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I Hong-Ou-Mandel with two beam splitters

  1. Mar 16, 2017 #1
    In http://ws680.nist.gov/publication/get_pdf.cfm?pub_id=104112 , Pittman et al write:
    Is it possible to make a version where the two photons pass through two different beam splitters altogether, but are brought together finally at the detectors? For example by slightly tilting the mirrors upwards and downwards, we can send one photon to a beam splitter that is located just above the central plane, while the other photon passes through another splitter just below the first. After exiting the beam splitters we can have some extra optics that bring the beams back into alignment.

    As long as the phase relationships are maintained, HOM interference should still occur -- is this true?
  2. jcsd
  3. Mar 22, 2017 #2


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    This is interesting experiment, but note that in experiment it is SA interference that is observed not HOM interference. While SA and HOM interference experiments are very similar, demonstration that SA interference does not happen at beam splitter does not prove that HOM interference does not happen at beam splitter.
    Actually when I first saw this experiment some time ago I got impression that it falsifies Bohmian mechanics exactly because I initially did not noticed this difference between SA and HOM setups.

    That's the tricky part. You can't join two beams into one. You can mix two beams into other two beams using beam splitter, but then the HOM interference would happen in this later beam splitter.
  4. Mar 22, 2017 #3
    Thanks for the reply. I need to read more about SA interference -- at the moment I'm not very clear about how it relates to HOM. (There's a lot more written about HOM than SA, it seems).

    Have you seen this paper:
    "Hong-Ou-Mandel interference without beam splitters" --- https://arxiv.org/abs/1508.01639
    Is this experiment really equivalent to a HOM setup?
  5. Mar 22, 2017 #4
    I'm trying to understand this question of, how do we say that two modes are distinguishable or indistinguishable from the detector's point of view? Let's say two beams (modes) are hitting a detector surface, and the angle between them is [itex]\Delta \theta[/itex]. If [itex]\Delta \theta[/itex] = 0, we have 100% visibility of interference. At [itex]\Delta \theta = 0.1\deg[/itex] we can expect, say, 99.9% interference visibility. At a [itex]\Delta \theta[/itex] of 10 degrees, maybe much less.

    But how do we calculate the function relating visibility to [itex]\Delta \theta[/itex]? What factors enter into it? For example, in a photoemulsion film, does each grain of emulsion "know" which beam the photon came from, even if the directions differ by a large [itex]\Delta \theta[/itex] like 20 degrees?
  6. Mar 23, 2017 #5


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    It seems more like HBT setup. I don't understand why it should be compared to HOM setup.

    I would say that angle between beams affect variation of phase across the surface of detector. So if collection pinhole of detector is very small you can get good visibility with larger angles. And that's it as I see.
  7. Mar 23, 2017 #6
    Thanks again. The phase variation thing is clear and convincing. But maybe, sometimes, authors think that "welcher weg" is more impressive. :)
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