D.64 Angular momentum uncertainty

Suppose that an eigenstate, call it $\big\vert m\big\rangle$, of ${\widehat J}_z$ is also an eigenstate of ${\widehat J}_x$. Then $[{\widehat J}_z,{\widehat J}_x]\big\vert m\big\rangle$ must be zero, and the commutator relations say that this is equivalent to ${\widehat J}_y\big\vert m\big\rangle$ $\vphantom0\raisebox{1.5pt}{$=$}$ 0, which makes $\big\vert m\big\rangle$ also an eigen­vector of ${\widehat J}_y$, and with the eigenvalue zero to boot. So the angular momentum in the $y$-direction must be zero. Repeating the same argument using the $[{\widehat J}_x,{\widehat J}_y]$ and $[{\widehat J}_y,{\widehat J}_z]$ commutator pairs shows that the angular momentum in the other two directions is zero too. So there is no angular momentum at all, $\big\vert m\big\rangle$ is an $\big\vert\:0\big\rangle$ state.