I have read for example here that the single slit experiment can be seen as a visualisation(?) of Heisenberg's Uncertainty principle. Basically, the photons passing through the slit are given a fixed value for the standard deviation of position. Therefore, the standard deviation for momentum is "fixed" and, given a statistically large amount of photons, the standard deviation for their momentum is made "visible" as the width of the central fringe, where the central fringe is the expected value of momentum for the photons.

If this is correct, why isn't the intensity pattern on the screen distributed according to the normal distribution, but shows individual fringes?

  • $\begingroup$ What you state is a first order approximation to the quantum mechanical level. The wave nature of the probability distribution in a quantum mechanical experiment "particle scattering off speccific slit" is seen in the fringes, which would not exist for classical particles. $\endgroup$ – anna v Apr 9 '18 at 15:55
  • $\begingroup$ Sorry, but I have no idea what the 'first order approximation' is. (And what are other approximations?) $\endgroup$ – FizzleDizzle Apr 9 '18 at 16:00
  • $\begingroup$ what about rule of thumb? rough approximation? that is what heisenberg's principle is about, not exact numbers which need solutions of quantum mechanical equations to describe data $\endgroup$ – anna v Apr 9 '18 at 16:13
  • $\begingroup$ So the reason why you get a normal distribution when using the uncertainty principle as an argument is that it only describes the 'things going on' in the experiment very imprecise (or wrongly)? $\endgroup$ – FizzleDizzle Apr 9 '18 at 16:24
  • $\begingroup$ But, within the logic of the argument I made with the uncertainty principle, should the experiment yield a normal distribution? $\endgroup$ – FizzleDizzle Apr 9 '18 at 16:26

Standard deviation is a general property of a data set, regardless of its probability distribution. For a random variable $X$, standard deviation is defined as

$$\sigma(X)=\sqrt{\operatorname E[X^2]-(\operatorname E[X])^2},$$

where $E[\cdot]$ is the expected value. As you see, nowhere here is the exact probability distribution used. It just so happens that normal distribution has standard deviation as a simple parameter.


There's no contradiction here. What happens fits both a classical and a QM interpretation.

For the QM, here's the best I can say:

Each photon can cross the slit at any location along the slit, and the probability for each location is more likely uniform, not gaussian. It's just as likely to be anywhere along there. Maybe it makes sense to say it's at all the locations. But sometimes it won't be uniform. Depending on how close a point-source is, the distribution could fit a slice from a sphere.

There is something about the photon which varies over time (and distance, at constant speed both will vary together) which fits a sine wave. I don't know what it is.

To get the chance that the photon will interact with a detector at a particular angle and distance from the slit, we look at all the straight-line paths it could take and their distances, and integrate our sine wave over all of them. The probability is proportional to the absolute value of that integral.

Some places the probability comes out to zero, places where the sine wave cancels. Other places it isn't zero. The maxima get smaller at bigger angles because the integral includes more whole cycles that cancel out.

What is it about light quanta that fit a sine wave? I dunno. I don't think anybody knows. But the math works, so we can use it.


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