In quantum physics, you can find commutators of angular momentum, L. First examine L* _{x}*, L

*, and L*

_{y}*by taking a look at how they commute; if they commute (for example, if [L*

_{z}*, L*

_{x}*] = 0), then you can measure any two of them (L*

_{y}*and L*

_{x}*, for example) exactly. If not, then they’re subject to the uncertainty relation, and you can’t measure them simultaneously exactly.*

_{y}Okay, so what’s the commutator of L* _{x}* and L

*? Using L*

_{y}*= YP*

_{x}*– ZP*

_{z}*and L*

_{y}*= ZP*

_{y}*– XP*

_{x}*, you can write the following:*

_{z}[L* _{x}*, L

*] = [YP*

_{y}*– ZP*

_{z}*, ZP*

_{y}*– XP*

_{x}*]*

_{z}You can write this equation as

But

So L* _{x}* and L

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

_{y}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

*, and L*

_{y}*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 |*

_{z}*l*, m >?

The usual trick here is that the square of the angular momentum, L^{2}, is a scalar, not a vector, so it’ll commute with the L* _{x}*, L

*, and L*

_{y}*operators, no problem:*

_{z}[L

^{2}, L] = 0_{x}[L

^{2}, L] = 0_{y}[L

^{2}, L] = 0_{z}

Okay, cool, you’re making progress. Because L* _{x}*, L

*, and L*

_{y}*don’t commute, you can’t create an eigenstate that lists quantum numbers for any two of them. But because L*

_{z}^{2}commutes with them, you can construct eigenstates that have eigenvalues for L

^{2}and any one of L

*, L*

_{x}*, and L*

_{y}*. By convention, the direction that’s usually chosen is L*

_{z}*.*

_{z}