What do physicists mean when they say photons have a "path"?

[andrew s 1905]

Light rays, i.e., directions of wave vectors or in this case, because single photons (in entangled biphoton states) are prepared, of photon momenta.

For most, but the very pedantic, that would seem to be what might be termed a "path" not "the path" of course. Regards Andrew

 

[vanhees71]

I always thought that "path" is inanimously defined as a trajectory of a classical particle in configuration space.

 

[andrew s 1905]

I always thought that "path" is inanimously defined as a trajectory of a classical particle in configuration space.

That may well be its meaning in classical mechanics but why can't it not have a use out side of that. Many terms in physics are like that. Langauge is intended to aid communication. If you insist on a rigorous precision no internal entity's term in QED has any meaning outside of the mathematical formalism and you should reject spin, momentum and others that have classical mechanics definitions and invent new ones.

Take an electron we still use the term but its "meaning" is very different now compared to that when JJ Tomson discovered it. It seems to me people still talk about electron tracts (paths) in a cloud chamber.

Regards Andrew
.

 

[vanhees71]

That may well be its meaning in classical mechanics but why can't it not have a use out side of that. Many terms in physics are like that. Langauge is intended to aid communication. If you insist on a rigorous precision no internal entity's term in QED has any meaning outside of the mathematical formalism and you should reject spin, momentum and others that have classical mechanics definitions and invent new ones.

That's not what I mean, as should be pretty clear from the specific topic we discuss here. The problem is that in the case of the discussed entangled photon pairs there is not specific paths for either of the two photons even in the loose meaning it may have for a single photon, and even for the single photon it must be understood in the correct way, as it is used in the textbooks and papers quoted so far in this thread (as far as I'm aware of them).

Take an electron we still use the term but its "meaning" is very different now compared to that when JJ Tomson discovered it. It seems to me people still talk about electron tracts (paths) in a cloud chamber.

Regards Andrew

It's well understood since Mott's famous paper, why one discovers "tracks" or "paths" of charged particles in a cloud chamber and that this of course doesn't contradict in any way the fundamental principles of quantum mechanics:

N. Mott, The Wave Mechanics of alpha-Ray Tracks,
Proceedings of the Royal Society of London. Series A 126, 79
(1929), https://doi.org/10.1098/rspa.1929.0205

Here a single ##\alpha## particle originating from a radioactive source. Quantum mechanically the ##\alpha## particles are described as a spherical wave, and indeed if you look at the cloud chamber you see many ##\alpha## particles being produced with momenta in all directions. On the other hand any single ##\alpha## particle makes a nice straight track (or a curved one when an em. field is applied). That's the issue nicely treated in Mott's paper: It's sufficient to have aview interactions of the ##\alpha## particle with a atom/molecule of the vapour in the cloud chamber to determine the momentum and position of the particle. This is of course by far less accurate than allowed by the uncertainty relation. So there is no contradiction with the fact that initially the ##\alpha## particle has no determined (direction of) momentum but nevertheless you see a single track (or "path") in the cloud chamber.

 

[DrChinese]

a. Once more, the correct description of the idealized entangled-photon pair is (here for the polarization-singlet state as an example):
$$|\Psi \rangle = \frac{1}{\sqrt{2}} [\hat{a}^{\dagger}(\vec{k}_1,H) \hat{a}^{\dagger}(\vec{k}_2, V) - \hat{a}^{\dagger}(\vec{k}_1,V) \hat{a}^{\dagger}(\vec{k}_2,H)] |\Omega \rangle.$$
In words: It describes the superposition of two possible "paths" (if you insist on using this word) of creation of this state: The creation of a photon with momentum ##\vec{k}_1## and polarization H and a photon with momentum ##\vec{k}_2## and polarization V is superimposed with the situation that a photon with momentum ##\vec{k}_1## and polarization V and a photon with momentum ##\vec{k}_2## and polarization H are created

a. And once again, your science is flat out wrong. There is absolutely nothing that requires an entangled photon pair to be momentum entangled, or path entangled, or to have ever been such. It might be, or it might not be. So the *correct* description of an idealized spin/polarization entangled photon pair DOES NOT INCLUDE a momentum term at all (you seem to think there is something special about photon momentum as an observable, and there isn't).

