Lamb 1969 states,

A misconception which most physicists acquire in their formative years is that the photoelectric effect requires the quantization of the electromagnetic field for its explanation. [...] In fact we shall see that the photoelectric effect may be completely explained without invoking the concept of "light quanta."

The paper gives a description in which an atom is ionized by light, with the atom being treated quantum-mechanically but the light being treated as a classical wave.

Is it true that all the standard treatments in textbooks are getting this wrong?

Lamb and Scully "The photoelectric effect without photons," in "Polarization, Matière et Rayonnement," Volume in Honour of A. Kastler (Presses Universitaires de France, Paris, 1969) -- can be found online by googling

  • $\begingroup$ see also the discussion in physicsforums.com/threads/… $\endgroup$ Commented Oct 15, 2014 at 11:19
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    $\begingroup$ At least one standard introductory physics book does not get this wrong -- Physics by Resnick, Halliday and Krane; 5th ed, Wiley, NJ. $\endgroup$ Commented May 31, 2018 at 15:56
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    $\begingroup$ A closely related question is whether lasers can be explained without photons. It us still open: physics.stackexchange.com/questions/408539/… $\endgroup$
    – my2cts
    Commented Jul 22, 2019 at 19:49

9 Answers 9


Yes, the photoelectric effect can be explained without photons!

One can read it in

L. Mandel and E. Wolf, Optical Coherence and Quantum Optics, Cambridge University Press, 1995,

a standard reference for quantum optics. Sections 9.1-9.5 show that the electron field responds to a classical external electromagnetic radiation field by emitting electrons according to Poisson-law probabilities, very much like that interpreted by Einstein in terms of light particles. Thus the quantum detector produces discrete Poisson-distributed clicks, although the source is completely continuous, and there are no photons at all in the quantum mechanical model. The state space of this quantum system consists of multi-electron states only. So here the multi-electron system (followed by a macroscopic decoherence process that leads to the multiple dot localization of the emitted electron field) is responsible for the creation of the dot pattern. This proves that the clicks cannot be taken to be a proof of the existence of photons.

An interesting collection of articles explaining different current views is in

The Nature of Light: What Is a Photon?
Optics and Photonics News, October 2003

Further discussion is given in the entry ''The photoelectric effect'' of my theoretical physics FAQ at http://arnold-neumaier.at/physfaq/physics-faq.html . See also the slides of my lectures http://arnold-neumaier.at/ms/lightslides.pdf and http://arnold-neumaier.at/ms/optslides.pdf .

QED and photons are of course needed to explain special quantum effects of light revealed in modern experiments (discussed in the Optics and Photonics News issue cited above) such as nonclassical states of light or parametric down conversion, but not for the photoelectric effect.

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    $\begingroup$ +1 because this was the kind of thing I was expecting as an answer to this question, rather than the self-answer at the top. I can't believe this is answer is sitting so low down. $\endgroup$
    – 299792458
    Commented Sep 16, 2014 at 13:47
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    $\begingroup$ @BenCrowell: Energy is never conserved in interactions with external fields. Hence this is no valid counterargument. $\endgroup$ Commented May 23, 2017 at 8:01
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    $\begingroup$ @BenCrowell: Compared to you I am considering a slightly different scenario where one cannot ask for energy conservation because the external field provides the energy. For a forced pendulum (a completely analogous situation), the enetgy is not conserved either. $\endgroup$ Commented May 24, 2017 at 18:42
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    $\begingroup$ +1 the quantum mechanical property of light is only revealed if one prepares light in certain special states. And studying those effects is a separate endeavour. Not the photoelectric effect. $\endgroup$ Commented Aug 31, 2020 at 18:52
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    $\begingroup$ Can you name some phenomenons/experiments that absolutely need the photon idea? Because till now, I thought the black-body radiation, photoelectric effect and Compton scattering were conclusive results. But they also seem to be explained via semi-classical arguments too (as people here are discussing). $\endgroup$ Commented Jul 9, 2021 at 16:41

The Lamb-Scully paper is a good example of how even a Nobel Prize winner can occasionally write a bad paper.

