# If $L_z$ has a $0$ eigenfunction, since $[L_x, L_y] = i\hbar L_z,$, then can $L_x, L_y$ have a simultaneous eigenfunctions?

In the lecture Quantum Mechanics by Dr. Adams in ocw.mit.edu, in the 16th lecture at 7:11, it is stated that since $$[L_x, L_y] = i\hbar L_z,$$ there is no state s.t it is eigenfunction of both $$L_x, L_y$$. In fact, this is stated whenever the commutator of any two operators is nonzero.

However, if the commutator, in this case $$L_z$$, have eigenvalue $$0$$ for some eigenfunction $$\phi$$, then can these two operators whose commuter is computed, in this case $$L_x, L_y$$, have a simultaneous eigenfunctions ?

You are right. $$|l=0,m=0\rangle$$ (i.e., $$Y^0_0$$ ) is a simultaneous eigenvector of $$L^2$$ and $$L_x,L_y,L_z$$ with eigenvalue $$0$$. What is impossible is the existence of a common basis of eigenvectors of $$L_x$$ and $$L_y$$: the two operators would be diagonal with respect to that common basis and therefore they would commute in contrast to the commutation relation you wrote since $$L_z\neq 0$$.
With the same argument you see that $$L_z\psi=0$$ if $$\psi \neq 0$$ is a simultaneous eigenvector of $$L_x$$ and $$L_y$$. So $$\psi$$ is also an eigenvector of $$L_z$$ with eigenvalue $$0$$. Permuting the axes, the same argument proves that $$L_x\psi=L_y\psi=0$$. Notice that such eigenvector is also an eigenvector of $$L^2$$. As a consequence, there are no common eigenvalues when $$L^2$$ assumes semi-integer values. In fact, it is known that the eigenvalues pf $$L^2$$ (I am referring to it as a general angular momentum observable, not necessarily the orbital momentum) are of the form $$\hbar l(l+1)$$, where $$l$$ is integer $$l= 0,1,2, \ldots$$ or semi integer $$l= 1/2, 3/2, 5/2,\ldots$$. Correspondingly, the eigenvalues $$m$$ of $$L_z$$ take values $$-l, -l+1, ..., l-1, l$$. Therefore, $$m$$ may be $$0$$ only if $$l$$ is integer.
Summing up, assuming that $$L_x,L_y,L_z$$ represent the total angular momentum of a given physical system (including the spin if there is),
1. There are no common (orthonormal) basis of eigenvectors of $$L_x,L_y$$;
2. If a common eigenvector exists for $$L_x,L_y$$ it must have eigenvalue $$0$$ simultaneously for $$L_x,L_y,L_z$$ and also $$L^2$$;
3. A common eigenvector of $$L_x,L_y$$ exists if and only if the operator $$L^2$$ -- whose eigenvalues are always in the set $$\hbar l(l+1)$$ for $$l \in \mathbb N/2$$ -- has eigenvalues of the form $$l=k/2$$ with even $$k$$.