Particle in 1D ring is a textbook problem, but there is one thing I don't understand – if the ground state is considered to have zero angular momentum, then its energy is also zero. And the wavefunction is constant in the ring. Isn't is unphysical? It means that there is equal probability to find particle in any position in the ring – so it's rotating, right? Even with $m=0$.
We can probably get rid of this state, and start with the first 'rotating' state with $m=\pm 1$, but in magnetic field, for example, we can't just ignore the $m=0$ state, because it's the only one that's not degenerate, and its existence is more or less confirmed by experiments.
And while 1D rings don't exist, the case of a finite width ring is absolutely the same. Both analytically solvable case and any other ring-like potential (see figure) have a ground state with zero angular momentum, but with 'stretched out' wavefunction maximum (it's a circle instead of a single point), so the particle effectively 'rotates'.
There is no such problem in most potentials. In a quantum box (rectangular, cylindrical or spherical) or in a hydrogen atom or in a harmonic oscillator there is only one maximum, so there is a single point where we can find the particle with most probability.
We can construct potentials with two or more 'central points', but the ring (or spherical layer which has the same property) is an extreme case, and I'm not sure what to think about its ground state.
Best Answer
This state is not a eigenstate of either the position or the momentum operator. That means it doesn't have a definite value for either of those variables, so the question doesn't actually make sense.
The question you are asking is essentially a classical one, but this is a quantum system. The right answer is "the system has zero angular momentum, so the wavefunction has no angular dependence on time", and that is all you can say about it.
It is worth noting that the s-states of the hydrogen atom (or indeed any central potential) have this same property, and this is one of the several things we mean when we describe the sates as "orbitals" not "orbits".