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Quantum imaging with undetected photons - adding of states

  1. Jan 7, 2015 #1
    I have a question concerning the paper "Quantum imaging with undetected photons".
    In the schematic (Fig. 1) a photon (idler) is created at NL1 and passing the object at O to be reflected further to NL2.
    It is then stated in the paper
    "By reflection at dichroic mirror D2, the
    idler from NL1 aligns perfectly with idler amplitude produced at NL2,
    [tex]|d\rangle_i \rightarrow |f\rangle_i[/tex]"
    However, if the incoming green laser beam at NL2 also creates an identical photon, why is the resulting state not a two-photon state? Why is it permissible to simply add the amplitude of the two idler photons
  2. jcsd
  3. Jan 7, 2015 #2


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    On page 2, they write "The probability that a down-conversion occurs at each crystal is equal and very low so the chance that more than one pair of photons is produced at a given time can be neglected."

    This doesn't mean only that cases where two or more pairs are produced in a single crystal can be neglected, but also cases where pairs are created in both crystals simultaneously. So you never have a two photon idler state but only single idler photons and you don't know whether a photon comes from one crystal or the other because both crystals produce idler photons in state |f>.

    That's at least the impression I got from skimming the relevant part of the paper.
    Last edited: Jan 7, 2015
  4. Jan 8, 2015 #3
    Thanks a lot for finding this - I was reading the nature version, where this sentence is missing (nasty nature's space restrictions). I think I get how it works now.
  5. Jan 9, 2015 #4
    O.k.,, after thinking again, I don't get it at all.
    If you never have simultaneous production of photons at NL1 and NL2, how can the states |c⟩ and |e⟩ interfere at BS2 (i.e., how can there ever be two photons at the dashed vertical line in Fig. 1)?
  6. Jan 10, 2015 #5


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    Let's look at the most simple case first. Remove all crystals and dichroic mirrors. Also let's assume that the green laser is a single photon source (which it isn't because regardless of intensity, lasers produce coherent states).

    So we only have a single photon, a beam splitter and a beam combiner. How would you describe the situation at the dashed vertical line now? What exactly goes wrong if you apply the same description to the more complicated case?

    (Also an answer to your last question is that signal and idler photons come in pairs. So if a down-conversion event is triggered in one of the crystal, there will always be two photons at the dashed line. But I don't think that this is the issue.)
  7. Jan 11, 2015 #6
    Thanks for the help. I think I got it now, here is my current understanding:

    "if a down-conversion event is triggered in one of the crystal, there will always be two photons at the dashed line"

    If down-conversion is triggered at, say NL1, I will have a photon at c and one idler f, but no photon e.
    If down-conversion is triggered at NL2, I'll have a photon e and f, but not c. This is what I also see in eq. (1) of the paper - the f-state occurs together with either c or e.

    So the system is actually in a superposition of states, one with a signal photon at c and one with a signal photon at e, because we do not know which of the two ways actually happened. (The same as in a standard double-slit.)

    And then I get interference between these two possibilities to get to eq. (2).

    Is this the trick?
  8. Jan 11, 2015 #7


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    Yes, this sounds good to me.
  9. Jan 11, 2015 #8
    Thanks a lot.
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