The historical context is important. Einstein hypothesized the photon in 1905, but his paper was ahead of its time and was not widely accepted. For decades afterward, even once the quantum-mechanical nature of the atom was assumed by all physicists, the quantum-mechanical nature of light was considered suspect. Bohr was influential in pushing a theory in which atoms were quantized, but the light they absorbed and emitted was classical. Lamb began his career during this era.

If you read the Lamb-Scully paper, the first thing you notice is that they explicitly state that photons are absolutely necessary in order to explain phenomena such as blackbody radiation, Compton scattering, spontaneous emission, and the Lamb shift. Any internet kooks who are trying to quote Lamb and Scully as authorities against quantization of light are way off base.

As in Bohr's old-fashioned dead-end approach, they then treat the atom as a quantum-mechanical system and the electromagnetic field as a classical one. They are able to reproduce the Einstein relation $E=hf-W$, where $E$ is the maximum energy of the electron once it leaves the cathode, $h$ is the quantum-mechanical Planck's constant, $f$ is the frequency of the light, and $W$ is the energy required for the electron to escape through the surface of the cathode. This is not particularly surprising or impressive in a bastardized quantum/classical calculation like this one; essentially it just says that the light wave has to have the energy taken out of it at a resonant frequency of the atom, that frequency has to match its own frequency.

They also show that the transition rate is nonzero even when the light is first turned on, saying that their result "certainly does not imply the 'time delay' which some people used to expect for the photoelectrons produced by a classical e.m. field." This result is not as impressive as they make it sound, since the classical prediction is what one expects for a classical light wave impinging on classical atoms.

In fact, the transition rate they derive shows the real problem with their calculation. Their calculation treats every atom as independent of all the other atoms. Therefore if a classical flash of light with energy $W$ illuminates the cathode, it may ionize more than one atom, violating conservation of energy. This unphysical result shows the opposite of what they claim; it shows that their mixed quantum-classical Frankenstein fails to provide a physically acceptable explanation of the photoelectric effect. What they really need is a quantum-mechanical entanglement between the different parts of the photon's wave packet, so that if the photon is observed at atom A, it is guaranteed not to be observed at atom B. Without this quantum-mechanical "spooky action at a distance," their theory violates conservation of energy.

This issue was recognized very early on in the development of the "old" quantum theory, and it led to the Bohr-Kramers-Slater (BKS) theory, in which energy and momentum were conjectured to be conserved only on a statistical basis. Experiments as early as Bothe 1925 falsified the BKS theory by showing that when x-rays were emitted in a spherical wave into two hemispherical detectors, the two detectors were completely anticorrelated.

A modern discussion of these issues is given by Greenstein 2005. In section 2.1, they first present a summary of the Lamb-Scully argument, and then discuss the experimental verification of the existence of the anticorrelations required in order to maintain conservation of energy (Grangier 1986). The fact that this anticorrelation was not successfully observed with visible light until 1986 was due to technical limitations on the ability to produce sources of light that were eigenstates of photon number. However, the equivalent anticorrelation result with x-rays had already been demonstrated by Bothe in 1925.

One could therefore argue that the observations of the photoelectric effect were not enough to establish the existence of photons without the further verification of anticorrelations some years later. This would be misleading, however. From the point of view of physicists reading Einstein's 1905 paper, before the quantum-mechanical nature of the atom had been established, a hybrid model such as Lamb's or the BKS theory was unavailable, and therefore the photoelectric effect really did require quantization of light. One could argue that, in the historical context of the period from 1913 (the Bohr model) to 1925 (Bothe), there was a viable BKS theory that avoided quantization of the electromagnetic field, but this is extremely misleading when modern authors such as Lamb fail to admit that nonconservation of energy was an ingredient.

Similar difficulties arise if one attempts to construct a consistent theory in which the gravitational field simply isn't quantized, unlike the other fundamental forces (Carlip 2008).

