# 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.

1. 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?

2. 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?

• More on Holstein-Primakoff. Apr 20 '19 at 12:03
• $S$ is a number, NOT an operator or a vector. Apr 20 '19 at 15:44

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 $$|0\rangle$$!