 Solution

Under the bipartite spin rotation, the staggered magnetization becomes the uniform transformed $S^z$. In the Bogoliubov transformation vacuum ${\alpha }_k|0_{\alpha }⟩=0$,The quantum depletion of Néel order therefore gives, with the zero modes regulated,Near each zero of $\sin k$, the integrand behaves as $1/(2|k-k_0|)$, producing a logarithmic divergence. With a finite-size cutoff of order $N^{-1}$, the depletion grows as ${\pi }^{-1}\log N+O(1)$. Thus the large-$S$ expansion about a state with finite Néel order is not self-consistent in the infinite one-dimensional chain. The divergent expression is not a negative physical magnetization; it signals breakdown of that ordered approximation. It does not determine whether the exact excitation spectrum is gapped.