Past exam of the mathematics course of the University of Cambridge / 2009 / ii / Paper 4 / 33D (Applications of Quantum Mechanics)

For potential period $a$, a Bloch state has the form ${\psi }_k(x)=e^{ikx}{u}_k(x)$, with $u_k(x+a)=u_k(x)$, or equivalently ${\psi }_k(x+a)=e^{ika}{\psi }_k(x)$. Quasimomentum is defined modulo the reciprocal-lattice vector $2\pi /a$. A Brillouin zone is one fundamental interval for that identification, usually $[-\pi /a,\pi /a)$. Here $a=2\pi$, so it is $[-1/2,1/2)$.

For small $V_0>0$, the first free-particle degeneracy occurs at the zone edge $k=1/2$, between plane waves $e^{ix/2}$ and $e^{-ix/2}$ of energy $1/4$. In this two-dimensional subspace, the diagonal matrix elements of the potential vanish. Its Fourier coefficient at wave number one is $-V_0/2$; the $\cos 2x$ term cannot connect these two waves at first order. Thus degenerate perturbation theory gives the matrixwith energies $1/4\pm V_0/2+O(V_0^2)$. Near the edge the two uncoupled energies have opposite slopes, and diagonalization gives the avoided crossing, so the lowest band gap is

For large $V_0$, expand about each deepest minimum $x=2\pi j$:The local quantum harmonic oscillator $-d^2/d{x}^2+(5V_0/2)x^2$ has levels $(2n+1)\sqrt{5V_0/2}$. Its localization width is $O(V_0^{-1/4})$, so the quartic term changes these low levels by $O(1)$. Tunnelling between successive deep wells gives exponentially narrow energy bands. The secondary minima at $x=(2j+1)\pi$ have potential zero, and their levels lie far above the first two deep-well levels, whose energies are $-2V_0+O(\sqrt{V_0})$. Therefore the gap between the first and second bands is the first oscillator spacing to leading order:The $O(1)$ anharmonic shifts and exponentially small bandwidths do not change this leading estimate.