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Sorry I know the title of this question is kind of weirdly worded, but it kind of takes context to understand what I'm asking better.

So my current understanding of quantum superposition is this: So when an electron is unobserved it occupies a superposition of all possible position states, essentially meaning it is everywhere it can be at the same time. (See: first few sentences of https://en.wikipedia.org/wiki/Quantum_superposition#Hamiltonian_evolution) However when observed it collapses into a single position with a degree of uncertainty (so still fuzzy where it actually is, just much less so.).

So imagine this scenario. An incoming photon approaches the electron. It can be said that when this photon interacts with the electron then the electron has been 'observed' and collapses into a single fuzzy position state. In some cases the photon is absorbed and the electron jumps up an energy level, and in others the photon is absorbed and then immediately rejected (non-resonant absorption). But either way most photons coming in are absorbed. But why does the electron take up a position by the photon, doesn't it have an equally likely chance of being on the other side of the atom, and thus not able to absorb the photon? But if that were the case the vast majority of incident photons would simply pass through materials. Most things would be transparent, but they're not.

So if the photon causes the position wave to collapse, then why doesn't it cause the vast majority of photons to not be absorbed.

Or I am completely misunderstanding the topic at hand?

Thank you!

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  • $\begingroup$ However when observed it collapses into a single position with a degree of uncertainty (so still fuzzy where it actually is, just much less so.). This part loses me, does not superposition immediately recommence? $\endgroup$ – user163104 Aug 11 '17 at 1:30
  • $\begingroup$ Once was no longer being 'observed' the superposition would recommence. But for the duration of the interaction the state would be collapsed (at least that's how I understand it) $\endgroup$ – Robotic_Cow Aug 11 '17 at 1:39
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    $\begingroup$ If you just have a one-atomic-layer thick film, most of the photons probably will go through. But if you have 1000-atomic-layer-thick film, then doesn't each photon get 1000 chances to be absorbed by an atom? So wouldn't the chance of a photon making it all the way through be much smaller? And most materials we see are much thicker than 1000 atoms. $\endgroup$ – Peter Shor Aug 11 '17 at 1:52
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"But why does the electron take up a position by the photon, doesn't it have an equally likely chance of being on the other side of the atom, and thus not able to absorb the photon" .

The basic misunderstanding lies in confusing classical frameworks with quantum mechanical frameworks.

Electrons, photons and atoms are quantum mechanical entities following quantum mechanical equations, not classical ones.

The electron is not in an orbit around a nucleus. It is in an orbital, a quantum mechanical probability locus. When a photon hits an atom, it is the whole atom that interacts and changes energy level, not the electron by itself. The photon is absorbed by the collective and the wave function giving the probability loci , orbitals, changes. The electron goes to a higher orbital and the photon disappears. If the photon scatters off, it scatters quantum mechanically from the residual field outside the atom, and the atom reacts as a whole if free, if in a lattice it transfers some momentum to the lattice.

The term "collapse" is a confusing way of saying "this single scatter has provided a point in the probability density distribution of photon atom scattering".

The answer to:

Why doesn't electron-photons interactions and the principle of quantum superposition cause most things to be transparent?

is that the photons encounter a quantum mechanical barrier in the solids and liquids and can only pass through them in lattices which allow total through scattering, retainining its energy and momentum, the transparent materials. Most materials are not transparent and the photon loses its energy in the first layers of the lattice.

In gases they scatter off the residual electric field or tranfer their energy to the atom by lifting an electron to a higher orbital, or losing some of their energy in transferring the electron out of the potential well binding it to the nucleus. It is the total atom, or lattice that interacts with the photon.

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Not every interaction/measurement has a position state as outcome - nothing about position is special, we can have many other observables whose eigenstates then form the outcomes of them being "observed". The interaction between electrons and photons starts and ends with the electron in an energy eigenstate (or atomic orbital), which is sort-of localized in the region we conventionally call "the atom", but which is far from the (fuzzy) position state you seem to be thinking of.

So yes, you are misunderstanding the topic at hand because nowhere in the quantum mechanical interaction between electron and photon is either of them forced to occupy a state of definite position, so thinking about the photon "missing" the electron because it's on the other side of the nucleus is plainly non-sensical.

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  • $\begingroup$ The interaction between electron and photons does not start and end with the electron in an energy eigenstate unless somebody or something "observes" the electron. And it's not the interaction that "observes" it; it's generally an "observer" that comes along after the interaction. $\endgroup$ – Peter Shor Sep 5 '17 at 19:51

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