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I have read some papers about experimental proof of non-locality involving a laser that goes through a beam-splitter and then each "half" goes to an observer (traditionally "Alice" and "Bob"). It has been shown Bob can manipulate Alice's results by making certain observations on his side. The idea is that what Bob does has a non-local, instantaneous affect on Alice's photon stream, due to entanglement between his photons and hers. If he makes a certain tweak to a beam that intersects on his side, it can instantly alter the probability that Alice's results come out a certain way.

However in all these experiments, Alice and Bob are stationary relative to one another. But what if Bob was on a spaceship, going in a circle at 0.5c, while Alice was in a stationary spaceship in the middle of that circle, and the laser came from a satellite some distance from them? Or something like that?

Einstein tells us that Alice and Bob can never agree on what counts as a "simultaneous" event in this circunstance because time is moving at a different speed for each observer. If two lightning strikes happen at different points, it's possible for Alice to see them as simultaneous while Bob sees them as sequential.

So if Bob tweaks his side of the split beam when he sees the sequential strikes one second apart, will Alice notice the non-local effect of his modulation as two separate events or just one? If two, will they be one second apart? It does not seem possible that they could be the same time apart from each other, because Bob's time is moving at a different rate than Alice's. But for non-locality to be true, they would have to match, would they not?

Since spacetime is one thing (time and space are all part of the same four-dimensional structure), how can an events be linked non-locally without also being linked non-temporally?

Update: I was basing my question on this paper: http://arxiv.org/pdf/1412.7790.pdf

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  • $\begingroup$ If Bob tweaks his side of the split beam when he sees the sequential strikes one second apart, Alice will not notice anything. Why should she? In fact (assuming you mean that Alice does her noting at a time that anyone views as simultaneous with Bob's tweak), how could she possibly? (Hint: She couldn't.) $\endgroup$ – WillO Nov 12 '15 at 21:25
  • $\begingroup$ If Alice were capable of observing such hypothetical modulation, then faster than light communication would be possible. $\endgroup$ – user83548 Nov 12 '15 at 22:32
  • $\begingroup$ Well, there is no proof that FTL communication is impossible. The no-communication theorem applies only to a single quantum state, not a series of them; i.e. the quantum state itself cannot be used to send information, but modulating the collapse of wavefunctions over time might :D Or not. I guess we'll see. I am skeptical but it's interesting. "Alice can steer Bob’s possible conditioned states by her choice of measurement setting θ" arxiv.org/pdf/1412.7790.pdf $\endgroup$ – CommaToast Nov 13 '15 at 3:59
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Alice makes the observation of her choice. Bob makes the observation of his choice. The pair of observations has an outcome with probability distribution determined by the initial joint state of the two particles.

The spacetime locations of the two observations, and the states of motion of the observers (relative to each other or anything else) have nothing to do with any of this. The probability distribution for a given pair of measurements is fully determined by the initial state and nothing else.

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  • $\begingroup$ I was basing my question on this paper: arxiv.org/pdf/1412.7790.pdf. It makes it sound like the probability can be affected by what Bob does, and is not deterministic, as you suggest. Or am I misreading this? $\endgroup$ – CommaToast Nov 13 '15 at 3:35
  • $\begingroup$ "Alice can steer Bob’s possible conditioned states by her choice of measurement setting θ" from that paper. $\endgroup$ – CommaToast Nov 13 '15 at 3:56
  • $\begingroup$ Without looking at the paper, I can assure you that either you are misreading it or it is wrong. Nothing Bob does can change the probabilities of the outcomes of any measurement Alice wants to make. $\endgroup$ – WillO Nov 13 '15 at 3:57
  • $\begingroup$ Yes, Alice can steer Bob's possible conditioned states. That doesn't mean she can steer the probability distribution of the outcomes of his measurements. $\endgroup$ – WillO Nov 13 '15 at 3:58
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    $\begingroup$ This isn't difficult: If the initial state is UU-DD, and if Alice makes a U/D measurement, then the state of Bob's electron is either U or D, whereas if she makes a U+D/U-D measurement, then the state of his electron is either U+D or U-D. But either way, if he makes, say, U/D measurement, he's got a 50% chance of U and a 50% chance of D. (And likewise for any other $\theta$ that Alice might pick.) $\endgroup$ – WillO Nov 13 '15 at 3:59

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