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Despite being presented as one of the fundamental results of Quantum Mechanics in practically every textbook, I realized this morning that I don't understand deeply how Quantum Mechanics predicts the double slit experiment.

The best explanation I have found seems to be closely related to the path integral formulation (e.g., since there are to a good approximation only two paths that a particle can take, the amplitudes due to these two paths add/interfere).

However, I am interested in how one could go about deriving the result using the methods taught in an introductory quantum mechanics course: defining a potential, solving the Schrodinger equation, using the generalized statistical interpretation, etc. Is there a simple way to see it from this angle? Or is this a result best left to understanding via the path integral?

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  • $\begingroup$ You're right! I find this unsatisfying as well, I'm very interested in the answers. $\endgroup$ – TheoreticalMinimum Aug 6 at 18:50
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I did once look at path integrals, but they are really just a complicated way of doing wave mechanics and I never found them useful. The double slit is much easier using basic methods, and I am not sure why your books would approach it any other way. There is no potential (or constant potential if you prefer), so the solution of the Schrodinger equation is a simple wave. The wave functions from the two slits sum, so one gets exactly the same pattern as for the classical Young's slit experiment.

enter image description here

In the figure, where wave crests (light grey) meet crests and troughs (dark grey) meet troughs, the amplitude of the wave increases (white crests, black troughs) creating a bright area on the screen. Where crests meet troughs, the waves cancel out (mid-grey), leading to a dark region.

The only difference between the classical interpretation and quantum mechanics is that in quantum mechanics the wave function is simply a way of calculating the probability for where a particle will be observed on the screen, as shown when particles pass through the slits one at a time.

single particle slits experiment

Result of the Young’s slits experiment using individual electrons, as carried out by Dr. Tonomura in 1989, showing the build-up of an interference pattern of single electrons. Numbers of electrons are 200 (b), 6000 (c), 40000 (d), 140000 (e).

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    $\begingroup$ But why must the particle diffract the way it does at the slit? Surely this must be caused by a potential (say, the "slit potential") on some level, right? I've seen classical explanations for this diffraction, but they don't seem to translate easily to the quantum case. $\endgroup$ – Uyttendaele Aug 6 at 19:30
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    $\begingroup$ This can be understood from Huygens principle, en.wikipedia.org/wiki/Huygens%E2%80%93Fresnel_principle, but it is simply the solution of a wave equation. Think of a point source (the slit). How else would waves flow from a point source, except radially? (the cause is the boudary condition of point source, not a potential). There is no difference in the solution of equations for quantum and classical wave mechanics. The difference is that qm gives probabilities, not actualities. $\endgroup$ – Charles Francis Aug 6 at 20:36
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    $\begingroup$ @Uyttendaele yes, as you very well point out there must be a potential that specifies the slit geometry. But the potential is infinite, so only affects the boundary conditions at the slits. $\endgroup$ – Wakabaloola Aug 8 at 8:47

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