# Angular Momentum Operators - Commutation Relations

I was going over past PGRE exam questions, and came across this one.

The components for the angular momentum operator $\mathbf{L}=(L_x,L_y,L_z)$ satisfy the following commutation relations. \begin{align*} [L_x,L_y]&=i\hbar L_z\\ [L_y,L_z]&=i\hbar L_x\\ [L_z,L_x]&=i\hbar L_y \end{align*} What is the value of the commutator $[L_xL_y,L_z]$? (The answer was (D), $i\hbar(L_x^2-L_y^2)$)

This is easily solved by expanding the commutator, but I was very interested in one of the answers given on this website.

A quick way to do this: As the commutator is $[L_xL_y, L_z]$, Use the right hand rule to point first in the $+X,+Y$ (diagonal) direction, and then curl up to the $Z$ direction. Your thumb will be pointing in the $+X,-Y$ direction, so thus choice (D).

How did the reasoning in that person's answer work? Is there more (than tricky coincidence) behind what he said?

• I think the reasoning is based on the analogy with classical angular momentum. – VictorSeven Jul 10 '15 at 18:17
• Could you expand on that? How does one end up with the difference in the squares of the operators? – Arturo don Juan Jul 10 '15 at 18:19

Let's calculate $[L_x L_y, L_z]$. I'm going to use the property $[AB,C]=A[B,C]+[A,C]B$.

If you apply the property to our case, you obtain $L_x[L_y,L_z]+[L_x,L_z]L_y$. Now you can substitute the value of the commutators and find the correct answer.

Note that the quantum commutation relations are pretty similar to vector product in cartesian coordinates. For example, $[L_x,L_y]=i\hbar L_z$ is analogous to $\hat{x} \wedge \hat{y} = \hat{z}$. I'm not saying that this is the same, but the math is really similar.

So if you look the commutators above, and the property that I've used, you find that you're doing something similar to $(\hat{x}+\hat{y})\wedge\hat{z} = \hat{x} \wedge \hat{z} + \hat{y}\wedge\hat{z} = -\hat{y} + \hat{x}$. So from here you infere that the answer should be something related with $L_x -L_y$. (D) is the only option which has this, so this must be the correct answer.

As a little disclaimer, I find this really strange. If I were you, I would use only the commutators and forget about that "classical tricks". They will often confuse you in QM.

I think this is more or less the reasoning, but maybe somebody has a more complete answer.