Past exam of the thermal and statistical physics course of the University of Cambridge / 2017 / 1

Put $\xi =ϵ-{ϵ}_F$ and $\Delta =|\Delta |>0$. The BCS quasiparticle spectrum has $E=\sqrt{{\xi }^2+{\Delta }^2}$, so on either inverse branchConservation of states on one branch gives $D_{\mathrm{branch}}(E)dE=g({ϵ}_F)|d\xi |$. Consequently the single-branch expression requested in the source isThere are no such states for $0<E<\Delta$ in this ideal fully gapped model. Near the edge the density of states diverges as $(E-\Delta {)}^{-1/2}$, an integrable singularity; at large $E/\Delta$ it approaches its normal-state value.

Counting qualification. The energy map in the source is two-to-one. If $g(ϵ)$ literally counts all normal-state levels both above and below ${ϵ}_F$, the unweighted positive-energy quasiparticle density from a folded spectrum must sum both preimages, and is insteadAlready at $\Delta =0$, folding $\xi$ into $|\xi |$ doubles the positive-axis density. Thus the printed formula is not the unweighted count of all states under the literal mapping without an additional normalization convention.

There is also a standard physical quantity with exactly the printed normalization: the BCS electronic density of states, measured for example by single-electron superconducting tunneling. At positive $E$, the electron spectral weights of the two roots are the BCS coherence factors $u_{\pm }^2=(1+{\xi }_{\pm }/E)/2$. They sum to one, so the weighted sum is $g({ϵ}_F)E/\sqrt{E^2-{\Delta }^2}$. This supplies a valid electronic-DOS interpretation, while keeping it distinct from counting every positive-energy excitation. Spin degeneracy must be included consistently in $g$ and the transformed density; it does not remove the branch-counting issue.

Repeated letters. There are six positions, with two copies of each of two letters. Counting labeled arrangements and dividing by the permutations of each repeated pair gives

Single-occupancy cells. For $N$ indistinguishable particles, specifying which $N$ of the $M$ distinct cells are occupied completely specifies a configuration. ThusFor distinguishable particles, each chosen occupied set admits $N!$ assignments of particle labels:Equivalently, successively place labeled particles in $M,M-1,\dots ,M-N+1$ available cells. The single-occupancy constraint is essential: this is not unrestricted stars and bars. Both counts give one empty configuration at $N=0$; at $N=M$ they give one unlabeled occupation pattern and $M!$ labeled assignments.

In the grand canonical ensemble, the system exchanges both energy and particles with a reservoir at fixed temperature $T$ and chemical potential $\mu$. The probability of a particular many-body microstate $i$, with energy $E_i$ and particle number $N_i$, is its Boltzmann weight $e^{-\beta (E_i-\mu N_i)}$ divided by the grand canonical partition function $\Xi$, where $\beta =1/(k_{B}T)$. The denominator normalizes ${\sum }_i{p}_i=1$; a degeneracy is accounted for by counting distinct microstates, rather than changing an individual state's probability.

For noninteracting particles, specify a microstate by the occupation numbers $n_r$ of one-particle levels of energy ${ϵ}_r$. Then $E-\mu N={\sum }_r({ϵ}_r-\mu )n_r$, so the Gibbs probability factorizes between levels. Put $x_r=e^{-\beta ({ϵ}_r-\mu )}$.

For fermions, the Pauli exclusion principle permits $n_r=0,1$ in each complete one-particle state, giving ${\Xi }_r=1+x_r$. Its mean occupancy isthe Fermi-Dirac distribution.

For bosons, $n_r=0,1,2,\dots$, so the geometric sum is ${\Xi }_r=1/(1-x_r)$ when $x_r<1$. Differentiating that sum givesthe Bose-Einstein distribution. Finite-system convergence requires $\mu <{ϵ}_{\min }$; condensation is the limiting approach to the lowest energy with macroscopic ground-state occupation. Both quantum distributions reduce to $e^{-\beta ({ϵ}_r-\mu )}$ when occupation is small. The quantities derived here are mean occupancies of states, rather than normalized probabilities over the set of one-particle energies.