The speed of one photon through a transparent medium

[PeroK]

So, basically, we don't know what a photon "does" between emission and detection

It's more fundamental than that. Take an atom itself as an example. It's not that we don't know what the electrons are doing. The atom exists as a bound energy state. That is a QM description of nature. It's not that there is a classical (locally realist) description under the covers that we don't know about. It's that atoms cannot be described in classical, locally realistic terms.

The QM description of light involves a quantized EM field (see post #32, for example). This description does not involve photons "doing" things. And, indeed, since the photon is a product of this quantum theory, it makes no sense to attribute classical behaviour to it.

I have one last question about photons in a medium: force carrier photons are always "traveling" with the speed c, regardless of the medium?

Those are virtual photons, which is a whole different subject. They are really just an aid to computation in perturbative QED - quite explicitly they do not have a speed; and, they don't even obey the energy-momentum relationship of a massless particle.

 

[vanhees71]

The input state to parametric down conversion is a coherent state from a laser. Does the BBO crystal somehow convert this into a pair of Fock states?

The laser emits of course a coherent state of high intensity. The spontaneous parametric downconversion is a pretty rare non-linear process, where one photon of the laser field is "split" into two photons (usually of half the frequency, but also other phase-matching conditions can occur). Due to the birefringence the photon momenta lie on two cones with axis in different directions, depending on the polarization. If you take photon pairs on the intersection of these two cones you get polarization- (and also momentum-) entangled photon pairs. The Wikipedia article explains it quite well (although you should, of course read, "momenta" instead of "trajectories" ;-)).

https://en.wikipedia.org/wiki/Spontaneous_parametric_down-conversion

 

[vanhees71]

So, basically, we don't know what a photon "does" between emission and detection, but we also don't need to, because, using QM, we

OK, I should probably buy Feynman's book. Thank you!I have one last question about photons in a medium: force carrier photons are always "traveling" with the speed c, regardless of the medium?

As I said before, you should not think of photons as classical particles. You cannot even define a position observable in the literal sense. That's because the em. field is a massless field with spin 1. As in the classical case also in the quantum description of the (quantized) electromagnetic field the field modes in the medium change due to the interaction of the em. field with the charged particles the medium is made of. This is described by the in-medium polarization (or "self-energy") of the electromagnetic field (aka photons).

It's also important to realize that there are different notions of speed/velocity of waves. There's the phase velocity, described by the relation between the (angular) frequency and the wave vector, $\omega=\omega(\vec{k})$ and the phase velocity then is $c_{\text{phase}}=\omega/|\vec{k}|$. There's nothing preventing this from being larger than $c_{\text{vac}}$, the speed of light in the vacuum. If you have, e.g., a dielectric, i.e., a medium where there are no (quasi-)free charge carriers but all the electrons are bound within the atoms/molecules making up the medium, you have resonances of the corresponding vibrations of these charges excited by an electromagnetic wave within the medium. You also have some "friction" in the effective description of the collective vibrational motion and also some resonances, and if the frequency of the incoming em. wave is close to such a resonance frequency the dispersion relation indeed becomes such that $c_{\text{phase}}=c_{\text{vac}}/n(\omega)>c_{\text{vac}}$ (where $n(\omega)$ is the refractive index of the medium at the frequency of the em. wave). That's the region of "anomalous dispersion", and this well-known fact gave rise to objections against the special theory of relativity, but the phase velocity has nothing to do with causal signal propagation, and thus it's not a violation of relativistic causality that $n(\omega)<1$ in the region of anomalous dispersion.

Another velocity of wave propagation is the socalled group velocity, $c_g=|\partial \omega/\partial \vec{k}|$. In the case that a certain approximation (saddle-point approximation) of the Fourier integral from the frequency to the time domain is valid, it describes the speed of the center of a wave packet. Now also this velocity can exceed $c_{\text{vac}}$, particularly in the region o anomalous dispersion, and again this doesn't mean any violation of relativistic causality. That's because in such cases the saddle-point approimation is no longer applicable, and the shape of the wave packet gets drastically deformed. That's why also the group velocity looses it's intuitive physical meaning.

What's really restricted to the speed of light, $c_{\text{vac}}$, is the "front velocity", i.e., if you have a wave pulse that's restricted to a finite volume in space, then the boundary of this volume must not spread faster than the speed of light, because then you'd really have faster-than-light signal propagation, i.e., you could always find a reference frame, where you get a signal even before the em. wave has been switched on, which is of course nonsense, and indeed there are (quite weak) constraints on the dispersion relation that guarantees causality and that the front velocity never exceeds the speed of, and these constraints are indeed fulfilled for the usual theory of dispersion (linear-response theory). In these models the front velocity turns out to be $c_{\text{vac}}$.

