Past exam of the mathematics course of the University of Cambridge / 2015 / iii / Paper 81 / 2 / d / Solution

In each factor of the BCS ground state, the BCS anomalous average is $⟨b_k⟩=u_k{v}_k={\Delta }_k/(2E_k)$. Self-consistency therefore gives the zero-temperature BCS gap equationPut ${\Omega }_D=\hslash {\omega }_D$. The specified constant attractive interaction makes ${\Delta }_k$ independent of $k$ inside the energy shell and zero outside it:The printed ${\delta }_{k,0}$ is incompatible with this interaction: $k$ labels relative pair momentum, not the total momentum of a Cooper pair. Every pair here has zero total momentum, while many relative momenta contribute.

For the nonzero solution, the constant-shell BCS gap equation becomes $1=(V/L^3){\sum }_{|{\xi }_k|<{\Omega }_D}(2\sqrt{{\xi }_k^2+{\Delta }^2}{)}^{-1}$. Let ${\nu }_s$ denote the approximately constant single-spin density of states per unit volume at the Fermi level. ThenIn weak coupling, $\Delta \simeq 2\hslash {\omega }_D{e}^{-1/({\nu }_{s}V)}$. The displayed answer in the question uses $\nu ={\nu }_s$. If “total electronic density of states” includes both spin species, ${\nu }_{\mathrm{tot}}=2{\nu }_s$, the argument instead reads $2/({\nu }_{\mathrm{tot}}V)$. If the density counts the whole box rather than unit volume, divide it by $L^3$ before using this formula.