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

[Paul Colby]

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

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

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

 

[Paul Colby]

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

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

 

[PeterDonis]

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

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

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

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

 

[Paul Colby]

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

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

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

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

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

 

[PeterDonis]

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

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

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

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

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

 

[Paul Colby]

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

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

 

[PeterDonis]

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

Why? Please show your work.

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

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

 

[Philip Koeck]

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

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

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

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

 

[vanhees71]

You surely meant a single electron ;-)

Ok course ;-(.

 

[vanhees71]

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

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

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

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

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

 

[Paul Colby]

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

Yes, I'm in error.

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

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

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

 

[DrChinese]

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

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

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

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

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

https://arxiv.org/abs/0809.3991

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

 

[vanhees71]

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

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

https://arxiv.org/abs/0809.3991

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

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

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

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

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

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

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

 

[DrChinese]

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

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

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

https://arxiv.org/abs/0809.3991

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

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

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

 

[neilparker62]

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

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

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

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

 

[vanhees71]

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

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

https://arxiv.org/abs/0809.3991

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

Obviously we don't understand each other. What I always stress is that the "paths" are indistinguishable if you have entangled photon pairs. The point is that you choose the pairs, where you can't know which photon ("labelled" by its momentum) has which polarization (H and V), as becomes very clear from this figure from Wikipedia's article on SPDC:

Taking the photons with momenta in the directions determined by the intersection of the two cones makes it impossible to know the "specific paths" of the H-polarized and the V-polarized photon. That's why you have a superposition of two paths and that's why you have the entanglement, discussed in the experiments, including the one about entanglement swapping in the paper you quote.

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

It has all to do with the quantum nature of how you treat the photons 2 and 3, enabling to "project" to either of the four possible Bell states. Again it's crucial that there is no determined paths at the moment you detect these two photons after running through the PBSs to get the entanglement of photons 1 and 4 when looking at either of the four subensembles, where photons 2 and 3 are selected to be in either of the corresponding Bell states.

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

The indistinguishability of the two "paths" is the key feature of the entanglement of the two photons we discuss here.

 

[vanhees71]

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

Well, I guess @DrChinese misunderstood me (on purpose?) as if I was advocating to avoid the words "photon" or even "paths". I never said we should follow Lamb in abandoning the word "photon" for good, because that's impossible given the ubiquitous use of this terminology. I'm only insisting on using it exclusively in the modern from as defined by QED. Otherwise there's no chance to get things right, particularly when it comes to specific extremes of the quantum behavior of the electromagnetic field. Also "paths" should be understood in terms of QED and in no other way, i.e., what has to be abandoned is a naive photon picture as proposed by Einstein in 1905. Einstein was the one who never was satisfied with this notion of his "heuristic picture" but also not with the modern notion of relativistic QFT/QED. For the em. field as a zero-mass spin-1 field there are no states resembling a classical, localizable point particle. Rather the classical limit are high-intensity coherent states, which resemble classical electromagnetic fields in the sense that the quantum fluctuations are very small compared to the mean values of the corresponding classical wave observables.

 

[CoolMint]

 

[DrChinese]

1. Well, I guess @DrChinese misunderstood me (on purpose?) as if I was advocating to avoid the words "photon" or even "paths".

2. I never said we should follow Lamb in abandoning the word "photon" for good, because that's impossible given the ubiquitous use of this terminology. I'm only insisting on using it exclusively in the modern from as defined by QED. ... Rather the classical limit are high-intensity coherent states, which resemble classical electromagnetic fields in the sense that the quantum fluctuations are very small compared to the mean values of the corresponding classical wave observables.

1. Vanhees71, post #15 of original thread: "Forget about paths here! It's really crucial to understand in this context that there are no paths and that photons are indistinguishable bosons!" Ah, I don't think I misunderstood anything - and certain not on purpose.

Vanhees71, post #106 above: "The indistinguishability of the two "paths" is the key feature of the entanglement of the two photons we discuss here." No significant issue with the science of that post, a couple of points to tidy up.

a. When discussing entangled photon pairs, and a diagram specifies a PDC source for the sake of discussion: the mechanics of PDC are not relevant. And yes, it is a factual (but completely technical) statement that basic entangled PDC requires indistinguishability of the 2 photons' sources. Specifically, in your diagram, you must not be able to determine if one particular photon came from the V cone or from the H cone (and vice versa). You could choose to pick up the photon pair such that one is known to be from the V cone - such pairs will not be polarization entangled.

b. Again, in general, polarization entangled photon pairs have no path indistinguishability requirement. They don't need to be path entangled, they don't need to be momentum entangled, or indistinguishable on either of those bases.

