Holstein Primakoff - Showing Spin-Relation $S^2 = S(S+1)$

Question

I'm trying to prove that Holstein-Primakoff-Transformation \begin{align} S_i^{+} &= \sqrt{2S-n_i}b_i \\ S_i^{-} &= b_i^{\dagger} \sqrt{2S -n_i} \\ S_i^{z} &= S-n_i \end{align}

holds the condition $$ \vec S^2=S(S+1)$$ but I get the following (with $S_x=\frac{1}{2}( S^+ + S^-) $ und $S_y=\frac{1}{2i} (S^+-S^-)$): \begin{align*} S^{2} &= S_x^{2} + S_y^{2} + S_z^{2} = \frac{1}{4} \left( S^{+} + S^{-}\right)^{2} - \frac{1}{4} (S^{+}- S^{-})^{2} + (S-n_i)^{2} \\ &= \frac{1}{4} (2 S^{+} S^{-} + 2 S^{-} S^{+}) + S^{2} - 2Sn_i + n_i^{2} \\ &= \frac{1}{2} \left( \sqrt{2S-n_i} b_ib_i^{\dagger} \sqrt{2S-n_i} + b_i^{\dagger}(2S-n_i) b_i \right) + S^{2} - 2Sn_i + n_i^{2} \\ &= \frac{1}{2} \left( \sqrt{2S-n_i}(1+n_i) \sqrt{2S-n_i} + b_i^{\dagger}(2S-n_i) b_i \right) + S^{2} - 2Sn_i + n_i^{2} \\ &= \frac{1}{2} \left( n_i (2S-n_i) + 2S -n_i + b_i^{\dagger}(2S-n_i) b_i \right) + S^{2} - 2Sn_i + n_i^{2} \\ &= S^{2} + \frac{n_i^{2}}{2} + S - \frac{n_i}{2} - \frac{1}{2} b_i^{\dagger} b_i^{\dagger} b_i b_i \end{align*} So my question is, whether the Holstein-Primakoff satisfies this condition or not? - It should satisifies the condition, because otherwise it would be unphysically to use this transformation.

Answer 1

What you've done is treat $n_i$ as a number, which is not completely honest, since it really an operator $\hat n$. Thus for instance \begin{align} S^+ \vert n\rangle &= \sqrt{2S-\hat n}\, b\vert n\rangle = \sqrt{2S-\hat n} \sqrt{n}\vert n-1\rangle = \sqrt{(2S-n+1)n}\vert n-1\rangle\\ S^- \vert n\rangle &= b^\dagger \sqrt{2S-\hat n} \vert n\rangle = \sqrt{(2S-n)(n+1)} \vert n+1\rangle \end{align} so that \begin{align} S^+S^-\vert n\rangle &= \sqrt{(2S-n)(n+1)} S^+ \vert n+1\rangle =\sqrt{(2S-n)(n+1)} \sqrt{(2S-n)(n+1)} \vert n\rangle \\ &=(2S-n)(n+1) \vert n\rangle \\ S^-S^+\vert n\rangle &=(2S-n+1)n \vert n\rangle\, . \end{align} If my algebra is right this should take care of it.