"Yet conceptually, the simplest examples of entangled states of two photons are the polarization entangled 'Bell states': HV ± VH and HH ± VV." See (1) from Kwiat et al (1999). Of course, the Zeilinger reference says precisely the same thing, that the photon pair is polarization entangled (no mention of momentum). Now it is true that PDC produced pairs are usually momentum entangled initially, but that is usually just a temporary state of affairs as is the case in the Zeilinger experiment. More importantly (and you will probably miss this point completely): In general, spin entangled photon pairs need never have been momentum entangled. Let me know if you need a reference on this point.

b. (even you must admit that the formula is much more clear than this description).

c. In the parametric down conversion it is impossible to predict which of the "two paths" is realized, and that's why we have this superposition and thus the entanglement...

b. Of course, my words were more clear than your (unnecessarily convoluted) formula to a B or I level reader.

c. This too is wrong. In PDC: there is a requirement of path indistinguishability, but it is NOT for the photon's path out of the crystal. For them to be entangled, it is ONLY a requirement that you don't know whether it emerges from the V cone or the H cone. (Note: formation of the V and H cone varies for different PDC crystal setups, and is not relevant for my answer.) Once you grab a pair of photons from the intersection of the V and H cones, you can do whatever you like to their path and momentum; they stay polarization entangled until a measurement of same occurs.

d. It's pretty ironic that you quote this paper against my opinion although it's another quite convincing example for it!
https://arxiv.org/abs/quant-ph/0607182

d. And once again you dismiss my specific refutation of your words, and even have the nerve to provide my reference to support your position without quoting a single word or otherwise explaining. Here's the black and white on this:

You said: "With a setup that establishes classical paths you destroy the entanglement..."

Zeilinger et al says: [In my words, read the full setup for a proper description of the 144 km path from the source to Bob] We did everything possible to locate a direct line of sight - a classical path that is as straight a line as is possible - so we could send a polarization entangled photon to Bob. That included use of classical optical devices to route the photon to the open atmosphere for 89 miles, and to collect the photon upon arrival at the destination island. That's how we established the classical path to Bob to confirm entanglement.

Conclusion: No quantum preparation was involved in establishing the photon path, which is as close to classical as is feasible. Entanglement was preserved and demonstrated, in contradiction to your statement. The photon going to Alice was measured as Bob's was just getting on its way. Therefore it did not exist during most of its partner's journey on its near perfect classical path. (Again, I am not saying entangled photons travel on a classical path. They can't, because they are quantum particles.)

 

[vanhees71]

a. And once again, your science is flat out wrong. There is absolutely nothing that requires an entangled photon pair to be momentum entangled, or path entangled, or to have ever been such. It might be, or it might not be. So the *correct* description of an idealized spin/polarization entangled photon pair DOES NOT INCLUDE a momentum term at all (you seem to think there is something special about photon momentum as an observable, and there isn't).

Can you write down the two-photon state you want to discuss?

I wrote down the (idealized) description of a state created in SPDC. That's what you read all the time in the papers describing the experiments using them, including those you quote yourself all the time. Often one writes something like
$$|\Psi \rangle=\frac{1}{\sqrt{2}} (|H_A V_B \rangle-H_B V_A \rangle),$$
where A and B refer to the positions of the detectors used to measure the single-particle polarizations. That's of course (together with the known frequencies/energies of the photons) equivalent to use the momenta of the photons.

 

[andrew s 1905]

@vanhees71 I was in Motts department at Cambridge and well understand his paper on alpha ray tracks! Although he was working on amorphous soilds when I was there.

Obviously, by your classical mechanic's definition it is not a "path" but a your "Light ray" definition well be considered to be!

However, no point in going round in circles.

Regards Andrew

 

[DrChinese]

I think it might help if the type of experiment fluidfcs seemed to refer to was discussed. There's a big choice of experiments including the very famous one one carried out by Kim et al. Patrick Edwin Moran prepared a nice schematic of the apparatus used and I think this is a good representation of the apparatus actually used. I'm guessing Kim et al approve of it. With reference to this schematic we could ask questions such as the following:
1. What is the purpose of the prisms the lens and the mirrors?
2. What do the red and green lines represent?

Good point.

The setup was presented in a diagram in OP's post #3. It is just a generic diagram of a Bell test using polarization, PDC source. Because he was asking basic questions about "paths", I answered using "paths". No reason to explain to him that some physicists don't believe in the Photon (see Lamp's Anti-photon), or to others that Photons are "excitations of the quantized electromagnetic field" (QFT), or that photons don't move precisely on classical paths. All of which are technically correct but not important to the OP - whose level of knowledge likely would not appreciate the nuance of those points.

