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When a particle is on a ring, we know that the commutator $[L_z,x]=i\hbar y$ and $[L_z,y]=-i\hbar x$. So we can't measure $L_z$ and $x$ & $L_z$ and $y$ simultaneously with infinite accuracy.

When $m=0$, then the energy of the particle is 0. So we can say that $L_z$(angular momentum in z-direction) is 0 as we consider potential energy to be 0 on the ring. This also implies $p_z$ is also zero as $L_z=rp_z$.

My question is that if the particle is in $m=0$ state on the ring, then if we find its position $x$ and $y$ simultaneously {as $[x,y]=0$}, we are confident that $L_z=0$ on the ring. Then doesn't the commutator relation for $[L_z,x]=[L_z,y]=0=[p_z,z]$ (as $z=0$ is considered on ring) for $m=0$?

Here we see that commutator relation and uncertainty principle is violated. I know that it can't be the case as they are generalized. Please tell me what is wrong in my reasoning.

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    $\begingroup$ Your definition of $L_z = rp_z$ isn't quite right. $$\vec{L} = \vec{r} \times \vec{p} \implies \pmatrix{L_x \\ L_y \\ L_z} = \pmatrix{x \\ y \\ z} \times \pmatrix{p_x \\ p_y \\ p_z} = \ldots?$$ $\endgroup$ – orthocresol♦ Jun 16 '20 at 7:22
  • $\begingroup$ Oh, I see. But the total energy is 0 in m=0, doesn't this imply that energy by virtue of any translation and rotational motion is 0 and L and p=0 because potential energy is considered to be 0 on ring? $\endgroup$ – Manu Jun 16 '20 at 7:43
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    $\begingroup$ The commutators you need are $[\hat L_x,\hat L_y]=i\hat L_z$ and cyclic variants on $x,y,z$. When the particle has zero energy its orientation is unknowable meaning that it can be anywhere on the ring which is accord with the uncertainty principle. $\endgroup$ – porphyrin Jun 16 '20 at 10:52
  • $\begingroup$ Ok, now I have understood $\endgroup$ – Manu Jun 16 '20 at 14:11

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