I Feynman path integral and events beyond the speed of light

kurt101
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Do events that can't actually happen because of the speed of light limitations contribute to probabilities that are calculated using the Feynman's path integral method?
In Richard Feynman's book "The Strange Theory of Light and Matter", in chapter 2, he explains how to calculate the probability that light from some source will be reflected by a mirror and be detected at some location. He explains how you sum up all of the probability amplitudes (represented as vectors) for each way light can reflect off of a mirror and end up at the detector and then square this result to get the final probability. Here is the illustration of this:
feynman_reflection.PNG
Feynman then explains how the probability amplitudes of reflections around A and B tend to cancel out and thus don't contribute very much to the final probability. Here is the illustration of this:

feynman_ab_cancel.PNG


Feynman then explains that if you remove sections of the mirror at A and B that you can change the contributions of the probability amplitudes around A and B so that they do contribute significantly to the final probability. Here is the illustration of this:

feynman_ab_contribute.PNG


My Question:
What happens if you turn on the light source for a small amount of time and only detect the light that could have reflected from the middle of the mirror (around say E through I) in the amount of time that it takes light from the source traveling at the speed of light to reflect and arrive at the detector. In this case would you eliminate the contributions to the final probability from the reflections around A and B because they could not have possibly contributed because of the speed of light?
 
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There's no certainty in when a photon is emitted, so you can't have QED on the one hand and classical precise events in spacetime on the other.

In fact, the arguments in that book generally rely on this uncertainty in order to have the interference in the first place.

Remember also that interference is generally a photon interfering with itself and not different photons interfering with each other. Although the full picture is more complicated than that.
 
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PeroK said:
There's no certainty in when a photon is emitted, so you can't have QED on the one hand and classical precise events in spacetime on the other.

In fact, the arguments in that book generally rely on this uncertainty in order to have the interference in the first place.
But you can know that the amount of time the light source was turned on is well between the times it took for reflection in the middle of the mirror (say source -> G -> detector) and the time it took for reflection at the end of the mirror (say source -> A -> detector). So you can run the experiment with these constraints and see if having the missing sections of mirror at A and B has any change on the probability. And so does having or not having the sections with this constraint change the probability?

And maybe there would be some variation in light detection based on the uncertainty, but if you ran enough trials the uncertainty should wash out.
 
If you manipulate the light in time as you describe it will necessarilly not be monochromatic and you will wash out potential interference patterns. In rereading this I am uncertain as to what the actual question is.
 
kurt101 said:
But you can know that the amount of time the light source was turned on ...
The emission of a photon, as with practically everything in quantum theory, is a probabilistic event. If you try to have a photon at spacetime coordinates ##(t, x, y, z)## moving with speed ##c## in some direction ##\vec n##, then that is not quantum theory in the first place. That's a classical view of the photon as a massless particle with a classical trajectory.

QED by Feynman, if it does anything, is trying to get you NOT to think of light like that!
 
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kurt101 said:
Summary:: Do events that can't actually happen because of the speed of light limitations contribute to probabilities that are calculated using the Feynman's path integral method?
In the path integral method photons do not have a well-defined "speed"; paths in which photons travel at different speeds all contribute to the path integral. In more technical language, worldlines which are timelike and spacelike, as well as worldlines which are null, make nonzero contributions to the path integral.

In many cases (such as light traveling over long distances in free space), the contributions to the path integral from any worldlines other than the classical null worldline are negligible; in these cases, we can view the light as simply traveling at the speed of light through free space. But there are also many cases where this is not true. The general rule is that you must include in the path integral all paths that are not forbidden by some condition of the problem (such as the presence of an opaque screen), even if those paths would make no sense classically.
 
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One should take the "null world lines" as what they really are, namely the characteristics of the eikonal equation, and this is just an approximate solution of the wave equation of the electromagnetic field. The phasor approach in Feynman's popular book is nothing else than the classical refraction theory in Fraunhofer approximation, i.e., it solves the wave equation for the electromagnetic field, and thus it is consistent with the propagation of these waves with the speed of light.
 
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vanhees71 said:
One should take the "null world lines" as what they really are, namely the characteristics of the eikonal equation, and this is just an approximate solution of the wave equation of the electromagnetic field. The phasor approach in Feynman's popular book is nothing else than the classical refraction theory in Fraunhofer approximation, i.e., it solves the wave equation for the electromagnetic field, and thus it is consistent with the propagation of these waves with the speed of light.
For the specific examples Feynman discusses, which all involve light propagation (reflection off a mirror, diffraction gratings, bending when passing from one medium to another), yes, he is only considering null paths through spacetime (different paths through space, but which are all traversed at the speed of light and therefore have different travel times from source to receiver). But the path integral formulation is much more general; for example, when modeling a static force between charged objects, the path integral will have significant contributions from timelike and spacelike paths through spacetime.
 
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hutchphd said:
If you manipulate the light in time as you describe it will necessarilly not be monochromatic and you will wash out potential interference patterns.
Why does turning on and off a monochromatic light source make the light from that source NOT monochromatic?

hutchphd said:
In rereading this I am uncertain as to what the actual question is.
I will try phrasing it again.

To make my question as clear as possible I am using the same context as Feynman used in chapter 2 of his book "The Strange Theory of Light and Matter" where he describes how to calculate the probability of light from a monochromatic source reflecting off a mirror to a detector. After he explains the basic procedure he goes on to explain how removing sections at the end of the mirror can alter the probability that light reaches the detector. Ok with this part?

My question involves taking the Feynman scenario and putting the following constraints on the light source and the detector. For clarity sake I will call this my experiment and to make the sequence of events clear, I will put in some time values.

Lets say the following:
It takes roughly 5 seconds for light to travel from the light source to the middle of the mirror (point G in the diagram) to the detector.
It takes roughly15 second for light to travel from the light source to the end of the mirror (point A in the diagram) to the detector.

In my experiment:
At 0 seconds, I turn on the light source and start recording at the detector.
At 10 seconds, I turn off the light source and stop recording at the detector.

In the first version of my experiment, I make no changes to the mirror. The mirror at point A and B look like:
feynman_ab_cancel.PNG


In the second version of my experiment, I remove sections at the end of the mirror. The mirror at point A and B now looks like:

feynman_ab_contribute.PNG


I run many trials of both versions of my experiment. Over all my trials I compare the average amount of light I measured in the first version of my experiment to the average amount of light measured in the second version of my experiment. Is the amount of light measured between the first version of my experiment the the same or different than the second version of my experiment?
 
