Can the wavelength of a single photon be measured?

In summary, there is no way to accurately measure the wavelength of a single photon, but it can be measured indirectly.
  • #36
The argument in the last several posts suggests a new question
which I think should be in a new thread. I will title it:

"Can anything be measured with a single measurement?"

-- Jeff, in Minneapolis
 
Science news on Phys.org
  • #37
sophiecentaur said:
This is making assumption that a photon has more parameters to describe it than just its energy. You can't be seriously suggesting that the photons an Astronomer measures can be selected in terms of their source (in some way rejecting other inputs).

This response puzzles me a bit, so I am not sure whether we talk about the same thing. The title of this topic was about single photons, so I assume we are talking about light fields with a photon number of 1. We do not necessarily know the energy: the wavelength uncertainty of a single photon taken from a femtosecond light pulse may as broad as tens of nm, while the wavelength uncertainty of a single photon originating from a narrow atomic transition may be as narrow as a few pm. I just wanted to emphasize that these photons do not have an undefined extent.
Historically speaking, a theoretical treatment using the Fourier decomposition of the light field into states of well defined energies of course yields modes of precise energy and undefined extent, but although unfortunately these modes have sometimes been termed photon modes, they are not at all related to any physically existing thing that could be considered as a single photon.

With respect to the astronomy question: I am not sure what you mean exactly. Of course there is no source-o-meter, which allows photons from Sirius to pass, but not those from the sun. However, the light beams still carry information about their source in terms of their coherence. Early intensity interferometers used this to determine the angular diameter of stars. However, this is getting a bit off-topic, I guess.
 
  • #38
The more accurately you know a particles position the less accurately you know it's speed. And the more accurately you know a particles speed, the less accurately you know it's position.
 
  • Like
Likes sophiecentaur
  • #39
slappmunkey said:
The more accurately you know a particles position the less accurately you know it's speed. And the more accurately you know a particles speed, the less accurately you know it's position.

Is that the apostrophe uncertainty principle?
 
  • Like
Likes Klystron and sophiecentaur
  • #40
PeroK said:
Is that the apostrophe uncertainty principle?
Heisenberg uncertainty principle
 
  • #41
Jeff Root said:
The argument in the last several posts suggests a new question
which I think should be in a new thread. I will title it:

"Can anything be measured with a single measurement?"

-- Jeff, in Minneapolis

Let me answer your original wavelength question instead: Send your photon through a spectrophotometer whose detector is a sensitive linear array capable of counting individual photons. (Incidentally you don't need a femtosecond laser here...just a very low intensity source and some time) The position of the detected photon will indicate its most probable wavelength.
The uncertainty in this measurement is determined by careful characterization of the apparatus (either before or after the event and either experimentally or theoretically!). No measurement is exact. If you need better numbers, send more photons.
There is an issue here with throughput (do you actually see any result at all?)...but that gets us into the weeds.

So yes a single measurement is useful information (and no measurement result is really useful without an explicit or implicit knowledge of certainty)
 
  • #42
slappmunkey said:
The more accurately you know a particles position the less accurately you know it's speed. And the more accurately you know a particles speed, the less accurately you know it's position.
Yes - that applies to a single photon, which is a quantum particle. So you can only know its Energy of you are prepared not to know where it is or how 'big it is'. This contrasts with a wave, which lasts forever and is a vast enough number of quantum particles to know both its dimensions and its frequency to as accurately as your equipment will allow.
 
  • Like
Likes Mister T
  • #43
sophiecentaur said:
Yes - that applies to a single photon, which is a quantum particle. So you can only know its Energy of you are prepared not to know where it is or how 'big it is'. This contrasts with a wave, which lasts forever and is a vast enough number of quantum particles to know both its dimensions and its frequency to as accurately as your equipment will allow.
I thought that time was the conjugate of energy. That is, if you measure energy very accurately, what you cannot know is when the measurement was obtained.

##\Delta E\ \Delta t > \frac \hbar 2##

But I just tried to resolve this notion for myself and found that things are rather more complicated than expected. http://math.ucr.edu/home/baez/uncertainty.html
 
  • Like
Likes hutchphd
  • #44
jbriggs444 said:
if you measure energy . . . . .etc . . .
I thought that referred only to quantum objects??
 
  • #45
sophiecentaur said:
I thought that referred only to quantum objects??
Ummm, since when is a photon anything other than a quantum object?
 
