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Since university I've always accepted without question, that electromagnetic radiation consists of photons; but is it true?

I believe it was Einstein who demonstrated that light comes in small packets, 'photons', and since light is EM radiation, it is tempting to conslude that all such radiation consists of photons. On the other hand, most (all?) light originates from discrete processes in atoms, so perhaps that is why it comes in discrete packets?

To take an extreme example, if I charge something, say a metal sphere, and then 'wave it about', I will generate EM radiation - will that radiation consist of photons?

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  • $\begingroup$ I mean, purely because it's warm, it will be emitting infrared photons, but I think this is orthogonal to what you are asking. $\endgroup$ – IronEagle Jan 10 at 18:33
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    $\begingroup$ EM radioation is always photons, there are other types of radioation that are not EM and therefore not made of photons. E.g. neutrino radioation $\endgroup$ – Christian Jan 10 at 21:48
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    $\begingroup$ Could you do some basic research, then re-phrase that Question? How would your being tempted make it reasonable to conclude that since light is EM, all such radiation consists of photons… or that if light originates from discrete processes, that's why it comes in discrete packets? If you charge a metal sphere and wave it about, what does your prior reading say the radiation generated will consist of? $\endgroup$ – Robbie Goodwin Jan 10 at 22:23
  • $\begingroup$ “light originates from discrete processes in atoms, so perhaps that is why it comes in discrete packets?” Or the processes in the atoms are discrete, because they can change energy states only by emitting discrete photons. $\endgroup$ – Holger Jan 11 at 9:04
  • $\begingroup$ @RobbieGoodwin I think what you are saying is, that I should go and look up the textbook answer and be grateful :-) I've done that - I did study maths and physics at university - but to put it a bit colloquially, I'm a little tired of variations of 'Because of Quantum'. QM, despite its many successes, remains unintuitive and unapproachable; one has to simply accept it on faith and live with not understanding, and I would like to feel that I understand. Hence my slightly odd-ball question - it is also meant to be a gentle provocation, as well as being seious. $\endgroup$ – j4nd3r53n Jan 11 at 14:09
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Generally speaking, all electromagnetic radiation can be reduced to being a combination of "photons."

When the light can be represented as a clean wave, $\mathbf{E}(t) = \mathbf{E}_0 \cos{\omega t}$, as we expect classically, then the light is a "coherent state" which has a Poissonian distribution in photon number (and technically as @aekmr points out, for it to be a coherent state the "noise", ie fluctuations, of the electric field needs to be normally distributed without changing as it propagates). So if you put a detector that could measure exactly the total number of photons for this coherent state, you would find that the probability of measuring a certain photon number is:

$$P(n)=\frac{\bar{n}^{n} e^{-\bar{n}}}{n !}$$

with $\bar{n}$ being the mean photon number (which doesn't need to be an integer). This visually looks like this:

enter image description here

If you notice it is a bit counter intuitive, basically any classical light is uncertain in its photon number. It's possible to get more "sharp" distributions with more photon number certainty called "sub-Poissonian", but these require special quantum effects.

I should also point out that (mentioned by @aekmr) classically you can also get "thermal light" (in the class called "super-poissonian" statistics) which has a different fatter distribution in photon number - and it looks like the one labeled here as a Bose-Einstein distribution:

enter image description here

To take an extreme example, if I charge something, say a metal sphere, and then 'wave it about', I will generate EM radiation - will that radiation consist of photons?

It certainly will radiate "photons," and in principle we could measure how many photons it has either by using (a lot of) number resolving detectors or by detecting the electric field and looking at the statistics of its values.

(Also, I'm pretty sure that these photons will be "coherent states," but I'm not exactly sure how to prove it though.)

EDIT:

Maybe this answer is a bit too simple, as @aekmr suggests, so I'll try to answer some more technical points about it.

Generally speaking, the quantum state of light of an electric field can be broken down into the "Fock state basis" (which is basically the basis that tells you how many photons your state is in). For example, your state might be a superposition of being in vacuum, one photon, and two photons, (like this $|\psi\rangle = c_1|0\rangle + c_2|1\rangle +c_3|2\rangle$).

Combinations of these superpositions produce both the average value of the electric field that you will measure and the statistics that you observe (so if you were to take the same state and measure the electric field multiple times, you'll observe a particular distribution for the values of the electric field). And also as a technical point, sometimes, like in the case of thermal light, the photons are not actually in "superpositions" but are just a mix of different photons that are entering your detector with certain probabilities.

