Code "dynmat.x" should do what you need

Paolo

On 1/2/2026 12:29 AM, Tang, Weilun via users wrote:
Hi Quantum Espresso users:

I am studying the example 5 from PHonon. Below is the snippet of the output from ph.x.

These Raman tensors correspond to the atom and different polarization. Is there a way to find the Raman tensor corresponds to the vibrational modes?

Thanks for your time,

Weilun Tang

Department of Materials Science and Engineering
Iowa State University


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Raman tensor (au^-1) in cartesian axis           atom      1  (      -0.000000000      -0.000000000       0.000000000 )          (    -0.000000000      -0.000000000      -0.784974138 )          ( 0.000000000      -0.784974138       0.000000000 )          (  -0.000000000       0.000000000      -0.784974138 )          ( 0.000000000       0.000000000      -0.000000000 )          (  -0.784974138      -0.000000000      -0.000000000 )          ( 0.000000000      -0.784974138       0.000000000 )          (  -0.784974138      -0.000000000      -0.000000000 )          ( 0.000000000      -0.000000000       0.000000000 )           atom      2          (      -0.000000000      -0.000000000       0.000000000 )      (      -0.000000000      -0.000000000       0.793396465 )  (      -0.000000000       0.793396465       0.000000000 )          (     0.000000000      -0.000000000       0.793396465 )          (  -0.000000000      -0.000000000       0.000000000 )          ( 0.793396465      -0.000000000       0.000000000 )          (  -0.000000000       0.793396465      -0.000000000 )          ( 0.793396465      -0.000000000       0.000000000 )          ( 0.000000000       0.000000000      -0.000000000 )



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Paolo Giannozzi, Dip. Scienze Matematiche Informatiche e Fisiche,
Univ. Udine, via delle Scienze 206, 33100 Udine Italy, +39-0432-558216

_______________________________________________________________________________
The Quantum ESPRESSO Foundation stands in solidarity with all civilians 
worldwide who are victims of terrorism, military aggression, and indiscriminate 
warfare.
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