= Solution
Choose the positive-$x$ polarized ground state selected by $B>0$, or its symmetry-selected $B\to0^+$ limit. It is exactly the boson vacuum in the <Holstein–Primakoff transformation>. Thus $n_j|\mathrm{GS}\rangle=0$ and $S_j^x=S-n_j$ give
$$
\boxed{\langle S_j^x\rangle=S,\qquad\langle(S_j^x)^2\rangle=S^2.}
$$
For 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 is
$$
\langle\tilde S^+\tilde S^-\rangle=2S,
$$
which follows using the \b[full ordered square-root operators], or directly from $\tilde S^-|S,S\rangle=\sqrt{2S}|S,S-1\rangle$. Therefore
$$
\boxed{\langle S_j^y\rangle=\langle S_j^z\rangle=0,\qquad
\langle(S_j^y)^2\rangle=\langle(S_j^z)^2\rangle=S/2.}
$$
An independent exact spin argument starts from $S_j^x|S,S\rangle_x=S|S,S\rangle_x$ 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 $\langle S_j^x\rangle=0$. The displayed local moments refer to the explicitly specified polarized member, not to every superposition or rotationally invariant mixture of zero-field ground states.
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