Past exam of the mathematics course of the University of Cambridge / 2021 / ii / Paper 4 / 34B (Applications of Quantum Mechanics)

$E_0(k)={\hslash }^2{k}^2/(2m)$. Periodic boundary conditions give $k=2\pi n/L$. Orthogonality of plane waves gives $⟨k|U|k\prime ⟩=U_l$ when $k\prime -k=2\pi l/a$, and zero otherwise.

Nondegenerate perturbation theory fails where free energies related by a reciprocal vector coincide, at $k=l\pi /a$. Diagonalizing the two-state block places the gap centre at ${\hslash }^2(l\pi /a{)}^2/(2m)+\lambda U_0$ and gives width $2\lambda |U_l|+O({\lambda }^2)$.

For $V=2\lambda \cos (2\pi x/a)$, only $U_{\pm 1}=1$ are nonzero. To first order there is one gap of width $2\lambda$ at the first Brillouin-zone boundary $k=\pm \pi /a$; higher gaps vanish at this order. The allowed bands are the folded free-electron parabolas with that interval removed.

For the Dirac comb, Poisson summation gives $U_l=1$ for every integer $l$. Hence every crossing at $k=l\pi /a$ opens a gap of width $2\lambda$ to first order, producing an infinite sequence of allowed bands between consecutive gaps.

Filled bands separated by gaps explain the distinction between electrical insulators and conductors: a completely filled band cannot change its net crystal momentum under a weak field, whereas a partially filled band supports current.