I Are EM fields clouds of photons?

w5dxp

Personally, I have trouble visualizing a flock of birds without the birds. I have the same trouble trying to visualize an EM field without the photons. More than once, I have been hounded off of the ham radio forums for mentioning the word "photon". As one ham radio operator put it, "We don't need no stinkin' photons!" Others conveyed the same feelings with more polite words like, "You've imbued photons with magic properties ('photons go where electrons cannot go') and made assumptions ('The velocity of a photon is the speed of light in the medium')...".
As Feynman implied in "QED", I picture photons entering a coaxial transmission line at the source and being transferred by the free electrons in the coax conductors to the load at the speed of light in the medium, i.e. at the velocity factor of the coax dielectric, and not necessarily in a perfectly straight line. I am told that before I am allowed to visualize an EM field as a cloud of massless photon particles, I must calculate the path of each individual photon which of course is intended to prohibit me from visualizing an EM field in that manner.
My question is: What is so wrong with visualizing EM fields as a cloud of photons?

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Nugatory

Mentor
My question is: What is so wrong with visualizing EM fields as a cloud of photons?
Because they aren't even a little bit like that.

When you hear that a photon is a "particle", the temptation to imagine that photons are to light as water molecules are to a river, or as birds are to a flock, is almost irresistible... but those analogies are terribly misleading.

A good I-level starting point for understanding the relationship between photons and EM fields might be http://www.physics.usu.edu/torre/3700_Spring_2015/What_is_a_photon.pdf

(PF mentor @bhobba first pointed me at that write up)

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w5dxp

Thanks, I already read that paper and I am not saying that photons are anything like birds. I know that photons are pure energy with no rest mass and the only reason they are called particles is that they possess momentum when traveling at the speed of light. The question remains: Do EM fields consist of photons or not?

PeterDonis

Mentor
Do EM fields consist of photons or not?
From a quantum field theory perspective, it's the other way around: "photon" is the name we give to a particular kind of state of the quantum EM field. Although even that understates the vagueness and ambiguity of the term "photon", since different textbooks and papers use the term to refer to different quantum EM field states with very different properties. A more detailed discussion of that would really require an "A" level thread. At the "I" level, the simplest answer to your question as you ask it is "no".

weirdoguy

I know that photons are pure energy
No they are not. Energy is a property of photons, not photons themselves.

Demystifier

2018 Award
My question is: What is so wrong with visualizing EM fields as a cloud of photons?
In a sense, it can be said that macroscopic EM field "contains" a large number of photons, but here "contains" does not have a usual meaning. It means that if you measure the number of photons you will probably find a lot of them, but you cannot know in advance how many. The number of photons in a macroscopic EM field is uncertain. There is even a small probability that the number of photons will be zero.

DarMM

Gold Member
There are two levels of problems with thinking of the EM field being made of photons.

The first is that for the free electromagnetic field (idealized case where it does not interact with other fields) states that correspond most closely to classical fields, coherent states, don't possess any well-defined photon number. They're a superposition of states with arbitrarily high photon count. So there's a probability to detect $n$ photons for virtually any $n$.

Secondly once the electromagnetic field is coupled to other fields in no longer possesses $n$ photon states for $n > 1$ and so even this decomposition in terms of a superposition of photon states becomes impossible.

It's just as @PeterDonis says, the quantum electromagnetic field possesses states that are like classical fields and states that are like classical relativistic lightlike particles. However the former isn't made of the latter and when I say they are "like" their classical counterparts one has to remember there are limits to the similarities (e.g. uncertainty principle, non-commuting observables)

vanhees71

Gold Member
In a sense, it can be said that macroscopic EM field "contains" a large number of photons, but here "contains" does not have a usual meaning. It means that if you measure the number of photons you will probably find a lot of them, but you cannot know in advance how many. The number of photons in a macroscopic EM field is uncertain. There is even a small probability that the number of photons will be zero.
For a corherent state, the probability to find the number $N$ is given by the Poisson distribution
$$P_{\lambda}(N)=\frac{\lambda^N}{N!} \exp(-\lambda),$$
where $\lambda$ is the average number of photons in the state, which is a measure for the intensity of the macroscopic em. field described by such a coherent state.

If you "dim down" this kind of light to $\lambda \lesssim 1$, indeed the most probable photon number is 0 :-)).

Demystifier

2018 Award
For a corherent state, the probability to find the number $N$ is given by the Poisson distribution
$$P_{\lambda}(N)=\frac{\lambda^N}{N!} \exp(-\lambda),$$
where $\lambda$ is the average number of photons in the state, which is a measure for the intensity of the macroscopic em. field described by such a coherent state.

If you "dim down" this kind of light to $\lambda \lesssim 1$, indeed the most probable photon number is 0 :-)).
Yes, but for macroscopic fields $\lambda$ is in fact much larger than 1.

Cthugha

If you "dim down" this kind of light to $\lambda \lesssim 1$, indeed the most probable photon number is 0 :-)).
Personally, I find it even more impressive that thermal light sources have a photon number distribution given by:
$$P_{\lambda}(N)=\frac{1}{\lambda+1}(\frac{\lambda}{\lambda+1})^N,$$
which always yields probabilities that decrease with n, so 0 is always the most probable photon number, irrespective of the mean photon number of the light field.

People are always somewhat surprised when I tell them that the sun is actually dark most of the time, but our eyes are way too slow to notice that. ;)

"Are EM fields clouds of photons?"

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