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

[vanhees71]

Good point AndresB, I wouldn't always take this position. But these are entangled particles. They do not produce double slit interference.

It depends, of course, on the setup. Recently we have discussed a nice double-double-slit experiment with entangled photons, demonstrating the original EPR debate about entanglement in momentum/position space (open-access article):

Kaur, M., Singh, M. Quantum double-double-slit experiment with momentum entangled photons. Sci Rep 10, 11427 (2020). https://doi.org/10.1038/s41598-020-68181-1

But even in the double slit, with a normal photon, I would say it took a "path". Being a quantum particle, it did not take one specific path in an experiment designed to highlight this quantum behavior. I point out another such in post #41 above, point 4: where light does not take classical paths.

Come on! That's at the very fundamentals of all quantum theory that there is no classical path. You cannot understand the diffraction pattern in experiments like that with single particles nor single photons. It's fasterfully explained in the introductory chapter of the Feynman lectures vol. 3.

But in an instance where such properties can be neglected - which was the case in the thread where this originated - of course I would talk about a photon's path. Virtually all photons that arrive anywhere do so upon what is almost perfectly a classical path - that is easily demonstrated by blocking anywhere along the most likely path. Should we need factor in the effect of gravity on light when we discuss how light moves? I would not mention that gravity bends light when discussing photon path unless there are large objects involved. To recap my position (from another post of mine above):

Entangled photons travel on paths whether or not they are in fiber. They almost exactly follow classical trajectories. They are NOT classical paths however, for a lot of reasons. The main reason is that photons are quantum particles, not classical particles. I don't know if an individual photon travels on one path, many paths (path integral concept), different paths in different MWI worlds, exact Bohmian trajectories, are continuous or not, etc. They can do lots of things when not being observed. (Nobody I aware of on this planet has any superior understanding of the "truth" of what happens.)

Photons don't take paths in the sense of classical particles. The classical limit of the quantized electromagnetic field is not a point-particle theory but classical Maxwell theory. AFAIK there is no satisfactory Bohmian interpretation of relativistic QFT.

And yet: every experimentalist does all test calibration as if they are observing entangled photons moving on their precisely desired "classical" path with a classically expected arrival time relative to the entangled partner. Let's just call that a path, like everyone else does.

No! When taking the effort to use entangled photon pairs usually researchers are after quantum effects, where a classical particle picture is inapplicable. Particularly photons don't even have a position observable. All that is physical are detection probabilities at positions defined by the detector at times measured by a clock.

Guess what? All of the above is also true about momentum, position, etc. of any quantum particle. And yet we use the words momentum and position to discuss such particles. Obviously, if I choose to place these attributes in superposition, those words can become meaningless. So I would say there is context to consider, just as when answering a question on any subject.

For massive particles there exists a position observable, a non-relativistic approximation, and a classical point-particle limit, but not for photons. It is important to use the words "position" and "momentum" in the right way, when it comes to quantum properties of particles and photons, particularly in cases, which are not describable in terms of the classical point-particle picture, and particles or photons in entangled states are the extreme example for a situation, where the classical point-particle picture and even the classical theory of "local realism" a la EPR fails.

Parametric downconversion to prepare entangled photon pairs is also a characteristic example: The entanglement comes from the fact that you select such pairs via phase matching that you can't know in principle which "path" each of the photons took, and that's why you have a superposition that is not a product state, like the type-II case
$$|\Psi_{\pm} \rangle=\frac{1}{\sqrt{2}} [\hat{a}(\vec{k}_1,H) \hat{a}(\vec{k}_2,V) \pm \hat{a}(\vec{k}_1,V) \hat{a}(\vec{k}_2,H)]|\Omega \rangle.$$
As an example, see the paper with the theory about it you quoted yourself yesterday, where the state of the created pairs is treated in great detail including all the deviations from the above quoted idealized example:

https://arxiv.org/abs/quant-ph/0103168v1

 

[vanhees71]

What exactly is the word photon supposed to refer to in this context? Is it just any state of the quantized EM field?

A photon is a single-photon Fock state by definition.

 

[Demystifier]

AFAIK there is no satisfactory Bohmian interpretation of relativistic QFT.

Define "satisfactory"!

Certainly, there are versions of Bohmian relativistic QFT that make the same measurable predictions as standard relativistic QFT. If that's not satisfactory enough, then you should say what more do you expect from a satisfying theory.

 

[vanhees71]

Is there a causal theory of classical point-particle trajectories?

 

[Demystifier]

Is there a causal theory of classical point-particle trajectories?

Yes, it's called classical mechanics. But I think you meant something else.

Anyway, nobody said that Bohmian QFT must be about point-particles. There are versions of Bohmian QFT with fields instead of particles.

