Are photons really particles or just a misconception?

[jorgdv]

Hello everyone, thanks for reading

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

 

[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.

 

[Nugatory]

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?

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.

 

[lightarrow]

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?

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

 

[bhobba]

I think you might find Feynmans classic illuminating:
https://www.amazon.com/dp/B00BR40XJ6/?tag=pfamazon01-20

Thanks
Bill

 

[jorgdv]

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.

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?

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?

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

 

[Nugatory]

n 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.

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/

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?

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.

 

[jerromyjon]

Well, maybe I'm totally wrong here and this is just speculation.

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.

 

[lightarrow]

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.

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

 

[Avodyne]

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.

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.

 

[microsansfil]

Professional physicists have been using the word "particle" to MEAN "quantized excitation of the field" for many decades.

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

Patrick

 

[lightarrow]

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?

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?

--
lightarrow

 

[jorgdv]

Sorry for the delay.

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?

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 modeled as classical potential in the Hamiltonian, so what's happening here?

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?
--
lightarrow

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.

 

[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.

 

[lightarrow]

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.

What does it mean?

The thing that could not be explained with classical EM radiation together with ordinary QM is photon momentum and scattering,

Are you referring to Compton effect?

--
lightarrow

 

[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.

 

[DrChinese]

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 modeled 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.

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

 

[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".

 

[DrChinese]

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".

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.

 

[ZapperZ]

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".

Read the paper!

As Dr. Chinese has pointed out, they used the classic photoelectric effect (i) in the historical context and (ii) they never claim that it is the definitive evidence! That is why they performed this which-way experiment in the first place! Their argument was that this experiment is a lot stronger in favor of the photon picture than that classic photoelectric effect experiment.

I think you need to include the word "suggest" in your list, because you have a misuse and misunderstanding of it yourself.

To come back to the original question, as has been pointed out, the misconception of "photons" only occurs when people think that it is a classical "particle". These which-way experiment, and photon anti-bunching experiments are clear evidence of the validity of the photon picture. They are NOT misconceptions!

Zz.

 

[vanhees71]

Yes, I take it all back. The paper is excellent, in fact the best, I've read in a long time about photons on the undergrad level. That's how it should be taught! Of course, one has to work out the theoretical details much better, particularly with beginners, but their main topic was the experiment, which is of course marvelous in its simplicity and very impressive concerning the accuracy reached.

 

[harrylin]

[..]
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

Depends who you ask!

For the no-particle view as explained by A.Neumaier in this group, see:
https://www.physicsforums.com/threads/what-exactly-is-a-photon.457930/#post-3060510
And also:
https://www.physicsforums.com/threads/are-clicks-proof-of-single-photons.474537/page-7#post-3180396

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?

I think that that was answered here in this thread in which you also participated (answering "how would the screen 'know' how to get the stats right"
https://www.physicsforums.com/threads/are-clicks-proof-of-single-photons.474537/page-8#post-3187214

But it's not easy stuff, don't ask me to explain it further!

 

[lightarrow]

I think that that was answered here in this thread in which you also participated (answering "how would the screen 'know' how to get the stats right"
https://www.physicsforums.com/threads/are-clicks-proof-of-single-photons.474537/page-8#post-3187214

I remember, it was one of the most interesting threads for me. But here I am not talking of the exact case I discussed there: in this thread I assume we know to have sent exactly one single photon before the subsequent detection of it, and to repet the same experiment a great number of times; it's more difficult for me to understand how to invoke statistics in this case.

--
lightarrow

 

[jorgdv]

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

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

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

Excellent article and claim, that's exactly the point of my initial question!: The photoelectric effect can be entirely explained "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". With that in mind, I asked to myself, why on Earth do we need photons as particles of light to explain the photoelectric effect, if this view is completely reasonable? I see now that I'm not the only one with that concern, and that as you said nowadays is understood much better with that interpretation instead of the "bullet of light" concept.

