Speed of individual photons in a vacuum?

[LarryS]

TL;DR Summary

Does speed of individual photons in a vacuum vary?

Is there experimental evidence that confirms that the speed of individual photons in a vacuum never varies, even slightly?

Thanks in advance.

 

[Dale]

Summary: Does speed of individual photons in a vacuum vary?

Is there experimental evidence that confirms that the speed of individual photons in a vacuum never varies, even slightly?

Thanks in advance.

I cannot think of how that could be tested even in principle.

 

[Vanadium 50]

never

Never? Never ever? Not even when we aren't measuring it?

even slightly

Even below the ability to measure?

By those requirements, I couldn't even prove reindeer can't fly.

 

[LarryS]

Alright, maybe "never" or "slightly" were not the best choice of words.

But, are you aware of any experiments to measure the speed of individual photons?

 

[PeroK]

Summary: Does speed of individual photons in a vacuum vary?

Is there experimental evidence that confirms that the speed of individual photons in a vacuum never varies, even slightly?

Thanks in advance.

There is something here on cosmological theories that involve a varying speed of light:

https://cds.cern.ch/record/618057/files/0305457.pdf

 

[PeterDonis]

are you aware of any experiments to measure the speed of individual photons?

Individual photons don't even have a well-defined "speed" in the first place. Strictly speaking, no quantum particles do, but for quantum particles like electrons, with nonzero rest mass, you can reasonably define a Hermitian "velocity operator" that you can physically realize, at least to a reasonable approximation, in a measurement. But no such operator can be defined for a photon, or for any massless particle, so there is no way to even define what you mean by the "speed" of a photon.

 

[Vanadium 50]

First, even revised, the scenario is goofy. Are you telling me that the speed of photons depends on how many there are nearby?

Second, one can measure the speed of gammas from individual nuclear disintegrations, and it's c. Is it known to be c to a zillion decimal places? Nope. Certainly to a percent, perhaps better. You can do something similar with accelerator based experiments.

 

[Delta2]

Individual photons don't even have a well-defined "speed" in the first place. Strictly speaking, no quantum particles do, but for quantum particles like electrons, with nonzero rest mass, you can reasonably define a Hermitian "velocity operator" that you can physically realize, at least to a reasonable approximation, in a measurement. But no such operator can be defined for a photon, or for any massless particle, so there is no way to even define what you mean by the "speed" of a photon.

very interesting. Then what do exactly scientists measure when they measure the speed of light? The average speed of a "collection" of photons?

 

[PeterDonis]

what do exactly scientists measure when they measure the speed of light? The average speed of a "collection" of photons?

No. They measure the speed of a beam of light which the measurement does not even try to resolve into photons. This is not the same as the average speed of the photons in the light because such a thing is not even well-defined to begin with.

 

[Vanadium 50]

While I absolutely agree you can't say "out of the zillion photons", this one was here at time t1 and that one was there at time t2, you certainly can look at pulse speed vs. intensity. This is a hard measurement because of something called slewing, but is at least defined.

Or you could go to single, energetic photons as I discussed.

Or - and bnest of all - you can create a model where this speed varies and looks for an effect. Otherwise this kind of question degenerates quickly: "Ah, but was it ever done on a Thursday!", "Ah, but was it ever done facing Fresno?" "Ah, but was it ever done on Mars?"

 

[Nugatory]

Then what do exactly scientists measure when they measure the speed of light? The average speed of a "collection" of photons?

When we first hear about photons being light particles, our experience with small classical objects like grains of sand leads us to assume that a flash of light is a collection of photons the same way that a beach is a collection of grains of sand or a river is is a collection of water molecules moving along the riverbed. That's not how photons work and not what it means to say that photons are quanta of light.

When we measure the speed of light, we are measuring the speed at which classical electromagnetic waves propagate, and photons don't come into the analysis at all.

 

[Delta2]

When we measure the speed of light, we are measuring the speed at which classical electromagnetic waves propagate, and photons don't come into the analysis at all.

Yes I was afraid this was the case, I see also @PeterDonis updated his answer, originally he had agreed with me.

 

[CoolMint]

Fast forward to 04:20

 

[PeterDonis]

I see also @PeterDonis updated his answer, originally he had agreed with me.

No, I mistyped and hit Post by mistake before I could correct it, so I had to edit.

 

[LarryS]

Question: Say our source of light is from a laser. I have read that the longitudinal position (same direction as the beam) of the photons is completely undefined/unknown. I assumed that was due to the HUP, that the longitudinal momentum of the photons was precisely defined. Is that correct?

