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

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.