Past exam of the condensed matter physics course of the University of Oxford / 2016 / 4 / c / ii / Solution

Differentiate the positive-frequency branch on either side of the origin:It tends to $+4.0$ km/s as $k\to 0^{+}$ and to $-4.0$ km/s as $k\to 0^{-}$, and vanishes at either zone boundary. A positive-$k$ wavepacket therefore transports energy to the right, while a negative-$k$ one transports it to the left. At the zone edge, neighboring atoms oscillate out of phase and the dispersion is flat, giving zero group velocity even though the frequency is nonzero.

The graph has separate one-sided limits at zero because $\omega (k)$ has a cusp there. Assigning one differentiable slope at exactly $k=0$ would erase the distinction between the two propagation directions; a precisely uniform translation has no preferred propagation direction. Plotting only $0\le k\le \pi /a$ is also legitimate if the missing negative branch is stated. Group velocity and phase velocity $\omega /k$ are different except in the long-wavelength linear limit.