Bothe and Geiger, "Experimentelles zur Theorie von Bohr, Kramers und Slater," Die Naturwissenschaften 13 (1925) 440. The experiment is described in Bothe's 1954 Nobel Prize lecture.

Carlip, "Is quantum gravity necessary?," http://arxiv.org/abs/0803.3456

Grangier, Roger, and Aspect, "Experimental evidence for a photon anticorrelation effect on a beamsplitter," Europhys. Lett. 1 (1986) 173 -- can be found online by googling

Greenstein and Zajonc, "The quantum challenge: modern research on the foundations of quantum mechanics," Jones and Bartlett, 2005.

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    $\begingroup$ note that a semi-classical treatment of Compton scattering (which Schrödinger derived in 1927) does get parts of the physics right - but not all of it (as with the photoelectric effect, you need QED for that); also note that a semi-classical approach that attributes clicks of a detector in optical experiments to the quantum nature of the detector (instead of the field) can be closer to reality than the naive picture of photon particles hitting the detector - a QED photon is inherently delocalized and shouldn't be confused with localized wave packets (which we sometimes call photons as well) $\endgroup$
    – Christoph
    Commented Jun 15, 2013 at 17:51
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    $\begingroup$ From what I remember, it's certainly the case that, in the context of the photoelectric effect, textbooks generally give the impression that it's just the discreteness of the photodetections which provide evidence for the quantum nature of light - it would be more illuminating (sorry) if they mentioned that discreteness occurs also in semiclassical theories, but that it leads to difficulties with energy conservation. $\endgroup$
    – twistor59
    Commented Jun 17, 2013 at 7:51
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    $\begingroup$ @twistor59: it would be more illuminating (sorry) if they mentioned that discreteness occurs also in semiclassical theories, but that it leads to difficulties with energy conservation. Such an approach would beg the question of why semiclassical theories are even worth considering. The reasons they were considered are IMO obscure and historical and not of much interest to modern students. There are infinitely many wrong theories we could set up for our students and then shoot down. How much of that to do is a matter of taste. $\endgroup$
    – user4552
    Commented Jun 17, 2013 at 14:20
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    $\begingroup$ @BenCrowell I disagree with your last comment. To follow you, we should never teach Newton's law, because it's of no much interest for modern students ? Your answer is definitely great by the way, even if I would erase the first sentence. I also subtly modify the interpretation usually when discussing with friends. I prefer to argue that neither the atom nor the light field is quantised (in a provocative way), but it's rather the exchange energy between the two sub-systems which is quantised. $\endgroup$
    – FraSchelle
    Commented Jun 19, 2013 at 5:15
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    $\begingroup$ -1: It is easy to dismiss a 50 ears old paper. But try to do the same with the 2003 collection of articles cited in my answer! $\endgroup$ Commented Aug 18, 2014 at 12:56

Yes, the textbooks are getting it very wrong.

The common narrative on these things is best summarized by the "three nails in the coffin" approach: the dead body being the wave theory of light, and the three nails being the blackbody spectrum, the photo-electric effect, and the Compton effect. Whatever difficulties the wave theory may or may not have with modern anti-correlation experiments, they are completely wrong in the arguments they present for rejecting the wave theory on the basis of the "three nails".

The reason textbooks and physicists in those days accepted those wrong arguments is that until 1926 there was no viable theory which allowed people to do wave-on-wave calculations. Once Schrodinger discovered the wave equations, clear explanations were available for all three phenomena. I will describe them briefly here.

First, the photo-electric effect. Even today modern textbooks make much of the frequency threshold, as though that were inexplicable by classical waves. The Schrodinger theory made it immediately clear that states of different energy levels are coupled only when excited by frequencies corresponding to the difference in those levels. Yet the textbooks continue to profess bafflement at the frequency effect.

The other glaring error of the textbooks is in using the physical cross-section of a single atom to calculate the absorption cross-section. Even Scully is guilty of this in a paper as recent as 2002 (if I remember the year). The physical cross-section is completely wrong, even in antenna theory; if it were true, a crystal radio could never collect enough energy to drive even the tiniest of earphones. I explain this in my blogpost on the crystal radio. (And I don't think anyone wants to argue that you need photons to explain the crystal radio.)

