Past exam of the quantum condensed matter course of the University of Cambridge / 2015

In the isotropic parabolic-band approximation implied by the single supplied curvature, the hole mass from a valence-band maximum isThe hole has charge $+e$, so its cyclotron angular frequency isIf “frequency” means cycles per second, $f_c={\omega }_c/(2\pi )=5.59\times 10^{10}\mathrm{Hz}$. The curvature unit is $\mathrm{J}{\mathrm{m}}^2$, so ${\hslash }^2/E^{\prime \prime }$ has mass units. The external constants used are $\hslash =1.054571800\times 10^{-34}\mathrm{J}\mathrm{s}$, $e=1.6021766208\times 10^{-19}\mathrm{C}$ and $m_e=9.10938356\times 10^{-31}\mathrm{kg}$.^[ref] A genuinely anisotropic band would require the cyclotron mass for the orbit plane, not a single arbitrary directional curvature.

Let the atomic spacing be $a$ and define the nearest-neighbour matrix elements $t_{\alpha }=⟨j,p_{\alpha }|H|j+1,p_{\alpha }⟩$, using the same translated orbital convention on every site. ThenFor the opposite sigma and pi hopping signs in a p-orbital chain, consider $p_z$ first: neighbouring lobes above the chain have the same sign, so the in-phase Bloch sum is bonding. Thus $t_{\pi }<0$ andFor $p_x$, the lobes facing one another have opposite signs for equal site coefficients. Bonding therefore needs alternating coefficients, giving $t_{\sigma }>0$ andEnd-on $\sigma$ overlap is normally stronger than side-on $\pi$ overlap, so the $p_x$ band is wider, with bandwidths $4|t_{\sigma }|$ and $4|t_{\pi }|$. The curve orientations require a stated hopping-sign convention; a Hamiltonian written with an explicit minus sign in front of the hopping uses correspondingly opposite parameter signs.^[ref]

A hydrogenic donor has the effective Bohr radius of a hydrogenic donorusing the external $a_0=0.52917721067\times 10^{-10}\mathrm{m}$.^[ref] Increasing donor density causes donor states to overlap and screens their binding, allowing a metallic impurity system rather than isolated frozen-out donors.

Using the standard approximate Mott criterion, $n_D^{1/3}{a}_B^{\ast }\simeq 0.25$, givesThe factor 0.25 is an externally documented empirical criterion, not supplied exam data or an exact universal threshold.^[ref] An order-of-magnitude overlap argument without that prefactor would give $(a_B^{\ast }{)}^{-3}\simeq 1.3\times 10^{25}{\mathrm{m}}^{-3}$. Compensation, disorder and multivalley anisotropy affect a real material's transition; this estimate uses the scalar mass and permittivity actually supplied, without substituting other germanium parameters.

Heavy fermion materials are correlated metals whose low-energy quasiparticles have exceptionally large effective masses. Partly localized $f$-electron moments hybridize with broader conduction states. The Kondo effect and coherence below the Kondo-lattice coherence temperature produce a narrow low-energy band, so a small energy change corresponds to a large momentum-space state density.

For the electronic heat capacity of a heavy Fermi liquid, the resulting density of states gives a large low-temperature electronic specific-heat coefficient,and often an enhanced magnetic susceptibility. Here $g$ includes both spins and must be expressed per sample, mole or volume consistently with $\gamma$. A coherent Fermi-liquid regime often has $\rho ={\rho }_0+A{T}^2$ with an enhanced coefficient $A$. Above coherence, scattering by local moments can instead give a Kondo-like resistivity upturn.

The competition between moment screening and intersite magnetic interactions can yield a heavy paramagnetic metal, magnetic order, quantum-critical behaviour or unconventional superconductivity. Ce- and U-based compounds supply familiar examples. “Heavy” describes the emergent electronic quasiparticles, not the ionic mass; the simple heavy-band picture need not survive at a quantum critical point.^[ref]

The Hubbard model combines hopping with a local repulsion,Hopping favours a broad itinerant band; the interaction penalizes opposite-spin double occupancy. For uniform Stoner decoupling of the Hubbard interaction, let $s=n_{\uparrow }-n_{\downarrow }$ be the spin imbalance per site. The interaction is $U(n^2-s^2)/4$, so polarization lowers it by $U{s}^2/4$, at the expense of kinetic energy.

