 Solution

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.