Today in class we learned that the commutator $[L_x, L_y] = i\hbar L_z$ Where $L_x, L_y, L_z$ are operators. A consequence of this seems to be according to Heisenberg's uncertainty relation $\Delta L_x \Delta L_y \geq \frac{1}{2}\cdot |<[L_x, L_y]>| $ that the uncertainty in $L_x$ and $L_y$ is dependent on $L_z$.

Now I'm considering the case where $L_z$, which is the value, not the operator, is exactly zero. This means that the mean value of $L_z$ is also zero. Hence the right hand side is zero. Suddenly it seems like I can measure both $L_x$ and $L_y$ exactly. But this contradicts what I have learnt, which is that you can only measure one of the components exactly.

Edit: I have an idea that $L_z=0 => L_x=L_y=0$. But I don't know if this is correct.

  • $\begingroup$ Read up. $\endgroup$ Mar 7, 2017 at 14:42

2 Answers 2


You're confusing operators with their eigenvalues. When you say $L_z=0$, you're not saying the operator is the null operator. You're saying we are considering a state which is an eigenstate of $L_z$ with eigenvalue zero.

Think about this for a moment, and you should understand all the rest.

Anyway, the above says nothing about the commutator of $L_x$ and $L_y$, which is still an operator, and not a number. Finally, your idea that $L_z = 0$ implies $L_x = L_y = 0$ is wrong. $L_x$ and $L_y$ don't have well defined values when $L_z=0$.

EDIT: it is indeed possible that they are, namely when $L^2=0$: all components are null.

  • 1
    $\begingroup$ I agree that the operator Lz is not zero. But when the wave equation of my particle is in the eigenstate so that the value of Lz is zero, then the mean value: integral("ksi* Lz ksi") is zero. Which is what is used in Heisenbergs uncertainty relation. I.e. <[Lx,Ly]> is zero even though [Lx,Ly] is not. $\endgroup$
    – B. Brekke
    Mar 7, 2017 at 12:31
  • $\begingroup$ Indeed, this is answered here: physics.stackexchange.com/questions/210464/… and in user123's answer. $\endgroup$
    – Martino
    Mar 7, 2017 at 16:36
  • $\begingroup$ I have edited my answer. $\endgroup$
    – Martino
    Mar 7, 2017 at 16:37

I find something here (Phys. Rev. Lett. 113, 260401 (2014)) that might be of interest to you. This is quite similar to what have been answered by Bzazz.

The Heisenberg-Robertson uncertainty relation bounds the product of the variances through the expectation value of the commutator \begin{equation} \Delta A \Delta B\geq \frac{1}{2}|\langle[A,B]\rangle|^{2}\tag1 \end{equation} where the expectation value and the variances are calculated on the state of the quantum system $|\psi\rangle.$ It was strengthened by Schröodinger who pointed out that one can add an anti-commutator term, obtaining \begin{equation} \Delta A \Delta B\geq \frac{1}{2}|\langle[A,B]\rangle|^{2}+|\frac{1}{2}\langle\{A,B\}\rangle-\langle A\rangle \langle B\rangle |^{2}\tag2 \end{equation} Both these inequalities can be trivial even in the case in which $A$ and $B$ are incompatible on the state of the system $|\psi\rangle$, e.g. if $|\psi\rangle$ is an eigenstate of $A$, all terms in (1) and (2) vanish.


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