I The speed of one photon through a transparent medium

[isotherm]

TL;DR Summary

I want to learn if the speed of one photon in a transparent medium is c/n or not and, of course, why.

We know from double-slit experiments that singular photons behave like waves, so I expect that one photon would undergo refraction when entering, with an angle different than 90 degree, into water, glass or other transparent material. Is that true?

If the refraction occurs, than the speed of singular photons though the transparent material is c/n? If this is true, how it is explained? I ask this because the explanation from wikipedia with the charges in the material "shaken" back and forth and radiating their own electromagnetic wave doesn't sound right. I don't see how from one photon you can obtain more (their own electromagnetic wave). If it is correct, we may use it to produce energy

So, what is the speed of one photon through a transparent medium and why?

 

[PeterDonis]

"One photon" is not what you think. The concept of "speed" does not even make sense for one photon. Nor does the concept of having a definite path through space, which is what you would need for "refraction" to make sense. Nor does the concept of "electromagnetic wave". (The "waves" that explain interference in the double slit experiment for single photons are quantum waves, which are not the same thing as the "electromagnetic wave" that appears in Wikipedia's explanation.)

In short, none of your questions are even well defined for "one photon".

 

[PeroK]

TL;DR Summary: I want to learn if the speed of one photon in a transparent medium is c/n or not and, of course, why.

We know from double-slit experiments that singular photons behave like waves, so I expect that one photon would undergo refraction when entering, with an angle different than 90 degree, into water, glass or other transparent material. Is that true?

If the refraction occurs, than the speed of singular photons though the transparent material is c/n? If this is true, how it is explained? I ask this because the explanation from wikipedia with the charges in the material "shaken" back and forth and radiating their own electromagnetic wave doesn't sound right. I don't see how from one photon you can obtain more (their own electromagnetic wave). If it is correct, we may use it to produce energy

So, what is the speed of one photon through a transparent medium and why?

The photon is, technically, the quantum of the quantized EM field. And a single photon is a particular state of the EM field. Feynman's book "The Strange Theory of Light and Matter" gives a basic QM description of the phenomena of reflection, refraction and diffraction.

There are no photons in classical EM, where light is modelled as an EM wave obeying Maxwell's equations. It's a common misconception to confuse the two theories of light. It doesn't make any sense to try to apply the classical explanation in Wikipedia to a photon.

 

[Dale]

As @PeterDonis and @PeroK said, your question actually is quite problematic. A lot of the things you like to think about for massive classical particles don’t apply for massless quantum particles. At least, not without a lot of caveats.

However, if you have a single photon source of optical photons, and you send them through a lens one at a time, then the density of photon detections will be very high at exactly the same place that classical light will focus.

Make what you will of that. To me it is a “yes with lots of aforementioned caveats”

 

[Vanadium 50]

You can have source emitting a photon at a certain place and time (sort of). You can have detector receiving a photon at a certain place and time (again, sort of). But there is no way to say these are the same photon, much less say anything about a path from source to detector.

 

[isotherm]

The concept of "speed" does not even make sense for one photon.

If I search "speed of photon" I get:

Photons are massless, so they always move at the speed of light when in vacuum, 299792458 m/s

It is true? And "always" means also in a medium?

 

[isotherm]

The "waves" that explain interference in the double slit experiment for single photons are quantum waves, which are not the same thing as the "electromagnetic wave" that appears in Wikipedia's explanation.

Ok.

It doesn't make any sense to try to apply the classical explanation in Wikipedia to a photon.

Ok, thank you.

 

[isotherm]

You can have source emitting a photon at a certain place and time (sort of). You can have detector receiving a photon at a certain place and time (again, sort of). But there is no way to say these are the same photon, much less say anything about a path from source to detector.

So, if you "shoot" singular photons, with an angle, through water, and you put detectors in the water, both straight ahead and where a light beam would go, you should determine if the refraction occurred. It was done?

 

[hutchphd]

h

So, if you "shoot" singular photons, with an angle, through water, and you put detectors in the water, both straight ahead and where a light beam would go, you should determine if the refraction occurred. It was done?

How does this differ from taking an underwater photograph? Source is the sun, CCD in camera has many detectors (in an array). I have seen such photographs.
What does "shoot" singular photons mean???

 

[isotherm]

How does this differ from taking an underwater photograph? Source is the sun, CCD in camera has many detectors (in an array). I have seen such photographs.
What does "shoot" singular photons mean???

It is sending/emitting one photon at a time, like in the double-slit experiment.

 

[PeterDonis]

It is sending/emitting one photon at a time, like in the double-slit experiment.

Again, you have to be careful. In double slit experiments where the intensity of the source is turned down very low, the source is not emitting "one photon at a time". The state that the source emits is not a "one photon" state (Fock state); it is a coherent state. Thinking of it as one little billiard ball moving at the speed of light is not correct. Neither is thinking of it as "one photon's worth of a wave".

