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I guess the following obvious question is answered by any flavor of relativistic Quantum Mechanics, but I just wanted to check whether I understand correctly:

Is it correct that nonrelativistic QM violates locality (allows "statistical superluminal communication") in the following way:

Let Alice and Bob be far away from each other (and at relative rest). Assume we have a particle determined at $t=0$ to be in a "small" region around Alice (and therefore with quite undetermined momentum, but not so undetermined that it is possible to reach Bob in a "very small time"). Alice and Bob agreed that Alice would at $t=0$ measure the momentum with "extremely high precision" iff she wants to send Bob a signal. (This would make the position very undetermined, and thus make it possible for the particle to be at Bob's position). At $t=0$ (or a "very small time afterwards") Bob tries to find the particle at his position. In the unlikely event that he succeeds, he knows that Alice must have tried to send the signal. (If he does not find it, he doesn't know anything.)

A weak point of this example might be that it is probably (?) not possible to have wave function with compact support in position space ("close to Alice") as well as in momentum space (not able to reach Bob "instantaneously") if you look at the Fourier transform. However, if you look at the Schrödinger Equation, it seems to be that case that a free particle cannot "instantaneously" enter a region separated from the support of the wave function (position space) at a given time? I have to admit that this confuses me and I cannot come up with reasonable examples (the Gauss curve being the only normalized example for a free particle I have seen thus far, which obviously does not have a compact support). But I would be surprised if the non-locality effect above would depend on such technical issues?

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    $\begingroup$ Indeed a function and its Fourier transform cannot both have compact support. This is implied by the Paley-Wiener theorem (en.wikipedia.org/wiki/Paley%E2%80%93Wiener_theorem) $\endgroup$ – doetoe Dec 16 '14 at 22:11
  • $\begingroup$ Thanks for pointing that out, @doetoe . But as I said, I would be surprised if that was very relevant to the question; I assume it should be possible to at least set up some wavefunction that avoids a small (far away) region for a free particle at t in [0,epsilon] for a tiny epsilon; but such that an "exact" measurement of momentum will make some part of this region have nonzero probability... (But again, I have no example for that) $\endgroup$ – Jakob Dec 16 '14 at 22:23
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    $\begingroup$ It may be possible to adapt your argument, but not in this way: in fact the Fourier transform of a compactly supported function is analytic, which implies it cannot be identically 0 an any set of positive volume. Note however that just a slight (predictable) alteration of the probability due to a measurement by Alice would constitute a transfer of information (a very small amount). I am confident that that would fail, but I cannot tell you where, and would also be interested in an explanation. $\endgroup$ – doetoe Dec 16 '14 at 22:41
  • $\begingroup$ I suggest to leave Fourier transforms and compact supports. Not because I am against them, but because the problem is simpler - see my answer. Not in the Fourier transform s=is the answer, but in probabilities. The experiment goes so that Bob won't be able to distinguish whether Alice sent him a 1 or a zero. Please tell me if you agree with my answer. The non-signaling (no FTL communication) is such a basic issue that one doesn't need much math. for finding mistakes in proposals. $\endgroup$ – Sofia Dec 16 '14 at 22:53
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    $\begingroup$ It kind of bugs me that nobody ever bothers to define "locality" when they talk about this topic :\ $\endgroup$ – DanielSank Dec 17 '14 at 1:31
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The Schroedinger equation is non-relativistic and it propagates effects at an infinite velocity to begin with. It is thereof nonsensical to even talk about "locality". Schroedinger's equation doesn't describe local physics any more than a first order diffusion equation describes the speed of sound. There is no technical issue here, at all, you are simply using the wrong equation for the purpose.

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    $\begingroup$ Isn't that beside the point? The non-locality stems from the wavefunction collapse, not from its unitary time evolution, and is not governed by the Schrödinger equation. $\endgroup$ – doetoe Dec 17 '14 at 9:28
  • $\begingroup$ The problem is that any part of the potential impacts every part of the non-relativistic wave function immediately. So in that sense it is always non-local. That impact can be very small for parts of the potential that are far away from the center of the wave packet, but it's always there, except for packets with compact support... and I think those should spread out instantaneously. This is not the case in relativistic quantum fields, where the speed of light is the limit for causal effects. The "wave function collapse" is not even described by the Schroedinger equation. $\endgroup$ – CuriousOne Dec 17 '14 at 10:36
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    $\begingroup$ It would of course not be surprising that the non-relativistic QM is not compatible with locality. But I do not think that the Schrödinger Equation ist so relevant here (maybe I should not have stressed it in my question): I am interested in the intantaneous collapse of the wavefunction (which has nothing to to with Schrdinger Equation, but with the QM framework). Usually people claim that such a collapse cannot transmit information (and I am sure they are all correct in some clever way), but in the setting described this seems not the case (as described)? $\endgroup$ – Jakob Dec 17 '14 at 11:25
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    $\begingroup$ I repeat: There are two different things: 1) QM is non-local. 2) We simply CANNOT use the non-locality of QM to send FTL messages or backward in time (BIT) messages, because we cannot RE-WRITE the past. $\endgroup$ – Sofia Dec 17 '14 at 12:39
  • $\begingroup$ @CuriousOne Could please expand a bit on the points you've brought up? very hard to understand, although I'm sure the answer lies in there, I cannot decipher it yet :( $\endgroup$ – user929304 Dec 18 '14 at 16:28
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The no-communication theorem is a no-go theorem from quantum information theory which states that, during measurement of an entangled quantum state, it is not possible for one observer, by making a measurement of a subsystem of the total state, to communicate information to another observer. The theorem is important because, in quantum mechanics, quantum entanglement is an effect by which certain widely separated events can be correlated in ways that suggest the possibility of instantaneous communication.

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  • $\begingroup$ Does the no-communication theorem deal with multiple measurements of single-particle states, though? From descriptions I've read it sounds like it might just be about how measuring one part of an entangled system can't change the probability of measurement results for another part of the same entangled system. $\endgroup$ – Hypnosifl Dec 17 '14 at 1:30
  • $\begingroup$ Yes, I believe the theorem includes both cases, but I might be wrong. I am gonna check the demonstration and update. Thanks! $\endgroup$ – Wolphram jonny Dec 17 '14 at 1:41
  • $\begingroup$ @Hypnosifl Just checked, and the theorem becomes trivial for single particle states. $\endgroup$ – Wolphram jonny Dec 17 '14 at 1:48
  • $\begingroup$ Hmm, obviously there are some examples of pairs of measurements on single-particle states where the second measurement gives information about the first, for example in the double-slit experiment, if you measure a particle at the position of a minima of a double-slit interference pattern, that can tell you that someone earlier measured which slit the particle went through. So what kind of information transmission are they ruling out, if it's non-relativistic and has nothing to do with light cones? Is it similar to the def. of "non-local" I suggested in my comment to DanielSank above? $\endgroup$ – Hypnosifl Dec 17 '14 at 1:54
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    $\begingroup$ @Jakob I agreed with Hypnosifl that the theorems doesn't not seem to include the simple particle case. I trying to figure it out if the theorem becomes trivial for a single particle, or if at least it is simple to extend it to it. $\endgroup$ – Wolphram jonny Dec 17 '14 at 13:15

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