A long-wavelength, e.g. radio frequencies, of say, 1 km, has a period lasting about 1/300000th of a second.

So for an imaginary fixed observer watching the incoming wave, it takes some time to go from one peak to the next (not saying this measurement is actually feasible).

Also, I would believe the generation itself of the wave was not instantaneous but rather took at least the duration of one cycle.

In other terms, I would believe, with my limited knowledge, that generating a very low energy photon is a very slow process.

And building on that, I am wondering how this generation would always lead to a perfect photon which' waveform is defined only by one energy, one frequency, and not a more complex waveform (which would introduce other frequencies, Fourier style).

I mean, if the electrons are accelerated in complex ways instead of a perfectly periodic motion, does that produce a spectrum of photons with unique frequencies each?

If the very long wavelength is distorted during its creation, how can energy still be quantized and produce one frequency per photon?


3 Answers 3


The essence of this is owing to the time--frequency relationship that is involved in Fourier analysis.

Suppose we have a process that can in principle put energy into the electromagnetic field. In photon language, this is a process that can create photons. If the process can move the field from its ground state to a state corresponding to a single photon of frequency $\omega$, then it is supplying energy $\hbar \omega$. Such a process has some sort of timescale associated with it, such as the time since a current started to oscillate or something like that. But by Fourier analysis, any process that has only lasted a duration $\tau$ cannot be monochromatic; it must have a range of frequencies $$ \Delta \omega \ge 1 / \tau. $$ This means the process cannot guarantee to only create a photon of one frequency. It may create photons of a range of frequencies. As $\tau$ gets larger this range can in principle get smaller. So you are basically correct to say that if we want a process which only creates low-energy photons, then the process must be slow. The same can be said at the detector end. In order that a detector can determine that the frequency was indeed $\omega$ to within some precision $\Delta \omega$, it will have to accumulate the signal for a time of at least $\tau \ge 1 / \Delta \omega$.

Now the question was, how can such a slow process result in a perfect photon? The answer is that, in general, it will not result in a perfect photon, but in a quantum superposition of perfect photons. For the term "photon" simply refers to a state which is by definition a state of perfectly precise energy. No process can be guaranteed ahead of time to realise that perfection, but we can analyse whatever does happen by describing it as a sum of photons with different weights, just like in Fourier analysis.

Finally, the state of the electromagnetic field having a precise number of photons, such as 1 photon, is not quite as envisaged in the question. Such a state does not have a phase of oscillation. The situation is similar to the first excited state of a quantum harmonic oscillator: the oscillator is not "moving too and fro". To get an oscillation having a phase you need to superpose two or more energy states. The classical limit corresponds to superposition of a large number of such states, where both the mean number of photons is large, and so is the variance of the distribution of the number of photons.


You are simply assuming that radiation (production of "light" by accelerating charges) produces point-like particles. This is not the case. When we have this sort of radiation the picture that best describes the output is a wave, having said that, this implies there is no localization, the wave itself is an EM wave as described by Maxwell´s equations and follows the geometry of the "antenna" that produced it, the wavefront itself is not necessarily spherical. Under this conditions it is more useful to think of a continuous process, that is if you accelerate a charge a little you get a little bit of radiation, whose power and other features can be computed. For an observer it will not look as a single wavelength wave (or even a periodic wave), specially if the acceleration was too short, instead he/she will see a superposition of waves corresponding to the specific shape of the wave. (All of this can be found in any standard course or book on Classical electrodynamics)

The phenomena of quantum transitions is perhaps what you had in mind, which does involve single photons (the energy scales are very different) of very specific wavelengths which correspond to the energy gap of the transition. This is more accurately described by thinking of light as a discrete package, photon. (More on this pertains courses/books concerning Quantum mechanics)

Current physics tries to view both pictures above by modelling the light as a field, whose modes (delocalized excitations, with precise wavelengths) as photons, this leads to the field of quantum field theory (QFT).

  • $\begingroup$ Wait a minute. This is the first time someone mentions QFT as a means to unify Maxwell and QM. I thought Maxwell was a limit of QM. If I want to use the photon picture only, how can I describe these deformed waves? $\endgroup$
    – Winston
    Mar 22, 2021 at 13:34
  • 1
    $\begingroup$ That is mistaken again. Quantum mechanics on its own does not say anything about Maxwells equations or quantization of electromagnetism. It comes later in the context of QFT and QED. Maxwell alone is a field theory, it is not quantized. One needs to move on to QFT to understand the details, as I tried explaining it boils down to which phenomena in which regime you are describing $\endgroup$
    – ohneVal
    Mar 22, 2021 at 13:52

It is simply wrong that during one half-period of the antenna generator, the accelerated electrons emit only one photon. They emit photons x times and the resulting radio wave is nothing more than the periodic change in the intensity of the emitted radiation.

The reason why radio waves are perceptible is that the polarisation of the photons is able to induce an electric current in the conductor. A single photon is converted into heat or is reflected from the surface of the receiver. A bunch of polarised photons is a concerted action on the conductor. Many electrons at once are under the influence of the incoming energy and begin to move synchronously. A single photon only slightly disturbs the receiving electron and the surrounding area weakens it.

  • $\begingroup$ Are you saying that the frequency/energy of single photons is actually irrelevant in radio wave communication? $\endgroup$
    – Winston
    Mar 23, 2021 at 7:25
  • $\begingroup$ Exocytosis, exactly that is the statement. $\endgroup$ Mar 23, 2021 at 16:54
  • $\begingroup$ I am sorry if this makes you repeat it one more time, but why is a certain range of EM frequencies are called radio waves if any photon energy can be used for, say, MHz radio transmission? $\endgroup$
    – Winston
    Mar 25, 2021 at 7:48
  • $\begingroup$ The number of photons emitted is entirely irrelevant to the fact that the photons produced by an antenna of course have "the frequency" of the radio wave within the natural line broadening as explained in Andrew's answer. This is misleading at best. $\endgroup$
    – ACuriousMind
    Mar 25, 2021 at 14:37
  • $\begingroup$ So you say that, if one electron moves to and fro in one second, not one ultra-long-wavelength-photon (c meter) is produced, but rather many (x)? $\endgroup$ Mar 26, 2021 at 10:53

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