= Solution
For real $\Delta_k$ and $\xi_{-k}=\xi_k$, introduce the particle-hole column $\Psi_k=(c_{k\uparrow},c_{-k\downarrow}^\dagger)^T$. Its quadratic block is the <Bogoliubov--de Gennes Hamiltonian>
$$
K_k=\Psi_k^\dagger\begin{pmatrix}\xi_k&-\Delta_k\\-\Delta_k&-\xi_k\end{pmatrix}\Psi_k+\xi_k.
$$
The <BCS coherence factors> can be chosen to obey
$$
u_k^2=\frac12\left(1+\frac{\xi_k}{E_k}\right),\quad v_k^2=\frac12\left(1-\frac{\xi_k}{E_k}\right),\quad 2u_kv_k=\frac{\Delta_k}{E_k},\qquad E_k=\sqrt{\xi_k^2+\Delta_k^2}.
$$
With $u_k\ge0$, choose the sign of $v_k$ to match $\Delta_k$. The rotation diagonalizes the matrix to $\operatorname{diag}(E_k,-E_k)$. Reordering $\gamma_{-k\downarrow}\gamma_{-k\downarrow}^\dagger=1-\gamma_{-k\downarrow}^\dagger\gamma_{-k\downarrow}$ and restoring the mean-field constant gives
$$
\boxed{K=\sum_{k,\sigma}E_k\gamma_{k\sigma}^\dagger\gamma_{k\sigma}+\sum_k\left(\xi_k-E_k+\Delta_k\langle b_k^\dagger\rangle\right)}.
$$
The inverse transformation is $\gamma_{k\uparrow}=u_kc_{k\uparrow}-v_kc_{-k\downarrow}^\dagger$ and $\gamma_{-k\downarrow}=v_kc_{k\uparrow}^\dagger+u_kc_{-k\downarrow}$. Both annihilate $(u_k+v_kb_k^\dagger)|0\rangle$. Since every positive-energy <quasiparticle> mode is empty, the normalized <BCS ground state> is
$$
\boxed{|\mathrm{g.s.}\rangle=\prod_k\left(\cos\theta_k+\sin\theta_kc_{k\uparrow}^\dagger c_{-k\downarrow}^\dagger\right)|0\rangle}.
$$
Each momentum label refers to the pair $(k\uparrow,-k\downarrow)$ once; distinct labels use disjoint single-particle spin states.
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