# Are photons really particles or just a misconception?

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1. Sep 1, 2015

### jorgdv

I'll explain my question. At first, light was described as electromagnetic waves, until Einstein proposed the photoelectric effect and thus creating the concept of photon, a particle of light with momentum and energy, but no mass. It could explain why the atoms become excited at discrete levels of energy, and why they emit only in specific frequencies knowing basics of quantum mechanics.

Nevertheless, quantum mechanics could explain successfully why electrons get excited to the next level in a resonance-like phenomenon, viewing light as a periodic potential in time and the electron as a wave function, without the need for light particles. If we observe the electron and we find it in the next eigenfunction of the unperturbed Hamiltonian, we can say that the electron has absorbed one photon of the electromagnetic field. However, that's a quantum of energy, an energy unit, not a particle.

So after all this, and without entering quantum field theory, there goes my question: If the quantum excitation of an electron can be explained treating light as a classical field, why do we need photons? Do they really exist as particles or it is just a misconception?

Thank you

2. Sep 1, 2015

### phinds

Photons are not particles (in the classical sense of that word) and they are not waves (in the classical sense of that word). They are quantum objects. If you measure for wave characteristics, you can see wave characteristics but photons are not waves. If you measure for particle characteristics, you can see particle characteristics but photons are not particles. Photons are quantum objects so worrying about whether they are particles or waves is a misguided effort.

3. Sep 1, 2015

### Staff: Mentor

Photons really are particles, but in this context the word "particle" means something quite different than it does in general English usage - we are not talking about something like a tiny little billiard ball, a solid object that just happens to be very small. It's mostly a historical accident that the word "particle" is used to describe photons and other quantum objects.

Much of the behavior of light and other electromagnetic interaction can be explained by treating light as a classical (actually, we should say "non-quantum" as you do need special relativity) electromagnetic wave - but not everything. Indeed, the classical wave model can be derived from quantum electrodynamics, the same way that special relativity reduces to Newtonian mechanics as long as you're only working with low speeds.

Last edited: Sep 1, 2015
4. Sep 1, 2015

### lightarrow

A plane, monochromatic light wave having low intensity hits a screen with a small hole, it diffracts and then it's detected, point after point, on another screen.
If light is a classical field, why do we detect single points on the second screen?

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lightarrow

5. Sep 1, 2015

6. Sep 1, 2015

### jorgdv

Thank you for your answer. So you are saying that it is not whether the classical description or quantum description are the good ones, but both are just an approximation of quantum field theory, making it impossible to proceed without it in this matter? In the description I put "without entering quantum field theory" because what seemed to be the most fundamental interaction of photons with matter, the excitation of electrons, seemed to have a very coherent description with ordinary quantum mechanics and classical electromagnetic fields (perturbation theory with periodic potentials), without the need of Einstein's paradigm to view the photons as particles.

I've read most of Zee's book about QFT and I was fascinated. However, in the view of QFT photons are not particles, but disturbances in the quantum field that propagate through it. So if neither QFT nor QM need to see photons as particles to explain things such as electron excitation, being much more related to fields, why are they considered as such?

Well, maybe I'm totally wrong here and this is just speculation. If I'm correct a photoelectric detector detects light by measuring the electrons excited by the "photons" that make it. Well, light could be a classical field entering the quantum Hamiltonian as a periodic perturbation. Then, when you measure it, you could measure discrete excitations produced by continuous fields. I mean, if you have the system at the fundamental eigenstate $\psi_0$ of the Hamiltonian $\hat{H}_0$, then in a simple approximation, a small electromagnetic field with electric field pointing in the $z$ direction and magnetic field pointing in the $x$ direction could be treated as a perturbation $\hat{W}=(|e|E_0\hat{z}+\mu_B B_0 \hat{\sigma}_x)sin(\omega t)$, so that the total hamiltonian after the field "hits" the system would be $\hat{H}(t)=\hat{H}_0+\hat{W}(t)$. After a while the state of the system would be a linear combination of eigenstates $\psi(T)=\sum_i c_i \psi_i$, where the probabilities $|c_i|^2$ could be approximated with Fermi's golden rule. Then each "point" could just be the measurement of an excited electron produced by a continuous field in the previous fashion. Please feel free to correct any mistakes that I could have made.

Thanks to everyone

7. Sep 1, 2015

### Staff: Mentor

Such a description does indeed work - in fact, ZapperZ just published an Insights article touching on this point: https://www.physicsforums.com/insights/violating-einsteins-photoelectric-effect-model/

When you see someone referring to photons as particles, there are two possibilities. One is that you're reading a popularization written by an author who does not care overmuch about accuracy. The other is that they're using the word "particle" to MEAN "quantized excitation of the field", in which case my earlier answer applies - in this context the word "particle" means something quite different than it does in general English usage.

In any case, if you're feeling that the word "photon" is overused and too often understood by laypeople to mean something more akin to a little tiny "bullet of light" than a quantized excitation of the field.... you're right.

8. Sep 1, 2015

### jerromyjon

You are missing the classic-quantum distinction, you can know how big of a "chunk" you will have and exactly that size "chunk" or any other word for particle would mean the same thing, where it winds up is quantum and strange.

9. Sep 1, 2015

### lightarrow

I know nothing of QFT so you certainly have more knowledge than me, but it's not clear to me how your model explain the fact that if you send one photon only, the detecting screen will register one only "click". How does it "know" that it have to obey to energy conservation?

