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Triple Slit Experiment?

  1. Sep 26, 2010 #1
    Suppose that instead of two slits for the classical experiment, we use three slits.

    First what pattern would appear? I tried drawing it out, but without a protractor to draw proper curves, it's hugely inaccurate and I'm too lazy right now.

    Anyway, to the actual meat of the question, I was wondering. If you were to put a detector at only one of the slits, what would happen to the resultant pattern on the screen detector?

    With two slits, detecting one meant detecting the other (because we're only given two choices), and so for whatever reason, the wave function of the particles collapse (correct me if I'm wrong).

    With a detector at only one of the three slits however, there would still be an uncertainty as to which of the two remaining holes the particle could go through. So would the final product be a "wave collapse" from 3-wave interference to a 2-wave interference superimposed on a single slit pattern?

    Or would the whole thing just collapse to a particle "pattern" just like the collapse of the 2 slit experiment?
     
    Last edited: Sep 26, 2010
  2. jcsd
  3. Sep 26, 2010 #2
  4. Sep 29, 2010 #3
    Thanks for the reply. I wonder if they've ever try to run the "measuring" experiment with diffraction grating? Would it affect the resultant spectrum? It's a lot of new information to take in though, so I'll have to take some time to digest it all.

    No other reply's here huh?... I would have thought the triple slit experiment would have been of some interest.

    btw Dr Lots, if you don't mind me asking, what is your interpretation of the phenomena?
     
  5. Sep 29, 2010 #4

    Cthugha

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    The triple slit gets some sort of interest. Earlier this year there even was a paper in Science on this topic:

    U. Sinha et al., Ruling Out Multi-Order Interference in Quantum Mechanics, Science 329, 418 (2010).

    It is also available on Arxiv. Basically I am a bit surprised that this paper made it to Science because the result is not surprising at all and the upper bound they get for possible multi-order interference is not too impressive, but nevertheless the experiment was done very well as far as one can judge from the publication.
     
  6. Sep 29, 2010 #5
    Would you possibly mind summarizing their experiment and results? That is, if it's not too long/complex/troublesome, in which case I'll read the paper myself. (I'm going to do it myself eventually, just swamped by homework right now).
     
  7. Sep 29, 2010 #6

    jtbell

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  8. Sep 29, 2010 #7
    Yes yes... quite right :redface:.

    Though my question still stands.
     
  9. Sep 29, 2010 #8

    Cthugha

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    Ok, basically the interesting result you get from the two-slit interference experiment is the difference between the two single slit diffraction patterns and the double slit diffraction pattern. So basically the probability to detect a photon at some position r behind the two slits A and B is:
    [tex]P(r)=\left|\psi_A(r)+\psi_B(r)\right|^2 =\left|\psi_A(r)\right|^2 + \left|\psi_B(r)\right|^2 +\psi_A^\ast(r) \psi_B(r) + \psi_B^\ast(r) \psi_A(r)[/tex]

    Or in easier terms: you get the two single-slit diffraction patterns and have to add or subtract the interference term at the corresponding position:
    [tex]P(r)=P_A(r)+P_B(r)+I_{AB}[/tex].

    Applying this to a three slit scenario with slits A, B and C is straightforward and gives aprobability that looks as follows:
    [tex]P(r)=P_A(r)+P_B(r)+P_C(r) +I_{AB}+I_{AC}+I_{BC}[/tex].

    Now all they did in this paper is check, whether also terms of the form [tex]I_{ABC}[/tex] exist. So they checked whether the distribution seen for the three-slit experiment is the same as the sum of the three single-slit diffraction patterns and the three two-slit interference terms for several light sources. It was always the same and it seems like there are no such third-order interferences and if there is one it must be abt least two orders of magnitude smaller than the common second-order interference. They advertise it as a test of Born's rule and therefore also as a test of the absolute square of the wave function being the probability density. These results are of course somewhat important, but as the result is not surprising at all, I would have liked to see a lower upper bound on the possible third order interference especially as the authors claim that already very small deviations from Born's rule could have drastic consequences.
     
  10. Sep 29, 2010 #9
    Wow thank you for the detailed response. Though the complexity is going way over my head currently, I'll take some time over the weekend to try to understand what it all means. Anyway, I just wanted to give my thanks in the mean time.
     
  11. Sep 29, 2010 #10
    Call the three slits A,B and C and let them be evenly spaced with B in the middle.A and B will form a two slit interference pattern as will B and C and the two patterns will overlap so closely that in effect they appear to form a single two slit pattern with a greater intensity variation than that formed by one set of two slits only.That leaves A and C.The separation of these two slits is twice that of A and B and B and C and the result is that these two slits form an interference pattern whose intensity variation is less than that of the two slit pattern and whose fringe separation is half that of the two slit pattern.The overall pattern is one of bright principle maxima due to the two close slit combinations separated by less bright subsidiary maxima due to the single more widely separated slit combination.
    If more slits are added the principle maxima get brighter and sharper and the subsidiary maxima get less bright and less pronounced.
     
    Last edited: Sep 30, 2010
  12. Sep 29, 2010 #11
    I'm mostly not fond of the term "collapse". I try not to get too distracted by impressive terminology, and keep distinguishing math models (such as QM) from actual experimental events. So currently, I can't help thinking that if a photon "collapses", then so does a tossed coin when it lands. It doesn't sound right. I still have much to learn though, and I won't debate with any QM professor who may be reading.
     
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