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Picture a situation where we have two observers, $A$ and $B$, and a system in a certain quantum state. If $B$ makes a measurement of some observable, say energy for example, the state will collapse to one of the possible energy states with a definite probability. If $A$ makes a measurement after this has happened he will observe a precise energy given by the state to which our system has collapsed after the the measurement made by $B$.

Pretty straightforward till now. Now, if both observers make a measurement at the same time they will both measure the same value. But we know from special relativity that simultaneity is relative to the observer, so we may think that from some observer this measurements won't be simultaneous and actually for him $A$ will have made the measurement before $B$.

My question is obvious then, how do measurements and relativity of simultaneity marry?

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marked as duplicate by Ben Crowell, Brandon Enright, Waffle's Crazy Peanut, Chris White, twistor59 Jun 7 '13 at 12:26

This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.

duplicate of – Ben Crowell Jun 3 '13 at 21:16

The relativity of simultaneity relies on spatial separation of the events. If two measurements are made at the same system, there will be no ambiguity of the order, because there is no spatial separation.

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The loophole you're missing is that while $B$ can decide when he measures and what observable he measures, he has no control over what measurement outcome he gets. Thus if they repeat this many times, $A$ will observe some probability distribution over energies, which is exactly what she would get if she had measured first, or even if $B$ hadn't measured at all. With this simple setup, there is no way that $A$ and $B$ can communicate at all, let alone faster than light, and it is faster-than-light communication that sits ill with special relativity, because only that can break causality.

There is still something weird going on, though. If $A$ and $B$ step up their game a little bit, then they can play Bell inequality or CHSH games, in which the correlations between their measurement outcomes are greater than they could possibly be (for a hidden-variable theory) unless they were communicating faster than light.

However, these games are always symmetric in $A$ and $B$. The really weird thing is that whatever entangled state $A$ and $B$ share, they cannot use it to communicate faster than light, because the local outcomes are always random. It's the correlations that are weirdly nonlocal.

As another example, take quantum teleportation. Here $A$ has some quantum state $\psi$ and shares some entangled state $\Psi$ with $B$. By entangling $\psi$ with her half of $\Psi$ and performing a measurement, she can collapse $B$'s half of $\Psi$ into a copy of $\psi$ - and she can do so instantly. Unfortunately, though, $B$'s teleported copy of $\psi$ is scrambled by some unitary operation which depends on $A$'s measurement outcome. To unscramble that unitary, $A$ needs to communicate with $B$ classically, which is at the speed of light or slower, and only then can $B$ get a trustworthy copy of $\psi$. Once again: instant collapse, but no way to use it to communicate.

That said, it is even weirder that this is nonrelativistic quantum theory we're talking about. By taking the usual Schrödinger equation, we're married to a specific observer, and there isn't a formal requirement that the resulting theory not allow instantaneous communication. To do this properly, you really should be using relativistic QM to do quantum information, which is an active field of research (see e.g. RQI-N).

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Here, you suppose that the sign of $t_A - t_B$ can change, but that would mean that A and B are separated by a space-like interval.

But that means that the "state" which comes from A to B, or from B to A, in a space-like interval way.

So it is not a physical acceptable state.

If, now, your state is a acceptable physical state, then A and B are separated by a time-like interval, then the sign of $t_A - t_B$ is always the same

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