There are many situations where a photon is emitted from a charged particle because it accelerates. This includes bremsstrahlung, which is usually a charged particle scattering of a nucleus and losing energy through radiation. And we have an electron radiating photons in a magnetic field because the electron is accelerated by the magnetic field. The latter is the situation I'm particularly interested in - in case it matters.

When a photon is emitted from an electron that was accelerated - for whatever reason. Let's say that electron had a well-defined spin before the photon was emitted (ex. it's spin up in the magnetic field, or its spin is parallel to its motion in the lab frame). Does the spin ever flip because of angular momentum carried away by the photon?

The very fact that there are spin polarized electron beams in accelerators proves that this isn't happening all the time. In Feynman diagrams that show this process, usually they depict a photon being absorbed from the nucleus/magnet and another one being emitted as the bremsstrahlung, which I think is why it typically doesn't flip the spin (the angular momentum of the emitted photon is then transferred to the nucleus or the magnet that caused the scattering?). But I'm wondering if its possible - or if I've misinterpreted something about how this process works.

  • $\begingroup$ When you say angular momentum I take it you mean orbital angular momentum and not the total one right?, $\endgroup$
    – schris38
    Mar 24, 2023 at 21:19
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    $\begingroup$ Sorry I don't follow the question. For the angular momentum of the light, I meant the spin. For the electron, I also meant the spin. It's not in an atom but it is in a magnetic field, so it definitely has angular momentum from its velocity within the magnet - but it's well known that that is lost to cyclotron radiation, and that's most of what cyclotron radiation does. As for how the angular momentum of the light could be transferred to a magnet - I guess I mean in the form of changing the $E\times B$ angular momentum of the field. $\endgroup$
    – AXensen
    Mar 24, 2023 at 22:27
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    $\begingroup$ You appreciate interaction with a photon preserves chirality, no? $\endgroup$ Mar 25, 2023 at 0:23
  • $\begingroup$ Does it? It preserves some kind of total chirality or it never changes the chirality of the electron? Anyway if there's some kind of answer from that fact then I wasn't previously aware of it. $\endgroup$
    – AXensen
    Mar 25, 2023 at 9:26
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    $\begingroup$ The photon couples equally to both L and R-chiral fermions, which preserve their helicity if ultra-fast. There are recondite complications/exceptions, but these are outside the routine scope of your question, here... $\endgroup$ Mar 25, 2023 at 14:12

1 Answer 1


QED is a vectorlike interaction, so it preserves chirality, i.e., it couples equally to L and R-chiral fermions. To the extent the fermions are fast, so their mass is not dispositive in the particular problem, chirality is close to helicity, and is also preserved.

So, for high energy physics events, bremsstrahlung preserves helicity, but low energy physics is full of complications of that picture, as linked in a comment. But I'm not versed in the minutiae of such processes...


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