Can the wavelength of a single photon be measured?

[PeroK]

Mentor's note: A thread derailment based on a misunderstanding of elementary quantum mechanics has been removed from this thread.

Although the main thread based on a misunderstanding of elementary quantum electrodynamics continues!

 

[hutchphd]

There are plenty of states of the quantum electromagnetic field that are not eigenstates of any "wavelength" observable, which means they have no definite wavelengths.

There are plenty of states of the classical Electromagnetic Field that we would not call "light" as well! We would not usually (at least colloquially) associate either the term "photon" or "light" with these states. The far field states are what we usually call "light" and I think that they usually do have such an observable in the quantum case. Or am I misunderstanding here??.

 

[PeterDonis]

There are plenty of states of the classical Electromagnetic Field that we would not call "light" as well!

Depending on how your intuition interprets "light", yes, this could be true.

The far field states are what we usually call "light"

I'm not sure what you mean by "far field states". The states that are closest to our intuitive idea of a classical "electromagnetic wave" are the coherent states. But those are certainly not "photon" states, as I've already pointed out, and the OP was asking about "single photon" states.

that they usually do have such an observable in the quantum case.

The issue is not "having an observable". You can apply an observable to any state. The issue is whether the states the OP is thinking of as "single photon" states are eigenstates of a "wavelength" observable. Which requires, at least, specifying which states, and which observable.

 

[hutchphd]

I'm not sure what you mean by "far field states". The states that are closest to our intuitive idea of a classical "electromagnetic wave" are the coherent states. But those are certainly not "photon" states, as I've already pointed out, and the OP was asking about "single photon" states.

Is there not an energy eigenstate in free space corresponding to one quantum of momentum p ?

 

[PeterDonis]

Is there not an energy eigenstate in free space corresponding to one quantum of momentum p ?

This would be a "photon state" (Fock state) with photon number 1 and an appropriate energy for the given momentum. But such a state is not anything like a classical electromagnetic wave or a classical electromagnetic field.

 

[Nugatory]

Although the main thread based on a misunderstanding of elementary quantum electrodynamics continues!

That phrase might be an oxymoron.

 

[hutchphd]

But such a state is not anything like a classical electromagnetic wave or a classical electromagnetic field.

Will it have a wavevector?

 

[PeterDonis]

Will it have a wavevector?

That would be an observable, and I'm not sure which observable it would be.

 

[Dale]

The issue is not "having an observable". You can apply an observable to any state. The issue is whether the states the OP is thinking of as "single photon" states are eigenstates of a "wavelength" observable.

Hmm, I didn’t read the OP that way. I understood he just wanted to know how you would measure the wavelength of a single photon (Fock state). I did not unterstand him to be asking if the wavelength was a definite value (ie I didn’t think he needed it to be an eigenstate)

 

[hutchphd]

This would be a "photon state" (Fock state) with photon number 1 and an appropriate energy for the given momentum. But such a state is not anything like a classical electromagnetic wave or a classical electromagnetic field.

If it has momentum p will it not also have a corresponding wavelength??

 

[PeterDonis]

If it has momentum p will it not also have a corresponding wavelength??

First, it's not clear what state you are describing when you say "has momentum p", since momentum is a 3-vector and a state can have a definite energy (which Fock states do) without having a definite momentum 3-vector. Furthermore, eigenstates of 3-momentum in ordinary non-relativistic QM are already non-normalizable, and ordinary non-relativistic QM can't be used for photons, or more generally for the quantum EM field, anyway.

In short, there are a lot of complexities lurking underneath the simple-sounding ordinary language that you and other posters are using in this thread, and those complexities mean that a lot of simple-sounding questions don't have well-defined answers, and aren't really even well-defined questions. What is needed to make the questions well-defined is an explicit specification of the math: what states and what observables are we talking about?

 

[hutchphd]

First, it's not clear what state you are describing when you say "has momentum p", since momentum is a 3-vector and a state can have a definite energy (which Fock states do) without having a definite momentum 3-vector. Furthermore, eigenstates of 3-momentum in ordinary non-relativistic QM are already non-normalizable, and ordinary non-relativistic QM can't be used for photons, or more generally for the quantum EM field, anyway.
In short, there are a lot of complexities lurking underneath the simple-sounding ordinary language that you and other posters are using in this thread, and those complexities mean that a lot of simple-sounding questions don't have well-defined answers, and aren't really even well-defined questions. What is needed to make the questions well-defined is an explicit specification of the math: what states and what observables are we talking about?

So put a vector sign on p and normalize periodically in a very large box. How is the question ill-defined? If it has a complicated answer, so be it.

 

[PeterDonis]

So put a vector sign on p and normalize periodically in a very large box.