All this is known since the early history of the theory of relativity. It has been clarified by Sommerfeld in 1907 in an answer to W. Wien, whether there's a contradiction between the known fact that in the region of anomalous dispersion phase and group velocity can exceed $c_{\text{vac}}$. Later Sommerfeld and Brillouin studied the propation of the head of a wave train, when it enters from the vacuum the medium with interesting results on the wave shape (Sommerfeld and Brillouin precursors) and confirming the analytical result on the front velocity being just $c_{\text{vac}}$ in all calculational detail.

You find a good treatment of this in Sommerfeld, Lectures on Theoretical Physics, vol. 4 (Optics) or also in Jackson, Classical Electrodynamics (where also an experiment with microwaves confirming Sommerfeld's and Brillouin's calculations is quoted).

 

[isotherm]

As in the classical case also in the quantum description of the (quantized) electromagnetic field the field modes in the medium change due to the interaction of the em. field with the charged particles the medium is made of. This is described by the in-medium polarization (or "self-energy") of the electromagnetic field (aka photons).

First, I want to thank you for the entire post. Of course, I knew about phase, group and front velocities, but I enjoyed reading your presentation.

About the quote above, this is the quantum description of what the microscopic explanation in wikipedia is trying to say? I must confess that I don't quite understand your explanation.

And also I'm not quite satisfied with the one in wikipedia, because if the "shaken" charges radiate their own electromagnetic wave that is at the same frequency, it would mean that energy/photons is/are produced without any energy being taken from the incident wave (no photons absorbed). By the way, such an explanation can be tested with low intensity waves, where

you can actually see individual dots on the detector screen

by using parametric down-conversion, which creates entangled photon pairs by shining a laser on a birefringent crystal (like -barium borate, BBO) (see post #27). One can use the "idler" photon to "herald" the presence of the other photon ("signal") going through the medium. In this way we can see if any new/extra photon is emitted by the "shaken" charges.

But PeterDonis wrote, in post #32:

For a high intensity source (what you are calling a "beam of light"), it is no longer possible to see the individual dots; all you see is the interference pattern. The classical wave model used in the Wikipedia article can explain the interference pattern, but only if you ignore what the same experiment with a low intensity source is telling you about the individual photon detections. If you want to have a single model that explains both experiments, the classical wave model does not work. Only the quantum field theory model does.

so, as expected, no extra photons would be produced.

It is a bit confusing, at least for me, to promote the "shaken" charges explanation when you know that it does not work for low intensity sources. On the other hand, the "quantum field theory model" is harder to understand, so maybe this is the reason for the "classical wave model" to be still presented.I kind of knew that force carrier photons

are virtual photons ...

but I was not sure. I'm not very happy when virtual something is introduced in a theory ...
Anyway, thank you PeroK for your entire post.One more thing:

causal signal propagation

(for EM forces) is with/at speed c, right? If something changes, say an electron is annihilated, the charges nearby would react with a delay, the time needed for the information to reach them. Or they react instantly?! Or there is another speed/speeds for propagation? And this speed/speeds depend on the medium or it is always c?

 

[PeterDonis]

The spontaneous parametric downconversion is a pretty rare non-linear process, where one photon of the laser field is "split" into two photons

This still doesn't answer my question: are the "two photons" that come out of the BBO crystal Fock states? Or are they still coherent states?

Experimentally, the way to tell the difference would be to see if the output photons exhibited antibunching.

 

[PeterDonis]

if the "shaken" charges radiate their own electromagnetic wave that is at the same frequency, it would mean that energy/photons is/are produced without any energy being taken from the incident wave

It means no such thing. The "shaken" charges move because they are absorbing energy from the incident wave. The radiation they emit comes from that energy.

by using parametric down-conversion

You don't need to use parametric down conversion to see individual dots on the detector screen. That happens in any experiment involving low intensity light, no matter what kind of source it comes from.

In this way we can see if any new/extra photon is emitted by the "shaken" charges.

I don't know what you're talking about here. Do you have any kind of reference for an experiment?

as expected, no extra photons would be produced

Still don't know what you're talking about or what you think I was saying.

It is a bit confusing, at least for me, to promote the "shaken" charges explanation when you know that it does not work for low intensity sources.

