Let's say we have two Hilbertspaces $H_1$ and $H_2$, with angular momentum operators $j_1$ and $j_2$. On their product space $H_1 \otimes H_2$ we define the total angular momentum operator as:
$J = j_1 \otimes id^{(2)} + id^{(1)} \otimes {j_2}$
Now i tried to calculate the commutator in the product space $[J_x,J_y]$ (it should equal to $i \hslash J_z$):
$$[J_x,J_y] = [j_{1x}\otimes id^{(2)} + id^{(1)} \otimes j_{2x},j_{1y}\otimes id^{(2)} + id^{(1)} \otimes j_{2y}]\tag{1}$$
This splits up into a sum of 4 commutators, the first being:
$$[j_{1x}\otimes id^{(2)},j_{1y}\otimes id^{(2)}] = (j_{1x}\otimes id^{(2)})(j_{1y}\otimes id^{(2)}) - (j_{1y}\otimes id^{(2)})(j_{1x}\otimes id^{(2)})\tag{2}$$.
Now lets multiply a vector $|\psi \rangle |\phi \rangle \in H_1 \otimes H_2$ onto this commutator:
$$((j_{1x}\otimes id^{(2)})(j_{1y}\otimes id^{(2)}) - (j_{1y}\otimes id^{(2)})(j_{1x}\otimes id^{(2)})) |\psi \rangle |\phi \rangle = (j_{1x}j_{1y}-j_{1y}j_{1x})|\psi \rangle (id^{(2)}id^{(2)}-id^{(2)}id^{(2)})|\phi \rangle = i\hslash j_{1z}|\psi \rangle \ 0|\phi \rangle = 0\tag{3}$$
And i think by the same logic the complete commutator equals to zero, which clearly can't be true.
Can somebody identify the mistake or explain how the commutator on the product space is defined? Thank you.
