Can A White Photon Exist?not so easy to Answer

[ZapperZ]

What do you understand by "the Fourier components of that pulse" else than a superposition of momentum states (which are nothing else but these Fourier components) ?

But these are the "wave" descriptions! No where in such a description is there anything that invokes "a photon". I haven't seen a connection between the presence of a superposition of Fourier components in a pulse of light with superposition of wavelengths in a single photon. This is what I stated earlier, that such an argument is missing.

Zz.

 

[Gonzolo]

Single photon sources do not operate via such truncation in time.

Would you consider Fourier transforming a single-photon pulse? If not, why? If so, what explains the linewidth?

Femtosecond pulses do not operate via truncation either, they are a superposition of coherent harmonics in a laser cavity. If you have such a pulse transmit through an absorber until one photon is left, what energy does this last photon have?

And now, assuming both are centered at the same frequency, is the last photon from an attenuated fs pulse any different than the one produced by a single-photon source?

 

[ZapperZ]

Would you consider Fourier transforming a single-photon pulse? If not, why? If so, what explains the linewidth?

Remember what "single-photon sources" are.. they are (simplistically) "plane wave" sources, but at an emission rate that's so small that on average, you get only one photon at any given time going through a particular length or volume. It isn't that you open and close a shutter so fast that you manage somehow to get only one photon.

I don't know how one does a "fourier transform" of a single photon. I can, however, consider a "fourier transform" pair quantity as determined via the HUP, i.e. I measure its location via a single slit, and then see where it hits the screen and thus, measure the corresponding momentum. If I do this enough number of times on the identically prepared photons, I get a "spread" in momentum, depending on the size of my slit. This would qualify as Fourier transforming the slit width, but I don't think this is the typical Fourier transform of the time domain signal into the freq. domain signal that I believe is what you're getting at.

Femtosecond pulses do not operate via truncation either, they are a superposition of coherent harmonics in a laser cavity. If you have such a pulse transmit through an absorber until one photon is left, what energy does this last photon have?

But that's what I meant as "truncation". You need a sum of various harmonics to create "wave packets". One does this by physically opening-shutting apatures, or have the superposition naturally within a cavity by exciting those higher modes.

And now, assuming both are centered at the same frequency, is the last photon from an attenuated fs pulse any different than the one produced by a single-photon source?

Based on what I said above about the nature of single-photon source, yes.

Zz.

 

[Gonzolo]

[...] a "fourier transform" of a single photon. [...] the typical Fourier transform of the time domain signal into the freq. domain signal that I believe is what you're getting at.

Indeed, my doubt lies in what I'm getting at. I am not convinced that we are forbidden to Fourier tranform a pulse of a single photon. I would tend to believe that it would somehow merge into a delta E delta t (vs delta p delta x) version of the HUP. But that is where my knowledge is limited on the matter at the moment. I agree with everything else in your post.

 

[ZapperZ]

Indeed, my doubt lies in what I'm getting at. I am not convinced that we are forbidden to Fourier tranform a pulse of a single photon. I would tend to believe that it would somehow merge into a delta E delta t (vs delta p delta x) version of the HUP. But that is where my knowledge is limited on the matter at the moment. I agree with everything else in your post.

But the HUP itself, to show the "spread" in any of the conjugate observable, requires MANY observations. A single photon, passing through a single slit, and making a dot somewhere on a screen, does NOT give you the HUP.

I also don't know what a "pulse of a single photon" mean. A wave packet is not a photon. Doing a Fourier transform on a single photon (presuming this operation has any valid meaning here considering it is a discrete signal) would just give me a delta function corresponding to the freq. of that photon. But this is where it makes no sense because an inverse FFT of this is a continuous, infinite time-domain wave, which we don't have when all we originally detected was just one single photon. Thus, my argument earlier that such a thing makes no sense. So now it is up to you to show me why you think this is still possible.

Zz.

