# Simple question on Angular Momentum

Need to know why $$L^2$$ and ONLY ONE of $$L_x$$, $$L_y$$, $$L_z$$ are constants of motion. Main problem arrives when $$V = f(r, \theta, \phi)$$ causing none of the $$L_x$$, $$L_y$$, $$L_z$$ to commute with Hamiltonian

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– user191954
Commented Feb 8, 2019 at 8:53

Need to know why $$L^2$$ and ONLY ONE of $$L_x$$, $$L_y$$, $$L_z$$ are constants of motion.
We say an observable $$A$$ is a constant of the motion if it commutes with $$H$$: Then $$\langle A\rangle$$, computed on a state whatever, doesn't depend on time and the same for all matrix elements. Moreover if $$A$$ is a constant of the motion it shares a base of eigentunctions with $$H$$.
As a special case: if potential energy is centrally symmetrical ($$V=f(r)$$) then every component of angular momentum is a constant of the motion, in particular $$L_x$$, $$L_y$$, $$L_z$$. And the same is true for $$L^2=L_x^2+L_y^2+L_z^2$$. But $$L_x$$, $$L_y$$, $$L_z$$ don't commute with one another, so you can't find a common base of eigenfunctions. Since $$L^2$$ does commute with each component, there is a common base of eigenfunctions of $$L^2$$ and - say - $$L_z$$. This is independent of their being constants of the motion or not.
If $$L^2$$ and $$L_z$$ are constants of the motion, then there is a common base of $$H$$, $$L^2$$, $$L_z$$. This is why - for instance - we're used to classify the eigenstates of energy for the hydrogen atom with the three quantum numbers $$n$$, $$l$$, $$m$$: $$n$$ for $$H$$, $$l$$ for $$L^2$$, $$m$$ for $$L_z$$.
If $$V$$ is a function of $$\theta$$ and $$\phi$$ as well as $$r$$, then there is no symmetry and none of the $$L_{x,y,z}$$ will be constants of motion. Re-wording your initial statemnt: $$L^2$$ and AT MOST ONE of $$L_x,L_y,L_z$$ MAY be constants of motion.