I don't think a previous question How are photons created? answers my question, which is.

As far as I know there are at least three ways to create photons:

  1. Electrons jumping between levels in an atom.
  2. Charged particles accelerating, even a proton by itself in intergalactic space with no electron in sight.
  3. Matter-antimatter annihilation, including proton-antiproton annihilation again with no electron in sight.

Is there a unified explanation for these three seemingly distinct cases?

Secondly, is there evidence that photons are quantized in all three cases? From what I read, Planck's quantization hypothesis was based on black body radiation, which only applies to scenario 1 above.


P.S. One main reason for my question is the role of protons and oher particles. QED and most of the literature only talks about electrons.

  • $\begingroup$ I'm not sure scenario 1 accounts for Blackbody radiation. Electron transitions in atoms generate the discrete emission spectrum. Blackbody is a thermal radiation, which does not involve electron transitions. $\endgroup$ Jul 24, 2021 at 1:45
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    $\begingroup$ What would you even mean by a non-quantized photon? The photon is the quanta of EM radiation. What is a non-quantized quanta? $\endgroup$
    – nasu
    Jul 24, 2021 at 1:59
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    $\begingroup$ Not sure why you think QED discusses only electrons, the QED Lagrangian includes ALL the charged fermion fields, and through loop effects you can even add in the uncharged neutrinos to the picture. $\endgroup$
    – Triatticus
    Jul 24, 2021 at 2:18
  • $\begingroup$ @LucasBaldo what is the mechanism for thermal radiation? does it not involve electron transitions at the fundamental level? would a gas of neutrons emit black body radiation? $\endgroup$
    – Jay
    Jul 24, 2021 at 6:34
  • $\begingroup$ @nasu at the simplest level, consider two photons with energy hv and h(v+1). There is no energy value possible between these two values. If someone were to find a situation where, for example, a charged particle lost momentum corresponding to such a forbidden energy value, that would be a problem. $\endgroup$
    – Jay
    Jul 24, 2021 at 6:42

1 Answer 1


A quantum of a quantity means that it comes in countable steps. It does not mean that the value given is not continuous. A number of bricks describes how many there are. It does not mean that the weight of a brick cannot be any number within the continuum of weight.

Is there a unified explanation for these three seemingly distinct cases?

All can be modeled by quantum field theory, and at the limit of h_bar compatible with zero (macroscopic), Maxwell's equations.

Secondly, is there evidence that photons are quantized in all three cases?

Photons are countable in all three cases, their their energy is $hν$ where $ν$ is the frequency of the electromagnetic wave they will belong to when in large numbers.

In transitions between bound states they have they come in quanta defined by the energy differences between states.

Particle interactions as annihilations and decays, are modeled by the QFT of standard model.

In two body annihilation to two photons just the kinematics and conservation laws are enough to define the energy of the photons.

  • $\begingroup$ Thanks but it seems my original wording was sloppy. What I meant was non-quantized energy. I have explained further in the comments to the original question. Basically, I can see that the energy difference between electron levels in an atom can be quantized, so any emission by electrons changing levels is quantized. But is there any such restriction on an accelerating proton out by itself in the middle of nowhere? Why must it only emit photons with quantized energy values? $\endgroup$
    – Jay
    Jul 24, 2021 at 6:59
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    $\begingroup$ "Why must it only emit photons with quantized energy values? " The bremsstrahlung can have any energy compatible with the charged particle energy and the probability distribution, certainly not quantized but countable number of photons. Photons can have any energy compatible with energy and momentum conservation. en.wikipedia.org/wiki/Bremsstrahlung $\endgroup$
    – anna v
    Jul 24, 2021 at 8:59
  • $\begingroup$ OK, so if I understand correctly, the energy 'content' of a matter particle can be any continuous value, but it can only be emitted in multiples of 'h' (since a photon's energy must satisfy E = hv and v is an integer). It's still not clear what constrains a proton to do so unless it involves some property of quarks. $\endgroup$
    – Jay
    Jul 24, 2021 at 11:23
  • $\begingroup$ On second thought, if energy is always transported in multiples of h (because E = hv), including the creation of matter-antimatter from a gamma ray photon, then I don't see how the energy content of a matter particle can ever by anything other than a multiple of h. $\endgroup$
    – Jay
    Jul 24, 2021 at 11:45
  • $\begingroup$ h is a very small irrational number though, so it is really a continuum . The constraint is experimental . see planck's radiation formula which fits the black body data hyperphysics.phy-astr.gsu.edu/hbase/mod6.html $\endgroup$
    – anna v
    Jul 24, 2021 at 12:07

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