Undergrad Speed of individual photons in a vacuum?

[PeterDonis]

I had assumed that SR required the photon to travel at exactly the speed of light and that this caused, via the HUP, to make the position unknown.

SR is a classical theory, not a quantum theory, and has no concept of "photons" (although unfortunately many SR textbooks misleadingly use the term "photon" instead of something like "light pulse" or "light ray"--Taylor & Wheeler is one textbook that, IIRC, actually explains why "photon" is not a good term in a classical theory).

Quantum field theory is our best current quantum theory that takes SR effects into account, but in QFT photons do not always travel at exactly the speed of light; they have nonzero probability amplitudes for traveling faster or slower than light. Also, in QFT, the HUP doesn't work the way you were thinking it does; photons still do not have a well-defined position operator so there is no way to relate their momentum uncertainty to any position uncertainty.

 

[PeterDonis]

I had assumed that SR required the photon to travel at exactly the speed of light and that this caused, via the HUP, to make the position unknown.

Even leaving aside all the other issues that have been raised, and assuming for the sake of argument that we could formulate a HUP for photons using some kind of position operator, this reasoning would still be invalid. The HUP is between position and momentum, not position and velocity, and even if photons in QM traveled at exactly the speed of light, they still do not all have the same momentum and they can still have momentum uncertainty. (Their momentum uncertainty would be related to uncertainty in their energy.) So it still would not follow that photons being forced to travel exactly at the speed of light would have to have completely uncertain position.

 

[Vanadium 50]

@LarryS have you considered trying to learn physics sequentially? This is kind of like having a blind man paint your house with paintballs. I mean, sure, eventually the job will get done, but it may not be the most efficient way.

 

[LarryS]

Good suggestion but I'm having too much fun bouncing around from subject to subject. I retired a few years ago after spending 50 years in IT. This is all for pure enjoyment. Physics Forums allows me to pick the brains of those with more formal education and experience in physics than myself.

 

[Vanadium 50]

Yes, but you are making the people who are helping you work harder than necessary as well. A more methodical approach will make their lives easier, and you will learn more and faster. Everybody wins.

There's a reason we don't paint houses with the blind paintball method.

 

[vanhees71]

Well, I looked for such references, and to my surprise I could not find any. The photon does not have a well-defined position, but for reasons other than I thought. I had assumed that SR required the photon to travel at exactly the speed of light and that this caused, via the HUP, to make the position unknown.

Thank you all for your input.

The photon has not even a position observable, and thus there's no HUP to begin with.

 

[bob012345]

The photon has not even a position observable, and thus there's no HUP to begin with.

And why is that?

 

[vanhees71]

http://arnold-neumaier.at/physfaq/topics/position.html

 

[Delta2]

@vanhees71 that page is too old, I think it was written before Latex was invented. This doesn't mean it is wrong just very hard to read

 

[Delta2]

em sorry @gentzen does my browser has a problem and doesn't display the page correctly or what?

 

[gentzen]

em sorry @gentzen does my browser has a problem and doesn't display the page correctly or what?

I have now read the content of that page. Of course I did not understand everything, but that was totally unrelated to that page not using LaTeX. And even before I read it, it was clear that its readability was unaffected by its ascii representation of the formulas.

 

[Delta2]

I have now read the content of that page. Of course I did not understand everything, but that was totally unrelated to that page not using LaTeX. And even before I read it, it was clear that its readability was unaffected by its ascii representation of the formulas.

Ok, well for me clearly its readability is problematic (remembers pre 2000 pages, circa 1995-1999) but anyhow at least I know its not my browser.

 

[bob012345]

Going back to the spirit of the OP, first, can an experiment be set up that produces what the consensus would agree can be considered a single photon? If so, does that photon, have measurable physical properties such as motion in space over time or extension over space that are measurable within experimental limits? If so, can such things as the movement through space be measured without significantly changing such properties? If so, could the same photon be measured twice at two different locations in its path and thus a speed experimentally defined? If the properties are changed at the first measurement, nevertheless could the differences between the two measurements still be used to define a photon speed and compare events? Thanks.

 

[Lord Jestocost]

Photons are defined by the three quantum numbers associated with momentum, energy and polarization; position and time do not enter into consideration. (Arthur Zajonc in “The nature of light: what is a photon?”)