 

[PeterDonis]

I have read

Where? Please give a reference.

 

[Delta2]

Where? Please give a reference.

I have also read (in Wikipedia I think) that the photon doesn't have a well defined position operator.

 

[PeterDonis]

I have also read (in Wikipedia I think) that the photon doesn't have a well defined position operator.

That's not what post #15 says. Post #15 says the position is completely uncertain by the HUP because the momentum is precisely defined (i.e., the photon is in a momentum eigenstate). Such a statement only makes sense if the photon does have a well-defined position operator.

 

[PeterDonis]

I have also read (in Wikipedia I think)

You've been here long enough to know that Wikipedia is not a valid reference.

 

[Delta2]

That's not what post #15 says. Post #15 says the position is completely uncertain by the HUP because the momentum is precisely defined (i.e., the photon is in a momentum eigenstate). Such a statement only makes sense if the photon does have a well-defined position operator.

Somehow I read post #15 in a "reverse" way. That the position of the photon is completely unknown, hence by HUP its momentum is precisely defined.

 

[PeterDonis]

Somehow I read post #15 in a "reverse" way. That the position of the photon is completely unknown, hence by HUP its momentum is precisely defined.

The HUP logic works fine either way, since "the position is completely unknown" and "the momentum is precisely defined" are logically equivalent by the HUP. You just failed to realize that "the position is completely unknown" is not the same thing as "the photon doesn't have a well-defined position operator".

 

[Delta2]

You just failed to realize that "the position is completely unknown" is not the same thing as "the photon doesn't have a well-defined position operator".

Hmm if the photon doesn't have a well defined operator, then its position is completely unknown at all times at all cases. Right?

However the reverse doesn't necessarily hold. Right?
By reverse I mean that if for some cases, the position of a photon is completely unknown then this doesn't imply that the photon doesn't have a well defined position operator

 

[PeterDonis]

if the photon doesn't have a well defined operator, then its position is completely unknown at all times at all cases. Right?

Wrong. If the photon doesn't have a well-defined position operator, then the whole concept of "position" doesn't even make sense for it, not even to say that its position is completely unknown.

if for some cases, the position of a photon is completely unknown then this doesn't imply that the photon doesn't have a well defined position operator

For the statement "the position of a photon is completely unknown" to even make sense, the photon must have a well-defined position operator.

In short: without a well-defined position operator, you can't even apply the position-momentum HUP to a quantum system.

 

[vanhees71]

Somehow I read post #15 in a "reverse" way. That the position of the photon is completely unknown, hence by HUP its momentum is precisely defined.

The HUP refers to statistical properties of observables, independent of the state the system is prepared in. In the case of photons there's no position observable, and thus there cannot be uncertainty relations for it.

I'd say the most convincing evidence for photons being massless is the upper limit of the mass of the em. field, given in the particle data booklet. There you can also find the experimental papers, this limit is extracted from:

https://pdg.lbl.gov/2022/web/viewer.html?file=../listings/rpp2022-list-photon.pdf

 

[Delta2]

In the case of photons there's no position observable

Yes isn't this statement equivalent to the statement "the photon doesn't have well defined position operator"?

I'd say the most convincing evidence for photons being massless is the upper limit of the mass of the em. field, given in the particle data booklet. There you can also find the experimental papers, this limit is extracted from:

Sorry I lost you, what's got to do here the fact that the photon is massless?

 

[weirdoguy]

what's got to do here the fact that the photon is massless?

For massive particles there is no problem with defining position operator.

 

[Dale]

one can measure the speed of gammas from individual nuclear disintegrations, and it's c

How can you do that? Even if it is a gamma, once you detect it to start your speed measurement, it is absorbed and cannot be detected to stop the speed measurement. Is there some non-destructive way to measure a single photon?

 

[Vanadium 50]

Didn't I say it was hard?

There a a few ways to handle it. Perhaps the easiest is to look at decays involving multiple particles, including recoil. Then t0 happens when the nucleus decays, and (at least) one particle is detected, and t1 happens later when the photon is detected a distance x away. Then c is x/(t1-t0).

⁶⁰Co might be a good candidate. It decays vie eγγ most of the time, with an angular correlation between the two gammas. You could vary the positions of the two detectors and fit for c, using distances, angles and energies as inputs: This has the advantage of relative insensitivity to background processes.

 

[Dale]

including recoil

Ah, of course. Then it does matter that it is a gamma so that you get a good amount of recoil.

 

[LarryS]

Where? Please give a reference.

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.

 

[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.