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When single photons are repeatedly fired through a double slit apparatus, eventually an interference pattern emerges on the detector, suggesting that each individual photon actually propagates as a wave. Because the presence of the double slit wall changes the distribution of locations where each photon actually lands, it seems fair to say that the wave must interact with the double slit wall in some manner.

If the photon doesn't "hit" the double slit wall, but instead makes it to the detector, it seems the only interaction that could have occurred between the wall and the photon is an exchange of momentum. That is, the wave and the wall exchange momentum, thereby changing the direction of the wave, and ultimately impacting the distribution of arrival locations at the detector, but this interaction falls short of forcing the wave to collapse into a particle. I.e., the photon wave doesn't give up any energy to the wall, but does change direction when it interacts with the wall.

Is this a fair description of what happens physically, or theoretically, in the double slit experiment? If so, then doesn't this suggest that a quantum wave can exchange momentum without collapsing into a particle?

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  • $\begingroup$ Why not go one step further and think about a quantized wall-photon interaction that causes a swelling deflection? The intensity distribution of photons is the result of the quantized force field between these photons and the (surface electrons of the) wall. $\endgroup$ – HolgerFiedler Sep 5 '18 at 18:48
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Yes what your saying is fair but a couple of points. When the photon interacts with the wall it is interacting with the electric (EM) field of the wall which can extend some small distance from the wall and is somewhat random and quantum in nature. Since the photon hasn't changed colour (energy) but has changed direction the wall must have lost some energy.

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  • $\begingroup$ In the case of Lorentz force the external magnetic field does not weaken, it works more like a spring. The electrons radiating and are loosing energy. In the case with the photons indeed the electrons of the wall have to be accelerated from the photons (which are indivisible) and have to emit photons. But since they are not driven from an electric voltage (like in an antenna rod), perhaps the most of the emitted photons get absorbed by other electrons of the wall? $\endgroup$ – HolgerFiedler Sep 5 '18 at 18:58
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Also look up "Quintuple slit experiment" on this site and see a mathematical solution presented by levitopher based on Feynman paths. (Ignore the quintuple part and just read the explanation)

Photons like to travel a path in "n" multiples of its wavelength where n is an integer. Based on the geometry of the situation certain paths are very probable and others are not (dark spots have no photons).

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