Are quantum measurements on two different particles always commutative?

Question

If we have some quantum system of two entangled particles -- let's call them $a$ and $b$. Let $A$ and $B$ be some arbitrary observables we can measure on the two particles, respectively.

Is it necessarily true that the observables commute (i.e., $[A, B] = 0$)?

It makes sense that they should because why would the order of measurement matter when conducted on two different particles. However, I've never seen an explicit theorem or statement that ever claimed this. Is there a neat way to show this in a proof?

Answer 1

Yes, they commute. However, it's not correct to talk about the commutator $[A,B]$ because these are operators defined on different Hilbert spaces.

When you have a system of two particles, the state of each particle lives in a Hilbert space $\mathcal{H}$. So, we say the state of the first particle lives in $\mathcal{H}_A$ and the second in $\mathcal{H}_B$. The overall Hilbert space of the two particle system is given by $\mathcal{H}_A \otimes \mathcal{H}_B$.

Now, an observable on particle $A$ is a Hermitian operator on $\mathcal{H}_A$, and the same it true for operators on particle $B$.

Since we're describing the system now as a two-particle system, we need to talk about operators as acting on the larger Hilbert space $\mathcal{H}$. How do we do that?

The operator corresponding to a measurement on only particle $A$ is given by $A \otimes I_{\mathcal{H}_B}$ where $I_{\mathcal{H}_B}$ is the identity operator on $\mathcal{H}_B$. The operator corresponding to a measurement on only particle $B$ is $I_{\mathcal{H_A}} \otimes B$. Then, it's easy to check that these commute:

$$\begin{align}[A \otimes I_{\mathcal{H}_B}, I_{\mathcal{H_A}} \otimes B] &= \left(A \otimes I_{\mathcal{H}_B}\right) \left( I_{\mathcal{H_A}} \otimes B\right) - \left( I_{\mathcal{H_A}} \otimes B\right)\left(A \otimes I_{\mathcal{H}_B}\right) \\ &= (AI_{\mathcal{H_A}} \otimes I_{\mathcal{H}_B} B) - (I_{\mathcal{H_A}} A \otimes B I_{\mathcal{H}_B})\\ &= (A\otimes B) - (A \otimes B)\\ &= 0 \end{align}$$

So the operators indeed commute, provided that each one is really a measurement on a single particle.

Answer 2

I haven't seen it explicitly stated either, but it follows directly from definitions.

If $A$ is an observable associated with only the first particle, then by definition, it actually takes the form $O_a \otimes I_b$ where $I_b$ is the identity operator acting on the Hilbert space for the second particle. Likewise, $B$ by definition should take the form $I_a \otimes O_b$. And again, by definition, $$AB = (O_a \otimes I_b) (I_a \otimes O_b) = O_a \otimes O_b$$ which is equal to $BA$, so the operators commute.