Dear develpoer
I tried to calculate exciton strength using yanbopy in two different ways:
1. l_residual*r_residual
2. starting from exciton eigenvector and IP dipole using formula abs(\Sigma_kcv A_kcv&\lambda d_cvk)**2
I got quite consistent results using both approaches, the relative strength of exciton is the same, however strength obtained using this two methods differ by an overall factor of 1E10 for all exciton state. So I suppose it is because of the different unit . However, since strength is in unit of the of area (I suppose). If the former method adpot international unit, while the latter adpot ang. The overall factor should be 1E20 instead of 1E10. This confuses me. What is the possible origin of the factor 1E10? What units these two methods adopt respectively?
exciton residual from two different methods
Moderators: palful, amolina, mbonacci

 Posts: 15
 Joined: Thu Dec 01, 2022 12:53 pm
exciton residual from two different methods
Yunfei Bai
Institute of Physics Chinese academy of sciences, Beijing
Institute of Physics Chinese academy of sciences, Beijing
 palful
 Posts: 68
 Joined: Tue Jan 26, 2016 11:23 am
 Location: Modena and Milan
Re: exciton residual from two different methods
Dear Yunfei,
Is your system a 3D bulk or 2D? I suspect this is related to having a 2D system.
Cheers,
Fulvio
Is your system a 3D bulk or 2D? I suspect this is related to having a 2D system.
Cheers,
Fulvio
Dr. Fulvio Paleari
S3CNR Institute of Nanoscience and MaX Center
Modena, Italy
S3CNR Institute of Nanoscience and MaX Center
Modena, Italy

 Posts: 15
 Joined: Thu Dec 01, 2022 12:53 pm
Re: exciton residual from two different methods
Dear palful:
Yes, my system is 2D. Do you mean that when I calculated exciton dipole using l_residual*r_residual it may include another extra factor in unit of length apart from exciton dipole ? So maybe the second way :abs(\Sigma_kcv A_kcv&\lambda d_cvk)**2 should be correct ? Is the exciton dipole obtained in the second way in unit ang^2 ?
Best wishes!
Yes, my system is 2D. Do you mean that when I calculated exciton dipole using l_residual*r_residual it may include another extra factor in unit of length apart from exciton dipole ? So maybe the second way :abs(\Sigma_kcv A_kcv&\lambda d_cvk)**2 should be correct ? Is the exciton dipole obtained in the second way in unit ang^2 ?
Best wishes!
Yunfei Bai
Institute of Physics Chinese academy of sciences, Beijing
Institute of Physics Chinese academy of sciences, Beijing
 palful
 Posts: 68
 Joined: Tue Jan 26, 2016 11:23 am
 Location: Modena and Milan
Re: exciton residual from two different methods
Dear Yunfei,
I've looked better into this and I don't think it is related either to dimensionality or to units. It depends on the definition of the residuals.
The residual is defined including the scalar product between the momentum q and the dipole matrix element d:
abs(\Sigma_kcv A_kcv&\lambda q \cdot d_cvk)**2
where q and d_cvk are to be considered both vectors (d_cvk = <ckrvk> in the length gauge). However, in optical absorption we are always taking the limit q>0, and furthermore, we know that in the full dielectric function 1vX the q^2 dependence coming from the residual squared (i.e. from X) is cancelled by the 1/q^2 dependence of the Coulomb potential. Therefore, in the q>0 case yambo defines the residual by multiplying the dipole to a purely formal "small" q0=1e5 (called q0_def_norm in the yambo source code).
Therefore, you have
l_residual*r_residual = q0^2 * abs(\Sigma_kcv A_kcv&\lambda d_cvk)**2,
or equivalently
dipole in the excitonic basis = residual/q0.
Since q0^2=1e10, I believe this could be the reason for the difference you find, meaning there is nothing wrong after all.
Cheers,
Fulvio
I've looked better into this and I don't think it is related either to dimensionality or to units. It depends on the definition of the residuals.
The residual is defined including the scalar product between the momentum q and the dipole matrix element d:
abs(\Sigma_kcv A_kcv&\lambda q \cdot d_cvk)**2
where q and d_cvk are to be considered both vectors (d_cvk = <ckrvk> in the length gauge). However, in optical absorption we are always taking the limit q>0, and furthermore, we know that in the full dielectric function 1vX the q^2 dependence coming from the residual squared (i.e. from X) is cancelled by the 1/q^2 dependence of the Coulomb potential. Therefore, in the q>0 case yambo defines the residual by multiplying the dipole to a purely formal "small" q0=1e5 (called q0_def_norm in the yambo source code).
Therefore, you have
l_residual*r_residual = q0^2 * abs(\Sigma_kcv A_kcv&\lambda d_cvk)**2,
or equivalently
dipole in the excitonic basis = residual/q0.
Since q0^2=1e10, I believe this could be the reason for the difference you find, meaning there is nothing wrong after all.
Cheers,
Fulvio
Dr. Fulvio Paleari
S3CNR Institute of Nanoscience and MaX Center
Modena, Italy
S3CNR Institute of Nanoscience and MaX Center
Modena, Italy

 Posts: 15
 Joined: Thu Dec 01, 2022 12:53 pm
Re: exciton residual from two different methods
Dear palful:
That answers all my questions, thanks a lot for your reply!
That answers all my questions, thanks a lot for your reply!
Yunfei Bai
Institute of Physics Chinese academy of sciences, Beijing
Institute of Physics Chinese academy of sciences, Beijing