The speed of one photon through a transparent medium

vanhees71 said:

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.

vanhees71 said:

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?

PeterDonis said:

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.

PeterDonis said:

Yes.

Ok, thank you.

PeterDonis said:

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.

Vanadium 50 said:

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:

isotherm said:

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

isotherm said:

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.

isotherm said:

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.

isotherm said:

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.

Thread closed for moderation.

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