Commuting angular momentum

Question

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.

Answer 1

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.

Answer 2

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.