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kurt101 said:
Lets say the following:
It takes roughly 5 seconds for light to travel from the light source to the middle of the mirror (point G in the diagram) to the detector.
It takes roughly15 second for light to travel from the light source to the end of the mirror (point A in the diagram) to the detector.
Light can travel from Earth four times further than the Moon in 5 seconds.
 
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  • #11
kurt101 said:
Why does turning on and off a monochromatic light source make the light from that source NOT monochromatic?
Because one of the requirements for a source to be truly monochromatic is that it emits light continuously forever, infinitely into the past and infinitely into the future. Which means, of course, that no real light source is ever truly monochromatic.
 
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  • #12
PeroK said:
The emission of a photon, as with practically everything in quantum theory, is a probabilistic event. If you try to have a photon at spacetime coordinates ##(t, x, y, z)## moving with speed ##c## in some direction ##\vec n##, then that is not quantum theory in the first place. That's a classical view of the photon as a massless particle with a classical trajectory.

QED by Feynman, if it does anything, is trying to get you NOT to think of light like that!
I am trying to understand how QED applies to the real world. I think that was one of the major reasons Feynman wrote the book. I think that is the major purpose of quantum mechanics. In the real world, when I turn on the light, I expect to detect photons in the detector at some later time that is roughly the distance of the shortest path of light divided by 3 x 10^8 m/s. I don't know how the light actually traveled from the source to the detector and I have tried to avoid any kind of interpretation in the presentation of my question.

PeroK said:
Light can travel from Earth four times further than the Moon in 5 seconds.
Ha, ha, I guess the mirror is very large.
 
  • #13
PeterDonis said:
In the path integral method photons do not have a well-defined "speed"; paths in which photons travel at different speeds all contribute to the path integral. In more technical language, worldlines which are timelike and spacelike, as well as worldlines which are null, make nonzero contributions to the path integral.

In many cases (such as light traveling over long distances in free space), the contributions to the path integral from any worldlines other than the classical null worldline are negligible; in these cases, we can view the light as simply traveling at the speed of light through free space. But there are also many cases where this is not true. The general rule is that you must include in the path integral all paths that are not forbidden by some condition of the problem (such as the presence of an opaque screen), even if those paths would make no sense classically.
I have done a lot of searching for answers related to my question prior to posting it. I get the sense that there is a lot of uncertainty on what actually contributes to the probability in the path integral formulation. While I believe what you told me is correct, you left it sufficiently vague (i.e. "The general rule is that you must include in the path integral all paths that are not forbidden by some condition of the problem") and so I not able to answer my question with this rule. In other words I don't know what those conditions are.
 
  • #14
kurt101 said:
I get the sense that there is a lot of uncertainty on what actually contributes to the probability in the path integral formulation.
No, there is no uncertainty at all. The math is perfectly clear.

For you, the problem is that you aren't looking at the math. You're looking at various people's attempts to describe what the math is saying in ordinary language. And there is no way to do that with 100% accuracy because our ordinary language simply doesn't have the words or the concepts required. That's why physicists don't use ordinary language to do physics; they use math.

kurt101 said:
I don't know what those conditions are.
There will be some paths through spacetime that are forbidden because of the presence of something (such as an opaque screen, which was the example I gave) that absorbs light. (You could, if you really wanted to, include the details of the absorption of the light by that something in your analysis, but all that would do is complicate the math even more to get the same answer--that none of that light makes it to your detector.)
 
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  • #15
PeterDonis said:
For the specific examples Feynman discusses, which all involve light propagation (reflection off a mirror, diffraction gratings, bending when passing from one medium to another), yes, he is only considering null paths through spacetime (different paths through space, but which are all traversed at the speed of light and therefore have different travel times from source to receiver). But the path integral formulation is much more general; for example, when modeling a static force between charged objects, the path integral will have significant contributions from timelike and spacelike paths through spacetime.
But of course the path integral for relativistic QT is not a path integral over paths in phase space of point particles (or after integrating out the momenta, if they appear only quadratically in the action, over paths in configuration space) in non-relativistic QT ("1st quantization") but a path integral over field configurations.
 
  • #16
PeterDonis said:
No, there is no uncertainty at all. The math is perfectly clear.
If it is so clear to you then help me understand how to apply the algorithm Feynman describes to answer the question I am asking.
 
  • #17
@vanhees71 I have read many of your discussions around microcausality in the context of QFT. For example you wrote:

vanhees71 said:
This together with the fact that a local relativistic QFT cannot describe any faster-than-light signal propagation (due to the microcausality built in this kind of relativistic QFTs) one must conclude that the correlation is not caused by the local measurements on each photon at far distant places but it is due to the preparation in the entangled state.

Does the microcausality condition require local (in a spatial sense) preparation?

In the scenario I am discussing in this thread where removing sections of a mirror (i.e. the grating) affect the probability of the light being detected; what is the (local?) preparation between the light source and the grating?

Ultimately in this thread, I am trying to understand how something distant affects where the light ends up. I am trying to understand this in the context of the QED algorithm Feynman describes.

For someone who truly understands QFT, my question about whether having the grating or not has an effect on the detection probability with the time window constraint I have added, should be trivial to answer.
 
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kurt101 said:
If it is so clear to you then help me understand how to apply the algorithm Feynman describes to answer the question I am asking.
The problem here is that we can’t do anything with a description of an algorithm. To do something we need the real thing, and “QED: The strange theory….” is not that; it’s an extended multi-page analogy that offers non-specialists some intuition for how local interference can yield many of the macroscopic behaviors of light. You should not be surprised to find that its math-free toy model becomes vague and mushy when you push it.

One particularly important limitation of the model is that it does not work for light that is switched on and off (more precisely, is not kept on for a time that is long compared with the time it takes for light to reach all parts of the setup). Thus it‘s not a good place to start when you’re trying to understand which paths to include and why some seem to be arbitrarily excluded. The real theory, expressed mathematically, does not have this limitation.
 
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  • #19
kurt101 said:
If it is so clear to you then help me understand how to apply the algorithm Feynman describes to answer the question I am asking.
hutchphd anwered your intial question: "If you manipulate the light in time as you describe it will necessarilly not be monochromatic and you will wash out potential interference patterns." You slightly modified your question in response to his answer, but basically his answer still applies. If you don't understand his answer in the context of your initial question, then you will understand it even less in the context of your modified question.

In QED, Feynman describes how coherent superposition works. But what happens in your setup is partial coherent superposition, i.e. an incoherent average over coherent superpositions (what hutchphd described as "... will wash out ..."). Perhaps trying to understand the picture and explanations on the last slide of this extract will help you a bit.

To make everything even more confusing, partial coherence can be "reduced" to full coherence by going to a (church of the) larger Hilbert space. But if you don't already understand partial coherence, this construction won't help you either.
 
