Spin operators commutation Why do the spin operators $ S_{x1}$ and $S_{x2}$ of two particles along the $x$-axis  commute i.e $S_{1x}S_{x2}-S_{2x}S_{1x}=0 $ ?
 A: Let $\mathcal{H}$ be the Hilbert space for one particle. Then, $S_{x}\in\mathcal{B}(\mathcal{H})$ is a bounded, self-adjoint operator. Now, if you want to have the Hilbert space for two particles, remember that this is the tensor product, i.e. $\mathcal{H}=\mathcal{H}_1\otimes \mathcal{H}_2$ (where $\mathcal{H}_1$ is the Hilbert space of the first and $\mathcal{H}_2$ the Hilbert space of the second particle). 
What then is the operator? Let's consider the spin operator $S_{x}^{(1)}\in\mathcal{B}(\mathcal{H}_1)$ and $S_{x}^{(2)}\in\mathcal{B}(\mathcal{H}_2)$, then the operator for the first particle is $S_{1x}:=S_{x}^{(1)}\otimes 1$ - because it doesn't do anything on the second particle. For the second particle, we have $S_{2x}:=1\otimes S_{x}^{(2)}$, because the second spin operator, only acts on the second particle. As you can easily show, those two commute, since 
$$S_{1x}S_{2x}=(S_{x}^{(1)}\otimes 1) \cdot (1\otimes S_{x}^{(2)})=S_{x}^{(1)}\otimes S_{x}^{(2)}=(1\otimes S_{x}^{(2)})\cdot (S_{x}^{(1)}\otimes 1)=S_{2x}S_{1x}$$
In general, every two operators that act on different parts of the system should commute. One simple reason is that the operators should be jointly measurable (because I can just do the measurement on the one particle, while you measure the other).
