Photon states should not evolve?

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SUMMARY

The discussion centers on the concept that photons do not evolve in their quantum mechanical state due to the absence of proper time in their reference frame. Participants argue that since photons travel at the speed of light, they lack a valid reference frame, leading to the conclusion that their states remain unchanged. The conversation also touches on the implications of General Relativity and the semantics surrounding the "experience" of time by photons. Key points include the distinction between different reference frames and the role of quantum electrodynamics (QED) in understanding photon behavior.

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  • #61
PeterDonis said:
Carrock said:
Saying that the concept of "time" does not apply to them' also often leads to incorrect inferences
How so?
Hard to find a reference, but I've seen claims that photons experience change in some way, but time does not pass for them. It's usually harder to understand when time is implicitly avoided but a vaguely similar concept is used.

PeterDonis said:
Carrock said:
I was simply trying to indicate that it's never necessary to have photons' energy, momentum etc change.
It is if you want to analyze actual experiments, since in actual experiments the energy relative to the source is often different from the energy relative to the detector.
Often isn't necessarily. There's no problem e.g. in moving a clock downwards to compensate for gravitational redshift from a lower clock to compare them.
PeterDonis said:
Carrock said:
Since an object can always be chosen such that the absorbed photon is in the same state as when it was emitted

No, this is not correct. For example, if you are here on Earth, and you are looking at photons coming from a distant galaxy, you don't get to choose the state of motion of the source or the detector.

Without doing the maths, I'd think a suitable particle at CERN could be created as a detector with an appropriate state of motion.

PeterDonis said:
Also, your use of the word "state" is incorrect here. The energy of the photon is an inner product of the photon's 4-momentum and the object's 4-velocity; that's not just a matter of the photon's state. (And the photon's state is not just its 4-momentum; it also includes polarization, which doesn't affect the photon's energy relative to an object, but does affect other measurements.)
I'm rather rusty on 4-momentum and 4-velocity as you may have guessed...
However I don't see my imprecision/error negating anything else in this post. From your quote, an object with a suitable 4-velocity is all that's required.

PeterDonis said:
Carrock said:
A particle's state 'really' evolving during flight, like a neutrino, is generally regarded as proof mass is associated with it.
Do you have a reference for this?
From https://www.nobelprize.org/nobel_prizes/physics/laureates/2015/advanced-physicsprize2015.pdf
The discovery that neutrinos can convert from one flavour to another and therefore have non-
zero masses is a major milestone for elementary particle physics.
Do you have a reference refuting
Carrock said:
A particle's state 'really' evolving during flight, like a neutrino, is generally regarded as proof mass is associated with it.
More precisely, refuting "...an object can always be chosen such that the absorbed photon (and absorber I suppose) is in the same state as when it was emitted."
 
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  • #62
PeterDonis said:
One of those mass eigenstates could still have zero mass (i.e., be massless), but the others could not.

Ahhh... Yes. I think I see. So the solar neutrino problem of 25 years ago was about missing electron neutrinos. For them to be missing because of a flavor change they would have to have mass, be traveling at speeds less than ##c##, and not have a zero arc length (or has been described loosely as not being able to experience time). Do I at least have that part right?
 
  • #63
Carrock said:
Hard to find a reference, but I've seen claims that photons experience change in some way, but time does not pass for them.

Um, have you read this thread? We've already discussed such claims and why they're not valid.

Carrock said:
From https://www.nobelprize.org/nobel_prizes/physics/laureates/2015/advanced-physicsprize2015.pdf

I didn't ask for a reference to show that neutrinos oscillate. I asked for a reference supporting your claim that it is the neutrinos "really evolving" that allows them to oscillate. I already explained in post #58, in response to @Mister T , why that claim is not correct (because the inference from neutrino oscillations to them having to have mass does not depend on the worldlines of massless particles having zero arc length). The reference you give is consistent with what I said in post #58.

Carrock said:
refuting "...an object can always be chosen such that the absorbed photon (and absorber I suppose) is in the same state as when it was emitted."

I didn't say this statement was wrong. I just said it wasn't useful, and I explained why. Did you read my post?
 
  • #64
Mister T said:
Do I at least have that part right?

Not quite. Neutrino oscillations imply that some neutrinos must have mass. They do not require that all neutrinos have mass. I explained why briefly in post #58. For a more detailed explanation, the PDF that @Carrock linked to gives a good discussion in the section "Theory of neutrino oscillations", starting at the bottom of p. 13.
 