As to your Kim diagram: the red and green lines represent the paths of the A and B photons...

 

[Dadface]

https://www.physicsforums.com/attachments/298157

Do you have a reference, where this schematic drawing is explained in more detail? Is it referring to the quantum-eraser experiment (at least it looks like a colored version of Fig. 2):

https://arxiv.org/abs/quant-ph/9903047v1
https://doi.org/10.1103/PhysRevLett.84.1ad 1. They have the usual purposes as optical elements and are described as in classical electrodynamics with an index of refraction.

Light rays, i.e., directions of wave vectors or in this case, because single photons (in entangled biphoton states) are prepared, of

The schematic I posted is basically the same as the schematic in Kim etal's paper ( Arxiv version) but is slightly more detailed in that idler photons from both semi silvered mirrors a can be detected.

 

[Dadface]

Good point.

The setup was presented in a diagram in OP's post #3. It is just a generic diagram of a Bell test using polarization, PDC source. Because he was asking basic questions about "paths", I answered using "paths". No reason to explain to him that some physicists don't believe in the Photon (see Lamp's Anti-photon), or to others that Photons are "excitations of the quantized electromagnetic field" (QFT), or that photons don't move precisely on classical paths. All of which are technically correct but not important to the OP - whose level of knowledge likely would not appreciate the nuance of those points.

As to your Kim diagram: the red and green lines represent the paths of the A and B photons...

Agreed.

 

[vanhees71]

What's the merit to call the lines "paths" rather than (directions of) photon momenta?

 

[Philip Koeck]

There's actually something I've been wondering about in connection with an electron microscope.
If you image very thin specimens, which are almost pure phase objects, the only way you can get an image is by interference between scattered and unscattered electrons using something like holography, a phase plate or just defocusing (Fresnel imaging).
An unscattered electron travels along the optical axis, whereas a scattered one leaves the specimen at some small angle.
Now the strange thing is that you can reduce the current of the electron beam so much that there is, on average, only 1 electron in the vicinity of the specimen (a few mm) at the same time, and you still get an image after a sufficient exposure time.

Does that mean the electrons are interfering with themselves and that each electron is both scattered and unscattered and therefore follows many paths?
Or am I overinterpreting?

 

[Paul Colby]

There is no doubt that experimentalists measure something that they call photon paths

I’ve done photon counting experiments in the past and I’ve never measured a photons path. I’ve certainly inferred a path or line of sight using slits baffles and lens systems. Calling this inference a measurement of the photon path seems strained.

 

[vanhees71]

There's actually something I've been wondering about in connection with an electron microscope.
If you image very thin specimens, which are almost pure phase objects, the only way you can get an image is by interference between scattered and unscattered electrons using something like holography, a phase plate or just defocusing (Fresnel imaging).
An unscattered electron travels along the optical axis, whereas a scattered one leaves the specimen at some small angle.
Now the strange thing is that you can reduce the current of the electron beam so much that there is, on average, only 1 electron in the vicinity of the specimen (a few mm) at the same time, and you still get an image after a sufficient exposure time.

Does that mean the electrons are interfering with themselves and that each electron is both scattered and unscattered and therefore follows many paths?
Or am I overinterpreting?

Although I'm not very familiar with the theory of the electron microscope from your description it's clear that in this case the wave aspects of electrons are needed to describe the phenomena.

The point of QT is that you cannot describe electons as classical point particles nor as classical (wave) fields. A single electron [edit: corrected in view of #85] at not too high energies can be described in non-relativistic QT by a wave function. It's meaning is that of a "probability amplitude", i.e., ##|\psi(t,\vec{x})|^2 \mathrm{d}^3x## is the probability to find the electron in a small volume ##\mathrm{d}^3 x## around ##\vec{x}##, i.e., you register a single electron always as a pointike object, but the probability distributions show the typical wave properties like interference, diffraction, refraction, etc.

 

[mattt]

Although I'm not very familiar with the theory of the electron microscope from your description it's clear that in this case the wave aspects of electrons are needed to describe the phenomena.

