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The assumptions are:

  • Alice and Bob have perfectly synchronized clocks
  • Alice and Bob have successfully exchanged a pair of entangled photons

The idea is simply to have Alice and Bob perform the Quantum Eraser Experiment (doesn't need to be the delayed choice).

Alice and Bob agree on a specific time when Bob's photon will be between the "path marker" (which is usually just after the slits) and the detector.

If Alice acts collpasing the wave-function on her photon, the interference pattern will disappear. If not it won't.

Alice and Bob can be spatially separated...

What am I misunderstanding?

The only meaningful difference from this spatially separated quantum eraser experiment to one done on tabletop is that you won't be able to use a coincidence detector, but that is not impeditive to identifying the interference pattern, just will make errors more probable. Which we should be able to deal with a appropriate protocol...

There is a experimental paper with a small amount of citations pointing out to the breaking of complementarity in a very similar setup: http://www.pnas.org/content/early/2012/05/23/1201271109

Flávio Botelho
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    If Alice and Bob are spatially separated, you cannot force the premise that they have perfectly synchronized clocks. Well, you can but that would be unphysical. – Jim Jun 20 '14 at 17:42
  • Why shouldn't they be able to compensate the time difference? Is some kind of uncertainty involved? I can't remember anything (from my limited knowledge as I'm an engineer). – WalyKu Jun 20 '14 at 18:51
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    @Jim One usually imagine that we're setting up a transmitter between Alice and Bob which functions as a continuous source of entangled particles on which the participant will act in an attempt to achieve super-luminal communications (once the system is established, which implies a one-time set-up delay in excess of $d/c$). If we also assume that the parties are at relative rest, then it is easy enough for the source to also transmit a clock signal to both parties and achieve synchronization. – dmckee --- ex-moderator kitten Jun 20 '14 at 20:36
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    Since I am out of my depth here I won't post this as an answer. I notice that all the descriptions of this experiment I read have the two separated detectors talking to a coincidence detector. If that thing is needed for the experiment to work, then the issue is resolved because you can't get the coincidence news any faster than a lightspeed signal. No, I don't understand why it would be needed, but I figure some bright bulb would have done the experiment without if it were possible... – dmckee --- ex-moderator kitten Jun 20 '14 at 20:38
  • @dmckee I have looked upon a lot of papers with experiments of delayed choice quantum eraser and in most of the setups the interference pattern only shows up in the coincidence detector. So you seem close enough. The only paper with disagreeing results was the one mentioned. The question that remains to me: if we were using one photon of the entangled pair for a local double slit experiment would it display the interference pattern or not? If it would, I cannot see why my experiment wouldn't work. – Flávio Botelho Jun 20 '14 at 21:01
  • Probably, the reason that you are confused is that you don't understand the quantum eraser experiment. However, I don't think you have explained your setup in sufficient detail for anybody to figure out the error. – Peter Shor Sep 13 '14 at 23:02
  • @PeterShor I am very probably wrong. But just noticed there is an easy simplification. Alice entangles 10k photons and send them 1 of each pair to Bob. Bob is 1 light year away. Bob will simply do the (non-delayed) quantum eraser experiment with his wave-function part of the entangled photons. For example, put a 45o polariser before the slits, horizontal and vertical polariser as which path marker after the 2 slits. And 1 minute later pass the photons thru another 45o polariser to erase which path information. Alice chooses to measure her photon or not exactly during that 1 minute window. – Flávio Botelho Dec 01 '14 at 03:56
  • If Alice doesn't do anything, an interference pattern should appear for Bob (without any coincidence counter, consider a 100% bright source of entangled photons). But if Alice measures the polarisation of her wave-function part will Bob see an interference pattern or not? If he does see it then complementarity is broken, because after 1 year Alice will be able to pinpoint the slit that the photon passed although for Bob there was an interference pattern pointing towards wave-like behaviour. – Flávio Botelho Dec 01 '14 at 04:00
  • People have thought about this experiment. It doesn't work. Bob can only see the interference pattern if he conditions on the results of Alice's measurements. – Peter Shor Dec 01 '14 at 04:03