  • #65
Carrock said:
Often isn't necessarily.

You claimed that it is never necessary to have the photon's energy/momentum change. One counterexample, which I gave (and it's not the case of gravitational redshift, which you mention), is sufficient to refute such a strong claim.

Before making even a weaker claim, I strongly suggest that you spend some time reviewing the kinds of actual experiments in which photon energies are relevant, so you can see how often actual experimenters have to use models in which photon energies and momenta are allowed to change in order to explain their results.

Carrock said:
Without doing the maths, I'd think a suitable particle at CERN could be created as a detector with an appropriate state of motion.

See what I just said above.

Carrock said:
From your quote, an object with a suitable 4-velocity is all that's required.

Sure, but you can't always guarantee that such an object exists, or make one if it doesn't. To do so, you would have to already know the photon's 4-momentum, and in many, if not most, cases of interest, you don't. That's why you have to make measurements on it--to find out what its 4-momentum is.

Furthermore, even if you could construct such an object, you might not care what its measurements are telling you.
 
  • #66
PeterDonis said:
No, the arc length along a null worldline in relativity is zero. But "proper time" is an incorrect term to use to describe that arc length.
Why? It's even written down as ##d \tau^2 = \ ... \ ##. How can it not be the proper time?
PeterDonis said:
to properly do a comparison here, you should be looking, not at non-relativistic QM, but at quantum field theory. Which is built on relativity.
Is there any difference as to how time is treated in Quantum Field Theory and non-relativistic Quantum Mechanics?
 
  • #67
Tio Barnabe said:
It's even written down as dτ2=

So what? ##d\tau^2## is a symbol, not physics. I could just as easily write the same timelike interval as ##ds^2## (in fact, many textbooks do exactly that); would that make it a proper length?

Tio Barnabe said:
Is there any difference as to how time is treated in Quantum Field Theory and non-relativistic Quantum Mechanics?

Most definitely.
 
  • #68
PeterDonis said:
So what? ##\tau^2## is a symbol, not physics. I could just as easily write the same timelike interval as ##ds^2## (in fact, many textbooks do exactly that); would that make it a proper length?
It's just a symbol, yes. But it's just a symbol with a well defined meaning. As long as we define ##d \tau^2## as the proper time squared, so it is the proper time squared.
 
  • #69
Tio Barnabe said:
it's just a symbol with a well defined meaning.

Yes, but not the meaning you think it has. See below.

Tio Barnabe said:
As long as we define ##d \tau^2## as the proper time squared, so it is the proper time squared.

"Proper time squared" is not the definition of ##d\tau^2##. The definition of ##d\tau^2## is "the square of the infinitesimal spacetime interval". There is nothing in the definition that tells you the physical meaning of that spacetime interval. For that you have to look at its actual value. "Proper time" is only the physical meaning of the interval if the interval is timelike. That is true regardless of what symbol you use.
 
  • #70
Hmmm, good to know. I thought ##d \tau^2## could always be regarded as a proper time, because in the reference frame of the "particle" there's no spatial displacement. So ##d \tau^2 = dt^2## in the particle reference frame.

PeterDonis said:
"Proper time" is only the physical meaning of the interval if the interval is timelike
because if it is space-like the "proper time" would be negative?
 
  • #71
Tio Barnabe said:
I thought ##d \tau^2## could always be regarded as a proper time, because in the reference frame of the "particle" there's no spatial displacement.

This will be true if the particle has nonzero rest mass and ##d\tau^2## is an interval along its worldline. But the definition of ##d\tau^2## makes no such assumptions. What you're describing is just one particular special case.

Tio Barnabe said:
because if it is space-like the "proper time" would be negative?

The square root of ##d\tau^2##, assuming you were using a timelike signature convention, would not be negative for a spacelike interval, it would be imaginary.

However, this is still focusing on superficial features instead of the fundamental definition. The fundamental definition just says that timelike, null, and spacelike intervals are physically different. How that difference is modeled in the math depends on your choice of signature convention--note that I specified a timelike signature convention above, which means that timelike intervals have positive ##d\tau^2## and spacelike intervals have negative ##d\tau^2##. Conversely, the spacelike signature convention means that timelike intervals have negative ##d\tau^2## (the symbol ##ds^2## is normally used in this case--but as I said before, it's often used in the timelike signature case as well) and spacelike intervals have positive ##d\tau^2##. Either way, null intervals have zero ##d\tau^2##.
 
