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.