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lightarrow

10. Sep 1, 2015

### Avodyne

Professional physicists have been using the word "particle" to MEAN "quantized excitation of the field" for many decades. The photon is certainly considered to be a particle. For example, it appears on the Particle Listings page of the Particle Data Group, which is recognized worldwide as the authoritative source for information on particle physics:
http://pdg.lbl.gov/2014/listings/contents_listings.html

As you say, this usage of the word particle does not necessarily coincide with common English usage (e.g., "particle of dirt"), but that is true of many other words such as "force", "energy", "work", etc etc.

11. Sep 1, 2015

### microsansfil

Could there not have been confused, in the common English usage, with the term of corpuscle ?

Patrick

12. Sep 2, 2015

### lightarrow

Here I was criptic because jorgdv was talking of classical em field interacting with the screen.
Let's say something as follows.
An excited atom emits an em pulse of (average, but we can assume it's well defined) frequency $$\nu$$and then comes back to the fundamental level. The atom's recoil allow us to measure momentum and energy of the pulse and it's found the first and the second are, respectively: $$\hbar\nu/c;$$ $$\hbar \nu$$ Later this em pulse interacts with a distant screen. Why it interacts with one point only of the screen, all the times we repeat the experiment?

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lightarrow

Last edited: Sep 2, 2015
13. Sep 8, 2015

### jorgdv

Sorry for the delay.

Well that's the thing, it's not QFT but ordinary Quantum Mechanics. In this context there's no need for "one photon", you have classical EM radiation interacting with a quantum system. Then, you measure consequences of the quantum system, such as the wavefunction of an excited electron. The conservation of energy is a good question, since here the EM field enters the quantum system via potential energy of the electron. So if you measure an excited electron, then that would mean that the EM wave has lost one photon of energy. But I guess that means that the wave should be in a way entangled with the electron wavefunction, even being modelled as classical potential in the Hamiltonian, so what's happening here?

Well, in the context of our discussion you would have an EM pulse with energy $h \nu$. Due to the conservation of energy, you can measure only one excited electron of the screen, because it has not enough energy to excite more. However lots of points in the screen would be excited and not excited simultaneously until you measure.

The thing that could not be explained with classical EM radiation together with ordinary QM is photon momentum and scattering, I guess in this point we would need QFT.

Thank you everyone.

14. Sep 9, 2015

### vanhees71

First, I'm self-advertising my Insights article concerning one aspect of your question:

https://www.physicsforums.com/insights/sins-physics-didactics/

To answer the question in the title of this thread: Photons are neither particles in any common sense of this word nor are they a misconception. The only way to get a correct idea about what a photon is, is to study relativistic quantum field theory and (fortunately the most simple part) of the Standard model of elementary particle physics, Quantumelectrodynamics (QED). The most common mistakes in starting to teach quantum theory with photons are to assume

(a) photons are little bullet-like particles, knocking out electrons from an atom ("explaining" the photoelectric effect). As detailed in my Insights article, in modern terms the full content of Einstein's 1905 article on the photoelectric effect is now understood by using time-dependent first-order perturbation theory with classical electromagnetic waves as the perturbation and any model for a bound electron in quantum theory. This shows that this heuristics is misleading and, in addition, cannot be established within modern quantum theory. There's not even a position observable for a photon in the strict sense.

(b) Very-low-intensity laser light represent photons. In the correct sense of QED, photons are (asymptotically) free excitations of the electromagnetic quantum field with a determined photon number. Particularly a single photon is represented by a one-photon Fock state. Very-low-intensity laser light is not represented by such a state but by a very-low-intensity coherent state. It consists of a superposition of all photon-number Fock states. At low intensity the largest component is the vacuum. Note that the photon number of a coherent state is Poisson distributed.

(c) You come pretty far with classical electromagnetic waves. The most simple case, where you need quantum theory is the case of thermal photon radiation, i.e., to derive the black-body spectrum. This can be done easily with a naive notion of photons, which is however correct also in the view of modern QED. You just use the occupation-number basis for free massless bosons and take into account that each momentum eigenmode comes in two polarization states. You end up with Bose statistics for the photon-phase space distribution, and thus immideately at Planck's radiation law. This can of course be derived also from QED with a bit of work, solving the problems due to gauge invariance, which is of course the key issue of modern QFT of the Standard Mode.

15. Sep 9, 2015

### lightarrow

What does it mean?
Are you referring to Compton effect?

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lightarrow

16. Sep 9, 2015

### vanhees71

Classical electromagnetic waves carry energy, momentum, and angular momentum. You don't need photons for that either. The Compton effect, i.e., elastic scattering of photons on charged particles, is however indeed an example, where you need the photon picture, i.e., QED.

17. Sep 9, 2015

### DrChinese

As already mentioned, if you equate "photon" with QFT excitation: you get the right answer and can stop there. But I think this will address your comment above: this experiment shows that there are fixed photon number states that cannot be explained by a classical analogy. Thus talking about "entanglement" with an electron (i.e. a superposition of excited/not excited) doesn't really make sense.

http://people.whitman.edu/~beckmk/QM/grangier/Thorn_ajp.pdf

18. Sep 10, 2015

### vanhees71

Hm, already the abstract makes the wrong statement that the photoeffect suggest the existence of photons. I'll have a look at the rest of the paper this evening.

Again: On the level of naive description of the photoeffect this is not justified, because it's entirely explained by classical electromagnetic waves interacting with a bound quantum electron. The Compoton effect is more suggestive, and I guess that's why Einstein got his Nobel at the time he got it, because of the discovery of the Compton effect, which made the existence of photons much more "suggetive".

19. Sep 10, 2015

### DrChinese

Considering that these experiments are more definitive proof of the existence of quantum photons, "suggestive" is appropriate for the context. I don't think they were making a deeper point. Hopefully you will approve of the content and the conclusion.

20. Sep 10, 2015

### ZapperZ

Staff Emeritus