  • #46
jbriggs444 said:
Ummm, since when is a photon anything other than a quantum object?
What I meant is that it's always a quantum object and, as such, it can't be treated like a wave which is not a quantum object and a parameter like wavelength basically refers to a classical wave. It refers to vast numbers of photons and not to what individuals may be doing.
 
  • #47
sophiecentaur said:
This contrasts with a wave, which lasts forever and is a vast enough number of quantum particles to know both its dimensions and its frequency to as accurately as your equipment will allow.
This seems a rather idiosyncratic definition of a "wave". Certainly a plane harmonic wave of single wavelength is only possible in an infinite space. It is, however, a good enough approximation to use in almost any macroscopic empty space. And it describes one particle or many particles equally well.
The measurement process necessarily involves individual particles and of course more measurements gives better accuracy. The dipole radio antenna still absorbs (many) individual (radiofrequency) photons one at a time. I guess I don't understand the point you are trying to make.
 
  • #48
hutchphd said:
And it describes one particle or many particles equally well.
But does it? In which way do you mean that? All the EM Wave model tells you about the photons that are involved is their mean energy and where they are most likely to be detected.
Lurking at the back of peoples' minds will always be tiny squiggles when they think of Photons. What evidence is there that photons are at all like that? Are they small sections of a larger squiggle? I thought the wave packet description was dropped a long time ago. The current model (afaik) of photons is that they can be regarded as point particles. How does that imply a wavelength?
hutchphd said:
The measurement process necessarily involves individual particles
That's true but the nature of the wave is based on the statistics of the particles acting as a group.
 
  • #49
sophiecentaur said:
The current model (afaik) of photons is that they can be regarded as point particles. How does that imply a wavelength?
As does the theory of electrons...do they also not have wavelength?? How can you create spatial interference without wavelength? I am frankly mystified here.
sophiecentaur said:
That's true but the nature of the wave is based on the statistics of the particles acting as a group
The interference pattern can be revealed by repeated isolated measurement of individual particles.
 
  • #50
sophiecentaur said:
But does it? In which way do you mean that? All the EM Wave model tells you about the photons that are involved is their mean energy and where they are most likely to be detected.

The difference between classical wave models and quantum optics models lies in the fact that there are commutation relations in the latter case, which make life more complicated and lead to effects such as antibunching or the Hong-Ou-Mandel effect. However, this also means that for every quantity of interest that involves only few operators and does not need any normal-ordering of operators, the classical and the quantum model give exactly the same predictions. This includes: measurements of intensity, first-order spatial coherence, first-order temporal coherence and similar properties. The two models give different predictions for physical quantities depending on more operators: coincidence measurements, two-photon interference and similar stuff.

sophiecentaur said:
Lurking at the back of peoples' minds will always be tiny squiggles when they think of Photons. What evidence is there that photons are at all like that? Are they small sections of a larger squiggle? I thought the wave packet description was dropped a long time ago. The current model (afaik) of photons is that they can be regarded as point particles. How does that imply a wavelength?

Photons are considered as point particles only in a very limited sense that is not faithful to what most people would naively consider as point particles: If they get absorbed, this happens locally and only at one position. Still, the whole light field is "updated" instantaneously, which gives rise to antibunching and similar effects. However, any single photon state has a finite extent in space and time, where the probability to detect it is non-zero and this extent is given by its coherence volume, which can be small or huge depending on the light source used. The current model of photons is rather that you get probability amplitudes to calculate events such as absorption processes. These probability amplitudes behave pretty much like classical electromagnetic wave packets and thus may also be well described by a wavelength (or a superposition of several of them). The absorption process itself is then localized. However, if you are interested only in things such as simple absorption measurements without taking correlations with measurements at other positions or times into account, the probability amplitudes are sufficient to describe an experiment and you are back at a wave description.
 
