Past exam of the general relativity and cosmology and condensed matter physics paper of Part B of the physics course of the University of Oxford / 2011 / Section VI / 7

Normalize the plane waves in a periodic crystal of area or volume $\mathcal{V}$. The Fourier coefficients of the periodic potential obeyTranslation of the integral by any lattice vector $\mathbf{R}$ leaves it invariant but multiplies it by $e^{-i(\mathbf{k}-{\mathbf{k}}^{\prime })\cdot \mathbf{R}}$. A nonzero result therefore requires $\mathbf{k}-{\mathbf{k}}^{\prime }$ to be a reciprocal lattice vector.

Near an isolated Bragg scattering degeneracy, ordinary nondegenerate perturbation theory fails because the free energies coincide. Retain the resonant plane waves $|\mathbf{k}⟩$, $|\mathbf{k}+\mathbf{G}⟩$ and apply degenerate perturbation theory. Absorb the constant $V_0$ into the energy origin. Since a real potential has $V_{-\mathbf{G}}=V_{\mathbf{G}}^{\ast }$, the projected Schrödinger equation has coefficient matrixA nonzero eigenstate requiresOff-resonant plane waves give higher-order energy shifts omitted in this leading nearly-free electron model.

Let $u=V_{10}<0$ andAt $X=(\pi /a,0)$ the two degenerate plane waves have momenta $(\pi /a,0)$ and $(-\pi /a,0)$ and coupling $u$. ThusTheir symmetric and antisymmetric combinations are the lower and upper states respectively. The local band gap is $2|u|$.

Put $w=V_{11}<0$. At $M=(\pi /a,\pi /a)$, the free energy is $2E_X$. The four cyclic combinations in the supplied secular equation give the four-plane-wave splitting at a square-zone cornerFor example, the normalized coefficient vectors in the source's cyclic basis are $(1,1,1,1)/2$, $(1,0,-1,0)/\sqrt{2}$, $(0,1,0,-1)/\sqrt{2}$, and $(1,-1,1,-1)/2$. Because $u<w<0$, these are ordered as written: a lower singlet, a central doublet and an upper singlet. The sketch puts the $X$ and $M$ energies on one common scale; its representative negative couplings are stated in the caption.

There are two electrons per primitive unit cell; a spin-degenerate energy band accommodates exactly that number. A semiconductor results when the first band is full and does not overlap the second. In the leading two-wave/four-wave approximation, compare the first-band maximum at $M$ with the second-band minimum at $X$:This is an indirect band gap; a nonzero local splitting at $X$ alone is insufficient. At the onset,With $a=0.3\mathrm{nm}$, ${\hslash }^2/(2m_e)=0.03810\mathrm{eV}\mathrm{n}{\mathrm{m}}^2$ gives $E_X\simeq 4.178\mathrm{eV}$, henceFor $V_{10}$ more negative than this value the estimated indirect band gap is positive; at equality the bands just touch. This is the onset estimate within the source's nearly-free electron model truncation, not an exact critical potential after coupling all plane waves.