# Spherical symmetry and mean of angular momentum

I have the following problem.

Consider a 3 dimensional system with spherical symmetry. Consider a state $|\psi \rangle$ such that the possible results of a measure of operator $L^2$ (square of angular momentum) are $0$, $2\hbar^2$ and $6\hbar^2$. Let all possible results for simultaneous measurements of $L^2$ and $L_z$ (third component of angular momentum) be equiprobable. Evaluate the average of $L_z$ and give a lower bound for the product $\Delta L_x \Delta L_y$.

Here's my attempt. $|\psi\rangle = \sum_{l,m} c_{l,m} |l,m\rangle$ for some coefficients. Due to the equiprobability request, $|c_{l,m}|^2=|c|^2$ for some $c$. Therefore, $c_{l,m} = c e^{i \theta_{l,m}}$. So, $\langle L_z\rangle=\hbar |c|^2 \sum_{l,m} \sum_{l',m'}e^{-i \theta_{l',m'}}e^{i \theta_{l,m}} \langle l',m'| m |l,m\rangle$, since $L_z |l,m\rangle = \hbar m |l,m\rangle$.

Now, $m=-l\dots l$ and $L=0,1,2$, so $\sum_{m=-l}^{l}m=0$ and $\langle L_z\rangle=0$. But, from Heisenberg's principle, $\Delta L_x \Delta L_y \geq \frac{1}{2} |\langle[ L_x, L_y]\rangle|$, so this would be $0$ aswell.. which is incorrect, from my understanding. What am I doing wrong?

• Please, consider separating your question in paragraphs. Also, I don't see the problem with Heinsenberg's principle. – FGSUZ Jul 20 '18 at 13:54
• Added a few paragraphs. Shouldn't I obtain a strictly positive quantity on the RHS, for the indetermination principle? Doesn't "0" as a result tell that there is no uncertainty at all? – user3461126 Jul 20 '18 at 13:58
• $\Delta L \neq \langle L\rangle$, plus, there is an inequality. – FGSUZ Jul 20 '18 at 14:01

## 1 Answer

Heisenberg's inequality says that $\Delta L_x \Delta L_y \geq 0$; this is not false! It tells you that if $\langle L_z \rangle = 0$, uncertainty doesn't forbid a state of definite $L_x$ or $L_y$.