 

[Lord Jestocost]

With respect to terms like “path“ in our language which is based on classical notions:

“I deduce two general conclusions from these thought-experiments. First, statements about the past cannot in general be made in quantum-mechanical language. …. As a general rule, knowledge about the past can only be expressed in classical terms.”

Freeman Dyson in „Thought Experiments in Honor of John Archibald Wheeler“ in “Science and Ultimate Reality”, Cambridge University Press, New York, 2004

 

[DrChinese]

1. That's at the very fundamentals of all quantum theory that there is no classical path.

2. You cannot understand the diffraction pattern in experiments like that with single particles nor single photons.

1. For the Nth time, read the words I write. I never said any photon takes a classical path. What I have said is:

a) Photons are quantum particles, demonstrate quantum observables, and do quantum things.
b) Entangled photons generally take a *near* classical path. They lack precise origin/creation times when created via PDC, which is what we were discussing.
c) The experimentalist, setting up a Bell test, acts as if it takes a classical path; and essentially ignores the contributions of paths outside of that as being grouped with the classical path.
d) There is nothing particularly different about the quantum nature of an entangled photon's path as compared to its momentum, wavelength, etc. A PDC entangled photon does not have a single well defined momentum or wavelength any more (or less) than it has a single well defined path. Yet we talk about photon momentum (you did anyway) and wavelength without feeling the need to consult a 664 page textbook when we discuss that.
e) Authors of scientific papers on the subject almost universally talk about photon path (or momentum, or wavelength), just as I did. Call it a "path" as it is commonly used in scientific writing. If it makes you feel better or it's more accurate: call it a "quantum path".

2. Importantly: the discussion topic was the entangled photons' paths labeled A and B. It had nothing to do with an experiment specifically designed to highlight quantum diffraction. Referring to entangled photon "path" makes perfect sense in this and most other contexts, just as you might refer to photon "wavelength".

 

[PeterDonis]

A photon is a single-photon Fock state by definition.

Is that the kind of state that is produced by the sources in the experiments we have been discussing. My understanding is that the answer to that is "no": the sources in these experiments produce coherent states, not Fock states.

 

[HomogenousCow]

I think that's the central issue here, the typical coherent state produced by a classical source is not an eigenstate of particle number. In fact I suspect you can't make a spatially localized state without superimposing states with arbitrarily-high numbers of photons.

 

[vanhees71]

Is that the kind of state that is produced by the sources in the experiments we have been discussing. My understanding is that the answer to that is "no": the sources in these experiments produce coherent states, not Fock states.

Parametric down conversion is today the standard method to produce true two-photon states,from which you can prepare proper single-photon states by using one of the photons in the pair to indicate that the other an only the other photon is there and not just a very dim coherent state.

 

[vanhees71]

1. For the Nth time, read the words I write. I never said any photon takes a classical path. What I have said is:

a) Photons are quantum particles, demonstrate quantum observables, and do quantum things.

Why then don't you stick to this obvious statement?

b) Entangled photons generally take a *near* classical path. They lack precise origin/creation times when created via PDC, which is what we were discussing.

No, they don't. To the contrary, as I stressed several times, that's the main misunderstanding when talking about paths. When it comes to entangled two-photon states you have rather "superpositions of two paths". That's what makes these particular "maximally entangles Bell states" so special and "most quantum".

c) The experimentalist, setting up a Bell test, acts as if it takes a classical path; and essentially ignores the contributions of paths outside of that as being grouped with the classical path.

No he doesn't but he uses a setup to demonstrate the entanglement. With a setup that establishes classical paths you destroy the entanglement, as is exemplified by the which-way experiments like, e.g., in the quantum erasure experiment discussed in the other thread.

d) There is nothing particularly different about the quantum nature of an entangled photon's path as compared to its momentum, wavelength, etc. A PDC entangled photon does not have a single well defined momentum or wavelength any more (or less) than it has a single well defined path. Yet we talk about photon momentum (you did anyway) and wavelength without feeling the need to consult a 664 page textbook when we discuss that.

There is an entangled-photon pair, not two single photons with well determined single-particle properties. That's the entire point of EPR, Bell and all that. I also don't understand why you critisize that I quoted a textbook I consider a very good one after you (sic!) asked for a reference.

e) Authors of scientific papers on the subject almost universally talk about photon path (or momentum, or wavelength), just as I did. Call it a "path" as it is commonly used in scientific writing. If it makes you feel better or it's more accurate: call it a "quantum path".

In the papers it's usually made clear what's meant by using the mathematical notation of QED, making it very clear what's meant and not a naive photon picture of 1905!

2. Importantly: the discussion topic was the entangled photons' paths labeled A and B. It had nothing to do with an experiment specifically designed to highlight quantum diffraction. Referring to entangled photon "path" makes perfect sense in this and most other contexts, just as you might refer to photon "wavelength".