 

[PeterDonis]

can an experiment be set up that produces what the consensus would agree can be considered a single photon?

Sort of. You can do experiments with very faint light sources in which individual dots show up on a detector; the dots in such experiments can be interpreted as the detection of single photons by the detector. In most cases, however, such light sources do not emit a state of the quantum electromagnetic field that would normally be interpreted as a "single photon" state. Such a state would be a Fock state, and sources that emit Fock states are extremely hard to make. Most light sources emit coherent states, which have no definite photon number.

If so, does that photon, have measurable physical properties such as motion in space over time or extension over space that are measurable within experimental limits?

No. Photons are not little billiard balls. They aren't even close.

 

[vanhees71]

Sort of. You can do experiments with very faint light sources in which individual dots show up on a detector; the dots in such experiments can be interpreted as the detection of single photons by the detector. In most cases, however, such light sources do not emit a state of the quantum electromagnetic field that would normally be interpreted as a "single photon" state. Such a state would be a Fock state, and sources that emit Fock states are extremely hard to make. Most light sources emit coherent states, which have no definite photon number.

Such "faint light sources" are no single-photon sources but just coherent states of very low intensity, i.e., it's mostly the vacuum state but contains also $N$-photon states for all $N \in \mathbb{N}_0$.

True single-photon states are not so simple to have, but for some decades one has such sources. One standard way is to use parametric down-conversion, where you use a laser and certain sorts of birefringent crystals (like beta-barium borat, BBO), where for certain directions behind the crystal you get entangled photon pairs, i.e., one photon from the laser interacts with the crystal in such a way that two photons are created such that energy and momentum is conserved ("phase matching") as well as the polarization of the two photons are entangled. Depending on the type of conversion you get singlet or triplet polarization states. You can use this as a "heralded single-photon source", i.e., you use one photon as a "trigger", announcing the other photon (the "idler") to be present in the corresponding other direction determined by the phase-matching condition. If you also measure the polarization state of the trigger photon you also know the polarization state of the idler photon. In this way you get a true one-photon source.

No. Photons are not little billiard balls. They aren't even close.

I couldn't agree more. Photons do not even have a position observable in the usual sense. To get an intuition for photons you should just think of them as electromagnetic waves and the intensity of the electromagnetic wave as the probability density to detect a photon at a place determined by the position of the detector. The most distinct property of a single-photon state in comparison to a classical em. wave is that it can be absorbed as a whole or not at all. There's no way to absorb "half a photon". That's the only sense you have "particle-like features" of single-photon states, but it's not possible to define a position of a single photon.

Another case are "cavity photons". There you have a box made of conducting walls. As classical electrodynamics tells you, in this case you have some boundary conditions for the electromagnetic field to fulfill, leading to discrete frequencies and wave numbers, out of which the most general radiation field can be built. These are all "standing waves", due to the boundary conditions. A photon is now a certain state of the corresponding quantized field. This photon is, of course "localized" within the cavity, i.e., the probability to detect it outside the cacity is 0, but it has no position within the cavity either. It's just within the cavity but not in any sense like a point particle, bouncing back and forth between the walls of the cavity.

 

[bob012345]

Sort of. You can do experiments with very faint light sources in which individual dots show up on a detector; the dots in such experiments can be interpreted as the detection of single photons by the detector. In most cases, however, such light sources do not emit a state of the quantum electromagnetic field that would normally be interpreted as a "single photon" state. Such a state would be a Fock state, and sources that emit Fock states are extremely hard to make. Most light sources emit coherent states, which have no definite photon number.

Thanks. FYI here is a talk about making a Fock laser

No. Photons are not little billiard balls. They aren't even close.

And I was not suggesting they are. I was carefully trying to ask about what can and can't be measured in the spirit of the OP.

 

[bob012345]

Another case are "cavity photons". There you have a box made of conducting walls. As classical electrodynamics tells you, in this case you have some boundary conditions for the electromagnetic field to fulfill, leading to discrete frequencies and wave numbers, out of which the most general radiation field can be built. These are all "standing waves", due to the boundary conditions. A photon is now a certain state of the corresponding quantized field. This photon is, of course "localized" within the cavity, i.e., the probability to detect it outside the cavity is 0, but it has no position within the cavity either. It's just within the cavity but not in any sense like a point particle, bouncing back and forth between the walls of the cavity.

