Past exam of the condensed matter physics course of the University of Oxford / 2017 / 2

The density of states $g(E)$ counts one-electron states per energy interval, including both electron spin orientations. It may be a total or a per-volume quantity; $U$ must have the matching normalization. The Fermi-Dirac distributionis the occupation probability of such a state. In the fixed-volume free electron model, the integration variable $E$ and the density of states do not depend on $T$. The occupation depends explicitly on $T$ and implicitly through the chemical potential $\mu (T)$ needed to preserve electron number. Thermal expansion or a changing electronic band structure would require additional effects outside this model.

At $T\ll T_F$, only states within order $k_{B}T$ of the Fermi energy change their occupation. There are order $g(E_F)k_{B}T$ such states, each carrying an excitation energy of order $k_{B}T$. Thus $U(T)-U(0)\propto g(E_F)(k_{B}T{)}^2$ and the electronic heat capacity is proportional to $g(E_F)k_B^{2}T$. Using the supplied coefficient,The shell argument fixes the scaling, not the numerical coefficient; the controlled Sommerfeld expansion fixes that coefficient at fixed electron number. Filled states far below $E_F$ cannot accept arbitrary thermal excitations because of the Pauli exclusion principle.

At fixed volume and particle number, reversible heating obeys $dS=C_{V}dT/T$. Integrating the electronic heat capacity from zero temperature givesWith no extensive residual ground-state degeneracy, the third law of thermodynamics sets the electronic $S(0)=0$. ThereforeThe entropy normalization follows that of $\gamma$: total, per volume or per mole. Lattice contributions are additional and are not part of this electronic result.

Let $g_v(E)$ denote the spin-summed density of states per volume. An electron moment with magnitude the Bohr magneton ${\mu }_B$ has Zeeman energies $\mp {\mu }_{B}B$. At low temperature, the two spin populations areThe common chemical potential changes only at order $B^2$, so it does not affect this linear imbalance. The magnetization is $M={\mu }_B(n_{+}-n_{-})={\mu }_B^2{g}_v(E_F)B$. Define SI volume magnetic susceptibility by $M={\chi }_{p}H$ and take the weak applied field as $B\simeq {\mu }_{0}H$. Then Pauli paramagnetism givesThe last equality uses the three-dimensional parabolic free electron model, $g_v(E_F)=3n/(2E_F)$. This susceptibility is nearly independent of temperature for $T\ll T_F$, unlike the Curie law for localized independent moments. The displayed result excludes orbital diamagnetism and interaction enhancements. If susceptibility instead denotes $\partial M/\partial B$, omit ${\mu }_0$; confusing that convention with $\partial M/\partial H$ changes the units.

In the one-band electron Hall effect model, $R_H=-1/(ne)$, with $e>0$. The measured Hall coefficient therefore givesA monovalent body-centered cubic lattice has two conduction electrons per conventional unit cell. Thus $n{a}^3=2$, andCounting two spin states within the Fermi sphere gives $k_F=(3{\pi }^{2}n{)}^{1/3}=9.0418\times 10^9{\mathrm{m}}^{-1}$. Using the specified effective mass in the dispersion,There is one mole of conduction electrons per mole of sodium. Its molar density of states at $E_F$ is $3N_A/(2E_F)$, givingUsing the volume density of states in c givesThe effective mass changes the dispersion and density of states; the electron Bohr magneton used for the spin coupling retains the bare electron mass. Replacing it by $e\hslash /(2m^{\ast })$ would be an additional, unjustified magnetic-moment assumption. A Hall factor or a multiband metal would also invalidate the simple density inference; the stated calculation uses the requested one-band approximation.