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What’s wrong with this EPR-Bell FTL Gedankenexperiment?

  1. Apr 22, 2013 #1

    DevilsAvocado

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    I truly believe Einstein was right; Faster-than-light (FTL) communication is not possible.

    And this far, nobody has proven Einstein wrong, not even “Italian Massive Mountains” could do it.

    There are hypothetical particles like the Tachyon that (in theory) always moves faster than light, but I don’t like it, and the obvious skeptic reason is unsolvable and chaotic Causality paradoxes that comes by default with FTL. Furthermore we have the No-communication theorem.

    In Bell test experiments no real usable information is propagated, only random “quantum information” and later (< c) established correlations between Alice & Bob, as a consequence of the shared wavefunction between the two entangled photons A & B.

    Therefore I know something must be wrong with my thought experiment.

    Could someone please tell me what it is?


    * EPR-Bell FTL Gedankenexperiment *

    Typical Bell Test Setup
    This is the typical setup for Bell test experiments, which can be performed in the undergraduate lab:

    sketch.jpg
    The yellow symbols are photon detectors

    For additional info, watch this short movie on the basic setup for testing Bell inequalities:

    https://www.youtube.com/watch?v=c8J0SNAOXBg
    http://www.youtube.com/watch?v=c8J0SNAOXBg&hd=1

    A complete description can be found in the paper Entangled photons, nonlocality and Bell inequalities in the undergraduate laboratory by D. Dehlinger and M. W. Mitchell.


    The Measurement Problem
    This Gedankenexperiment has the potential to goof right here, so watch out for errors. My assumption is that the shared wavefunction between the two entangled photons A & B does not decohere/collapse (or split the MWI-universe) when passing the Polarizing Beam Splitters (PBS).

    AFAIK most will agree. As an example; one prominent SA in this forum once quoted Professor Joseph H. Eberly, pointing out:

    The global quantum object characterized by one state vector (i.e. entanglement), is finally and irreversibly affected once the photon enters the detector and is actually measured. The PBS is not acting as a choice-device for “this or that way” – the single photon (wavefunction) takes both paths with its [itex]\left|H\right\rangle[/itex] and [itex]\left|V\right\rangle[/itex] superposition components.

    An argument for this line of reasoning is the good old Double-slit experiment. We know that the two slits do not destroy the (single) wavefunction, hence interfering with itself behind the slits:

    DoubleSlitExperiment_secondspace_2013-01-12.gif

    Other beamsplitter examples are the Mach–Zehnder interferometer and the Delayed choice quantum eraser.


    The EPR-Bell FTL Setup
    Please don’t laugh. This will not work, but I’m just too ignorant to find out why. :smile:

    6oh8ac.png

    There’s actually nothing strange about this setup. In fact it’s almost identical to the undergraduate lab setup, except for the extended (wind up & asymmetrical) optical fibers (red).

    Bob is an old fashion guy, he always calls the cards first (10 km to detectors D3 & D4 vs. 15 km to detectors D1 & D2) and thereby imposing Alice a restricted “quantum freedom”, and in some cases like perfect correlation – she has no choice at all – if Bob get 1V Alice will get 1V as well.

    Alice is deeply dissatisfied with current chauvinistic situation and has therefore forced Bob out of the lab – 10 km out in the wilderness – with only one detector D3 (i.e. a highly realistic quantum drama :smile:).

    Despite these “quantum-drama-modifications”, the setup ought to generate exactly the same result as a standard Bell test in the undergraduate lab.

    But the excitement doesn’t end here – Bob feels lonely & grumpy – and starts disconnecting detector D3 every now and then, to let photon B pass without any detection. Payback time! :grumpy:


    The Big Question
    What will Alice see in the lab? Will she notice anything? My claim is that she will definitely get the message – there’s a “Bob Revolution” going on out there in the wilderness.

    Why??

    Well, let’s do the math. In the setup we use a SPDC Type I down converter that along orthogonal axes will always result in the same outcome (i.e. perfect correlation).

