Past exam of the mathematics course of the University of Cambridge / 2020 / ii / Paper 2 / 35C (Applications of Quantum Mechanics)

For a lattice of period $a$, wavevectors differing by the reciprocal-lattice vector $2\pi /a$ describe the same translation eigenvalue. The first Brillouin zone may therefore be chosen as

Bloch theorem states that every energy eigenstate in a periodic potential may be chosen in the formTo prove it, let $T_a$ be translation by one period:Periodicity gives $[T_a,H]=0$. The translation operator is unitary, so it can be diagonalized within each energy eigenspace. Choose a common eigenstate,Unitarity gives $|\lambda |=1$, so $\lambda =e^{ika}$ for some real $k$. Thus $\psi (x+a)=e^{ika}\psi (x)$. Defining $u_k(x)=e^{-ikx}\psi (x)$ giveswhich proves the theorem.

The potential has period $2\pi$, so the first Brillouin zone is $-1/2\le k<1/2$. At its boundary the free-particle statesare degenerate, with energyThe relevant Fourier coefficient is supplied by $V_0\cos x$:The term $-V_0\cos 2x$ changes wavevector by two and has no matrix element within this degenerate pair. Thus degenerate perturbation theory gives the matrixand the two band-edge energiesThe nearly-free electron model therefore predicts the lowest band gap

The minima occur at $x=(2j+1)\pi$, where $V=-2V_0$. Writing $x=(2j+1)\pi +\xi$ givesThe harmonic approximation near a potential minimum therefore has frequencybecause $\frac{1}{2}m{\omega }_0^2{\xi }^2=\frac{5}{2}{V}_0{\xi }^2$. The two lowest localized energies are

In the tight-binding model, tunnelling between neighbouring wells broadens each localized level into a band,For large $V_0$, the hopping amplitudes $J_n$ are exponentially small compared with the oscillator spacing. Hence the gap between the first two bands isup to exponentially small tight-binding bandwidth corrections.