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  • #20
kurt101 said:
If it is so clear to you then help me understand how to apply the algorithm Feynman describes to answer the question I am asking.
You can't apply the specific algorithm Feynman describes to the specific case given in your OP, because, as has already been pointed out, his algorithm doesn't work for light sources that are switched on and off. More precisely, it doesn't work for light sources that are switched on and off on a time scale comparable to the time scale of the experiment you are doing.

You can use the math of QED directly to answer a question like the one you ask in your OP, but, as has already been pointed out, the math of QED is not the same as the heuristic ordinary language description that Feynman gives. If you really want to learn how to apply QED to general cases, not just the specific cases Feynman describes in his book, you will need to learn how to use the math of QED. There are no shortcuts that let you use vague ordinary language to actually do physics.
 
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  • #21
PeterDonis said:
You can use the math of QED directly to answer a question like the one you ask in your OP, but, as has already been pointed out, the math of QED is not the same as the heuristic ordinary language description that Feynman gives.
Note that the last paragraph of my post #14 was an attempt to at least convey a more general sense of what the math of QED is doing, beyond what Feynman says in his book. There are other possible reasons besides the presence of an opaque screen for particular paths through spacetime being forbidden: one such reason is that the light source is turned off, so paths through spacetime that originate from the source when it is turned off will not have any nonzero amplitude associated with them. But you can't analyze such cases using the heuristic ordinary language picture Feynman gives in his book.
 
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  • #22
kurt101 said:
Is the amount of light measured between the first version of my experiment the the same or different than the second version of my experiment?
1. You already know that if a diffraction grating is placed at appropriate spots on the mirror, there will be no contribution to the final result from that section. Any way that contribution is alternatively excluded gives the same results as if a diffraction grating were present. There is no faster-than-light effect present in any version of this scenario.2. As a practical matter, using some classical reasoning to consider your version of the experiment: In a femtosecond (10^-15 seconds), light travels about about 333 nanometers (nm) - little shorter than a wavelength of visible light. The difference X in distance light travels when it is reflected from the middle (your B) versus when it is further away by a single wavelength (i.e. ever so slightly closer to your A or C) is extremely small. Using some "reasonable" parameters for visible light, approximately 45 degree angle of incidence, 1000 wavelength distance to mirror, back of a napkin calculations... that X works out to be on the order of magnitude of 1 nanometer (nm). That translates to a time window of roughly 0.003 femtoseconds.

In other words, we would need to nail down the precise time of emission AND detection to less than approximately .003 femtoseconds to be able to perform this experiment. The best detectors today have a comparable resolution of no better than 1 picosecond (1000 femtoseconds), or nearly a million times longer that what you'd need to even think about running your experiment. And that doesn't begin to consider a variety of quantum effects and uncertainty you'd encounter.3. Think of the Feynman paths as follows: Because of inherent uncertainty in photon emission times: a detection at experimentally precise (if there was such a thing) time T3 could result from a slightly shorter path traversed by a photon B emitted at time T2, or a slightly longer path traversed by a photon A emitted slightly earlier at time T1. Both of those possibilities contribute to the result. You could perform this experiment a photon at a time, but it is not possible to force (or restrict) a single photon to be emitted at a specific time with sufficient resolution to distinguish between scenarios A and B. To the extent that you could perform this impossible task, you would only succeed in demonstrating what we already know would occur. And that would not show any FTL effect at all.
 
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  • #23
It's best not to try to imagine the Feynman paths as real processes or worry about FTL paths and so on. Most of the "paths" contributing to the path integral aren't even functions but distributions and so in no sense describe actual paths, i.e. assign a position for each time.
 
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  • #24
gentzen said:
hutchphd anwered your intial question: "If you manipulate the light in time as you describe it will necessarilly not be monochromatic and you will wash out potential interference patterns." You slightly modified your question in response to his answer, but basically his answer still applies. If you don't understand his answer in the context of your initial question, then you will understand it even less in the context of your modified question.
I added more detail to my question, but did not change it. In Feynman's book, chapter 2 that I used as the basis for my description, the light source is described as monochromatic and I left that detail out. Here is the actual quote from the book:

"At S we have a source that emits light of one color at very low intensity (let’s use red light again). The source emits one photon at a time." - Feynman, Richard P.. QED: The Strange Theory of Light and Matter (Princeton Science Library) (p. 38). Princeton University Press. Kindle Edition.

Also I do not say in my question that the light source emits one photon at a time, but I do not think that is relevant to my question. Is it important to the result at the detector that the light source emits 1 photon at a time?

gentzen said:
In QED, Feynman describes how coherent superposition works. But what happens in your setup is partial coherent superposition, i.e. an incoherent average over coherent superpositions (what hutchphd described as "... will wash out ..."). Perhaps trying to understand the picture and explanations on the last slide of this extract will help you a bit.

I understand the light source in Feynman's example is not coherent; at least in the sense that he does not describe it as a laser where light is in phase. That said, I don't understand how turning on the light makes the light any different (i.e. more or less coherent) than how Feynman already describes it. Unless I am mistaken about Feynman's example and he does mean to imply that is light source is fully coherent even though he never states it. I would be surprised if this is the case.
 
  • #25
kurt101 said:
I added more detail to my question, but did not change it. In Feynman's book, chapter 2 that I used as the basis for my description, the light source is described as monochromatic and I left that detail out. Here is the actual quote from the book:

"At S we have a source that emits light of one color at very low intensity (let’s use red light again). The source emits one photon at a time." - Feynman, Richard P.. QED: The Strange Theory of Light and Matter (Princeton Science Library) (p. 38). Princeton University Press. Kindle Edition.

Also I do not say in my question that the light source emits one photon at a time, but I do not think that is relevant to my question. Is it important to the result at the detector that the light source emits 1 photon at a time?
I understand the light source in Feynman's example is not coherent; at least in the sense that he does not describe it as a laser where light is in phase. That said, I don't understand how turning on the light makes the light any different (i.e. more or less coherent) than how Feynman already describes it. Unless I am mistaken about Feynman's example and he does mean to imply that is light source is fully coherent even though he never states it. I would be surprised if this was the case.
So, why don't we just stick with one photon? The whole point of that section is to describe how the probability of the photon being detected depends on the diffraction grating.

Why have you brought all these added compexities into the equation?
 
  • #26
PeterDonis said:
There are other possible reasons besides the presence of an opaque screen for particular paths through spacetime being forbidden: one such reason is that the light source is turned off, so paths through spacetime that originate from the source when it is turned off will not have any nonzero amplitude associated with them.
This sounds promising to helping me answer my question, but in the context of everything else told to me, I am far from certain.

