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Consider the Bell state $|\psi\rangle =\frac{|00\rangle +|11\rangle}{\sqrt{2}}$.

When measuring one of the individual systems, the superposition is projected onto one of the eigenstates $|00\rangle$ or $|11\rangle$.

Imagine we measured the entangled systems at the same time. Which part will project the state? That is, at measurement one part of the system determines the outcome of the other. But if measured at the same time, which determines the outcome?

For the moment I neglect special relativity and think the observer to be in the rest frame of the experiment.

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  • $\begingroup$ I've deleted an off-topic comment thread. $\endgroup$
    – David Z
    Commented Jul 6, 2016 at 17:34

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The outcome is determined (probabilistically) by the state, and is either $|00\rangle$ or $11\rangle$ (assuming that's the observation you're making) equiprobably.

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  • $\begingroup$ No, the state doesn't determine the outcome. Before a measurement we have superposition. That's what the Bell test proves: The outcome is not determined before the measurement is made and thus rules out all hidden variable theories. (Given there are no loopholes and c is constant) $\endgroup$
    – Zetaman
    Commented Jul 6, 2016 at 14:23
  • $\begingroup$ @JohannesFankhauser: It rules out some classes of hidden variable theories. It doesn't seem to rule out all hidden variable theories. Not that this matters. That's not a question to nature, anyway. $\endgroup$
    – CuriousOne
    Commented Jul 6, 2016 at 14:25
  • $\begingroup$ I'm not familiar with these theories. Which are they? $\endgroup$
    – Zetaman
    Commented Jul 6, 2016 at 14:29
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    $\begingroup$ @JohannesFankhauser : *"No, the state doesn't determine the outcome". * Perhaps you overlooked the word "probabilistically". $\endgroup$
    – WillO
    Commented Jul 6, 2016 at 14:30
  • $\begingroup$ @JohannesFankhauser: You have to ask the quantum philosophers about that, it's certainly not a physics questions. $\endgroup$
    – CuriousOne
    Commented Jul 6, 2016 at 14:33
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To add on to WillO's answer, I think the key in entanglement is not "which measurement happens first" or "which measurement affects the other", but rather that $|00\rangle$ and $11\rangle$ are the only possible outcomes! No matter what frame you switch to, you will NEVER get $|01\rangle$ or $|10\rangle$.

Even if you consider the different frames in which the two events can have different chronological order, the Bell test still rules out local hidden variable theories because the two events are causally disconnected regardless of which frame you look at it from. So the causal disconnection is not affected by the chronological ambiguity.

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    $\begingroup$ I really believe the concept of 'locality' or 'causality' need to be reconsidered when entanglement is involved. Since entanglement is related with the spacetime geometry, on which our concept of 'locality' is built. $\endgroup$
    – XXDD
    Commented Jul 6, 2016 at 14:44
  • $\begingroup$ @X.Dong: Locality is well defined and it works just fine. Causality is also well defined and it is not violated by entanglement. Entanglement simply doesn't exchange anything, even though it looks that way, and that's what throws people's minds off. $\endgroup$
    – CuriousOne
    Commented Jul 6, 2016 at 14:47
  • $\begingroup$ I do not think so. Before we have a clear picture of the structure of spacetime, we can not have the correct concept of 'locality'. If Preskill is correct, the AdS/CFT is related with quantum error correcting code, then we have to reconsider these concepts. $\endgroup$
    – XXDD
    Commented Jul 6, 2016 at 14:50
  • $\begingroup$ There are recent developments relating entanglement and wormholes by Maldacena and Susskind, but I don't understand the details of their argument at all, so I will not comment... Maybe you want to look at this SE post: physics.stackexchange.com/questions/34650/… ? $\endgroup$
    – Zhengyan Shi
    Commented Jul 6, 2016 at 14:54
  • $\begingroup$ At this point it seems that whenever Susskind doesn't understand something, he confabulates a wormhole into it. Don't fall for that stuff. $\endgroup$
    – CuriousOne
    Commented Jul 6, 2016 at 14:58

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