    Let’s assume we measure 100 entangled photons with a 90° relative angle between Alice & Bob:

    sin^2(90°) = 100% = 100 correlated matches, i.e. [1V, 1V] or [1H, 1H]

    We derive the result from:

    • 50% of the time photon B is detected at D4 and Alice will then detect photon A at D2.
    • 50% of the time photon B is not detected at D4 and Alice will then detect photon A at D1 (and we assume Bob detected photon B at D3).
    If Bob decides to disconnect detector D3 before measurement, we get:

    • 50% of the time photon B is detected at D4 and Alice will then detect photon A at D2.
    • 50% of the time photon B is not detected at all.
    • When photon B is not detected, Alice measurement will be 100% random, i.e. 50/50 chance for photon A to be measured as [itex]\left|H\right\rangle[/itex] or [itex]\left|V\right\rangle[/itex].
    And we get:
    sin^2(90°) = 100% x 50 = 50 correlated matches
    + 50% x 50 = 25 = A total of 75 matches

    This is clearly enough to confirm there’s a “Bob Revolution” going on 10 km out in the wilderness.
    (In fact, as soon as Alice detects A at D2 and B is not detected at D4 we know something is going on.)


    Where’s FTL!?
    Well, once we established that Bob could use his detector D3 to signal a change to the lab 10 km away, it’s easy to build a system for Morse code, or better, a binary transmitter. In the movie above on the undergraduate lab we get 500 entangled measurements per second. If 100 photons is enough to confirm that Bob did a change, that mean we could send 5 bit/s.

    Ha, 5 bit/s! Grandpa’s bicycle is faster!!

    True, but the communication is instantaneous. Change km to ly and it will definitely be FTL.


    Wait! Something Is Missing!
    I know some of you are ROFL by now. There’s one “little” thing missing... do you know what?

    We need a tool to separate the rare entangled photons (1 pair in every 10^12) from the noise, and the only way to do this is timing and coincidence counting, meaning Bob has to transmit this data thru classical channels (< c) back to the lab. There goes FTL down the drain. Period. Case closed. End of story.

    But still, I can’t believe that our current inability to generate entanglement in a deterministic process is the only thing between us and FTL communication?? This can’t be true?? (How are we going to build quantum computers without 'controlled' entanglement?)

    AFAICT we now have the following “non FTL” options:

    • Entanglement is (by theory?) forever ‘doomed’ to be generated only from completely random sources like quantum fluctuations, and this will “save us” from FTL.
    • The Polarizing Beam Splitters are not that innocent, they do act as a “this or that way” measurement after all.
    • The Polarizer acts as a measurement.
    • The combination Polarizer + PBS act as a measurement.
    • I missed something else in the Gedankenexperiment ...
    Any clarification is much appreciated. Thanks.

    Regards
    DA


    UPDATE:
    If I’ve done the thinking right, we don’t need perfectly controlled/deterministic entanglement. All that’s needed is improved efficiency. With an efficiency of 1 entangled pair in every 10 it would be possible to see a statistical difference when “FLT” Bob is on/off.

    Is there anything in QM preventing us from higher efficiency?

    UPDATE2:
    I know some of you will dismiss the whole “FLT” setup by relativity, and argue that it’s impossible to say that Bob always does his measurement first because; to a moving observer Alice will be the one ‘breaking’ the shared wavefunction/entanglement. This is of course correct; however remember detectors D1, D2 and D4 are all at the same location in the “lab frame”, where D4 will always “flash” before D2, no matter how fast or in which direction an observer travels (or?). Since the wavefunction always takes both paths to D3 & D4 with its (decomposed) superposition components, it will probably be a quite advanced relativity assignment to explain how the same QM superposition could dissolve/exist in different frame of reference. Maybe it’s possible... I have absolutely no idea...
     
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  3. Apr 22, 2013 #2

    DrChinese

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    The Devil is back!

    :biggrin:
     
  4. Apr 22, 2013 #3

    DrChinese

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    You know that identical measurements on A and B are redundant. So Alice sees a random 50/50 pattern. She will see D1 go off 1/2 the time, and D2 1/2 the time. When she sees D1 go off, she knows Bob will see D3 go off. Of course, Bob is too lazy to bother to connect his D3 sometimes. But that makes no difference, the photon that approaches D3 is H>. Neither does the ordering.

