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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?

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2 Answers 2

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Is it always the case here that operators acting on different properties?

If different operators act on different spaces then the operators do for sure commute. I think it's not so hard to see that if you make an operator operate on two distinct spaces, one after another, then it doesn't matter which operator is applied first. If the operation of one of them takes place in a space different from the space in which the other operates then the wavefunction corresponding to the total space is acted upon in two independent ways. The wavefunction's part in one space doesn't change the part of the wavefunction belonging to the other space when acted upon by a corresponding operator.
If they act in the same space then it depends.

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.

So, if operators act on different spaces they certainly commute but when they act on the same space this remains to be seen.

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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".

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