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Is Faster-than-light Communication Possible?

  1. Jan 25, 2005 #1
    Dear group, recently I have written a paper on the possiblity of sending faster-than-light signals using entangled photons. I have shown it to several people and I have not yet received any feedback on it. I also sent a copy to arxiv.org but it was rejected and no reason was given. I am looking for some feedback on it before I try and get it published. If you have the time please have a look and give me some feedback. You may find it worthwhile. You can access the file here:

    http://www.nd.edu/~rjensen2/article...050113FTLC2.pdf

    I understand the topic is controversial, however my idea is simple: you have a source which emits entangled photons, one to each "end." On one end there's a receiver, and the other, a transmitter (of the FTL signals, not the photons). To send signals, the transmitter either performs a momentum, or position measurement on his/her photon. Say he/she chooses to do a momentum measurement, using a double slit, on N entangled photons. On the other end, the receiver has a double slit also. The receiver notices an interference pattern formed from the N photons he/she receives. Now, say the transmitter does a position measurement, on the next N photons. This time, the receiver sees no interference pattern, even though the photons go through a double slit. This is all elementary quantum mechanics here- nothing fancy. Now, think about how you can transmit signals from the transmitter to the receiver by causing the receiver to either see an interference pattern, or not. And remember, one photon of a correlated pair influences the other "instantaneously," so it is possible to send FTL signals in this manner, provided that the distance between sender and receiver is great.

    Again, feedback is welcome.

    Sincerely, Ray Jensen
     
  2. jcsd
  3. Jan 25, 2005 #2

    Hans de Vries

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  4. Jan 25, 2005 #3
    Hans, (and others) please try this link:

    http://tinyurl.com/4chaq
     
  5. Jan 25, 2005 #4

    Doc Al

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    I haven't read your paper, but I'm pretty confident that this won't work. Nothing you can do to one particle (A) of an entangled pair can be detected soley by measurements made on the other particle (B). Yes, measured values are correlated, but this correlation is only evident after comparing the measurements made of both particles. Make any measurement you wish (or none at all) on particle A: the experimenters measuring particle B will never know it until they all get together and compare lab notebooks. No FTL communication. (I don't think this topic is controversial.)
     
  6. Jan 25, 2005 #5
    Lets say both observers A and B have a double slit, and each receive one photon from each entangled pair emitted by the source. Now, if N photon pairs are emitted and received, and both A and B leave their slits open, then both A and B observe interference patterns. Now suppose A closes one of his slits, and another N photon pairs are emitted and received. Then both A and B see no interference pattern, even though B left both slits open.

    There is nothing fancy in this, it is just Heisenberg's uncertainty principle at work.

    Conclusion: A can control whether B sees an interference pattern, or not, depending on whether A covers one of her slits or not. Herein lies a basis for FTL communication. I hope that you can see it now.
     
  7. Jan 26, 2005 #6

    vanesch

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    Again and again and again. That's the fault of all these "collapse at a distance" and "quantum erasure" and other papers, with their fancy, misleading terms.

    Let's first fix a working hypothesis: we work within the framework of QM, ok ?

    If that's the case, then consider the following theorem:

    "If a system is made up of two subsystems, described respectively by 2 hilbert spaces H1 and H2 (so that the total hilbert space is H = H1 x H2 tensor product), and if the state of the overall system is described by a density matrix rho, then all expectation values of all operators A1 which only work on H1 can be found by using the local density matrix rho1, where rho1 is obtained from rho by tracing out the H2 states ; <A1> = Tr(A1.rho1)"

    So your local observations are completely independent of what is observed at the other side.

    cheers,
    patrick.
     
  8. Jan 26, 2005 #7
  9. Jan 26, 2005 #8
    Vanesch, what I have described has already been shown to be true in an experiment. See Zeilinger, Rev. Mod. Phys. vol. 71(2) 1999, in particular, read the part on an experiment done by his graduate student Dopfer on page S290 and take a look at the apparatus at the bottom of the page.
     