Second, the Compton Effect. When I figured out a semi-classical explanation of the Compton Effect, I thought I would win the Nobel Prize. So I was disappointed to find that Schrodinger had published exactly the same explanation in 1927. You take the light and the electron in a center-of-mass system, and you consider the system at the midpoint of the interaction... when the electron is in a superposition of states, half moving to the left, and half moving to the right. You can see right away that this superposition sets up layers of charge equally spaced at a distance of $\frac{1}{2}\lambda$, creating a perfect diffraction grating for total reflection.

Of course, Compton couldn't have come up with this explanation because he didn't know about electron waves. His "proof" debunking the wave theory of light treated the electron as a tiny charged ping-pong ball.

Finally, the black-body spectrum is an interesting case. Oddly enough, it is known that Planck's Law must prevail even if electromagnetism didn't exist, as exemplified by the low-temperature specific heat of solids. The deviation from the law of Dulong and Petit was (I think) recognized by Einstein in a 1905 paper. But it is hard to argue that it is caused by "photons". Surely we must believe that the suppression of the high-frequency modes is here just a mechanical consequence of Schrodinger's Equation.

And if that is so, then there is no need to invoke "photons" to explain the extension of Planck's Law to the electromagnetic spectrum, because a careful classical argument shows that the energy per mode at any given frequency of the classical e-m field must be equal to the energy per mode of the mechanical oscillators at that same frequency. I show how this calculation works in a series of articles culminating here.

For good measure, I also explicitly show in a later series of blog posts that the Copenhagen "quantum leap" between eigenstates gives the same radiation field as the Schrodinger continuous transition model with the atoms radiating semi-classically.

Thanks to Helder Velez for flagging some of my articles. Yes, I am the kook identified as such by Ben Crowell, so feel free to ignore my post.

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    $\begingroup$ Nobody disputes that semi-classical theories such as Bohr-Kramers-Slater (or rehashes of it by Lamb or you) can explain some aspects of these phenomena. A viable theory needs to explain all of the observations. Whatever difficulties the wave theory may or may not have with modern anti-correlation experiments [...] "Modern" is misleading. BKS was proposed in 1924, then disproved in 1924-1925 in a series of experiments by Bothe and Geiger, one of which observed exactly the anticorrelations predicted by the photon theory. $\endgroup$
    – user4552
    Commented Jun 16, 2013 at 16:11
  • $\begingroup$ The question is specifically about the photoelectric effect, so the material about Compton scattering and black-body radiation is off topic. However, a semi-classical picture of Compton scattering can't explain (1) a change in wavelength in the limit of low-intensity incident radiation, or (2) the results of the 1924 Bothe-Geiger electron-x-ray coincidence experiment. $\endgroup$
    – user4552
    Commented Jun 16, 2013 at 16:16
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    $\begingroup$ Ben, the textbooks dispute precisely those aspects of these phenomena which can be explained by a good semi-classical approach. And I'm not talking about BKS...you obviously haven't had time in the 16 minutes since I posted to read the articles where I show how it's done. $\endgroup$ Commented Jun 16, 2013 at 16:16
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    $\begingroup$ Ben, you're the one that introduced Compton and Black Body into the discussion, not me.So you shouldn't criticize me for going "off-topic". There's nothing in the original post about anti-correlation either. The question was about whether the "textbooks" have it wrong, meaning I would think the common undergrad textbooks with their "three nails" narrative. I think I answered the question. $\endgroup$ Commented Jun 16, 2013 at 17:08

I disagree with OP in that I don't consider energy conservation as a fatal flaw.

If one lets $t\to\infty$ in the perturbative calculation, one gets a nice delta function $\delta(\epsilon_f-\epsilon_i-\hbar\omega)$ but in such case the external energy supply is infinite and no meaningful energy conservation argument can be formulated, so I guess OP must be talking about the finite-time result, so let's focus on this.

