Inspired by this question, I have slightly different question.

One way how photons are produced is through quantum oscillator transition from upper state $E_2$ to lower state $E_1$. The photon energy, and its wavelength is described as $$E_2-E_1=E_{ph}=h\nu_{ph}=\frac h \lambda_{ph}.$$

Suppose there is no "source" of photons from nothing.

  • Are there limits for the stransition steps, leading to limits for the photon wavelengths?
  • Are there othe sources of photons, that does't have such limits?
  • Is the presumption of nonexistence of photon "source" false?
  • 1
    $\begingroup$ Bands in a solid are, essentially, a set of continuum states, so there is no lower limit on a transition energy. One could argue, perhaps, that they are discrete at a small enough energy scale, but since that is perhaps ~$10^{-22}$eV I'm not going to worry about it. $\endgroup$
    – Jon Custer
    Commented Oct 6, 2016 at 19:55

2 Answers 2


As the comments mention, there is no practical lower limit on the energy. As to upper limits:

Photons can be produced by atomic transitions, and here, the drop of an electron from free space to the $1s$ orbital in a hypothetical stripped nucleus would set an upper limit. But then you can go to photons emitted from nuclear interactions (X-rays to Gamma rays) and those are much more energetic. Past that, at accelerators like Fermilab and Cern heavy particles created for an instant of time decay, often emitting "hard" gamma rays on the order of tens of GeV.

Cosmic events can yield even more energetic gammas than anything man has produced. A practical limit is set by the energy at which the gamma-gamma interaction (picture a 1-loop Feynman diagram with an electron mediating between two photons) of a hard gamma against a $3^\circ$ radiation photon becomes significant. For photons travelling intergalactic distances, this imposes a limit of about $10^{16}$ to $10^{17}$ eV. And we actually do detect events explained by such high-energy "gamma ray bursts"; in fact, some pretty much have to have originated in our own galaxy because otherwise they would have lost too much energy in collisions aginst the $3^\circ$ background.


(This is point number 3) I'm not sure about what you mean with "source of photons from nothing": you can't create energy, so for example you need to provide the energy needed to your quantum system for going to $E_2$ first. Said that, the photon wasn't "inside" the system, it was created, 'from nothing' if you want.

  1. See the comment for lower energy limit. For an upper limit, if you give too much energy to your oscillator (how much depends on the system) you'll end up destroying it (tearing apart its costituents), so any theoretical upper limit (see the question you linked) is not very relevant.

  2. Yes, for example using Compton backscattering en.wikipedia.org/wiki/Compton_scattering. In this process, low energy photons scatter on high energy charged particles (electrons are best suited) and some of them gain a lot of energy. Notice that there you don't create photons 'from nothing', but instead transform already existent photons (altough is most correct to state that you are destroying the old photon and creating a new one).

    The limit here is basically the speed your electrons can reach (so it's about particle accelerators max energy, which grows as technology improves): for backscattering (the new photon goes in opposite direction than the old one, with the biggest possible energy), given the electron $\beta = v/c$, the new photon energy $E'_\gamma$ relates to the old $E_\gamma$ as $E'_\gamma = \frac{(1 + \beta)^2}{1 - \beta^2} E_\gamma$ (the formula is approximated but very good in this setting). E.g., 1.5 GeV electrons scattering on 2.5 eV photons yelds photons up to 100 MeV, so GeV photons and beyond should be achievable today.

  • $\begingroup$ Formally all non-trivial photon scattering processes destroy the incoming photon and replace it with a different outgoing photon. Even if you ignore that and take some poetic license and describe the outgoing photon as a transformation of the incoming one "accelerate" is probably not the word you want. $\endgroup$ Commented Oct 6, 2016 at 20:49
  • $\begingroup$ Indeed. It seemed to me that the OP would be interested in the fact that he needs a pre-existent photon for making that work, and that specifying 'destroy and create' would just mess around. I will change 'accelerate' to 'transform', peraphs? $\endgroup$ Commented Oct 6, 2016 at 20:54
  • $\begingroup$ @dmckee I corrected and expanded the Compton part, hope that now it is formally correct while being clear for everyone and not too boring. Thank you for the feedback. And, aehm, I can't list point .3 well formatted without the next one becoming point 4 instead of 1: is that possible? $\endgroup$ Commented Oct 7, 2016 at 13:48

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