We don't know any such thing. See above.

On the other hand, the "quantum field theory model" is harder to understand, so maybe this is the reason for the "classical wave model" to be still presented.

The classical wave model, as I've already said, works fine as long as you don't have to account for individual photon detections. Which is the case for many experiments. You are just focusing on experiments where we do have to account for individual photon detections, so the classical wave model doesn't work.

If something changes, say an electron is annihilated, the charges nearby would react with a delay, the time needed for the information to reach them.

Yes, as I've already said, the "speed of causality" is $c$. (Strictly speaking, causal influences don't always travel at $c$ because they don't always travel at the maximum possible speed for causal influences. But in the case we are discussing, they do.) However, as I've also already said, the "speed of causality" is not the same as either the phase velocity or the group velocity of EM waves in the case under discussion (where a medium is present). So thinking of EM waves "propagating at $c$" in this case is not correct.

 

[PeroK]

And also I'm not quite satisfied with the one in wikipedia, because if the "shaken" charges radiate their own electromagnetic wave that is at the same frequency, it would mean that energy/photons is/are produced without any energy being taken from the incident wave (no photons absorbed). By the way, such an explanation can be tested with low intensity waves ...

It's absurd to continue to confuse the classical and quantum theories of light.

 

[vanhees71]

This still doesn't answer my question: are the "two photons" that come out of the BBO crystal Fock states? Or are they still coherent states?

They are Fock states. That's the point, and that's why you can do all the Bell experiments with these true photon pairs from spontaneous parametric downconversion.

Experimentally, the way to tell the difference would be to see if the output photons exhibited antibunching.

Yes, this was among the first things done with SPDC by, among others, Mandel, Hong, Ou et al. You use one of the photons in the pair to "herald" the 2nd photon, which you use in, e.g., an HBT experiment, clearly demonstrating antibunching.

For a nice review on SPDC, see

https://arxiv.org/abs/1809.00127
https://doi.org/10.1080/00107514.2018.1488463

 

[Dale]

They are Fock states. That's the point, and that's why you can do all the Bell experiments with these true photon pairs from spontaneous parametric downconversion.

I didn't know that. So the laser comes in as a coherent state and the BBO crystal, when it produces a spontaneous parametric downconversion, produces a 2-photon Fock state.

 

[vanhees71]

Exactly. That's why SPDC is such a breakthrough for all these quantum-optics experiments (and applications), where you need true Fock states. In the very early days of experiments concerning entanglement, Bell's inequalities, and all that, they had to use atomic "cascades" as entangled-photon-pair sources, which was very inefficient and time-consuming.

For the historical development of the technical side of the "quantum-foundation revolution", see

Climerio Paulo da Silva Neto, Materializing the Foundations of Quantum Mechanics, Springer (2023)
https://doi.org/10.1007/978-3-031-29797-7

 

[PeterDonis]

They are Fock states.

Ok, thanks for the clarification and the references.

 

[isotherm]

The "shaken" charges move because they are absorbing energy from the incident wave.

Without photons from the incident wave being absorbed? How? The photoelectric effect showed that:

The experimental results disagree with classical electromagnetism, which predicts that continuous light waves transfer energy to electrons, which would then be emitted when they accumulate enough energy. An alteration in the intensity of light would theoretically change the kinetic energy of the emitted electrons, with sufficiently dim light resulting in a delayed emission. The experimental results instead show that electrons are dislodged only when the light exceeds a certain frequency—regardless of the light's intensity or duration of exposure. Because a low-frequency beam at a high intensity does not build up the energy required to produce photoelectrons, as would be the case if light's energy accumulated over time from a continuous wave, Albert Einstein proposed that a beam of light is not a wave propagating through space, but a swarm of discrete energy packets, known as photons—term coined by Gilbert N. Lewis in 1926

You don't need to use parametric down conversion to see individual dots on the detector screen. That happens in any experiment involving low intensity light, no matter what kind of source it comes from.

Yes, but by using it you know exactly how many photons entered the medium, not only how many photons got through it and hit the detector.

thinking of EM waves "propagating at c" in this case is not correct.

I didn't say or imply that. I only implied that EM forces are propagating/transmitted with the speed c, not instantly, nor slower.

 

[PeterDonis]

Without photons from the incident wave being absorbed?

How do you know photons from the incident wave are not being absorbed? You're not detecting any photons inside the medium. Your quote about the photoelectric effect is irrelevant since we are not discussing the photoelectric effect here. The photoelectric effect is not the only way a medium can absorb light.

by using it you know exactly how many photons entered the medium

How?

