 Solution

Write $a_k=g-\cos k$, $s_k=\sin k$ and $r_k=\sqrt{a_k^2+s_k^2}$. The Ising-chain Bogoliubov diagonalization uses the eigenvalues $\pm r_k$ of the two-by-two Nambu matrix. For distinct paired modes $k\ne -k$, choose $\cos {\theta }_k=a_k/r_k$, $\sin {\theta }_k=-s_k/r_k$ and ${\theta }_{-k}=-{\theta }_k$. The rotation $U_k=e^{-i{\theta }_k{\sigma }_y/2}$ satisfies $U_k^{†}{M}_k{U}_k=r_k{\sigma }_z$, and its first transformed component isThe opposite signs of the paired angles preserve the canonical anticommutation relations. Self-paired spinless fermion modes have zero pairing and are treated directly as occupied or empty number levels; they do not require this paired-angle formula.

Both $k$ and $-k$ occur in the sum, so the physical quasiparticle coefficient is $2J{r}_k$, not $J{r}_k$. ConsequentlyThis is the bulk quadratic result under the stipulated boundary simplification. Restoring the finite-chain parity sectors adjusts allowed modes and global excitation constraints. In the thermodynamic limit, the ground-state energy per site isFor $g\ge 0$, the bulk quasiparticle gap is $2J|g-1|$. The requested three regimes are shown in the original sketch and discussed separately below.