Past exam of the quantum condensed matter course of the University of Cambridge / 2022 / Section A

The drift velocity is the small ensemble-average velocity acquired by charge carriers in an applied electric field. It is distinct from their much larger random thermal or Fermi velocity. For conduction Electrons of charge $-e$, the isotropic Drude model giveswhere $\tau$ is the momentum relaxation time.

The positive electron mobility ${\mu }_e=|{\mathbf{v}}_d|/|\mathbf{E}|$ measures the ease of this drift in linear response. In the same approximation,Its units are ${\mathrm{m}}^2{\mathrm{V}}^{-1}{\mathrm{s}}^{-1}$; in an anisotropic crystal the mobility is a tensor.

The band effective mass describes the acceleration implied by the curvature of the dispersion relation. The group velocity is $\mathbf{v}={\hslash }^{-1}{\nabla }_{\mathbf{k}}E$, while $\hslash \dot{\mathbf{k}}=-e\mathbf{E}$. ThusFor an isotropic parabolic minimum this reduces to $E=E_c+{\hslash }^2{k}^2/(2m^{\ast })$ and $m^{\ast }\dot{\mathbf{v}}=-e\mathbf{E}$. It incorporates the periodic crystal potential and need not equal the bare electron mass. Negative electronic curvature near a band maximum is conveniently described by positive-mass electron holes.

Work at zero temperature and fixed total carrier density. Write $n_{\uparrow }=n/2+\delta$ and $n_{\downarrow }=n/2-\delta$. Let $g_s(E_F)$ be the single-spin density of states per unit volume in the unpolarised state. Since $d{\mathcal{E}}_s/d{n}_s=E_{Fs}$ and $d{E}_{Fs}/d{n}_s=1/g_s$, the linear kinetic-energy shifts cancel and the quadratic cost isThe repulsive interaction instead decreases under spin polarization:Hence the Stoner criterion for an opposite-spin contact repulsion isThe two forms differ only in whether the stated density of states counts one spin or both. For the three-dimensional free electron model, $g_{\mathrm{tot}}(E_F)=3n/(2E_F)$, soEquivalently, for fractional polarisation $\zeta =(n_{\uparrow }-n_{\downarrow })/n$, the kinetic-energy density is $\frac{3}{10}n{E}_F[(1+\zeta {)}^{5/3}+(1-\zeta {)}^{5/3}]$; its quadratic coefficient is $n{E}_F/3$, while the interaction contributes $-U{n}^2/4$. With number densities, $U$ has units of energy times volume and the displayed interaction is an energy density; multiplying by the sample volume does not change the threshold.

An angle-resolved photoemission spectroscopy intensity map locates occupied electronic band structure through its bright dispersing ridges. For a nearly isotropic graphene cone,Fit straight lines to the reasonably linear portions of the ridges, avoiding the broad central crossing, and extrapolate them to locate the Dirac point energy $E_D$. The PDF places the crossing near $k_y=0$ and plotted energy roughly $1.4$–$1.5\mathrm{eV}$. On a lower branch, a representative change from approximately $(|k_y|,E)=(0.10{\mathring{\mathrm{A}}}^{-1},0.9\mathrm{eV})$ towards $(0,1.5\mathrm{eV})$ gives a slope near $6\mathrm{eV}\mathring{\mathrm{A}}$. Allowing for the ridge width gives about $5$–$7\mathrm{eV}\mathring{\mathrm{A}}$, henceHere $1\mathrm{eV}\mathring{\mathrm{A}}/\hslash =1.52\times 10^5\mathrm{m}{\mathrm{s}}^{-1}$.

To determine the carrier Fermi energy, identify the Fermi level from a calibrated photoemission cutoff, then measure its separation from the Dirac point:The separation of the two Fermi crossings gives $2k_F$ for a cut through the cone centre. This determines the doping energy relative to the Dirac point; the absolute plotted energy origin is not itself a Fermi energy.

The reproduced panel has no marked Fermi level or stated energy-zero calibration. Its upper ridges end near $1.9\mathrm{eV}$, but that alone does not prove that the endpoint is a Fermi cutoff. If it is the intended cutoff, the estimate is $E_F^{\mathrm{cone}}\simeq 0.4$–$0.5\mathrm{eV}$ and $k_F\sim 0.06$–$0.08{\mathring{\mathrm{A}}}^{-1}$. If instead the plotted numbers are binding energies measured from an unstated Fermi zero, the inferred separation would be roughly $1.4$–$1.5\mathrm{eV}$. The supplied figure securely supports the slope estimate, but it does not distinguish these energy-reference conventions; an unconditional numerical Fermi energy requires the missing calibration.