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Quantum eraser experiment

  1. May 15, 2009 #1
    I have read a description of a Quantum eraser experiment: http://www.bottomlayer.com/bottom/kim-scully/kim-scully-web.htm

    I don't understand the technicalities, like the process how the entangled photon pair is generated. So things like that may solve my problem, or may not. What I'm not getting is this: why do we need to include both a part with "which path" information and one without, simultaneously in the same experiment. This makes the final correlated information seem like a sampling of the distribution at D0. Whereas, making the idler side always erase path information (or always keep it) would show a 100% picture.

    It seems reasonable to assume, that this simpler experiment has been tested before inventing the one in the link, with that labyrinth of mirrors and splitters and coincider circuit. Based on what I understand, I would expect the following to happen (which Im sure is wrong somehow):

    If we just remove the idler side of the experiment completely, we should not see any interference pattern, due to the path information being observable (?). Even if Im wrong about this, we can then put there detectors that tell us the path information, no problem here.

    However, if we put the eraser in the idler side, it removes the path, and the interference pattern should appear at D0. It worked when applied to only D3 and D4, so why not now?
    I understand that in the original experiment there was no interference at D0, somehow I feel that should be important for this?

    If this actually worked, it would mean that my choice of setting up the device determines if there will be path information. However, that in turn determines the distribution pattern at D0. Just for visibility, lets say that the idler part of the experiment and D0 detector are in different rooms (they could be galaxies apart), then me changing the setup in one room would change the interference pattern in the other room.

    Surely this cant be true. Not because it would be counterintuitive, Im ready to accept that. But this would be a relatively simple experiment (compared to the one in the link), and it would completely disprove local hidden variables, much more obviously than Bell's inequality, making that obsolete. So the pure existence of Bell's theorem indicates that it wont work like I described. But then how?
  2. jcsd
  3. May 15, 2009 #2


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    No, the important point is that the quantum eraser is about two-photon interference. You NEVER see an interference pattern at D0 alone. The interference pattern is only present in the correlation counting data. The reason for that is, that the phase of the two photon state is well defined. This means, that while the phase of the photon states after the down conversion is random, the relative phase between them is not.
  4. May 18, 2009 #3
    Thanks. It took me quite some time to understand, but I think I got it now.

    So if my understanding of what you said is correct, the distribution is always the same at D0 regardless of what happens at the idler side. But this aggregated data can be grouped into subsamples, depending on the setup at the idler part. By correlating the entangled pairs, we basically cut up this distribution. So depending on the setup, we can get 2 patterns with interference and 2 without, (as in the original experiment), with some changes we can get 4 with interference, or 4 normal distribution patterns. But if we add these together, their summed distribution is always the same.

    Please correct me I misunderstood something... but it actually sounds right to me now.
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