Following OP's argument, actually, we don't even need two atoms to see energy is not "conserved" - one atom is enough. The result of harmonic perturbation gives the probability of transition from ground state $|g\rangle$ to the $k$th excited state $|k\rangle$ as (quoting Lamb & Scully equation (13)),

$$4\left|\frac{\langle k|\hat x|g\rangle E_0}{\hbar}\right|^2\frac{\sin^2\{(\epsilon_k\hbar^{-1}-\nu)t/2\}}{(\epsilon_k\hbar^{-1}-\nu)^2}$$

where $E_0$ is the external EM wave E-field strength. Normally the matrix element $\langle k|\hat x|g\rangle $ can be non-zero up to $|k\rangle$ with arbitrarily high energy. For classical light, we can make $E_0$ arbitrarily close to $0$, this means in a finite $t$ the energy supply can be arbitrarily small, yet the probability is non-zero for the transition to a $|k\rangle$ with very high energy (i.e., $\epsilon_k-\epsilon_g>\text{ external energy supply}$). If after a measurement, the atom indeed ends up at $|k\rangle$, then energy conservation is violated.

However, what is the reason for this violation? It is because our energy measurement for external EM wave is classical while the energy measurement for the atom is quantum mechanical. In other words, we are comparing initial energy with some eigenvalue $\epsilon_k$ of the quantum Hamiltonian. In a fully quantum mechanical (i.e., not semi-classical) system, this is exactly what we should not do; what we should compare are the energy expectation values, that is, something like initial $\langle i|H|i\rangle$ and final $\langle f|H|f\rangle$, but never just some eigenvalues (unless both are eigenstates). So if we do the same in the semi-classical treatment of photo-electric effect, we see energy conservation is satisfied qualitatively, because from equation $(13)$ we see the energy expectation value will be proportional to $|E_0|^2$. I believe the same argument applies to OP's two-atom experiment.

I must say OP's argument is justified for a semi-classical system, because it's certainly operationally possible. But my point is that this is a generic problem of all semi-classical systems (in fact there has been similar argument showing that if light is treated classically then the uncertainty principle for the electron can be violated. See Sakurai's "Advanced quantum mechanics" page 34~35). So I think it's good enough that Lamb and Scully could reproduce $E=\hbar\omega-\phi$ and the no-delay emission of electrons. If one wants to use energy conservation as an objection, one might as well just say quantum-classical coupling is impossible, there's no need to assign any special significance to the photo-electric effect.

I'd like to move my last comment to the main text for the sake of completeness. The energy conservation difficulty is only conceptual not experimental, because the very original photoelectric effect could only measure energy expectation values, and from my above analysis we see energy expectation values are conserved. Even on a conceptual level, there's still a way out, that is, take energy conservation to be true only on a statistical level (which of course needs an experimental test, and indeed there were as Ben mentioned), and this was exactly what Bohr proposed, due to exactly the same reason. In a word, I believe Lamb & Scully did explain all experimental aspects of the photoelectric effect.

