Can we define operators like $\dfrac{1}{a^\dagger a}$? I was recently reading this paper on Enhancement of Few Photon Optomechanical Effects and could not quite understand eq.(2). The author has written an operator like this:
$$\hat \xi=\dfrac{g_oa^\dagger a}{w_m-g_{cK}a^\dagger a}$$
I don't understand how I am supposed to interpret that number operator in the denominator. My guess is that the operator $\hat \xi$ is defined such that its product with $(w_m-g_{cK}a^\dagger a)$ gives $g_oa^\dagger a$. But I am not sure.
Any help is appreciated.
 A: To supplement Andrew's nice answer, note that vectors in the occupation number basis are eigenvectors of $a^\dagger a$, i.e.
$$a^\dagger a |n\rangle = n|n\rangle$$
As a result, the action of your operator $\hat \xi$ on such a vector becomes
$$\hat \xi|n\rangle = \frac{g_0 n}{w_m - g_c n}|n\rangle$$
which extends by linearity to any state $|\psi\rangle = \sum_n c_n|n\rangle$.  This basis can also be used to rewrite the operator as
$$\hat\xi = \sum_n \left(\frac{g_0 n}{w_m - g_cn}\right) |n\rangle\langle n|$$
This so-called spectral decomposition is the means by which most "functions of operators" are actually defined.

As a technical note, the trouble with power series expansions of operators is that they generally only converge properly when the operator in question is bounded.  As an example, one is tempted to write
$$\frac{1}{w_m - g_c a^\dagger a}\simeq \frac{1}{w_m} \sum_k \left(\frac{g_c}{w_m}a^\dagger a\right)^k = \frac{1}{w_m}\left(1+\frac{g_c}{w_m}a^\dagger a + \ldots \right)$$
The problem with this expansion should be clear if we try to apply it to the vector $|n\rangle$, at which point we have
$$\frac{1}{w_m - g_c a^\dagger a}|n\rangle = \left[\frac{1}{w_m} \sum_k \left(\frac{g_c}{w_m}n \right)^k\right]|n\rangle$$
The prefactor clearly goes to infinity whenever $n> \left| w_m/g_c\right|$.
If instead we use the spectral definition, we simply obtain
$$\frac{1}{w_m - g_c a^\dagger a}|n\rangle = \frac{1}{w_m - g_c n}|n\rangle$$
which (a) is perfectly well-defined for all $n$, and (b) can easily be shown to agree with the power series expansion, provided the latter converges.
A: In general, $\frac{A}{B}$ is lazy physicist shorthand notation for $B^{-1}A$. You might rightly complain that there is an ordering ambiguity and the expression could also mean $A B^{-1}$. That's completely correct, and this notation is only meaningful if $[A,B^{-1}]=0$.
If we take $A=g_o a^\dagger a$ and $B = w_m - g_{cK} a^\dagger a$, then indeed $[A, B^{-1}]=0$ so there is no ordering ambiguity. To see this, define $\epsilon = g_{cK}/w_m$ and write $B=w_m C$, with $C=1-\epsilon a^\dagger a$. Then (assuming the Taylor series converges -- see J. Murray's answer for more details on this assumption and ways to remove the need for it) we can use the Taylor series (specifically it is a geometric series) to write
\begin{equation}
B^{-1} = \frac{1}{w_m} C^{-1} = \frac{1}{w_m} (1-\epsilon a^\dagger a)^{-1} = \frac{1}{w_m} \sum_{n=0}^\infty \epsilon^n (a^\dagger a)^n
\end{equation}
Since $[a^\dagger a, a^\dagger a]=0$, we have that $A$ commutes with every term in the Taylor series, so $[A,B^{-1}]=0.$

Here's another argument that $[A,B^{-1}]=0$, which doesn't rely on the Taylor series.
First, we note that if $[A,B]=0$, and if $B^{-1}$ exists, then $[A,B^{-1}]=0$. Here's a proof:
\begin{eqnarray}
0 &=& BA - AB\\
&=& A - B^{-1}A B\\
&=& AB^{-1} - B^{-1} A\\
&=& [A, B^{-1}]
\end{eqnarray}
Since it's easy to see that for $A=g_o a^\dagger a$ and $B=w_m - g_{cK} a^\dagger a$, that $[A,B]=0$, it follows that $[A, B^{-1}]=0$.
