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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.

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  • $\begingroup$ I don't quite understand your question from where you say "...we can write the effect of the measurement of Alice and Bob on the state as...". Measurement cannot be represented by unitary transformations like $\sigma_3^A \otimes I^B$ or $I^A \otimes \sigma_3^B$. $\endgroup$ – Sanchayan Dutta May 29 at 18:34
  • $\begingroup$ Uh, okay, trying really hard here but, on a one qubit state you apply $\sigma_3$ to measure the spin on the z-axis? So if you would try to measure the spin on the z-axis of spin A of a two qubit system, I would assume you use $\sigma_3^A \otimes I^B$? How do you mean that measurement cannot be represented by unitary transformations? Okay, maybe I get that, but what does "apply $\sigma$" mean then if we don't let the operators work on our state? My question remains that $\sigma_3^A \otimes I^B$ and $ I^A \otimes \sigma_3^B $ "applied" consecutively comes down to $\sigma_3^A \otimes \sigma_3^B$ $\endgroup$ – CFRedDemon May 29 at 18:49
  • $\begingroup$ So I don't really get the difference between applying $\sigma_3^A \otimes I^B$ and $ I^A \otimes \sigma_3^B $ and $\sigma_3^A \otimes \sigma_3^B$ then? So maybe you can help me understand the difference? What does the measurement do then? Or how do you depict it, is it because the post measurement state is completely different? Or what happens? $\endgroup$ – CFRedDemon May 29 at 18:51
  • $\begingroup$ Preskill probably means that you're carrying out the measurements in the eigenbases of the operators. Measurements themselves cannot be represented as operators. $\endgroup$ – Sanchayan Dutta May 29 at 18:51
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    $\begingroup$ Playing loose with terminology will only confuse you. POVM is a measure (in the measure theoretic sense); it's not an operator. Sure you can describe it as a set of positive semidefinite operators, but it's not an operator in itself. $\endgroup$ – Sanchayan Dutta May 29 at 19:11
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  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.

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