 

[selfAdjoint]

Heh heh! chemistryknight's title of this thread was right. His question is truly not so easy to answer!

Question: apart from visual sensation, is QED capable of answering the question about ultra-short pulses of light? Or is this phenomenon beyond the reach of valid perturbation analysis? Or can we set up a series of Feynman diagrams featuring just ONE photon in and out which will superpose different wave states on it?

 

[Gonzolo]

But the HUP itself, to show the "spread" in any of the conjugate observable, requires MANY observations. A single photon, passing through a single slit, and making a dot somewhere on a screen, does NOT give you the HUP.

I also don't know what a "pulse of a single photon" mean. A wave packet is not a photon. Doing a Fourier transform on a single photon (presuming this operation has any valid meaning here considering it is a discrete signal) would just give me a delta function corresponding to the freq. of that photon. But this is where it makes no sense because an inverse FFT of this is a continuous, infinite time-domain wave, which we don't have when all we originally detected was just one single photon. Thus, my argument earlier that such a thing makes no sense. So now it is up to you to show me why you think this is still possible.

Zz.

Either it makes no sense to Fourier tranform a single photon pulse, either it makes no sense that they have discrete energy.

With discrete energy, you have to accept discrete frequency, which means infinite lifetime and measurement. Accepting discrete energy means having to explain the time-energy version of the HUP.

I believe it takes a small amount of time to measure the frequency of a photon. Excitation of an atom is not instantaneous. Otherwise, I don't see how two-photon absorption would be possible. Two-photon absorption needs a slight time window for the second one to join in.

 

[ZapperZ]

Either it makes no sense to Fourier tranform a single photon pulse, either it makes no sense that they have discrete energy.

With discrete energy, you have to accept discrete frequency, which means infinite lifetime and measurement. Accepting discrete energy means having to explain the time-energy version of the HUP.

I honestly don't know how this fits into this whole thread.

I believe it takes a small amount of time to measure the frequency of a photon. Excitation of an atom is not instantaneous. Otherwise, I don't see how two-photon absorption would be possible. Two-photon absorption needs a slight time window for the second one to join in.

I'm not sure why a "time" factor here is involved in measuring the energy of a photon. The "time" factor in your example is the life of the excited state of the atom. It has nothing to do with the photon, or its own lifetime. So the HUP in your example is for the atom. But again, if you consider it carefully, you cannot get any kind of HUP relationship with just ONE photon, or just ONE measurement, or just one transition. This has never been what the HUP is describing.

Zz.

 

[masudr]

Zz asks if the photon can be in a superposition of different wavelengths. Can we not say that it can be in a superposition of different energy eigenstates, each one corresponding to some energy, each energy level corresponding to some frequency this corresponding to some wavelength? Or have I missed something?

In any case, we have no reason to call a photon in a superposition white, because it will never be seen as white. Instead, the better way to describe it would be "a superposition of several eigenstates". And Zz is also right in that the last few posts have strayed far from the original thread.

 

[Gonzolo]

As I have said when I divided the question in two, "white" either means :

1. How our brain interprets photon absorption. In this case, details of how retinal sensors work, answers the question, as masudr and self-Adjoint have discussed. The answer is no with little doubt.

2. Although it's an abuse of language, the only other possible meaning of "white" photon is that it's in a superposition of energy eigenstates, as has been said.

"So now it is up to you to show me why you think this is still possible."

I did what what I was asked to do, how can it not fit the thread? The original question has been answered beyond reasonable doubt long ago. Either we talk of 1. of 2. or we close it.

"The "time" factor in your example is the life of the excited state of the atom"

Not exactly, it's the lifetime of a virtual state between two actual states. I used this example to suggest absorption isn't instantaneous, thus nothing is. Maybe I can't get a HUP relationship with only one photon, I'll check around. Though HUP is derived from experiments with many photons, I don't see why I can't extrapolate to the case of a single photon, if only theoretically.

