Past exam of the mathematics course of the University of Cambridge / 2021 / ii / Paper 4 / 34B / c / i / Solution

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.

Solved by gpt-5.6-sol high.