Is the classical EM wave a single photon?

[S Beck]

Does a single photon travel in two different waves at once? If photons are particles like the Photoelectric Effect, Compton Scattering, and Blackbody radiation all suggest, how do polarizing filters block light completely? Is a particle from a radio antenna actually that large in size?

 

[PeterDonis]

@S Beck the answer to the question in the title of this thread is: Obviously not, because the concept of "photon" only applies in quantum mechanics, not classical EM, and the concept of "classical EM wave" only applies in classical EM, not quantum mechanics.

Does a single photon travel in two different waves at once?

I'm not sure what this even means, but you can probably figure out an answer from the answer to the title question that I gave above.

If photons are particles like the Photoelectric Effect, Compton Scattering, and Blackbody radiation all suggest, how do polarizing filters block light completely?

Um, by stopping the particles from going through? Why is this even a problem?

Is a particle from a radio antenna actually that large in size?

A photon does not have a "size". Also, a real radio antenna does not emit single photons; it emits huge numbers of them (if you insist on trying to describe the emission using quantum electrodynamics instead of just using the classical EM wave model).

 

[S Beck]

I'm not sure what this even means

Familiar with this picture?

Does this complex wave explain a single particle? Say, a red photon or a gamma ray photon?

However I assume you have already answered this with a no. It arises from a number of photons. I don't know what QED says about EM waves but I just ordered a copy of Feynman's book, QED.

 

[PeterDonis]

Familiar with this picture?

I've seen pictures like it. Where did you get it from and what do you think it describes?

Does this complex wave explain a single particle? Say, a red photon or a gamma ray photon?

Why do you think it's a complex wave? The E and B fields are real.

Every time I've seen a picture like this, it's describing a classical EM wave.

I assume you have already answered this with a no. It arises from a number of photons.

No, it arises from a classical EM wave.

I don't know what QED says about EM waves but I just ordered a copy of Feynman's book, QED.

That's a good layman's treatment of QED, but it's not a textbook.

 

[S Beck]

I've seen pictures like it. Where did you get it from and what do you think it describes?

An image search for EM wave. It even shows up when searching "photon."

Why do you think it's a complex wave? The E and B fields are real.

Every time I've seen a picture like this, it's describing a classical EM wave.

If this is what a single photon is like, a particle with no size, then it seems rather complex and of size. A single water molecule isn't a water wave.

No, it arises from a classical EM wave.

And this EM wave is what? A single photon? Many photons?

That's a good layman's treatment of QED, but it's not a textbook.

I wouldn't do well with a text without first studying Special Relativity and a few other large subjects. The text would be full of advanced mathematical stuff.

 

[PeterDonis]

An image search for EM wave. It even shows up when searching "photon."

Which certainly does not make this a reliable way of finding out what "photon" means. You need to be looking at textbooks or peer-reviewed papers, not trying to learn science from random image searches.

If this is what a single photon is like

It isn't. I already told you what the picture is describing: a classical EM wave. A classical EM wave is not a photon; I've already answered that too.

And this EM wave is what? A single photon? Many photons?

No and no. See above. A classical EM wave is a classical EM wave. It's what you get when you solve Maxwell's Equations, the equations of classical electrodynamics, for the case of no sources (zero charge and current density).

I wouldn't do well with a text without first studying Special Relativity and a few other large subjects. The text would be full of advanced mathematical stuff.

Then it sounds like you have a fair bit of work to do if you really want to understand the concepts you are asking about. There is no short cut.

 

[S Beck]

Which certainly does not make this a reliable way of finding out what "photon" means. You need to be looking at textbooks or peer-reviewed papers, not trying to learn science from random image searches.

Actually I have read science articles explaining photons as electromagnetic waves. I should've bookmarked them and linked them here, but this is why I asked such a question.

It isn't. I already told you what the picture is describing: a classical EM wave. A classical EM wave is not a photon; I've already answered that too.

Fair enough. This is all I asked if an EM wave is a photon. I got my answer.

Then it sounds like you have a fair bit of work to do if you really want to understand the concepts you are asking about. There is no short cut.

To ask a seemingly simple question? Don't think so. I don't need to know programming language if I want to know what the computer software is or what it does, or what it doesn't, but to program one. Likewise, I don't need to learn engineering to understand what ceiling fans are. Finally, one can get by without learning what the cryptic mathematical symbols mean if they want to understand what a quasar is, or isn't. All this machinery is central to scientific work, not explaining what a neutron star is to the public.

This is what was great about Richard Feynman, Albert Einstein, and Issac Newton. Simplification was important to them.

 

[PeterDonis]

I have read science articles explaining photons as electromagnetic waves.

Some pop science articles do try to use such language. It's not really accurate.

To ask a seemingly simple question?

Not if you just want to ask the simple question and get a simple answer. As you note, I already gave you the answer to your simple question.

But when you start going into more detail, like posting an image you got from a random Internet search, then you're not just asking a simple question.