  • $\begingroup$ Nice answer, +1. I think we only differ in emphasis. If one wants to use energy conservation as an objection, one might as well just say quantum-classical coupling is impossible, there's no need to assign any special significance to the photo-electric effect. I agree. The photoelectric effect is just one example that demonstrates the general impossibility of quantum-classical coupling. But historically it was one of the first and most important such examples in the development of quantum mechanics, and pedagogically it's a nicer intro than the Compton effect or black body radiation. $\endgroup$
    – user4552
    Commented Jun 17, 2013 at 14:08
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    $\begingroup$ The problem with all of your "conservation of energy" arguments is that they can never be verified experimentally. The device which measures the supposedly early electron is inevitably a tube with a 1200-volt power supply or something of that kind, which is clearly capable of supplying the energy necessary for the detection event. There is no reason to think the detection energy had to come from the "photon". $\endgroup$ Commented Jun 17, 2013 at 15:14
  • $\begingroup$ @MartyGreen: Nonsense. For example, in a Geiger-Mueller tube, the photon has to ionize a molecule initially. Until that initial ionization occurs, you don't get an avalanche or amplification, because the unionized gas is an insulator. Statistical versus absolute conservation of energy is exactly the issue tested by the 1924-1925 Bothe-Geiger experiments, for which Bothe received the Nobel Prize. $\endgroup$
    – user4552
    Commented Jun 17, 2013 at 18:03
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    $\begingroup$ Ben, it is hard to see how you can be absolutely sure that some of the energy for that ionization didn't come from the 1200-VDC power supply. You should read my article on Quantum Siphoning, linked elsewhere on this page by Helder Velez, where I explain how those kinds of processes can work. In my article I explain how it works for the case of the photographic plate, but I think the Geiger Counter is the same in principle. $\endgroup$ Commented Jun 17, 2013 at 18:16
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    $\begingroup$ @BenCrowell: But the difficulty is only on conceptual level, at least when only the original photo-electric effect experiment is under consideration, because in the original set up only energy expectation values can be measured. Since statistical versus absolute conservation of energy was not tested before, absolute energy conservation was better not to be taken as true unconditionally. I still hold the opinion that Lamb & Scully's explanation is good enough for the historical photo-electric effect experiment. $\endgroup$
    – Jia Yiyang
    Commented Jun 21, 2013 at 4:58

The photoelectric effect can be explained without photons. First, let us define “photon.” It is not a thing like a wave packet. Please, it is not a detection click either. It is a dualistic phenomenon. A photon will go one way or another at a beam-splitter, but if you re-converge the beam you will develop an interference pattern. That is a rough quote from Bohr (ref 1) explaining Einstein's view of a photon. This encapsulates the way quantum mechanics (QM) works: a probability-wave guides an absorption event. The key feature of this model is that an emission of one quanta will end with a total of one quanta. If the quanta splits in half it can produce two half-quanta. QM is usually handled in this one-to-one sense. There is a poorly understood alternative to this model. It has been called the accumulation hypothesis, and the loading theory. We call it the threshold model (TM). To understand the photoelectric effect without photons is to see the flaw in QM and understand TM. We show by analysis of past experiments, and by new experiments, how QM fails. Please look at what experiments are saying, not people. These experiments include the photoelectric effect, element of time in the photoelectric effect, Compton effect, charge diffraction, black body tests, giant molecule diffraction, and beam-split coincidence tests (see website, ref 2).

The most important of these tests is the beam-split coincidence test. This test is famous for showing the particle property of going one way or another at a beam-splitter, just like in Einstein's definition of the photon. This test has been described with visible light, but now we do it with Gamma-rays. Newly shown is how the coincident detector click-rate past the beam-splitter can substantially exceed the accidental chance rate. Quantum mechanics predicts chance. Chance is easily determined by (time window)(singles rate of detector 1)(singles rate of detector 2) = (chance rate), see ref (3). Cadmium-109 emits only one gamma in spontanious decay. We know this from the sandwich test: a coincidence test with its pair of detectors on opposite sides of the radioisotope, in close range (see Knoll ref 3). Cd-109 also emits an x-ray, but we eliminate that with electronic pulse-height discriminators. We set the discriminators to read pulses above two thirds the characteristic height assigned to its 88 KeV gamma-ray photo-peak. Previous tests have shown that pulse-height is proportional to electromagnetic frequency, and it is also proportional to their so-called photon energy, in electron-volts. Here we use eV only for convenience because we do not believe in photons.