    So the point is that context (which includes the beam splitter) is what governs. You cannot say specifically that the collapse occurs at the beam splitter or when detected. In a sense, it is both.
     
  5. Apr 22, 2013 #4

    DevilsAvocado

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    Hey DrC, long time no see! Good to be back @home! :smile:

    Many thanks for your response. Probably I’ve gone (more) sloppy in my time out of PF...

    Aha! This is what I suspected! But what happens if we recombined B’s H> & V> after the PBS... Is the “partial measurement” undone then??
     
  6. Apr 22, 2013 #5

    DevilsAvocado

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    You know I’m a green little Devil (with never ending questions). :smile:

    Are you sure about the ordering? If we rig the setup to be asymmetrical when it comes to the apparatus having an influence on the shared wavefunction – are we not allowed to say that one thing happen before the other (since photons A & B travels at c)??

    And in this case, what happens in the setup below. If the PBS is the one “making the decision”, photon A will be there first, but photon B will nevertheless be measured first?

    How is one supposed to handle this...?? :rolleyes:

    257ne6g.png
     
  7. Apr 22, 2013 #6

    DrChinese

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    This is a thorny question: is the wave function split/collapsed at the PBS or not? There are interpretations both ways. But I think the fact that the PBS outputs can be re-merged (restoring entanglement, presumably) tips the scales in favor of NO.

    Imagine there was no PBS on Bob's side, and Alice sees a click at D1. What is the polarization of Bob? :smile: It must be H>. That is the EPR conclusion. So the PBS on the right is superfluous. Of course, the same is true vice versa. So go figure.
     
  8. Apr 22, 2013 #7

    DrChinese

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    The usual view is: you must look at the entire context of a measurement to properly estimate the results. Ordering definitely does not matter, and there are no relativistic effects to consider here.
     
  9. Apr 23, 2013 #8

    DevilsAvocado

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    I love these words, makes life worth living! https://www.physicsforums.com/images/icons/icon14.gif [Broken] (:smile:)

    Agree, but your first answer seems to put some ‘restriction’ on this conclusion...

    Back to square one. :bugeye:

    This is the amazing thing about QM/EPR: You think you’ve learned something and then you realize you don’t understand anything.

    I have to reset the EPR-neurons in my left hemisphere, and see what comes out... I’ll be back.

    Many thanks for the stimulating responses until then!
     
    Last edited by a moderator: May 6, 2017
  10. Apr 23, 2013 #9

    DevilsAvocado

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    This was unexpectedly quick. I was drinking coffee it just struck me that there might be a way to tie up some loose ends a bit:

    i5p4wi.png

    In this setup old Bob owns the game. :smile: No matter what we are talking about, PBS or else, Bob will be done a long time before (5 km ≈ 16.5 µs) Alice’s photon A has chance to do any choice.

    Now, the question remains if the PBS qualifies as a measurement (in case of D3 disconnected), and I think there might be a chance to figure this out as well...

    We let Bob’s polarizer stay fixed and start rotating Alice’s polarizer.

    Now, when photon A enters the PBS it has to obey the predictions of QM + Bob’s result. If we get a sinusoidal curve with D3 disconnected, somehow Bob’s PBS (+ D4) is enough information for photon A to know what to do. Otherwise we get no sinusoidal.

    I don’t know... something tells me there will be no sinusoidal with D3 disconnected... how on earth is photon A going to know that D4 did not register and therefore it has to act as if D3 did...??

    The only information available to A should be the status of shared wavefunction.

    Or, I’m totally wrong... weird...
     
  11. Apr 23, 2013 #10

    DrChinese

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    Ha ha. You are going around in circles, DA. :smile:

    I guess in a sense everything is available to A when D4 is +, and equally available to A when D4 is -. And vice versa. Reversing the time ordering yields the same results.