An actual reference (as in, a published peer-reviewed paper) for the kind of experiment you are describing would be helpful.

 

[isotherm]

How do you know photons from the incident wave are not being absorbed?

So photons from the incident wave are absorbed, in order to "shake" the charges inside the medium? Interesting, especially when photons are "traveling" one at a time, as described here (thank you renormalize for the link in post #18).

 

[PeterDonis]

So photons from the incident wave are absorbed, in order to "shake" the charges inside the medium?

If the charges are "shaken" by the incoming wave, then energy from the incoming wave must be absorbed by those charges. To describe this as "photons being absorbed" is not really a good description and I don't know why you insist on continuing to use it even though you have been repeatedly been told it is not a good description. But if you do insist on using it, then yes, if any energy is absorbed from the incoming wave then photons must be absorbed from the incoming wave; there is no way to shake the charges inside the medium without doing so.

especially when photons are "traveling" one at a time, as described here (thank you renormalize for the link in post #18).

Not all of the source photons make it through the medium. Figures 4 and 5 clearly show that at almost all angles the fraction of light transmitted is less than 1.

That said, even in these experiments, where the light passing through the medium is in a Fock state, it is still not a good description to think of what happens in the medium as "photons being absorbed". But again, if you insist on using that description, then the fact that the fraction of light transmitted is less than 1 means that some incoming photons are absorbed.

 

[vanhees71]

More formally what is discussed here is the self-energy of the photon (i.e., the em. field) in a medium. An example for a corresponding one-loop result within a model for propagation in a thermalized hadronic medium (simplified to a $\rho \pi$ gas) can be found here:

C. Gale and J. I. Kapusta, Vector dominance model at finite
temperature, Nucl. Phys. B 357, 65 (1991),
https://doi.org/10.1016/0550-3213(91)90459-B

 

[isotherm]

More formally what is discussed here is the self-energy of the photon (i.e., the em. field) in a medium. An example for a corresponding one-loop result within a model for propagation in a thermalized hadronic medium (simplified to a $\rho \pi$ gas) can be found here:

C. Gale and J. I. Kapusta, Vector dominance model at finite
temperature, Nucl. Phys. B 357, 65 (1991),
https://doi.org/10.1016/0550-3213(91)90459-B

I can only see the abstract and, maybe because of that, I don't understand how is the article relevant for this discussion.

But if you do insist on using it, then yes, if any energy is absorbed from the incoming wave then photons must be absorbed from the incoming wave; there is no way to shake the charges inside the medium without doing so.

Ok, so photons must be absorbed from the incoming wave in order to "shake" the charges.

How about the radiation emitted (with a delay) by the "shaken" charges? Is this a normal radiation or virtual? I ask this because if it's "normal" it must "contain" photons, meaning that photons are emitted by the "shaken" charges, and this explanation with the "shaken" charges sounds like absorption/re-emission, especially when we have single photons passing through the medium.

 

[vanhees71]

I can only see the abstract and, maybe because of that, I don't understand how is the article relevant for this discussion.

The photon propagator, which is given by the photon self-energy in the medium, describes how a photon "moves through this medium". So it's exactly what you asked for, i.e., how a photon behaves in the medium.

Ok, so photons must be absorbed from the incoming wave in order to "shake" the charges.

How about the radiation emitted (with a delay) by the "shaken" charges? Is this a normal radiation or virtual? I ask this because if it's "normal" it must "contain" photons, meaning that photons are emitted by the "shaken" charges, and this explanation with the "shaken" charges sounds like absorption/re-emission, especially when we have single photons passing through the medium.

That's what's described by the in-medium photon propagator. The photon self-energy takes into account precisely the interaction of the photon (or rather the em. field) with the charged particles making up the medium.

 

[PeroK]

The photon propagator, which is given by the photon self-energy in the medium, describes how a photon "moves through this medium". So it's exactly what you asked for, i.e., how a photon behaves in the medium.

That's what's described by the in-medium photon propagator. The photon self-energy takes into account precisely the interaction of the photon (or rather the em. field) with the charged particles making up the medium.

You may be taking at cross purposes. You are describing the scenario using QED. The OP is describing the scenario based on a single photon as a classical particle that moves through space at $c$ and also represents an EM wave.

 

[vanhees71]

The only way you can describe photons is in terms of QED. There is no classical particle of light in any sense. Then the question is simply not making any sense, i.e., you cannot describe something in physics which cannot exist on the very principles the physical theory you need to describe this situation (in this case the special theory of relativity).

