 Solution

The Holstein–Primakoff transformation maps the finite spin basis $|S,m⟩$ to bosonic Fock states $|n⟩$ with $n=S-m$. Thus $n=0$ is the fully polarized reference state, and creating a boson lowers $S^z$ by one. The physical subspace has the Holstein–Primakoff occupation constraint $0\le n\le 2S$, not the whole infinite oscillator Hilbert space.

The specified ordering of the square root givesThese are the exact spin ladder operator matrix elements; in particular, the lowering amplitude vanishes at $n=2S$. The underlying oscillator obeys its usual canonical commutation relation on the enlarged space, while the spin operators preserve the constrained physical subspace. Projecting the oscillator operators themselves to a finite space would change their commutator at its upper endpoint.

Expanding $\sqrt{2S-n}=\sqrt{2S}(1-n/(4S)+\cdots )$ converts an ordered spin problem to approximately free magnons, with interactions in higher orders of $1/S$. Two complications are the nonpolynomial, order-sensitive square roots, which generate interacting bosons, and loss of the occupation bound when a truncated expansion uses the full bosonic Fock space. Large depletion, high temperature, or soft modes near a phase transition can invalidate the assumption $⟨n⟩\ll S$ even when $S$ is large.