 

[ZapperZ]

"So now it is up to you to show me why you think this is still possible."

I did what what I was asked to do, how can it not fit the thread? The original question has been answered beyond reasonable doubt long ago. Either we talk of 1. of 2. or we close it.

That question from me was asked in the context of your claim that we can do a Fourier transform on a single photon. It came from:

"I also don't know what a "pulse of a single photon" mean. A wave packet is not a photon. Doing a Fourier transform on a single photon (presuming this operation has any valid meaning here considering it is a discrete signal) would just give me a delta function corresponding to the freq. of that photon. But this is where it makes no sense because an inverse FFT of this is a continuous, infinite time-domain wave, which we don't have when all we originally detected was just one single photon. Thus, my argument earlier that such a thing makes no sense. So now it is up to you to show me why you think this is still possible."

I believe I haven't seen any evidence that this is possible, or even if such a thing as any meaning.

"The "time" factor in your example is the life of the excited state of the atom"

Not exactly, it's the lifetime of a virtual state between two actual states. I used this example to suggest absorption isn't instantaneous, thus nothing is. Maybe I can't get a HUP relationship with only one photon, I'll check around. Though HUP is derived from experiments with many photons, I don't see why I can't extrapolate to the case of a single photon, if only theoretically.

A photon cannot be absorbed by an atom, for example, if there are no states that allow for such a transition. Your example only has in it the lifetime of that excited state. That's where a 2-photon absorption can come in, in which the excited state absorbs another photon, and gets promoted to an even higher state. It isn't the lifetime of the transition process, it's the lifetime of the excited state, that allows for such an event. The "Delta(t)" in the HUP refers specifically to such a lifetime. The longer it lives, the sharper the state becomes when we look at a collection of such transition (that's how we view their spectra). So in the example that you brought up, this is how the HUP is applied, not through the "transition time" of the absorption, and not with just one process by one atom due to one photon.

I would again refer to the example that I've already given elsewhere on a photon passing through a single slit as an illustration of the HUP. I will maintain that it is only through repeated measurement of identically prepared system does one obtain such observation, and not simply from one single event.

Zz.

 

[vanesch]

But these are the "wave" descriptions! No where in such a description is there anything that invokes "a photon". I haven't seen a connection between the presence of a superposition of Fourier components in a pulse of light with superposition of wavelengths in a single photon. This is what I stated earlier, that such an argument is missing.

That's not true: there is a 1-1 link. Single-wavelength coherent states are the QED equivalent of strictly monochromatic classical waves (fourier components). If you then want to describe Fourier superpositions of classical waves, you simply use superpositions of the coherent states, which lead you to have superpositions of 1-photon states with exactly the same phase factors, together with equivalent superpositions of 2-photon (same momentum) states etc...

So the QED description of non-monochromatic, classical EM waves ARE superpositions of photon states of different momentum.

cheers,
Patrick.

 

[vanesch]

Heh heh! chemistryknight's title of this thread was right. His question is truly not so easy to answer!

Question: apart from visual sensation, is QED capable of answering the question about ultra-short pulses of light? Or is this phenomenon beyond the reach of valid perturbation analysis? Or can we set up a series of Feynman diagrams featuring just ONE photon in and out which will superpose different wave states on it?

I thought it was very simple to describe this in QED (quantum optics, if you like).

Consider fock space:
we have:
1) a vacuum state |0>
2) single photon, pure momentum states: |k>
3) two-photon states |k1,k2>
...

This is the difference with nonrelativistic QM:
you only have line 2) because there's no creation or destruction.
If we talk about "single photons" we limit ourselves to the subspace spanned by the states in 2).

A single-photon "localized" state would then be exactly like a "localized" position state in NR QM: |x> = integral dk exp(i k x) |k>

(ok, the normalization is different in QED but let's forget all these details)
I honestly don't see why nobody complains about such a momentum superposition in NRQM, and why it is a problem in QED.

