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Photon shot effect

  1. May 24, 2013 #1
    1. I have never understood Purcell's explanation of the Hanbury Brown Twiss effect saying that the correlation they measured is only due to the clumping of the bosons; that the pure shot effect would not yield any correlation. Why not?
    2. I was reminded of this by Purdy's recently announced observation of the Heisenberg uncertainty principle on a macroscopic object. He only talks about shot effect fluctuation. Shouldn't he see a slightly larger fluctuation due to the contribution of photon clumping?
     
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  3. May 24, 2013 #2

    Cthugha

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    Did he use that wording in his comment in response to Ferguson/Brannen about the HBT-effect? Basically, the HBT-effect is a question of the variance and mean of the underlying photon number distribution. As long as that is equal to the variance and mean of a Bose-Einstein distribution, the coincidence count rates will match the ones expected in the HBT effect. Photons will not necessarily show bunching. If you start from a nonequilibrium situation, you may get other results. For example laser light does not bunch. It follows a Poissonian distribution.

    The experiments in the Regal lab were done using laser light which does not show bunching. Laser light is coherent and therefore in a minimal-uncertainty state. If you look at its Wigner function you will see that the expectation values of quadrature variance equal that of the vacuum state, which means that you are left with shot noise.

    One should indeed see larger noise if one used a filtered light bulb or a laser below threshold as a light source. These sources will indeed show photon bunching.
     
  4. May 24, 2013 #3
    I understand your answer to 2 but not that to 1. If I put laser light thru a half-silvered mirror will I not find correlation between the intensity fluctuations (due to pure Poisson shot effect) of the 2 beams? If not, why not?
     
  5. May 24, 2013 #4

    Cthugha

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    To check whether there are correlations present, one will typically compare the joint detection rates to the joint detection rate expected if there are no correlations present - the latter will just be the product of the mean detection rates n. Assume for simplicity that we simply sample the same field twice. Then this ratio will be:

    [tex]g^{(2)}=\frac{\langle: n^2 :\rangle}{\langle n \rangle^2}.[/tex]

    This is the correlation function. A value of 1 will indicate the absence of correlations. The thing you measure in the HBT-effect is the normal-ordered correlation function (the two ":" are there to indicate that). Normal-ordering of the underlying photon operators just means that you correctly incorporate the effect of the detection of the first photon: the light field will now have one photon less:

    [tex]g^{(2)}=\frac{\langle n (n-1) \rangle}{\langle n \rangle^2}.[/tex]

    Now the instantaneous photon number will be the mean photon number with some deviation added:

    [tex]g^{(2)}=\frac{\langle (\langle n \rangle +\delta) (\langle n \rangle +\delta -1) \rangle}{\langle n \rangle^2}.[/tex]

    Now you can evaluate all these terms. The mean value of the deviation should vanish. The expectation value of the square of the deviation survives. This is the variance of the photon number distribution. That leaves us with three surviving terms:

    [tex]g^{(2)}=\frac{\langle n \rangle^2 +\langle \delta^2 \rangle- \langle n\rangle) }{\langle n \rangle^2}=1+\frac{\langle \delta^2 \rangle}{\langle n \rangle^2}-\frac{1}{\langle n \rangle}.[/tex]

    Now one can perform a sanity check and evaluate this result for three typical states of the light field:

    a) A single photon. A non-classical single photon state has no noise at all and a photon number of 1. This will leave us with a g2 of 0. The correlation is negative. If you have exactly one photon and detect it, it is gone and you will not be able to detect another photon afterwards.

    b) Thermal light. Thermal light follows the Bose-Einstein distribution. For a mean of <n> it has a variance of n^2 +n. That leaves us with a value of 2. The detection events are correlated.

    c) Coherent light. This follows a Poissonian distribution. For a mean value of <n>, this distribution has a variance of sqrt(<n>). Inserting that into g2, you will find that the last two terms cancel exactly and you are left with a value of 1. The shot noise contributions exactly cancel the contribution you get by destroying a photon with the first detection. This is also the reason why coherent states are eigenstates of the photon annihilation operator. They are immune to loss. So the detections will be completely uncorrelated for laser light.
     