So, is there such a thing as a single photon propagating in space? I'm hearing no, don't think that way. If it is a pure wave it surely can't fill the universe (cavity) at once so it must have some kind of leading edge. Why can't that edge be measured as it passes a detector? It seems photons as real moving and interacting objects were fundamental to early Quantum theory and effects such as the photoelectric effect and Planck's blackbody radiation.

 

[PeterDonis]

is there such a thing as a single photon propagating in space? I'm hearing no, don't think that way.

You're hearing correctly.

Why can't that edge be measured as it passes a detector?

Any such measurement does not detect an "edge" of a wave. It detects a particle--for example, a dot on a detector screen. Wave patterns seen in measurements of light are obtained from measurements of large numbers of photons, either because the light source is high intensity, or because the experiment is run for a long time in order to detect many photons one at a time over a long period (for example, a double slit experiment run with a very faint light source, in which individual dot impacts on the detector gradually build up an interference pattern).

It seems photons as real moving and interacting objects were fundamental to early Quantum theory and effects such as the photoelectric effect and Planck's blackbody radiation.

Photons as a concept associated with quantum properties of light were fundamental to these things. But photons as "moving and interacting objects" in the sense you mean were not. The early theorists didn't know that because they didn't know very much about photons and didn't have a theory for them; they just had some heuristic ideas. Now we have a theory of photons, the quantum field theory of the electromagnetic field, and we know that all the effects you describe can be accounted for just fine even though photons are not "moving and interacting objects" in the sense you mean.

 

[bob012345]

Are you trying to say something like it's the field, always the field and only the field that exists and any manifestation of what one would label a photon is merely a local quantum state of the field?

 

[PeterDonis]

Are you trying to say something like it's the field, always the field and only the field that exists and any manifestation of what one would label a photon is merely a local quantum state of the field?

Yes. That is the basic idea of quantum field theory and is true for any field, not just the quantum electromagnetic field.

 

[DrChinese]

1. So, is there such a thing as a single photon propagating in space? I'm hearing no, don't think that way. If it is a pure wave it surely can't fill the universe (cavity) at once so it must have some kind of leading edge. Why can't that edge be measured as it passes a detector?

2. Are you trying to say something like it's the field, always the field and only the field that exists and any manifestation of what one would label a photon is merely a local quantum state of the field?

With all of this, there are so many nuances that the language and specific words can get in the way. We tend to debate this stuff around and around, as within the scientific community there are a variety of opinions.

1. You can have single photons propagating in space. One, and only one (as you mention above, a Fock state). And you can localize it to block of spacetime. You can manipulate it in many ways, and then detect it where you would expect to find it. See details here, for one example which goes far past the photoelectric effect:

http://people.whitman.edu/~beckmk/QM/grangier/Thorn_ajp.pdf2. With due respect to PeterDonis' always correct answers: a photon may or may not be considered a local object depending on the experimental setup. And also depends what you call "local".

The main idea of this thread is that photons don't have something that corresponds to a position in the same manner that an electron does. There are a number of reasons for this, and generally within QFT the explanation is not dependent on the Uncertainty Principle.

One problem is that single photons cannot be easily be created on demand in the very specific sense that emission/absorption times are precise and knowable. In an attosecond (10^-18 seconds), a photon will travel roughly 5 times the radius of a hydrogen atom (just to put in perspective). There are not many experiments running at that level of accuracy, and in fact single photon experiments often resolve photons not much better than to the femtosecond to nanosecond scales (which are a thousand to a billion times less resolution). You also have the problem that the wavelength of the photon may be larger than the spacetime block you are trying to localize the photon to, which leads to more difficulty. So you can't push a button and know the extremely precise time of photon creation. Ergo, you could never use a standard approach to assessing its position at some precise point in time.

So if it cannot have a position, how can it be considered local to anything? Again, this is a point that is hotly debated. I think the best one can say is that QFT makes predictions as to what a specific experimental setup will demonstrate about a photon, and the results match the predictions. That's impressive. I don't think the English language conveys the issues well in the context of some of the thread answers.

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

Does a photon have a position, or can it be considered localized in a meaningful manner? Apparently, it depends.

