Dispersion relation diagrams, phonons

[_Andreas]

In dispersion relation diagrams, where omega is plotted against k, omega is sometimes nonzero at k=0. How is this possible? I thought a wave had to have a nonzero wavenumber

 

[Dr Transport]

Optical phonons have non-zero components at the center of the Brillouin zone.

 

[Worzo]

The reason this is possible (as Dr T says, the optical branch has non-zero energy at k=0) is because k is not really a wavenumber.

p = [hbar]k is the 'crystal momentum', which is not a real momentum.

 

[ZapperZ]

Also, you could easily be reading a reduced zone scheme, in which the band from the next zone is folded back into the first zone.

Zz.

 

[kanato]

Optical phonons occur in crystals which have more than one atom per unit cell. If you have a phonon with k=0 that means the displacement of atoms is the same in every cell. When you have only one atom per cell, then a k=0 displacement is just a shift of the whole crystal, so there can't be a restoring force (hence, $$\omega=0$$). But if you have more than one atom per unit cell then the atoms could displace relative to one another (eg. like a bond-stretching mode). Then you can have a k=0 wave, where the displacement is the same in each cell, but the atoms in the cell move relative to one another. Then you will have a restoring force, and have $$\omega > 0$$ for this type of phonon.

 

[_Andreas]

Wow, thanks guys!