But the link with classical EM is a bit more subtle. Classical EM waves are indeed no single-photon states (but can approach them - except for one detail - in the low-intensity limit), but "coherent states".

Pure harmonic, classical EM waves are described in QED by a superposition of states with different numbers of photons, but with all the same momentum:

|EM harm wave> = Sum_n alpha^n/n! |k,...k>
where |k...k> is an n-photon state, with all of them momentum k.
(if I'm not mistaking).

Alpha is the complex amplitude of the EM harmonic wave.

So if we now consider an EM wave which has a non-trivial Fourier decomposition g(k), this is represented in QED by:

|EM "g(k)" field> = Integral dk |EM harm wave (k) with alpha = g(k)>

If the amplitude is low (low intensity) we can neglect the 2-photon and higher order n-photon states, and our state is essentially:

|0> + Integral dk g(k) |k> + O( g^2)

So it is |0> + |white single photon> + peanuts.

where |white single photon> = Integral dk g(k) |k>, for instance |x>, in the case of an ultrashort pulse (at given t, we know where it is, IF it is there).

The difference is a big contribution of the vacuum state: most of the time, in an attenuated beam, we don't have anything, and "single photons" are rare.

There is a way to make *pure* single-photon states, and that is by entangling them with something else (another photon): upon detection of that other photon, we then know that there is ONE photon (and not ZERO photon) in the other branch.

cheers,
patrick.

 

[vanesch]

Zz asks if the photon can be in a superposition of different wavelengths. Can we not say that it can be in a superposition of different energy eigenstates, each one corresponding to some energy, each energy level corresponding to some frequency this corresponding to some wavelength? Or have I missed something?

I don't see any reason why you can't have such a superposition.

In any case, we have no reason to call a photon in a superposition white, because it will never be seen as white. Instead, the better way to describe it would be "a superposition of several eigenstates". And Zz is also right in that the last few posts have strayed far from the original thread.

I don't think so. It goes to the heart of the problem.
Lots of "white photons" in identical, pure states will be seen as white, just as a statistical mixture of red, green, blue, yellow, etc.. photons will give you an impression of "white". There's no way to make the difference, because you cannot distinguish between a mixture and a superposition if you stay in the basis you're talking about (here, the momentum, or energy basis).

But you CAN make the difference if you go to another basis, for instance the position basis. Now, how do you determine (or prepare) photons in a "position" (knowing that they always go at lightspead) ? Well, you could think of a mechanical shutter, but it is not fast enough.
So a femto second laser can do, because you know where they are when.
And you can only do that through a superposition, not a mixture.
A mixture of momentum states, in this basis, will ALSO give you a mixture of position states, so they will be "everywhere". A superposition CAN make them be in a confined position.

cheers,
Patrick.

 

[ZapperZ]

That's not true: there is a 1-1 link. Single-wavelength coherent states are the QED equivalent of strictly monochromatic classical waves (fourier components). If you then want to describe Fourier superpositions of classical waves, you simply use superpositions of the coherent states, which lead you to have superpositions of 1-photon states with exactly the same phase factors, together with equivalent superpositions of 2-photon (same momentum) states etc...

So the QED description of non-monochromatic, classical EM waves ARE superpositions of photon states of different momentum.

cheers,
Patrick.

But this isn't what we have here. 1-photon states are not equivalent to "measurement of one photon". Of course there are single-photon states - if not, I have no business in pointing out the existence of single-photon sources. But I question in the ability of having just one photon and writing it as a superposition of freq. (or energy, or wavelength, etc.) in such a way that when we "view it", it is "white". We seem to be forgetting that we NEVER observe, upon measurement, all the superposition components. We just measure ONE! That is why I argued earlier that if you do a Fourier transform on a single photon (if such an operation actually has any meaning), then no matter how one argues how many freq. that photon could have, one only gets ONE delta function. Remember, a photon is not a "wave packet" or a "pulse" in the typical sense of the word. It cannot contain Fourier components that the sum of which produces a wave packet. If it is, then those of us doing photoemission measurements would have been producing nonsensical results the past 60 years or so.