 

[Vanadium 50]

@DrChinese , all of what you wrote is true. But a lot of those problems go away when you're working with gamma rays. These can be placed in a state very close to a Fock state and have small wavelengths. Indeed, this is as close to a "billiard ball" as you're likely to find.

Those photons, we know travel at c. (There is a 1951 paper by Cleland and Jastram where they measure their speed to be 0.995 +/- 0.005 c,) I guessed a percent - they did half a percent.

 

[PeterDonis]

You can have single photons propagating in space. One, and only one (as you mention above, a Fock state). And you can localize it to block of spacetime.

In a block of spacetime, yes. But in general you can't assign a photon a particular trajectory within that block of spacetime. The best you can say in general is that, for example, in the experiment in the Thorn et al. paper you referenced, the photon travels from the down conversion crystal through the beam splitter to the detectors; but you can't say that any particular curve in spacetime is its trajectory between those points.

An item of interest about this particular experiment is that, if I'm understanding it correctly, it allows you to test specifically for a Fock state--i.e., it distinguishes a Fock state from other possible states such as coherent states, by the vanishing coincidence count (a coherent state would not have a coincidence count of zero, within error bars, in this experiment).

 

[vanhees71]

With all of this, there are so many nuances that the language and specific words can get in the way. We tend to debate this stuff around and around, as within the scientific community there are a variety of opinions.

1. You can have single photons propagating in space. One, and only one (as you mention above, a Fock state). And you can localize it to block of spacetime. You can manipulate it in many ways, and then detect it where you would expect to find it. See details here, for one example which goes far past the photoelectric effect:

http://people.whitman.edu/~beckmk/QM/grangier/Thorn_ajp.pdf

Of course one can prepare single photons, e.g., by using parametric down-conversion an using one photon as a "signal photon" to "herald" the other single "idler" photon. You can even determine the polarization state of the idler photon when doing a polarization measurement on the signal photon thanks to the entanglement of the photon pair.

It's, however indeed not simply a massless point-like classical particle. It's not only not localizable but doesn't even have a position observable at all. All you can know, given its state ("preparation procedure") are the probabilities for detecting it at a given time and at the position of the detector. This is encoded in the autocorrelation functions ("Green's functions") of the em. field's energy density. It's nicely described in the quoted paper.

2. With due respect to PeterDonis' always correct answers: a photon may or may not be considered a local object depending on the experimental setup. And also depends what you call "local".

Not again ;-). But here I agree. Photons are not localizable. All you can observe are "detection events" at a certain time and position of the corresponding detector. Of course, the QED as the paradigmatic example of a local relativistic QFT is, well, local in the very clear and specific sense it is by construction:

(a) The local observables are built by field operators that transform locally under Poincare transformations
(b) The Hamilton density commutes with all local observables at space-like distances (microcausality)
(c) This implies that the S-matrix (transition-probability densities) obeys the cluster-decomposition principle, is unitary and Poincare invariant.

The main idea of this thread is that photons don't have something that corresponds to a position in the same manner that an electron does. There are a number of reasons for this, and generally within QFT the explanation is not dependent on the Uncertainty Principle.

Of course, the uncertainty principle holds also in QFT. It's a mathematical property following from the very basic rules of any QT. If there is a position and a momentum observable they fulfill the corresponding uncertainty principle $\Delta x_j \Delta p_k\geq \hbar/2 \delta_{jk}$. Photons have no position observable though. "Localizability" of photons must thus be a derived concept, i.e., rely on the observables, like detection probabilities.

One problem is that single photons cannot be easily be created on demand in the very specific sense that emission/absorption times are precise and knowable. In an attosecond (10^-18 seconds), a photon will travel roughly 5 times the radius of a hydrogen atom (just to put in perspective). There are not many experiments running at that level of accuracy, and in fact single photon experiments often resolve photons not much better than to the femtosecond to nanosecond scales (which are a thousand to a billion times less resolution). You also have the problem that the wavelength of the photon may be larger than the spacetime block you are trying to localize the photon to, which leads to more difficulty. So you can't push a button and know the extremely precise time of photon creation. Ergo, you could never use a standard approach to assessing its position at some precise point in time.

That precisely goes in the direction I mentioned above.