If $D(E_F)$ is the noninteracting density of states per spin per site, small polarization costs $s^2/[4D(E_F)]$. Thereforeis the mean-field ferromagnetic instability criterion. The exchange splitting is $\Delta =Us$, and the paramagnetic response is enhanced as $\chi ={\chi }_0/[1-UD(E_F)]$. A density of states including both spins changes the written factor in the criterion.

The model explains how itinerant bands can polarize without imposing pre-existing localized atomic spins. It is not a claim that every large-$U$ metal becomes ferromagnetic: near half filling the Hubbard model can instead form a Mott insulator with antiferromagnetic superexchange of scale $4t^2/U$. Band filling, bandwidth and fluctuations matter.^[ref] The interaction-energy and kinetic-cost expansion above is the stated mean-field derivation of the enhancement.^[ref]

For an electron wavepacket in one isolated band under a uniform electric field,Because $ϵ(k+2\pi /a)=ϵ(k)$, acceleration through successive Brillouin zones makes the velocity periodic rather than increasing indefinitely. For $ϵ={ϵ}_0-2J\cos (ka)$,for a positive field along the chosen axis. Integration givesThe excursion amplitude of a Bloch-oscillating wavepacket is $2|J|/(e|E|)$ about its mean and peak-to-peak extent $4|J|/(e|E|)$. The turning of velocity corresponds semiclassically to Bragg reflection near a zone boundary.

Coherent Bloch oscillations require scattering/dephasing times long compared with $T_B$ and a field weak enough to avoid substantial interband tunnelling. Ordinary metals usually scatter too rapidly; semiconductor superlattices and cold atoms in periodic optical potentials provide cleaner realizations. A completely filled band has canceling velocities, so its net current does not oscillate simply because individual wavepackets can.^[ref]

For a spin-degenerate band, each unit cell contributes two available states. An even electron count per cell can fill an integer number of bands; a band insulator additionally needs a global gap between the highest occupied and lowest unoccupied bands. Even filling alone is not sufficient when bands overlap.

For a free two-dimensional gas with spin degeneracy two,The reciprocal vectors of the periodic potential are $Q\hat{\mathbf{x}}$ and $Q\hat{\mathbf{y}}$. Its first Brillouin zone is the square $-Q/2\le k_x,k_y\le Q/2$, of side $Q$ and area $Q^2$; the real-space cell has side $a=2\pi /Q$.

At zero potential, the extended-zone Fermi surface is the circle of radius $k_F$. Since $\sqrt{2}<\alpha <2$, it crosses every square face but misses the corners. For Fermi-surface reconstruction in a weak square potential, weak positive potential opens gaps at the Bragg crossings: the second band has electron pockets at the two inequivalent edge centres $X,Y$, while unfilled regions of the first band form a hole pocket at the zone corner $M$. Opposite edges are identified: four edge half-pockets are two electron pockets, and four corner pieces are one hole pocket.

For the circular-cap carrier count in a square Brillouin zone, put $R=k_F$ and $s=Q/(2R)=\alpha /2$. Each circular cap outside a square face has areaNo two such caps overlap in the specified interval. The total second-band electron area is $A_e=4C$. The first-band hole area is the square area minus the occupied central-circle area, $A_h=Q^2-\pi R^2+4C$. Including spin,Thus at $\alpha =\sqrt{3}$,The carrier densities are the electron- and hole-pocket counts, not the total electron density. The carrier compensation and even band filling in a square lattice follows because their difference obeys $n_e-n_h=n-Q^2/(2{\pi }^2)$, so compensation occurs at $\alpha =\sqrt{\pi }$. These area formulas are the $V_0\to 0^{+}$ limit, before finite-gap reconstruction changes the pocket shapes.^[ref]

The filling per real-space cell is $\nu =n{a}^2=2\pi /{\alpha }^2$. In the given interval, the only even integer filling is $\nu =2$, again requiring $\alpha =\sqrt{\pi }$. At any other fixed filling a full gap cannot yield an ordinary spin-degenerate band insulator in this unchanged periodicity. In particular, the $\alpha =\sqrt{3}$ system retains excess electron carriers even after its holes disappear.

