Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

FTL communication

  1. Mar 14, 2008 #1
    Hi,

    I stumbled upon this page about a double slit quantum eraser experiment. Reading it I got the idea that it could be used for faster-than-light communication. The problem is that I couldn't figure out in which way this method would fail (other than being a rather tough nut to crack engineering wise). I would appreciate it if someone could point out where it crashes with theory.

    Here's the idea:

    Alice generates a several batches of entangled photons and gives half, one from each pair, to Bob via conventional means. Bob then travels to a nearby star, where he wants to communicate back his findings to Alice as soon as possible.

    So, at a predetermined time after the Bob arrives, Bob begins to send photons from the first batch through the p path of the detector (as described in the eraser experiment). To encode a zero, he would insert the polarizer, and to encode a 1 he would remove it.

    Simultaneously, Alice sends her first batch trough the s path of the detector. If she notices an interference pattern, she writes down a 1, else a 0.

    The problems I see are:
    a) Keeping the photons entangled and preventing loss, so that the batches are "in sync".
    b) Some way of counting the number of photons sent into the detector device, so that Alice knows when to move Ds. Or could she just use a regular double slit setup (photographic plate or similar)?

    Are there any theoretical issues with the above? Or are they just (very hard) engineering challenges?

    Although the photons have to get to Bob via conventional means, if it could work, wouldn't this for all intents and purposes be a FTL communication device?
     
  2. jcsd
  3. Mar 14, 2008 #2

    Doc Al

    User Avatar

    Staff: Mentor

    I haven't studied the Double-Slit Quantum Eraser Experiment very closely, but it seems to me that all measurements are of coincidence counts between the two separated detectors. Alice can only notice the presence or absence of an interference pattern after she correlates her data with Bob's via the coincidence counter. Now if Alice could detect the presence or absence of an interference pattern without any interaction with Bob or his data--that would be something!
     
  4. Mar 14, 2008 #3
    Hmm yes, I tried to find some more info on this coincidence counter... From what I read it appears that the only thing it does is allow one to easily measure the ratio of sent photons vs detected photons (at Ds). However, if you knew you sent N photons to the s detector, and you measured n, couldn't you use that information to build up the detection pattern in the same way? Or is that where it all breaks apart?
     
  5. Mar 14, 2008 #4

    Doc Al

    User Avatar

    Staff: Mentor

    That doesn't make sense to me. The reason for coincidence counting is to correlate measurements from each detector. (To make sure that measurements are being done on the same entangled pair, for instance.)

    I don't see how Alice can detect anything done by Bob (or even if Bob is still alive) by looking at her data alone.
     
  6. Mar 14, 2008 #5

    DrChinese

    User Avatar
    Science Advisor
    Gold Member

    Entangled photons do not exhibit the kind of interference pattern you describe. This is a specific physical and detectable difference between entangled photons and "normal" photons. Please note that this is NOT the same setup as what is sometimes looked at in the Quantum Erasers.

    You can see this in an enlightening article by Anton Zeilinger, p. 290, Figure 2.

    Experiment and the foundations of quantum physics

    "FIG. 2. A source emits pairs of particles with total zero momentum.
    Particle 1 is either emitted into beams a or a' and
    particle 2 into beams b or b' with perfect correlations between
    a and b and a' and b', respectively. The beams of particle 1
    then pass a double-slit assembly. Because of the perfect correlation
    between the two particles, particle 2 can serve to find
    out which slit particle 1 passed and therefore no interference
    pattern arises."

    Besides, I do not believe that a momentum measurement and a polarization measurement are non-commuting anyway. So the polarizer would not be a factor. I think... :)
     
    Last edited: Mar 14, 2008
  7. Mar 14, 2008 #6
    If you only send one pair of photons into the apparatus, you don't need the coincidence counter do you? Wouldn't the s photon use a certain amount of time traveling to the detector, so that you can use the time to register a hit / no hit?
     
    Last edited: Mar 14, 2008
  8. Mar 14, 2008 #7
    Thanks for the pdf, looks very good, will read it carefully :)
     
  9. Mar 14, 2008 #8

    Doc Al

    User Avatar

    Staff: Mentor

    I think you are missing my point about coincidence counting, which correlates measurements made by both detectors. For every entangled pair, both Bob and Alice make measurements. Just by looking at her own data, Alice will not be able to detect anything done by Bob. (No communication whatsoever.)

    But if Alice selects only her data that matches a certain result that Bob has measured, then she may well observe a different pattern than if she did not select data based on Bob's results. But that can only be done after the fact, once Bob and Alice get together and compare data--or at least transmit results to each other in the usual non-FTL manner. (No FTL communication.)
     
  10. Mar 14, 2008 #9

    JesseM

    User Avatar
    Science Advisor

    Lord Crc, you might want to take a look at this thread (I explain why the DCQE can't be used for FTL in post #2). Basically, the total pattern of signal photons never shows interference, only subsets of signal photons whose idlers went to a particular detector do. On the subject of entangled photons not creating interference patterns in their total pattern, you might also find this thread and this one interesting.
     
  11. Mar 14, 2008 #10
    Thanks for the pointers, I'll have to digest it a bit though, as right now I can't quite manage to see how the results you mention maps to the setup described in the article I linked.

    And I still can't see what makes the coincidence counter so special :)

    Perhaps a good nights sleep will help sort things out. In any case, cheers!
     
  12. Mar 14, 2008 #11

    JesseM

    User Avatar
    Science Advisor

    Well, note that the position of Ds is varied over many trials to produce an interference pattern, suggesting the range of positions it can register a photon on any given trial is fairly narrow, while Dp's position is held constant. If Dp's range is narrow as well, then that means that in a coincidence count you're only looking at a subset of photons at Ds, namely the ones whose entangled doubles went to just the right position to hit Dp; I would imagine that if you graphed all hits at Ds, regardless of whether Dp registered a hit, you'd see a non-interference pattern. Likewise, if you replaced Dp with a very wide detector which would intercept virtually all incoming photons, I would think that in this case the coincidence count with Ds wouldn't show interference, but then if you picked only the subset of photons at Ds whose twins ended up in some narrow range of positions at the wide detector in the position of Dp, you'd recover an interference pattern.
     
  13. Mar 15, 2008 #12
    Ah, good point. I thought perhaps it had to do with the polarization filter which would let only a subset of photons through, and only this subset produced the interference (similar to looking at only the photons that hit say D2 in the description you gave).
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook

Have something to add?