 

[Dale]

Ok, so photons must be absorbed from the incoming wave in order to "shake" the charges.

How about the radiation emitted (with a delay) by the "shaken" charges? Is this a normal radiation or virtual? I ask this because if it's "normal" it must "contain" photons, meaning that photons are emitted by the "shaken" charges, and this explanation with the "shaken" charges sounds like absorption/re-emission, especially when we have single photons passing through the medium.

I don't think that this is a good way to think about this.

 

[Orodruin]

The only way you can describe photons is in terms of QED. There is no classical particle of light in any sense. Then the question is simply not making any sense, i.e., you cannot describe something in physics which cannot exist on the very principles the physical theory you need to describe this situation (in this case the special theory of relativity).

Unfortunately, many SR textbooks will use ”photon” in the sense described by @PeroK . I usually make a point of rather using ”short light pulse” instead in my relativity classes because that is usually what you can substitute meaningfully.

 

[PeroK]

Unfortunately, many SR textbooks will use ”photon” in the sense described by @PeroK . I usually make a point of rather using ”short light pulse” instead in my relativity classes because that is usually what you can substitute meaningfully.

The difference is that Taylor and Wheeler and Helliwell, for example, use photon as a placeholder for the energy-momentum that is released in a decay or particle collision. They don't attempt to analyse the finer points of EM theory using this model.

A vague analogy would be to study geology using the model of the Earth as a perfect sphere or point mass, as it is modelled in basic solar system dynamics.

 

[Nugatory]

You may be taking at cross purposes. You are describing the scenario using QED. The OP is describing trying to describe the scenario based on a single photon as a classical particle that moves through space at c and also represents an EM wave and finding that it doesn't work.

Edits are my take on why this discussion between @isotherm and @vanhees71 is at cross purposes. The underlying problem is that photons aren't what people expect when they hear "particle" or "particle of light" or the like, but many relativity texts and posters here (myself included when I'm in a hurry or don't want to come across as a cranky old pedant) don't worry about this when it is otherwise irrelevant to the relativistic principle being discussed. However, in this discussion it very much does matter.

Unfortunately, many SR textbooks will use ”photon” in the sense described by @PeroK . I usually make a point of rather using ”short light pulse” instead in my relativity classes because that is usually what you can substitute meaningfully.

I have, with tongue in cheek, suggested we should install a bot that automatically substitutes "flash of light" for "photon" in all posts in the relativity forum.

 

[PeterDonis]

photons must be absorbed from the incoming wave in order to "shake" the charges.

Energy must be absorbed from the incoming wave to shake the charges. Thinking of this as "photons" being absorbed is not a good way to think about it, as has already been said multiple times.

How about the radiation emitted (with a delay) by the "shaken" charges? Is this a normal radiation or virtual?

Normal. It's just radiation emitted by accelerated charges.

I ask this because if it's "normal" it must "contain" photons

It must contain energy. As above, and as has been said multiple times, thinking of this as "photons" is not a good way to think about it. Certainly there is no way to measure individual photons in either the absorption of the incoming wave or the emission by the shaken charges.

this explanation with the "shaken" charges sounds like absorption/re-emission

Yes.

especially when we have single photons passing through the medium.

Once more, "single photons" is not a good way to think about what is going on in the case of very low intensity.

 

[Vanadium 50]

Once more, "single photons" is not a good way to think about what is going on in the case of very low intensity.

We've been saying this for three weeks. The OP hasn't budged.

Where I think he's going is that if a photon goes into the apparatus, and a photon of the same energy comes out of the apparatus, any explanation involving energy transferred to the medium cannot be right because there is no energy to do it. (If this is not the OP's position, I urge him to write more clearly and specifically)

Unfortunately, the OPs description is a hodgepodge of classical and quantum. It suffers from not proposing a specific configuration and a specific set of measurements (for example, one does not measure velocity so much a sposition and time.)

Without clarification, I can only speculate, but I am unconvinced that even with a single photon state in you get a single photon state out. Even in those case where you do, the medium can still transfer energy to and from the electromagnetic field without violating anything. What it might be doing whenn it is not being measured is not something addressibnle by QM, or arguably by science.

But let's hope we don't go three more weeks.

 

[PeterDonis]

I am unconvinced that even with a single photon state in you get a single photon state out.

So am I, since the experiments that have been referenced in this thread explicitly show less than 100% transmission of incoming light.