The point of QT is that you cannot describe electons as classical point particles nor as classical (wave) fields. A single photon at not too high energies can be described in non-relativistic QT by a wave function. It's meaning is that of a "probability amplitude", i.e., ##|\psi(t,\vec{x})|^2 \mathrm{d}^3x## is the probability to find the electron in a small volume ##\mathrm{d}^3 x## around ##\vec{x}##, i.e., you register a single electron always as a pointike object, but the probability distributions show the typical wave properties like interference, diffraction, refraction, etc.

You surely meant a single electron ;-)

 

[gentzen]

Now the strange thing is that you can reduce the current of the electron beam so much that there is, on average, only 1 electron in the vicinity of the specimen (a few mm) at the same time, and you still get an image after a sufficient exposure time.

Does that mean the electrons are interfering with themselves and that each electron is both scattered and unscattered and therefore follows many paths?
Or am I overinterpreting?

Yes, each electron is both scattered and unscattered, as long as certain conditions are satisfied: Your "electron source + electron optics" ensures that the electrons reach the surface of your specimen with a pretty well defined energy and momentum. Otherwise it doesn't even matter whether the electron interferes with itself, because the different interference patterns of the different electrons would just average out each other, so that no useful pattern would be left. You need more than one or two electrons to be able to recognize a useful interference pattern.

Overall, this condition restricts how your electron optics can look like, how much you can focus your beam, and which electron energies you can use. Another condition is that scattering events that change the energy of the electron by a random amount should be rare, at least if you are working in transmission. This basically means that (elastic) scattering at the nuclei should be much more common than (inelastic) scattering the atomic electrons. This implies that the incident electron energy should not be too small.

The thought experiment to reduce the current until on average the electrons are isolated suggests a misleading image of the electron source as simply single electrons. The correct image of the source includes the entire preparation procedure, including the electron optics and the energy spead of the emitted electrons.

 

[DrChinese]

I’ve done photon counting experiments in the past and I’ve never measured a photons path. I’ve certainly inferred a path or line of sight using slits baffles and lens systems. Calling this inference a measurement of the photon path seems strained.

No one called our discussion a measurement of a path - we were simply talking about what happens with entangled photon pairs. Specifically, we were referring to one taking a path to detector A and another to detector B. The presented diagram (by the OP) using labeling as A and B. What you call it - path or whatever - had absolutely no relevance to the question. So all of this is a giant sidetrack, .

------------------------------

On the other hand, there are plenty of entangled photon experiments where path measurement (or path timing - essentially the same thing for our purposes) IS a precise factor. It varies according to the experiment, and it is not unusual for the path measurement to be relative (path to Alice vs path to Bob). For example, that is what is done with observation of H-O-M dip ("Hong-Ou-Mandel interferometry relies on spatial overlap of the photons requiring balancing of path lengths traversed by them."). Here is an experiment which appeared where there is extensive theoretical and technical discussion on the topic of photon path length.

Weiner et al (2022) Nonlocal subpicosecond delay metrology using spectral quantum interference
https://arxiv.org/abs/2202.11816

"Path" and "path length" are mentioned 17 times. Not once did they feel the need to deny that photons don't have paths. (Again, no one is claiming photons are classical particles or take classical paths.) As you say, as a practical matter you have inferred a path without calling it a measurement of same. But there are times when it is, such as this experiment: when high resolution timing measurements are made and compared to theory.

 

[Paul Colby]

Specifically, we were referring to one taking a path to detector A and another to detector B.

Sorry, I’m going off the post #1 which is about photons having a path. When one reduces the discussion to just detector A and B, things seem quite well defined. What I’m sticking on has little to do with entanglement per say.

Consider two detectors, A and B, viewing a monochromatic point source located at O. The detectors are sufficiently spaced so that interference isn’t a factor. What I’m hearing is that a photon detected in detector A follows a path OA while a count in B followed the path OB. To me it seems reasonable (well, it’s reasonable but completely false) that if a photon followed path OA it did not follow path OB and, likewise, if it followed OB it did not follow OA. The problem with this statement is it implies an anti correlation between the two detectors. Turns out for all light sources I can think of the counts are completely uncorrelated. How does this square with the photon having a path?

 

[PeterDonis]

Turns out for all light sources I can think of the counts are completely uncorrelated.

No, they're not. If you have a source emitting single photons, for each emission, you either get a detection at A or a detection at B. You never get both.

