# [Pw_forum] Identifying the phonon mode from eigen vector

Ajit Vallabhaneni avallabh at purdue.edu
Wed Feb 8 04:00:47 CET 2012

```Dear users and developers,

I am trying to understand how the atoms will be displaced for each of the six phonon modes in graphene at various q's from the eigen vectors. At gamma point (0,0,0), the eigen vectors are symmetric with the real part being of the same sign for acoustic modes and of opposite signs for optical modes with zero imaginary part for all of them. But for a different q like the one showed below, I can tell which are long, transv and out of plane (z) modes, but i want to know if it is possible to distinguish optical and acoustic modes  (assuming frequencies and dispersion curves are not known). On a relevant note, i read some previous archives of 2009 discussing the possibility of identifying the frequencies by it's branch, not just from lowest to highest, i was wondering if that is possible at all without looking at the eigen vectors at each q.

Diagonalizing the dynamical matrix

q = (    0.000000000   0.384900179   0.000000000 )

**************************************************************************
omega( 1) =       8.644497 [THz] =     288.349368 [cm-1]
(  0.000000  0.000000  0.000000  0.000000 -0.111751  0.698190 )
(  0.000000  0.000000  0.000000  0.000000 -0.707137  0.000000 )
omega( 2) =      16.684631 [THz] =     556.539398 [cm-1]
(  0.369888  0.602593  0.000000  0.000000  0.000000  0.000000 )
( -0.687258  0.166557  0.000000  0.000000  0.000000  0.000000 )
omega( 3) =      23.786354 [THz] =     793.427359 [cm-1]
(  0.000000  0.000000  0.000000  0.000000 -0.111760  0.698249 )
(  0.000000  0.000000  0.000000  0.000000  0.707077  0.000000 )
omega( 4) =      29.776017 [THz] =     993.221002 [cm-1]
(  0.000000  0.000000  0.667916  0.231960  0.000000  0.000000 )
(  0.000000  0.000000  0.111765  0.698277  0.000000  0.000000 )
omega( 5) =      42.320122 [THz] =    1411.647329 [cm-1]
( -0.312873  0.634172  0.000000  0.000000  0.000000  0.000000 )
(  0.507076  0.492758  0.000000  0.000000  0.000000  0.000000 )
omega( 6) =      45.622781 [THz] =    1521.812160 [cm-1]
(  0.000000  0.000000  0.668026  0.231999  0.000000  0.000000 )
(  0.000000  0.000000 -0.111746 -0.698162  0.000000  0.000000 )
**************************************************************************

Thanks
Ajit

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