What about when the the light source is turned on? Do paths that don't yet exist through spacetime (outside of the light source's light cone), have a nonzero amplitude?
 
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  • #27
kurt101 said:
What about when the the light source is turned on? Do paths that don't yet exist through spacetime (outside of the light source's light cone), have a nonzero amplitude?
You may have completely lost the point of this treatment of QED. I think you should ask yourself whether you are intent on studying the material in Feynman's book, or whether you'd rather let your mind wander onto vague and superfluous questions.

Perhaps I'm too harsh. If so I apologise. But, if you were my student I'd tell you to knuckle down and concentrate.
 
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  • #28
kurt101 said:
I understand the light source in Feynman's example is not coherent; at least in the sense that he does not describe it as a laser where light is in phase.
He describes the light source as monochromatic. This is what makes it coherent. To get quasi-monochromatic light from a non-laser source, you have to filter out most frequencies, and only keep a small portion of the frequency spectrum. That is typically still wider than for light from a laser, but the coherence length can be made longer than the distance-differences in your optical instruments. And that is enough for treating the light as quasi-monochromatic.

kurt101 said:
That said, I don't understand how turning on the light makes the light any different (i.e. more or less coherent) than how Feynman already describes it.
Let me first descibe it in Feynman's photon picture. Assume that each photon would have exactly one frequency, and nicely form a coherent superposition with itself. If different photons have slightly different frequencies, then their interference patterns are slightly different. And the sum of those interference patterns no longer shows sharp interference peaks, but appears washed out. But to see an interference pattern at all, you must look at the image produced by many photons, so all you can see is the washed out pattern.

If your light has the frequency f, and you switch it on for t seconds, then you end up with an frequency distribution in the range f +- 1/t. If f is big compared to 1/t, then it makes sense to talk of quasi-monochromatic light. And if c*t is big compared to the distance-differences in your optical instruments, then treating the light as monochromatic should work well.
 
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  • #29
DrChinese said:
1. You already know that if a diffraction grating is placed at appropriate spots on the mirror, there will be no contribution to the final result from that section.
It is actually the opposite in the example, the diffraction grating placed at the end of the mirror adds a contribution, but I could see where one placed at the middle of the mirror counteracts the contribution, and so I get what you are saying.

DrChinese said:
Any way that contribution is alternatively excluded gives the same results as if a diffraction grating were present.
I am not sure if I follow what you are saying. Are you saying that given the constraints of my experiment, you will see the same results with or without the diffraction grating? (i.e. were you giving a direct answer to my question?).

DrChinese said:
There is no faster-than-light effect present in any version of this scenario.
Ultimately this is what I want to understand. I have had trouble finding clear statements in any of the descriptions of the Fenman path integral (both with math and without) that clearly say contributions prohibited by the speed of light don't count.

That said, I still don't understand if contributions that are prohibited by the speed of light in my example are from the light sources perspective or say the mirror's perspective. Hopefully you understand what I am saying here and this would be a part 2 question for this thread and so I am reluctant to get into it too much before I am comfortable that I understand my current question.
DrChinese said:
2. As a practical matter, using some classical reasoning to consider your version of the experiment: In a femtosecond (10^-15 seconds), light travels about about 333 nanometers (nm) - little shorter than a wavelength of visible light. The difference X in distance light travels when it is reflected from the middle (your B) versus when it is further away by a single wavelength (i.e. ever so slightly closer to your A or C) is extremely small. Using some "reasonable" parameters for visible light, approximately 45 degree angle of incidence, 1000 wavelength distance to mirror, back of a napkin calculations... that X works out to be on the order of magnitude of 1 nanometer (nm). That translates to a time window of roughly 0.003 femtoseconds.

In other words, we would need to nail down the precise time of emission AND detection to less than approximately .003 femtoseconds to be able to perform this experiment. The best detectors today have a comparable resolution of no better than 1 picosecond (1000 femtoseconds), or nearly a million times longer that what you'd need to even think about running your experiment. And that doesn't begin to consider a variety of quantum effects and uncertainty you'd encounter.
I know my example is not a practical one, at least the numbers I gave, but it is the principle of the example that is important and I still have the understanding that in principle you can run my experiment. If that is not the case, and I am still missing something, please make that clear to me.
DrChinese said:
3. Think of the Feynman paths as follows: Because of inherent uncertainty in photon emission times: a detection at experimentally precise (if there was such a thing) time T3 could result from a slightly shorter path traversed by a photon B emitted at time T2, or a slightly longer path traversed by a photon A emitted slightly earlier at time T1. Both of those possibilities contribute to the result. You could perform this experiment a photon at a time, but it is not possible to force (or restrict) a single photon to be emitted at a specific time with sufficient resolution to distinguish between scenarios A and B. To the extent that you could perform this impossible task, you would only succeed in demonstrating what we already know would occur. And that would not show any FTL effect at all.

If you changed the time constraint by adding additional detection time to be sure scenarios A and B (with and without the grating) should give a different result; then ran scenarios A and B many times; then took the averages; I would expect on average you would see a consistent difference between A and B, where scenario B with the grating that add contributions would on average detect more photons than A.
 
  • #30
PeroK said:
You may have completely lost the point of this treatment of QED. I think you should ask yourself whether you are intent on studying the material in Feynman's book, or whether you'd rather let your mind wander onto vague and superfluous questions.

Perhaps I'm too harsh. If so I apologise. But, if you were my student I'd tell you to knuckle down and concentrate.
That is an interesting perspective. I don't think my questions are vague or superfluous. I think understanding the constraints for any algorithm, formula, etc. is extremely important. So I don't know where you are coming from.

My goal is one of understanding out of curiosity and not one of learning QM so I can earn a living from it (at least not right now (no jokes here)) and so maybe that plays a part in the types of questions I ask. I am biased towards a causal deterministic perspective and tend to think the universe is not fundamentally statistical and so maybe that bias shows (certainly obvious in my earlier threads) and plays a part in the types of questions I ask. That said, I enjoy reading the different perspectives on this forum and I am open to all well supported positions.

You also said "knuckle down and concentrate". I do read carefully most everyone's responses and try very hard to understand what is being said to me. So I hope you are not saying this because you don't think I am not reading or not trying to understand what is being said to me. If I ask more questions or try to clarify where I am coming from that does not mean I am rejecting what others are telling me. It might mean I still don't understand and/or I think I have been misunderstood.
 
