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It is common wisdom - and mathematically proven - that quantum entanglement cannot be used to bypass the relativistic speed limit and transfer information faster than light. So there must be something wrong with the following gedankenexperiment, but I can't figure out what:

Let there be a device which (nearly) simultaneously creates lots of entangled, coherent photon pairs and sends them to Alice and Bob, each of them receiving one photon of each pair. Both are far away from the light source and have placed screens into the - highly focused - light beam, so that they detect some small area of light on the screen.

Now Alice wants to transfer one bit of information to Bob. If it is a 0, she does nothing. If it is a 1, she puts an appropriate double-slit plate into the light path, immediately before the photon bunch arrives. This will create an interference pattern on her screen, and entanglement will instantly replicate the photons' paths and such the pattern onto Bob's screen. So if he sees an interference pattern instead of a dot, he knows that it's a 1, and vice versa.

Of course Bob's screen will not show an exact copy of what Alice sees: it will be olverlayed by noise, e.g. from photons which bounced off Alice's double-slit plate or decohered through environment interaction. Worst case a statistical accident may create a pattern out of a 0-bit which looks like the result of interference. But we may optimize the signal/noise ratio through the experimental setup, and besides of that any real-world communication channel is subject to partial information loss (which must be compensated by error correction).

So it seems like this experiment transfers information faster than light. What did I miss?

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  • $\begingroup$ This is indeed similar to the "EPR-type experiments and faster-than-light communication using interference effects as signaling mechanism" thread, which unfortunately had flaws both in the question and in the final self-answer. Therefore I give it a new shot. $\endgroup$ – following Feb 25 '14 at 15:15
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    $\begingroup$ Why would Bob see an interference pattern though? Just because the particles are entangled doesn't mean that the interference experiment gets replicated on Bob's side. He still has to decide between putting a double slit down or not. $\endgroup$ – Raskolnikov Feb 25 '14 at 15:31
  • $\begingroup$ @Raskolnikov I think that is an answer. $\endgroup$ – innisfree Feb 25 '14 at 15:43
  • $\begingroup$ I think your confusion stems from the fact you seem to think that "interference" is a quantum state. It isn't. Interference is what happens when we make a measurement that's incompatible with the state we measure. $\endgroup$ – Raskolnikov Feb 25 '14 at 15:50
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Your error seems to be the misconception that entanglement will magically make the results of any experiment of photon B exactly mimic those of a similar experiment done on its entangled partner. Entanglement is more subtle than that and must be treated carefully.

In particular, there are many different types of entanglement. For example, photons may be entangled in their polarization (i.e. they are both $x$-polarized or both $y$-polarized), or in their path (they're both going on the right 'lane' or both of the left 'lane' of a two-channel waveguide), or in more complicated degrees of freedom such as their spatial profile. Mostly, though, they need to be entangled in some degree of freedom which can have more than one possible state.

In Kim and Shih's example, the photons are entangled in their momentum: if Alice measures her photon to have momentum $p$, Bob will always measure the opposite. However, this entanglement only ever shows up in correlation experiments where they both measure momentum and observe correlated results. This fact is made clear from the outset by Kim and Shih's reference for the "ghost image".

If Alice puts a double slit in front of her photons and Bob doesn't, Alice will observe interference and Bob will not, since there is nothing in Bob's beamline to cause interference; it's that simple. In fact, in this setup their measurements will not turn up any correlations that would indicate the entanglement, since they are measuring conjugate observables.

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  • $\begingroup$ Thanks for the extensive explanation, which also invalidates an similar FTL communication idea based on momentum measurement that I found elsewhere. So I misunderstood the Kim and Shih example because I overlooked the need for correlation and did not fully understand the entanglement mechanism. $\endgroup$ – following Feb 25 '14 at 21:47
  • $\begingroup$ No worries. To be honest, I too found it hard going with that paper; I don't think it's particularly clearly written. $\endgroup$ – Emilio Pisanty Feb 25 '14 at 21:49
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Your proposal is very similar to that of John Cramer and Raymond Jensen (resting on Dopfer experiment: see 1998 Dopfer thesis under the direction of A. Zeilinger). The mainstream point of view is that this shouldn't work. Nevertheless, the proof of this conclusion is probably very technical, doesn't seem to be easy to find and cannot rest on an handwaving evocation of the no-signalling theorem.

Have a look for instance at Why does the Dopfer EPR experiment require coincidence counting?

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The flaw in your argument is that the claim "entanglement will instantly replicate the photons' paths and such the pattern onto Bob's screen [sic]" is incorrect.

The statistics for the measurement outcomes of any experiment performed on one subsystem of a maximally entangled pair is independent of what goes on with the other subsystem.

In your case, from Bob's perspective, if Alice puts a double-slit cannot give rise to anything he can observe on his photon that would not have occurred otherwise. Formally, this is because the density matrix describing Bob's system remains unchanged despite Alice's intervention on her photon. Entanglement may give rise to correlations between the outcomes of both subsystems, but not to any change in the individual outcome statistics that could be detected in order to transmit information.

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