 Solution

In the position representation, the orbital angular momentum operator isOn a common smooth domain use $[x_a,p_b]=i\hslash {\delta }_{ab}$, $[x_a,x_b]=[p_a,p_b]=0$. Expanding the product commutator givesContracting with the two alternating tensors gives the orbital angular momentum commutation relations:For $L_{\pm }=L_1\pm i{L}_2$, these imply $[L_3,L_{\pm }]=\pm \hslash L_{\pm }$. Therefore $L_3\psi =0$ yieldsThusThe nonzero qualification is necessary: a rotationally invariant wavefunction is annihilated by all three components, and the zero vector is not an eigenvector.