  • Like
Likes sophiecentaur and PeroK
  • #51
Cthugha said:
Photons are considered as point particles only in a very limited sense that is not faithful to what most people would naively consider as point particles: If they get absorbed, this happens locally and only at one position. Still, the whole light field is "updated" instantaneously, which gives rise to antibunching and similar effects. However, any single photon state has a finite extent in space and time, where the probability to detect it is non-zero and this extent is given by its coherence volume, which can be small or huge depending on the light source used. The current model of photons is rather that you get probability amplitudes to calculate events such as absorption processes. These probability amplitudes behave pretty much like classical electromagnetic wave packets and thus may also be well described by a wavelength (or a superposition of several of them). The absorption process itself is then localized. However, if you are interested only in things such as simple absorption measurements without taking correlations with measurements at other positions or times into account, the probability amplitudes are sufficient to describe an experiment and you are back at a wave description.
That makes sense and I notice you only used the word 'wavelength' once then it includes the superposition of a number of wavelengths - as with the classical Fourier transform of a signal burst of finite length. A problem that I have is that this model seems to apply differently from photon to photon - depending on the way it happens to be interacting with matter, each end of its journey - but doesn't that go against the idea that photons have only one parameter (energy) to describe them?
"Still, the whole light field is "updated" instantaneously" and "depending on the light source used." are difficult concepts and to me it seems an artificial mechanism to explain away some of the paradoxes that seem to go go with quantum mechanics. If more than just the energy of a photon is involved, is the idea that you could somehow filter out photons of one kind from another kind (not just the energy involved)?
 
  • #52
sophiecentaur said:
That makes sense and I notice you only used the word 'wavelength' once then it includes the superposition of a number of wavelengths - as with the classical Fourier transform of a signal burst of finite length. A problem that I have is that this model seems to apply differently from photon to photon - depending on the way it happens to be interacting with matter, each end of its journey - but doesn't that go against the idea that photons have only one parameter (energy) to describe them?
"Still, the whole light field is "updated" instantaneously" and "depending on the light source used." are difficult concepts and to me it seems an artificial mechanism to explain away some of the paradoxes that seem to go go with quantum mechanics. If more than just the energy of a photon is involved, is the idea that you could somehow filter out photons of one kind from another kind (not just the energy involved)?

I find this whole discussion strange. A photon field can be decomposed into modes of definite wavevector, or, to be more precise, into wavepackets localized very narrowly around a certain wavevector. These modes are eigenstates of some operator (the momentum operator). Hence, when we measure the momentum of a photon, we also determine its wavelength. As always in quantum mechanics, wavelength is not a property a photon always haves, but only when we measure it.
E.g. the photons making up a spherical wavefront are a superposition of many wavevectors and frequencies.
 
  • #53
slappmunkey said:
It applies to everything sir. No matter how big or small.
Atomoc particles to pitchers throwing a ball...
EVERY calculation has uncertainties.

The Heisenberg Uncertainty Principle does not relate to calculations. It applies to the state of a system. Uncertainty in calculations is something very different.
 
  • #54
sophiecentaur said:
That makes sense and I notice you only used the word 'wavelength' once then it includes the superposition of a number of wavelengths - as with the classical Fourier transform of a signal burst of finite length. A problem that I have is that this model seems to apply differently from photon to photon - depending on the way it happens to be interacting with matter, each end of its journey - but doesn't that go against the idea that photons have only one parameter (energy) to describe them?

Well, do they have only energy to describe them? In most quantum treatments of the electromagnetic field, people start by mapping the problem onto the harmonic oscillator. Mathematically, you end up with the field quadrature (amplitude) along some phase as the equivalent of position in the harmonic oscillator and the amplitude along the orthogonal phase as equivalent to momentum. Then one considers the whole field as an infinitely-numbered ensemble of such oscillator modes each with different energy and essentially applies the physics of the harmonic oscillator to problems in optics. These photon modes have well defined energy (and polarization and wavevector). However, just like a real harmonic oscillator allows for quite complex oscillations which are a superposition of its possible modes, so does the light field. What people mean when discussing "single photons" is that the total number of excitations in the field is 1. Still, this single excitation may be a superposition in the sense that the probability amplitudes for several "modes" may be non-zero and we still have a single-photon state if the sum over all modes yields an occupation number of 1. These two concepts of photon number states and photon modes of well defined energies are sometimes mixed up in old books, but they are very different concepts. Also, in principle you can decompose the light field in many other ways, not only into modes of well-defined energy.

sophiecentaur said:
"Still, the whole light field is "updated" instantaneously" and "depending on the light source used." are difficult concepts and to me it seems an artificial mechanism to explain away some of the paradoxes that seem to go go with quantum mechanics. If more than just the energy of a photon is involved, is the idea that you could somehow filter out photons of one kind from another kind (not just the energy involved)?