  • #72
PeterDonis said:
This will be true if the particle has nonzero rest mass and ##\tau^2## is an interval along its worldline.
This is going against what is said in Weinberg's book. I remember a part of the book that says for light the LHS of the equation ##d \tau^2 = \ ... \ ## is equal to zero.
 
  • #73
Tio Barnabe said:
This is going against what is said in Weinberg's book.

No, it isn't. I said "if the particle has nonzero rest mass". For something with zero rest mass, like light, the interval along the worldline is zero, yes, but the term "proper time" is not a proper description of it, for that very reason--because the worldline is null, not timelike.
 
  • #74
PeterDonis said:
No, it isn't. I said "if the particle has nonzero rest mass". For something with zero rest mass, like light, the interval along the worldline is zero, yes, but the term "proper time" is not a proper description of it, for that very reason--because the worldline is null, not timelike.
So, also, is the statement that times stop at the speed of light incorrect?
 
  • #75
Tio Barnabe said:
is the statement that times stop at the speed of light incorrect?

I would say it's "not even wrong", because it assumes that the concept of "proper time" or "elapsed time" is even meaningful for light, or anything that moves at the speed of light.
 
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  • #76
Tio Barnabe said:
Hmmm, good to know. I thought ##d \tau^2## could always be regarded as a proper time, because in the reference frame of the "particle" there's no spatial displacement. So ##d \tau^2 = dt^2## in the particle reference frame.?
If ##d\tau^2=0## then there is no particle reference frame.
 
  • #77
Carrock said:
Since an object can always be chosen such that the absorbed photon is in the same state as when it was emitted, and emission and absorption are the only times a photon's state can be observed, there is no requirement for a photon's state to evolve over time.
This is wrong. The wave function representing the photons state evolves over time according to the Schrödinger equation. At least in standard QFT it is required.
 
  • #78
Dale said:
The wave function representing the photons state evolves over time according to the Schrödinger equation. At least in standard QFT it is required.

Just to clarify: the Schrödinger equation is non-relativistic, and you can't really apply it to photons/light. (You can apply it to atoms or particles interacting with an electromagnetic field, but the field itself will be modeled as an external potential.) A better way of expressing what I think you're trying to express here would be that, in QFT, you can't assume that, if the photon field (electromagnetic field) is in a given state at a particular event, it will be in the same state at any other event that is null separated from that event.
 
  • #79
Dale said:
This is wrong.

To clarify a bit further: I was interpreting @Carrock's use of the term "photon" classically, i.e., it should really be "very short pulse of EM radiation", which we can model as a "particle" with a well-defined 4-momentum vector which is null. Given a known null 4-momentum vector, and some chosen value for energy, we can always find a local inertial frame at any event on the null worldline of the photon in which the inner product of that 4-momentum vector and the timelike basis vector of the frame yields the chosen value. If we imagine a series of such local inertial frames all along the null worldline, we could say that, in these frames, the photon's energy is "unchanged" as it travels.

The problem, as I pointed out in response, is that in order to know which frames these are, you have to already know the photon's 4-momentum, and in most cases of practical interest, you don't. Also, even if you do, the frame given by the above prescription might not be the one you actually care about.
 
  • #80
PeterDonis said:
Carrock said:
Hard to find a reference, but I've seen claims that photons experience change in some way, but time does not pass for them.
Um, have you read this thread? We've already discussed such claims and why they're not valid.
Um, have you read the post you quoted from? ( https://www.physicsforums.com/threads/photon-states-should-not-evolve.925675/page-4#post-5847397 )

Carrock: Saying that 'the concept of "time" does not apply to them' also often leads to incorrect inferences.
PeterDonis: How So?
Carrock: Hard to find a reference, but I've seen claims that photons experience change in some way, but time does not pass for them.

So you request an example of an incorrect inference; when I provide it you infer I haven't read this thread.
PeterDonis said:
I didn't ask for a reference to show that neutrinos oscillate. I asked for a reference supporting your claim that it is the neutrinos "really evolving" that allows them to oscillate.
I never claimed that it is 'the neutrinos "really evolving" that allows them to oscillate.' Still, the time I wasted searching for a good reference has saved you the necessity of looking for yourself.

