Yambo Community Forum

Dear Daniele:
Thanks for your patience!
I just want to know the |<e|D|h>|^2,the combination of one dipole elements,could I get that value from the outfile of yambo?
I need the dipole which is composed of electron-hole pair to analyze my system.
I think that the |Acv|^2 as weight could be obtained，so the dipole|<e|D|h>|^2 should also could be printed,is it?
how can I got the value of the dipole element or the square of modules of transition matrix elements of each electron-hole pairs?
That is very important for me to analyze the exciton properties in my system,please give me more help.
Everything goes well with your work！
Best wishes！
Quxiao
BIT

Dera Quxiao,
you can read them form the ndb.dipoles* databases.
Here, an example python script to extract them. Of course you need to adapt to your needs.

Otherwise the easiest way use the yambo-py utility.
You can have a look to this post on how to use it:
viewtopic.php?f=15&t=1889&p=9564&hilit=python#p9564


Best,
Daniele

Dear Daniele:
Thanks very much for your help
I use the script written by palful to get the dipole elements because that is easy to learn.
However I use the yambo4.4.1,so the filename is ndb.dip_iR_and_P(not the ndb.dipoles)
I got the dipole elements,however that is complex number,which is beyond my expectation.
Besides,in the o-3D_BSE.exc_qpt1_weights_at_14,I get the Acv(Band_V=15,Band_C =17)is not zero,but the dipole elements<15|D|17> is 0j which printed by the script
Now,firstly.I would use your script to get the dipole elements again.Then I think that edition maybe a reason for that ,so I need to use the yambo5.0,is it?
Thanks for your kindness!

BIT
QUxiao

Dear Quixiao,

however that is complex number,which is beyond my expectation.

wfs are complex so dipole moments are.

Besides,in the o-3D_BSE.exc_qpt1_weights_at_14,I get the Acv(Band_V=15,Band_C =17)is not zero,but the dipole elements<15|D|17> is 0j which printed by the script

Acv are the eigenvector of the BSE and they are not related with the dipoles.

Then I think that edition maybe a reason for that ,so I need to use the yambo5.0,is it?

In any case I suggest you to use an updated version of the code.

Best,
Daniele

Hi Quxiao,

You can print the transition dipole for each exciton with modifying the ypp/excitons/excitons_sort_and_report.F. viewtopic.php?f=9&t=1742&start=20

Best,
Xiaoming

Dear xiaoming:
Thanks for your help,but I use the yambo4.4.1,so I find that ypp is not very satisfied with your script.
I am appreciate it for your kindness, and I will think how to solve that.
maybe refer to your script or change the yambo5.0.
Best Wishes
Quxiao
BIT

Hi Daniele,

I have a slight confusion.
You mentioned:

Daniele Varsano wrote: ↑Tue Jun 29, 2021 12:47 pm Note that in that file they are normalised to its maximum, the maximum is found at the beginning of the file "Maximum Residual Value=".

Does the Strength column in 'o.exc_qpt1_E_sorted' give μ_s or μ_s^2. And in that case, if I want to obtain the true oscillator strength, do I need to multiply the Strength column values with maximum residual value or the square root of the maximum residual value?

Thanks and Regards

Dear Harshita,
please sign your post with your name and affiliation, this is a rule of the forum, and it can be done once for all by filing the signature in your user profile.

The reported strengths are the squared modulus of the Residuals, where the residuals are obtained by multiplying the BSE eigenvector by the dipole matrix elements:
Residual= \sum_k BS_mat(k,i) x conjg( BSS_dipoles(k) )

Here k is a generalized index (k,c,v).

Best,
Daniele

Pardon, Daniel!
Thank you for the quick response.
Can you please elaborate your answer, I didn't get it. I understand it may be very basic to you but any help would be highly appreciated.

Regards,
Harshita

Dear Harshita,

you can have a look at the Yambo cheatsheet:
https://www.yambo-code.eu/wiki/images/c ... et-5.0.pdf

Slide 11a gives you the expression of the optical spectrum.
The strength is essentially the numerator of the expression of eps_M for each lambda excitations.
I hope that now it is more clear.

Best,
Daniele