4
$\begingroup$

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?

$\endgroup$
  • 2
    $\begingroup$ 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. $\endgroup$ – ACuriousMind Jul 1 '16 at 22:34
1
$\begingroup$

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.

$\endgroup$
  • $\begingroup$ I think he is asking about angular velocity and not angular momentum. Similar things, not the same thing. $\endgroup$ – J. Manuel Nov 17 '16 at 10:52

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.