I The speed of one photon through a transparent medium

[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".

 

[isotherm]

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.

Not exactly, because the abstract says that "The neutral ϱ-meson propagator is computed" ... and, as I already wrote, the abstract is all that I can read.

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.

Where I can read about this?

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

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.

Individual photons exiting the medium can be and are measured but, of course, there is no way to tell if the photon detected is the one that entered or one (re-)emitted by the shaken charges.

Yes.

Ok, thank you.

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

Why? Because the "shaken charges explanation" becomes awkward? I ask this because "single photons" is not only accepted, but very much used in Bell experiments.

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)

Yes and no. My point is that if one photon enters the medium and the same photon exit it, yes, energy cannot be transferred to the medium in order to shake the charges. But that doesn't mean that energy cannot be transferred. If the original photon is absorbed "in order to shake the charges" and than the "shaken charges" emit a similar photon, it would work. PeterDonis already admitted in post #60 that:

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

 

[PeterDonis]

Why? Because the "shaken charges explanation" becomes awkward?

No. The reasons have already been given repeatedly in this thread. I do not understand why you continue to ignore these repeated statements.

I ask this because "single photons" is not only accepted, but very much used in Bell experiments.

By "single photons" you mean Fock states (@vanhees71 has given references for the usage of those in optical experiments). But as has already been said (and as you appear to agree), the fact that you have a 1-photon Fock state going in and a 1-photon Fock state coming out does not mean you can reason based on a "single photon" traveling inside the medium. As has already been said repeatedly in this thread.

 

[PeterDonis]

if one photon enters the medium and the same photon exit it

Which, as has already been said, and as you appear to agree, is not the case. Even if you have a 1-photon Fock state going in and a 1-photon Fock state coming out, that does not mean the two are "the same photon". You are reasoning from a false premise.

 

[PeterDonis]

Thread closed for moderation.

 

[PeterDonis]

After moderator review, the thread will remain closed. The OP question has been answered. Thanks to all who participated.