Now for my test we use the same isotope, detectors, electronics, pulse-height settings as was used in the sandwich test, but we make two changes. 1) We put the two detectors to one side of the radioisotope such that the gamma needs to go through the first detector to be received by the second detector. 2) We make the first detector thinner so that the probability of reaching either detector is the same. This tandem geometry splits energy similar to the geometry of a beam-splitter, but it works better. The result is typically 15 times the accidental chance rate. The result is highly repeatable but depends on the details of the setup. This is not a special case. The effect works with other isotopes, other detectors (sodium-iodide, HpGe), and different geometries. If you believe in photons, this two-for-one effect seemingly violates energy conservation. Energy is still conserved, but now we understand how a pre-loaded state must exist from previous energy exchanges. If you do this test with visible light you are looking at noise. Doing the test with gamma-rays gives the punch to overcome the noise. If you do the test with a detector having Compton effect efficiency exceeding photoelectric efficiency, you will see noise and think QM is correct. The detectors and gamma-ray we used were chosen for dominant photoelectric response. The new tests with gamma-rays show there are no photons.

One might object by citing many similar tests that uphold photons. Examine them carefully. You will often see polarization trickery. A polarized atomic emission of an $hf$ of energy will be routed by a polarized beam-splitter to go one way or another, thereby making you think their test upholds the photon model. $E=hf$ is still true, but please let us not call it a photon. I like to call it an h-new in honor of Planck. TM is really an extension of Planck's second theory (4). Also, do any of those tests discuss how they set their pulse-height discriminators? I never see them show it. It turns out that the pulse-height distribution using monochromatic visible light with any detector is too wide to make the distinction between a QM or TM. If you set the discriminator too high you seemingly falsify QM; if you set it too low you seemingly prove QM.

TM calls for a pre-loaded state. In the photoelectric effect the pre-loaded state is the amount of electronic kinetic energy. This same two-for-one effect is evidenced in my similar beam-split coincidence tests with the alpha-ray. The atom splits like a wave. These tests do not simply describe wave properties; they reveal failure of tests that are famous for upholding the particle property. To see how such a pre-loaded state is possible requires a conceptual adjustment to our physical constants. Let us describe for the electron: charge constant $e$, mass constant $m$, and Planck's $h$. To see how a pre-loaded state and matter-waves are possible, we take these constants to express maxima, revealed in our experiments. In this theory we do not see sub-$e$ but we can understand that it exists nevertheless. In tests that display wave properties there are ratios of $e/m$, $h/e$, $m/h$. For example, if you see an $e/m$ ratio in an equation, it means the experiment related to that equation can have $e/2$ and $m/2$, but you will not know it. Equations that do not show these simple ratios are cases where the waves hold themselves together, we are dealing with real particles, and those systems will not diffract. Just described are the important points of my discovery toward removing wave-particle duality, and more specifically, how to see the photoelectric effect without photons. Writings on my website (2) show a derivation of the photoelectric effect, linking it to the deBroglie equation. To understand the photoelectric effect without photons is to transcend wave-particle duality and re-shape our most fundamental physics.

(1) Bohr, Atomic Physics and Human Knowledge, see Pg 50
(2) http://www.thresholdmodel.com
(3) Knoll, Radiation Detection and Measurement
(4) Kuhn, Black Body Theory and the Quantum Discontinuity 1894-1912


All the above explanations describe measurable effects at endpoints of energetic interaction, they do not demonstrate photons as anything other than a concept of pure convenience that derives historically from the dreaded billiard ball analogy. The "so-called" propagation of the interaction energy is observable only at the endpoints and the effect is associated with c (the so called speed of light), so at c time and space dilation make the endpoints essentially the same event. It's important to divest yourself of the anthropomorphic boundaries of observability. The "effect" you are measuring is not only loaded with your bias to a notion of time and distance but also to causality. The end point effects do not require there to be a photon. The very notion of one is an anachronism.


Any classical or quantum electrodynamics effect can be explained without photons.

Photons are not real, but simply a device for simplifying direct particle-particle interactions. Instead writing

$$\mathrm{e}_1 + \mathrm{e}_2 \rightarrow \mathrm{e}_1' + \mathrm{e}_2'$$

for a process where electron $\mathrm{e}_1$ is loosing energy, we subtract the environmental electron to get

$$\mathrm{e}_1 \rightarrow \mathrm{e}_1' + [ \mathrm{e}_2' - \mathrm{e}_2 ]$$

The term within brackets is what we call photon

$$\mathrm{e}_1 \rightarrow \mathrm{e}_1' + \gamma$$

All the properties of the photon (mass, energy, spin) can be derived from therein

Mass: $m_\gamma \equiv m_{\mathrm{e}_2'} - m_{\mathrm{e}_2} = 0$.

