# Does an electron's wave function collapse when it emits a photon?

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?

• 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. Jan 1 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? Jan 1 at 19:44
• For "electromagnetic interactions between electrons " see page two here 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 physics.stackexchange.com/questions/249057/… Jan 1 at 19:49
• This question is similar to question below where I gave an explanation.physics.stackexchange.com/questions/667799/… Jan 1 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? Jan 1 at 21:19

$$\newcommand\ket{\left|{#1}\right>}$$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.

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.