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.

That's right, sorry I meant conservation of momentum. So with phenomena such as the Compton effect it is clear that the classic EM picture is not enough, and you need something that conserves the concept of field while having particle-like phenomena such as scattering, that is precisely the view of QFT. But my concern was that the photoelectric effect did not proove the existence of photons as such, despite the claims of Einstein and the common "particle" conception, due to the view of time-dependent QM with bounded electrons. That made me think that if that if the classic proof of the existence of photons that gave Einstein the novel price does actually not proove it, then photons as such could be a misconception, since it's clear that they have much more of EM pulses than particles ("bullet" interpretation). But it's clear that these pictures with the quantum mechanical nature of light are only unified in QED.

Thank you everyone

 

[ZapperZ]

Excellent article and claim, that's exactly the point of my initial question!: The photoelectric effect can be entirely explained "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". With that in mind, I asked to myself, why on Earth do we need photons as particles of light to explain the photoelectric effect, if this view is completely reasonable? I see now that I'm not the only one with that concern, and that as you said nowadays is understood much better with that interpretation instead of the "bullet of light" concept.
That's right, sorry I meant conservation of momentum. So with phenomena such as the Compton effect it is clear that the classic EM picture is not enough, and you need something that conserves the concept of field while having particle-like phenomena such as scattering, that is precisely the view of QFT. But my concern was that the photoelectric effect did not proove the existence of photons as such, despite the claims of Einstein and the common "particle" conception, due to the view of time-dependent QM with bounded electrons. That made me think that if that if the classic proof of the existence of photons that gave Einstein the novel price does actually not proove it, then photons as such could be a misconception, since it's clear that they have much more of EM pulses than particles ("bullet" interpretation). But it's clear that these pictures with the quantum mechanical nature of light are only unified in QED.

Thank you everyone

As I've stated elsewhere, the naive and simplistic version of the photoelectric effect is certainly not the definitive experiment to show the existence of photons. But this phenomenon has evolved and expanded BEYOND just this simplistic experiment! From the way this is being discussed, it is as if angle-resolved photoemission, resonant photoemission, and multiphoton photoemission do not exist already! NONE of those have ever been described using anything else than the photon picture. Period! I've even asked several people who made claims of being able to describe the basic photoelectric effect using the classical EM wave picture to formulate the description of these experiments, and no one could. And this was at least 10 years ago!

The basic photoelectric effect is a "sphere" seen from very far away. It is only when you look at the other more advanced photoemission phenomena will you see the cow!

Zz.

 

[vanhees71]

Have a look at the following paper, suggested by DrChinese

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

It's precisely making the correct statement about the naive-photon picture, and it describes a very-well understandable most simple experiment to prove the necessity of the QFT picture, i.e., the existence of single-photon states of the em. field. The key issue is that nowadays the quantum optics people have "heralded single-photon sources", using parametric down conversion to create two entangled photons. Then detecting one of those (the "signal photon"), one knows one has another one (the "idler photon"). Then you are sure to have prepared a one-photon Fock state (which is not so easy without the discovery of parametric down conversion). Then you let run the idler photon through a beam splitter and measure the rate of coincidental detection of a signal at both "exits" of the beam splitter. If you had exactly always a single photon running through the beam splitter you'd never observe such a coincidental event, i.e., the 2nd-order coherence measure $g^{(2)}=0$. In reality you have not 100% always only a single-photon state, but in the mentioned paper the authors find $g^{(2)}=0.0177 \pm 0.0026$, while for classical em. waves one must have $g^{(2)} \geq 1$, where the value 1 occurs if you have a perfectly coherent plane wave and $g^{(2)}=2$ for perfectly incoherent ("thermal") light (the situation of the Hanbury-Brown Twiss measurement of the diameter of a far-distant star, which with some right can be seen as the historical birth of quantum optics). This result is indeed overwhelming evidence for the correctness of the photon picture of the electromagnetic field. The only point is, and also this is really very well explained in this paper, to have the right picture in mind when talking about photons, and this right picture is a mathematical one and called QED!