Hopefully, the science of these 2 points is not in question. 2. I agree with this, no issue. Hopefully we can leave it at this.

 

[vanhees71]

1. Vanhees71, post #15 of original thread: "Forget about paths here! It's really crucial to understand in this context that there are no paths and that photons are indistinguishable bosons!" Ah, I don't think I misunderstood anything - and certain not on purpose.

Vanhees71, post #106 above: "The indistinguishability of the two "paths" is the key feature of the entanglement of the two photons we discuss here." No significant issue with the science of that post, a couple of points to tidy up.

This is indeed crucial: The "paths" are indeed indistinguishable here, i.e., there is a coherent superposition of two possibilities, which you always insist to call "paths": "path 1": the photon with momentum $\vec{k}_1$ has polarization H and the photon with momentum $\vec{k}_2$ has polarization V and "path 2" vice versa. This clumsy verbal description is made clear by the math,
$$|\Psi \rangle=\frac{1}{\sqrt{2}} [\hat{a}^{\dagger}(\vec{k}_1,H) \hat{a}^{\dagger}(\vec{k}_2,V) - \hat{a}^{\dagger}(\vec{k}_1,V) \hat{a}^{\dagger}(\vec{k}_2,V)] |\Omega \rangle.$$

a. When discussing entangled photon pairs, and a diagram specifies a PDC source for the sake of discussion: the mechanics of PDC are not relevant. And yes, it is a factual (but completely technical) statement that basic entangled PDC requires indistinguishability of the 2 photons' sources. Specifically, in your diagram, you must not be able to determine if one particular photon came from the V cone or from the H cone (and vice versa). You could choose to pick up the photon pair such that one is known to be from the V cone - such pairs will not be polarization entangled.

No it's not simply a technical statement, it's what's defining entanglement in the first place! Of course, I could choose to pick up another photon pair as you describe, but now you yourself admit: "such pairs will not be polarization entangled." As soon as what you call "paths" are distinguishable, you don't have (complete) entanglement anymore and you prepare a different state, like the unentangled product state
$$|\Psi' \rangle=\hat{a}^{\dagger}(\vec{k}_1,H) \hat{a}^{\dagger}(\vec{k}_2 V) |\Omega \rangle.$$
I used PDC as an example, because it's pretty easy to understand, why one has entangled photons for this specific choice of the momenta, i.e., on the interaction points of the cones. The same basic "mechanism" is quite common to the production of entangled states though (e.g., the original cascade in Aspect's experiment).

b. Again, in general, polarization entangled photon pairs have no path indistinguishability requirement. They don't need to be path entangled, they don't need to be momentum entangled, or indistinguishable on either of those bases.

You contradict yourself, because two lines before you said it yourself, if you choose a situation, where you know on which cone the photon with momentum $\vec{k}_1$ is located (i.e., which polarization it has), you don't get an entangled state.

I still don't understand, what you mean by "path entangled". The photons are either in an entangled state or not!

Hopefully, the science of these 2 points is not in question.2. I agree with this, no issue. Hopefully we can leave it at this.

 

[andresB]

While answering another thread I came across these interesting words by Landau and Lifshitz (see the introduction)
https://www.amazon.com/dp/0750633719/?tag=pfamazon01-20

 

[throw]

While answering another thread I came across these interesting words by Landau and Lifshitz (see the introduction)
https://www.amazon.com/dp/0750633719/?tag=pfamazon01-20

One should compare those words to those in the first sections of the earlier QM volume too.

The idea that quantum objects have paths contradicts the very first thing QM says as one can see from the first section of both of these books.

 

[PeterDonis]

The idea that quantum objects have paths contradicts the very first thing QM says

Since this thread started because there is more than one possible meaning of the term "path", you should be more specific about which particular meaning is intended here. I believe a good alternative term for that meaning would be "trajectory", or the SR term "worldline", i.e., a single curve in spacetime that describes the object.

I don't think there are any physicists working on QM who would dispute the statement that quantum objects do not have trajectories in the above sense. However, I also don't think physicists who use the term "path" in a quantum context (and plenty of references have been given to such usage in the literature) are using it to mean trajectories in the above sense. I think they are using it to mean something else.