  10. Jan 26, 2005 #9

    vanesch

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    You mean figure 3 ?


    Ok, let us read carefully:

    So the interference pattern occurs when you SUBSELECT a sample of the photon 2 impacts, which are coincident with the photon-1 trigger (which, somehow, has to travel through the wire linking D1 with the COINCIDENCE COUNTER. So this is a correlation measurement. But if you would now trigger also the photon-2 events in anti-coincidence with photon 1 detection, you would find a COMPLEMENTARY interference pattern, in the complementary subset. And if you DO NOT have this coincidence information, but strictly look only at the photon-2 impacts, you would get the sum of the two subsamples of course, which hasn't got any interference pattern.

    cheers,
    Patrick.

    EDIT: :devil: :devil:

    In a way, these experiments allow the same kind of FTL communication as the following: Station S puts each time a red ball in one bag, and a black ball in another bag. One of these bags is randomly sent to Alice, and the other one is sent to Bob.
    Alice can now decide or not to throw the bag in the dustbin, or to open it. If she opens it, it turns out that each time she sees a red ball, Bob sees a black ball (that's the same kind of coincidence information as in the experiment above). If Alice doesn't look at the bags, then Bob sees, each time Alice puts a bag in the dustbin, a red ball in 50% of the cases, and a black ball in 50% of the cases. Question: can Alice send information to Bob by deciding, or not, to throw the bags in the dustbin ??
     
    Last edited: Jan 26, 2005
  11. Jan 26, 2005 #10

    Doc Al

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    link to Zeilinger paper

    FYI, that paper is available here: http://www.hep.yorku.ca/menary/courses/phys2040/misc/foundations.pdf
     
  12. Jan 26, 2005 #11

    DrChinese

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    Coincidentally, we had a discussion of almost the exact same setup last week here : https://www.physicsforums.com/showthread.php?t=59412. Vanesch gave some interesting analysis which I am still reviewing. My diagram of the setup is :

    EPR Paradox Using Double-slit Setup (and measuring position)

    I will review your paper more - I gave just a quick glance so far.

    Regardless of the theoretical issues, it seems that an experiment could be performed to test the idea regardless. Even a null result is useful.
     
  13. Jan 26, 2005 #12

    DrChinese

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    Oops!

    I see from Doc Al's reference above to the Zeilinger paper that this experiment HAS already been performed. No interference was detected, regardless of whether or not a measurement is performed on the "other" beam.

    Just as Vanesch had stated over in the other thread, entangled particles obey different state vectors than ones which are not entangled. I think he (or someone) said that the HUP does not apply the same way as normally expected in these situations. (Sorry if I have this way off, Vanesch!)
     
  14. Jan 26, 2005 #13
    At page 3 of the document it says (just opposite to equation 4.)

    "Therefore, a double-slit interference pattern for photon 2 is registered conditioned
    on registration of photon 1 in the focal plane of the lens."
    http://www.hep.yorku.ca/menary/courses/phys2040/misc/foundations.pdf

    The whole problem seems to be that not every "photon 1" is registered at the focal plane, but if the photon 1 was registered then the coincidence counter singles out thoose of photon 2 that can be used to build up a inteference pattern.

    vanesch seems to assume that only half of the "photon 1's" are registered at the focal plane. Do you know that thas it the case Vanesch? And if so is there a reason why we cant register more? (apart from allowing faster than light signals).
     
    Last edited: Jan 26, 2005
  15. Jan 26, 2005 #14

    vanesch

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    In figure 4, you see what is recorded, as a function of the position of the detector 1 (the "trigger" detector !). So at a fixed position, you do not record all of them (and it is at a fixed position that you get coincidences with detector 2 that generate an interference pattern at *detector 2*).

