Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Can one identify the vacuum state of a photon?

  1. Sep 7, 2011 #1
    If an attenuated laser pulse (with expected number of photons <1) is incident on a perfectly efficient detector, and the detector does not click, does it mean we know the |n=0> state was realized, and does it imply that all subsequent detectors measuring this pulse will also not click?
     
  2. jcsd
  3. Sep 7, 2011 #2

    Cthugha

    User Avatar
    Science Advisor

    No, the photon number output from a laser is random. All you have is the expected photon number per unit time interval and the photon number distribution per unit time interval will be a Poissonian distribution with that given mean.

    So in other words if an ideal detector does not click within that time interval that does not give you any information about what will happen in future measurements on time intervals of the same length as all detections in all time intervals are completely independent of each other. This is, however, slightly different for thermal light.

    edit: just a comment. I assume that you intend to have a streak camera or something similar as detectors so that you can time-resolve the pulse and different times within the pulse are effectively detected by different detectors. Otherwise the mixture of having a perfect detector and using subsequent detectors indeed makes no sense as xts pointed out.
     
    Last edited: Sep 7, 2011
  4. Sep 7, 2011 #3

    xts

    User Avatar

    Take it a bit practically...

    You speak about laser pulse, so we may assume you are speaking about visible light / near infrared / near ultraviolet, rather than about gamma rays.

    What kind of a nearly "perfectly efficient detector" for such wavelenghths you know, which is not "perfectly absorbing" simultaneously?

    So what do you mean by "subsequent measurements" of "this" pulse?
     
  5. Sep 7, 2011 #4

    Bill_K

    User Avatar
    Science Advisor

    I believe he is talking about a quantum nondemolition measurement. See http://www.physorg.com/news197873165.html, which describes an experiment that counts the number of photons in a microwave cavity without disturbing them.
     
  6. Sep 7, 2011 #5
    The "additional detectors" are there in case the first one does not click so whatever is now in the beam (vacuum state?) can reach the next detector. I also assume the detector is on during the whole time interval so we can ignore the temporal probability distribution.
     
  7. Sep 7, 2011 #6

    Cthugha

    User Avatar
    Science Advisor

    Ok, common detectors detect photons by destroying them. So if you have a perfectly efficient detector you will destroy all photons anyway and the subsequent detectors will not detect any photons in any case.

    If you want to perform QND measurements, it is more complicated. Here you use a series of QND measurements to nail down the photon number and you can predict the photon number distribution with increasing accuracy by increasing the number of subsequent measurements. However, this method is never ideal, but only non-demolition in a certain percentage of cases. The optimum I heard about is about 90%.

    In a nutshell, if you had a perfect and ideal detector for all wavelengths which is also otherwise ideal in all respects, the light field arriving at the detector could be in a vacuum state. However, it is not necessarily so. It could also just be a result of destructive interference at the detector position.
     
  8. Sep 7, 2011 #7
    "In a nutshell, if you had a perfect and ideal detector for all wavelengths which is also otherwise ideal in all respects, the light field arriving at the detector could be in a vacuum state. However, it is not necessarily so. It could also just be a result of destructive interference at the detector position."

    That part is clear to me. But if we modify the original setup to include a Mach - Zehnder interferometer, then wouldn't destructive interference at Detector 1 automatically mean constructive interference at Detector 2? For example, we split the pulse, phase-shift one half by 180, and recombine it to get a deterministic result. Wouldn't that distinguish with certainty between destructive interference and the vacuum ket? That is, when neither detector clicks in this scheme, wouldn't that tell us that the pulse was in a vacuum state?
     
  9. Sep 7, 2011 #8

    xts

    User Avatar

    Just a stupid question by an (ex)experimentalist, used to GeV rather than single eV...

    For gamma rays we may easily track a photon. It kicks electrons by Compton scattering, barely changing its original direction, and we may easily nail the scattered electrons (nice track left by gamma in a bubble chamber). We may detect it again, and again - many times - not destructively, just taking some part of its energy. So maybe the "perfect" but "non-absorbing" detector is possible?

    Is any fundamental (I am not asking about practical...) reason why similar mechanism can't be taken down to the visible light region? Is the only obstacle the rest mass of electron (the lightests possible charged "detector"), which is easily kicked by 1 GeV gamma, but rather not vulnerable to 1 eV kicks?
     
    Last edited: Sep 7, 2011
  10. Sep 7, 2011 #9

    Cthugha

    User Avatar
    Science Advisor

    Well, if you know the possible states and the setup, you can of course place the detectors in a manner that you get to know the state of the light field. I was just pointing out that this in necessarily true for a completely unknown state of the light field and a unknown setup.

    Good question. Compton scattering is basically a change of the photon mode to lower energy which leaves the momentum more or less unchanged, right?

    I suppose that it should be very difficult to find a process that does this at low photon energies - at least one that has a reasonable probability to occur. I guess you will not get electrons to move around quickly by scattering with low energy photons - I suppose you need a certain velocity to see the electron traces in the bubble chamber. Lighter particles might be able to scatter off with higher velocity, but I am not aware of any light particle with reasonable scattering cross-section and I suppose it should also be charged to leave a trace in bubble chamber. But if there was such a fitting particle I see no fundamental reason why it would not work.

    By the way, are these measurements in bubble chambers photon number sensitive? I mean could you distinguish between one or more photons entering at the same initial energy?
     
  11. Sep 7, 2011 #10
    "But if there was such a fitting particle I see no fundamental reason why it would not work."

    Suppose we found such a particle, or better yet, built a hypothetical gamma laser (a gedankenexperiment no doubt, but physics remains the same) so that a perfect detector would not need to absorb the photon but would simply scatter it. Then if we place several (perfect!) detectors (or in a more refined example, several Mach -Zehnder interferometers equipped with perfect detectors) one after another, and the first detector (or the detector pair in the first Mach-Zehnder) did not click, does it mean the state transmitted/scattered at this detector is now certainly a |n=0> state, and if so, can it still produce a noticeable outcome at the next detector? To put it differently, if there any physical difference between sending a vacuum state toward a detector, and not sending anything?
     
  12. Sep 8, 2011 #11

    xts

    User Avatar

    Yes, photon loses just small amount (10 keV or so) of energy, then you may see those electrons making small spirals. If you see multiple such spirals along straight line - that's a signature of gamma.
    Of course - Compton's scattering, and especially electron tracks in bubble chamber can't be used for single eV range. I gave it only as an example of non-destructive detection of photons. So I wonder if there is any mechanism allowing to pick 1meV from 1eV photon and thus detect its presence. HUP allow for it if the size of detector would be in order of mm. And, of course, it would have to be cooled closely to 0K, to eliminate thermal noise - if such 1meV are to be detected. Breaking of Cooper's pairs - is my first guess candidate to detect such 1meV energy transfers.

    Except that I never saw two 10GeV gammas running in the same direction, they should be - just producing twice bigger number of "spirals". Number of kicked electrons depends also on gamma energy, but it may be estimated from spirals size, so in principle you may measure both: energy of gamma and number of photons.
    And, of course, no one uses bubble chambers nowadays - last of them got closed in 1980 or so - I gave it only as a picture easier to imagine than modern electronic track chamber.
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook




Similar Discussions: Can one identify the vacuum state of a photon?
Loading...