Past exam of the condensed matter physics course of the University of Oxford / 2015 / 1

Take one conduction electron per atom and a three-dimensional free electron gas, with spin degeneracy two and electron mass $m_e$. State counting inside the occupied Fermi sphere givesThe highest occupied energy at zero temperature is $E_F={\hslash }^2{k}_F^2/(2m_e)$, soThis relates the atomic number density to the Fermi temperature only because monovalency and the free-electron dispersion have been assumed. A real energy band can require a different density of states and band effective mass.

For the isotropic Debye model, count three acoustic polarizations up to one spherical wavevector cutoff $q_D$. The number of vibrational modes per volume isWith $\omega =cq$ and $\hslash {\omega }_D=k_B{\theta }_D$,The effective $c$ is the Debye sound speed: matching the low-frequency mode density of one longitudinal and two transverse branches requires $3/c^3=1/c_L^3+2/c_T^3$. It is not generally the arithmetic average of their speeds. The cutoff preserves the mode count; it need not coincide with the actual nonspherical Brillouin zone boundary.

Pauli exclusion principle prevents most filled electronic states from changing occupation at $T\ll T_F$. Only a shell of width of order $k_{B}T$ around the Fermi energy participates. Since the density of states near that energy is of order $n/E_F$ per volume and energy, the fraction of active electrons is of order $k_{B}T/E_F=T/T_F$. Their typical change of energy is of order $k_{B}T$, so the thermal energy increment per volume scales asDifferentiating with respect to temperature gives the electronic heat capacityThe simple shell argument fixes the linear law, not its coefficient. A Sommerfeld expansion using the spin-summed free-electron density of states $g(E_F)=3n/(2E_F)$ gives $C_e=({\pi }^2/3)g(E_F)k_B^{2}T$, hence $A={\pi }^2/2$, consistent with the supplied value. The result excludes strong band-structure or interaction corrections and requires a degenerate electron population.

The Debye model treats the crystal as an isotropic elastic continuum with independent harmonic vibrational modes. There are three acoustic polarizations, approximated by linear dispersions and a common effective speed. A cutoff is chosen to retain $3N$ modes for $N$ atoms. The model neglects anharmonic interactions and the detailed high-wavevector dispersion. In a multi-atom primitive cell, optical branches require additional treatment; copper has one atom per primitive cell even though its conventional cubic cell contains four atoms.

A frequency interval contains, per volume,Each phonon mode has Bose-Einstein distribution occupation $(e^{\hslash \omega /(k_{B}T)}-1{)}^{-1}$. The fixed zero-point energy contributes no heat capacity. Its thermal energy density isFor $T\ll {\theta }_D$, extending the upper limit to infinity gives $u_{\mathrm{ph}}=(3{\pi }^4/5)n{k}_B{T}^4/{\theta }_D^3$. ThereforeThe cubic law reflects the three-dimensional acoustic mode density proportional to ${\omega }^2$. It is a low-temperature limit, not a valid extrapolation through the classical high-temperature regime.

The conventional face-centered cubic lattice cell contains four atoms, soThe preceding free-electron and Debye expressions giveAt low temperature the electronic and lattice heat-capacity crossover in a monovalent metal is found fromThus the electronic term dominates for $0<T\lesssim 3.4$ K within the normal-metal model. Both low-temperature approximations are well controlled at that crossover. At exactly zero temperature both heat capacities vanish; below sufficiently small finite-size energy scales a continuum model also needs revision.

Room temperature is not $T\ll {\theta }_D$: using the cubic formula at 300 K would incorrectly predict $C_{\mathrm{ph}}\simeq 142n{k}_B$, far above the harmonic high-temperature limit $3n{k}_B$. Instead use the Dulong-Petit law as an estimate:Keeping the finite cutoff in the Debye heat capacity givesAt 300 K this is $2.8006n{k}_B$, whereas $C_e=0.017994n{k}_B$, giving $\boxed{C_e/C_{\mathrm{ph}}=6.42\times 10^{-3}}$. Either appropriate room-temperature estimate shows that the lattice contribution dominates. The electronic approximation remains excellent because $300\ll T_F$.