# Uncertainity principle of a particle on a ring in $m=0$ state

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.

• Your definition of $L_z = rp_z$ isn't quite right. $$\vec{L} = \vec{r} \times \vec{p} \implies \pmatrix{L_x \\ L_y \\ L_z} = \pmatrix{x \\ y \\ z} \times \pmatrix{p_x \\ p_y \\ p_z} = \ldots?$$ – orthocresol Jun 16 '20 at 7:22
• 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? – Manu Jun 16 '20 at 7:43
• 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. – porphyrin Jun 16 '20 at 10:52
• Ok, now I have understood – Manu Jun 16 '20 at 14:11