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I simply don‘t understand why the assumption that light moves as a Wave makes more sense than that it moves as a Particle... (I know that the assumtion is more about light „sometimes“ being a Wave, but that just makes even less sense).

Wouldn‘t it be more appropriate to assume that the probability of the particle being at a certain place is based on the Wave function?

Now what causes this probability of it deviating from the straight path can be anything from an unstable setup that slighlty changes the direction of the particle being shot, to any outside force change, but that‘s Not the point now.

If we shoot many particles (in a Computer Simulation) towards 2 slits and slightly modify the angle for each this will result in a pretty perfect interference pattern... and we have no particle being a Wave, or having anything to do with one.

Now if we have the amount of deviation as being inverse to the probability (smaller deviations happen more often), then the interference pattern looks pretty much like the ones seen in Double-Slits experiments, being brighter in the center than towards the sides.

So how come that physicists say it behaves like a Wave? Am i completely missing something about this?

Btw, i made a simulation with this assumtion and it looks just how it should...

Edit : This is close to what i got when i did the simulation. The simulation takes a lot of time, so i just made an image of what i got

This is what it resembles (not 100%, but that's due to the quality of the simulation) enter image description here

Turns out, that it resembles the 2nd image less, the further the particles travel...

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  • $\begingroup$ So it will obey a wave equation but you will call it a particle? $\endgroup$ – Jon Custer Jan 6 '19 at 0:58
  • $\begingroup$ @JonCuster It is a particle and follows the path a particle should follow, but if you repeat it with slight deviations in its path (slightly changing the starting direction) the overall Image created is exactly what the Double slits experiment shows. $\endgroup$ – Lexyth Jan 6 '19 at 1:12
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    $\begingroup$ In your simulation, if you cover a slit what do you get? $\endgroup$ – Francesco Bernardini Jan 6 '19 at 2:48
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    $\begingroup$ @FrancescoBernardini if i Cover one slit, then it creates 2 equal spikes with the center of both being directly under the Open slit... which shouldn‘t be... ok seems like i made some mistakes 😅 $\endgroup$ – Lexyth Jan 6 '19 at 3:23
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    $\begingroup$ @FrancescoBernardini No, that‘s why i Said i made a mistake 😅 I get 2 beams with the „dark“ space between them being where the beam should be. Thus there has to be something wrong with my Simulation... $\endgroup$ – Lexyth Jan 6 '19 at 3:50
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Why can‘t the Double Slits Experiment be explained by a Particle?

In the standard model of particle physics the photon is a zero mass point elementary particle . These elementary particles obey quantum mechanical equations. When they interact it is the wave function that gives the probability for the path that the photon will follow . It is the probability that waves. In this double slit single photon experiment :

snglphot

Single-photon camera recording of photons from a double slit illuminated by very weak laser light. Left to right: single frame, superposition of 200, 1’000, and 500’000 frames.

the experiment is "photon interacting with two slits a fixed distance apart and a fixed width" for a given material whose atoms and molecules provide the potential with which the photon will interact.

The particle footprint is seen on the left the built up interference pattern , of the electromagnetic wave is seen on the right.

The standard model is a very successful model in describing existing data and predicting new set ups. Classical electromagnetic fields emerge in a complicated mathematically way from the quantum mechanical formulation with photons as can be seen here .

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Just so there is a clear record: what you're seeing is not an interference pattern - it is just the shadow of the two slits. If you cover one of the slits, the expected wave behaviour is a pattern which is equally broad, but without any nodes. As you noted in the comments, that's not what you observe in your simulations.

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  • $\begingroup$ It should be the real answer sought ....plus one $\endgroup$ – Alchimista Jan 6 '19 at 9:09
  • $\begingroup$ @Alchimista It is more of a record of this question and the answers and delivers an explanation to why i was wrong, but doesn't answer the direct question as precisely as the currently selected answer, at least in my opinion. $\endgroup$ – Lexyth Jan 6 '19 at 10:00
  • $\begingroup$ Yes that is possible. I have thought your simulation just challenged something that you were aware of. It was just an appreciation of the coincise answer. $\endgroup$ – Alchimista Jan 6 '19 at 10:18

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