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Young's Slits and "one-at-a-time" photons

  1. Jul 14, 2014 #1
    Hi Everyone

    There is a problem that I have with understanding what happens when photons are emitted "one-at-a-time" in the Young's Slits experiment. However, I think I may have come up with a solution to this problem and would be very grateful for your thoughts.

    Let's say a single photon is emitted from a photon emitter and let's assume that absolutely no electrical or magnetic disturbances whatsoever are emitted from the emitter before or after the emission of the photon until the next photon is emitted after a period of time which allows the current photon to reach the screen. As I see it, the photon's probability distribution which will emerge from the aperture of the emitter (assuming the photon has not already been absorbed by the physical structure of the emitter itself) will have the shape of a thin spherical shell fragment with perhaps some diffractive fringing around the edge of the fragment. This probability distribution moves forward until it meets the filter containing the Young's Slits. At this point, the photon may get absorbed by the filter itself but let's assume that this does not happen and the probability distribution continues on through the slits. If we imagine that there is an imaginary vertical plane which is perpendicular to the filter and is exactly half way between the two slits and we assume that the emitter is positioned exactly on this plane then this means that the probability distribution will pass through both slits at exactly the same time. The slits will convert the probability distribution into two probability sub-distributions. As I see it, as these two thin, shell-like sub-distributions move forwards towards the screen, they can only intersect and therefore interfere with each other at points which are on the imaginary vertical plane I mentioned earlier which is exactly half way between the two slits. This means that interference can only happen along a single line down the middle of the screen, not the whole screen.

    I have managed to think of a solution to the above problem which allows a proper interference pattern to emerge on the whole screen. The solution is that the shell of the probability distribution is not as thin as I have been assuming. If it is as thick as the distance between the slits or thicker then I can see that this will enable a proper interference pattern to emerge. I would be very grateful for the thoughts of the experts on this forum about my solution.

    Thank you very much.
  2. jcsd
  3. Jul 14, 2014 #2


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    Good question. Indeed, this "shell" is not as thin as one might think. The "thickness" is given by the coherence time of the source. It might sound strange, but photon emission is not taking place at a well fixed moment in time. There is some uncertainty to it. For example, if you consider a photon placed inside a resonator with good mirrors, it will go back and forth several times and you will typically not be able to tell whether it left the resonator after a few round trips more or less. For emission from an atom, you will get a superposition between the excited state and the ground state and the probability density to find an electron in one of this states will oscillate back and forth several times. So will the probability density to find a photon in the field.

    Accordingly, the uncertainty in photon emission roughly corresponds to how long this superposition is kept alive. If the photon could now take two possible paths to one spot on the screen and both are with a distance given by the speed of light times the duration of the superposition, interference is possible and it is impossible to find out in principle when exactly the photon was emitted.

    The exact timescale of the coherence time depends on your light source. It is around 200 fs for our sun and can go up to milliseconds for good lasers. Accordingly, the "thickness" of that "shell" can be anything from micrometers to kilometers depending on your light source.
  4. Jul 14, 2014 #3
    Hi Cthugha

    Thank you very much for your reply. You have confirmed what I thought.
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