  • @PeterShor If the interference pattern is intrinsically linked to Alice's measurements, there is a modification that can be made that should make this work: unlink the measurements. Changing the thought experiment so Alice decides whether to collapse or not the entanglement 1 minute before the photons arrive for Bob (at his first polariser before the two slits). If she collapses it, then the first 45o polariser realigns whichever measurement Alice did and makes them irrelevant. An interference pattern should appear just as if normal photons (non-entangled) were sent by Alice. – Flávio Botelho Dec 01 '14 at 04:42
  • @Flávio Botehlo - "If Alice doesn't do anything, an interference pattern should appear for Bob (without any coincidence counter, consider a 100% bright source of entangled photons)." That's incorrect, entangled photons never show interference in the total pattern without coincidence count, totally regardless of what happens to their entangled twins. – Hypnosifl Dec 02 '14 at 02:59
  • (continued) for example, see p. 305 of Meeting the Universe Halfway, which cites a paper that experimentally confirmed the Wootters/Zurek result, where the authors say it doesn't actually matter whether the idlers are measured in a way that determines the which-path information: "Whether or not this auxiliary measurement...is actually made...appears to make no difference. It is sufficient that it could be made, and that the photon path would then be identifiable, in principle, for the interference pattern to be wiped out." – Hypnosifl Dec 02 '14 at 03:05
  • @Hypnosifl If that is true, then my new experiment is the idea of Alice to "untangle" the photons right before the 1st 45o polariser. Alice makes the measurement and now knows the Bob's photon orientation, but right after, the 1st 45o erases whichever orientation the original photon had (making Alice's measurement irrelevant) and we get a traditional non-delayed quantum eraser being performed by Bob on a ordinary (non-entangled) photon now. This HAS to show the interference pattern... – Flávio Botelho Dec 02 '14 at 03:06
  • @Flávio Botelho - Are you saying Bob is the one who measures the signal photons that have gone through the double slit, and Alice measures the idlers, or vice versa? – Hypnosifl Dec 02 '14 at 03:10
  • @Hypnosifl The original quote you took from me, I am assuming the interpretation you guys have given. So I am assuming that to be false in this second experiment. All that is needed for FTL in this new setup is "entangled photons never show interference in the total pattern without coincidence count". All I do in this second experiment (which looks a lot like the first) is let Alice decide whether to break the entanglement right before it reaches Bob's apparatus. – Flávio Botelho Dec 02 '14 at 03:17
  • @Hypnosifl Alice keeps the idlers. Bob gets the signal. – Flávio Botelho Dec 02 '14 at 03:20
  • @Flávio Botelho - You can't really "break" entanglement in the way you're imagining. If particle 1 and particle 2 are created in an entangled state, then the first measurement on particle 1 will display whatever statistics are characteristic of entanglement (like non-interference in this case), regardless of what type of measurement has already been performed on particle 2. – Hypnosifl Dec 02 '14 at 03:22
  • @Hypnosifl If Alice measures the idler right before it reaches the first 45o polariser before de double slit, then an interference pattern will appear for Bob. If she doesn't then it won't as "entangled photons never show interference in the total pattern without coincidence count". The key here is the first 45o polariser before the slits, it will 'erase' the correlation between Alice's measurement and Bob's photon. – Flávio Botelho Dec 02 '14 at 03:24
  • Breaking entanglement is just the measurement. Collapsing the wave function... The key again is that the only measurement at the time Bob's photon hits the 45o polariser is Alice's in the idler photon. That 'broke' the entanglement, but yet no pattern would appear because we know what Bob's photon direction. But after it passes the 45o polariser, Alice's measurement becomes irrelevant and we once again don't know what Bob's photon measurement will be. – Flávio Botelho Dec 02 '14 at 03:30
  • To make it easier to understand, Bob's part of the apparatus will be just like Figure 12 of this pdf: David Ellerman, A Very Common Fallacy in Quantum Mechanics: Superposition, Delayed Choice, Quantum Erasers, Retrocausality, and All That – Flávio Botelho Dec 02 '14 at 03:35
  • @Flávio Botelho - "Breaking entanglement is just the measurement. Collapsing the wave function..." But if you have an entangled pair A and B and you measure particle B, the wavefunction for particle A always "collapses" in such a way that its probability distribution is identical to what it would have been if you hadn't measured particle B first. In technical terms, I believe this means that if you look at the total probability distribution for A if B had been measured first but you didn't know what the result was, this is exactly the same as the distribution you would get from the – Hypnosifl Dec 02 '14 at 03:56
  • (continued) "reduced density matrix" for A if it was still a member of an entangled 2-particle state with B (the 'reduced density matrix' gives you the probability distribution for individual elements of entangled systems). – Hypnosifl Dec 02 '14 at 03:58
  • Just found a paper that has exactly the same idea: Is Faster‐Than‐Light Communication Possible? by Raymond W. Jensen – Flávio Botelho Feb 09 '18 at 03:08

