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I have read this question where anna v says:

The photon is an elementary particle in the standard model of particle physics. It does not have a wavelength.

What exactly is meant by the wavelength of a photon?

And this one where Emilio Pisanty says:

Photon frequency and wavelength are the same as the corresponding classical mode. If the state of the field is such that there is, whenever you look, only one excitation present, then we say the field is in a single-mode, single photon state. This photon then has a well-defined frequency (ν=ω/2π) and wavelength (λ=2π/k).

Frequency and wavelength of photons

And if you look on this site, you find numerous occasions where people talk about the wavelength of a single photon.

Relation between radio waves and photons generated by a classical current

How many wavelengths does a single photon span?

Naively, I would think that a photon does have energy, and frequency associated with it, and since frequency and wavelength are inversely related (in vacuum), even a single photon could have a wavelength too. But if I interpret it as a point particle defined in the standard model, then the meaning of wavelength is not so obvious.

Question:

  1. Does a single photon have a wavelength or not?
Árpád Szendrei
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    write what a single photon is for me – user220348 Feb 08 '21 at 00:14
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    This is a duplicate of https://physics.stackexchange.com/q/267034/ where @AnnaV has given the correct answer. – my2cts Feb 08 '21 at 00:24
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    @my2cts as you can see Emilio Pisanty explains the opposite of that answer, and many others on this site clearly talk about the wavelength of a single photon. – Árpád Szendrei Feb 08 '21 at 01:28
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    @Dale I was understanding it too that way, until I read Emilio Pisanty's answer in the other link. And I read on many occasions on this site answers where people clearly talk about the wavelength of a single photon. – Árpád Szendrei Feb 08 '21 at 01:29

1 Answers1

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This is a long comment.

In the standard model of particle physics the photon is an axiomatic point particle,has spin +/-1 only , energy=hnu , mass zero and speed the velocity of light. Wavelength is meaningless for point particles. The standard model is continuously validated as far as photon behavior goes.

After a number of discussions on similar questions I have come to the conclusion that the term "photon" as defined in particle physics, is defined differently in quantum optics, and the use of the same term to describe two different things gives rise to the confusion .

To start with, one can have many useful and mathematically consistent field theories with creation and annihilation operators operating on fields. (I first learned of field theory in a nuclear physics calculation back in 1962.)

I believe that in quantum optics, they have a field theory that describes well the classical electromagnetic wave behavior in a quantum state, as a collective photonic field on which excitations , with creation and annihilation operators, are created. Those excitations they call "photons" , creating the semantic confusion.

These "photons" are composite states of the elementary particle photons, but are fundamental in their field theory, and can have a wavelength.

In this question here and discussions, including chat, I understood the semantic difference.

In my opinion, in physics it is wrong to have the same word defining two different entities, particularly confusing elementary particle ones of the standard model with other, very useful models in physics sub-disciplines.

All physics sub-disciplenes emerge from the elementary particle standard model and, imo, terminology should be consistent and not confusing.

