Past exam of the physics A course of the University of Cambridge / 2013 / Paper 2 / Section C

spin normalization in two-dimensional free-electron state counting. Treat the mobile electrons as independent particles in a constant interior potential, with parabolic dispersion $ϵ={\hslash }^2{k}^2/(2m)$, and impose confinement or periodic boundary conditions. electron–electron interactions and the detailed periodic ionic potential are neglected, while Fermi statistics are retained. Each spatial state occupies area $(2\pi {)}^2/A$ in two-dimensional wavevector space. For spin degeneracy $g_s$,The printed coefficient $Am/(2\pi {\hslash }^2)$ is the per-spin density of states. Unpolarized spin-one-half electrons have $g_s=2$, doubling it. Actual graphene near its Dirac points has approximately linear dispersion, so its energy-dependent density of states is not this parabolic-model result; the plane geometry alone does not establish the free-electron dispersion.^[ref]

Occupied disk and Fermi energy. At zero temperature fill the lowest states up to a disk of radius $k_F$. State counting givesThus the physical spin-degenerate result is ${ϵ}_F=\pi {\hslash }^{2}N/(Am)$. If $N$ instead counts one spin species, the per-spin convention gives $2\pi {\hslash }^{2}N/(Am)$. Each occupied wavevector can contain one electron of each allowed spin, never two in the same complete one-particle state.

Occupation probability and chemical potential of a two-dimensional ideal Fermi gas. A single fermion state permits occupation zero or one. In a grand canonical ensemble their relative weights are $1$ and $e^{-\beta (ϵ-\mu )}$. Normalizing givesThe chemical potential is fixed by the total particle number. Here the total density of states $D=g_{s}Am/(2\pi {\hslash }^2)$ is constant above the band bottom. Integrating the occupation gives $N=D{k}_{B}T\ln (1+e^{\mu /(k_{B}T)})$, and thereforeThe approach is exponentially close in this ideal two-dimensional constant-density-of-states model; a three-dimensional $T^2$ correction should not be inserted here. A narrow interval of width of order $k_{B}T$ around $\mu$ is thermally smeared.

One mass per cell. If $u_j$ is the displacement from equilibrium, the two springs give $m{\ddot{u}}_j=\alpha (u_{j+1}+u_{j-1}-2u_j)$. Substituting $u_j=U{e}^{i(qja-\omega t)}$ yieldsThe first Brillouin zone is $-\pi /a\le q\le \pi /a$; adding $2\pi /a$ leaves the sampled displacements unchanged. The branch is linear near zero, with sound speed $a\sqrt{\alpha /m}$, and reaches $2\sqrt{\alpha /m}$ at either zone edge.

Two masses per cell. Take a cell of length $2a$, with light displacement $u_j=U{e}^{i(2jaq-\omega t)}$ and heavy displacement $v_j=V{e}^{i[(2j+1)aq-\omega t]}$. The coupled equations reduce toA nonzero displacement requires determinant zero:Solving this quadratic in ${\omega }^2$ givesNow the first zone is $|q|\le \pi /(2a)$. The minus branch is acoustic: as $q\to 0$, both sublattices move together and ${\omega }_{-}\to 0$. Its sound speed is $a\sqrt{2\alpha /(M+m)}$. The plus branch is optical: at $q=0$ the atoms move oppositely with $mU+MV=0$, giving finite frequency $\sqrt{2\alpha (1/m+1/M)}$. This terminology describes relative motion; optical activity itself also requires coupling to the electromagnetic field.

diatomic-chain zone-edge polarizations and dimensionality. At the edge, $\cos qa=0$, so the heavy and light sublattices decouple: the acoustic edge frequency is $\sqrt{2\alpha /M}$ and the optical edge frequency is $\sqrt{2\alpha /m}$. HenceThe acoustic edge mode moves only the heavy sublattice; the optical edge mode only the light one. If $M=m$, the gap closes and the two branches are the zone-folded monatomic branch.

A three-dimensional crystal with two atoms per primitive cell has six branches: three acoustic and three optical. For propagation along a suitable symmetry direction these resolve into one longitudinal and two transverse branches of each type. The wavevector and Brillouin zone are three-dimensional; anisotropic force constants determine the dispersions, and polar crystals can have longitudinal–transverse optical splitting. A universal six-branch curve cannot be specified from masses alone.