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  • #31
PeroK said:
So, why don't we just stick with one photon? The whole point of that section is to describe how the probability of the photon being detected depends on the diffraction grating.
I am ok with 1 photon at a time versus many. I did not specify either way. I still don't think it matters for my question.
PeroK said:
Why have you brought all these added compexities into the equation?
Because I am trying to understand how distant objects (e.g. the grating) affect the result. Its kind of like what others are telling me, the ordinary language does not matter, you have to understand the math. However in my case I would say it is the experiment that ultimately tells me the truth and while I don't doubt the math is correct from some perspective, math obviously has its limitations on where and how it can be used, and so it is helpful for me to understand what the actual experiment says.
 
  • #32
kurt101 said:
I do not say in my question that the light source emits one photon at a time, but I do not think that is relevant to my question. Is it important to the result at the detector that the light source emits 1 photon at a time?
It is important to the result at the detector to know what state the light emitted from the source is in. Saying "the source emits one photon at a time" does not tell you that. That is one of the key limitations of trying to learn physics from a pop science book written in ordinary language--even one by Feynman, who was one of the best physicists I know of at minimizing the distortion inherent in any such presentation.

The issue here, which is not necessarily important to the exposition Feynman was trying to make, but is very important if you are actually trying to understand the underlying physics in detail, is that "monochromatic" light is not in an eigenstate of photon number. "Monochromatic" light (I put the term in quotes because of the issue already raised, that truly "monochromatic" light can only be emitted by a source that emits light forever), such as light emitted by a laser, so called because (at least to whatever level of approximation is appropriate for the specific problem) it has a well-defined frequency, is in a coherent state. A coherent state does not have a well-defined photon number, since it is not in an eigenstate of photon number. So it is really not correct to describe such a source as emitting "one photon at a time". A strictly correct statement would be something like "the intensity of the source is such that the expected number of clicks at the photon detector over some standard interval of time is one". (Note that it is the detector that produces discrete clicks, or some other discrete phenomenon, that we associate with the term "photon".)

An eigenstate of photon number is called a Fock state, and is a very different state that requires a very different type of source (one which is much more difficult to make than a laser; experiments have been done using such sources, but they are much less common). That is not the kind of source Feynman was talking about; he was talking about something like a laser.
 
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  • #33
PeroK said:
why don't we just stick with one photon?
Because, as I noted in my previous post just now, "one photon" is a misleading description of the actual state of the light emitted by the kind of source (something like a laser) that Feynman was talking about.
 
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  • #34
kurt101 said:
What about when the the light source is turned on? Do paths that don't yet exist through spacetime (outside of the light source's light cone), have a nonzero amplitude?
You are not thinking correctly about spacetime. Spacetime includes time. A spacetime model already includes all the information about when (i.e., at what events in spacetime) the light source is turned on. Only paths originating from those events in spacetime will have a nonzero amplitude. There is no such thing as "paths that don't yet exist through spacetime"; spacetime is not something that "changes" as things happen. A spacetime model already includes all the information about everything that happens.
 
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  • #35
kurt101 said:
Unless I am mistaken about Feynman's example and he does mean to imply that is light source is fully coherent even though he never states it.
If the light is monochromatic, it is in a coherent state, as I have already said. A source like an ordinary light bulb does not produce monochromatic light. You need something like a laser.
 
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  • #36
gentzen said:
He describes the light source as monochromatic. This is what makes it coherent. To get quasi-monochromatic light from a non-laser source, you have to filter out most frequencies, and only keep a small portion of the frequency spectrum. That is typically still wider than for light from a laser, but the coherence length can be made longer than the distance-differences in your optical instruments. And that is enough for treating the light as quasi-monochromatic.

If a laser would work better, than I am all for a better experiment. That said, it would now be interesting to me to compared the difference between using a laser vs. filtered light source.
gentzen said:
Let me first descibe it in Feynman's photon picture. Assume that each photon would have exactly one frequency, and nicely form a coherent superposition with itself. If different photons have slightly different frequencies, then their interference patterns are slightly different. And the sum of those interference patterns no longer shows sharp interference peaks, but appears washed out. But to see an interference pattern at all, you must look at the image produced by many photons, so all you can see is the washed out pattern.
I don't think it matters to your point, but just to be clear, I am not detecting an interference pattern in my experiment, but measuring the number of photons or accumulated intensity.

gentzen said:
If your light has the frequency f, and you switch it on for t seconds, then you end up with an frequency distribution in the range f +- 1/t. If f is big compared to 1/t, then it makes sense to talk of quasi-monochromatic light. And if c*t is big compared to the distance-differences in your optical instruments, then treating the light as monochromatic should work well.

So if I make a more practical version of my experiment and use a 200 m mirror and assume red light (700 nm, 430e15 cycles / second) and:
put the source in the middle of the mirror, 100 m from the mirror
put the grating at the far left end of the mirror
put the detector at the far right end of the mirror, 100 m from the mirror

Then as a rough calculation:
The time for light to reach the grating is about 470 ns.
The time for light to reflect off the grating and go past the source is about 840 ns.
The time for light to reflect off the grating and reach the detector is about 1200 ns.
The time for light to take the shortest path, reflect off the mirror and reach the detector is 750 ns.

In this scenario, I would turn on the light source and look at the data from detector for up to 840 ns or 1200 ns.

So f = 430 x 10^15 cycles / second is much greater than (1 / t) = 1.2 x 10 ^ 6 and by your classification would be considered quasi-monochromatic light.

And c*t would be 252 m which seems fairly large compared to what I imagine the size of the optical instruments are and so by your classification would also be considered quasi-monochromatic light.

So given your guidance, I believe it would be justified to treat the light in this more practical scenario as monochromatic.
 
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  • #37
Well, you can also have a single-photon source which is pretty much monochromatic. It can't be literally monochromatic, because this would refer to an energy-momentum eigenstate of a single free photon, which is like a plane wave in classical electrodynamics and this is not a proper state in Hilbert space but a distribution in a larger space. So a true single-photon Fock state has a certain width in energy and thus also in momentum. To get a coherent quasi-monochromatic state you need a state with a very small energy width (energy uncertainty). According to the energy-time uncertainty relation (which is a complicated topic of its own!) you need a very long-lived unstable state of, e.g., an atom, which can emit a pretty monochromatic photon.
 
  • #38
kurt101 said:
a laser is directional (right?)
No, not for sum-of-paths purposes. That tight directed beam that we expect from a laser is itself the result of adding the contributions from paths in all directions and having the amplitudes cancel everywhere except along the path of the beam.
 
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  • #39
kurt101 said:
Like I said to gentzen: Another reason I thought Feynman did not imply a laser in his example is because a laser is directional (right?) and his light source was not.
I wish you good luck trying to understand that stuff. I tried my best to explain it. As long as considerations when you turn the light on or off are relevant, assuming the light to be quasi-monochromatic is problematic. I won't try any further to explain it.
 
