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First off, I assume you accept that the probability density for the position of a particle is $|\psi(x)|^2$. Now, what roughly happens in real scattering is that a particle whose wavefunction looks like the below image comes in: Then, after interacting with a potential, two smaller packets come out. The packet that transmitted through to the other side ...

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You are actually not calculating the same thing. The momentum space Feynman of the propagator is used for internal momentum, which is off-shell. So your first calculation is correct. While in your second case, you are used the LSZ formula to force the momentum to be on-shell. One way to get the same result is as follow: Consider the following diagram: ...

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As pointed in the comment by @CuriousOne, there is indeed a paper providing such explanation ("The photoelectric effect without photons" by Lamb and Scully). It was however later found to be faulty. A very good review and references on this can be found in these answers to a different question posted on this website, but apparently there is no consensus on ...

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The space appearing black despite the abundance of stars is Olbers' paradox. Poet Edgar Allan Poe suggested that the finite size of the observable universe resolves the apparent paradox: because the universe is finitely old and the speed of light is finite, only finitely many stars can be observed within a given volume of space visible from ...

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The reason the sky is blue on Earth is because of the Earth's atmosphere. The molecules and gas in the atmosphere interact with solar light via Rayleigh scattering, which allows for blue light to be scattered more efficiently than lower frequencies. This results in an abundance of blue light, which makes the sky look blue. Actually it should be said that ...

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Some brief answers, and references for further reading below. 1) You are right to be concerned about your sample having crystals with a preferred orientation. A single crystal would produce points instead of circles. The points would fall somewhere on the circles. But the image you provided shows that it is possible for the point to be off the film. ...

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Most of the time, an (elastic) scattering problem can be reduced in : An incoming initial wave / quantum state $|\phi\rangle$, which most of the time is taken to be a plane wave / free state $|\textbf{k}\rangle$ eigenstate of the free hamiltonian :  ...

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Essentially, fluorescence "scattering" is inelastic - transferring some energy to the fluorphore and decreasing the energy / frequency of the emitted light relative to the excitation - and Rayleigh scattering is elastic - conserving energy and thus wavelength. Rayleigh scattering is also very fast: elastic scattering of photons off medium electrons and each ...

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In general you need both kinds of functions $H_n^{(1)}(k\,r)\,e^{i\,n\,\phi} = (J_n(k\,r) + i\,Y_n(k\,r))\,e^{i\,n\,\phi}$ and $H_n^{(2)}(k\,r)\,e^{i\,n\,\phi} = (J_n(k\,r) - i\,Y_n(k\,r))\,e^{i\,n\,\phi}$ in any superposition, where $(r\,\phi)$ are the polar co-ordinates and $k$ the wavenumber. If the homogeneous region in question contains the point $r=0$ ...

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