Particle in ring zero energy?

The ground state is going to look like:

$$ psi(theta) = frac 1 {sqrt{2pi}}e^{intheta} $$

with $n=0$, which is:

$$ psi(theta) = frac 1 {sqrt{2pi}} $$

which manifestly does not have a preferred coordinate.

But we can find the expectations values and uncertainties via:

$$ langletheta^nrangle = int_0^{2pi}psi^*theta^npsi dtheta$$ $$ langletheta^nrangle = frac 1 {2pi}int_0^{2pi}theta^ndtheta = frac 1 {n+1}theta^{n+1}|^{2pi}_0 = frac{(2pi)^n}{n+1}$$

so that:

$$ langlethetarangle = pi$$

and

$$ langletheta^2rangle = frac{4pi^2} 3$$

Hence:

$$ sigma_{theta} = (frac{4pi^2} 3 - pi^2)^{frac 1 2}= frac 1 {sqrt 3} pi $$

which one would expect, since a uniform random variable has standard deviation $1/sqrt{12}$.

To paraphrase Feynman, this is a "[just] calculate" situation.

One thing to note here, which is also true in central potentials (hydrogen atom), is that the energy eigenstates are also angular momentum eigenstates, that means the uncertainty in $J^2$ is zero, while the angular coordinate is distributed around the ring (or according to $Y_l^m(theta, phi)$ in the 3D case), which is of course uncertain.

When $J=0$, the solution is completely rotationally symmetric, so any rotation leaves the state unchanged (up to a possible global phase factor). That means the probability amplitude is uniform around the ring. This is the same as the free particle case with $vec p=0$: translations do not change the state, and the probability is uniform over all $vec r$. Any wave function with "clumping", so to speak, means there is non-zero angular/linear momentum in the wave function.

This behavior all derives from the fact that momenta in a coordinate $q$ is proportional to $-ipartial_q$, and that's because translations (rotations) are generated by linear (angular) momentum operators.

My guess is that by "violation of HUP" you mean that $$leftlangle theta^2rightrangleleftlangle L_z^2rightrangle=0$$

The reason for that is that while we can define the operator $L_z=-ipartial_theta$ we cannot define it's canonical conjugate $theta$ globally, as we get a winding around $2pi$. We could think of the ring as the line $[0,2pi]$ with the ends identified, and define the operator that sends $psi(theta)mapsto thetapsi(theta)$ which is a legitimate operator in the Hilbert state. We then get, however, that $$[partial_theta,theta]=1-2pidelta(theta)$$ (as a result of the jump) and so $$leftlangle[-ipartial_theta,theta]rightrangle=0$$ The uncertainty relation will then give $$leftlangle theta^2rightrangleleftlangle L_z^2rightranglege0$$ which is consistent with the ground state wavefunction.