For ordinary non-relativistic QM, yes, this addresses one of the complexities. Did you see the part where I said that ordinary non-relativistic QM can't be used for photons, or more generally for the quantum EM field?

 

[hutchphd]

So is your answer to the OP please don't ask the question? You can count quanta on a very modest photodetecting camera...one at a time ...are these not individual photons?
What then is the appropriate question?

 

[Herman Trivilino]

What then is the appropriate question?

Something about the phenomenon rather than an interpretation.

 

[hutchphd]

Something about the phenomenon rather than an interpretation.

The original question:

Is there any way to accurately measure the wavelength of a single,
individual photon?

This is sort of a binary question. If the answer is no: then why? If the answer is yes: then how and how accurately? But answer the question!

 

[PeterDonis]

? You can count quanta on a very modest photodetecting camera...one at a time ...are these not individual photons?

Go read my post #59.

This is sort of a binary question. If the answer is no: then why? If the answer is yes: then how and how accurately? But answer the question!

If you count quanta on a photodetecting camera, are you measuring their wavelength? No. Does that answer the question?

If not, then whoever is asking the question needs to explain how, experimentally, they are going to make "photons" and detect them, before we can answer questions about whether we can measure their wavelength or anything else.

 

[hutchphd]

If you count quanta on a photodetecting camera, are you measuring their wavelength? No. Does that answer the question?

If you first send them to a diffraction grating then yes you are (as discussed earlier).

 

[PeterDonis]

If you first send them to a diffraction grating then yes you are (as discussed earlier).

No, you're not, as discussed earlier, because the individual counts on the detector are not all in the same place. You only build up the diffraction pattern over many counts, but that fact means you can't interpret the spacings of the pattern as telling you "the wavelength of individual photons". (And that is even more reinforced by the fact that you can use the grating to make a diffraction pattern with light sources that don't even emit Fock states in the first place, so the state emitted by the source has no valid interpretation as a "single photon" state. Which is precisely the caution I gave in my post #59, which I asked you to read.)

 

[Cthugha]

This is self-contradictory. A coherent state, which is basically what you're describing in the second sentence, is not a "single-photon state", even if its expectation value of photon number is 1, and is not a "single excitation" of the field. The only kind of state that is a "single excitation" or "single-photon state" is a Fock state with $n = 1$.

That is unfortunately wrong. Most probably you misunderstand my scenario. So let me try to make it more clear. I am not talking about decomposing the Fock state into separable states in the energy basis. You rather get something similar to entangled states where you get 1 photon at a certain energy and 0 in all others and sum over all of these possible states. What I described is a single photon state then. There is no need for a single photon state to be spectrally pure. This is also very different from a coherent state, which has a photon number variance equal to the mean photon number, while a Fock state has zero variance in the chosen basis and this is the only criterium there is. A state that has a guaranteed photon number of 1 is never a coherent state in that basis. There are good books on quantum optics that discuss this in detail. Vogel/Welsch is good for theorists, Mandel/Wolf, if you like old-school books or Gerry/Knight for something in between.
As an example, consider Jaynes-Cummings and the simple example of a single atom inside a cavity. You can find an ortogonal set of eigenmodes of the cavity. However, they have a finite Q-factor and thus a finite linewidth. A single atom in the excited state in resonance with one of the cavity modes may emit a single photon into the cavity mode. After the emission process, one photon will be inside the cavity. We know that the occupation number of this cavity mode is 1. However, due to the finite line width of the transition, it is not pure spectrally. If you perform a spectral decomposition, you will find that the probability amplitudes for all the infinitely narrow spectral modes are non-zero and the total probability to find the photon within the spectral width of the transition is one. Still, this is a Fock state, although the energy is not well-defined as the probability to find a different photon number than 1 when summing over the whole line width is exactly 0.

The concepts of "Fock state" and "coherent state" are often mixed up, yes. But the former is both a "photon number state" and a "photon mode of well-defined energy", while the latter is neither.

If you think that I talked about the difference between Fock and coherent states than you are exactly a victim of the misconceptions I talked about. I was not talking about coherent states at all.

Your statement is incorrect(*). Of course you can build a Fock state on energy eigenstates, but you do not have to. You just want some orthonormal basis of states of the underlying single-particle Hilbert space and there are much more options than energy eigenstates (although this is of course a very common choice). According to your description any real atom would be unable to emit single photons because the transition necessarily has a finite spectral width, which necessarily means that there is a finite range of energies involved. Spectral purity is not a part of the definition of single photons. A well-defined exact photon number is sufficient.

(*) Unless you are purist in the sense that you consider only photons in vacuum that do not interact with anything as faithful representations of what is meant by the term photon - there it makes perfect sense to restrict the description to energy eigenstates.

