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Coherence time of parametrically downconverted entangled photons

  1. May 30, 2014 #1
    What is the coherence time for a pair of entangled photons produced in a nonlinear crystal? Is it related to coherence time of the pump photon? Also, if we say that the coherence time of the two photons is T, then does it mean that the two photons can interfere with each other even if their relative delay is T?
  2. jcsd
  3. Jun 2, 2014 #2


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    You need to consider two coherence times here. The coherence time of the two photon state is pretty long and may be given by the pump. The coherence time of the single photons is usually pretty short.

    It is easier to see when discussing transversal coherence length instead. The argument is similar. SPDC gives you a broad range of emission angles (and has to in order to violate Bell inequalities). This is the opposite of the requirement of spatial coherence (a narrow angular size of the light source). However, the total momentum (and therefore also the sum of the emission angles) is very well defined. That means that if you perform a measurement in coincidence you can use a small detector and therefore postselect on a narrow range of angles. The corresponding photons in the other arm must also necessarily come from a narrow range of emission angles due to momentum conservation. This subset has large spatial coherence. So the total state has well defined momentum, while you do not get well defined momentum looking at the single arms.

    Coherence time is instead related to the power spectral density of your light field. In SPDC, the total energy of the incoming photon must be conserved, but there is no need for the energy to be distributed 50/50 among the outgoing photons. All the combinations are allowed as long as they fulfill the phase matching condition. So in each arm you will get a spectrally broad distribution and a short coherence time. However, if you perform spectral filtering in one arm and do coincidence counting now, you postselect on an ensemble with well defined energy, so the corresponding photons in the other arm will have a narrow spectral distribution and a long coherence time. So here, the two photon state has a well defined sum of energies so you have a long coherence time for the two-photon state, but the individual single photon states have very short coherence time when looked at in total.
  4. Jun 2, 2014 #3
    Thanks a lot Marc for your detailed reply.
    I understand your point that individually the two photons have low coherence both spatially and temporally but due to entanglement of two photons, if you post-select one photon in a narrow momentum or spectral state, then the other photon will have high coherence. Can't we post-select both the photons in this way e.g. by using the same very narrow spectral filters in both the outgoing photon arms?

    What I was puzzled about was the fact that if the uncertainty in wavelength for the SPDC process is equal to the 10 nm for outgoing photons of wavelength 1060 nm, then the corresponding coherence time is of the order of ps. However in some of the experiments I saw in literature e.g. in Hong-Ou-Mandel experiments, the effect of two photon interference appears to be visible even in ns range. It may be due to detector jitters but I am not sure.

    So if the two photons are generated with 10 nm bandwidth at 1060 nm wavelength, then the two photons are generated with maximum of 1 ps delay between the two (This comes from the uncertainty principle). So, if I now send these two photons to the two input ports of a beam splitter, then the two photons should interfere with each other only if the delay between them is less than 1 ps. We can use spectral filters to increase the coherence time and in that case the maximum delay for which the two photons can interfere with each other will be equal to this new coherence time. Are these statements correct?
  5. Jun 3, 2014 #4
    If what I say above is correct, then another experiment can be imagined. Two photons generated by SPDC with broad spectral power density are made to fall on two input ports of a beam splitter with a delay between them which is much greater than the temporal coherence of two photons. Therefore, the two photons do not interfere and the probability of the two photons going in two output arms and the two photons going in same output port is 1/2 in both cases. A narrow wavelength filter (in a thought experiment can be assumed to be a delta function in wavelength domain) is placed in one output port (port A) of the beam splitter. If the detector in this output port A now makes a click, then the coherence time of the other photon (in output port B) becomes infinite. In such a case, photon of port B should have interacted with the photon that has been detected at port A. I understand that the question of interference seems impossible since one photon has already been detected. But then again, that raises the question for me: when does the coherence time of the second photon change to infinity? Exactly at the instant when the photon at output port B is detected? Or when this photon passes through the spectral filter? Or did this photon did did interfere with the other photon of same wavelength component and has been detected afterwards?
    This last question can be answered by looking at the statistics of coincidence counts with and without the spectral filter.
  6. Jun 4, 2014 #5


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    In principle yes, but you will not gain much from that. If you already picked a subset via postselection, the spectral distribution in the second arm will be narrow anyway.

    It would be helpful to see the exact paper in question. HOM measurements are routinely done to check whether single photons are indistinguishable. That may include photons created by consecutive pulses and whatever.

    Seems right to me. It should be noted that doing so "breaks" entanglement. You will not be able to violate Bell's inequalities using the filtered subset as the distribution is now too narrow.

    This sounds like a different take on quantum eraser experiments. It is well accepted that the photons do not even have to overlap at the beam splitter. The crucial thing is that you have two different pathways with two different processes leading to the same final result. If these processes are in principle indistinguishable, you will get interference of the two-photon probability amplitudes. A nice example was given in "Can Two-Photon Interference be Considered the Interference of Two Photons?" by T.B. Pittman et al., Phys. Rev. Lett. 77, 1917 (1996). http://journals.aps.org/prl/abstract/10.1103/PhysRevLett.77.1917
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