    The future affects the past in QM! (0r at least participates.) It may help on the time ordering side to look at pages 4-5 of this reference.

    http://arxiv.org/abs/quant-ph/0201134

    "A seemingly paradoxical situation arises — as suggested by Peres [4] — when Alice’s Bell-state analysis is delayed long after Bob’s measurements. This seems paradoxical, because Alice’s measurement projects photons 0 and 3 into an entangled state after they have been measured. Nevertheless, quantum mechanics predicts the same correlations. Remarkably, Alice is even free to choose the kind of measurement she wants to perform on photons 1 and 2. Instead of a Bell-state measurement she could also measure the polarizations of these photons individually. Thus depending on Alice’s later measurement, Bob’s earlier results either indicate that photons 0 and 3 were entangled or photon s 0 and 1 and photons 2 and 3. This means that the physical interpretation of his results depends on Alice’s later decision."
     
  12. Apr 23, 2013 #11

    DevilsAvocado

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    Oh god... I felt it in my backbone that something was terribly wrong after posting... there’s no technical way a dumb flat detector could tilt/set a “superposition-spinning-photon-ball”, right? It’s the polarizer who do the first ‘tilting’ and then it is ‘finalized’ in the PBS, right? It’s just that we humans have no idea what’s going on before we measure that little evil thing, right? And also kill the entangled magic in the same blow, right? But ‘kill’ is not the same as ‘set’, right?

    Okay, it’s late over here, sleep> + coffee> + reading> = hopefully better understanding tomorrow

    See ya!
     
  13. Apr 26, 2013 #12

    DevilsAvocado

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    Thanks for the link DrC.

    Well, I’ve been thinking> + coffee> + reading> + thinking> etc... and maybe “The Old Man” was right – “It’s hard to see the wood for the trees in QM”... :smile:

    I’ve been reading about Asher Peres, Yakir Aharonov and time-symmetric QM (Two-state vector formalism). I can’t say it was the ‘clearest’ day in my life... :smile: (what’s completely incomprehensible is that we are supposed to reject FTL but accept Retrocausality... and to me they are the same thing??)

    If we peel off all “experimental overload” and focus on the core, we are left with one entangled pair and two polarizers. That’s what it’s all about.

    We can discuss collapse/decoherence/pilot waves and split universes for the rest of our life, without guaranteed progress (of course possible brain-collapses will be included in that kit ;).

    When we run ‘normal’ unpolarized light thru a polarizer, it’s a no-brainer to understand what happens:
    400px-Wire-grid-polarizer.svg.png

    50% of the unpolarized light gets thru, and it doesn’t matter what angle we set.

    Entangled photons are in an indefinite state, i.e. they are not in a definite state of polarization (or maybe it’s better to say they’re in a superposition of all possible polarizations?). An example of a Bell state (Type I):

    [itex]\frac{1}{\sqrt{2}} \left( | \uparrow \rangle_A |\uparrow \rangle_B + |\rightarrow \rangle_A |\rightarrow \rangle_B \right)[/itex]

    And we all know what it means; if Alice get V> Bob will get V>, and if Alice get H> Bob will get H>, and there’s a 50/50 chance for either.

    Now, one could think that it’s all about perfectly horizontal/vertical polarization, right? But that’s of course wrong, we can choose any angle and still get the “orthogonal linked correlation” between A & B. An example of a diagonal Bell state (Type I):

    [itex]\frac{1}{\sqrt{2}} \left( | \nwarrow \rangle_A | \nwarrow \rangle_B + | \nearrow \rangle_A | \nearrow \rangle_B \right)[/itex]

    So, now we have this “orthogonal linked correlation” between the entangled pair A & B (without any ‘pre-fixed’ angle) and two ordinary polarizers, and it’s Time for the Big Question:

    What ‘stuff’ alters this fragile indefinite state of polarization into an irreversible definite state??

    Don’t need to answer, right? There’s only one possibility left... (thank god!) :smile:

    This also put the “who is first” and “in what order” in completely new light. On larger distances we have relativity stopping us from claiming that A was before B thru the polarizer, on small distances we have the uncertainty principle hindering us from perfect resolution.

    = It’s impossible to say that Alice was before Bob and vice versa!

    I stop here for now, and wait for your (“massacre”) response... :wink:
     
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