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The paper Gloge, Marcuse 1969: Formal Quantum Theory of Light Rays starts with the sentence

Maxwell's theory can be considered as the quantum theory of a single photon and geometrical optics as the classical mechanics of this photon.

That caught me by surprise, because I always thought, Maxwell's equations should arise from QED in the limit of infinite photons according to the correspondence principle of high quantum numbers as expressed e.g. by Sakurai (1967):

The classical limit of the quantum theory of radiation is achieved when the number of photons becomes so large that the occupation number may as well be regarded as a continuous variable. The space-time development of the classical electromagnetic wave approximates the dynamical behavior of trillions of photons.

Isn't the view of Sakurai in contradiction to Gloge? Do Maxwell's equation describe a single photon or an infinite number of photons? Or do Maxwell's equations describe a single photon and also an infinite number of photons at the same time? But why do we need QED then at all?

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Because photons do not interact, to very good approximation for frequencies lower than $m_e c^2 / h$ ($m_e$ = electron mass), the theory for one photon corresponds pretty well to the theory for an infinite number of them, modulo Bose-Einstein symmetry concerns. This is similar to most of the statistical theory of ideal gases being derivable from looking at the behavior of a single gas particle in kinetic theory.

Put another way, the single photon behavior $\leftrightarrow$ Maxwell's equations correspondence only holds if you look at the Fourier transform version of Maxwell's equations. The real space-time version of Maxwell's equations would require looking at a superposition of an infinite number of photons — one way to describe the taking an inverse Fourier transform.

If you want to think of it in terms of Feynman diagrams, classical electromagnetism is described by a subset of the tree-level diagrams, while quantum field theory requires both tree level and diagrams that have closed loops in them. It is the fact that the lowest mass particle photons can produce a closed loop by interacting with, the electron, that keeps photons from scattering off of each other.

In sum: they're both incorrect for not including frequency cutoff concerns (pair production), and they're both right if you take the high frequency cutoff as a given, depending on how you look at things.

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  • $\begingroup$ Isn't it remarkable that in tree level the description of a photon doesn't need Planck's constant at all? It seems to me that only when interactions with (fermionic?) matter are included the theory of photons needs Planck`s constant. $\endgroup$ – asmaier Nov 30 '16 at 16:44
  • $\begingroup$ @asmaier : the description of a photon needs Planck's constant! $\endgroup$ – John Duffield Dec 2 '16 at 13:09
  • $\begingroup$ @JohnDuffield Maxwells equations, which describe the wavefunction of a single noninteracting photon, don't need Planck's constant. I find that remarkable. $\endgroup$ – asmaier Dec 2 '16 at 14:16
  • $\begingroup$ Tree-level does not mean without interactions. It simply means there are no loops in the diagram. A tree-level diagram in QED has at least two vertices (unless you are thinking of the bare propagator). So I don't think classical electromagnetism is described by tree-level QED. Loop diagrams only serve to renormalize the tree-level diagrams. $\endgroup$ – flippiefanus Dec 3 '16 at 5:44
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    $\begingroup$ I do not agree with the Fourier transform part. The sharp-momentum states are only one basis we can choose in the Fock space. In the free theory we can take any of the solutions of free-field equations (vacuum Maxwell) as the single state in which the particle currently resides. Many people are presented with just a narrow exposition of QFT leading to S-matrix formalism, and this leads to ignorance about this freedom to choose the free states. $\endgroup$ – Void Dec 4 '16 at 0:27
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Do Maxwell's equation describe a single photon or an infinite number of photons?

Both.

(i) The single photon wave function is a solution of the free Maxwell equations in vacuum, and any nonzero solution of the free Maxwell equations in vacuum is a possible single photon wave function.

(ii) The mode of a coherent state of (arbitrarily many) photons is a solution of the free Maxwell equations in vacuum, and any nonzero solution of the free Maxwell equations in vacuum is a possible mode of a coherent state of photons. (The zero solution corresponds to the coherent state of zero intensity, usually called the vacuum state.)

For full details see the entry ''What is a photon?'' in Chapter B2 of my theoretical physics FAQ.

The above follows easily from the pure electromagnetic sector of QED (i.e., no matter present). The full strength of QED is needed to properly describe interactions of photons with electrons and other charged matter on a microscopic level.

More generally, for any free bosonic field theory (relativistic or not) there is a 1-1 correspondence between modes of coherent states and state vectors (wave functions) of the 1-particle Hilbert space of the theory.

