Torque-free precession

by espen180
Tags: precession, torquefree
espen180 is offline
Mar6-12, 07:44 AM
P: 836
In classical mechanics, an asymmetric rotating object will generally precess. Expressed in the body-fixed normal system of the object, we have [itex]I_i \dot{\omega_i}=(\vec{L}\times \vec{\omega})_i[/itex] where [itex]L_i=I_i\omega_i[/itex].

Choosing a simple example where [itex]I_1=I_2[/itex], we obtain [itex]\dot{\omega_3}=0[/itex] and, for [itex]\Omega=\frac{I_1-I_3}{I_1}\omega_3[/itex],
[itex]\dot{\omega_1}=\Omega \omega_2[/itex]
[itex]\dot{\omega_2}=-\Omega \omega_1[/itex]
describing the precession. Thus, [itex]\vec{\omega}(t)=(A\cos(\Omega t) , A\sin(\Omega t), \omega_3)[/itex].

My question is; can this motion be described quantum mechanically?

My first guess was to write the Hamiltionian as [itex]\hat{H}=\frac12 \hat{\vec{\omega}}I\hat{\vec{\omega}}[/itex] with [itex]I[/itex] being the inertia tensor. The difficulty is then to describe [itex]\hat{\vec{\omega}}[/itex] in terms of [itex]\hat{x},\hat{p_x}[/itex] 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.
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M Quack
M Quack is offline
Mar6-12, 08:36 AM
P: 635
This can happen for example when rigid molecules with an electric dipole moment are placed in an eletromagnetic field.

Check this out:
espen180 is offline
Mar6-12, 10:45 AM
P: 836
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
M Quack is offline
Mar7-12, 03:19 AM
P: 635

Torque-free precession

I see what you mean.

Try the following Hamiltonian:

[itex]\hat{H} = \frac{1}{2} \sum\limits_{ij} \hat{L}_i I^{-1}_{ij} \hat{L}_j[/itex]

where [itex]I^{-1}_{ij}[/itex] is the invserse of the inertia tensor. In the normal system [itex] I^{-1}_{ij} = \delta_{ij} \frac{1}{I_i}[/itex]

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 is offline
Mar7-12, 05:04 AM
P: 836
Great! I'll try it.
Thank you very much!

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