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In my answer to a recent question on World Building, I suggested that quantum-entangled particles would be a good way for ships traveling at relativistic speeds to communicate. My understanding is that quantum-entanglement allows for instantaneous action at a distance.

Is this accurate? Do quantum-entangled particles, when flipped, effect an immediate/instantaneous flip on the other side? Do relativistic speeds affect quantum-entangled particles in their speed of "communication?"

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marked as duplicate by ACuriousMind, Brandon Enright, BMS, Danu, Kyle Kanos Dec 12 '14 at 2:15

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    $\begingroup$ Quantum entanglement is the product of a local theory, and hence will not permit information to be transmitted faster than light. This is a duplicate of, e.g., physics.stackexchange.com/q/78118/50583 $\endgroup$ – ACuriousMind Dec 11 '14 at 23:15
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    $\begingroup$ Quantum entanglement is possibly the most extreme example of "correlation is not causation" that we know about. It produces pure correlation without any causal relationship, whatsoever. As such it plays very nicely into our sense for paradoxes, which is basically another way of saying that the human mind is very stubborn to accept facts that contradict "obvious" but fundamentally false lines of intuitive reasoning. $\endgroup$ – CuriousOne Dec 11 '14 at 23:35
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First of all I recommend you to read the answer of Hypnosifl to the question "Entangled photons never show interference in the total pattern without coincidence count" implies FTL.

Next, what flip you want to transmit? Flipping the spin of a particle is a local, unitary operation, not even a measurement. You pass your (spin endowed) through a spin-flipper, or, if it is a photon you can rotate its polarization if it is linear. It's local, completely controllable operation, with no implication on the other side. To the difference, if you introduce a certain decoherence between the states of one particle, that influences the entanglement as a whole, but flipping is not of this type.

To be clear, here is an example of introducing the decoherence that I am talking about: consider the very well-known singlet-state

$$\frac{\lvert x \rangle \lvert x \rangle + \lvert y \rangle \lvert y \rangle}{\sqrt{2}}$$

where the first vector in a product refers to a particle in the lab of one experimenter, and the second vector to a particle in the lab of the other experimenter. Let's name the two experimenters A, respectively B. If A measures the property of eigenstates |x> and |y>, and finds, say, x, that decoheres the whole wave-function. The topic of decoherence is widely covered in this forum - see related topics. But the experimenter B will get no information of what did A.

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