2 Answers2

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This page has a good simple summary of the experiment, they present the following simple schematic of a delayed choice quantum eraser (the same as Figure 1 in the original paper by Kim et al.):

enter image description here

In this case, if the entangled "idler photon" is detected by Alice at D3 or D4 then she will know whether the "signal photon" went through slit A or slit B, but if the idler is detected at D1 or D2, then the which-path information for the signal photon is "erased". Alice would be free to replace the beam-splitters BSA and BSB with mirrors to make sure all the idler photons were directed at the middle beam splitter BS, ensuring they all ended up at D1 or D2 and had their which-path information erased. She would also be free to simply remove BSA and BSB so all the idlers would end up at D3 and D4, so none of them would have their which-path information erased. And for a large enough setup, she can in principle make either choice after Bob has already observed the pattern of the signal photons that have passed through the double slit and been detected at D0 (D0's position is supposed to be varied to see how many signal photons it detects at different positions, but feel free to replace D0 by the more traditional "screen" behind a double-slit if it makes things more clear, the point is that this is where you expect to see either an interference pattern or a non-interference pattern depending on whether you know which slit the signal photon went through).

The key as to why this doesn't allow for FTL or backwards-in-time signalling is that Bob never sees any interference pattern in the total pattern of signal photons going through the slits, he can only see an interference pattern if he looks at the subset of signal photons that he knows were entangled with photons that ended up at one of the two which-path-erasing detectors D1 or D2. If you graph what the original paper calls the "joint detection" of signal photons with idlers that all went to D1, you see an interference pattern, likewise if you graph the joint detection of signal photons with idlers that all went to D2, you see an interference pattern. But even if Alice makes sure all the idlers go to D1 or D2 by replacing BSA and BSB with mirrors, the total pattern of signal photons that Bob sees at D0 will be the sum of signal photons entangled with idlers that end up at D1 and signal photons entangled with idlers that end up at D2--and the sum of the two "joint detection" patterns is actually not an interference pattern, because the peaks of the D1 interference pattern line up with the valleys of the D2 interference pattern and vice versa. This is illustrated in another diagram from the page I linked to earlier:

enter image description here

You can also see it in the graph of the R01 and R02 interference patterns on the wikipedia page:

enter image description here

The original paper by Kim et al. notes this issue on p. 2, saying "It is clear we have observed the standard Young’s double-slit interference pattern. However, there is a $ \pi $ phase shift between the two interference fringes."

So, regardless of what Alice does, Bob never sees an interference pattern if he just looks at the total pattern of signal photons arriving at D0, only after he has communicated with Alice and learned which idlers went to which detectors can he look at particular subsets of signal photons whose idlers all went to a particular detector, and see an interference pattern if that detector is D1 or D2.

Hypnosifl
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If I understand correctly, Alice and Bob generate an entangled pair of photons and each take one. Alice does something with hers (you specified a quantum eraser experiment at a particular time, but I won't make that assumption) while Bob does a standard double-slit experiment with his.

A minor point here is that with one photon you'll never get interference fringes, just a dot, but they can work around that by repeating the experiment many times, with Alice trying to send the same information each time, until a pattern has built up on Bob's screen.

The outcome of this experiment depends on whether any measurement of Alice's photon could reveal which slit Bob's photon went through. (A more precise description of the experimental setup would settle this question.) If it can, then Bob will not see an interference pattern no matter what Alice does. This is because for the interference pattern to disappear it's only necessary that which-path information be recorded somewhere and in principle accessible to measurement; it doesn't matter whether the measurement actually happens.

If it can't, then Bob will see an interference pattern no matter what Alice does, because any measurement Alice could perform will not collapse the part of the wave function related to the two slits. Collapse isn't all or nothing; only the part of the wave function associated with the measured quantity collapses.

These two cases are actually two ends of a continuum; if measurements on Alice's photon can reveal partial information about the path, then Bob will see a pattern intermediate between the fully interfering and non-interfering patterns. But in no case does the pattern depend on what Alice actually does with her photon, only on what it could tell her in principle.

In the paper "Time-resolved double-slit experiment with entangled photons" (mentioned in your answer), although the text for Fig. 4 says "Interference pattern fringes move as the phase is changed remotely by the QWP", it appears that they are talking about fringes that only appear after postselecting with D1 or D2, as in the usual quantum-eraser setup. Note that the fringes in Fig. 4(B/D) are labeled "heralded by D1/D2", and Fig. 4(A) shows a total detection envelope with no fringes.

I only read the abstract of the paper mentioned in your question, but it only alleges a violation of the principle of complementarity. Unlike the uncertainty principle, the complementarity principle isn't a foundational principle of quantum mechanics and never had a standard mathematical formulation. If a particular attempt to formalize it turns out to be wrong, that's not a problem for quantum mechanics.

benrg
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