This an example of the problem I have with this interface between classical and quantum. The word "instantaneously" is just one of the words which are used 'colloquially" but when are they literally right? If I used the term "instantaneously" when talking about travel or communications, you guys would all be down on me like a tin of bricks.
I think the whole problem is that there is too much common use of words in the two fields. To describe a "small region of perturbations" (eg a photon) in terms of wavelength, you would need to specify the spectrum unless the wave were continuous. etc. etc.. It's as if there is a very inadequate analogy being used to reconcile the two sides of our world

Sure, you are perfectly right. It is important that there is no transfer of information involved. Let me put it this way: As an experimentalist, it is perfectly sufficient to me that the math works out. You consider the initial state and the final state, add up all the probability amplitudes for indistinguishable ways to get from the former to the latter and take the modulus squared (and potentially consider any distinguishable ways as well), respect the operator algebra and things work out fine. There is little need to consider what actually happens in between. However, many people feel the need for some ontology here and this is where hell breaks loose. You get the typical discussion about interpretations which I tend to avoid. If you want some real physical field to travel from a to b and no superdeterminism, you need to introduce some non-local update mechanism as soon as a photon gets absorbed. If you can get along without real fields, you do not have to do that. These discussions are typically endless and basically everybody may pick an interpretation that suits his needs. Essentially, I think it is not a good idea to consider photons as important in the case of small perturbations. A coherent state may have smaller mean intensity than a single photon Fock state. It is important to consider photons when the discreteness matters, e.g. in antibunching experiments or other correlation experiments. For classical linear experiments (double slit, linear absorption, Michelson interferometer and so on) a coherent state and a Fock state of the same mean intensity, spectral composition, spatial and temporal coherence will just show exactly the same results as such experiments cannot be sensitive to photon statistics.

 

[hutchphd]

No, you're not, as discussed earlier, because the individual counts on the detector are not all in the same place. You only build up the diffraction pattern over many counts, but that fact means you can't interpret the spacings of the pattern as telling you "the wavelength of individual photons". (And that is even more reinforced by the fact that you can use the grating to make a diffraction pattern with light sources that don't even emit Fock states in the first place, so the state emitted by the source has no valid interpretation as a "single photon" state. Which is precisely the caution I gave in my post #59, which I asked you to read.)

Is my assertion in post #41 incorrect? If so, please explain why.

What if light from Sirius is put thru a bandpass (interference) filter and then a single event is recorded by a photomultiplier cascade. Is this not a wavelength measurement of single photon?

 

[Jeff Root]

The kind of observation hutchphd just described is familiar to me.
Apache Point Observatory sends large numbers of short bursts of laser
light at the Apollo and Lunokhod retroreflectors on the Moon, and
detects the returned light through a very narrow-band filter. Each
detection consists typically of only one photon per pulse. Previous
observations at other observatories were only able to detect one
photon per hundred pulses, yet they were able to discriminate the
signal, due in part to its specific wavelength. The exact wavelength
changes continuously as the distance and relative speed between
the observatory and the retroreflector on the Moon changes.

-- Jeff, in Minneapolis

 

[Jeff Root]

Is there any way to accurately measure the wavelength of a single,
individual photon? How precise could such a measurement be?

First, you need to say what you mean by "a single, individual
photon". Since there are lots of misleading descriptions in
ordinary language of what "a single, individual photon" is, the
best way to answer that is to describe precisely what kind of
source you are imagining that will produce a "single, individual
photon". Can you do that?

No, because I have no idea why the source would matter, unless,
as I suggested as a possibility, all that can be done is confirm
the wavelength of a known source.

I have read that a beam of visible light can be reduced in
intensity by filters until there is *probably* at most only
one photon of visible light in the darkened chamber at a time.

That's what I had in mind. But I don't see that the source
would make make any difference.

It should be obvious that I'm not talking about eliminating
*all* other photons, since low-energy photons are constantly
emitted by all matter as thermal radiation. For simplicity,
I would not be trying to detect those. The detector would
have at least a minimum cutoff so that only photons of some
given energy or greater would be detected. I expect it would
be limited to a moderately-narrow band such as visible light.

I have also read that monochromatic sources exist, and that
some monochromatic sources can also be coherent. If such a
source is useful or required for measuring the wavelength of
a single photon, then perhaps you can explain why, but I have
no reason to think it would make any difference.

-- Jeff, in Minneapolis

 

[Jeff Root]

A photon cannot change over time. Time does not exist in the
photon's reference frame. That means the wave nature of the
photon cannot be caused by the photon changing, such as pulsing
or moving up and down. Instead, the wave must be a fixed form
which moves as a whole, unchanging unit.