To repeat it ince more: that's the main misunderstanding when talking about paths. When it comes to entangled two-photon states you have rather "superpositions of two paths". That's what makes these particular "maximally entangles Bell states" so special and "most quantum".

 

[vanhees71]

I think that's the central issue here, the typical coherent state produced by a classical source is not an eigenstate of particle number. In fact I suspect you can't make a spatially localized state without superimposing states with arbitrarily-high numbers of photons.

Indeed that was the main challenge to experimentally realize Bell's ideas to test the EPR notion of "local realusm" against QT. Indeed classical sources produce (maybe pretty dim) coherent states which are not true Fock states, with which you can't do the corresponding experiments. Today one has parametric down conversion as a "heralded-single-photon source" as well as all kinds of entangled two-photon states.

 

[gentzen]

To repeat it once more: that's the main misunderstanding when talking about paths. When it comes to entangled two-photon states you have rather "superpositions of two paths". That's what makes these particular "maximally entangles Bell states" so special and "most quantum".

Indeed, this "superpositions of nearly classical paths" is what "experimentalist" mean by an entangled photon's path. But it wasn't clear to me whether you would not also object to this way of including classical paths in the description of a quantum experiment. And I guess DrChinese will also agree that the paths in quantum experiments can be in superposition. But the point is that using the word "path" is fine, there is no need to forbid this word.

 

[DrChinese]

1. Why then don't you stick to this obvious statement?

2. No, they don't. To the contrary, as I stressed several times, that's the main misunderstanding when talking about paths. When it comes to entangled two-photon states you have rather "superpositions of two paths". That's what makes these particular "maximally entangles Bell states" so special and "most quantum".

3. No he doesn't but he uses a setup to demonstrate the entanglement. With a setup that establishes classical paths you destroy the entanglement, as is exemplified by the which-way experiments like, e.g., in the quantum erasure experiment discussed in the other thread.

4. I also don't understand why you critisize that I quoted a textbook I consider a very good one after you (sic!) asked for a reference.

1. I did. It's called a path.2. Of course the entangled pair is a single quantum system consisting of 2 photons. And yes, you can correctly specify that the entangled pair has shared properties that are not separable. And you can, in the case of many PDC experiments, state that the paths are entangled and therefore in a superposition of paths.

Of course, they don't have to be path entangled. They might be just spin entangled (see reference in my 3. below for example). And their paths, regardless of whether those entangled photons are path entangled, can be distinguished and identified as 2 distinct paths. Which is of course how the experimentalist views them.3. This is flat out wrong science. There is nothing particularly "quantum" about the path the experimentalist directs entangled photons. They go in free air with no problem, as well as through classical optical filters, classical mirrors, classical beamsplitters, classical lens, etc. Knowing the path does not "destroy the entanglement" (your exact words).

Zeilinger et al (2006) Free-Space distribution of entanglement and single photons over 144 km
https://arxiv.org/abs/quant-ph/0607182

PS They use a laser pointer (from Bob to the source) to control the path of the entangled photons going from the source to Bob. So they go along an almost perfectly parallel path to each other. I guess we wouldn't call a laser path precisely classical either, but it sure makes it hard to say the entangled photons do much "quantum stuff" in a 144 km (89 mile) trip if they follow the straightest line known to man. The experimentalist plans every step with classical path in mind.4. One of the biggest criticisms I have of your posts is that you fail to provide references when challenged. A 664 page reference doesn't cut it. You may as well a) quote yourself; b) say it is obvious without explaining why; and c) say all of QFT supports you. (Well, actually you do all 3 of those. ) You would never do that in a paper you write, obviously PF posts are not papers.

I don't mind a reference to a entire paper if your basic point is in the abstract, or the paper is very short. But asking someone to search a book is preposterous. Further, you ignore/dismiss my direct quoted references with a flip of the hand, when the appropriate response is to refute it with an equally suitable reference.

 

[vanhees71]

Indeed, this "superpositions of nearly classical paths" is what "experimentalist" mean by an entangled photon's path. But it wasn't clear to me whether you would not also object to this way of including classical paths in the description of a quantum experiment. And I guess DrChinese will also agree that the paths in quantum experiments can be in superposition. But the point is that using the word "path" is fine, there is no need to forbid this word.

I don't want to forbid any word, but if it is used in a wrong way in a specific situation, I'm also allowed to point this out.

Imho have paths in QT a definite and generally correct meaning in terms of Feynman's path integrals. For photons the adequate paths in this sense are "field configurations" since in relativistic QFT we integrate over field configurtation not over point-particle paths.

 

[vanhees71]

1. I did. It's called a path.2. Of course the entangled pair is a single quantum system consisting of 2 photons. And yes, you can correctly specify that the entangled pair has shared properties that are not separable. And you can, in the case of many PDC experiments, state that the paths are entangled and therefore in a superposition of paths.

