What is the quotient of two quantum operators? It's probably useful to explain the context, which led me to this question. We were asked the following question:
By writing ${L}^2 = \sum_{ijklm}\epsilon_{ijk}{x}_j{p}_k\epsilon_{ilm}{x}_l{p}_m$ show that: $$p^2 = \frac{L^2}{r^2}+\frac{1}{r^2}\left\{(\textbf{r}\cdot \textbf{p})^2-i\hbar(\textbf{r}\cdot \textbf{p})\right\}$$
I got up to this point: $${L}^2 = {\textbf{r}}^2{\textbf{p}}^2-\left(\textbf{r}\cdot\textbf{p}\right)^2+i\hbar\textbf{r}\cdot\textbf{p}$$
However now my question is, am I allowed to just divide by $\textbf{r}^2$ and if yes what is the interpretation of a division of two quantum operators? After all $L,\textbf{r},\textbf{p}$ are all quantum operators so I'm quite worried about just applying "normal" Algebra rules and callying it a day.
 A: Yeah, it's a subtle issue. You can't just divide two non-commuting operators - you need to specify whether you're left-multiplying or right-multiplying the numerator by the reciprocal of the denominator. I would avoid ever using "division" notation and only multiply operators and their reciprocals, for clarity. You can left-multiply your operator expression by $\left({\bf r}^2 \right)^{-1}$ to get one particular quantization of the final result that you're supposed to show.
Strictly speaking, in $d$ spatial dimensions the domain of the operator $r^{-2}$ is the subset of the Hilbert space $L^2 \left( \mathbb{R}^d \right)$ which $r^{-2}$ takes to $L^2 \left( \mathbb{R}^d \right)$, i.e. the set of square-integrable functions $\psi({\bf r})$ such that
$$\left \langle \psi \middle| \left(r^{-2} \right)^\dagger r^{-2} \middle| \psi \right \rangle = \int d^dx\ \frac{|\psi({\bf r})|^2}{r^4} = \int d\Omega \int r^{d-1} dr \frac{|\psi(r, \Omega)|^2}{r^4}$$
is finite. This is the set of functions $\psi({\bf r})$ that go to zero at the origin at least as fast as $r^p$ for some power $p > 4 - d$. In three spatial dimensions, this means that $\psi({\bf r})$ must go to zero faster than $r$ near the origin.
