Past exam of the condensed matter physics major option course of the University of Oxford / 2016 / 6 / Solution

The conduction and valence dispersions are two cones touching at the Dirac point, with slopes $\pm \hslash v_F$. Here $\mathbf{k}$ is measured relative to one valley centre. Each cone has a density of states proportional to $|E|$, which vanishes at the touching energy. Gate-induced carriers satisfy approximately $n=C_g(V_g-V_D)/e$ per unit area. Negative and positive offsets from $V_D$ produce holes and electrons, respectively. Away from neutrality a simple charge carrier mobility model gives $\sigma \simeq e\mu |n|$, so resistance decreases on either side of a maximum near $V_D$. Disorder, electron–hole puddles, contacts and quantum transport round the peak; the cone dispersion alone fixes neither its height nor its width, nor does it require exact electron–hole symmetry in a real device.

A direct derivation of the magnetic spectrum starts from the single-valley Dirac Hamiltonian operator $H=v_F({\sigma }_x{\pi }_x+{\sigma }_y{\pi }_y)$. Introduce oscillator operators using the magnetic length ${\ell }_B=\sqrt{\hslash /(e|B|)}$. Up to exchanging sublattices or valleys,Squaring gives adjacent oscillator number operators, henceThe positive expression printed in the source describes the electron branch only; the negative-energy partners must also be included. The same square-root dependence follows from quantizing circular orbit area, with the Dirac Berry phase cancelling the usual half-integer offset.

Each spin and valley contributes $Ae|B|/h$ states in area $A$. Including two spins and two valleys gives $N_{\mathrm{LL}}=4Ae|B|/h$, or areal degeneracy $4e|B|/h$. The zero-energy level is half-filled at charge neutrality: the carrier density counts departures from that state.

For $n=10^{16}{\mathrm{m}}^{-2}$, the filling factor and fourfold graphene Hall sequence areIn a plateau, ${\rho }_{xx}\simeq 0$ and $|{\rho }_{xy}|=h/({\nu }_N{e}^2)$. The Hall sign reverses between electron and hole doping; the sketch uses its magnitude to avoid tensor-index sign conventions. Representative plateau-centre fields are $20.68$, $6.893$, $4.136$ and $2.954\mathrm{T}$ for $\nu =2,6,10,14$. Between plateaux, crossing extended states in broadened levels gives longitudinal peaks and Hall steps. Their ideal filling-centre positions have $\nu =4j$, namelyThus peaks near $10.34$, $5.17$ and $3.45\mathrm{T}$ separate the first prominent plateaux as field increases. Many weak oscillations accumulate toward zero field and are damped there. Finite-width plateaux require localized states between extended-state energies; an exactly clean Landau spectrum does not by itself prescribe plateau widths. The illustration uses an explicitly schematic broadening, not inferred experimental widths.

At $6.5\mathrm{T}$, $\nu =6.36$, close to the $\nu =6$ plateau. With $v_F=10^6\mathrm{m}{\mathrm{s}}^{-1}$, $E_1=92.5\mathrm{meV}$ and $E_2=130.8\mathrm{meV}$. The relevant adjacent gap near the chemical potential is therefore $\Delta =E_2-E_1\simeq 38.3\mathrm{meV}$, rather than the gap from the zero level. Resolving levels requires quantum-lifetime broadening $\hslash /{\tau }_q\lesssim \Delta$ and thermal broadening appreciably smaller than $\Delta$, giving the order-of-magnitude limitsThe upper temperature is an energy-scale estimate, not a guaranteed sharp-feature temperature. For oscillation amplitudes the thermal damping factor is $X/\sinh X$, $X=2{\pi }^2{k}_{B}T/(\hslash {\omega }_c^{\ast })$, with $\hslash {\omega }_c^{\ast }=\hslash eB{v}_F^2/E_F\simeq 36.7\mathrm{meV}$ and $E_F=\hslash v_F\sqrt{\pi n}$. Requiring $X\lesssim 1$ gives the more conservative scale $T\lesssim 22\mathrm{K}$ for little thermal damping. Visible weaker features may persist above that scale; their actual cutoff depends on resolution, disorder and temperature-dependent scattering. The transport lifetime and quantum lifetime also need not coincide.