This is what was great about Richard Feynman, Albert Einstein, and Issac Newton. Simplification was important to them.

Simplification, yes. Inaccuracy, no.

 

[king vitamin]

A plane electromagnetic wave may be understood as the classical limit of a "coherent state" of photons, the latter of which is made up of an indefinite number of photons. In particular, the number of photons contained in a coherent state obeys some probability distribution, but in the classical limit that distribution is peaked at some very large number with very little spread around it, so one can characterize the state by N, the average number of photons in the state.

Just like with the number of photons, the electric and magnetic fields are also uncertain, taking values from some probability distribution. However, if you take the expectation value of the electric and magnetic fields, you find that they are also very sharply peak about some mean values, and that mean value evolves in space and time just as in the picture you've posted. This evolution can be represented mathematically as
$$
<|\mathbf{E}(x,t)|> = \sqrt{\frac{N \hbar}{\omega \kappa}} \cos(\omega t - kx)
$$
where $\omega$ is the frequency of the photon, $k = \omega/c$, $\hbar$ is the reduced Planck's constant, and $\kappa$ stands for some mess of constants (depending on your choice of unit system) which I will not bother to figure out. I will also point out that the average energy of this state is given by the usual Planck formula formula
$$
<E> = N \hbar \omega.
$$

We now take the classical limit, which clearly requires $\hbar \rightarrow 0$. However, doing so naively results in the amplitude and energy of the wave going to zero. Instead, we must take the limit
$$
\hbar \rightarrow 0, \quad N \rightarrow \infty, \quad N \hbar \rightarrow \mathrm{constant}.
$$
The constant can be any real number. Therefore, in the classical limit we take the number of photons in the wave to be infinite. (This is as close as your question comes to having an answer.) With such a limit, we also see that energy is proportional to the square of the amplitude of the electric field. The constant which appeared is a tunable parameter in the classical theory (it just corresponds to boundary conditions when you solve Maxwell's equations).

Finally, let me stress that a single photon of a given frequency looks nothing like anything you've seen in classical physics. In particular, the electric and magnetic fields of a single photon are constant in time, and they are (as always in quantum electrodynamics) described by a probability distribution. The average fields of a photon are actually zero, but there is some finite spread in probabilities around zero. This is not something that looks like classical electromagnetism.

 

[PeterDonis]

n the classical limit that distribution is peaked at some very large number with very little spread around it, so one can characterize the state by N, the average number of photons in the state.

But note that this "N" does not have to be an integer; in fact, it most likely won't be.

a single photon of a given frequency

For clarity, what particular quantum state are you describing here?

 

[king vitamin]

But note that this "N" does not have to be an integer; in fact, it most likely won't be.

Very true, thanks for adding that!

For clarity, what particular quantum state are you describing here?

I'm not quite sure how to answer what exactly a single photon of a given frequency is without going beyond the "Basic" level (and I just noticed that's defined as high school, so my previous answer is a little on the advanced side too). So to go to the more advanced level, I am considering the exact energy eigenstates of non-interacting QED; the Hamiltonian is just a sum over oscillator modes, and I'm calling a photon of frequency $\omega$ the state which is created by $a^{\dagger}(\omega)$. In fact, my description above holds for a state with any number of photons as long as the photon-number is definite.

 

[PeterDonis]

I'm calling a photon of frequency $\omega$ the state which is created by $a^{\dagger}(\omega)$.

Ok, got it. Another way of describing this would be a Fock state with energy $\hbar \omega$.

 

[DrChinese]

Does a single photon travel in two different waves at once? If photons are particles like the Photoelectric Effect, Compton Scattering, and Blackbody radiation all suggest, how do polarizing filters block light completely? Is a particle from a radio antenna actually that large in size?

In the quantum physics forum, we get a lot of questions like yours. PeterDonis, a most respected mentor, has answered the questions well and accurately - as has King Vitamin. You can take either of two approaches in understanding QM: the popular view is basically that "a camel is a horse". What I mean be that is: the use of language is loose and precision is not important. Note that Camels and horses both have 4 legs, and are mammals. At some point, the similarities end.

PeterDonis is providing the other, more correct view: a camel is not a horse. Don't be frustrated by this approach, nor by the approach that popular articles may use. The use of the word photon means different things in different contexts, therefore the context is quite important. It should be specified to get a good answer, whether about photons or anything else quantum.

For example, you may have noticed PeterDonis' use of "average photon number" and that not being an integer. Just as the average family might have 2.5 children, you will never find a family with 2.5 kids. That's one kind of photon state. Another is a Fock state, with a specific number of photons (an integer). That's another photon state, and the properties are quite different. And so on.

The language of quantum mechanics is mathematical. Therefore, it is quite difficult to approach it without resorting to math. Shifting to descriptions using words will often lead to confusion unless the context is agreed upon. For example, it an appropriate context: a single photon could be considered the size of the entire universe. That context is not useful for most discussion, however.

Good luck!