## The oscillator strength and the imaginary part

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Quxiao
Posts: 22
Joined: Fri Mar 26, 2021 11:27 am

### The oscillator strength and the imaginary part

Dear all:
I am a novice. Before asking the question,I have learned some topics about the strength of excitonic.I still have something need to ensure.
Firstly,I want to ensure that if the exact expression of the oscillator strength which printed in the E_sorted is the |BS_R|^2(BS_R=rho_cv*Acv) ?
or the (|BS_R|^2)*8pi/(q^2\Omega Nq) is the strength of the excitonic?
Secondly, is the residuals equal to strength? What is the relationship between the residuals and the strength?
Finally,I plot the spectrum by using the imaginary part of dielectric function,then I find the second absorption peak is higher than the first,which means the imaginary part of the dielectric function of the second is larger than first.However, the oscillator strength of the second absorption peak is smaller than the first.I am confused that the imaginary part of the dielectric function is proportional to the strength.From the http://www.yambo-code.org/wiki/index.ph ... xcitons.So what cause that ,if (ω − Eλ),the energy at denominator may cause that?(at Eq.23 of the2009 Yambo paper.)
Have a good day!
Quxiao
BIT
Quxiao in BIT,calculate the exciton

Daniele Varsano
Posts: 2997
Joined: Tue Mar 17, 2009 2:23 pm
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### Re: The oscillator strength and the imaginary part

Dear Quixiao,
what is printed in th E_sorted file is:
(|BS_R|^2)*8pi/(q^2\Omega Nq)

Note that in that file they are normalised to its maximum, the maximum is found at the beginning of the file "Maximum Residual Value=".
.I am confused that the imaginary part of the dielectric function is proportional to the strength
What it is proportional to the strength is the areas beneath each peak, while is absolute height will depend on the considered damping value you can provide in the input.
The imaginary part of the dielectric function is build by summing up Lorentzian function centred around each excitation energies multiplied by its strength using the damping factor provided in the input.
You can have that a peak is higher than another if you have for example excitation energies which are close and the Lorentzian function add up to each other.

It could be instruct to plot the Im eps file together with the E_sorted file s impulse e.g. in gnuplot:

p "o.eps*" w l, "o*E_sorted" u 1:2 w i

Best,
Daniele
Dr. Daniele Varsano
S3-CNR Institute of Nanoscience and MaX Center, Italy
MaX - Materials design at the Exascale
http://www.nano.cnr.it
http://www.max-centre.eu/

Quxiao
Posts: 22
Joined: Fri Mar 26, 2021 11:27 am

### Re: The oscillator strength and the imaginary part

Dear Daniele Varsano
Thanks for your reply,"The imaginary part of the dielectric function is build by summing up Lorentzian function centred around each excitation energies multiplied by its strength using the damping factor provided in the input.",I want to know the exact expression of the imaginary part of the dielectric function,for example,what is the Lorentzian function?I know that is basic knowledge，however，I can not find that.
Thank you very much!
Best wishes!
Quxiao
BIT
Quxiao in BIT,calculate the exciton

Daniele Varsano
Posts: 2997
Joined: Tue Mar 17, 2009 2:23 pm
Contact:

### Re: The oscillator strength and the imaginary part

Dear Quxiao,
it comes from the imaginary part of the expression 23 of the yambo paper.
At 0 kelivn it should be a delta function centred around the excitation energy. In order to have a line shape to be compared with experiment an artificial damping (lifetime) is considered:
1/omega-E--> 1/(omega-E+i delta)
Here the definition of the Lorentzian function (or Cauchy distribution):
https://docs.mantidproject.org/nightly/ ... tzian.html

Best,
Daniele
Dr. Daniele Varsano
S3-CNR Institute of Nanoscience and MaX Center, Italy
MaX - Materials design at the Exascale
http://www.nano.cnr.it
http://www.max-centre.eu/

Quxiao
Posts: 22
Joined: Fri Mar 26, 2021 11:27 am

### Re: The oscillator strength and the imaginary part

Dear Daniele
I think maybe I got that.Is the delta represent the energy loss?
You mean that,when 0 k,the imaginary part of the dielectric function will be a delta function,I can learn that you want to construct the lorentzian function,is it?
the second question,at Eq.23 of the2009 Yambo paper，ρnm(k, q,G) = <nk|exp[i(q+G)·r]|mk−q>，however,BS_R= (ρcv*Acv)
so the ρcv is <v|r|c> or <v|exp(iqr)|c> ?
the oscillator strength is proportional to the Weighted transition strength，is it?
Thanks for your selfless help and patience ！
Quxiao
BIT
Quxiao in BIT,calculate the exciton

