The angular momentum operators $L_x = yp_z-zp_y$, $L_x = zp_x-xp_z$, $L_x = xp_y-yp_x$ do not commute. In fact, $[L_x,L_y] = i\hbar L_z$. We also know that $L_z |nlm\rangle = \hbar m$. So what if we have an eigenstate $\psi = |100\rangle$, this will make $L_z \psi = 0$. For this state, can we measure $L_x$ and $L_y$ simultaneously, in theory?

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    $\begingroup$ "$\psi = |100>$... For this state, can we measure $L_x$ and $L_y$ simultaneously, in theor?" Yes, they are also zero. $\endgroup$
    – hft
    Commented Nov 10, 2022 at 2:44
  • $\begingroup$ I think the title of this question has a typo. $\endgroup$
    – DanielSank
    Commented Nov 10, 2022 at 22:26
  • $\begingroup$ Um... Can we measure any two things simultaneously for a single object? I think the fact you have commutators here indicates you don't really mean simultaneously. Lx Lz is one then the other. $\endgroup$
    – Boba Fit
    Commented Nov 10, 2022 at 22:39
  • $\begingroup$ @BobaFit Yes, for example you can measure the $x$ and $y$ coordinates of a particle simultaneously. Also note that, following standard notation in quantum mechanics, the product $L_x L_y$ refers to multiplication of linear operators on Hilbert space, and not a sequence of measurements. $\endgroup$
    – Andrew
    Commented Nov 10, 2022 at 23:14

1 Answer 1


$L_x, L_y, L_z$ do commute on the subspace of states where $L^2=0$. So, if you prepare a state that lies in this subspace, you can measure all three simultaneously. (All three must be zero). That subspace includes the state $|100\rangle$ that you asked about explicitly.


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