Well-definedness of Holstein-Primakoff transformation In many-body physics, Holstein-Primakoff transformation is defined as follows:
\begin{align}
S_i^+ &= \sqrt{2S}(1-a_i^\dagger a_i/2S)^{1/2}a_i, \\
S_i^- &=\sqrt{2S}(1-a_i^\dagger a_i/2S)^{1/2},
\end{align}
where $S_i^{\pm}, S$ are spin operators and $a_i$ is the transformed operator.
I have two questions regarding this.


*

*Is the square root or division of an operator (not a number!) well-defined?
(Furthermore, $S$ is a vector operator!)
It seems that the square root is defined in terms of Taylor series, but how can we guarantee that the series converges?

*The transformed operator $a_i$ is defined implicitly by the above relation. Does there actually exist such $a_i$ that satisfies the above relation? If so, is it unique?
 A: Since you reference Wikipedia, I'll use its definition instead of your garbled one, so consider 
$$
S_+ = \hbar   \sqrt {2s- {a^\dagger a} }~~ a ~, \qquad
S_- = \hbar   a^\dagger\, \sqrt{2s-{a^\dagger a}} ~, \qquad 
S_z = \hbar(s - a^\dagger a) ~, $$
where capital S s are operators, $[S_+,S_-]=2\hbar S_z$; and $[a,a^\dagger ] =1$ ab initio.
Crucially, unlike what you seem to suspect, s is a parameter, such that, by straightforward plugin, (check!),
$$
S^2= S_z^2 + \tfrac{1}{2}(S_+ S_- + S_- S_+)=\hbar^2 s(s+1) 1\!\!1.
$$
The last equality holds, of course, formally, by dint of $a^\dagger a a = a a^\dagger a -a$.
Now to you question: The (number) operator $a^\dagger a$ is a diagonal matrix with eigenvalues 0,1,2,3,4,... so the diagonal matrix under the square root sign has eigenvalues 2s,2s-1,2s-3,... terminating with a 0 in its bottom rightmost component. That is, it has only 2s+1 components, and cannot allow $S_-$ to lower a state past $a^{\dagger ~~2s}|0\rangle$, nor $S_+$ to raise it past $|0\rangle$! 
The states connected by these S operators then comprise a 2s+1  multiplet: a spin s representation of angular momentum. So, the diagonal matrix under the square root sign is not even infinite dimensional.
How do you take a square root of a finite diagonal matrix D? It is but the diagonal matrix of the square root of each eigenvalue of the original matrix D. 
