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All you can do with your half of the pair is observe its state after the wave function collapsed. Whether it collapsed because you just poked it or because the other half of the pair was poked first can't be known without comparing notes afterward using classical means.

https://en.m.wikipedia.org/wiki/No-communication_theorem




An interferometer can differentiate that. If a photon is in superposition, it exits interferometer on a certain path, if it's collapsed, it exits on a random path. If you send 100 such photons, you can send 1 bit this way with good precision.


Although not informed enough to comment on this, I do know that if it was true, it would break relativity and that would have come across my feed at some point.

So I would say you are missing something.


I made it up. Any nonlocality breaks relativity, this doesn't break it further. You can still say the same thing - it's okay because why not. Or you mean it will be a paradox if they receive a message in the past? I guess it can be prevented if the entangled photon doesn't interfere with itself.


Information cannot move faster than the speed of light. If the non-locality has a random output, then it doesn't violate relativity because random noise cannot carry information (by definition it's no longer random if it does).


That's the trick - to replace random process with deterministic process which is still sensitive to collapse. Or rather make collapse switch from deterministic to random process.


Whether or not the photon is collapsed isn't the process that makes entangled particles "spooky" though. (Sorry I should have originally stated this)

Its that they have opposite states when you collapse them. So collapsing one doesn't collapse the other. Its just that when you collapse one (by measuring it) you can know the state of the other (even if it's on the other side of the universe).

The real "spookiness" is that the particles don't have a predetermined state "underneath" their superposition. When you collapse one its odds of being heads or tails are 50/50 up to the moment you collapse it. And when you collapse it you know what reading you will get when you collapse the other one.


Did you try to describe Everett's interpretation? There observation of one particle doesn't change the state of the other indeed. If everything is local, then no FTL is ever implied. In copenhagen collapse is nonlocal and happens instantaneously across the universe, so FTL actually happens.


>In copenhagen collapse is nonlocal and happens instantaneously across the universe, so FTL actually happens.

I don't know if this is a mathematical deduction or something empirically verified. Technically when you measure one particle the wave function of both collapses, because now you know what the properties of the other particle are.

But it could be like a pair of dice that always land on opposite faces. When the first die lands, the fate of the second die is sealed and it's no longer in a "superpostion" despite the second die still tumbling around in the air. I strongly suspect this is the case, because otherwise, like you said, you could transmit information faster than light, breaking relativity.


That's more or less right, as far as I know. Manually turning one of the dice doesn't manipulate the other one, and tossing your die only tells you what the other die did or will come up if it's ever thrown. It doesn't tell you if it was thrown.

Any superluminal signalling method proposed using entanglement has to explain which assumption of the no-communication theorem it's negating, otherwise it's breaking the known laws of quantum physics. Also, explain why nobody's noticed this effect in the lab before and sold it to high frequency traders and then won the Nobel prize.


wikipedia says: there can be also cases where is still possible to communicate through the quantum channel encoding more than the classical information

What I propose is to switch between deterministic and indeterministic process, the former encoding 0, the latter encoding 1. Maybe indeterministic process counts as more than the classical information. Also since it's indeterministic, it has a small chance of miscommunication. Since communication isn't precisely certain, maybe the theorem treats it as failure.


Wikipedia is referring to superdense coding, which is not superluminal and still needs you to physically send a particle from place to place to transfer the information.

"an entangled state (e.g. a Bell state) is prepared using a Bell circuit or gate by Charlie, a third person. Charlie then sends one of these qubits (in the Bell state) to Alice and the other to Bob. Once Alice obtains her qubit in the entangled state, she applies a certain quantum gate to her qubit depending on which two-bit message (00, 01, 10 or 11) she wants to send to Bob. Her entangled qubit is then sent to Bob who, after applying the appropriate quantum gate and making a measurement, can retrieve the classical two-bit message." (emphasis mine)

Alice physically transmits her qubit to Bob, so it's not superluminal and doesn't break the no-communication theorem.

https://en.m.wikipedia.org/wiki/Superdense_coding


Okay, I found my error: deterministic measurement I proposed is impossible, it works only for factorized state, but entangled state produces second spooky result with uniform distribution, so collapsed photon is indistinguishable from noncollapsed.


Mathematically speaking entanglement is local: in entangled state |11>|22>+|12>|21> the state |11> coexists with |22> and not with |21>, so when the particles fly away and meet again, |11> still meets with |22>, it's sealed at the time of creation of the state.




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