Does that mean I have influenced the measurement result of one member of the entangled pair by acting on the other? Can information be sent this way using entanglement?
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$\begingroup$ No, the other particle will not be influenced. The disentanglement is the main problem in quantum cryptography. $\endgroup$– HolgerFiedlerCommented Apr 7, 2016 at 5:04
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1$\begingroup$ I'd advise to read any of the multitude of threads about faster-than-light communication via entanglement or spooky action at a distance. For example: physics.stackexchange.com/questions/3158/… $\endgroup$– MartinCommented Apr 7, 2016 at 14:46
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1 Answer
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The answer to your first question is interpretation-dependent. In many-worlds, the other state is unaffected. In the Copenhagen interpretation, the other state is affected.
Since interpretations all give the same predictions, and they disagree about the answer to your first question, the answer can't be determined experimentally or mathematically.
What is agreed on is:
- You can use pre-shared entangled states to perform some space-like-separated coordination tasks better than would be possible classically.
- Any non-local effects are undetectable by the other side until they can condition on your measurement result.
In summary:
Does the measurement of one member of an entangled pair act on the other?
Can information be sent this way using entanglement?
No. (Caveat: superdense coding.)
answered Apr 7, 2016 at 16:24