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

[PeterDonis]

They almost exactly follow classical trajectories.

The above seems inconsistent with this:

They are NOT classical paths

The latter I would agree with; the former I would not. I don't see how both can be true (to be clear, I'm talking about the case where there are no fiber optic cables or other devices present, just free space between source and detector).

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.)

I agree with all of this.

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.

So what do you think justifies this, given the other statements quoted above?

 

[PeterDonis]

what do you think justifies this, given the other statements quoted above?

Let me give an example of a possible justification to see if it helps: suppose I assume that a photon is released from my photon source at time $t = 0$. Say the expectation value of the wavelength of photons from my source is $\lambda$. I know the distance from the photon source to the parametric down conversion crystal, and I know the distances from the crystal to detectors A and B, where the pair of down converted photons will be detected (assuming this run of the experiment produces such a pair). Could I compute the probability amplitudes for detecting photons at A and B as a function of time, and show that those amplitudes were sharply peaked around a time $t = T$, where $T$ is the classical light travel time over the sum of the relevant distances? Would the sharpness of the peak be a function of how large the distances were as compared to $\lambda$? Has any such computation been done in the literature?

 

[Demystifier]

Before saying that photons have or not have paths, one should first say what photon is.

So what is a photon? A state in the one-photon sector of the QED Hilbert space? A click in the photon detector? A pointlike object in the Bohmian inerpretation? Something else? If we first agree on that, I think it will be much easier to agree on existence or non-existence of photon paths.

There is no doubt that experimentalists measure something that they call photon paths. But also there is no doubt that theorists can describe those experiments without dealing with a notion of photon paths. So in a sense, both sides are right.

 

[CoolMint]

In double slits experiments, it is possible for classical-like paths of observed photons to be inferred/derived.
Unobserved photons don't appear to have classical-like paths.

 

[vanhees71]

Oops, forgot to post the reference itself (this is a just one I picked out of the blue):

Kim et al, 2001
https://arxiv.org/abs/quant-ph/0103168

As expected this paper uses the standard quantum (!) optics treatment of photons in terms of the quantized electromagnetic field. You claim one could describe parametric down conversion, entangled two-photon states and all that within the naive photon picture of 1905. IMO that's simply impossible, because these phenomena need the correct QFT treatment. That's what's in all introductory chapters of modern quantum-optics textbooks, you however don't seem to accept as valid sources.

 

[DrChinese]

You claim one could describe parametric down conversion, entangled two-photon states and all that within the naive photon picture of 1905.

Wow, where did you get that? What I actually said: "Conservation [of momentum, etc.] and basic QM is plenty enough to describe PDC as it relates to entanglement tests."

You don't need to know how to build a car to drive one. That's an analogy! You don't need to know how non-linear crystals generate entangled photon pairs to perform a Bell test. Or answer a post question.

Let me give an example of a possible justification to see if it helps: suppose I assume that a photon is released from my photon source at time $t = 0$. Say the expectation value of the wavelength of photons from my source is $\lambda$. I know the distance from the photon source to the parametric down conversion crystal, and I know the distances from the crystal to detectors A and B, where the pair of down converted photons will be detected (assuming this run of the experiment produces such a pair). Could I compute the probability amplitudes for detecting photons at A and B as a function of time, and show that those amplitudes were sharply peaked around a time $t = T$, where $T$ is the classical light travel time over the sum of the relevant distances? Would the sharpness of the peak be a function of how large the distances were as compared to $\lambda$? Has any such computation been done in the literature?

Why yes! That was essentially done in the reference - and quote - I placed in my post #12 in this thread. Note that the answer is expressed differently - it's expressed as the relative difference in the arrival times at the A and B detectors (labeled differently in the paper). That's because the pair creation time cannot be well constrained to an arbitrarily small time window. (After all, it's a quantum particle. ) But note that comparing the path length to the wavelength precision, it's about 12 billion to 1 in terms of the peak expectation values.

1. The meaning of the above: It is AS IF (even though it's not) each photon travels a path that is very very nearly a continuous classical path - and nothing else! To the extent that they DON'T travel such a continuous classical path: they do such similar things that they arrive extremely closely together.

2. Further, it should be obvious from comparing these results with other Bell tests with total path lengths on the order of a laboratory room (perhaps 2 meters as compared to 8 kilometers from city to city, a ratio of 1:4000) that there is no more "quantum discontinuous" action during their travel due to the total length traversed. If there were, the much longer travel time of the Tittel et al experiment would require a proportional larger coincidence window. But it doesn't.

3. The conclusion: to the extent that entangled photons do not travel in classical paths, it is not measurable as having any dependency on path length. Please do not quote me as saying entangled photons travel on classical paths, they are quantum particles and subject to quantum rules. But there is no clear measurable evidence that they don't travel on continuous paths with current experiments.

4. One of the great and easily understandable proofs that photons don't travel on a classical path is the one in which light is reflected from a mirror to a source that measures the intensity of reflected light. Let's call that amount i1. It is possible to place small etchings at precise spots AWAY FROM the primary reflection point such that some destructive quantum interference is eliminated (which should result in MORE transmitted light). Such spots are NOT in the classical path, but are in the quantum path. We measure the intensity with the etchings in place, and it is i2. If photons travel along classical paths, i1=i2. In actual experiments, i1<i2 - defying common "sense". I have not seen a reference where a similar setup used entangled photons as the light source, but I think that would be interesting. According to Kaur and Singh (2020): "Because of path revealing quantum entanglement of particles the single particle interference is suppressed." Would that mean that entangled photons wouldn't have something like i1<i2?

