# Torque-free precession

1. Mar 6, 2012

### espen180

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.

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.

2. Mar 6, 2012

3. Mar 6, 2012

### espen180

That would still be concidered precession by an external torque, which is not what I am interested in here. Diatomic molecules don't experience free precession. I am sorry if I worded the problem poorly.

What I am interested in is the kind of precession the rotational axis of the Earth experiences, but at the quantum level. For example, a free spinning molecule of white phosphorus (tetrahedral molecule) would experience precession.

4. Mar 7, 2012

### M Quack

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.

5. Mar 7, 2012

### espen180

Great! I'll try it.
Thank you very much!