Quantum entanglement is 2 atoms that are paired together and when you stop one from spinning the other also stops with the same spin. Can you use these pairs to have FTL communication between 2 computers?

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    $\begingroup$ I'm voting to close this question as off-topic because this question is about physics and better asked on Physics. There is no computation involved. $\endgroup$ – jknappen Mar 12 '18 at 18:12
  • $\begingroup$ @jknappen isn't the computation the using it for quantum computers? I.E if it works cool and all but can/will it be used for QCs? $\endgroup$ – Christopher Mar 12 '18 at 18:13
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    $\begingroup$ @jknappen I think communication is a pretty important part of computation. Obviously, the distinction between what belongs here or on the Physics site will be an ongoing point of discussion here. $\endgroup$ – DanielSank Mar 12 '18 at 18:27
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    $\begingroup$ This question is pretty clearly relevant to quantum information technology. While not strictly speaking "computation", it is certainly part of what one might imagine a quantum computer would be used for. Having said that, the answer is pretty clearly "no, entanglement cannot be used for FTL communication" (and the OP should be edited to actually make use of the word 'entanglement' rather than 'quantum pairs'). $\endgroup$ – Niel de Beaudrap Mar 13 '18 at 0:42
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    $\begingroup$ I think there should be a clear distinction between "this question is on-topic", "this question is clear", and "this question is good". While I understand why people think this might belong to physics (because it sounds foundations-y), it is very obviously on-topic as an application of quantum information technology, which seems to be on-topic generally for this site. The question was also pretty clear to start with, albeit it benefitted from some editing. The only problem is that (with apologies to Christopher) it was not a good question, except to allow us to dispel a popular myth. $\endgroup$ – Niel de Beaudrap Mar 13 '18 at 14:32

There seems to be some debate about wether or not this question should be allowed but as it's one of the first examples in Nielsen & Chuang I'll go ahead and type out their explanation here for anyone else that is interested in entanglement for faster than light communication.

The following is an abridged version of Nielsen & Chuang section 1.3.7 entitled 'Example: quantum teleportation'

Quantum teleportation is a technique for moving quantum states around, even in the absence of a quantum commincations channel linking the sender of the quantum state to the recipient.

Here’s how quantum teleportation works. Alice and Bob met long ago but now live far apart. While together they generated an EPR pair, each taking one qubit of the EPR pair (also known as Bell states) when they separated. Many years later, Bob is in hiding, and Alice’s mission, should she choose to accept it, is to deliver a qubit $\lvert \psi \rangle$ to Bob. She does not know the state of the qubit, and moreover can only send classical information to Bob. Should Alice accept the mission?

Intuitively, things look pretty bad for Alice. She doesn’t know the state $\lvert \psi \rangle$ of the qubit she has to send to Bob, and the laws of quantum mechanics prevent her from determining the state when she only has a single copy of $\lvert \psi \rangle$ in her possession. What’s worse, even if she did know the state $\lvert \psi \rangle$, describing it precisely takes an infinite amount of classical information since $\lvert \psi \rangle$ takes values in a continuous space. So even if she did know $\lvert \psi \rangle$, it would take forever for Alice to describe the state to Bob. It’s not looking good for Alice. Fortunately for Alice, quantum teleportation is a way of utilizing the entangled EPR pair in order to send $\lvert \psi \rangle$ to Bob, with only a small overhead of classical communication.

In outline, the steps of the solution are as follows: Alice interacts the qubit $\lvert \psi \rangle$ with her half of the EPR pair, and then measures the two qubits in her possession, obtaining one of four possible classical results, 00, 01, 10, and 11. She sends this information to Bob. Depending on Alice’s classical message, Bob performs one of four operations on his half of the EPR pair. Amazingly, by doing this he can recover the original state $\lvert \psi \rangle$!

Skipping some of the details...

First, doesn’t teleportation allow one to transmit quantum states faster than light? This would be rather peculiar, because the theory of relativity implies that faster than light information transfer could be used to send information backwards in time. Fortunately, quantum teleportation does not enable faster than light communication, because to complete the teleportation Alice must transmit her measurement result to Bob over a classical communications channel. The classical channel is limited by the speed of light, so it follows that quantum teleportation cannot be accomplished faster than the speed of light, resolving the apparent paradox.

  • $\begingroup$ "Fortunately, quantum teleportation does not enable faster than light communication, because to complete the teleportation Alice must transmit her measurement result to Bob over a classical communications channel. ", so basically, we can't beat light speed because measuring quantum states is hard and requires them to share information from classical channels, correct? $\endgroup$ – Discrete lizard Mar 22 '18 at 21:41
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    $\begingroup$ Not quite. Bob can measure his qubit easily. But to discern the wave function that Alice wants to send from the potential states Bob will get from his measurement, he needs that extra classical information from Alice. $\endgroup$ – Andrew O Mar 22 '18 at 21:43
  • $\begingroup$ Ah, I see. The information from the measurement is too 'little' (i.e. insufficient) to have certain communication, is that a better interpretation? $\endgroup$ – Discrete lizard Mar 22 '18 at 21:47
  • $\begingroup$ More like not useful. He can't learn anything without knowing what state Alice's qubit is in. $\endgroup$ – Andrew O Mar 22 '18 at 21:54
  • $\begingroup$ Just realized that you got this answer from Looking Glass Universe, it would be beneficial to all of us if you post your citations on how you came up with your answer. $\endgroup$ – Dylan Dodds Mar 30 '18 at 15:43

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