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I've been wondering how electromagnetic interactions between electrons are calculated if the position isn't determined until the wavefunction describing position collapses.

Does the wavefunction need to collapse in order to emit a photon? Otherwise the photon's path would be undetermined... or is the path of the photon linked to the uncollapsed wave function.

Is this related to the idea that the force between two electrons can be calculated by combining all possible paths the photon could travel weighted by the probability amplitude?

Iron Attorney
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    The photon is also a quantum "particle" with no specific position unless its wavefunction is collapsed for some reason. So you shouldn't in general expect the photon to have a determined path to wherever it goes to be detected. – The Photon Jan 01 '23 at 19:36
  • So does the electromagnetic interaction (like an electron emitting or absorbing a photon) not cause wavefunction collapse? And building on that, does that mean electrons can interact without their wavefunction collapsing? – Iron Attorney Jan 01 '23 at 19:44
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    For "electromagnetic interactions between electrons " see page two here http://hyperphysics.phy-astr.gsu.edu/hbase/Particles/expar.html ". The photons of interaction are virtual, mathematical constructs. For a single electron to emit a photon it has to interact with a field, see diagrams here https://physics.stackexchange.com/questions/249057/how-does-bremsstrahlung-occur-in-a-vacuumized-particle-accelerator – anna v Jan 01 '23 at 19:49
  • This question is similar to question below where I gave an explanation.https://physics.stackexchange.com/questions/667799/what-happens-to-the-wavefunction-in-an-allowed-transition/667817#667817 – emir sezik Jan 01 '23 at 19:50
  • @annav - so are you saying that the interaction between two electrons via a virtual photon does not cause the wavefunction to collapse for either electron? If so, are all possible positions the electrons could be in taken into account in the infinite set of paths the virtual photon can take? – Iron Attorney Jan 01 '23 at 21:19
  • @annav - and the virtual photon exchange is still an interaction with the electromagnetic field... so maybe my question is actually "does an interaction with the electromagnetic field cause an electron's wavefunction to collapse?" – Iron Attorney Jan 01 '23 at 21:29
  • The Feynman diagrams, which are iconic representations of the integrals that have to be performed in order to get the probability distribution for the reaction, give a continuity of values to the four vectors of virtual lines. A measurement happens when the real four vectors of the input and output particles is done. That is when the wavefunction "collapses" (though not a balloon). Within the integral there is no measurement. to the last question, yes, but not within an integral representation of virtual particles. – anna v Jan 02 '23 at 04:33
  • maybe this answer of mine will help https://physics.stackexchange.com/questions/622155/does-the-collapse-of-the-wave-function-happen-immediately-everywhere/622193#622193 – anna v Jan 02 '23 at 04:46

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$\newcommand\ket[1]{\left|{#1}\right>} $Electrons do not emit or absorb photons: this would violate conservation of momentum in the electron’s rest frame. Electrons can scatter from photons. And if an atom emits a photon, it is common to say “the electron has changed states from $\ket A$ to $\ket B$” rather than “the atom has changed changed state.” Putting all of the change on the electron is roughly like assuming the nucleus is infinitely massive — but that assumption is better in heavy atoms, where you have to talk about the evolution of the entire multi-electron wavefunction in any case.

Your final paragraph (v1) asks about “virtual photons,” which are a computational tool in relativistic quantum mechanics, or quantum field theory (QFT). Virtual photons aren’t needed to understand entanglement. You might be amused, however, to learn that one model of atomic emission in QFT is that virtual photons are “scattered” from the charged constituents of the atom into real photons.

As for your actual questions:

I've been wondering how electromagnetic interactions between electrons are calculated if the position isn't determined until the wavefunction describing position collapses.

The position of a bound electron famously isn’t determined, apart from the overall shape of the wavefunction. While an individual atom may be well-localized within a solid, photon emission in solids tends to come from the delocalized electrons which make up the electronic energy-band structure. One example is the ensemble of light-emitting diodes which are probably illuminating this text on your computer screen.

Does the wavefunction need to collapse in order to emit a photon? Otherwise the photon's path would be undetermined

It’s perfectly fine for a photon’s path to be undetermined. A historical non-photon example. It’s harder for a photon’s energy to be undetermined, so there may be a “collapse” of the atomic wavefunction. But then again, there may not. In laser emission, the photon number operator doesn’t commute with the Hamiltonian which means you can’t say “there are exactly $N$ photons in this laser cavity.” I suspect (but haven’t verified) this also means you can’t determine the number of ground-state versus excited-state atoms in your laser medium.

It’s totally reasonable to imagine a matter interferometer where a single excited atom takes both of two paths, and in which light emitted by those atoms would originate from both paths, even in the single-photon limit. I’m not sure whether that has been done or not; most of the gee-whiz interferometer experiments are done with photons, because making coherent photons is really easy.

rob
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