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  • $\begingroup$ I agree wholeheartedly with (ii), but am somewhat confused by your claim (i). How does the mapping from arbitrary classical field to single photon wave function work, given that a classical field can have a continuously varying intensity while a single photon cannot? $\endgroup$ – Rococo Nov 29 '16 at 5:02
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    $\begingroup$ The mapping is given by the Riemann-Silberstein vector. en.wikipedia.org/wiki/Riemann–Silberstein_vector The nonrelativistic analogue is that the solution of a Schroedinger equation has a continuously varying intensity (absolute value squared) which is reinterpreted as a probability density for position measurement. This reinterpretation works only in the nonrelativstic case. $\endgroup$ – Arnold Neumaier Nov 29 '16 at 10:04
  • $\begingroup$ @Rococo: ... but qualitatively, something similar happens in the relativistic case. in any case it shows that there is no contradiction between the two views. $\endgroup$ – Arnold Neumaier Nov 29 '16 at 11:31
  • $\begingroup$ Thought I would look at your FAQ, ( after your comment to my answer) and the link was not working so I took the liberty to correct it $\endgroup$ – anna v Nov 29 '16 at 16:08
  • $\begingroup$ Isn't it remarkable that in tree level the description of a photon doesn't need Planck's constant at all? It seems to me that only when interactions with (fermionic?) matter are included the theory of photons needs Planck`s constant. $\endgroup$ – asmaier Nov 30 '16 at 16:26
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I think that the authors are sloppily combining two quite distinct ideas into a single sentence.

The first half of the sentence, "Maxwell's theory can be considered as the quantum theory of a single photon," refers to the fact that the single tree-level vertex of QED involves one photon interacting with an electron/positron. So as Sean Lake said, tree-level QED is linear in photon amplitudes, in the sense that photons cannot scatter off each other (because doing so requires the exchange of closed loops of virtual electrons), and in this tree-level classical limit the photon dynamics are described by Maxwell's equations.

In the second half of the sentence, "geometrical optics [can be considered] as the classical mechanics of this photon," when the authors say "photon" they're really referring to the much older concept of a corpuscle. In the high-frequency limit - more precisely, the limit where the wavelength of light is much smaller than the size of the objects that it scatters off of - Maxwell's equations reduce to the much simpler theory of geometrical or ray optics, in which light rays simply travel in straight lines and can be thought of as being transmitted by discrete classical-mechanical particles called "corpuscles." These particles are very different from photons, as they are completely classical and have no wavelike nature whatsoever.

So indeed a single photon at very high frequency is technically formally similar to a Newtonian corpuscle. But this isn't very useful, because if there's even one other photon (which, realistically, there will be), then the photons will exchange virtual electrons and scatter in a very non-classical way. The more natural way to get geometrical optics is to get a huge number of photons propagating in a semiclassical coherent state (so that as Sakurai says, you can ignore their discrete nature), with a wavelength that is small compared to the size of the scattering objects (so that you can ignore the classical wavelike behavior of Maxwell's equations) but large compared to the electron Compton wavelength (so that you can ignore QED scattering effects).

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Do Maxwell's equation describe a single photon or an infinite number of photons?

When interpreted as a classical equation where the E and B are classical fields, they describe the emergent envelope from zillions of photons.

When E and B solutions of Maxwell's equations are considered as part of the imaginary wave function, then they describe the wave function of a single photon.

This explains why single photons in superposition , i.e. as superposition of wave functions, can build up the classical field since it is the same solutions, different interpretations.

When a single photon is detected , one gets the energy , the spin and the probability of being found at that (x,w,z). The ensemble of photons gives the classical E and B fields, as explained in the first link.

QED is necessary to describe photon particle interactions. (There are no photon photon interaction without photon particle interactions after all).

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    $\begingroup$ ''When a single photon is detected , one gets the energy , the spin and the probability of being found at that (x,w,z,t). '' No. The probability is ill-defined, and cannot be defined in a natural, rotation invariant way. $\endgroup$ – Arnold Neumaier Nov 29 '16 at 15:55
  • $\begingroup$ @ArnoldNeumaier Are you aware of this single photon accumulation of double slit ? sps.ch/en/articles/progresses/… . That looks like a probability distribution for the QM solution "photon scattering off two slits" . The experiment seems to define one, though I do not know about the "rotational invariant way" . $\endgroup$ – anna v Nov 29 '16 at 16:12
  • $\begingroup$ See my comments on Virgo's answer, and the entries ''Particle positions and the position operator'' and ''Localization and position operators'' in Section B1 of my FAQ. There is an extended literature about position operators. $\endgroup$ – Arnold Neumaier Nov 29 '16 at 17:44
  • $\begingroup$ A single photon is detected at a single point. To get the interference pattern one needs to look at a large number of photons. The number of single detections per area is proportional to the cumulative intensity of the incident field at the measurement surface, hence no events can appear where there is zero intensity, i.e., destructive interference, and events tend to appear first where there is high intensity. This explains the observed pattern. But note that this happens in 2 dimensions only, at the screen. $\endgroup$ – Arnold Neumaier Nov 29 '16 at 17:52
  • $\begingroup$ But your probability statement referred to a 4-dimensional distribution, which is a tempting illusion but a mathematical impossibility. The impossibility is due to the transversal nature of polarization; to construct a probability density that transforms correctly under the rotation group one would need the longitudinal modes, too. $\endgroup$ – Arnold Neumaier Nov 29 '16 at 17:52
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Here's one approach to visualizing why both assertions are valid.

