# Confused how to solve my equation of operators

I am trying to show $$[J_x,J_y] = i\hbar J_z$$ commutation relation for the operators.

I have so far expanded to get:

$$=(L_xL_y+L_xS_y+S_xL_y+S_xS_y)-(L_yL_x+L_yS_x+S_yL_x+S_yS_x)$$

Giving:

$$= [L_x,L_y]+[S_x,S_y] + (L_xS_y-S_yL_x)+(S_xL_y-L_yS_x) \\= i\hbar L_z + i\hbar S_z + [L_x,S_y] + [S_x,L_y] \\= i\hbar (L_z+S_z) + [L_x,S_y] + [S_x, L_y] \\= i\hbar J_z + [L_x,S_y] +[S_x,L_y]$$

Is it always the case here that operators acting on different properties (such as the last two terms here) always have a commutation relation that is zero? Or do they evaluate to something else?

For example. the spin operators and the angular momentum operators act on different spaces (the internal spin space and Hilbert space of state functions, which are functions of either space and time or momentum, depending on the representation) while the operators for $$L_x$$ and $$L_y$$ act on the same space and don't commute. The operators for angular momentum in the same direction do commute though.
The word is not "different" properties (observables), but "compatible". That is, properties that can be determined at the same time (admit eigenstates that are common to both). Generally there are two ways in which properties (observables: Hermitian operators) $$A$$ and $$B$$ can commute --and thus be compatible--; one is that $$A$$ is a diagonal function of $$B$$, $$A=f\left( B \right)$$ (although they "live" in the same subspace) and the other is that, say, one is "external" to the other, so that the complete basis is made up of tensor products of both eigenstates, $$\left|a\right\rangle \otimes\left|b\right\rangle$$ The latter is the case as concerns orbital and spin angular momentum. So, in your case, $$\left[L_i,S_j\right]=0$$ The same is the case with, e.g., $$x$$ and $$y$$ position variables. But this is more postulational than otherwise. That's, I think, the reason why you got stuck there. Spin is "internal".