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

Write $b_k=⟨b_k⟩+\delta b_k$. The mean-field approximation drops the quadratic product of the fluctuations:For a Hermitian pairing interaction, the two linear terms are Hermitian conjugates. Defining the BCS anomalous average and pairing field by ${\Delta }_k=-{\sum }_{k^{\prime }}{V}_{k,k^{\prime }}⟨b_{k^{\prime }}⟩$, and writing ${\xi }_k={ϵ}_k-\mu$, givesThe final constant corrects the double counting in the reduced BCS pairing Hamiltonian.

The canonical anticommutation relations are preserved precisely when the real matrix acting on $({\gamma }_{k\uparrow },{\gamma }_{-k\downarrow }^{†})$ is orthogonal. Directly,The other relations are identical or vanish between disjoint pairs. Hence $u_k^2+v_k^2=1$, and one may write $u_k=\cos {\theta }_k$, $v_k=\sin {\theta }_k$. This fermionic Bogoliubov transformation uses an ordinary rotation; bosonic mixing instead preserves a difference of squared coefficients.

For real ${\Delta }_k$ and ${\xi }_{-k}={\xi }_k$, introduce the particle-hole column ${\Psi }_k=(c_{k\uparrow },c_{-k\downarrow }^{†}{)}^T$. Its quadratic block is the Bogoliubov--de Gennes HamiltonianThe BCS coherence factors can be chosen to obeyWith $u_k\ge 0$, choose the sign of $v_k$ to match ${\Delta }_k$. The rotation diagonalizes the matrix to $\mathrm{diag}(E_k,-E_k)$. Reordering ${\gamma }_{-k\downarrow }{\gamma }_{-k\downarrow }^{†}=1-{\gamma }_{-k\downarrow }^{†}{\gamma }_{-k\downarrow }$ and restoring the mean-field constant givesThe inverse transformation is ${\gamma }_{k\uparrow }=u_k{c}_{k\uparrow }-v_k{c}_{-k\downarrow }^{†}$ and ${\gamma }_{-k\downarrow }=v_k{c}_{k\uparrow }^{†}+u_k{c}_{-k\downarrow }$. Both annihilate $(u_k+v_k{b}_k^{†})|0⟩$. Since every positive-energy quasiparticle mode is empty, the normalized BCS ground state isEach momentum label refers to the pair $(k\uparrow ,-k\downarrow )$ once; distinct labels use disjoint single-particle spin states.

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.