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Question on Double Slit experiment

  1. Aug 5, 2007 #1
    I have a simple question on double slit experiment,
    In double slit experiment, after the light rays pass through the slits, place two beam splitters (down converters) which split the light into two rays, one going towards the screen and other going orthogonal to it. Now my question is do we get interference on the screen?
    (Remember that I do nothing to the orthogonally travelling light. I leave it undisturbed)

    The simple answer seems to be 'Yes' because we are in no way trying to obtain 'which-path' information. We just split the beam. But I believe the answer is 'No'.
    I would like to know the correct answer.
    If I am correct, placing reflecting mirrors do not affect interference. Similarly placing beam-splitters also should not affect it (But this contradicts at times). But placing polarizers do affect interference because it gives an 'identity' to the photon.
  2. jcsd
  3. Aug 6, 2007 #2
    Just to clarify, are you placing a different beamsplitter behind each of the two slits? If so, how are the beams recombined toward the screen? I'm just trying to get the layout right before attempting to answer your question.
  4. Aug 7, 2007 #3
    Sorry, my intention is to place down-converters. Not beam-splitters. down-converters will make 2 photons out of one. So one of them always travel towards the screen. The other one will travel orthogonal to it. Like this I have placed down-converters in front of both slits. So two rays of down-converted photons still hit the screen (forming interference or not I am not sure). Only the idler photons travel orthogonal to the original beam.
    We have something like,

    A<--- []......[]---> B

    Ignore the dots. They just represent empty space.
    Here [] is the down converter. A and B are idler photons. I am not disturbing them in anyway. They just travel into empty space. Will I get interference on the screen?
    Last edited: Aug 7, 2007
  5. Aug 7, 2007 #4
    Answer is no. It doesn't matter whether you actually measure the which path information from the idler beam, or whether you just "could have" measured it, or even (as far as raw aggregate data is concerned) whether you at some point manipulated the idler beam so that you "can't in principle" measure it (eg. quantum eraser), the signal photons still produce no pattern on the screen.
  6. Aug 8, 2007 #5
    I'm assuming that in your diagram the slits are at the same locations as the down converters. The only pertinent question is whether the two beams going to the screen are coherently related. Are the down-converted photons in a fixed frequency and phase with respect to the photons incident on the slits? If so, there will be an interference pattern at the screen. If not, there won't.
  7. Aug 11, 2007 #6
    I think that the inerference will occur because it is a wave phenomenon.If two waves are reaching,the interfernce must occur.Although the intensity may be disturbed.
    One more thing,are photons really divisible?I think that as they are fundamental so they are not.
  8. Aug 11, 2007 #7
    What do you mean?
  9. Aug 15, 2007 #8
    [red]down-converters will make 2 photons out of one[/red]
  10. Aug 16, 2007 #9
    Ok, but if you send, one photon at a time, light against a screen with two slits and put two detectors behind the slits, you can't detect the photon simultaneously in the two detectors. Why?
    (it's just a question!)
  11. Aug 16, 2007 #10
    But this has nothing to do withthe photon nature (or ANY other elementary particle for that matter) itself. This is a measurement property !

  12. Aug 16, 2007 #11


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    And the reason for this is what you explained in our other thread, that the down converters introduce a random phase in the photons so that the signal photons are no longer coherent. Right?
  13. Aug 16, 2007 #12
    What does it mean that the photon is not divisible?
  14. Aug 16, 2007 #13
    Because the crystals work by a process of *spontaneous* (ie. random in time, like nuclear decay) emission.
  15. Aug 17, 2007 #14
    I refered to the measurement property because your question involved the breakdown of superposition upon measurement. That is why you detect the photon at one slit only. This has nothing to do with "the photon being divisible". Actually, for the above reason, the divisibility is a concept that does not apply to a photon.

  16. Aug 17, 2007 #15
    Ok, I just would like to understand better what exactly mean "if you send one photon at a time, you can't detect the photon simultaneously at the two detectors" as written in some QM books.
  17. Aug 17, 2007 #16
    Err, i could have sworn i just answered to that question : breakdown of superposition of the particle's wavefunction upon measurement. This is a basic ingredient of the QM formalism.

  18. Aug 17, 2007 #17
    Thanks. However my question was actually intended in this way: if the two detectors click simultaneously, it's said it's because there arrived two photons. How do I know that it was sent only one photon or two photons, apart from what the detectors registered?
  19. Aug 17, 2007 #18
    Because you are performing TWO measurements here. Again, once you perform ONE (thus this also happens with TWO) measurement, the wavefunction collapse has occured and you either detect the photon through slit 1 or slit 2.

  20. Aug 17, 2007 #19
    So, how can we say that the photon is not the click of the detector?
  21. Aug 17, 2007 #20
    What does this have to do with your previous question ?
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