Thank you that I'm allowed to use the scientifically correct terminology, though I still think "path" is a misleading wording. Why don't you want to allow me to use the correct expressions, i.e., quantum states, which define the situation correctly, and be it in the very simplified and idealized way (using idealized momentum-polarization generalized eigenstates)?

Of course, they don't have to be path entangled. They might be just spin entangled (see reference in my 3. below for example). And their paths, regardless of whether those entangled photons are path entangled, can be distinguished and identified as 2 distinct paths. Which is of course how the experimentalist views them.

Sigh. Once more, the correct description of the idealized entangled-photon pair is (here for the polarization-singlet state as an example):
$$|\Psi \rangle = \frac{1}{\sqrt{2}} [\hat{a}^{\dagger}(\vec{k}_1,H) \hat{a}^{\dagger}(\vec{k}_2, V) - \hat{a}^{\dagger}(\vec{k}_1,V) \hat{a}^{\dagger}(\vec{k}_2,H)] |\Omega \rangle.$$
Inwords: 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 (even you must admit that the formula is much more clear than this description). 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, i.e., the crucial point is that due to this preparation procedure (aka the two-photon state) the "two paths" can NOT be identified as distinct before measuring at least one of the photon. Only after measuring the polarization of the photon with momentum $\vec{k}_1$ (selected by positioning the detector in the corresponding direction relative to the source) you "identify a path" (it's strange wording for me, but ok). That's the crucial point of this kind of states and because it's specifically the indefiniteness of "which path was realized" in such situations that makes us debate about them still even after about 40 years of the definite experimental disproof of the "local-realism hypothesis" a la EPR using, e.g., the violation of Bell's inequality to test it with these particular entangled two-photon states.

3. This is flat out wrong science. There is nothing particularly "quantum" about the path the experimentalist directs entangled photons. They go in free air with no problem, as well as through classical optical filters, classical mirrors, classical beamsplitters, classical lens, etc. Knowing the path does not "destroy the entanglement" (your exact words).

Zeilinger et al (2006) Free-Space distribution of entanglement and single photons over 144 km
https://arxiv.org/abs/quant-ph/0607182

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

PS They use a laser pointer (from Bob to the source) to control the path of the entangled photons going from the source to Bob. So they go along an almost perfectly parallel path to each other. I guess we wouldn't call a laser path precisely classical either, but it sure makes it hard to say the entangled photons do much "quantum stuff" in a 144 km (89 mile) trip if they follow the straightest line known to man. The experimentalist plans every step with classical path in mind.

You obviously don't understand my point! Of course, if I measure a photon's polarization at A (determined by the direction from the source to this point, and of course you can use a laser to determine this position accurately, which does not contradict anything I'm saying, because the laser beam is a coherent state with a pretty well specified wave vector), knowing that the entangled pair was prepared in the above quoted state (according to the formula on p. 481 of the above quoted paper), I know which polarization the photon at A has and thus also which polarization the photon it B must have, but before I couldn't know that, i.e., before the measurement not one of the paths was specified.

4. One of the biggest criticisms I have of your posts is that you fail to provide references when challenged. A 664 page reference doesn't cut it. You may as well a) quote yourself; b) say it is obvious without explaining why; and c) say all of QFT supports you. (Well, actually you do all 3 of those. ) You would never do that in a paper you write, obviously PF posts are not papers.

That's really ridiculous. You asked for a reference, and I gave you one. You find this in the first few sections of the textbook. Why it should be forbidden to quote myself, I also don't understand. If you don't believe me, you can ask for other references supporting my point of view, and this I gave. I don't know, what you are criticizing.

I don't mind a reference to a entire paper if your basic point is in the abstract, or the paper is very short. But asking someone to search a book is preposterous. Further, you ignore/dismiss my direct quoted references with a flip of the hand, when the appropriate response is to refute it with an equally suitable reference.

All the references you gave so far support what I'm saying, not what you are claiming. To say an entangled two-photon state (even a Bell state) specifies a "path" contradicts the physical content this very state describes!

 

[gentzen]

Imho have paths in QT a definite and generally correct meaning in terms of Feynman's path integrals. For photons the adequate paths in this sense are "field configurations" since in relativistic QFT we integrate over field configurtation not over point-particle paths

Imho, paths can enter much earlier in the description of such experiments, namely as part of the geometrical optics approximation of what is going on.

 

[vanhees71]

That's an interesting point. Geometrical optics means the eikonal approximation of wave optics, which you can of course also use for the quantized Maxwell equations as long as you are in the linear regime. How to describe non-linear optics within the eikonal approximation I don't know.

 

[Dadface]

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?

 

[vanhees71]

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?

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.

2. What do the red and green lines represent?

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

 

[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".