<s>[INST] Holstein Primakoff - Showing Spin-Relation $S^2 = S(S+1)$ 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. 
 [/INST] 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.
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