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

The right spring changes length by $u_{n+1}-u_n$ and pulls atom $n$ to the right with force $C(u_{n+1}-u_n)$. The left spring contributes $-C(u_n-u_{n-1})$. Adding these restoring forces and applying Newton's second law givesUniform displacement makes both spring extensions vanish, as translational invariance requires. Search for a normal mode $u_n=u_0{e}^{i(kna-\omega t)}$. Substitution yieldsOne representative first Brillouin zone is $-\pi /a\le k\le \pi /a$. This is one positive-frequency acoustic branch: near zero wavevector, $\omega \simeq a\sqrt{C/m}|k|$, so the sound speed is $v_s=a\sqrt{C/m}$. Opposite signs of $k$ label opposite propagation directions, not a second polarization branch.

The model assumes small displacements, harmonic nearest-neighbor forces, one scalar displacement per atom and uniform equilibrium spacing. It excludes damping, anharmonic expansion and transverse polarizations. In a three-dimensional monatomic crystal there would normally be three acoustic polarizations; that counting cannot be imported into this explicitly one-dimensional scalar model.