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Just for a completeness from an experimentalist's pov: Is it true every sub-atomic particle can have a mathematical representation as a wave? Experiments have determined the fundamental particles that define the standard model of particle physics. These are mathematically represented as point particles, with the mass in the table and the other quantum ...

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The wavefunction associated to a free particle of momentum $p$ is $$\psi_p(x) = \mathrm{e}^{\mathrm{i}px/\hbar}$$ which is obviously just a plane wave with wavelength $h/p$, fully compatible with the deBroglie relation. Strictly speaking, this function itself is not an admissible wavefunction because it is not square-integrable and hence ...

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No. The general relation is given by $$v = \lambda\nu$$ Where $v$ is the velocity of the considered wave and $\lambda$ and $\nu$ its wavelength and frequency. Of course in the case of an electromagnetic wave which is traveling at the speed of light you gain $$c = \lambda\nu$$ If you're treating instead some massive particle, then thou have p = ...

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What we call reflection is in reality a more complicated process than bouncing a ball to a wall. For the part of the electromagnetic radiation that we call visual light and for low densities of this light the surface electrons are responsible for the absorption and re-emission of this photons. So yes, mirror will gain momentum and the photons will lose ...

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The trick is that because the "slit" is infinitely wide, you shouldn't work in the far-field approximation (Fraunhofer difraction integral which leads to fourier optics), but with distances computed to the square order (Fresnel diffraction integral). The results are appropriately named Fresnel integrals (this time this is a name of a special analytical ...

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Every time a photon passes two slits or one slit and is detected, you obtain a dot. Indeed that's what a photon defined to be: a particle of light. A single photon will always be measured as a point. You observe interference bands only when you make many many measurements of single photons crossing two slits. In fact, a laser is just that: many many ...

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My question is, why doesn't interference occur with the observer here? Aren't there still probability waves between the quantum objects going through each slit, and shouldn't these waves interfere and create fringes of some sort? The results of any experiment when modeled with mathematics is absolutely dependent on the boundary conditions, which pick up ...

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With an observer there are no longer two probability waves travelling through each slit. There is a particle exiting one slit, which can be described by having some probability of where it will hit on the other side but there is no "second wave" from the other slit for it to interfere with. Another way to make it a little easier. If you imagine the ...

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Here is an updated video concerning single-slit diffraction (functions). Waves: Light, Sound, and the nature of Reality https://www.youtube.com/watch?v=Io-HXZTepH4

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The drawing is a little bit confusing - but the key is that the source $S$ is a relatively long way from the slits (compared to their spacing), and at a height of $\frac{d}{2}$ (from "directly behind $S_1$"). This means you can draw a diagram to calculate the relative path distance between S and each of the two slits. The difference in path length results in ...

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The text states that the source $S$ is placed just behind the slit $S_1$, so it looks to me that the sources $S_1$ and $S_2$ already are out of phase by a factor $\exp(i k d)$. Thus the phase difference in $O$ is $\Phi_O = k d$. This is different from the usual experiment where $S$ is placed at equal distance from $S_1$ and $S_2$.

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A phase difference may also be zero. Whenever the difference of the distances the beams travel is not a multiple of the wavelength, there is a non-zero phase difference - as simple as that. So if the source is not equidistant from both slits, and the point of measurement is equidistant from the slits, mean you already have a pretty good chance of non-zero ...

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