Feynman showed in his version of QED that the the most fundamental way to assess where a photon (or any particle) will travel in the future is to calculate every possible path it could take, then add all these complex-phase future paths so that the phases along each path can add or subtract. This causes the paths that are most consistently in phase to end up dominating, while other paths fade out as their phases add up chaotically to more-or-less zero.

In the case of a photon, this sum of an infinite number of "could have been" photon paths gives, to a very good first approximation, a set of wave functions whose probabilities look very much like the field intensities you get by using Maxwell's equations.

That is what Gloge is talking about when he asserts that Maxwell's equations nicely summarize the results predicted by QED for a single photon. However, Maxwell's equations do not give the mechanism for why that should be. QED reveals this deeper level of through its integral of all possible histories, and so enables a better understanding of a much broader range of phenomena.

It's worth noting that if a single photon did not behave like an infinite set of photons exploring every possible future path in parallel, you would be utterly blind and unable to read these words. A truly particle-like photon that cannot perceive the larger space around it would be scattered like a pinball in the maze of atoms and molecules that is your cornea and eye lens. Without the ability to explore every possible path and thus to "see" the large-scale structure of your eye, a photon could not even enter your eye, let alone be focused by the cornea and lens down to a single point on your retina.

You can interpret this larger perspective in either of two ways: As a QED style integral of all possible photon-particle histories, or as the quantized outcome of an excruciatingly weak electromagnetic wave subject to Maxwell's equations, with his equation re-interpreted as a probability once the field becomes so weak that only individual quanta of energy can be formed from it.

As for Sakurai's perspective, his point is simply that you can sort of "fill up" the infinite number of possible photon paths with enough energy to make some subset of them into real photons. This is helped by the fact that photons are bosons, and thus able to share the same wavefunction.

The process of adding energy to a photon wavefunction is pretty open ended, but you do have a minimum limit: One photon. After that you can add enough emitted energy to create two photons on arrival, or three, or some enormous number such as the photons in a laser beam. It really makes little difference for ordinary cases, since they all end up following the same infinite bundle of possible future histories. More energy just enables more photon pops at the receptor end, until they all start to merge together into what we think of as a beam of light.

One last note, which was covered nicely in one of the other answers: There are limits at which some of the simplifying assumptions start to break down. It is, for example, not strictly true that photons never interact with each other. It's "true enough" for ordinary energy domains to make it OK to ignore the possibility. However, if you point sufficiently intense beams of light at each other, with at least one of them in the gamma range, you can get a few photons to annihilate each other and produce electron-positron pairs. Time reversal symmetry requires that, since you can do the opposite interaction of colliding an electron and a positron to create two gamma rays.

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As we know, Maxwell came up with his equations long before the advent of quantum mechanics. So Maxwell's equations were not intended to describe photons. It is in that sense by itself a purely classical theory. It would fail to describe scenarios that represent a truly quantum aspect of nature. However, now that we know about photons, it is not uncommon to find an overlap between Maxwell's equation and quantum physics. So what is the relationship?

The role of quantum mechanics is often somewhat over stated, in that one can use a quantum mechanical description for situations that is perfectly well described by classical physics. Or, stated differently, the mathematical formulations that we would use to describe quantum mechanics can sometimes also be used to describe classical scenarios. (This begs the question, when is something intrinsically quantum? However, I'm not addressing this question here.)

