6
$\begingroup$

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?
Improve this question
$\endgroup$
4
  • 1
    $\begingroup$ write what a single photon is for me $\endgroup$
    – user220348
    Commented Feb 8, 2021 at 0:14
  • 1
    $\begingroup$ This is a duplicate of physics.stackexchange.com/q/267034 where @AnnaV has given the correct answer. $\endgroup$
    – my2cts
    Commented Feb 8, 2021 at 0:24
  • 2
    $\begingroup$ @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. $\endgroup$
    – Árpád Szendrei
    Commented Feb 8, 2021 at 1:28
  • 1
    $\begingroup$ @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. $\endgroup$
    – Árpád Szendrei
    Commented Feb 8, 2021 at 1:29

1 Answer 1

Reset to default
2
$\begingroup$

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.

Improve this answer
$\endgroup$
23
  • $\begingroup$ 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. $\endgroup$
    – A. P.
    Commented Feb 8, 2021 at 9:12
  • $\begingroup$ @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 motls.blogspot.com/2011/11/… . I am an experimental physicist so can only hand wave. $\endgroup$
    – anna v
    Commented Feb 8, 2021 at 10:06
  • $\begingroup$ @A.P. this use of quantum field theory is indicative sciencedirect.com/science/article/pii/S0370157312002773 of what I means. . The particles created by the creation operators are called "quasi-particles" . $\endgroup$
    – anna v
    Commented Feb 8, 2021 at 11:31
  • 2
    $\begingroup$ 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. $\endgroup$
    – Emilio Pisanty
    Commented Feb 14, 2021 at 13:09
  • 2
    $\begingroup$ 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. $\endgroup$
    – Emilio Pisanty
    Commented Feb 14, 2021 at 13:10

Not the answer you're looking for? Browse other questions tagged or ask your own question.