Is there an angular velocity operator in quantum mechanics?

In classical mechanics we can write as velocity of a rotating object $\vec{v} = \vec{\omega} \times \vec{r}$ or in analogy the momentum $\vec{p} = m (\vec{\omega} \times \vec{r})$ using the angular velocity $\vec{\omega}$ and the (rotating) position vector $\vec{r}$.

In quantum mechanics we have a momentum operator $\hat{p}$ and a position operator $\hat{r}$, but I have never seen an angular velocity operator $\hat{\omega}$ . Is there an angular velocity operator in quantum mechanics and how is it defined? Can one write an equation like $\hat{p} = m (\hat{\omega} \times \hat{r})$ in quantum mechanics?

• You have the angular momentum operator $\vec p \times \vec x$. But since there is no notion of the moment of inertia, it's not apparent to me that the idea of "angular velocity" is meaningful at all - indeed, the notion of linear velocity too is only meaningful insofar as that you can write $p/m$ and define this to be the velocity operator. – ACuriousMind Jul 1 '16 at 22:34

The angular momentum is $\vec L~=~\vec r\times\vec p$ which according to a bulk system with a moment of inertial $I$ is also $\vec L~=~\hat n I\omega$. Here the unit vector $\hat n$ is normal to the plane of $\vec r$ and $\vec p$. In Hamiltonian mechanics we have $${\dot L}_i~=~I\dot\omega_i~=~\{H,~L_i\}_{pb}~=~0,$$ for $i$ the coordinate direction which pretty easily means that $H~=~\frac{1}{2}L^2/I$. Now let us write this as $H~=~\frac{1}{2}I\omega^2$ and consider the momentum of inertia as pertaining to a single particle, so $I~=~mr^2$. The Hamiltonian is then $$H~=~mr^2\omega^2.$$ Now consider your own form $p_i~=~\epsilon_{ijk}\omega_jr_k$ then $$p^2~=~m\epsilon_{ijk}\epsilon_{imn}\omega_jr_k\omega_mr_n~=~\left(\delta_{jm}\delta_{kn}~-~\delta_{jn}\delta_{km}\right)\omega_jr_k\omega_mr_n$$ $$=~m(\omega^2r^2~-~(\vec r\cdot\omega)^2).$$ for the rigid case of $I~=~mr^2$ the last term is zero. So this agrees with your definition.
In somewhat greater generality we consider the momentum of a particle in the plane with variables $r,~\theta$. A momentum vector is then $$\vec p~=~{\bf r}\frac{dr}{dt}~+~r\frac{d\theta}{dt}\bf\theta.$$ The square of the momentum is then $$p^2~=~\left(\frac{dr}{dt}\right)^2~+~r^2\left(\frac{d\theta}{dt}\right)^2~=~p_r^2~+~r^2\omega^2$$ The Lagrangian for this is ${\cal L}~=~\frac{1}{2}(p_r^2~+~r^2\omega^2)$ minus what ever radial potential there may be. It is not hard to show that $${\cal L}~=~\frac{1}{2}p_r^2~+~\frac{L^2}{2mr^2}~-~V(r)$$ This is in agreement with above. This is a general way that motion is set up, and it does agree with $\vec p~=~m\omega\times r$
Now for quantization. The Lagrangian (or its corresponding Hamiltonian) above is expressed according to the angular momentum operator. We might however consider the two momenta $p_r$ and $p_\theta~=~mr\omega$ which is your momentum operator for the moment arm perpendicular to the velocity. For the $p_r$ we have the standard rule $\hat p_r~=~-i\hbar\partial/\partial r$. For the $\theta$ momentum we consider a similar thing $$\hat p_\theta~=~-ir^{-1}\hbar\frac{\partial}{\partial\theta},$$ which is just the angular momentum operator divided by the moment arm $\hat L/r$ Now take $\frac{1}{2m}(\hat p_\theta)^2$ and find $$\frac{1}{2}(\hat p_\theta)^2~=~\frac{\hat L^2}{2mr^2},$$ and voila we have found the kinetic energy term for angular momentum operator in standard form
As a rule however, people do not want to quantize $\vec p~=~m\vec\omega\times\vec r$. It is not a terribly convenient way of working with these systems.