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Torquefree precession 
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#1
Mar612, 07:44 AM

P: 836

In classical mechanics, an asymmetric rotating object will generally precess. Expressed in the bodyfixed 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_1I_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. 


#2
Mar612, 08:36 AM

P: 660

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/Diatomi...ional_energies http://chemwiki.ucdavis.edu/Physical...omic_Molecules 


#3
Mar612, 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. 


#4
Mar712, 03:19 AM

P: 660

Torquefree 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. 


#5
Mar712, 05:04 AM

P: 836

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


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