 

[isotherm]

Again, you have to be careful. In double slit experiments where the intensity of the source is turned down very low,...

Ok, but if was done? We used the same source (or similar) to determine refraction?

 

[PeroK]

Ok, but if was done? We used the same source (or similar) to determine refraction?

A percentage of light will reflect from water. That's an interesting question. If you emit one photon at a time, why do some photons pass through the water (and refract) and some reflect?

 

[Dale]

So, if you "shoot" singular photons, with an angle, through water, and you put detectors in the water, both straight ahead and where a light beam would go, you should determine if the refraction occurred. It was done?

What do you think is different in this question than in what I already posted? If the probability of a single photon refracts through a lens, what makes you think it wouldn’t refract in water?

 

[PeterDonis]

We used the same source (or similar) to determine refraction?

Yes.

 

[Vanadium 50]

If I search

So why should I bother writing an answer if you're just going to disregard what people write and ask a search engine or CrapGPT?

 

[PeterDonis]

If I search "speed of photon"

This is not a good way of understanding physics.

It is true?

It's too ill specified to even be evaluated as "true" or "false".

And "always" means also in a medium?

Even leaving aside what I said above, read what you quoted again. Does it say "in a medium"? No; it says "in" something else. What is that something else?

 

[renormalize]

Ok, but if was done? We used the same source (or similar) to determine refraction?

You can without a doubt measure the refraction of single photons. Here's one experiment by Wang and Strausser:
Single-photon determination of transmission, index of refraction and material thickness

 

[PeterDonis]

You can without a doubt measure the refraction of single photons.

"Single photons" here refers to the detectors used, which are discrete; they record individual "photon detections", not a continuous measurement of something like an electromagnetic field.

The states of the quantum EM field involved in these experiments, however, are, as far as I can tell, not Fock states; they are coherent states (from a laser via down conversion in a BBO crystal). These states are not eigenstates of photon number and cannot be accurately described as "single photon" states. They are the closest quantum analogue to classical "electromagnetic wave" states.

The extremely low intensity of the source in this experiment can also be taken to mean that the expectation value of photon number inside the experiment is no more than one; that is another sense in which the description "single photon" is sort of applicable. But the expectation value won't be exactly one, and doesn't correspond to any kind of "single photon" in the actual EM field state anyway.

 

[Dale]

The states of the quantum EM field involved in these experiments, however, are, as far as I can tell, not Fock states; they are coherent states (from a laser via down conversion in a BBO crystal)

Are there readily available sources that do produce Fock states?

 

[PeterDonis]

Are there readily available sources that do produce Fock states?

Not readily available, no. My understanding is that, while experiments have been done with such sources, they are hard to build and hard to do experiments with.

 

[vanhees71]

Again, you have to be careful. In double slit experiments where the intensity of the source is turned down very low, the source is not emitting "one photon at a time". The state that the source emits is not a "one photon" state (Fock state); it is a coherent state. Thinking of it as one little billiard ball moving at the speed of light is not correct. Neither is thinking of it as "one photon's worth of a wave".

You can nowadays do the double-slit experiment with single photons, and what you get is when repeating the experiment many times just the expected interference pattern.

It's of course wrong to think of photons as localized massless particles. They don't even have a position observable and thus photons cannot be localized in any classical-particle sense. As an intuitive picture it's usually best to think in terms of electromagnetic waves.

 

[Dale]

You can nowadays do the double-slit experiment with single photons,

What is the name of a source that produces a single photon Fock state? Or are you talking about coherent states with low enough amplitude that the expectation of more than one photon is negligible.

 

[isotherm]

You can without a doubt measure the refraction of single photons. Here's one experiment by Wang and Strausser:
Single-photon determination of transmission, index of refraction and material thickness

Very interesting. Thank you!

What do you think is different in this question than in what I already posted? If the probability of a single photon refracts through a lens, what makes you think it wouldn’t refract in water?

Sorry, somehow I missed your confirmation of refraction (post #4).

So, single photons do refract when entering in another medium. Does it mean that there is a decrease in speed (for single photons) when entering (from vacuum) in a medium with refractive index greater than 1? How/why is this (both the refraction and the possible decrease in speed) happening?

So why should I bother writing an answer if you're just going to disregard what people write and ask a search engine or CrapGPT?

The information was from google and wikipedia. And I really doubt that the quote (see post #6) was wrong. Was it? How it is correct?

 

[Nugatory]

You can nowadays do the double-slit experiment with single photons,

”Single-photon” in the sense that the detection events happen one at a time, or nthe sense that single-photon state has been prepared?