Well, "updated instantaneously" is obviously colloquial. Please do not quote me on that. When you put a single photon on a beamsplitter, you can only detect it at one of the output ports. In a pure field/wave picture this is obviously hard to explain. When trying to stay inside the picture, you arrive at this updating terminology, which is not exactly adequate. If you consider it in some more detail, the "depending on the light source" bit is not so puzzling. Obviously, before a photon starts to exist, some emission process must have taken place. If photons indeed were point particles, this emission process must be instantaneous, which means that the transition must be incredibly broad spectrally - and the photon as well. So this picture of the emission process is obviously wrong. Instead, what usually happens is that a polarization (another really bad wording) is created. This means that a superposition is created between the emitter being in the excited state and the light field being empty on the one hand and the emitter in the ground state and one photon present in the light field on the other hand. This superposition is of course quite fragile and one may easily break it, e.g. by checking the state of the atom. However, if one checks the atom and it is in the ground state, the exact emission time is still not well defined, but has some uncertainty. This now directly carries over to the spatial extent inside which there is a non-zero probability to detect this photon and also via the Fourier limit to the spectral width of the single photon.

Now different light sources have different emission mechanisms: A single atom may only emit one single photon, while blackbody emitters tend to emit photons in bunches. This can be seen in the photon number statistics. Can you filter photons according to the source? Well, kind of. Consider a saturable absorber that saturates when absorbing already a single photon. It will absorb all of the photons from a single photon emitter, but a lot of the light from the blackbody source will go through. However, this is not really filtering. If you know that you have a "clean" light field just from one source, you can tell something about that source. This is sometimes done when testing lasers. However, you cannot put a mixed light field somewhere and keep the single photons while removing the blackbody radiation.
 
  • #55
Cthugha said:
Well, "updated instantaneously" is obviously colloquial.
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. :wink:
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
 
  • #56
Jeff Root said:
Is there any way to accurately measure the wavelength of a single,
individual photon?

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?
 
  • #57
Jeff Root said:
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. There is no such thing as "a photon's reference frame". You have a mistaken mental model of what a photon is.

Jeff Root said:
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.

Jeff Root said:
That's the idea I'm ultimately exploring

I don't see how such explorations will get you anywhere since they are based on a mistaken understanding.
 
  • Like
Likes Dale
  • #58
Mentor's note: A thread derailment based on a misunderstanding of elementary quantum mechanics has been removed from this thread.
 
  • #59
Jeff Root said:
An interference pattern can be made on a screen by passing a
beam of monochromatic light through a barrier with two parallel
slits of appropriate size and spacing for the wavelength of the
light. This works even if the beam is so sparse that only one
photon hits the screen at a time. I find this a convincing
argument that each individual photon has a wavelength.

You shouldn't. You can do this experiment with laser light, which is a coherent state of the quantum EM field. A coherent state does not have a definite photon number. The fact that you detect discrete dots on the screen, one at a time, does not mean that the thing that causes the dots can be accurately described as "one photon at a time traveling through space".
 
  • #60
Cthugha said:
What people mean when discussing "single photons" is that the total number of excitations in the field is 1. Still, this single excitation may be a superposition in the sense that the probability amplitudes for several "modes" may be non-zero and we still have a single-photon state if the sum over all modes yields an occupation number of 1.

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##.

Cthugha said:
These two concepts of photon number states and photon modes of well defined energies are sometimes mixed up in old books, but they are very different concepts.

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.
 
  • #61
Nugatory said:
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! :wink:
 
  • Like
Likes Klystron, davenn and Dale
  • #62
PeterDonis said:
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??.
 
  • #63
hutchphd said:
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.

hutchphd said:
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.

hutchphd said:
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.
 
  • #64
PeterDonis said:
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 ?
 
  • #65
hutchphd said:
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.
 
  • #66
PeroK said:
Although the main thread based on a misunderstanding of elementary quantum electrodynamics continues! :wink:
That phrase might be an oxymoron.:smile:
 
  • #67
PeterDonis said:
But such a state is not anything like a classical electromagnetic wave or a classical electromagnetic field.
Will it have a wavevector?
 
  • #68
hutchphd said:
Will it have a wavevector?

That would be an observable, and I'm not sure which observable it would be.
 
  • #69
PeterDonis said:
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)
 
  • #70
PeterDonis said:
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??
 

Similar threads

Replies
24
Views
3K
Replies
25
Views
1K
Replies
18
Views
3K
Replies
3
Views
1K
Replies
81
Views
6K
Replies
29
Views
2K
Replies
4
Views
1K
Back
Top