A science geek here with a question about QM.

I've been watching conferences and reading about Quantum Entanglement, and my doubt is whether information is or is not exchanged between two entangled particles.

What I think is let A and B be two particles with entangled spins. It is unknown what the spin of each particle is, but it is known that one is the "opposite" of the other. So when you measure A and see what is its spin you "instantly" know the spin of the other because particle B no longer exists in its superposition state and its spin is now defined by particle A.

Now, is really a message, information, you name it, sent or it is just something that happens from the point of view of the person doing the measurement?

My interpretation is that no information is sent and no FTL communication happens, as it is everything just the revealing of a previously-set property (in the moment of the entanglement) as seen by the viewer.


There are several misconceptions here:

  • everything is just the reveal of a previously set property (in the moment of the entanglement) as seen by the viewer

    What you're describing here is a local hidden variables model of the experiment, and these are known (through Bell's theorem) to be incompatible with quantum mechanics. Where local-hidden-variable theories conflict with QM, experiment has consistently sided with the quantum mechanical predictions.

  • That said, if you do have an entangled pair in an 'opposite-spins' state (technically, the singlet state $\left|\uparrow\downarrow\right>-\left|\downarrow\uparrow\right>$, but it's important to know that there are other entangled states that do not share the 'opposite-spins' property), you can try to transmit a message by measuring at A in the $\{\left|\uparrow\right>,\left|\downarrow\right>\}$ basis, and thereby "controlling" the measurements on B, which are constrained in this state to be completely anti-correlated with the measurements in A.

    That won't work, for a simple reason: you don't control what measurement outcome you'll get at A ─ you'll get an even mixture of ups and downs ─ and therefore you can't control what's seen at B. You have no way to dial in a message to begin with.

    You can try and get beyond that by controlling the type of measurement you apply at A. That also doesn't work, for much the same reasons.

  • More generally, entanglement cannot be used for communication, period; that is the content of the no-communication theorem. It can still be used to show correlations that go beyond what can be displayed by classical systems (due to the Bell-inequality violations mentioned above), but those only show up once you collate (at light-speed or slower) the results from measurements at the two locations.

  • $\begingroup$ I didn't realize I was thinking with "hidden variables", any good source (and not too technical) of information to understand more about Bell's theorem? $\endgroup$ – Gusdc Jun 26 '17 at 14:31

In addition to the explanation in Emilio Pisanty's answer:

Even if entanglement could be used to communicate, it is interesting to note that this communication will be limited as all others- by the speed of light.

Consider Alice and Bob taking two boxes containing two entangled spinors. Bob now travels to mars.

Now, Alice opens the box. She collapsed the wave function, but how can she use it to send information back to Bob? Will this information travell faster than light?

If Bob opens the box now, he finds the particle in a set state, but how can he know for sure that he wasn't the one to collapse the wave function? He can't open the box without risking "ruining" the experiment and collapsing the wavefunction himself- so Alice must give him a "clear" sign to let him know he can open the box- and that travels at the speed of light like every other form of information.

You can also try to understand it in the Everettian (many-worlds) frame- instead of the particles being in a superposition, the entire universe (described by a single wave function) is in a superposition, between a universe in which Bob's particle is UP and Alice's is DOWN to a universe in which Bob's particle is DOWN and Alice's is UP. When Alice opens her box, she entangles herself and finds (WLOG) UP, with the latter universe. Only when Alice can communicate with Bob to discuss the state of his spinor, will she be entangled to him. She has to be entangled to a Bob that has a DOWN spinor. The interesting thing is, it is allowed for Bob to open his box before being entangled to Alice and find UP- it just happens in a different "world", in which Alice will find DOWN.

So, to sum it up, information never travels faster than light.

  • $\begingroup$ I understand that information cannot travel FTL and this forbids any use of the entanglement fenomenon as a comunicational tool, if my question was interpreted that way maybe I expressed myself poorly. Having said that, what was my concern is how the wave function "knows" whether to collapse or not? @emilio-pisanty $\endgroup$ – Gusdc Jun 28 '17 at 15:41
  • $\begingroup$ @Gusdc This is very beautifully resolved in the many-worlds interpretation: the wave function collapse "progresses through space" at the speed of light. Bob and Alice can collapse the wave function and get an UP result, but this cannot happen at the same universe. When this two situations 'coexist' (before the "collapse progressions" of the measurements meet), we can't say that Alice and Bob are in the same "world" as they are space-like seperated. Until they can communicate, they can't discuss the experiment, but afterwards they are bound to be in a universe in which the spins are opposite. $\endgroup$ – A. Ok Jun 28 '17 at 17:34
  • $\begingroup$ @Gusdc it is important to stress that nothing in the collapse of the wave function really propogates in space- it is just that a measurement entangles an observer to a universe, and when Alice and Bob open their boxes they are both entangled to a world, but only when they are time-like separated they are entangled to each other- meaning the UP measured by one implies a DOWN measured by the other. $\endgroup$ – A. Ok Jun 28 '17 at 17:57
  • $\begingroup$ One final question, is it right to think in the entangled pair A and B as B = F(A) where F is the "relation between the mesurement", like one would define a value in function of another. $\endgroup$ – Gusdc Jun 29 '17 at 12:01

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