anna v
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  • This is a very helpful statement. I was thinking we were discussing about the same object in the linked question. Can you elaborate on what you mean by "photons" are composite states in quantum optics? I don't know if this is still within the scope of this question, but it would also be interesting to know how the two different meanings arose. Historically photons were theoretically discovered in the blackbody spectrum as quantized excitations of field modes and in the photoeffect as particle-like objects showing localized interactions with matter. – A. P. Feb 08 '21 at 09:12
  • @A.P. If the particles appearing by their creation and annihilation operators have a wavelength in space these particles cannot be elementary photons and they have to be composites of the elementary photons of QED on the lines given in this blog by Motl https://motls.blogspot.com/2011/11/how-classical-fields-particles-emerge.html . I am an experimental physicist so can only hand wave. – anna v Feb 08 '21 at 10:06
  • @A.P. this use of quantum field theory is indicative https://www.sciencedirect.com/science/article/pii/S0370157312002773 of what I means. . The particles created by the creation operators are called "quasi-particles" . – anna v Feb 08 '21 at 11:31
  • Thank you so much! – Árpád Szendrei Feb 08 '21 at 16:41
  • anna, any chance you could add this "comment" to your answer, at least if this question remains closed? – stafusa Feb 08 '21 at 18:08
  • @stafusa do you mean the answer in the duplicate? – anna v Feb 08 '21 at 18:31
  • Yes, exactly. Your remark on the inconsistent use of "photon", including the short comment on QFT are great contributions. – stafusa Feb 08 '21 at 18:33
  • @stafusa I will give a link to this answer, as otherwise it will create a confusion, not having the context of this closed question. – anna v Feb 08 '21 at 18:39
  • Nice. That's also a good way. I was afraid this question would end up deleted if not reopened, but now found out that its positive score should protect it. – stafusa Feb 08 '21 at 18:43
  • @stafusa . Duplicates are not deleted, they are just not available in searches. – anna v Feb 08 '21 at 18:47
  • Actually, the help states that "The system will automatically delete negatively-scored, non-migrated, unlocked, and unanswered questions (both open and closed, including as duplicates) that are older than 30 days." So not only the score, but also your answer prevent it from being automatically deleted. But, otherwise, they are deleted. – stafusa Feb 08 '21 at 18:51
  • @annav Thanks for linking Motl's blog. This was a really well-written summary of quantum-classical transitions for various situations. Unfortunately neither the blog nor the linked paper really answer why quantum optic's photons should be considered quasiparticles. In the case the photon propagates through matter this is widely accepted, because its energy is distributed between the EM-field and the movement and displacement of charged particles. But for a photon in vacuum I would disagree. If the answer to this is too long I can also ask that in a separate question. – A. P. Feb 08 '21 at 21:33
  • @A.P. Sorry, but I do not know whether the this photonic study works in vacuum. It is all about quantum computing and such, so it must happen in media, with fibers and conductors ..The introduction in wikipeida gives a clue that it happens in matter https://en.wikipedia.org/wiki/Photonics – anna v Feb 09 '21 at 05:26
  • @annav Photonics is what I would consider the engineering branch of (quantum) optics. And sure, manipulating light only works with charged particles around. But the quantum optical concept of a photon also works in vacuum. It's probably best to formulate a question on this rather than squeezing the discussion into 500 character comments. I will do this towards the end of the week. – A. P. Feb 09 '21 at 09:48
  • @A.P. I will read it with interest. Quantum Optics in wikipedia includes the original photon discovery observations,, thats why I went to photonics. – anna v Feb 09 '21 at 10:25
  • @annav I think this answer is relevant with respect to the way quantum opticians use the term, which may not be so far from the standard model interpretation after all: https://physics.stackexchange.com/a/122571/101770 – Wolpertinger Feb 10 '21 at 11:51
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    This answer is dead wrong. There is no "semantic confusion". The term "photon" means the same thing in quantum optics as it does in QFT. – Emilio Pisanty Feb 14 '21 at 13:09
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    The claim "in quantum optics they have a field theory that describes well the classical electromagnetic wave behavior in a quantum state, as a collective photonic field on which excitations , with creation and annihilation operators, are created", taken literally, is correct: that field theory is QED, same as in particle physics. QED perfectly allows for single photons both with and without a well-defined wavelength. This might seem unintuitive from a phenomenological particle-physics perspective, but the rigorous theory trumps the phenomenological intuition; always has and always will. – Emilio Pisanty Feb 14 '21 at 13:10
  • Anna somehow this answer relates with my question https://physics.stackexchange.com/questions/612110/is-it-possible-to-confine-a-photon-in-less-than-its-wavelength I do realise that the answer is something on these base. Indeed I've tried to formulated as to be difficult to be dismissed. I keep finding a lot of questions on this subject. – Alchimista May 08 '21 at 11:04
  • I even realised that the textbook explanation of the electron diffraction while clarifies the electron sides generates misconception on the photon side. In fact one can conclude that electrons propagate as wave by comparison to light. But the wave sketched would be that of light, not the wavefunction of a photon! Take this as a sort of discussion. You might further comment if you find a sense in my writing. – Alchimista May 08 '21 at 11:05
  • @Alchimista you may be interested in my answer here https://physics.stackexchange.com/questions/273032/what-exactly-is-a-photon/273180#273180 , this for electrons https://physics.stackexchange.com/questions/238855/is-it-wrong-to-say-that-an-electron-can-be-a-wave/238866#238866 and this for photons double slit https://physics.stackexchange.com/questions/267034/what-exactly-is-meant-by-the-wavelength-of-a-photon/267141#267141 – anna v May 08 '21 at 11:53
  • @annav above I meant electron interference in the double slit setup, otherwise my comment seems more non sense than as it is. – Alchimista May 08 '21 at 12:38
  • And thanks for the links of course. – Alchimista May 08 '21 at 12:48