    I think that the explanation is quite subtle, but must go like the following.
    In PDC, there is a strict relationship between the angular relation of the incoming and outgoing momenta, and the energy, because you have a monochromatic pump beam with fixed momentum, and you have to satisfy the 2 equations:
    E_pump = E_1 + E_2
    p_pump = p_1 + p_2

    This means that for a given energy of the photons, you have a fixed outgoing angle (you get a "rainbow" out of it).

    This explains you why, with one beam out of a PDC xtal, you can never obtain any interference with a double slit if the light is emanent from a "point" in the Xtal: indeed, the difference in angular position of the two slits means that we have light of slightly different momenta, which do not interfere.

    But the pump beam, although "parallel (p_pump = constant) is "wide" ; meaning you get conversions at different lateral points in the PDC Xtal.
    So it is envisable that light is generated at two different positions of the xtal (with a distance of the generation points exactly equal to the distance between the two slits). That light transports parallel to either of the slits, and is strictly monochromatic (same angular direction, so same momenta and same energy). That light can give you an interference pattern. But it will be washed out with similar light originating from two OTHER spots within the xtal, because these OTHER spots will give rise to another, parallel, direction, hence another angle and another energy.
    But if we could now SELECT one strict angle of the outgoing light, by looking at the _other_ photon and selecting a strict angle for THAT one, then we fix for that other photon the momentum, and hence the energy and momentum of the first photon that will go to the double slit. Because of the strict angle of that "double slit" photon, it comes down to selecting 2 spots on the XTAL in such a way that they get through the slits. And that is exactly what is done in this experiment:
    The lens + detector D1 in the focal plane projects all parallel beams of the other photon into a point ; if we trigger on that detector, on that spot, we select out one strict angle (and hence, geometrically, only 2 spots on the xtal can send out the first photon through the slits, and these spots are distant by exactly the slit distance). We now have selected parallel and strictly monochromatic light on the first beam, and hence we can expect an interference pattern at D2. This is what is observed.
    If we now move D1 a bit, we will select out ANOTHER angular relationship, and thus another energy, and 2 other spots. They will also give rise to an interference pattern at D2, but slightly shifted. If we move D1 again, we will again shift the interference pattern at D2. If we don't look at D1, then we have all these patterns together at D2, which wash out and give you a blob.

    But we now also understand figure 4: if we have a photon BEHIND the slits on a fixed spot at D2, this means that the light was emanating from two different spots on the XTAL, and they will produce an interference pattern at D1 if this time we scan D1. The monochromaticity condition is this time fixed by D2, because in order to arrive on a certain spot behind a 2-slit experiment, you have to satisfy also a frequency relation (in fact, the experiment has now in fact been turned around, in that D2 selects the energy, and D1 sees the 2-spot interference pattern).

    cheers,
    Patrick.
     
  16. Jan 28, 2005 #15
  17. Jan 28, 2005 #16

    selfAdjoint

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    Just one suggestion. Introduce your |, -, /, \ symbols right away when you first describe the polarization states. Then they won't be something else to deal with when you move on to describe the experiment. Generally I think this is handsome and clear, and should be convincing to any person who hasn't already made up their mind. Expect to get some flak from the other sort on why Bob has to choose a camera.
     
  18. Jan 28, 2005 #17
    thanks selfAdjoint... got it...

    you are close to becoming my personal mentor...

    regards
    marlon
     
  19. Jan 28, 2005 #18

    DrChinese

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    I like it! Even though I know there is no FTL signalling possible, it is always fun to see if you can "beat the system" anyway.

    -DrC
     
  20. Jan 28, 2005 #19
    thanks DrC...

    glad to see the specialists are liking it...please let me know if you think something should be added. I am planning to include a paragrafe on delayed chose applied on this situation...i will think it through

    regards
    marlon, the humble pupil
     
  21. Jan 28, 2005 #20
    besides DrC

    how do you think it can be beaten??

    marlon
     
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