Torque-Free Precession in Classical & Quantum Mechanics

[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.

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.

 

[M Quack]

This can happen for example when rigid molecules with an electric dipole moment are placed in an eletromagnetic field.

Check this out:
http://en.wikipedia.org/wiki/Diatomic_molecule#Rotational_energies

http://chemwiki.ucdavis.edu/Physical_Chemistry/Spectroscopy/Rotational_Spectroscopy_of_Diatomic_Molecules

 

[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.

 

[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.

 

[espen180]

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