PeterDonis said:
Carrock said:
refuting "...an object can always be chosen such that the absorbed photon (and absorber I suppose) is in the same state as when it was emitted."
I didn't say this statement was wrong. I just said it wasn't useful, and I explained why. Did you read my post?
I read your posts but don't recall which is the relevant one. I'm not going to trawl through this thread yet again and perhaps pick a post you didn't mean. Perhaps you should read my posts to avoid changing their meaning with misleading editing and inaccurate precis.

I see in post #79 most of your objections seem rather forgotten as well as the reason I introduced this rather old concept in the first place. If you'd posted something like this earlier I'd have clarified this concept; as it is, when I spend this amount of time on a post like this, I know it's time to quit.
 
  • #81
PeterDonis said:
we could say that, in these frames, the photon's energy is "unchanged" as it travels.
Sure, but the state is more than just the energy. In classical EM the fields evolve over time per Maxwell's equations. In QED the wave function evolves over time per Schrödinger's equation. Either way, the state evolves over time.
 
  • #82
PeterDonis said:
Just to clarify: the Schrödinger equation is non-relativistic
Hmm, I thought that the Schrödinger equation, in a suitably generalized form, still determines the evolution of the wavefunction. Is it called another name?
 
  • #83
Dale said:
In QED the wave function evolves over time per Schrödinger's equation.

Not really. See below.

Dale said:
Either way, the state evolves over time.

Agreed, although there are caveats when we are talking about QFT; a better way to put it in QFT would be the way I put it before, that if we know the state of the quantum EM field at a particular event, we cannot just assume that the quantum EM field will be in the same state at any other event that is null separated from the first one.

Dale said:
I thought that the Schrödinger equation, in a suitably generalized form, still determines the evolution of the wavefunction.

In QFT, there isn't a "wave function" in the sense that there is in non-relativistic QM. It's a very different framework conceptually. We should probably start a separate thread in the quantum forum if this needs to be discussed further.
 
  • #84
Carrock said:
you request an example of an incorrect inference; when I provide it you infer I haven't read this thread.

I asked for an example of "the concept of time does not apply to photons" leading to an incorrect inference. You gave me an example of "time does not pass for photons" leading to an incorrect inference. But "time does not pass for photons" implies that the concept of time does apply to photons. So your example was not responsive to my question.

Carrock said:
I never claimed that it is 'the neutrinos "really evolving" that allows them to oscillate.'

I suppose that's true; here's what you said:

Carrock said:
A particle's state 'really' evolving during flight, like a neutrino, is generally regarded as proof mass is associated with it.

Which is still wrong, as I have already explained in other posts.
 
  • #85
Carrock said:
I read your posts but don't recall which is the relevant one.

It was the exact same one you were already responding to in the quote you gave (your quote starting with "refuting..."), which is from your post #61. In other words, in your post #61, you were responding to the relevant post of mine, which gave the explanation you are looking for.
 
  • #86
phinds said:
That is not correct. "c" is taken as the universal speed limit and it light were found to have mass, "c" would remain the same but would no longer be the speed of light. The point, then, is that light does NOT travel at "c" in various media, so your statement "vacuum or not" is incorrect.

You're right, but think you should read why. By the same interaction light may go faster than light...I mean c.

"If light were found to have mass..." has nothing to do with it.
 
  • #87
nitsuj said:
You're right, but think you should read why. By the same interaction light may go faster than light...I mean c.

"If light were found to have mass..." has nothing to do with it.
I don't understand any of this. Can you expound on it please?
 
  • #88
phinds said:
I don't understand any of this. Can you expound on it please?
sure, for light going through something where it is not absorbed and emitted it easiest to think of it as a wave as opposed to a photon. the EM of the light wave when it begins to pass through the medium, such as glass induces a wave in the glass, the sum of these waves is what we see leave the glass, most often this amounts to the original light wave as slowing. This induced wave and original wave can also amount to the the phase velocity bring greater than c.

You should read up on it it's pretty interesting...
 
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  • #89
nitsuj said:
sure, for light going through something where it is not absorbed and emitted it easiest to think of it as a wave as opposed to a photon. the EM of the light wave when it begins to pass through the medium, such as glass induces a wave in the glass, the sum of these waves is what we see leave the glass, most often this amounts to the original light wave as slowing. This induced wave and original wave can also amount to the the phase velocity bring greater than c.

You should read up on it it's pretty interesting...
Ah. Phase velocity. You're right, I should read up on it, since I keep hearing about it here on PF.
 

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