2D rotation and angular momentum uncertainty

[hokhani]

TL;DR

How a quantum particle may be confined to a planar rotation?

In quantum mechanics, the uncertainty among the components of the angular momentum doesn't allow to find all the components together. However, when a quantum particle is confined to the $xy-$plane, it seems that $L_x$ and $L_y$ are both zeros. How about the uncertainty here?

 

[pines-demon]

What do you mean? What do you want to know? In 2D you only have one generation of rotations: $L_z$.

 

[hilbert2]

Even in 3D there can be state vectors, such as the hydrogenic atom s orbitals, where all of the angular momentum components have value 0. The uncertainty relation just says that you can't form a complete basis of $\mathcal{H}$ from only vectors that have simultaneously definite $L_x$, $L_y$ and $L_z$.

 

[PeterDonis]

when a quantum particle is confined to the $xy-$plane, it seems that $L_x$ and $L_y$ are both zeros.

No, when a quantum particle is confined to the $x y$ plane, its Hilbert space is restricted, and there aren't even any such operators as $L_x$ and $L_y$ on vectors in the restricted Hilbert space (there is only one angular momentum operator on that space, $L_z$). So it doesn't even make sense to talk about any uncertainty relation between $L_x$ and $L_y$.