 Solution

Choose the positive-$x$ polarized ground state selected by $B>0$, or its symmetry-selected $B\to 0^{+}$ limit. It is exactly the boson vacuum in the Holstein–Primakoff transformation. Thus $n_j|\mathrm{GS}⟩=0$ and $S_j^x=S-n_j$ giveFor the transverse components, $S^y=({\tilde{S}}^{+}+{\tilde{S}}^{-})/2$ and $S^z=({\tilde{S}}^{+}-{\tilde{S}}^{-})/(2i)$. In the vacuum, the only nonzero quadratic ladder-operator expectation iswhich follows using the full ordered square-root operators, or directly from ${\tilde{S}}^{-}|S,S⟩=\sqrt{2S}|S,S-1⟩$. ThereforeAn independent exact spin argument starts from $S_j^x|S,S{⟩}_x=S|S,S{⟩}_x$ and $(S_j^x{)}^2+(S_j^y{)}^2+(S_j^z{)}^2=S(S+1)$. Rotational invariance about $x$ makes the two transverse second moments equal, so each is $[S(S+1)-S^2]/2=S/2$. This agrees exactly with the full boson representation.

The first moments describe the local magnetization. Subtracting their squares from the second moments gives spin component variances: zero longitudinal variance and transverse variance $S/2$. Transverse quantum fluctuations persist in a perfectly polarized state even though its magnon occupation is zero. Their relative scale decreases as $S^{-1/2}$ in the classical limit.

At precisely $B=0$, “the ground state” requires a choice within the ferromagnetic multiplet. For example, when $NS$ is an integer, a maximum-total-spin state with total $x$ projection zero is also a ground state but has $⟨S_j^x⟩=0$. The displayed local moments refer to the explicitly specified polarized member, not to every superposition or rotationally invariant mixture of zero-field ground states.