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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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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 $L_z$.

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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.

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  • $\begingroup$ Thanks! I thought the same, now i have clarity $\endgroup$ – Aditya Singh Feb 7 at 14:17

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