# Single photon's effect on conservation of momentum?

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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"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. – Prathyush Apr 18 '13 at 21:26
@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. – Chris White Apr 18 '13 at 21:45
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. – Evan Mata Apr 29 '13 at 2:38

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.