The ground state of particle in a ring have zero energy. But doesn't that mean, position of the particle is precisely deterministic.
How do we reconcile HUP in this case?
And what is zero point energy here. Is it 0 or next higher energy state corresponding to n =+-1
P.S - Sorry don't know way to write maths here.
The ground state is going to look like:
$$ \psi(\theta) = \frac 1 {\sqrt{2\pi}}e^{in\theta} $$
with $n=0$, which is:
$$ \psi(\theta) = \frac 1 {\sqrt{2\pi}} $$
which manifestly does not have a preferred coordinate.
But we can find the expectations values and uncertainties via:
$$ \langle\theta^n\rangle = \int_0^{2\pi}\psi^*\theta^n\psi d\theta$$ $$ \langle\theta^n\rangle = \frac 1 {2\pi}\int_0^{2\pi}\theta^nd\theta = \frac 1 {n+1}\theta^{n+1}|^{2\pi}_0 = \frac{(2\pi)^n}{n+1}$$
so that:
$$ \langle\theta\rangle = \pi$$
and
$$ \langle\theta^2\rangle = \frac{4\pi^2} 3$$
Hence:
$$ \sigma_{\theta} = (\frac{4\pi^2} 3 - \pi^2)^{\frac 1 2}= \frac 1 {\sqrt 3} \pi $$
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 $-i\partial_q$, and that's because translations (rotations) are generated by linear (angular) momentum operators.
My guess is that by "violation of HUP" you mean that $$\left\langle \theta^2\right\rangle\left\langle L_z^2\right\rangle=0$$
The reason for that is that while we can define the operator $L_z=-i\partial_\theta$ we cannot define it's canonical conjugate $\theta$ globally, as we get a winding around $2\pi$. We could think of the ring as the line $[0,2\pi]$ with the ends identified, and define the operator that sends $\psi(\theta)\mapsto \theta\psi(\theta)$ which is a legitimate operator in the Hilbert state. We then get, however, that $$[\partial_\theta,\theta]=1-2\pi\delta(\theta)$$ (as a result of the jump) and so $$\left\langle[-i\partial_\theta,\theta]\right\rangle=0$$ The uncertainty relation will then give $$\left\langle \theta^2\right\rangle\left\langle L_z^2\right\rangle\ge0$$ which is consistent with the ground state wavefunction.
