Past exam of the mathematics course of the University of Cambridge / 2022 / ii / Paper 3 / 35A (Statistical Physics)

Bosons have symmetric many-particle states, so any number may occupy the same one-particle state. At low temperature, an ideal Bose gas therefore places a macroscopic population in its lowest one-particle state; below the critical temperature this is Bose-Einstein condensation.

Fermions have antisymmetric many-particle states and obey the Pauli exclusion principle, so each one-particle state can hold at most one fermion of each available internal label. At zero temperature, a noninteracting Fermi gas fills distinct states in increasing order of energy up to the Fermi energy. Its many-particle ground state is thus the filled Fermi sea.

For an ultrarelativistic particle, $\epsilon =pc$. Including the two spin states of an electron, counting wavevectors in a sphere gives the density of statesThe grand canonical partition function of the ideal Fermi gas is thereforeSince the grand potential satisfies $pV=k_{B}T\log \mathcal{Z}$, integration by parts givesThusThis is the equation of state of an ideal ultrarelativistic gas.

At zero temperature the Fermi-Dirac distribution is the indicator of $0\le \epsilon <E_F$. HenceSolving for the Fermi energy gives

The zero-temperature internal energy isUsing part (i),Consequently the adiabatic equation of state isThe pressure is degeneracy pressure: it remains nonzero at zero temperature because of the Pauli exclusion principle. By comparison, a classical ultrarelativistic ideal gas at fixed temperature obeys $pV=N{k}_{B}T$, so its corresponding volume exponent is one and its pressure vanishes as $T\to 0$.

Let $g(E)$ denote the total density of states, including both spin directions. The field lowers one spin energy by ${\mu }_{B}B$ and raises the other by the same amount. To first order in $B$, each spin population changes by the one-spin density $g(E_F)/2$ times the displacement of its Fermi edge. The population imbalance is thereforeMultiplying by the Bohr magneton carried by each excess aligned electron givesThus, with this total-density convention,The zero-temperature magnetic susceptibility isThis weak positive response is Pauli paramagnetism. If susceptibility per unit volume is required, both $M$ and $g$ are divided by $V$.