Torque-free precession

by espen180
Tags: precession, torquefree
 P: 836 In classical mechanics, an asymmetric rotating object will generally precess. Expressed in the body-fixed normal system of the object, we have $I_i \dot{\omega_i}=(\vec{L}\times \vec{\omega})_i$ where $L_i=I_i\omega_i$. Choosing a simple example where $I_1=I_2$, we obtain $\dot{\omega_3}=0$ and, for $\Omega=\frac{I_1-I_3}{I_1}\omega_3$, $\dot{\omega_1}=\Omega \omega_2$ $\dot{\omega_2}=-\Omega \omega_1$ describing the precession. Thus, $\vec{\omega}(t)=(A\cos(\Omega t) , A\sin(\Omega t), \omega_3)$. My question is; can this motion be described quantum mechanically? My first guess was to write the Hamiltionian as $\hat{H}=\frac12 \hat{\vec{\omega}}I\hat{\vec{\omega}}$ with $I$ being the inertia tensor. The difficulty is then to describe $\hat{\vec{\omega}}$ in terms of $\hat{x},\hat{p_x}$ etc. Am I going about this the wrong way? Is there any treatment of this problem available? I tried searching, but all the treatments of precession I found were related to magnetic moment precession. Any help is greatly appreciated.
 P: 660 Torque-free precession I see what you mean. Try the following Hamiltonian: $\hat{H} = \frac{1}{2} \sum\limits_{ij} \hat{L}_i I^{-1}_{ij} \hat{L}_j$ where $I^{-1}_{ij}$ is the invserse of the inertia tensor. In the normal system $I^{-1}_{ij} = \delta_{ij} \frac{1}{I_i}$ The angular momentum operator L is well defined, and the moment of inertia can be taken as constant. If I am not mistaken, then L does not commute with the Hamiltonian, so that you get precession.