Past exam of the mathematics course of the University of Cambridge / 2026 / ii / Paper 3 / 34E (Applications of Quantum Mechanics)

Bloch theorem follows because lattice translations commute with $H$ and with each other: simultaneous eigenstates obey $T_R\psi =e^{ik\cdot R}\psi$, so ${\psi }_k(r)=e^{ik\cdot r}{u}_k(r)$ with $u_k$ lattice-periodic. Conversely this form has the required translation character.

The listed points are precisely integer combinations of $a_1,a_2,a_3$ (the face-centred cubic lattice), as follows by sorting the parity classes of doubled Cartesian coordinates.

$|a_1\cdot (a_2\times a_3)|=a^3/2$.

Using $b_i=2\pi (a_j\times a_k)/V$ gives $b_1=(2\pi /a)(0,1,1)$, $b_2=(2\pi /a)(1,0,1)$, $b_3=(2\pi /a)(1,1,0)$. The reciprocal cell volume is $(2\pi {)}^3/V=16{\pi }^3/a^3$.

Bloch theorem $|k⟩\propto {\sum }_R{e}^{ik\cdot R}|R⟩$ are eigenstates with $E(k)=E_0-2\mu {\sum }_{i=1}^3\cos (k\cdot a_i)$.

The sum of three cosines ranges from $-3$ to $3$ on the Brillouin zone, so $\Delta E=12|\mu |$.

Near $k=0$, $E=E_0-6\mu +\mu {\sum }_i(k\cdot a_i{)}^2+O(k^4)$. The quadratic form has eigenvalues $a^2/4,a^2/4,a^2$ along its principal axes, so comparison with ${\hslash }^2{k}_j^2/(2m_j^{\ast })$ gives $m^{\ast }=2{\hslash }^2/(\mu a^2)$ twice and $m^{\ast }={\hslash }^2/(2\mu a^2)$ once (with signs interpreted at the selected band extremum).