Is it possible to calculate eph coupling per phonon mode on different q-points?

[CHILLL]

Dear Experts and users,

I am very sorry if this question has been asked before.
Could you please tell me if EPW can provide information about eph coupling per phonon mode at a certain q-point?
I am referring now to the isotropic coupling, e.g. as in QE that outputs something like:

Code: Select all

  Gaussian Broadening:   0.005 Ry, ngauss=   0
     DOS = 11.828320 states/spin/Ry/Unit Cell at Ef= -1.315632 eV
     lambda(    1)=  0.0000   gamma=    0.01 GHz
     lambda(    2)=  0.0000   gamma=    0.00 GHz
     lambda(    3)=  0.0000   gamma=    0.01 GHz
     lambda(    4)=  1.5756   gamma=   12.02 GHz
     lambda(    5)=  0.0349   gamma=    1.62 GHz
     lambda(    6)=  0.2431   gamma=   14.26 GHz
     lambda(    7)=  1.4495   gamma=   99.51 GHz
     lambda(    8)=  0.1898   gamma=   22.28 GHz
     lambda(    9)=  1.1763   gamma=  153.64 GHz
     lambda(   10)=  0.0651   gamma=   17.08 GHz
     lambda(   11)=  0.1034   gamma=   29.04 GHz
     lambda(   12)=  0.0504   gamma=   17.05 GHz

In general, it would be very useful to see the coupling on the phonon band structure(or the full BZ) to see what modes are responsible for the most of the coupling. Could you please tell me if something like this exists in the package?
Thank you very much!

Best regards, Mikhail

[hlee]

Dear Mikhail:

Are you referring to $\lambda_{q\nu}$?
If so, you can check the file of lambda.phself (See Page 8 of the slide in http://indico.ictp.it/event/8301/sessio ... al/0/0.pdf ).
Additionally, you can check the array of lambda_all, for instance, the subroutine selfen_phon_q in the file of selfen.f90.

Even if you want the different thing, you can easily extract this information.

For example,
you can check the input flag of prtgkk and the subroutine of print_gkk in the file of printing.f90.
Based on this, you need some additional evaluations.

Sincerely,

H. Lee

[CHILLL]

Dear H. Lee,

$\lambda_{q\nu}$ is exactly what I was looking for.
Thank you very much!

Best regards, Mikhail