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I was reading some notes and it says that $\langle L_z^2\rangle=\langle L^2\rangle$ IFF the system is radially symmetric. I can see that in order that the LHS of the statement implies that $\langle L_x^2\rangle=0=\langle L_y^2\rangle$. But I am not sure how to rigorously prove to myself that in order to make those 2 vanish we must have an isotropic system. Could someone please kindly explain this? Many thanks.

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I am assuming you're talking about the expectation values of these operators on hydrogenlike wave functions?

Consider the eigenvalues of $\hat{L_z}$ and $\hat{L^2}$ operating on some eigenstate $|\psi>$:

$L_z|\psi>=\hbar m|\psi>$ and $L^2|\psi>=l(l+1)\hbar^2|\psi>$. So applying $\hat{L_z}$ twice to get $\hat{L_z^2}$ gives $\hbar^2m^2|\psi>$.

Therefore, $<L_z^2>=<\psi|L_z^2|\psi>=\hbar^2m^2<\psi|\psi>=\hbar^2m^2$.

Similarly, $<L^2>=<\psi|L^2|\psi>=\hbar^2l(l+1)<\psi|\psi>=\hbar^2l(l+1)$

If we equate the two, so $<L_z^2>=<L^2>$, we get $\hbar^2m^2 = \hbar^2l(l+1)$.

Since the allowed values of $m$ are limited by $\pm l$, the only valid solution is the solution where $m=0$ and $l=0$.

This solution, yields the spherical harmonic $Y_0^0(\theta,\phi) = \sqrt{\frac{1}{4\pi}}$ which is spherically symmetric because there is no $\theta$ or $\phi$ dependence.

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This is true in general, it does not require spherical harmonics--- it works for spin too. The algebra here only requires the definition of angular momentum. – Ron Maimon May 11 '12 at 8:40
Can give us a idea how to proof this for general angular momentum algebras? – FlashTek Jun 19 at 19:59

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