 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 $A\cosh 2{\theta }_k-B_k\sinh 2{\theta }_k$ and $B_k\cosh 2{\theta }_k-A\sinh 2{\theta }_k$. The pair term vanishes whenFor $\alpha >0$ the transformation is real and finite for every $k$. Summing the two Nambu number operators over the full zone givesHere $E_0$ names the constant in the requested oscillator form; the actual quadratic ground-state energy is $E_{\mathrm{gs}}^{(2)}=E_0+\frac{1}{2}{\sum }_k{ϵ}_k=-J\Delta N{S}^2+JS{\sum }_k({\Omega }_k-\Delta )$. 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 $k\ne 0,\pi$, the same formula holds with ${\Omega }_k=|\sin k|$. The exact zero modes have $|\tanh 2{\theta }_k|=1$, 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 $\tanh 2{\theta }_k=\cos k/(2\Delta )$ and ${ϵ}_k^{\mathrm{print}}=2JS\sqrt{4{\Delta }^2-{\cos }^{2}k}$. That result remains gapped at $\alpha =0$ and is not the spin-wave spectrum of the Hamiltonian stated at the start.