 

[isotherm]

The "waves" that explain interference in the double slit experiment for single photons are quantum waves, which are not the same thing as the "electromagnetic wave" that appears in Wikipedia's explanation.

So the interference in the double-slit experiment for "single-photons" is different, in result, from one recorded with a beam of light (with the same frequency)? If not, why not?

 

[vanhees71]

”Single-photon” in the sense that the detection events happen one at a time, or nthe sense that single-photon state has been prepared?

In the sense that you prepare a true single-photon state. That can be done by using parametric down-conversion, which creates entangled photon pairs by shining a laser on a birefringent crystal (like $\beta$-barium borate, BBO). Then you can use the one photon ("idler") to "herald" the presence of the other photon ("signal"), which you let go through the double slit. What you get is of course a random dot at the screen for each photon. The pattern that forms when you store all these dots after repeating the experiment with many equally prepared photons is the interference pattern you expect from the classical theory. The meaning is that the detection probability distribution is proportional to the (time-averaged) energy density of the em. field.

This is not too exciting, but you can of course do things with single photons very much different from classical em. waves, which from the point of view of QED are coherent states, even with very much dimmed-down ones (where you have mostly vacuum). One example is to make use of both of the entangled photons from parametric down-conversion to realize things like the quantum eraser (also in the Wheeler delayed-choice setup).

 

[PeroK]

So, single photons do refract when entering in another medium. Does it mean that there is a decrease in speed (for single photons) when entering (from vacuum) in a medium with refractive index greater than 1? How/why is this (both the refraction and the possible decrease in speed) happening?

There is no locally realistic picture of quantum mechanical phenomena, such as the refraction of light. If you set up an experiment with a light source and a detector, then you can calculate the probability of detecting a photon with the detector in any given position. In this case, the probabilities are consistent with refraction.

You really ought to read Feynman's notes on this:

https://en.wikipedia.org/wiki/QED:_The_Strange_Theory_of_Light_and_Matter

It's becoming a little tiresome that you are still pounding us with your locallly realistic questions on QM. Questions like "what was the photon doing while it was moving through the water?" and "what speed was it moving at time $t$?" are not questions that can even be asked in QM.

If you don't want to think quantum mechanically, then you should stick to classical EM and forget about photons - as they are elements of a QM theory of light and not the classical particles, that you been have repeatedly insisting on!

 

[vanhees71]

Indeed "the in-medium photon" is characterized by its in-medium spectral function, which can be calculated in many-body (relativstic) QFT. For a application of this in contemporary heavy-ion research, see, e.g.,

https://itp.uni-frankfurt.de/~hees/publ/habil.pdf

 

[Dale]

Does it mean that there is a decrease in speed (for single photons) when entering (from vacuum) in a medium with refractive index greater than 1?

Again, that question is fraught with all of the issues and caveats identified earlier. Insofar as, in my opinion, it can be answered at all, I already answered it. If you perform the experiment I indicated you will get the results described.

Again, make what you will of that. To me it is a “yes with lots of aforementioned caveats”.

I am not going to exclude the caveats, because they are important. But I also don’t want to simply surrender and say that the question has no part that can be answered. So I answer it by reference to an experiment.

 

[PeterDonis]

The information was from google and wikipedia.

And, as you were told in post #17, that is not a good way to understand physics, and what you quoted is not reliable information.

And I really doubt that the quote (see post #6) was wrong.

As you were told in post #17, it's too ill specified even to be evaluated as "right" or "wrong".

 

[PeterDonis]

So the interference in the double-slit experiment for "single-photons" is different, in result, from one recorded with a beam of light (with the same frequency)? If not, why not?

Apparently you aren't even reading what others are posting in response to you. The term "single photons" is not a good description of the experiments you are referring to. The only difference between what you are calling "single photons" and "a beam of light" is the intensity of the source: the quantum state emitted by the source is the same in both cases.

For a low intensity source (what you are calling, mistakenly, "single photons"), you can actually see individual dots on the detector screen. These detections are often referred to as "photon detections", but that does not mean what you think it means. It just means that the detections are discrete dots. It doesn't mean that the light in between the source and detector is "single photons". It's just a very low intensity coherent state--the same state as with a high intensity source (see below). Over time, the individual dots build up an interference pattern on the detector. This behavior is not explainable by classical waves; the Wikipedia article's explanation using classical waves does not explain why the detections are individual dots.

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.

 

[PeterDonis]

In the sense that you prepare a true single-photon state. That can be done by using parametric down-conversion

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?

 

[isotherm]

Questions like "what was the photon doing while it was moving through the water?" and "what speed was it moving at time ?" are not questions that can even be asked in QM.

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

can calculate the probability of detecting a photon with the detector in any given position

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?

 

[PeterDonis]

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

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

You do realize that the first quote from you above, which is correct, means that the second quote from you above is a bogus question, right?

 

[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.