I understand that diffraction patterns are seen with single slits with multiple photons/electrons, due to Fraunhofer effects. But do multiple single photons/electrons produce the same pattern, as they would in the double-slit experiment? (i.e. single photon detections aggregated over time)

  • $\begingroup$ As far as I know this was one of the big flags for the wave-particle duality of electrons. They could shoot them one by one and still ended with a diffraction pattern. $\endgroup$
    – JMac
    May 3 '17 at 12:00
  • $\begingroup$ Hi. This was certainly the case with the double-slit experiment but was it also seen in the single-slit case? $\endgroup$
    – Trumpy
    May 3 '17 at 12:02
  • $\begingroup$ Do you have any reason to believe it would not? Just showing your thought process gives others a way to address the motivation for the question. $\endgroup$
    – JMac
    May 3 '17 at 12:14
  • $\begingroup$ here for photons, at the end. youtube.com/watch?v=RNfENiKFWag $\endgroup$
    – anna v
    May 3 '17 at 13:46

The diffraction pattern seen when you have two "real" slits is actually not a pure cosine function; since the diffraction pattern is the Fourier Transform of the aperture function, and since the aperture function is the convolution of a single ("wide") slit with two "infinitely narrow" slits, the actual pattern observed is the product of a single slit pattern (sinc function) with the cosine function of the "ideal" double slit.

In other words - the "double slit" experiment observed both the "single slit" diffraction pattern, and the "double slit" diffraction pattern.

So the answer is "yes, a single slit diffraction experiment works, even when you use a single photon/electron at a time".

For reference, this site shows the result of a single-photon-double-slit experiment, which clearly shows the envelope (due to the finite width of the individual slits):

enter image description here

  • $\begingroup$ Out of curiosity, how does one make sure that only one photon reaches the slit at a time? $\endgroup$
    – Apoorv
    May 3 '17 at 12:35
  • $\begingroup$ This is done by using a very low flux. Given the speed of light, if you are sending (say) 1000 photons per second through the aperture, and you consider "at the same time" to be "they are within 1 mm of each other when they go through", then they need to be arriving within 3 ps ($\tau$) of each other. The probability of that happening (if the N=1000 are randomly distributed) is $2\tau N^2$ - there are $N$ intervals of width $\tau$ that can overlap with each other. That is clearly a very small number... so the vast majority of photons making up the pattern came through "one at a time". $\endgroup$
    – Floris
    May 3 '17 at 12:49
  • $\begingroup$ There are also 'single photon sources' where the emitter can really only emit one photon at a time. Often this is a single defect that is optically active. One example is the N-V (nitrogen-vacancy) center in diamond. $\endgroup$
    – Jon Custer
    May 3 '17 at 12:55

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