So when we use Maxwell's equations, as opposed to quantum electrodynamics, to describe a situation that we would otherwise have considered to be a quantum mechanical scenario, then one may conclusion that such a scenario do not really represent the quantum aspects of nature. So what scenarios are like that? (Finally I come to the question):

There are two such scenarios that I can think of. The one is that of a single photon. But one needs to be careful. It is not the photon itself that is being described, but rather the wave function; in other words, the probability amplitude to find the photon at a particular point (or in a particular state, to be more general). So, if $|\psi\rangle$ is a single photon state, then one can expand it as $$ |\psi\rangle = \int |\mathbf{k}\rangle \psi(\mathbf{k}) d^3k , $$ where $\psi(\mathbf{k})$ is the Fourier domain wave function. In the classical sense the latter is interpreted as the angular spectrum, and can be used as such in calculations. However, more generally, one can also expand it as $$ |\psi\rangle = \sum_m |m\rangle \psi_m , $$ where $|m\rangle$ some discrete yet complete basis (such as the Laguerre-Gauss modes).

One also needs to be careful to exclude all interactions in this scenario, because interactions can introduce quantum phenomena for which Maxwell's equations are inadequate to deal with. In this sense Maxwell's equations describe the evolution of the wave function, of which a photon is considered to be a single excitation.

Having described it like this, we can allow more excitations, provided that we impose certain restrictions. The photons all need to be in the same state, which is in turn allowed by their bosonic nature. This restriction, however, makes it unfavourable to consider multiple photons. The reason is that superpositions of multiple photons can introduce the notion of non-local entanglement, which is a truly quantum aspect of nature and therefore cannot be represented in Maxwell's equations. This also reveals why interactions are to be excluded: they can (and generally do) lead to situations where one can find quantum entanglement.

Now for the other scenario. This one corresponds to the case where one wants to consider an infinite number of photons. It turns out that there is a type of quantum state that is called a coherent state which is considered to be the closest thing to a classical state. This state is a superposition of all the different number states (Fock states) all the way from a single photon state to a state that have an infinite number of photons (however, the latter comes with a coefficient that is practically zero). All the photons in a coherent state have the same properties in terms of their other degrees of freedom, thus avoiding the issue with quantum entanglement.

So, to summarize, I'd say that both Sakurai and Globe are correct. They just considered different scenarios.

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Do Maxwell's equations describe a single photon or an infinite number of photons?

Neither. Like I said, Maxwell's equations are really Heaviside's equations*. And they predate the photon, which arose due to the ultraviolet catastrophe. They do not describe the quantum nature of light. If they described a single photon, you'd be explain to me what a photon is, and you can't. And how can they possible describe an infinite number of photons? There aren't an infinite number of photons streaming into your eye, or flying around a magnet.

The paper Gloge, Marcuse 1969: Formal Quantum Theory of Light Rays starts with the sentence: Maxwell's theory can be considered as the quantum theory of a single photon and geometrical optics as the classical mechanics of this photon.

It can't. If it could, Planck wouldn't have needed to have come up with Planck's constant.

That caught me by surprise, because I always thought, Maxwell's equations should arise from QED in the limit of infinite photons according to the correspondence principle of high quantum numbers as expressed e.g. by Sakurai (1967): The classical limit of the quantum theory of radiation is achieved when the number of photons becomes so large that the occupation number may as well be regarded as a continuous variable. The space-time development of the classical electromagnetic wave approximates the dynamical behavior of trillions of photons.

I'm sorry, but that's wrong too. An electron in an orbital emits a photon. Not trillions of photons. This photon has an E=hc/λ nature. It has a wavelength. It is an electromagnetic wave. So an electromagnetic wave is not trillions of photons.

Isn't the view of Sakurai in contradiction to Gloge?

It is. But that's not to say either of them were right.

Do Maxwell's equation describe a single photon or an infinite number of photons?

The former, but they don't describe it enough because they don't cover the quantum nature of light. See this answer of mine for something about that.

Or do Maxwell's equations describe a single photon and also an infinite number of photons at the same time?

No.

But why do we need QED then at all?

Because Maxwell's equations don't describe the photon adequately. Or the electron. Can you explain how a magnet works? No. Because Maxwell's equations don't describe enough. But there again, QED doesn't describe the photon either, see this question of mine. Nor does it describe how photons interact with photons. It isn't the full story either.

  • See Wikipedia : "The four modern Maxwell's equations can be found individually throughout his 1861 paper, derived theoretically using a molecular vortex model of Michael Faraday's "lines of force" and in conjunction with the experimental result of Weber and Kohlrausch. But it wasn't until 1884 that Oliver Heaviside, concurrently with similar work by Josiah Willard Gibbs and Heinrich Hertz, grouped the twenty equations together into a set of only four, via vector notation.[3] This group of four equations was known variously as the Hertz–Heaviside equations and the Maxwell–Hertz equations, but are now universally known as Maxwell's equations".
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