Daniele Varsano
Posts: 2997
Joined: Tue Mar 17, 2009 2:23 pm
Contact:

### Re: The oscillator strength and the imaginary part

Dear Quxiao,
s the delta represent the energy loss?
No, the delta is the resonance you have at the excitation energy, Fermi golden rule!
You mean that,when 0 k,the imaginary part of the dielectric function will be a delta function,I can learn that you want to construct the lorentzian function,is it?
The result of the calculation is a discrete excitation energy (eigenvalues of the of the excitonic matrix), i.e. a delta function. When adding a damping (lifetime) you have a Lorentzian function.
ρcv is <v|r|c> or <v|exp(iqr)|c> ?
in optics you are looking at the q-->0 limit so exp(iq.r) is ~ 1+iq.<v|r|c> (dipole approximation).
the oscillator strength is proportional to the Weighted transition strength，is it?
Yes the dipole moments are wighted with eigenvector of the BSE Hamiltonian.

Best,
Daniele
Dr. Daniele Varsano
S3-CNR Institute of Nanoscience and MaX Center, Italy
MaX - Materials design at the Exascale
http://www.nano.cnr.it
http://www.max-centre.eu/

Quxiao
Posts: 22
Joined: Fri Mar 26, 2021 11:27 am

### Re: The oscillator strength and the imaginary part

Dear Daniele：
Thanks for your reply! I have acquired quite a lot of knowledge！
I have the last question need your help，if I just want to get the dipole(<e|r|h>)of the specific excitons,what could I do in yambo?or I just could get the BS_R if I could not change the source code？
Your help was very much appreciated！
Quxiao
BIT
Quxiao in BIT,calculate the exciton

Daniele Varsano
Posts: 2997
Joined: Tue Mar 17, 2009 2:23 pm
Contact:

### Re: The oscillator strength and the imaginary part

Dear Quxiao,
what do you mean exactly by "the dipole(<e|r|h>)of the specific excitons".
A specific exciton will have a linear combination of dipole elements weighted by the Acv solution of the BSE.
You can look at the weight by using ypp (see here http://www.yambo-code.org/wiki/index.ph ... 021_school)

Next you can get the single particle dipoles using python scripts.

Best,
Daniele
Dr. Daniele Varsano
S3-CNR Institute of Nanoscience and MaX Center, Italy
MaX - Materials design at the Exascale
http://www.nano.cnr.it
http://www.max-centre.eu/

Quxiao
Posts: 22
Joined: Fri Mar 26, 2021 11:27 am

### Re: The oscillator strength and the imaginary part

Dear Daniele:
I know that "a specific exciton will have a linear combination of dipole elecments weighted by the Acv solution of the BSE":
http://www.yambo-code.org/wiki/images/e/e8/Weights.png
The strength is defined ashttp://www.yambo-code.org/wiki/images/f/ff/Strengh.png
So for an exciton,o-3D_BSE.exc_qpt1_weights_at_*reports the weights defined as http://www.yambo-code.org/wiki/images/e/e8/Weights.png,
so I guess that the Energy column in the o-3D_BSE.exc_qpt1_weights_at_* is defined as |<e|D|h>|^2,is it?
if not that,I think the energy is also related to dipoles element, is it?
the Column in the o-3D_BSE.exc_qpt1_weights_at_* :
# Band_V Band_C Kv-q ibz Symm_kv Kc ibz Symm_kc Weight Energy
Have a good day !
Thanks
BIT
Quxiao
Quxiao in BIT,calculate the exciton

Daniele Varsano
Posts: 2997
Joined: Tue Mar 17, 2009 2:23 pm
Contact:

### Re: The oscillator strength and the imaginary part

Dear Quxiao
the energy column is simply Ec-Ev,
note that this is the KS energy and not the QP energy.

Best,
Daniele
Dr. Daniele Varsano
S3-CNR Institute of Nanoscience and MaX Center, Italy
MaX - Materials design at the Exascale
http://www.nano.cnr.it
http://www.max-centre.eu/