So if it cannot have a position, how can it be considered local to anything? Again, this is a point that is hotly debated. I think the best one can say is that QFT makes predictions as to what a specific experimental setup will demonstrate about a photon, and the results match the predictions. That's impressive. I don't think the English language conveys the issues well in the context of some of the thread answers.

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

Does a photon have a position, or can it be considered localized in a meaningful manner? Apparently, it depends.

As you say it depends on what you mean by "localized". It cannot be in the sense of a position observable as for massive particles but only in terms of detection at a certain time and space with a detector, and these quantities have of course a finite resolution only, where particularly the temporal resolution is also to be considered separately, because the related "energy-time uncertainty relation" is not of the usual kins of uncertainty relations between observables since time never is an observable (i.e., neither in non-relativistic QM nor in relativistic QFT), but that's another story.

 

[vanhees71]

In a block of spacetime, yes. But in general you can't assign a photon a particular trajectory within that block of spacetime. The best you can say in general is that, for example, in the experiment in the Thorn et al. paper you referenced, the photon travels from the down conversion crystal through the beam splitter to the detectors; but you can't say that any particular curve in spacetime is its trajectory between those points.

Indeed there cannot be trajectories of photons in the sense of point paricles since photons don't have a position observable to begin with. A much better intuition to think about "propagation" is indeed in the sense of field theory. Heuristically it's almost always safe to say that the detection probability of one (or here when dealing with two-photon states the joint detection probability of these two) photons is proportional to the energy density (or the energy flux) at the place of the detector (or in the case of the two-photons the two-particle correlation operator of the energy density).

An item of interest about this particular experiment is that, if I'm understanding it correctly, it allows you to test specifically for a Fock state--i.e., it distinguishes a Fock state from other possible states such as coherent states, by the vanishing coincidence count (a coherent state would not have a coincidence count of zero, within error bars, in this experiment).

Yes, these kind of experiments are indeed the only clear demonstration that the electromagnetic field must be quantized. For a long time, i.e., before it was possible to prepare true photon Fock states in a reliably efficient way, the only (indirect) hint that the em. field has to be quantized was spontaneous emission. Otherwise you get very far with the semiclassical approximation, i.e., quantizing only the "matter" (particularly the electrons in atomic, molecular and condensed matter physics) and keeping the em. field as "classical": That explains, on the tree level, among other things, the photoelectric effect and Compton scattering, which are often falsely quoted as evidence for em-field quantization ;-).

 

[bob012345]

Indeed there cannot be trajectories of photons in the sense of point particles since photons don't have a position observable to begin with. A much better intuition to think about "propagation" is indeed in the sense of field theory. Heuristically it's almost always safe to say that the detection probability of one (or here when dealing with two-photon states the joint detection probability of these two) photons is proportional to the energy density (or the energy flux) at the place of the detector (or in the case of the two-photons the two-particle correlation operator of the energy density).

So we say photons travel but don't have point particles like trajectories. Does this have anything to do with Feynman's multiple paths? Your safe heuristic above suggests skirting the issue altogether.

 

[vanhees71]

Photons cannot be described by point-particle path integrals either, because photons have no paths to begin with. Of course, it can be described by path integrals over field configurations, and that's even a very elegant method particularly for gauge theories as QED. As I said, you can safely forget about the point-particle picture for photons (if not even for all particles in the relativistic realm).

 

[Nugatory]

So we say photons travel but don't have point particles like trajectories.

You can say that, but it's unlikely to be helpful. What does the idea of "travel" add to our understanding if there is no path along which the travel happens? What happens is that in an interaction at point A some amount of energy is transferred to the electromagnetic field; we can then calculate the probability that the electromagnetic field will transfer that amount of energy to a detector at point B at some later time. Those are statements about what happens at points A and B - nothing stops you from saying that the energy "travels" from A to B but saying that doesn't tell you anything new or useful.

Does this have anything to do with Feynman's multiple paths?

No. Multiple paths is a technique for calculating these probabilities, and the paths aren't even paths through space.

Your safe heuristic above suggests skirting the issue altogether.

That's the whole point. The issue should be skirted, for about the same reason that if you asked me how much February weighs I would refuse to consider the issue of how to weigh a month.

 

[Vanadium 50]

<bracing for the inevitable followup> "Yeah, but if you could weigh February, what would it weigh?"