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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?

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

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  • $\begingroup$ Actually, it was other person who linked the Wikipedia article. Anyway, thank you! $\endgroup$
    – eigenvalue
    Apr 26 '19 at 12:18

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