Past exam of the quantum condensed matter course of the University of Cambridge / 2019 / Section B / 4

In the nearly-free electron model, a weak periodic potential has Fourier components $V_{\mathbf{G}}$ that couple plane waves at $\mathbf{k}$ and $\mathbf{k}-\mathbf{G}$. Far from a degeneracy their admixture gives small perturbative energy shifts. At a Bragg plane, however,so degenerate perturbation theory is required. In one dimension the first such crossing is at $k=\pm G/2$, the boundary of the first Brillouin zone.

Within the degenerate pair the eigenvalues are ${ϵ}_{G/2}\pm |V_G|$. ThusThe eigenstates are standing-wave combinations of counterpropagating waves. Their charge densities sit differently relative to the ionic potential, giving different potential energies and removing the free-electron degeneracy. The resulting avoided crossing produces continuous allowed energy bands separated by a forbidden interval. A Fourier component or symmetry-allowed coupling is needed: geometry alone does not require every possible gap to be nonzero.

Write $U=V_g$ for the coupling Fourier coefficient, with $|U|=|V_0|$ for the supplied cosine potential. Projecting the Schrödinger equation onto the two retained plane waves givesThe diagonal potential average is zero. The characteristic equation isso the nearly-free electron dispersion near a one-dimensional band gap isOnly the coupling magnitude enters the energies; its sign changes the standing-wave eigenvectors. The two-wave truncation is controlled near this crossing when $|V_0|\ll {ϵ}_{g/2}$, since the other plane waves are separated by much larger kinetic energies.

Let ${ϵ}_B={\hslash }^2{g}^2/(8m)$ and take $V_0>0$ as the coupling magnitude. The dispersion becomesExpanding the square root at the boundary givesUsing $1/m^{\ast }={\hslash }^{-2}{d}^{2}E/d{q}^2$, the band-edge effective mass in a weak periodic potential isThe requested expression is the leading weak-potential limit. The upper band has positive band curvature and small positive effective mass; the lower band has negative curvature at its maximum and can instead be described by positive-mass electron holes. The small-$q$ expansion additionally requires ${\hslash }^{2}g|q|/(2m)\ll V_0$; its validity interval shrinks with the gap.

An acoustic phonon branch approaches zero frequency linearly, ${\omega }_{\mathrm{ac}}\simeq v_s|q|$, because a long-wavelength translation has no restoring force. An optical phonon branch has nonzero frequency at $q=0$ and represents relative motion of atoms within a primitive unit cell. Such an optical branch requires more than one independent atom per primitive cell; a truly monatomic simple-cubic model has only the acoustic branches.

The Bose occupation is $N_q=[e^{\hslash {\omega }_q/(k_{B}T)}-1{]}^{-1}$. At low temperature, appreciably populated acoustic modes haveAs temperature falls, the available phonon wavevectors concentrate near the zone centre and optical modes freeze out when $k_{B}T\ll \hslash {\omega }_{\mathrm{op}}(0)$. These are thermal-occupation restrictions, not a change in the allowed geometric Brillouin zone.

For a spherical Fermi surface inside a primitive cubic lattice zone, the shortest reciprocal vector has magnitude $G=2\pi /a$. The nearest translated Fermi sphere is separated by the electron-phonon Umklapp thresholdA phonon of this wavevector can connect nearly opposite-velocity Fermi states using $\mathbf{k}+\mathbf{q}={\mathbf{k}}^{\prime }+\mathbf{G}$, transferring reciprocal-lattice momentum to the crystal and efficiently relaxing electrical current. With the supplied sound speed, its thermal scale isFor $T\ll T_U$ such phonons have exponentially small occupation. As they become populated, Umklapp scattering can make the electrical conductivity fall rapidly. Lower-energy normal scattering and residual disorder still give resistance below this scale; this is a crossover estimate, not a sharp transition. The numerical estimate extrapolates the linear acoustic dispersion to $q_U$ and assumes the spherical Fermi surface implicit in giving one $k_F$.

Using $2k_F$ alone as the phonon wavevector misses this mechanism: direct backscattering would need $1.4\pi /a$, beyond the first-zone edge along this direction, while the reciprocal-lattice transfer permits it with $0.6\pi /a$.