Past exam of the physics A course of the University of Cambridge / 2022 / Paper 2 / Section C

For a plane wave wavefunction ${\psi }_k(x)\propto e^{ikx}$, impose the cyclic or periodic boundary condition ${\psi }_k(x+L)={\psi }_k(x)$. It requiresAdjacent allowed wavenumbers on the full signed $k$ axis are spaced by $\Delta k=2\pi /L$. Ignoring spin, the density of states per unit $k$ is consequentlyBoth positive and negative $k$ are included in this density. It is not yet the density per unit magnitude $|k|$, which has an extra factor of two away from zero. In $d$ independent dimensions the same periodic boundary conditions give one spatial state per $k$-space cell of measure $(2\pi /L{)}^d$.

To preserve the convention of part (a), first count one spin species. Define $g_d^{(1)}(ϵ)dϵ$ as the number of spatial states in the whole specimen in that energy interval. PutIn one dimension there are intervals near both $+k$ and $-k$, so the one-dimensional free-electron density of states isThus $g_1^{(1)}\propto {ϵ}^{-1/2}$. The factor of two here comes from opposite momenta, not spin. For physical Electrons with two spin states, the spin-summed density of states is $g_1=2g_1^{(1)}=L\sqrt{2m}{ϵ}^{-1/2}/(\pi \hslash )$. These expressions are continuum approximations away from individual discrete levels.

In two dimensions the density of states in $k$ space per spin species is $(L/2\pi {)}^2$. A annulus of radius $k$ and thickness $dk$ has area $2\pi kdk$. ThereforeThe two-dimensional free-electron density of states is constant in energy, $g_2^{(1)}\propto {ϵ}^0$, above the band bottom. Including both spin states gives $g_2=L^{2}m/(\pi {\hslash }^2)$. Dividing by $L^2$ gives the density per unit physical area, rather than the total specimen's density.

In three dimensions use the $k$-space density of states $(L/2\pi {)}^3$ per spin species and the spherical shell volume $4\pi k^{2}dk$:Thus the three-dimensional free-electron density of states grows as ${ϵ}^{1/2}$. For physical Electrons, including both spin states givesAll three total density of states functions have units ${\mathrm{J}}^{-1}$ when energy is in joules; they count states per energy interval. The paper explicitly suppresses spin in (a) but does not repeat that convention in (b), so giving both conventions avoids a hidden factor-of-two ambiguity. Parts (c)--(f) below use the physical two-spin-state electron gas.

Let $d=1,2,3$ and take a specimen of length, area or volume $L^d$. One monovalent atom per unit cell suppliesElectrons. At zero temperature, the Pauli exclusion principle fills the lowest energy states, with two spin states per spatial wavevector. The occupied region is an interval, disk or ball of radius $k_F$. Using the $k$-space density of states gives, in one dimension,In two dimensions,In three dimensions,Each Fermi energy follows from the free electron model dispersion relation $E_F={\hslash }^2{k}_F^2/(2m)$. The Fermi wavevectors have units ${\mathrm{m}}^{-1}$ and the Fermi energies have units joules for $a$ in metres and $m$ in kilograms. No numerical lattice spacing is supplied, so numerical energies in electronvolts cannot be uniquely calculated. The spin factor of two is necessary here for actual monovalent crystals; the spinless auxiliary count of part (a) was not an instruction to remove electron spin from these crystals.

For each primitive linear, square or cubic Bravais lattice of spacing $a$, the reciprocal lattice has spacing $2\pi /a$. The first Brillouin zone is consequently the interval, square or cube with faces at components $k_i=\pm \pi /a$. The nearest zone boundary is a distance $k_B=\pi /a$ from the centre. The ratios areAll are below one: the ideal free-electron Fermi surface lies wholly inside the first Brillouin zone. Its nearest approach to the boundary is in three dimensions, where it lies only about $1.53\%$ short of the nearest face. A lattice-induced band gap at a zone face therefore does not by itself make these half-filled systems insulating in the stated free-electron model. Interactions, structural changes and actual electronic band structure can alter that model prediction.

The length, area and volume of a first Brillouin zone are $(2\pi /a{)}^d$. The corresponding occupied $k$-space measures areHenceThis is state counting, independent of the interval/disk/ball geometry. One Brillouin zone contains one spatial state per unit cell in a single energy band, and that band can hold two Electrons per cell because of spin. One Electron per cell therefore occupies half of it. The “volume enclosed” is the occupied Fermi sea, not the lower-dimensional area of the Fermi surface itself. Its half-volume does not mean half the distance to the zone boundary in two or three dimensions.

Keep the Electron number fixed, soand $\mu (0)=E_F$. Heating smooths the Fermi-Dirac distribution across the Fermi energy, creating holes below $\mu$ and particles above it. The sign of the low-temperature adjustment of chemical potential follows from the local slope of the density of states.

In one dimension, $g\propto {ϵ}^{-1/2}$ decreases with energy. At fixed $\mu$, smoothing removes more Electrons below it than it adds above it, so $\mu$ initially rises. In three dimensions, $g\propto {ϵ}^{1/2}$ increases, so $\mu$ falls. In two dimensions the constant density of states makes the leading changes cancel, so $\mu$ stays extremely close to $E_F$ in the degenerate regime.

For quantitative checks, a Sommerfeld expansion gives, for $k_{B}T\ll E_F$,For a constant two-dimensional density of states above an unbounded parabolic band bottom, the number integral can be done exactly:soIt actually decreases, but its low-temperature correction is exponentially small rather than quadratic. “Constant in two dimensions” is therefore a low-temperature approximation, not an exact statement for every $T$.

The one-dimensional rise is also only an initial trend. In the high-temperature Maxwell-Boltzmann distribution limit, the ideal $d$-dimensional electron gas has $\mu \simeq k_{B}T\ln [n_d{\lambda }_T^d/2]$, where $n_d=N_e/L^d$ and ${\lambda }_T=\sqrt{2\pi {\hslash }^2/(m{k}_{B}T)}$ is the thermal de Broglie wavelength. It eventually becomes negative and decreases in every dimension. The one-dimensional chemical potential thus passes through a maximum. This high-temperature statement refers to the stipulated unbounded free-electron dispersion relation, not a claim that a real finite-width crystal band remains parabolic indefinitely.