wrong.
Note the values reported are normalised to the maximum value so the relative intensity for each q is correct.
The absolute value instead should be modified as S=S*10^-10/|q|^2

Best,
Daniele

Dear Daniele,
Thank you for solving my doubts very quickly！

Thanks!
Pangrt

Dear Pangrt,
please note I wrote the wrong sign in the exponent, now corrected.

Best,
Daniele