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Single photon, timing jitter, reduced indistinguishability?

  1. Jan 1, 2015 #1
    I saw in a recent review paper- Engineered quantum dot single-photon sources (Rep. Prog. Phys. 75 (2012))- a discussion about how a "timing jitter" problem lead to reduced indistinguishability of single photons, which I find very hard to understand.

    According to the paper, for quantum-dot-based single photon sources that use an incoherent pumping (basically, pump the quantum dot to a high excited state, and then it will relax to the first excited state quickly, and decay from this state produces a single photon), a non-infinite relaxation rate from the higher order excited states to the first excited state (from which the single-photon pulse is emitted) leads to a jitter in the arrival time of the single-photon wavepacket, and this will lead to reduction in indistinguishability.

    The relevant formula is Eq.(12) of the paper.
    According to this formula, a non-infinite relaxation rate will lead to an reduction in the indistinguishability.
    Also according to this formula, if relaxation rate is infinite, and if dephasing (another possible source of reduction of indistinguishability) is not present, the indistinguishability can reach 100 percent.

    However, I cannot understand this. After all, after the quantum dot reaches its first excited state, it does not emit a photon right away, but will do this at a random time within several lifetimes, which are actually much longer than the relaxation delays. Why this does not create jitters in the arrival times and reduce the indistinguishability?

    Two photon emitted by identical decaying atoms should be indistinguishable, whether they are emitted at the same time or not-- This I feel reasonable. But why if there is some random relaxation delay and they are not? (since they are delayed to the random decay times anyway?)

    I am sure the paper is correct of course. I must have misunderstood something, but what is it that I have misunderstood?
    Last edited: Jan 1, 2015
  2. jcsd
  3. Jan 2, 2015 #2


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    The exact time of a photon emission event is not defined to a better degree than the coherence time of the transition (which is under ideal circumstances given by twice the spontaneous decay time of the QD transition in question). So,loosely speaking, a single photon is stretched out in time over such a duration. You can see that, if you perform a Michelson interferometer experiment on single photons from QDs. Even if you change the delay between the two arms, you will still get the typical interference pattern even for single photons for temporal delays smaller than the coherence time.

    Now have a look at how one measures indistinguishability. Inthe Hong-Ou-Mandel experiment you take two consecutively emitted photons, insert a delay in the path of the first emitted photon and put the two photons on different entrance ports of a 50/50 beam splitter. If they are indistinguishable, they will leave through the same exit port because of two-photon interference. Otherwise, they will not. Now, the prerequisite for observing two-photon interference is of course that the two photons must overlap at the beam splitter. As a single experiment does not tell you much, you need to repeat the experiment several thousand times. If you revisit the Michelson interferometer experiment I outlined before,it becomes clear that having them overlap is not actually trivial. There is a non-zero probability for detecting a photon only during a time window of the length of the coherence time, so the delay between the two photons must be shorter than the coherence time in order to see two-photon interference events. While the mechanical delay set for the first photon compensates the time delay between two excitation cycles, it cannot account for the relaxation rates of free excited carriers towards the QD states as these change from shot to shot. So if the shot-to-shot fluctuations in the time these relaxation processes take is large than the temporal width of the photon, the photons just do not meet at the beam splitter and are of course not indistinguishable as you could tell which photon was emitted in which excitation cycle.
  4. Jan 19, 2015 #3
    Cthugha, thank you for your answer. They are very clear though I find the picture that follows from it are really amazing. I can understand that "the prerequisite for observing two-photon interference is of course that the two photons must overlap at the beam splitter". Also, I can understand that coherence time of the transition is "under ideal circumstances given by twice the spontaneous decay time of the QD transition in question". Thus, I picture the two photons as two very long snakes- as long as speed of light plus twice the spontaneous decay time. And if the two snakes overlap at the beam splitter for a large fraction of their length, a Hong-Ou-Mandel dip will be formed (after many repetitions of experiments of course).

    This is really against my intuition though. As I undertand it, if one put a photon detector before a single emitter, the time it detects a photon is random and subject to the exponential decay law. Thus before I see the your explanation, I thought since the two photon are emitted at random times, even if at time zero they are at exactly the same state- 100% inside the emitter, they generally cannot meet at the beam splitter if some miraculous timing coincidence did not happen.

    Thus I think the picture that follows from your explanation is very amazing: if one put two identical emitter at time zero at two arms of a Hong-Ou-Mandel interferometer, they always exit from the same output port, in spite of the fact that, if one put two single photon detectors right before the two emitters, they seem to emit photons at random, uncorrelated and in general quite different times. If I try to be a little more concrete, I would think of two identical emitter that has a lifetime in the range of minutes. It is just hard to accept that the photons that they emit will have any appreciable chance to meet at the beam splitter. But maybe if the condition is ideal enough that the photons they emitted have a coherence time that is indeed as long as approximately twice the spontaneous emission lifetime, and the detectors are far away enough that the photons have enough space to stretch out their whole length, a Hong-Ou-Mandel dip- a perfect one- will nevertheless appear?
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