Lattice Vibrations and Phonons Questions and Answers

Answer: A

The density of states is proportional to \(\omega\).

In two dimensions the number of modes with wave vector less than \(k\) is proportional to the area of a circle in \(k\)-space, \(N \propto \pi k^{2}\). With the linear (acoustic) dispersion \(\omega = v k\), \(N \propto \omega^{2}\), and

\[ g(\omega) = \frac{dN}{d\omega} \propto \omega \]

\(\omega^{2}\) is the Debye result for three dimensions. \(\sqrt{\omega}\) is the 3D density of states for free particles with quadratic dispersion, and \(1/\sqrt{\omega}\) belongs to one dimension with quadratic dispersion.

Answer: C

The resulting phonon has wave vector ⟨0.48 Å^-1, 0.23 Å^-1⟩, the only option inside the first Brillouin zone.

Three phonons combine to \(3 \times \langle 1.2, 0.6 \rangle = \langle 3.6, 1.8 \rangle\) Å^-1. This lies outside the first zone, which runs over \(|k_{x}| \le \pi/a = 0.52\) Å^-1 and \(|k_{y}| \le \pi/b = 0.79\) Å^-1, so the process is an umklapp process and a reciprocal lattice vector is subtracted to bring the result back in. The reciprocal lattice spacings are \(2\pi/a \approx 1.05\) and \(2\pi/b \approx 1.57\) Å^-1.

\[ \begin{aligned} k_{x} &= 3.6 - 3(1.05) \approx 0.46 \\ k_{y} &= 1.8 - 1.57 \approx 0.23 \end{aligned} \]

With \(2\pi/6\) rounded to 1.04 the \(x\) component comes to 0.48, the value printed in the keyed option. The other options lie outside the zone: ⟨3.60, 1.80⟩ is the unreduced sum, and ⟨2.56, 0.23⟩ subtracts only one \(2\pi/a\).