 

[PeterDonis]

That was essentially done in the reference - and quote - I placed in my post #12

First, that experiment uses fiber optic cables, which, as I've already noted, physically constrains the "path"; it's not the same as having free space between the source and the detector.

Second, I don't see how any calculation of the sort I described is made in the paper. The authors assert that the relative path lengths will affect the relative arrival times at the detectors in a particular way, but they don't calculate it. If they are relying on such a calculation done elsewhere in the literature, they don't give a reference to it. I'm wondering if there is any paper in the literature that actually calculates the probability of arrival at the detector as a function of time and shows that it is sharply peaked around the expected classical light travel time from the source given the classical path length, at least for path lengths that are large compared with the wavelength of the light.

 

[PeterDonis]

One of the great and easily understandable proofs that photons don't travel on a classical path is the one in which light is reflected from a mirror to a source that measures the intensity of reflected light. Let's call that amount i1. It is possible to place small etchings at precise spots AWAY FROM the primary reflection point such that some destructive quantum interference is eliminated (which should result in MORE transmitted light). Such spots are NOT in the classical path, but are in the quantum path. We measure the intensity with the etchings in place, and it is i2. If photons travel along classical paths, i1=i2. In actual experiments, i1<i2 - defying common "sense".

Yes, IIRC these experiments were discussed by Feynman in his physics lectures, and also in the popular book of his "QED", where he gives them as an example of light not behaving classically.

 

[DrChinese]

First, that experiment uses fiber optic cables, which, as I've already noted, physically constrains the "path"; it's not the same as having free space between the source and the detector.

Second, I don't see how any calculation of the sort I described is made in the paper. The authors assert that the relative path lengths will affect the relative arrival times at the detectors in a particular way, but they don't calculate it. If they are relying on such a calculation done elsewhere in the literature, they don't give a reference to it. I'm wondering if there is any paper in the literature that actually calculates the probability of arrival at the detector as a function of time and shows that it is sharply peaked around the expected classical light travel time from the source given the classical path length, at least for path lengths that are large compared with the wavelength of the light.

1. I don't think that would make any measurable difference at all (by not constraining the path using fiber). I don't have any relevant references that would answer that either way. That is what I was mentioning in my post #41, point 4. You'd need to do a specific experiment to discern either way.

2. I am not sure anyone if working like that. There must be a lot of tuning going on in these experiments. As the photons go through varying materials, the speed through that medium varies from others. I will keep my eyes open for anything that seems to calculate that. One of the problems being that the photon creation times are random.

 

[andresB]

I'm surprised by this tread and the position of DrChinese, I thought it was consensus that in general there is no such thing as a particle path in the usual interpretation of QM (no Bohmian mechanics or anything like that).

In a standard single particle, two-slit experiment, given the original position of the source and the position of the measurement when the particle hit the detector: (1) what method from QM can we use to define a path? I don't think it exists, but even if it exists, (2) how can we be experimentally sure that was the actual path taken by the particle?

 

[DrChinese]

I'm surprised by this tread and the position of DrChinese, I thought it was consensus that in general there is no such thing as a particle path in the usual interpretation of QM (no Bohmian mechanics or anything like that).

In a standard single particle, two-slit experiment, given the original position of the source and the position of the measurement when the particle hit the detector: (1) what method from QM can we use to define a path? I don't think it exists, but even if it exists, (2) how can we be experimentally sure that was the actual path taken by the particle?

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

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.

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.)

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.

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.

 

[HomogenousCow]

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

 

[andresB]

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.

Well, I agree that If we don't look too deep into it, if we don't try to zoom too much into the particle position over time, then we can use the word "path" without any danger. That is compatible with the fact that at the most fundamental level in (the standard interpretation of) QM, particles don't have trajectories.

 

[CoolMint]

A tennis ball is a collection of quanta and is generally assumed to have a path whether it is observed or not. In practice, all attempts to derive a classical path are successful and virtually everyone assumes an unobserved tennis ball has a definite well-established path. In QM terms, the tennis ball 'path' is the average of many trajectories that peak and average out mostly around the classical 'path'. Thus I suggest that QM treatments put commas around 'path' while practical treatments treat the path in question without commas for all practical purposes. Theory meeting practice is the unresolved issue of single outcomes in QM.
For ultimate exactness and truthfullness with theory however, the path should always be denoted as 'path'.

 

[PeterDonis]

One of the problems being that the photon creation times are random.

True, but since the arrival times at the detectors are recorded (within a fairly narrow window), one can just do the calculation in reverse and look at the amplitude for emission from the source as a function of time, given a detection within the known time window, and see if it is sharply peaked about the expected classical emission time given the total path length from source to detector.

The quantum optics literature might not have such a calculation, but that could be because earlier literature on QED more generally did a calculation something like this. Unfortunately I'm not familiar enough with the QED literature to know where to look.

 

[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.