In the case where you get a clean sine wave and when the distribution (in the electric field) as it propagates looks is a gaussian distribution, then you have a coherent state as your underlying state. As a visualization, this is what a coherent state looks like in time:

enter image description here

Then you can use the strength of the amplitude relative to the noise to infer the mean photon number for this coherent state.

In the case where your light does not follow this expected classical behavior, then you might have an unusual superposition of photons - and these can give you all sorts of funky statistics in the strength of the electric field. For instance while a single photon is just a single "click" in a "single-photon-detector" - if you were to look at the strength of the electric field for a photon, you would see that it has an unusual statistical distribution:

enter image description here

Which strangely shows that a single photon's average E-field value is zero! I should add here that while it seems trivial to make photons from your example, creating a perfect single photon is very challenging and is an active field of research.

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  • $\begingroup$ Is the mean photon number always an integer? $\endgroup$ – PyRulez Jan 10 at 19:08
  • $\begingroup$ The mean photon number can be any real number (and does not have to be an integer). It's the average photon number that is expected to be measured. $\endgroup$ – Steven Sagona Jan 10 at 19:37
  • $\begingroup$ I'm sorry but there are at least two points wrong in this answer: 1) The monochromaticity of the radiation ("When the light can be represented as a clean wave...") has nothing to do with its photon number distribution. You can have a "clean wave" with a non-Poissonian photon number distribution, just like you can have a wave-packet with a strange shape in which the number of photons is Poisson-distributed. 2) In a thermal state, the number of photons is not Poisson-distributed. The distribution is exponential: $P(n) = \exp(-\beta\hbar\omega n)/Z$. $\endgroup$ – aekmr Jan 10 at 23:08
  • $\begingroup$ @aekmr, thanks for the feedback. I always thought thermal states were just coherent states but in a mixed state form where the coherence was lost. I removed that part entirely. Additionally, I added a technical note about your first point. If an E-field has a clean sine wave AND it doesn't have statistical noise (in the strength of the electric field) vary as its phase changes, then it is a coherent state. I left the second part out because I think it's a more nuanced point that is highly technical. $\endgroup$ – Steven Sagona Jan 10 at 23:32
  • $\begingroup$ @StevenSagona Glad I could help to improve your answer :). $\endgroup$ – aekmr Jan 11 at 8:05
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Photons are what light is and they can have frequencies anywhere from radio to gamma. Billions of coherent photons resemble a wave.

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    $\begingroup$ I understand what you are saying, but I think we should be a little bit careful with making absolute statements like that. According to our best theory, light may be explained this way, but even the best theory is limited in scope and may one day be replaced. $\endgroup$ – j4nd3r53n Jan 10 at 9:40
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    $\begingroup$ @j4nd3r53n there are many absolutely statements in science that have not been proven or completely explained $\endgroup$ – Bill Alsept Jan 10 at 9:52
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TL;DR: Yes: every kind of electromagnetic radiation is made by photons.

In Quantum Mechanic and in Quantum Field Theory forces, such as the electric force are carried through space by quantum particles called bosons. (The adjective quantum in the term quantum particle is not a decorative one, it is important: in fact quantum particles are almost nothing like classical particles and you should keep it in mind; picturing photons as little balls could rapidly lead you to mistakes). An electromagnetic wave is nothing more that the manifestation of the electric and magnetic force acting at a distance, so our theory states that it is made up by bosons, specifically photons since we are dealing with electromagnetism.

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The question of whether or not a field “consists” of its corresponding particles is perhaps more interpretational than physical. That is, sure, we observe that whenever light is emitted, the emitter loses a discrete amount of energy, momentum, and angular momentum, and the absorber gains a discrete amount as well (always, regardless of the mechanism of emission/absorption). But since light seems to behave like a wave “in transit”, how do we know that it really is photons? Could the “reality” be that light is simply a wave subject to a law of quantum emission/absorption events? If so, would you say that the light “consists” of photons? Is there any meaningful difference??

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  • $\begingroup$ The name that we have given to these discrete amounts that you mention is "photon". $\endgroup$ – fishinear Jan 11 at 21:48
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Yes, all EM radiation consists of "photons", ex definition, because whatever EM radiation consists of, that's what the word "photon" refers to.

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