How to Find Commutators of Angular Momentum, L

In quantum physics, you can find commutators of angular momentum, L. First examine L_x, L_y, and L_z by taking a look at how they commute; if they commute (for example, if [L_x, L_y] = 0), then you can measure any two of them (L_x and L_y, for example) exactly. If not, then they’re subject to the uncertainty relation, and you can’t measure them simultaneously exactly.

Okay, so what’s the commutator of L_x and L_y? Using L_x = YP_z – ZP_y and L_y = ZP_x – XP_z, you can write the following:

[L_x, L_y] = [YP_z – ZP_y, ZP_x – XP_z]

You can write this equation as

But

So L_x and L_y don’t commute, which means that you can’t measure them both simultaneously with complete precision. You can also show that

Because none of the components of angular momentum commute with each other, you can’t measure any two simultaneously with complete precision. Rats.

That also means that the L_x, L_y, and L_z operators can’t share the same eigenstates. So what can you do? How can you find an operator that shares eigenstates with the various components of L so that you can write the eigenstates as | l, m >?

The usual trick here is that the square of the angular momentum, L², is a scalar, not a vector, so it’ll commute with the L_x, L_y, and L_z operators, no problem:

• [L², L_x] = 0

• [L², L_y] = 0

• [L², L_z] = 0

Okay, cool, you’re making progress. Because L_x, L_y, and L_z don’t commute, you can’t create an eigenstate that lists quantum numbers for any two of them. But because L² commutes with them, you can construct eigenstates that have eigenvalues for L² and any one of L_x, L_y, and L_z. By convention, the direction that’s usually chosen is L_z.