Yambo Community Forum

Hello,
I would like to know the exact meaning of the strength (residuals?) printed in the sort and report.
Looking at the code it reports sum_c,v | A_c,v < c| r | v> |^2 times a number of prefactors I assume have to do with the numerics?
In this case I would assume, that the units are charge^2 length^2 (or just length^2). But in the forum these values are also referred to as oscillator strength, which have no unit. And if I'm not mistaken these strengths divided by energy go into the dielectric constants, which mean the strength has a unit of energy?
Maybe in atomic units it's all the same, but I need to the dipole moment in SI units so C^2m^2 (or m^2), and I'm really not sure which unit conversion to use.

Also, would I need to get rid of some of the prefactors to get the proper dipole moment?

Thanks for your hard work and making Yambo available freely.
Cheers,
Robert

Dear Robert,
maybe it is useful to look at Eq.23 of the 2009 Yambo paper.

There you can see the dielectric matrix expression. Following that notation in Yambo what is called BS_R is defined the
BS eigenvector multiplied by the dipole moment (\rho_cv*Acv).
The Residuals are defined as |BS_R|^2 multiplied by the factor you see in Eq.10 (all beside the energy at denominator).
Residuals*8pi/(q^2\Omega Nq).
So, the whole expression as the unit of energy. Note that in the code before dumping it in the output it has been converted from Hartree to eV (note HA2EV).

It is important to note that the strengths reported in the output are normalized to their max value so in order to look at their absolute value you need to rescale them. The scaling factor is in the output as: If you want to look at a different quantity and you need to look at dipole moments only \rho_cv, there are available tool to extract the values from the computed databases ndb.dipoles.

Best,
Daniele

Dear Daniele,
thanks for the reply. Unfortunately, I am still a confused as before.

So, I was thinking of Eq.23 before and thought the residuals might have energy units, but with atomic units one never knows. There might be a hidden hbar or whatever somewhere.
F.e.: If the residuals are in Ha (or eV later), and \Omega = (2\pi)^3/volume and q^2 is 1/length^2 then |BS_R|^2 must be energy times over length^5 ? Or something weird like that.

Originally, I thought |BS_R|^2 would be the quantity I want, and it should be in units of bohr radius squared, I hoped. I used GaAs as a test system, where I get |BS_R|^2 about 1e-7 (a_0^2 ?) (I modified the code to print it directly in ypp) , which yields about 4e-28 (m^2 ?), which is far too small (should be about 1e-20 m^2) for the transition dipole matrix element of the that specific exciton.

I checked the yambopy source and there isn't any part dealing with BSE dipoles explicitly, only the normal ones, and the get_chi in excitondb.py uses the same expressions as yambo, and doesn't offer further inside.

I'm sorry if I'm dense or overlooked some obvious information.

Cheers,
Robert

Dear Robert,

1. Note that \Omega here is the direct lattice volume and not the BZ volume.
2. I'm not sure BS_R is in ao unit as there should be a volume to cell to normalize as <psi|r|psi> is integrated in the unit cell.
3. What you can do is to extract is the <v|r|c> dipoles, which are not summed up and weighted for a particular exciton.
4. About the discrepancy, you find for GaAs I cannot tell much, I hope someone else can give you some hints.

Best,
Daniele

Dear Robert

regarding your last question, I do not have an answer,
but I can suggest you this paper, done with Yambo,

https://arxiv.org/abs/1908.09962

where they calculated the radiative lifetime of excitons that is related to the exciton dipole matrix elements.
In such a way you have a system, the GaN, to compare with, and all the relative formulas.

I hope it will be useful to you

best regards
Claudio Attaccalite

Dear Robert,
let me correct with respect to the previous post:

2. I'm not sure BS_R is in ao unit as there should be a volume to cell to normalize as <psi|r|psi> is integrated in the unit cell.

besides that: the BS_R is defined as <c|e^iq.r|v> which for q=0 is then treated as q0.<c|r|v> where q0 is a small number (10^-5).
Thus the dipole is BS_R/qo

This could exaplain why you find such a small number for the GaAs case.

Best,
Daniele

Hi Robert

not completely sure if there is some prefactor missing but I agree with Daniele I think the excitonic dipole you are looking for is BS_R/q0_def_norm
This should be in length units (bohr) .

Dear Claudio, Daniele, Palummo,

thank you all for your help.
I completely did not think about the 1/q, although it's so obvious now that you spelt it out.
Sorry for not seeing it myself.

Cheers,
Robert

Daniele Varsano wrote: ↑Wed Apr 22, 2020 1:36 pm
The Residuals are defined as |BS_R|^2 multiplied by the factor you see in Eq.10 (all beside the energy at denominator).
Residuals*8pi/(q^2\Omega Nq).

Dear Daniele,
The Residuals are 8pi/(q^2\Omega Nq) in your explanation. It should be 8pi^2 or just 8pi? And for loe dimentions system like 2D, the omega is volume of the supercell aor the area of the xy plane, say z along the vauum.

Thanks!

Xiangtian

Dear Xiangtian,

I think it is 8pi, have a look to Eq.23 of the 2009 Yambo paper (Computer Physics Communications 180 (2009) 1392–1403):
Omega is the volume of the supercell, even in 2D systems, the BZ is 3D.

Best,
Daniele