= Solution
For the <easy-axis XXZ antiferromagnetic spin-wave spectrum>, the consistent quadratic <Hamiltonian> in part (d) has $A=\Delta$ and $B_k=\cos k$. The inverse transformation acts on the Nambu matrix by a congruence. Its diagonal and off-diagonal entries become $A\cosh2\theta_k-B_k\sinh2\theta_k$ and $B_k\cosh2\theta_k-A\sinh2\theta_k$. The pair term vanishes when
$$
\tanh2\theta_k=\frac{\cos k}{\Delta},\qquad \Omega_k=\sqrt{\Delta^2-\cos^2k}.
$$
For $\alpha>0$ the transformation is real and finite for every $k$. Summing the two Nambu <number operators> over the full zone gives
$$
\boxed{H=E_0+\sum_k\epsilon_k(b_k^\dagger b_k+\tfrac12)+O(S^0),\quad E_0=-J\Delta NS(S+1),\quad\epsilon_k=2JS\sqrt{\Delta^2-\cos^2k}.}
$$
Here $E_0$ names the constant in the requested oscillator form; the actual quadratic ground-state energy is $E_{\rm gs}^{(2)}=E_0+\tfrac12\sum_k\epsilon_k=-J\Delta NS^2+JS\sum_k(\Omega_k-\Delta)$. This distinguishes the normal-ordering constant from the total zero-point contribution. <bosonic Bogoliubov diagonalization> is a canonical, not an ordinary unitary, diagonalization of the Nambu matrix.
At $\alpha=0$ and $k\ne0,\pi$, the same formula holds with $\Omega_k=|\sin k|$. The exact <zero modes> have $|\tanh2\theta_k|=1$, so no finite squeezing angle diagonalizes them as positive-frequency oscillators. Treat collective rotations separately or retain a small anisotropy regulator before taking the limit. Finally, if one instead diagonalizes the literal erroneous printed matrix, it gives $\tanh2\theta_k=\cos k/(2\Delta)$ and $\epsilon_k^{\rm print}=2JS\sqrt{4\Delta^2-\cos^2k}$. That result remains gapped at $\alpha=0$ and is not the spin-wave spectrum of the <Hamiltonian> stated at the start.
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