Energy: $E_\gamma \equiv E_{\mathrm{e}_2'} - E_{\mathrm{e}_2}$.

Spin: $S_\gamma \equiv S_{\mathrm{e}_2'} - S_{\mathrm{e}_2} = (\pm1/2 - \pm1/2) = (-1,0,0,+1)$.

Instead postulating them, as field-theory does.

Basic review to electrodynamics without photons can be found in next two works:

Classical Electrodynamics in Terms of Direct Interparticle Action. 1949, Rev. Mod. Phys. 21(3), 425--433. Wheeler, John Archibald; Feynman, Richard Phillips

Cosmology and action-at-a-distance electrodynamics. 1995, Rev. Mod. Phys. 67(1), 113--155. Hoyle, F.; Narlikar, J. V.

  • $\begingroup$ You mean that virtual photons are not real, right? Because photons are definitely real. $\endgroup$ Commented Jul 9, 2021 at 16:38
  • $\begingroup$ All photons are virtual. Check the references: "The question is whether there are any real photons at all. As Feynman put it all photons if observed over a sufficiently long time scale are virtual." $\endgroup$
    – juanrga
    Commented Jul 11, 2021 at 10:27
  • $\begingroup$ but what about the energy-time uncertainty principle? A photon is an excited state of the electromagnetic field, whose energy is well defined. Hence, it can exist for a long enough time. And if one claims that that field excitation doesn't exist as a 'real' particle, then by the same logic any matter particle too is just a 'virtual' particle after all. Rather, all we have are fields whose excitation are the particles that we see. $\endgroup$ Commented Jul 11, 2021 at 14:07
  • $\begingroup$ The fact that one can explain classical electrodynamics via an action at a distance formulation doesn't remove the overall need of a field theory. We know that Quantum Mechanics holds and we have observed several phenomenons that are explained only by QFT, for instance decay of massive particles. $\endgroup$ Commented Jul 11, 2021 at 14:15
  • $\begingroup$ @quirkyquark In the action-at-distance formalism, the particles of matter are not the excitation of any field. There are no fields at all! Not only can classical electrodynamics be explained, quantum electrodynamics can also be explained (check the second reference I gave). I don't know of a single phenomenon that cannot be explained by action-at-a-distance. $\endgroup$
    – juanrga
    Commented Jul 16, 2021 at 10:02

Ben Crowell's answer contains the seed to a different answer, where he wrote that Lamb & Scully would need to add non-local spooky "self-entanglement" to their wave-only model. Okay, let's add that self-entanglement feature. In other words, absorption of quantum waves is quantized and implies non-absorption elsewhere... even if the absorbing detector is placed far from the rest of the experiment (which suggests a backward-in-time effect in order to prevent absorption at the detectors reached earlier by other parts of the wave).

Note: Experiments that falsify "classical waves" do not necessarily falsify non-classical (quantum) waves, and thus do not prove particles. So we should disregard answers and cited papers that argue for quantum particles by arguing against classical waves.


No! In fact, the existence of photons is crucial to the photoelectric effect. To understand why this is so, think of the billiard ball collision, Photons collide with an electron in a metal with a specific work function, the electron in turn absorbs the energy of the photon; instead of reflecting the light completely, according to the classical wave theory of light.

  • 7
    $\begingroup$ You've just said that photons are a sufficient condition in order to explain the photoelectric effect. But you haven't explained if they are a neccesary condition. $\endgroup$
    – jinawee
    Commented May 29, 2014 at 8:18
  • $\begingroup$ That isn't entirely true... Please read the last sentence of my comment... $\endgroup$
    – user34793
    Commented May 29, 2014 at 14:34

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