 

[throw]

By a path is meant a curve which can be determined by a position vector at each instant, which can equivalently be given by knowing an initial position, a tangent vector (velocity/momentum) at that point, and a law which tells us how they evolve (with respect to e.g. time, or some other parameter like proper length etc). We can do this in Euclidean space parametrized by time of Minkowski space etc so yeah it agrees with your sense.

The first thing QM says (see the references above) is that we simply can't simultaneously specify a position vector and a tangent vector simultaneously (at the same 'instant'), otherwise there would be no reason why we couldn't have a path for a 'quantum' particle and so find some classical description, it's only one or the other.

It's a very poor choice of language to use the word path in a quantum context (at least in the sense of minimizing confusion) unless one is using a quasi-classical approximation in which the classical notion of a path becomes more relevant, and it seems that is what is being argued for in this thread:

Paths only exist in a classical sense, i.e. the less accurately we measure the more accurate the notion of a path becomes (basically the eikonal approximation), I can see this sense being argued for e.g. in post 64 ("There is nothing particularly "quantum" about the path the experimentalist directs entangled photons") which seems to be the sense argued for at the start of the thread and seems to agree with post 46 too in arguing there is a difference between 'classical paths' and 'quantum paths'.

The only/main place I think see this not really being argued for is in that post 46 where it is argued a 'quantum' particle followed a path, but just not one specific path, which doesn't make sense - but neither does the idea that a particle doesn't follow a path i.e. it was likely meant in some vague path integral sense - instead of no path it travels all possible paths, which of course is just an equivalent starting point for QM.

In all cases, I don't see anything other than the usual notion of a path being used, where in a quantum context we do this via a quasi-classical approximation, which seems to be how it's being used at least in my reading, maybe the references meant something different.

 

[PeterDonis]

By a path is meant

No, by a "path" you mean what you said. You can't dictate to other people what they mean by "path". You have to actually go and look at what they mean.

 

[throw]

You have to actually go and look at what they mean.

That's what I did when I cited multiple posts and argued they seem to be using only one notion of a 'path', you seem to be the only person here arguing there are multiple notions of a path (without any citations to justify this claim I might add).

It's absolutely not an accepted fact that there is any than one notion of a path - that in QM we can only invoke this single idea of a path in a quasi-classical sense, it does not mean there are multiple notions of the idea of a path, and even the path integral invokes this single idea of a path, even this MWI notion of multiple paths is still just using the single idea of a path, nothing else, this is getting absurd.

 

[PeterDonis]

I cited multiple posts and argued they seem to be using only one notion of a 'path'

I don't think you're reading very carefully. In this thread overall, I see at least three notions of "path":

(1) The classical approximation notion that you describe;

(2) The "path" defined by an actual physical device such as a fiber optic cable (e.g., post #12) or a series of detectors with a particular layout in space (e.g., post #22);

(3) The "path" defined by a source at a particular location in space and a detector at a particular location in space (e.g., post #2, with "source" CRT and "destination" eye).

A possible fourth notion would be "path" as in "path integral", which you also refer to.

 

[throw]

I've already pointed out that only one idea of a path is being used in this thread, to see you argue that these four manifestations+examples of the (simple) idea of a path in a quantum and/or a classical context are all somehow radically different things that encode different fundamental ideas of a path and somehow result in different definitions of (the simple idea of) what a path is, which is just jaw-dropping ridiculous if we take it seriously, it at least explains why you'd think

I don't think you're reading very carefully

 

[PeterDonis]

I've already pointed out that only one idea of a path is being used in this thread

You've given that as your opinion, yes. I don't agree with your opinion.

I get that you don't agree with my opinion either, which is fine. My main point is to emphasize that both of us are giving opinions. There is no law of physics that says unambiguously what the term "path" means, or what the author of any particular reference must have meant by"path". But you are talking as if there is; you are claiming it is "jaw-dropping ridiculous" to interpret the ordinary language word "path" any other way than your particular preferred way. That claim is nonsense.

Ordinary language is vague. There is no way around that. That's why, if we really want to do physics, we don't use ordinary language; we use math. We don't try to build physical theories in terms of vague ordinary language words like "path"; we build them using precise mathematical equations.

 

[jedishrfu]

Since this thread has pretty much run its course and the OPs questions have been answered as best as can be, it's time to close this thread and to say thanks to everyone who have contributed to it.

Take care,
Jedi