Proving Commutation of an Operator with Rotation Generator Components

[MathematicalPhysicist]

Homework Statement

Prove that if A is an operator which commutes with two components of the rotation generator operator, J, then it commute with its third component.

Homework Equations

$$[A_{\alpha},J_{\beta}]=i \hbar \epsilon_{\alpha \beta \gamma} A_{\gamma}$$
(not sure about the sign of this commutator it might be minus.

The Attempt at a Solution

Ok, then I am given:
$$[A,J_{\alpha}]=[A,J_{\beta}]=0$$
thus also $$[A^2,J^2]=[A^2_{\alpha}+A^2_{\beta}+A^2_{\gamma},J^2_{\alpha}+J^2_{\beta}+J^2_{\gamma}]=[A^2,J^2_{\gamma}]=0$$ I used the above relevant equations to get to the last equality, but here is where I am stuck, I also know that $$[A^2,J_{\gamma}]=0$$, but I don't seem to get to the last punch line which is $$[A,J_{\gamma}]=0$$.

Any hints?

 

[George Jones]

$$[A_{\alpha},J_{\beta}]=i \hbar \epsilon_{\alpha \beta \gamma} A_{\gamma}$$
(not sure about the sign of this commutator it might be minus.

Did you really mean to write this?

All directions are equivalennt, so, WLOG, you can pick J_1 and J_2 as the components that commute with A.
Use [J_1 , J_2] = i \hbar J_3 in [A , J_3], and maybe use the Jacobi identity.

 

[MathematicalPhysicist]

OK thanks, I glossed over Jacobi identity in Wiki, and it does the job.

 

[George Jones]

A more transparent way might be to expand completely (i.e., get rid all commutators) [A, [J_1, J_2]], and use the fact that A commutes with J_1 and J_2 to move all the A's completely to right (or left) of all the terms.