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  • #40
PeterDonis said:
You are not thinking correctly about spacetime. Spacetime includes time. A spacetime model already includes all the information about when (i.e., at what events in spacetime) the light source is turned on. Only paths originating from those events in spacetime will have a nonzero amplitude. There is no such thing as "paths that don't yet exist through spacetime"; spacetime is not something that "changes" as things happen. A spacetime model already includes all the information about everything that happens.
Maybe I don't understand, I am saying it wrong, or something. When I draw a space time diagram (e.g. y-axis c*t and x-axis position), as y increases I add events that will occur. If I consider time 0, position 0, on the spacetime diagram as the start of the experiment at the light source, then the detector will be say at position 100 m away and outside of the light cone of the light source. The event where a photon from the light source reflects off of the grating and reach the detector will not happen until say y is c * 1200 ns (using the example I gave to gentzen). So if I am considering the experiment, up to only 840 ns, this event will not yet have happened. If the event has not happened, do I include it in the path integral?
 
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  • #41
gentzen said:
I wish you good luck trying to understand that stuff. I tried my best to explain it. As long as considerations when you turn the light on or off are relevant, assuming the light to be quasi-monochromatic is problematic. I won't try any further to explain it.
I am not sure why you sounded frustrated, but based on the context of your reply, it sounded like it was on my question about lasers being directional. Something I google confused me. Anyways it was a stupid thought and I am going to just delete that I ever said anything about it.
 
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  • #42
kurt101 said:
I am not sure why you sounded frustrated
I didn't intent to sound frustrated. I am not frustrated. Everybody here knows that this stuff is hard to understand. Don't worry. I "unwatched" this thread, and simply decided to also explicitly say "goodbye".
 
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  • #43
kurt101 said:
Maybe I don't understand, I am saying it wrong, or something. When I draw a space time diagram (e.g. y-axis c*t and x-axis position), as y increases I add events that will occur. If I consider time 0, position 0, on the spacetime diagram as the start of the experiment at the light source, then the detector will be say at position 100 m away and outside of the light cone of the light source. The event where a photon from the light source reflects off of the grating and reach the detector will not happen until say y is c * 1200 ns (using the example I gave to gentzen). So if I am considering the experiment, up to only 840 ns, this event will not yet have happened. If the event has not happened, do I include it in the path integral?
Let me try to fight fire with fire here. Let's look at the Lagrangian formulation of classical mechanics for a particle in a force field (gravity or EM or whatever).

You could describe this model as "the particle tries all possible paths and picks the one that minimises the action integral". Now we ask: while the particle is trying paths in one direction, what happens if we very quickly change the potential when the particle isn't looking? Does the particle take account of that change in the potential or not?

It's obviously absurb, because you're not supposed to take the idea that the particle is trying all paths and calculating the action integral literally.

It's the same with QED: you're not supposed to be thinking that the photon is literally traveling all these paths simultaneously - so that you can fool the particle by stopping the experiment before it's had a chance to try out all the paths. Which is effectively what you are asking.

You're not supposed to be overlaying a classical model of precise spacetime events at every location for every photon path in the calculation. Not least because in QED you have the uncertainty principle undepinning everything in any case.

Your whole line of thought is inappropriate to studying QM.
 
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  • #44
PeroK said:
Let me try to fight fire with fire here. Let's look at the Lagrangian formulation of classical mechanics for a particle in a force field (gravity or EM or whatever).

You could describe this model as "the particle tries all possible paths and picks the one that minimises the action integral". Now we ask: while the particle is trying paths in one direction, what happens if we very quickly change the potential when the particle isn't looking? Does the particle take account of that change in the potential or not?
I would suspect that any change you made quickly is limited by the speed of light. So this would not bother me because I can rationalize the cause.

I suspect in the path integral approach it is the same and it is limited by the speed of light. I thought Dr. Chinese more or less said this, but I am waiting for his confirmation before I am comfortable with that. I would also like to get a sense of how it is limited in the path integral approach. If someone told me we simply don't count the contributions from paths that can't happen because they violate the speed of light that would make sense to me, but so far nobody has told me this is what is done.

I understand these paths are not thought of as real paths, but used for calculation, but at the same time if the paths make a real contribution to the probability calculation then they have some aspect to them that has a real effect. So it is important to understand how that contribution is added or removed when the path integral approach is applied to an actual scenario (e.g. like my experiment).

PeroK said:
It's obviously absurb, because you're not supposed to take the idea that the particle is trying all paths and calculating the action integral literally.

I would tend to agree that a particle trying all paths and calculating the action seems like a crazy way for the universe to work. Like I said above, I have never thought of the idea that the particle takes all paths as actually what it is doing.

PeroK said:
It's the same with QED: you're not supposed to be thinking that the photon is literally traveling all these paths simultaneously - so that you can fool the particle by stopping the experiment before it's had a chance to try out all the paths. Which is effectively what you are asking.

I am more or less trying to ask my question in two ways:
I am trying to understand in an actual experiment when and where contributions from phenomena such as a grating begin and end.
I am trying to understand in the path integral approach how faster than light contributions from phenomena such as the grating are handled (i.e. do you remove them?).
It is my understanding that both the experiment and the path integral model should give the same result.
PeroK said:
You're not supposed to be overlaying a classical model of precise spacetime events at every location for every photon path in the calculation. Not least because in QED you have the uncertainty principle undepinning everything in any case.

Your whole line of thought is inappropriate to studying QM.

It is not clear to me whether you are saying my experiment would not tell me anything because of the uncertainty principle or that the uncertainty principle is just something that if I am not careful could skew the results of my experiment. I thought you and everyone else were saying the later, but maybe I have misunderstood.
 
  • #45
kurt101 said:
I suspect in the path integral approach it is the same and it is limited by the speed of light. I thought Dr. Chinese more or less said this, but I am waiting for his confirmation before I am comfortable with that. I would also like to get a sense of how it is limited in the path integral approach. If someone told me we simply don't count the contributions from paths that can't happen because they violate the speed of light that would make sense to me, but so far nobody has told me this is what is done.
You definitely must include all paths. You cannot exclude paths that "exceed the speed of light". One way to think about this is again the UP (Uncertainty Principle). There is no definite time when the photon is omitted - that is subject to quantum uncertainty. There's also no definite time when the photon was detected - for the same reason. So, you cannot say definitely that the photon traveled faster than light - but, again, that presupposes we are talking about definite photon paths, rather than contributions to a calculation.

