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I know there are already some similar questions like Amplitude of an electromagnetic wave containing a single photon but I can't get my head around it since the answers there seem not to fit very precisely to the question: What is the amplitude of a single photon? I know the energy of a photons depends strictly on its frequency and light is described by electromagnetic waves. Are (single) photons also seen as electromagnetic waves? If so, what does it amplitude then tell? Thank you!

Ben
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    The intensity tells how many photons are in a beam of light. As intensity is the energy per unit area and unit time it makes sense as average for dealing with light beam rather than single photons. The latter are counted, in practice. – Alchimista Jan 31 '18 at 14:36
  • This is a very big question, with lots of discussion, as you noted. Briefly: a photon are not seen as a wave itself, but rather as an excitation of a wave. Roughly speaking amplitude is related to the number of photons. – garyp Jan 31 '18 at 14:37
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    I see you perhaps mean if there is an electromagnetic wave for a single photon? The substance is the same but I cannot answer to this way of seeing your question. Surely duplicate in both viewpoints – Alchimista Jan 31 '18 at 14:40
  • @garyp But what is the amplitude in case of a single photon since photons are electromagnetic waves (?) – Ben Jan 31 '18 at 16:24
  • @Alchimista Yes, you're right. Can a photon be described by electric and magnetic fields/waves? – Ben Jan 31 '18 at 16:24
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    It is clear for me and as from other answers till we get to few or a single photon. In the latter case forget a wave train. The answer is tricky to me but should be here in SE. These next days I will reformulate the Q to get kind of strict yes or no. For the details we can then dig. – Alchimista Jan 31 '18 at 16:34
  • @Ben Photons themselves are not waves. – garyp Jan 31 '18 at 16:43
  • The term wave is only useful for a flux of photons? – Ben Jan 31 '18 at 16:48
  • Ben I would say that a single photon propagate as a wave EM too. At the end think of a trace at a point x. You get an oscillation whose amplitude is the smallest you can have at that frequency, ie 1 over the period (total duration of the em signal). – Alchimista Jan 31 '18 at 17:21
  • Perhaps related https://physics.stackexchange.com/questions/287394/amplitude-of-light-waves/287502#287502 – HolgerFiedler Jan 31 '18 at 20:32
  • https://physics.stackexchange.com/questions/278307/how-does-the-size-of-the-magnetic-field-vary-with-the-wavelength-of-a-photon/278325#278325 – HolgerFiedler Jan 31 '18 at 20:37
  • You should mean this https://physics.stackexchange.com/questions/174446/is-a-single-photon-also-a-maxwellian-wave – Alchimista Feb 01 '18 at 16:34
  • Look at the accepted answer here https://physics.stackexchange.com/questions/47105/amplitude-of-an-electromagnetic-wave-containing-a-single-photon – Alchimista Feb 02 '18 at 13:43
  • Maybe I'm not smart enough.. I don't see the point there. I see it's (very) related to the question but I don't see the answer..? Just a formula of a photon. – Ben Feb 02 '18 at 14:43

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The energy of light wave is not simply it's frequency component - it is both its frequency and amplitude.

But as with any wave, amplitude is a different quality than frequency, and the two are not completely interchangeable in their effects (even though a wave with high-amplitude and low-frequency may carry the same or more energy as a wave with low-amplitude and high-frequency) because of the effects of resonance (which is somewhat related to inertia).

Consider when your car bumps over a pothole - even a relatively shallow pothole might break wheels and almost knock your fillings out. That's a high-frequency, low-amplitude shock.

And yet you may drive up and down a mountainside comfortably (even though the amplitude of that movement involves orders of magnitude more energy being borne by the car through its wheels and suspension, it is so diffuse over time that it is insufficient to disrupt the physical integrity of the car or your body, which simply rides the wave rather than being shattered by it).

There is no way to substitute amplitude to achieve the same effect that frequency has - excessive amplitude would compromise the superstructure of the car rather than shocking it's components or passengers.

