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The treatment of EM radiation without including the topic of photons is successful. Just as the description of thermal processes works without the inclusion of the atomic level. Or the gas theory. But no one will deny the existence of atoms because of this.

Not so with EM radiation. W.E. Lamb has propagated the anti-photon (pdf) and the breaking down of physical phenomena of EM radiation to the level of photons thus seems to be unnecessary.

Hence my question. Does every phenomenon of EM radiation have to be explainable at the level of photons? And more in detail, are there other sources of EM radiation besides the emission of photons by excited subatomic particles?

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    $\begingroup$ I have given the article a quick look. To me it seems that he does not propagate the "anti-photon", he merely does not like the term and concept "photon". He states in the summary "It is high time to give up the use of the word "photon", and of a bad concept which will shortly be a century old. Radiation does not consist of particles...". I can only disagree and have not met or heard of any other physicist making such a claim. But maybe you can clarify what he means if I misunderstood. $\endgroup$ – Cream Apr 8 at 7:00
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    $\begingroup$ Sorry but it seems the question is contradictory or unclear as written: how is “Not so with EM” radiation related to “The treatment of EM radiation without including the topic of photons is successful”? $\endgroup$ – ZeroTheHero Apr 8 at 13:17
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    $\begingroup$ Concerning your question: Do you mean something like this as an example where light is not emitted by subatomic particles? Can you shake a charged object fast enough to create light? One could break it down and say that even there it is the subatomic particles, because they are what makes up the macroscopic charge. $\endgroup$ – A. P. Apr 8 at 19:43
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    $\begingroup$ HolgerFiedler, as you are still interested in an answer to your question (you placed a bounty on it, after all) could you clarify the points that have been adressed in the comments? Otherwise, every answer will contain a speculation on what you were asking. $\endgroup$ – Cream Apr 12 at 8:15
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    $\begingroup$ @HolgerFiedler The sentence is not clear. $\endgroup$ – my2cts Apr 12 at 10:54
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I see two questions. Q1. Does every phenomenon of EM radiation have to be explainable at the level of photons? All electromagnetic radiation involves photons. However, often it is enough to just consider the classical em field. This is the case when the em field intensity is high so that many photons are involved and no shot noise occurs. For low frequencies at reasonable intensity, such as in radio, this is mostly the case. Here other sources of noise dominate. For high frequencies such as EUV radiation, photon shot noise becomes observable. Q2. are there other sources of EM radiation besides the emission of photons by excited subatomic particles? In principle any charged particle can emit radiation, such as ions or even dust particles. In practice subatomic particles, mainly electrons, are nearly always involved.

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  • $\begingroup$ The question is closed, having 3 answers, an award and an accepted answer. Strange. $\endgroup$ – HolgerFiedler Apr 18 at 4:21
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Does every phenomenon of EM radiation have to be explainable at the level of photons?

No. In quantum field theory, the electromagnetic field is still a field, albeit a quantum one. We call it "quantum" because it uses non-commuting operators. It's not made of particles. Particles (photons) are one of its many possible manifestations, but most of its manifestations are nothing like particles. Trying to describe every state of the EM field in terms of photons is about as productive as trying to describe every state of the atmosphere in terms of tornadoes. Even if we can do it, we certainly don't have to do it, and we're probably better off if we don't do it.

...are there other sources of EM radiation besides the emission of photons by excited subatomic particles?

I'm not sure what this means. It could be interpreted very broadly, because all matter is made of subatomic charged particles. However, transitions between excited/deexcited states of an electron bound to a molecule are only one special case. An unbound accelerating electron also produces EM radiation, and there's no good reason to describe that radiation in terms of photons — unless we're describing an experiment that is designed to detect photons. Even then, we can only detect photons whose energy is above some threshold (depending on the detector), so we're not detecting all of the radiation that is actually being emitted. That doesn't mean we're not detecting all of the photons, though, because "all of the photons" is just plain meaningless. The state of the EM field produced by an accelerating electron does not consist of any well-defined number of photons. If that field washes over a photon detector, then sure, the detector will count photons, because that's all it can do. If that field washes over a field-amplitude-measuring device, then that device will report the field's amplitude (which might not be very meaningful if the amplitude is very small, because of the relevant "uncertainty principle"). Those are two different observables, and they don't commute with each other, so we cannot interpret the measurement outcomes as merely revealing pre-existing properties of the EM field itself. Quantum theory doesn't work that way, and neither does the real world.

There is one good cure for the confused-about-photons ailment, and that's to study the math. The quantum theory of a single harmonic oscillator is a good place to start. It has "photons" (discrete quanta of energy), but most of its states don't have any well-defined number of photons. It also has states with relatively well-defined "field amplitude" (like the so-called coherent states), for which the number of photons is not well-defined at all. Any state with a relatively well-defined large field amplitude is necessarily practically orthogonal to every state with a well-defined number of photons. The field-amplitude observable does not commute with the number-of-photons observable, and most states don't have well-defined values of either.

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    $\begingroup$ Superb answer. I wish more people in schools and universities taught photons with this level of understanding. $\endgroup$ – Ján Lalinský Apr 12 at 0:56
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    $\begingroup$ A calm weather is just a bunch of virtual tornadoes popping in and out of existence $\endgroup$ – Ruslan Apr 13 at 8:16
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No. There are other sources of electromagnetic radiation :

  1. Antennas, for instance: The simplest embodiment is perhaps the one used in the early Marconi experiments: Appling pulses of current from a battery to a wound (insulated) copper coil

Marconi’s device

More generally, according to Maxwell’s equations, setting any electrical charged body in motion generates an electromagnetic field.

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