Angular momentum uncertainty principle and the particle on a ring

By considering a particle on a ring, the eigenfunctions of ##H## are also eigenfunctions of ##L_\text{z}##:

$$\psi(\phi) = \frac{1}{\sqrt{2\pi}}e^{im\phi}$$

with ##m = 0,\pm 1,\pm 2,\cdots##. In polar coordinates, the corresponding operators are

$$H = -\frac{\hbar^{2}}{2I}\frac{d^{2}}{d\phi^{2}}$$ $$L_\text{z} = -i\hbar\frac{\partial}{\partial \phi}$$

The Robertson version of uncertainty principle, valid in a statistical sense, tell us

$$\Delta L_{x}\Delta L_\text{y} \geq \frac{1}{2}\left | \int \psi^{*}[L_{x},L_{y}]\psi d\tau\right | $$ where usually ##[L_{x},L_{y}] = i\hbar L_{z}##. The operators for ##L_{x}## and ##L_{y}## are

$$L_{x} = i\hbar \left ( sin(\phi)\frac{\partial}{\partial\theta} + cot(\theta)sin(\phi)\frac{\partial}{\partial\phi}\right )$$

$$L_{y} = -i\hbar \left ( cos(\phi)\frac{\partial}{\partial\theta} - cot(\theta)sin(\phi)\frac{\partial}{\partial\phi}\right )$$

Since the ring plane is located on xy plane, i.e ##\theta = \pi/2##, we obtain

$$L_{x}\psi = L_{y}\psi = 0$$

This make sense, since the particle is rotating around z-axis. Then, ##[L_{x},L_{y}]\psi = 0##? Thus, in this specific case, doesn't the relation ##[L_{x},L_{y}] = i\hbar L_{z}## hold? The interpretation of uncertainty principle is valid for a complete set of eigenfunctions of two operators, but can exist some eigenfunctions of ##L_{z}## which are eigenfunctions of ##L_{x}## and ##L_{y}## (I only know the ##Y_{0}^{0}## case until now). Does it means that ##[L_{x},L_{y}] = i\hbar L_{z}## should be use with care? Why? Is it valid only when ##[L_{x},L_{y}]\psi \neq 0## ? Even expanding the ##L_{z}## eigenfunctions in spherical harmonic, I can't obtain a satisfactory answer. Angular momentum is a vector quantity. Could you help me?

If the particle is confined to a ring in the ##x y## plane, what physical meaning do the operators ##L_x## and ##L_y## have? Nothing can rotate about the ##x## or ##y## axis.

PeterDonis said:

If the particle is confined to a ring in the ##x y## plane, what physical meaning do the operators ##L_x## and ##L_y## have? Nothing can rotate about the ##x## or ##y## axis.

@PeterDonis, classically, when a particle is rotating about z-axis, the x,y components of angular momentum are zero, since L is a vector quantity and there is no rotation around such axes. But, in Quantum mechanics, there is an uncertainty relation which cannot allow this in principle (except when we have ##Y_{0}^{0}##). But, in the case above, the uncertainty relations seems to be in contradition, no?

physical_chemist said:

@PeterDonis, classically, when a particle is rotating about z-axis, the x,y components of angular momentum are zero, since L is a vector quantity and there is no rotation around such axes.

In three dimensions, yes. But you have constrained the problem by confining the particle to a ring in the ##xy## plane. That means the problem is no longer three dimensional and what you are calling "the x,y components of angular momentum" are no longer even physically meaningful, since you have made it physically impossible for anything to rotate except in the ##xy## plane. Or, to put it another way, in two spatial dimensions, which is what you've confined this problem to, there is only one angular momentum component, the one you are calling ##L_z##.

physical_chemist said:

But, in Quantum mechanics, there is an uncertainty relation which cannot allow this in principle (except when we have ##Y_{0}^{0}##). But, n the case above, the uncertainty relations seems to be in contradition, no?

No, because everything I said above about the problem being confined to two spatial dimensions is still true when we pass over from classical physics to quantum mechanics. There is still only one angular momentum component. It's now an operator instead of a number, but it's still just one operator, so the question of uncertainty relations for angular momentum operators doesn't even arise.