Simultaneous eigenstate of commuting observables and their tensor product

Question

So this is about something from Preskill's notes on Quantum Computation and Information, Chapter 4, page 3.

Imagine we have a maximally entangled state (Bell state). We can identify the Bell state by measuring the phase and parity bit with the commuting operators

$$\sigma_1^A \otimes \sigma_1^B \\ \sigma_3^A \otimes \sigma_3^B$$

Now they want to show that you cannot do this locally, so when Alice and Bob both are spacelike separated and can only do measurements on their own qubit A and B.

Following Preskill, they can start by measuring $\sigma_3^A$ and $ \sigma_3^B$ each on their own, and because this commutes with $\sigma_3^A \otimes \sigma_3^B$, this does not disturb the parity bit. So they prepared a simultaneous eigenstate of $\sigma_3^A$ and $ \sigma_3^B$. These eigenstates do not commute with $ \sigma_1^A \otimes \sigma_1^B$ so they disturbed the phase bit so they cannot measure it. Conclusion: they cannot read the information that is in the entanglement.

But my real question now is: why do the first two observables $ \sigma_1^A \otimes \sigma_1^B$ and $ \sigma_3^A \otimes \sigma_3^B$ commute, when $\sigma_3^A$ and $ \sigma_3^B$ will not commute with $ \sigma_1^A \otimes \sigma_1^B$. Because in my opinion, we can write the effect of the measurement of Alice and Bob on the state as;

$$\sigma_3^A \otimes I^B \\ I^A \otimes \sigma_3^B$$

which, when measured after each other will have the effect on $|\psi\rangle$:

$$(\sigma_3^A \otimes I^B) (I^A \otimes \sigma_3^B) |\psi\rangle = \sigma_3^A \otimes \sigma_3^B |\psi\rangle$$

which is the original measurement that can be done when both qubits are at the same place? So what is the difference, why is this not the same eigenstate? I must be missing something, but I don't see where I make an error. Can someone please explain.

Answer 1

1. Measuring an observable does not mean applying the observable operator to a quantum state but rather measuring the state in the eigenbasis of the operator. A measurement will basically produce an eigenvalue of that observable operator and the system will collapse to a subspace corresponding to all states having that eigenvalue. For example, if you're measuring in the $\sigma_1$ basis, then you'll get either $+1$ corresponding to the eigenstate $|+\rangle$ or $-1$ corresponding to the eigenstate $|-\rangle$. Similarly, if you measure in the $\sigma_3$ basis then you'll either get $+1$ corresponding to the eigenstate $|0\rangle$ or $-1$ corresponding to the eigenstate $|1\rangle$.

2. The fact that $\sigma_3^A$ and $\sigma_3^B$ do not commute with $\sigma_1^A\otimes \sigma_1^B$ is a purely mathematical artefact and it's easy to prove along these lines. There's also a technical mistake in your sentence: "These eigenstates do not commute with $\sigma_1^A\otimes \sigma_1^B$ so they disturbed the phase bit so they cannot measure it."; it's meaningless to say that a state doesn't commute with an operator — commutators are only defined for operators.