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 ?


1 Answer 1


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 $k$.

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.