 

[ZapperZ]

Have a look at the following paper, suggested by DrChinese

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

It's precisely making the correct statement about the naive-photon picture, and it describes a very-well understandable most simple experiment to prove the necessity of the QFT picture, i.e., the existence of single-photon states of the em. field. The key issue is that nowadays the quantum optics people have "heralded single-photon sources", using parametric down conversion to create two entangled photons. Then detecting one of those (the "signal photon"), one knows one has another one (the "idler photon"). Then you are sure to have prepared a one-photon Fock state (which is not so easy without the discovery of parametric down conversion). Then you let run the idler photon through a beam splitter and measure the rate of coincidental detection of a signal at both "exits" of the beam splitter. If you had exactly always a single photon running through the beam splitter you'd never observe such a coincidental event, i.e., the 2nd-order coherence measure $g^{(2)}=0$. In reality you have not 100% always only a single-photon state, but in the mentioned paper the authors find $g^{(2)}=0.0177 \pm 0.0026$, while for classical em. waves one must have $g^{(2)} \geq 1$, where the value 1 occurs if you have a perfectly coherent plane wave and $g^{(2)}=2$ for perfectly incoherent ("thermal") light (the situation of the Hanbury-Brown Twiss measurement of the diameter of a far-distant star, which with some right can be seen as the historical birth of quantum optics). This result is indeed overwhelming evidence for the correctness of the photon picture of the electromagnetic field. The only point is, and also this is really very well explained in this paper, to have the right picture in mind when talking about photons, and this right picture is a mathematical one and called QED!

I am not sure to whom you are replying to. If this is for me, please note that I made a reference to this very paper from way back in 2004 already!

https://www.physicsforums.com/threads/particle-wave-duality.41838/#post-304909

So I was well-aware of it, and in fact, even asked you to actually read it first before making a previous comment.

Furthermore, this is not about a which-way experiment, but rather the photoelectric effect experiment. There are many other more definitive experiments that point to the validity of the photon picture, but since the photoelectric effect experiment is the topic in question, then I am pointing out that if one simply look at the simple, classic photoelectric effect, then one is missing out a whole zoo of other photoemission phenomena beyond just that simple experiment.

Zz.

 

[carllooper]

I remember, it was one of the most interesting threads for me. But here I am not talking of the exact case I discussed there: in this thread I assume we know to have sent exactly one single photon before the subsequent detection of it, and to repet the same experiment a great number of times; it's more difficult for me to understand how to invoke statistics in this case.

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lightarrow

Making the assumption that we have sent exactly one single photon, prior to subsequent detection of it, is not really possible. For how do we know a single photon has been sent if it hasn't been detected? While the rate of detections give us a rate of emission, and possible way of knowing a future not-yet-detected photon has been sent, this rate is only a statistical average of the rate (computed from previous detections), so we're not really any wiser at any given moment with respect to whether a single not-yet-detected photon has been sent. In other words we can only have a statistical picture of emission.

The problem is assuming an exact (or mathematical) picture of something prior to statistical evidence of it. In a sense it is the statistical picture which becomes more "exact" and the mathematical picture which becomes the "approximation".

C

 

[lightarrow]

Making the assumption that we have sent exactly one single photon, prior to subsequent detection of it, is not really possible. For how do we know a single photon has been sent if it hasn't been detected?

Measuring the atom's recoil. I described a situation where the single photon was emitted by an atom's de-excitation.

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lightarrow

 

[vanhees71]

...or by using parametric downconversion to have a heralded-single-photon source by measuring (and absorbing) one of the entangled photons. See, e.g., this thesis

https://www2.physics.ox.ac.uk/sites/default/files/2013-11-08/mosley_2007_pdf_12235.pdf