Moreover, the photon paths must be a heuristic calculation that replaces a full QFT calculation involving a quantised EM field and photons as excitations of that field. I.e. not the point particles of the path integral formulation.

kurt101 said:
I understand these paths are not thought of as real paths, but used for calculation, but at the same time if the paths make a real contribution to the probability calculation then they have some aspect to them that has a real effect. So it is important to understand how that contribution is added or removed when the path integral approach is applied to an actual scenario (e.g. like my experiment).
Your proposed experiment changes the scenario to one that is no longer modeled by the simple path integral approach. The EM field behaves differently; the QFT calculations are different; the simple scenario presented by Feynman no longer applies; and, the simple path integral calculations no longer apply.

Your analysis depends on the paths being "real" (even though you say they are not).

kurt101 said:
I am trying to understand in the path integral approach how faster than light contributions from phenomena such as the grating are handled (i.e. do you remove them?).
No, you do not remove them. They are not real paths.

kurt101 said:
It is not clear to me whether you are saying my experiment would not tell me anything because of the uncertainty principle or that the uncertainty principle is just something that if I am not careful could skew the results of my experiment. I thought you and everyone else were saying the later, but maybe I have misunderstood.
The way you have described and analysed your experiment ignores the concept of uncertainty in QM. You say things like:

kurt101 said:
When I draw a space time diagram (e.g. y-axis c*t and x-axis position), as y increases I add events that will occur. If I consider time 0, position 0, on the spacetime diagram as the start of the experiment at the light source, then the detector will be say at position 100 m away and outside of the light cone of the light source. The event where a photon from the light source reflects off of the grating and reach the detector will not happen until say y is c * 1200 ns (using the example I gave to gentzen). So if I am considering the experiment, up to only 840 ns, this event will not yet have happened. If the event has not happened, do I include it in the path integral?
Where you make no concessions towards uncertainty. The photon is at point ##(x_0, y_0)## at time ##t = 0##; the photon is at point ##(x_1, y_1)## at time ##t = 1## etc. That is classical physics. If you mix that in with QED, you get something that is simply a wrong analysis.

There are no "events" here; there are simply contributions to a probability amplitude. You cannot talk about these virtual paths as comprising spacetime events with well-defined coordinates.

In general, you appear to have tried to pick up some QM/QED concepts, but are generally not prepared to relinquish any notions from classical physics.
 
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  • #46
You shouldn't take this addition over "all photon paths" literary. In relativistic QT you deal necessarily with a QFT, particularly for massless particles. The path integrals for relativistic QFT are over field configurations rather than single-particle paths, which works for nonrelativistic QM.
 
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  • #47
PeroK said:
You definitely must include all paths. You cannot exclude paths that "exceed the speed of light". One way to think about this is again the UP (Uncertainty Principle). There is no definite time when the photon is omitted - that is subject to quantum uncertainty. There's also no definite time when the photon was detected - for the same reason. So, you cannot say definitely that the photon traveled faster than light - but, again, that presupposes we are talking about definite photon paths, rather than contributions to a calculation.
I still don't understand the impact of the uncertainty principle on light emission in my experiment. I am ok with light emission being random in my experiment as long as this random emission does not affect the average intensity measured at my detector. In the Feynman scenario without my constraints the grating clearly affects the probability outcome regardless of the uncertainty principle. So clearly it is my constraints that you are claiming are a problem for the experiment, but in principle I can modify my experiment so it is more and more like the Feynman scenario. I can make the path to the grating much much longer than the shortest path to the detector and thus make the uncertainty principle less and less of a factor, right?
PeroK said:
Moreover, the photon paths must be a heuristic calculation that replaces a full QFT calculation involving a quantised EM field and photons as excitations of that field. I.e. not the point particles of the path integral formulation.Your proposed experiment changes the scenario to one that is no longer modeled by the simple path integral approach. The EM field behaves differently; the QFT calculations are different; the simple scenario presented by Feynman no longer applies; and, the simple path integral calculations no longer apply.

Your analysis depends on the paths being "real" (even though you say they are not).

No, you do not remove them. They are not real paths.
Ok, I can accept that the simple Feynman path approach described in Feynman's book" The Strange Theory of Light and Matter" can not model my experiment. Also I am taking from what you are telling me that the approach that can successfully model my experiment is sufficiently different that it would not be easy for you or anyone to explain how FTL effects from say the grating in my experiment are removed from the calculation.
 
  • #48
kurt101 said:
1. It is actually the opposite in the example, the diffraction grating placed at the end of the mirror adds a contribution, but I could see where one placed at the middle of the mirror counteracts the contribution, and so I get what you are saying.

2. I am not sure if I follow what you are saying. Are you saying that given the constraints of my experiment, you will see the same results with or without the diffraction grating? (i.e. were you giving a direct answer to my question?).

3, Ultimately this is what I want to understand. I have had trouble finding clear statements in any of the descriptions of the Fenman path integral (both with math and without) that clearly say contributions prohibited by the speed of light don't count.

4. That said, I still don't understand if contributions that are prohibited by the speed of light in my example are from the light sources perspective or say the mirror's perspective. Hopefully you understand what I am saying here and this would be a part 2 question for this thread and so I am reluctant to get into it too much before I am comfortable that I understand my current question.

5. If you changed the time constraint by adding additional detection time to be sure scenarios A and B (with and without the grating) should give a different result; then ran scenarios A and B many times; then took the averages; I would expect on average you would see a consistent difference between A and B, where scenario B with the grating that add contributions would on average detect more photons than A.
1. A grating prevents a reflected contribution from the section where the grating exists. It can be a positive contribution, a negative contribution, or no contribution depending on size and positioning.

2. There are many "alternative" ways to prevent contributions from a section of the mirror. If you blocked it, for example. Or in your version of the experiment, which in my opinion is impossible to implement, there wasn't a wide enough time window for the path to be traversed, that would be an alternative method.

3. I think everyone is saying the same things: there is no FTL effects. So another issue here is that reflection of light is a more complicated process than we are describing. You can't really draw a straight line from a source to a point on the mirror and then to the detector and say "it went this way". As a result, there may appear to be paths that loosely appear to be FTL. But no such path can be demonstrated as such experimentally, because the true picture is quite different.

4. I don't follow you here. There are no FTL contributions if you get strict enough. You would be violating the uncertainty principle if you tried to assert you know a particle's position at 2 precise points in time, and momentum in between.

5. There isn't a scenario where you can distinguish A and B. There is always quantum uncertainty in addition to technological constraints. At some point as the time window shrinks, there just won't be any photons emerging from the small time window.