A rough gap threshold comes from comparing different symmetry directions, rather than only a direct zone-face gap. Define $E_Q={\hslash }^2{Q}^2/(2m)$. The zone-face two-plane-wave gap for a cosine potential follows because each cosine couples plane waves by $-V_0/2$, giving the one-dimensional zone-edge levels $E_Q/4\mp V_0/2$. The four-plane-wave splitting at a square-zone corner is the sum of two such pairs. The potential is separable in $x,y$, so to first orderThe indirect gap of a separable square cosine potential is consequentlyAt the compensated filling, this is $V_{0,\min }\sim (\pi /6)E_F^{(0)}$, with $E_F^{(0)}={\hslash }^2{k}_F^2/(2m)$. An insulator needs both the even filling and a positive indirect gap. At the touching threshold the gap is zero; a positive direct gap at every Bragg crossing is not by itself sufficient. The estimate neglects higher-plane-wave corrections and is not an exact critical amplitude.

Assume three-dimensional parabolic band edges, with density-of-states masses $m_c,m_h$ and one valley per edge, each with spin degeneracy two. Their effective densities of states areFor the semiconductor mass-action law, integrating the square-root density of states against the Boltzmann tail gives $n=N_c{e}^{(\mu -E_c)/(k_{B}T)}$ and $p=N_v{e}^{(E_v-\mu )/(k_{B}T)}$. Multiplication cancels the chemical potential, givingIf there are additional valley degeneracies, multiply by their product; anisotropic bands use the appropriate density-of-states masses. In the later equal-mass model, $A=4m^3/(2\pi {\hslash }^2{)}^3$. No numerical mass or valley multiplicity has been invented.

When $\mu =E_c=0$, the conduction-edge occupation is exactly $1/2$, so the nondegenerate approximation for electrons fails at every positive temperature. Define the Normalized complete Fermi-Dirac integralFor the carrier densities with chemical potential pinned to a band edge, with the two specified equal masses, put $N(T)=2[m{k}_{B}T/(2\pi {\hslash }^2){]}^{3/2}$. Direct Fermi-Dirac integration then givesThus $n\propto T^{3/2}$ exactly within the parabolic model. At low temperature, $p\simeq N(T)e^{-E_g/(k_{B}T)}$ is thermally activated; at $k_{B}T\gg E_g$, $p/n\to 1$ from below. Both vanish at zero temperature and increase at fixed chemical potential. The other states that pin $\mu$ can supply the compensating charge; one must not additionally impose $n=p$.

For the Pauli reduction of the pinned-edge mass-action product, the exact product relative to the classical law isIt approaches $f_0$ at low temperature and $f_0^2\simeq 0.58545$ in the formal high-temperature limit. The low-temperature activation factor is unchanged but its prefactor is reduced; the classical mass-action formula overestimates the product. The high-temperature limit here refers to the given unbounded parabolic bands, not a claim that real semiconductor band structures remain unchanged at arbitrarily high temperature.^[ref]

Finally, for a fixed positive $\mu$, the zero-temperature Fermi sphere has radius $k_{\mu }=\sqrt{2m\mu }/\hslash$, soThe low-temperature expansion at fixed $\mu$ gives the fixed-chemical-potential Sommerfeld density increase:The monotonic density at fixed chemical potential in a square-root density of states initially rises quadratically from $n(0)$. For $k_{B}T\gg \mu$, its leading behaviour crosses to $f_{0}N(T)\propto T^{3/2}$. The fixed chemical potential is what allows the particle number to change.