The same is true of atoms, that at too low a frequency of light, they may be temporarily perturbed by such light but it is insufficient to affect their integrity and cause (for example) the photoelectric effect, and at too high an amplitude, the solidity of the superstructure which is composed of the atoms would be compromised before the individual atoms were.

Steve
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  • Thanks for the vivid explanation regarding frequency and amplitude! But still the question remains: What is the amplitude of a single photon? The energy of a photon is E=hv, therefore its energy should only depend on the frequency? Where's the amplitude? – Ben Jan 31 '18 at 16:28
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    Steve this is clear. The juice of the Q is more or less "is a single photon described by a electromagnetic wave? An oscillator absorbing it it is supposed to do it by virtue of frequency. By is seem more a kind of solitonic wave, being alone. Tricky. Naively i would take the wave of 1 mole of photons and divide by Na....... – Alchimista Jan 31 '18 at 16:41
  • This answer is ok for the classical picture of EM radiation, but not the quantum mechanical picture, in which one has to carefully distinguish between energy, power, and intensity. – garyp Jan 31 '18 at 16:45
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    @Ben, the answer is that the Planck constant stands in for the "amplitude" when the photon is conceived as a particle (i.e. all photons have a constant amplitude, and the only variable about them is their frequency). Obviously, the aggregate amplitude can be increased by increasing the number of photons involved, but what this does say is that the amplitude of light must always be an integer multiple of the Planck length (i.e. it "steps"), whereas the frequency of light is infinitely variable. – Steve Jan 31 '18 at 17:00
  • I suppose what this also shows is that a light wave cannot radiate out infinitely with equal intensity at all points on a uniform wave front, since to maintain a minimum, integer amplitude, the wave front eventually has to disintegrate into moving points that have an intensity of one, and interstitial ranges between each point that have an intensity of zero. – Steve Jan 31 '18 at 17:09
  • @Steve your last is simply that it deparrs as photons. – Alchimista Jan 31 '18 at 18:16
  • @Steve up voted your comment on Planck. It is really the amplitude (given taken a bit maths). – Alchimista Jan 31 '18 at 18:18
  • Thank you! Just to make it explicit: The planck quantum is already the (fixed) amplitude? – Ben Feb 01 '18 at 12:15
  • @Ben, the amplitude is not "fixed" - it is stepwise. The Planck constant determines the stepping value (and by implication the minimum value, beneath which the amplitude will be zero). On review I muddled my own thinking somewhat by adopting the particulate conception of the photon to describe it, but it would be false and ultimately misleading to describe them strictly in that manner - it is closer to the known truth to stay with a wave-based conception of them, but accept that their amplitude is stepwise. – Steve Feb 01 '18 at 15:06
  • Sorry, I don't get that. Given a fixed frequency, is a photon forced to have always the same energy? – Ben Feb 01 '18 at 15:17
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    @Ben, there isn't such a thing as a "photon", in the sense of a little particle with certain properties. That's why I say I've fallen into error by adopting that approach in order to explain amplitude. What you really have is a wave-front, but instead of a wave-front with an even amplitude at all points (which continuously decreases as the wave propagates out), you actually have localised peaks (with an amplitude that is an integer multiple of the Planck constant), and as the wave-front spreads, these peaks do not decrease (below 1), but instead the empty space between them merely increases. – Steve Feb 01 '18 at 15:44
  • Am I making sense when I describe it like that? So if the amplitude at a point was 2, and the wave propagated some more, then at some point it may divide into two separate localised peaks of 1 (with space between), but having reached 1, a peak will never divide further, and further propagation of the wave-front will only increase the space between the peaks (it will not reduce the amplitude of the peaks that are on the wave-front, or extinguish them spontaneously). – Steve Feb 01 '18 at 15:45
  • @Steve if you were to visualize that "What you really have is a wave-front, but instead of a wave-front with an even amplitude...." description in your comment above for a single photon, what would that look like? – Dunois Aug 01 '20 at 16:02