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

As you continue to battle about conducting your experiment, you are actually attempting - whether you know it or not - to say that you can perform an experiment that would yield complete certainty in measurements of non-commuting observables. You either already accept that as impossible per QM, or you are tilting at a windmill using the mirror as a disguise in the process. I don't think any of us can help you much further, as it has already been demonstrated no FTL events are occurring. Once you accept that, your question is fully answered.
 
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  • #49
kurt101 said:
I still don't understand the impact of the uncertainty principle on light emission in my experiment.
When you say:

kurt101 said:
When I draw a space time diagram (e.g. y-axis c*t and x-axis position), as y increases I add events that will occur. If I consider time 0, position 0, on the spacetime diagram as the start of the experiment at the light source, then the detector will be say at position 100 m away and outside of the light cone of the light source. The event where a photon from the light source reflects off of the grating and reach the detector will not happen until say y is c * 1200 ns (using the example I gave to gentzen). So if I am considering the experiment, up to only 840 ns, this event will not yet have happened. If the event has not happened, do I include it in the path integral?
Then that is a classical, not a QM analysis. You cannot analyse QM experiments in those terms. It cannot be said any simpler.
 
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  • #50
DrChinese said:
2. There are many "alternative" ways to prevent contributions from a section of the mirror. If you blocked it, for example. Or in your version of the experiment, which in my opinion is impossible to implement, there wasn't a wide enough time window for the path to be traversed, that would be an alternative method.
I am fine with the alternative of blocking the grating if that helps anything in this discussion.

I am not sure what you mean by impossible. If you mean my initial example that used times of 10 seconds, that was clearly an in principle experiment not meant to be practical. If you mean your back of an envelope calculation where you used a distance to the mirror of 1000 wavelengths and came up with 0.003 femtoseconds then that was your choice of experimental setup. I came up with some more reasonable numbers in my discussion with gentzen with a difference of 90 nanoseconds. That said, I don't see why in principle you can't make the difference in paths as long as you want.

DrChinese said:
3. I think everyone is saying the same things: there is no FTL effects. So another issue here is that reflection of light is a more complicated process than we are describing. You can't really draw a straight line from a source to a point on the mirror and then to the detector and say "it went this way". As a result, there may appear to be paths that loosely appear to be FTL. But no such path can be demonstrated as such experimentally, because the true picture is quite different.
If that is the truth that there are no FTL effects, I certainly can accept that.

I don't know what you mean by loosely appear FTL.

DrChinese said:
4. I don't follow you here. There are no FTL contributions if you get strict enough. You would be violating the uncertainty principle if you tried to assert you know a particle's position at 2 precise points in time, and momentum in between.
I have not asserted I know anything about a particle position or momentum with any certainty in anything I have presented in this thread. I have asserted that the grating has an effect on the probability at the detector and I have theorized that there is something that goes from the source to the grating to the detector that can't go faster than light. By making this "something" go on a longer path, I was thinking my experiment would demonstrate that this "something" only has an effect on the probability at the detector when allowing for enough time in this experiment for this "something" to transverse its path and make it to the detector.

DrChinese said:
5. There isn't a scenario where you can distinguish A and B. There is always quantum uncertainty in addition to technological constraints. At some point as the time window shrinks, there just won't be any photons emerging from the small time window.

I am confused what you mean that there is no scenario where you can distinguish A and B. First of all that is my goal, to show that I get the same result for scenario A and B. However if you let A and B run for long enough time, I expect it to go to the scenario that Feynman presented which he clearly says there is a difference! So clearly we are not on the same page unless you are agreeing with my expectation which it does not sound like it.

To be clear we are on the same page, let's use the more practical version of my experiment that I presented to gentzen and use a 200 m mirror and assume red light (700 nm, 430e15 cycles / second) and:
put the source in the middle of the mirror, 100 m from the mirror
put the grating at the far left end of the mirror (Scenario B only)
put the detector at the far right end of the mirror, 100 m from the mirror

Then as a rough calculation:
The time for light to reach the grating is about 470 ns.
The time for light to reflect off the grating and go past the source is about 840 ns.
The time for light to reflect off the grating and reach the detector is about 1200 ns.
The time for light to take the shortest path, reflect off the mirror and reach the detector is 750 ns.

In this experiment, I would turn on the light source and look at the data from detector for up to 840 ns.

In scenario A there is no grating. We simply turn on the light source and record the accumulated intensity at the detector for 840 ns. We run this scenario many times to get an average accumulated intensity.

In scenario B there is a grating at the far left end of the mirror. We simply turn on the light source and record the accumulated intensity at the detector for 840 ns. We run this scenario many times to get an average accumulated intensity.

If I compare scenarios A and B, my expectation is that I will NOT see any difference in the average accumulated intensity (i.e. the intensity I averaged over many trials).

If I run scenarios A and B for longer than 840 ns they will become closer and closer, the longer I run them, to the scenarios Feynman described and I will see a difference in accumulated intensity.

So now that you clearly understand what I am comparing. Can you say whether you agree or disagree with my expectations and explain why?

DrChinese said:
------------------------------------

As you continue to battle about conducting your experiment, you are actually attempting - whether you know it or not - to say that you can perform an experiment that would yield complete certainty in measurements of non-commuting observables. You either already accept that as impossible per QM, or you are tilting at a windmill using the mirror as a disguise in the process.

I don't understand what you mean when you say "whether you know it or not - to say that you can perform an experiment that would yield complete certainty in measurements of non-commuting observables",
That said if you are implying I am trying to say I can perform an experiment that would yield complete certainty in my measurements. I would not say that at all. I am collecting averages over many trials. I am not expecting individual trials to measure the same intensity, but I do expect given enough trials that I will converge on some consistent average value. Even if I am measuring random noise, I will converge on some consistent average value.

So can you explain what the non-commuting observables are in my experiment?

DrChinese said:
I don't think any of us can help you much further, as it has already been demonstrated no FTL events are occurring. Once you accept that, your question is fully answered.

My expectation has always been that there would be no FTL events and no FTL effects in this scenario. So I have no problem accepting it and a little confused that you think I would.

In my mind an FTL effect would be like the EPR scenario, but in my scenario there is no local preparation (as vanhees71 has described in other threads) between the source and the grating and so that has been my rationale for expecting no FTL effect.

So if there are clearly no FTL effects then I am mostly satisfied. I would still appreciate a definitive answer on my experimental discrepancies that I list above. If you clearly understand what I am doing in the experiment, you should be able to definitively say what the result of the experiment will be. You can tell me that I can't conclude anything useful because my result but I would still like you to tell me what the result will be.

I look forward to completing this thread as soon as I can, but I would like to end on some clarity.
 
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