What will be true (at least, assuming the experiment is set up to not favor either detector over the other) is that, for each emission, you cannot predict which detector, A or B, will detect a photon; that will be completely random. So of course if you turn up the intensity of the source so that you can no longer distinguish single photon emissions, it will look like the readings at the detectors are completely uncorrelated. But that won't remain the case as you turn the intensity down to the point where you can distinguish single photon emissions again.

 

[DrChinese]

Sorry, I’m going off the post #1 which is about photons having a path. When one reduces the discussion to just detector A and B, things seem quite well defined. What I’m sticking on has little to do with entanglement per say.

Consider two detectors, A and B, viewing a monochromatic point source located at O. The detectors are sufficiently spaced so that interference isn’t a factor. What I’m hearing is that a photon detected in detector A follows a path OA while a count in B followed the path OB. To me it seems reasonable (well, it’s reasonable but completely false) that if a photon followed path OA it did not follow path OB and, likewise, if it followed OB it did not follow OA. The problem with this statement is it implies an anti correlation between the two detectors. Turns out for all light sources I can think of the counts are completely uncorrelated. How does this square with the photon having a path?

Entangled pairs will essentially go both OA and OB (one each path). The A and B detectors will click near simultaneous (when any path length difference is considered). I think that squares nicely with photons having a path.

Of course, you can choose to partially block any hypothetical alternate path and see if that changes the click coincidences. I have never seen that done, but it could be.

 

[Paul Colby]

No, they're not. If you have a source emitting single photons, for each emission, you either get a detection at A or a detection at B. You never get both.

This is very true, however the vast majority of light source[1] are poisson distribute. In these cases there is exactly no correlation for quantum particles. On the other hand a poisson distributed source of classical particles will be anti correlated because they have a trajectory or path. To convince I’d really need to write out the classical case.

[1] I’m not aware of any sources of single photon sources that don’t use entangled partners. If there are, I’d be interested.

 

[Paul Colby]

Entangled pairs will essentially go both OA and OB (one each path). The A and B detectors will click near simultaneous (when any path length difference is considered). I think that squares nicely with photons having a path.

Yes, but in the unentangled case I was discussing, I don’t believe this is the case.

 

[PeterDonis]

the vast majority of light source[1] are poisson distribute. In these cases there is exactly no correlation for quantum particles.

Most light sources can't even operate at intensities low enough where you could try to distinguish individual photon emissions. But for sources where you can get the intensity down that low--for example, highly attenuated lasers--if you have two detectors and both detect a photon, the explanation is that the source emitted two photons, not that you had one photon going both places. Which means this actually works the same as your description of the classical case:

a poisson distributed source of classical particles will be anti correlated because they have a trajectory or path.

Each individual particle does: but if the source has a nonzero probability of emitting two particles at once (just as a laser will have a nonzero probability of emitting two photons at once), the directions of the two particles will have zero correlation. So I don't see how the classical and quantum cases are any different in this respect. (Which should not be surprising since a coherent state, which is what a laser emits, is the closest we can get in quantum field theory to a classical state of the electromagnetic field.)

 

[Paul Colby]

Most light sources can't even operate at intensities low enough where you could try to distinguish individual photon emissions. But for sources where you can get the intensity down that low--for example, highly attenuated lasers--if you have two detectors and both detect a photon, the explanation is that the source emitted two photons, not that you had one photon going both places. Which means this actually works the same as your description of the classical case:

The experiment runs as follows. Photons are counted for an interval of time, ##\Delta T##. This results in ##A_1## counts in A and ##B_1## in B. This is then repeated generating count pairs, ##A_k## ##B_k##. The correlation,

##\sigma = \langle AB\rangle - \langle A\rangle \langle B \rangle##

the intensity in the source is not an issue other than it needs to be stable.

edit. I’m discussing the experiment given in #88 second paragraph. It involves photon paths for non entangled particles.

 

[PeterDonis]

The experiment runs as follows. Photons are counted for an interval of time, ##\Delta T##. This results in ##A_1## counts in A and ##B_1## in B. This is then repeated generating count pairs, ##A_k## ##B_k##. The correlation,

##\sigma = \langle AB\rangle - \langle A\rangle \langle B \rangle##

the intensity in the source is not an issue other than it needs to be stable.

edit. I’m discussing the experiment given in #88 second paragraph. It involves photon paths for non entangled particles.

So how is this different from an experiment where classical particles are emitted and counted?

 

[Paul Colby]

So how is this different from an experiment where classical particles are emitted and counted?