 

[vanhees71]

How do you come to that conclusion? With a single photon going into a medium you can have much more than simply a kind of "elastic scattering". There can come out any number of other particles, more photons and whatever else is "allowed" by the conservation laws (energy, momentum, electric charge).

 

[PeterDonis]

How do you come to that conclusion?

By reading the paper referenced in post #18.

With a single photon going into a medium you can have much more than simply a kind of "elastic scattering".

Yes, but all those other phenomena appear to be outside the scope of this thread. The OP is having enough trouble understanding the simplest case.

 

[renormalize]

So am I, since the experiments that have been referenced in this thread explicitly show less than 100% transmission of incoming light.

To the extent that the refractive medium has negligible loss (absorption), isn't $T<100\text{%}$ just because the experiment I cited didn't measure the single photons that are reflected? In other words, for a lossless medium, shouldn't single photons in give single photons out, either transmitted or reflected?

 

[PeterDonis]

To the extent that the refractive medium has negligible loss (absorption), isn't $T<100\text{%}$ just because the experiment I cited didn't measure the single photons that are reflected?

As far as I can tell, that experiment did not measure reflected photons, so yes, at least some of the less than 100% transmission could be due to reflection.

In other words, for a lossless medium, shouldn't single photons in give single photons out, either transmitted or reflected?

I think the general answer here is still no, because "photons" aren't trackable; even if you have a 1-photon Fock state going in and a 1-photon Fock state coming out, you can't say the photon that came out is "the same photon" as the one that went in. You just happened to have a process happening inside the medium that output a 1-photon Fock state from a 1-photon Fock state input. (This is the case whether the photon coming out is transmitted or reflected.) And as @vanhees71 has pointed out, there are lots of other processes that could also happen, even if we eliminate absorption in the medium.

 

[Vanadium 50]

So am I, since the experiments that have been referenced in this thread explicitly show less than 100% transmission of incoming light.

This may actually be required by the boundary conditions.

I've been thinking about the reverse situation. Do the experiment under water. A single photon source could be built, but of course photons in water are not the same as photons in air or vacuum. OK, now put in the medium, which is vacuum. Do you still have one photon in the vacuum? I don't think you do. I can't see matching boundary conditions and still being in an eigenstate of photon number.

I don't think the boundary condition of perfect transmission works here. At some angles you will have total external reflection. Since a photon's path will include such angles, well there you go.

 

[renormalize]

I think the general answer here is still no, because "photons" aren't trackable; even if you have a 1-photon Fock state going in and a 1-photon Fock state coming out, you can't say the photon that came out is "the same photon" as the one that went in.

I'm not claiming that the single photon-out is the same photon that came in, merely that a single photon-in results in a single flash in either the transmit- or reflect-detectors (but never both!) with a probability given by the Fresnel transmission and reflection coefficients (assuming no absorption).

And as @vanhees71 has pointed out, there are lots of other processes that could also happen, even if we eliminate absorption in the medium.

This statement I need help to understand. If, as @vanhees71 says, a single photon in can give rise to "any number of other particles, more photons and whatever else is "allowed" by the conservation laws (energy, momentum, electric charge)", why doesn't this "non-Fresnel" behavior persist as we increase the light intensity to "classical" levels? Simply put, why does an EM wave transiting a lossless medium obey the Fresnel equations if single photons do not?

 

[Dale]

why does an EM wave transiting a lossless medium obey the Fresnel equations if single photons do not?

The classical equations are obtained from the expectation values of the quantum equations. In other words, they are what an average photon does. Individual photons are not strictly average.

 

[PeterDonis]

I'm not claiming that the single photon-out is the same photon that came in, merely that a single photon-in results in a single flash in either the transmit- or reflect-detectors (but never both!) with a probability given by the Fresnel transmission and reflection coefficients (assuming no absorption).

Ok.

If, as @vanhees71 says, a single photon in can give rise to "any number of other particles, more photons and whatever else is "allowed" by the conservation laws (energy, momentum, electric charge)", why doesn't this "non-Fresnel" behavior persist as we increase the light intensity to "classical" levels? Simply put, why does an EM wave transiting a lossless medium obey the Fresnel equations if single photons do not?

The amplitudes for the other processes @vanhees71 refers to would be expected to be very small, small enough that even "classical" light intensities would not trigger them to a measurable extent. But if one is trying to understand the underlying theory, one has to at least be aware that these other processes exist and that their existence complicates any attempt to construct a simple picture in terms of "photons".