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I'm guessing the answer is no ;) but figured I'd ask anyway. Suppose I have one partner in an entangled pair and my friend on Alpha C has the other. I measure an observable (spin?) and the wave function collapses to a single eigenvalue. Does this cause the wave function for that observable to collapse for the remote partner, and if so, can the wave function collapse event itself be used as a superluminal signal? I assume the answer is no simply because he can't tell if the wave function has collapsed without making a measurement . . . which would cause the wave function to collapse. The whole idea falls apart if wave function collapse per se is not something that can be observed directly but is only inferred from measurement, I suppose.

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PS it also occurs to me I am not sure if my measurement collapses the wave function of the remote partner, or simply creates a condition such that when the remote partner is measured the necessary correlation is seen. –  JForster Sep 24 '13 at 22:26
Your basic problem here is that you can't observe a wavefunction collapse. You can just measure or not measure. When you measure you get a result, but you can't tell if it was required because the measured system was in a pure state or just came up by chance from among the allowed results. –  dmckee Sep 25 '13 at 1:28
@dmckee thanks very much; this is what I suspected. This clarifies the problem. –  JForster Sep 25 '13 at 2:15
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1 Answer

In the 2 slit experimentsuppose Bob is located at hole 2 with a mechanism that can instantly close the hole or open it. Sue is located on the screen plane a large distance, D , from the slits and at the position of the first interference fringe towards hole 1, where the probability of particle detection is zero when both holes are open. Bob closes hole 2, so the probability distribution for detecting a particle at Sue’s position changes from p1*p1 +p2*p2 +2p1*p2(=0) to p1*p1 > 0 . Does this change occur instantly? or does the effect propagate from Bob to Sue with the velocity of the particles? taking a time equal to the time of flight t = mD/p where m= particle mass, p= particle momentum?

If the change in probability distribution at Sue’s location is instantaneous ( as delayed choice experiments confirm that it is) can Bob use this to instantly modulate the number of particles per second received by Sue and hence send a signal faster than light? Why not?

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This is not an answer, just a replication of the OP's question, which moreover has been asked and answered many, many, many times on this site. –  Mark Mitchison Oct 17 '13 at 13:21
Not quite! No one has explained why Bob cannot send a message faster than light using the instant change of the wavefunction which occurs when he opens or closes hole 2. This action does not involve Bob observing a particle so the wave function collapse associated with a specific observation appears to be irrelevant here. Suppose the intensity of the particle beam through the apparatus is high enough to continuously produce an interference pattern (like that on the wall by a laser passing the two slits). What stops instant messaging in this case? Sue sees different results for Bob open/closed –  user31182 Oct 18 '13 at 8:49
Sue can tell if Bob's state is open or closed without reference to any measurement by Bob.The answer would involve considering the shot noise in the beam due to it being made up of particles and not a continuous wave. Sue would have to collect data for a minimum time T for each bit of information, to prevent errors due to misidentifying hole2 closed with open state. This will limit the rate at which information can be transferred from bob to sue. Has anyone reported this analysis and calculated if this effect is large enough to limit information transfer to slower than the speed of light? –  user31182 Oct 18 '13 at 9:16
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