Well, the claim is ##\sigma \ne 0## for poisson distributed classical particles while for poisson distributed monochromatic light ##\sigma = 0##. Its been a while since I’ve looked at this so I need to step through the classical argument again. I recall posting this a while back.

 

[PeterDonis]

the claim is ##\sigma \ne 0## for poisson distributed classical particles

Why? Please show your work.

for poisson distributed monochromatic light ##\sigma = 0##.

Yes. But the reasoning that leads to this conclusion applies equally well to the classical case: that each particle emission is independent of the others. In other words, the same property that justifies using a Poisson distribution to model the emissions.

 

[Philip Koeck]

Yes, each electron is both scattered and unscattered, as long as certain conditions are satisfied: Your "electron source + electron optics" ensures that the electrons reach the surface of your specimen with a pretty well defined energy and momentum. Otherwise it doesn't even matter whether the electron interferes with itself, because the different interference patterns of the different electrons would just average out each other, so that no useful pattern would be left. You need more than one or two electrons to be able to recognize a useful interference pattern.

Overall, this condition restricts how your electron optics can look like, how much you can focus your beam, and which electron energies you can use. Another condition is that scattering events that change the energy of the electron by a random amount should be rare, at least if you are working in transmission. This basically means that (elastic) scattering at the nuclei should be much more common than (inelastic) scattering the atomic electrons. This implies that the incident electron energy should not be too small.

The thought experiment to reduce the current until on average the electrons are isolated suggests a misleading image of the electron source as simply single electrons. The correct image of the source includes the entire preparation procedure, including the electron optics and the energy spead of the emitted electrons.

I believe an actual experiment was carried out (somewhere in England) quite recently, but I can't find the paper anymore.
I'm pretty sure they used a TEM with a cold field emission gun so the electron wave hitting the specimen ought to be quite (spatially) coherent and monochromatic. Energy is probably 200 or 300 keV.
I wish I could find the paper.

 

[vanhees71]

You surely meant a single electron ;-)

Ok course ;-(.

 

[vanhees71]

Entangled pairs will essentially go both OA and OB (one each path). The A and B detectors will click near simultaneous (when any path length difference is considered). I think that squares nicely with photons having a path.

Of course, you can choose to partially block any hypothetical alternate path and see if that changes the click coincidences. I have never seen that done, but it could be.

This is the misconception! All you can say is that there are two photons being produced in O and being registered in A and B (supposed both detectors registered a photon). You cannot say each one took a certain path but the two possibilities (H-polarized photon going to A V-polarized photon going to B and vice versa) are not distinguished in this state. That's the whole fascinating issue with entanglement.

It's also important to keep in mind that you consider only situations which are really measured. All you can say are the probabilities for photon-detection events given a specific setup and the given preparation of the two photons in the entangled state. Of course, the coincidences will change, depending on the experimental setup. E.g., if you put a polarization filter letting through H-polarized photons before the detector at A and one letting through V-polarized photons (with ideal detectors) you'll always get either a coincidence registration of both photons or none. That will of course change if you rotate on of the polarization filters to another angle. You can calculate the corresponding coincidence probabilities from the state.

I don't really know, what you mean by "blocking any hypothetical alternate path".

 

[Paul Colby]

Yes. But the reasoning that leads to this conclusion applies equally well to the classical case: that each particle emission is independent of the others. In other words, the same property that justifies using a Poisson distribution to model the emissions.

Yes, I'm in error.

If the emission rate is ##\lambda## then the joint classical probability works out to

##P(A,B) = \frac{\lambda^{A+B}e^{-\lambda}}{2^{A+B}A!B!}##

which clearly factors. ##A## and ##B## are independent classically as well.

 

[DrChinese]

1. This is the misconception! All you can say is that there are two photons being produced in O and being registered in A and B (supposed both detectors registered a photon). You cannot say each one took a certain path but the two possibilities (H-polarized photon going to A V-polarized photon going to B and vice versa) are not distinguished in this state. That's the whole fascinating issue with entanglement.

2. It's also important to keep in mind that you consider only situations which are really measured.

3. I don't really know, what you mean by "blocking any hypothetical alternate path".

1. Of course, with polarization entangled photons: there is no H and there is no V prior to a polarizer (you said this, but I wasn't sure it was clear). But when discussing the paths, it matters not that they are indistinguishable.