  6. May 25, 2013 #5
    Thanks. I think I understood most of what you said. Also may have come up with a simpler explanation, of which I would like your opinion. Consider a Poisson process with an average pulse rate f. Model the half-silvered mirror by a switch which randomly directs half the pulses into each of two channels. (Without loss of generality this can be simplified to directing every other pulse into each of two channels.) Lowpass filter each channel to a bandwidth << f, then correlate. My naive expectation was that, during a time interval having a pulse density higher than average, this would show up in both channels, but this must be wrong. The emission of a pulse is a random process whose probability is fixed and independent of when any other pulses were emitted. The two channels will, therefore, each receive pulses from a Poisson process with average pulse rate f/2 and these two processes are independent (uncorrelated). This doesn't even involve QM. Depending on your answer, I may write a simulation to demonstrate this.
     
    Last edited: May 25, 2013
  7. May 28, 2013 #6

    Cthugha

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    Absolutely correct. The Poisson distribution is a distribution describing statistically independent events. So your approach should work. You rather need qm to assure that you can use a Poisson distribution, but once you accept that, this classical reasoning will work well.
     
  8. Aug 13, 2013 #7
    To Cthugha:
    1. I made a bad mistake in my post of 5/25. I said that a half-silvered mirror would either transmit or reflect a photon. That's wrong; (the wave function of) each photon follows both paths. Otherwise we could not get phase interference in dim light where the photons are far apart. That destroys my argument.
    2. I don't understand how (5/24) you got from your penultimate to your last equation. I have a leftover term (2n-1)delta/n^2.
     
  9. Aug 13, 2013 #8

    Cthugha

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    The delta is the deviation from the statistical mean value that occurs in a single detection run. In the statistical average over many runs the mean deviation from the mean value must disappear (otherwise the mean value is obviously not the mean value), so any term linear in delta must be 0 in the statistical average.
     
  10. Aug 14, 2013 #9
    OK. Now I can follow the math, with one minor clarification: do the angle brackets mean 'expected value'?

    But I still don't understand the physics. If I send laser light into a Mach-Zehnder interferometer, I can observe phase interference at the combining plane. But if I intercept the two beams before combining with a pair of PMTs, their outputs will not show correlation of the intensity fluctuations? There will be intensity fluctuations because the light is a poisson process. And they will be synchronized because (the wave function of) every photon goes through both beams. Well, then, why no correlation?
     
  11. Aug 14, 2013 #10

    Cthugha

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    Yes, this is the expectation value or the ensemble average.

    These are two different things. One is taking place at the field level. The other is taking place at the intensity level. Even if you do not intercept the beams in the interferometer, but instead look at its output ports, you will find no correlation (if the setup is stationary and therefore phase stable).

    But there is some correlation term. It is the second of the three terms in the final result. However, for a coherent state, this is exactly as large as the anticorrelation you always get when destroying a photon while measuring it. If you have exactly three photons in a state and detect one of them and destroy it, this reduces the probability of detecting another photon compared to the initial state as you only have two photons left . This represents an anticorrelation. To get no correlation in total (neither positive nor negative), you need a small amount of synchronized fluctuations in both beams to cancel that anticorrelation.

    The interesting thing about coherent states is that these terms cancel exactly.
     
  12. Aug 14, 2013 #11
    3 photons is a convenient, small, number. Suppose the video integration time allows 600 photons, on average.
     
  13. Aug 14, 2013 #12

    Cthugha

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    Going to larger photon numbers does not change much, but the numbers get a bit smaller. The anticorrelation term is
    [tex]-\frac{1}{\langle n \rangle}.[/tex]
    For a mean photon number of 600, this is only -1/600 and pretty small.

    The correlation term is:
    [tex]\frac{\langle \delta^2 \rangle}{\langle n \rangle^2},[/tex]
    which is just the variance of the photon number divided by the squared mean of the photon number. Now a Poissonian process has the interesting property that the variance is exactly equal to the mean (or equivalently the standard deviation follows the famous sqrt(n)-law). This leaves us with 600/600^2 =1/600.

    The result is completely independent of the mean photon number as long as you have a process for which the photon number variance equals the mean - which is a Poissonian process.
     
  14. Aug 15, 2013 #13
    The light is beginning to dawn over my thick head. When a PMT detects a photon, that photon is exclusively assigned to that path; the other PMT cannot detect it. So, while, for phase interference, each photon follows both paths and interferes with itself, this is not true for intensity interferometry. Detecting any correlation here requires that there be correlation between different photons, and that requires bunching. Have I got it, at last?
     
  15. Aug 15, 2013 #14

    Cthugha

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    Leaving all the points aside which people have been fighting over how to interpret them correctly now for decades (What does it mean for photons to be different? Do photons really follow both paths at once and if not, what does?) and interpreting your wording in an intuitive sense, this is about right.

    Bunching is about intensity correlations and therefore this is indeed about two-photon correlations (or anticorrelations).
     
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