 Solution

For $\alpha >0$, the dispersion is periodic and has minima at $k=0,\pi$ in the rotated full-zone convention, withThe maximum is $2JS(1+\alpha )$ at $k=\pm \pi /2$. The easy-axis state has no continuously broken symmetry, so a gap is allowed. Near a minimum the energy behaves as $2JS\sqrt{\alpha (2+\alpha )+(\delta k{)}^2}$. At $\alpha =0$ it becomes ${ϵ}_k=2JS|\sin k|$, linear near its zeros, with small-$k$ slope $2JS$ in the chosen units. Restoring lattice spacing $a$ and $\hslash$ gives spin-wave speed $2JSa/\hslash$. Zone folding reorganizes the same modes into the magnetic Brillouin zone.

These are linear spin-wave predictions, not a statement that the exact one-dimensional quantum chain always has long-range Néel order. At the isotropic point the zero-point spin reduction contains $N^{-1}{\sum }_k(|\sin k{|}^{-1}-1)/2$, which diverges logarithmically as the infrared cutoff vanishes. This failure of the assumed ordered reference must not be mistaken for a calculation of the exact interacting spectrum.