There is NO requirement that entangled photons are indistinguishable for them to be polarization entangled. For example, they can be from different PDC sources - that would distinguish them nicely.

https://arxiv.org/abs/0809.3991

CHSH measured at S=2.37 (where S<2 is required by local realism).2. I've referenced multiple papers where the relative path / path length / path timing are measured as accurately as current technology allows. Above reference does this too.3. That would be some other path than the classical one. Since entangled photons take almost precisely the classical path, you could block that path (with a small object precisely positioned) and see the coincidence count changes. That would demonstrate the path taken was quantum (non-classical). You would expect that occasionally, since photons are quantum particles. (I am not aware any similar test has ever been published.) I wonder how big that object would been to be, to block close to 100% of the coincidences?

 

[vanhees71]

1. Of course, with polarization entangled photons: there is no H and there is no V prior to a polarizer (you said this, but I wasn't sure it was clear). But when discussing the paths, it matters not that they are indistinguishable.

There is NO requirement that entangled photons are indistinguishable for them to be polarization entangled. For example, they can be from different PDC sources - that would distinguish them nicely.

https://arxiv.org/abs/0809.3991

CHSH measured at S=2.37 (where S<2 is required by local realism).

But the entanglement swapping also is precisely possible, because you don't have classical paths. The selection via projecting the two photons in directions 2 and 3 into one of the Bell states (ensuring in fact maximum indeterminacy of the single photon "paths") lead to the entanglement of photons in directions 1 and 4 although indeed these never interacted with each other before. In other words, again you need the specifically quantum feature of quanta to do this, and the violation of the "CHSH inequality" demonstrates right this!

2. I've referenced multiple papers where the relative path / path length / path timing are measured as accurately as current technology allows. Above reference does this too.

No paths are measured. There are only coincidence registrations of several photons.

3. That would be some other path than the classical one. Since entangled photons take almost precisely the classical path, you could block that path (with a small object precisely positioned) and see the coincidence count changes. That would demonstrate the path taken was quantum (non-classical). You would expect that occasionally, since photons are quantum particles. (I am not aware any similar test has ever been published.) I wonder how big that object would been to be, to block close to 100% of the coincidences?

Again, what do you mean by path here and what do you mean in saying they were taking almost precisely the classical path? Classically we have light waves and the matter (various optical elements used in the experiment) in the area where they propagate.

Just explain what you really want to do, i.e., which optical elements you want to place where in the above setup and which coincidence measurement you want to perform. I hope I can then figure out what is expected concerning these measurements.

 

[DrChinese]

But the entanglement swapping also is precisely possible, because you don't have classical paths. The selection via projecting the two photons in directions 2 and 3 into one of the Bell states (ensuring in fact maximum indeterminacy of the single photon "paths") lead to the entanglement of photons in directions 1 and 4 although indeed these never interacted with each other before. In other words, again you need the specifically quantum feature of quanta to do this, and the violation of the "CHSH inequality" demonstrates right this!

Once again, not a single word of your response has the slightest connection to what I said. Which was:

"There is NO requirement that entangled photons are indistinguishable for them to be polarization entangled. For example, they can be from different PDC sources - that would distinguish them nicely.

https://arxiv.org/abs/0809.3991

CHSH measured at S=2.37 (where S<2 is required by local realism)."

How the entangled pair 1 & 4 got entangled is not really relevant to my point, it's just another of your rabbit trails. But it certainly has nothing to do with the quantum nature of the paths of 2 and 3 (and yes, they are quantum objects). In fact, they too trace an almost perfect classical path to the Bell State Analyzer in the experiment. Of course, as I get tired of saying, entangled photons are not classical particles. And in fact their very entanglement is proof of same, with S=2.37 and therefore ruling out local realism (where S must be less than 2).

So I will repeat: Indistinguishability is not necessary for entanglement, in contradiction to your statement. And in general, not referring to photons at all, entanglement is not even required to be between like quantum objects. Different quantum objects, of course, being very distinguishable as well.

 

[neilparker62]

Number of arxiv hits on "Photon" in the Abstract: 64,980
Number of arxiv hits on "excitations of the quantized electromagnetic field" in the Abstract: 0

Usually, voting doesn't make it so in science. But in this case, it actually does. It's "Photons" and "Photon Path" by a landslide!

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"Why 100? If I were wrong, one would have been enough." - Albert Einstein, when told of publication of the book One Hundred Authors Against Einstein.