Past exam of the quantum condensed matter course of the University of Cambridge / 2013 / 1

Use the $e^{-i\omega t}$ convention, so passive photon absorption has positive imaginary permittivity. For a single Lorentz oscillator model resonance, write the photon energy as $E=\hslash \omega$ andHere $S$ has units of energy squared, $\Gamma =\hslash \gamma$ is a damping energy, and ${ϵ}_b$ is an approximately constant background. The corresponding real and imaginary parts areThus the real part starts above the background, rises on the low-energy side of the resonance, drops rapidly through ${ϵ}_r^{\prime }={ϵ}_b$ at $2\mathrm{eV}$, and approaches the background from below at larger energy. The imaginary part starts at zero, has a positive photon absorption peak near $2\mathrm{eV}$, and decays above it. For weak damping its peak is close to the resonance, but not exactly fixed there for arbitrary damping. At asymptotically large energy the corrections scale as ${ϵ}_r^{\prime }-{ϵ}_b\sim -S/E^2$ and ${ϵ}_r^{\prime \prime }\sim S\Gamma /E^3$.

The resonance position is fixed at 2 eV; peak height, width and background are not specified. The figure chooses an explicit finite damping and oscillator strength only to illustrate the shape over 0–5 eV. In particular, a negative real permittivity above resonance is possible for suitable strength and damping, but is not forced by the resonance energy alone. Setting damping exactly to zero would give a singular idealized response instead of a finite-width photon absorption peak.

In the nearly-free electron model, the free parabola ${\epsilon }_k^0={\hslash }^2{k}^2/(2m)$ is coupled to copies shifted by reciprocal wavevectors $G=2\pi /a$. The first degeneracy is at $k=\pm \pi /a$, where the counterpropagating waves have equal kinetic energy. Retaining their coupling gives two branchesConsequently a nonzero Fourier component $V_G$ opens a gap $\boxed{\Delta E=2|V_G|}$ at the zone edge. The dispersion remains approximately free-electron-like away from the crossing, but the branches avoid crossing and their slopes flatten at the band edges. The horizontal axis can be labeled either wavevector $k$, with edges at $\pm \pi /a$, or physical momentum $p=\hslash k$, with edges at $\pm \pi \hslash /a$.

At the edge, an appropriate choice of spatial origin makes $V_G$ real. The eigenstates are equal-weight standing waves. In a normalization length $L$ containing an integer number of periods, representative real forms areThey come from the sum and difference of $e^{i\pi x/a}$ and $e^{-i\pi x/a}$, up to a global phase. Their densities have maxima and nodes displaced by $a/2$ relative to one another. Both wavefunctions change sign under translation by $a$, as required by their zone-edge Bloch phase, while their densities are lattice-periodic.

For the figure's choice $V(x)=2V_G\cos (2\pi x/a)$ with $V_G<0$, the potential minima occur at $x=na$. The cosine state puts density there and is lower, while the sine state puts a node there and is higher. Reversing the potential's phase relative to the coordinate origin exchanges that assignment. Since the question supplies only the spatial period, it fixes the zone-edge locations and qualitative standing-wave structure, but not a numerical gap or an origin-independent choice of which drawn sine/cosine is lower.

For one isotropic carrier species in the Drude model, the longitudinal electrical resistivity at zero transverse electric current is ${\rho }_{xx}=1/(ne\mu )$, and the Hall electrical resistivity has magnitude $|{\rho }_{yx}|=B/(ne)$. Equivalently, the electric field balances the magnetic deflection through $E_y=v_{x}B$, while $|v_x|=\mu |E_x|$. Therefore the magnitude of the Hall angle obeysThe required ratio givesThe electron density cancels. With electric current along $+x$ and field along $+z$, Electrons have negative Hall coefficient and hence $E_y<0$; the requested one-percent condition concerns the magnitude. This calculation assumes Electrons dominate transport and the transverse contacts draw no electric current, so a Hall field can develop.