This is an accident of the fact that when you are given a quantum system and its solution, the classical limit must have been integrable, because you could solve it. In the integrable case, there are action variables $J_n$ which fill up the whole phase space with invariant tori, defined by fixing the action variables, and letting the canonically conjugate angle variables vary over the full range The action variables $J_n$'s are conserved quantities by definition. Integrability means that any point in the space is labeled by the $J$'s and the conjugate $\theta$'s.

In the semiclassical limit of large number labels for the energy levels, the energy is given by setting the angle variables to integers times $2\pi$ (in units where $\hbar=1$)

$$ J_k = 2\pi n_k $$

and near large J_k,
$$ {\partial H\over \partial J_k} = {1 \over T_k} $$

Where $T_k$ is the classical period of the action variable. This makes the energy levels spacing a multiperiodic projected lattice near any one energy $E_0$

$$ E = E_0 + \sum_k {2\pi \Delta n_k\over T_k} = \sum_k \omega_k \Delta n_k $$

You can extract the classical periods of the angle variables from the energy level diagram. The energy level diagram near any (large) energy will look periodic with multiple incommensurate periods.

If you want a quantum system where the energy states are not labelled by classical conserved quantities, at least not by useful ones, take any classically chaotic system and consider the quantum version. You need at least two degrees of freedom, so the simplest example is the quantum double pendulum, where there are two angles $\theta$ and $\phi$, and the Lagrangian is given by

$$L = {1\over 2} (\dot{\theta}^2 + \dot{\phi}^2) - \cos(\theta) - A cos(\phi)$$

If you choose an A so that the system is classically chaotic, the eignestates of this are going to have discrete energy levels, but the labels cannot correspond to action variables, because there are none. The energy levels in a chaotic quantum system obey random matrix statistics in the semiclassical limit, they do not form multiperiodic sequences locally, like the integrable case.

This post imported from StackExchange Physics at 2014-03-22 17:30 (UCT), posted by SE-user Ron Maimon