When your looking at basic Compton theory you find that if you shoot a stream of photons at a particle (usually atoms or electrons), then you have the basic laws of conservation of momentum. The photon acts like a particle, like a "billiard ball." The photon interacts with the said electron and the photon goes off in a new path described by $h/\lambda_2$ ($h$ being Planck's constant and $\lambda_1$ being the original wavelength of the photon). This wavelength is increased. Using a basic vector diagram with $h/\lambda_1$, $h/\lambda_2$ and $m\vec{v}$ for the particle you get basic conservation of momentum.

My question is basically what happens when you are only using a single photon? I'm not aware of any experiments done with a single photon, so far this concept (Compton experiments) have only been done with multiple photons. The reason a single photon is important is because the energy of a photon is inversely proportional to the wavelength. The problem that I have with this is that $E = h\nu$. This is how we get our inverse wavelength in the formula (wavelength and frequency being inversely proportional). With a single photon you have no frequency, since you only have 1 event/photon. Thus how can you place Planck's constant over $\lambda$ to represent the energy of the photon, since there is no frequency?

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    $\begingroup$ "With a single photon you have no frequency, since you only have 1 event/photon" this sentence does not make any sense, The frequency of the photon can be inferred from the preparation procedure used,(example if a photon was produced using an excited atom,) I think you are confusing between frequency as in number of samples taken and frequency as in number of periods in a unit interval of time. $\endgroup$
    – Prathyush
    Apr 18, 2013 at 21:26
  • $\begingroup$ @Evan I never remember how to type UTF-8 symbols an any OS - if needed I usually just copy from, say, the wikipedia article. But in any event, we support Latex-style markup on this site, which is preferable to inserting special characters. A nearly comprehensive list of things you can do with it can be found here. $\endgroup$
    – user10851
    Apr 18, 2013 at 21:45
  • $\begingroup$ You can tell the frequency of a single photon from where your getting that photon, whichever energy level etc. But I'm trying to reason what occurs if you have a single photon of an unknown frequency. This means you'd have to take frequency as number of periods in a unit interval of time, but if its only a single period then how do you measure frequency? Since you can't measure the distance between peaks as there is no 2nd peak to measure against. As to number of samples taken I'm referring to that as I'm essentially wondering about frequency when you have a single sample, a specific instance. $\endgroup$
    – Evan Mata
    Apr 29, 2013 at 2:38

1 Answer 1


The frequency of light is not a property of many photons but a property of a single photon. (This is also strictly inaccurate, since we should think about a field with ripples in it; we then call a little clump of waves a photon.)

Anyway, let's imagine a photon/wavepacket as having a typical wavelength $\lambda$ given by the peak-peak spacing in this diagram:


This is a single photon, with the colour given by wavelength $\lambda$. Then this whizzes past you at the speed of light, $c$. But then you see the peaks go past at a rate of 1 per $\lambda/c$. Thus there is a sensible definition of frequency as $f=c/\lambda$. Then we let $E=hf$ and carry on happily.

This is all essentially an example of wave-particle duality. If you want to try to understand this better, try reading about the photoelectric effect. The key idea is that there is a wave-like phenomena (light) which we perceive as interacting with the rest of the world in discrete, particle-like events (photons).

  • $\begingroup$ Unfortunatly I don't think I'm being very clear with my question. Your response (Sharkos) is when you have multiple photons all adding up to a single larger wavepacket. In that instance you have frequency because you have other photons to compare to. I'm wondering what occurs if you have just a single photon making up the packet. You have no other peaks to compare with, thus you can't measure frequency. This is my question, what occurs if you have a single discreet photon that has no other peaks. Without other peaks you can't measure the rate of the peaks passing you by. You have no frequency. $\endgroup$
    – Evan Mata
    Apr 29, 2013 at 2:31
  • $\begingroup$ I did draw a single photon actually. This, in simple QM, is the wavefunction for one photon, describing (e.g.) the probability of it having various possible momenta. It is a superposition of pure momentum states for one photon. If you want a pure, non-superposition momentum state then you get a plane wave with infinitely many evenly spaced peaks and a unique well-defined frequency. $\endgroup$ Apr 29, 2013 at 8:21
  • $\begingroup$ For the wavepacket case, when it hits something and its momentum is 'measured' it collapses into a pure momentum state, and that is the amount exchanged in the collision $\endgroup$ Apr 29, 2013 at 8:23

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