None of this is correct.

I'm sure at least some of it must be correct. I've read it
enough times from multiple different physics professors and
people very knowledgeable about physics.

There is no such thing as "a photon's reference frame".

I very rarely complain about anyone arguing semantics.
Agreement on the meaning of terms and expressions is
crucial, but I think I have complained about this sort of
thing only once before in the last ten or twenty years.
I complain again now.

I agree that there is no such thing as "a photon's reference
frame". It is another way of saying that time does not exist
in a photon's reference frame. Nothing does. In a photon's
reference frame, the photon comes into existence and goes out
of existence simultaneousy and instantaneously. So there is
no time for it to do anything, or "experience" anything, or be
affected by anything-- in its own reference frame. Others, real
observers, not in a photon's reference frame, can see that
things sometimes happen to photons while they are in flight--
photons can be lensed, for example-- but the photons can't
"know" that because they don't even exist in their own
reference frame.

I think your apparent disagreement on this point is just a
semantic disagreement. But if it is something more than that,
I'll be happy to read your explanation.

You have a mistaken mental model of what a photon is.

Probably. But if so, I don't have a better one to replace it.
Until I get something better, I think I'll have to continue
building on the mental model I have.

... since we know the photon must have a wavelength ...

This is not correct either. Again, you have a mistaken mental
model of what a "photon" is. There are plenty of states of
the quantum electromagnetic field that are not eigenstates
of any "wavelength" observable, which means they have no
definite wavelengths.

As you quoted me in your next post, I am convinced that an
individual photon must have a wavelength by the fact that
one-by-one, individual photons can paint an interference
pattern that represents a specific wavelength.

I do not know what a "state" is in this context. I do not
know what a "quantum electromagnetic field" is. I have no
idea what an "eigenstate" is. I have no idea why all that
jargon implies that photons have no definite wavelengths.

I don't even know what you mean by "definite" here.

My interpretation, based on everything I do know, is that a
photon has a wavelength that can be measured with a maximum
precision whose limit is described by Heisenberg uncertainty,
but which, in a practical measurement, might be measured to
have *any* value in the range of the detector. However, that
value is much more likely to be close to the wavelength of
the monochromatic light source than far from it. Most of the
individual photons will be measured to have wavelengths close
to the overall wavelength of the source, while only a few
will be measured to have wavelengths that are very different.

I believe this was suggested by another poster early in this
thread, but I didn't understand what he or she meant until
just this moment. Individual photons in an extremely sparse
beam from a monochromatic source can be diffracted from a
grating onto an array of calorimeters. The angle at which
they are diffracted gives one measure of the wavelength, and
the energy they impart to the detectors gives a second,
independent measure of the wavelength. Both measurements
will be limited by Heisenberg uncertainty and by the physical
limits of the detectors, but they should always agree to some
extent. If they disagree often, then I might have reason to
doubt that individual photons have definite wavelengths.

-- Jeff, in Minneapolis

 

[PeterDonis]

I am not talking about decomposing the Fock state into separable states in the energy basis.

I guess I don't understand what state you're talking about. It would be helpful if you would use math instead of ordinary language to describe it. (This thread in general could use more math and less ordinary language.)

 

[PeterDonis]

I do not know what a "state" is in this context.

Then you don't have the proper background to even be talking about "photons" at all, let alone whether their wavelengths can be measured and under what circumstances. As I have already said, your reasoning is based on misunderstandings. You need to fix those misunderstandings before you can even begin to try to answer your original question. The best way to fix those misunderstandings is to work through a good textbook on quantum electrodynamics.

Another example of misunderstanding:

Each
detection consists typically of only one photon per pulse.

Nope. It has already been explained that laser light does not consist of "single photon" states. The fact that many sources describe laser light pulses of low intensity as "single photon" pulses is an indication that those sources are using sloppy ordinary language.

 

[PeterDonis]

I agree that there is no such thing as "a photon's reference
frame".

You think you do, but you don't. I know you don't because everything else you say in that same paragraph is wrong, yet you think you are agreeing with me.

So in addition to working through a good textbook on quantum electrodynamics, you need to work through a good textbook on special relativity. You might also want to read the PF FAQ article on why there is no "rest frame of a photon".

 

[PeterDonis]

Is this not a wavelength measurement of single photon?

Not if the light coming from the source is not in a state that can be described as a "single photon" state, as I've already explained. Which it certainly won't be if the light is from Sirius. Your post #41 doesn't specify a single light source, but you mention a laser, which, as I've already explained, also does not produce "single photon" states.

 

[PeterDonis]

The OP of this thread is based on misunderstandings, which do not appear to be corrected by the explanations already given. Thread closed.