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