Zz.

 

[ZapperZ]

Ah, after reading this thread further, I believe I have discovered a major miscommunication in the meaning of "single-photon".

Single-photon states that vanesch describes are not what I was trying to describe. Those single-photon states are states that describe photons produced with practically no degree of entanglement with others. The equivalent picture of this in condensed matter are single-particle excitation, where you have no particle-particle interaction. I now see why 2-photon states were invoked. Photon sources, specially those produced by parametric down converters, typically do not produce single photon states, because often more than 1 photon is created - which is why they are a popular source of EPR-type experiment.

This isn't what I have been talking about. What I have been describing are not "single-photon", but more accurately, INDIVIDUAL photons. Unless I have completely misread and misunderstood the original question in this thread, this is what was being asked : can an individual photon be "white" the same way a photon that make up a HeNe laser, for example, be "red". In a HeNe laser, can one argue against the fact that a photon from it, hitting a surface, imparts a well-defined energy and not a superposition of energy?

Zz.

 

[vanesch]

But this isn't what we have here. 1-photon states are not equivalent to "measurement of one photon". Of course there are single-photon states - if not, I have no business in pointing out the existence of single-photon sources. But I question in the ability of having just one photon and writing it as a superposition of freq. (or energy, or wavelength, etc.) in such a way that when we "view it", it is "white".

Ah, I see your point. You mean that even in order to qualify the spectrum, and be able to say that you measured "different frequencies", you have to do an energy measurement, and so that upon measurement, the particle has to choose an energy (according to the Born rule).

We seem to be forgetting that we NEVER observe, upon measurement, all the superposition components. We just measure ONE! That is why I argued earlier that if you do a Fourier transform on a single photon (if such an operation actually has any meaning), then no matter how one argues how many freq. that photon could have, one only gets ONE delta function.

IF YOU DO AN ENERGY MEASUREMENT, yes. But if you do a position measurement ; or a position preparation ?

What is so fundamentally different between a superposition of momentum states, for, say, an electron, when you know it is localized in a certain position, and the superposition of momentum states for a photon ?

Remember, a photon is not a "wave packet" or a "pulse" in the typical sense of the word. It cannot contain Fourier components that the sum of which produces a wave packet. If it is, then those of us doing photoemission measurements would have been producing nonsensical results the past 60 years or so.

I don't understand this. It is very well possible that most of the time you naturally work in the momentum basis (for instance because there are natural transitions between energy levels in the material you're considering, so that you do preparations and measurements in that momentum basis), but I really don't see why a photon cannot be in a "wave packet state" any less than an electron can be.


cheers,
Patrick.

 

[ZapperZ]

What is so fundamentally different between a superposition of momentum states, for, say, an electron, when you know it is localized in a certain position, and the superposition of momentum states for a photon ?

Because an electron is NOT defined by it's momentum. An individual photon is defined by it's energy (freq. and wavelength) and it's spin. That's about it! We never define an electron with a superposition of charge (let's not get into fractionalize charges here)! It can be in superposition of other observables that are not part of the definition of what an "electron" is.

I don't understand this. It is very well possible that most of the time you naturally work in the momentum basis (for instance because there are natural transitions between energy levels in the material you're considering, so that you do preparations and measurements in that momentum basis), but I really don't see why a photon cannot be in a "wave packet state" any less than an electron can be.


cheers,
Patrick.

If a photon is a "wave packet state", meaning it has a sum of various other Fourier components (which is the ONLY way to make a wave packet), then it's energy can be any of that Fourier component, and the photoelectric effect is bogus. This is because even when I use a filter to only select a "monochromatic" source, you are telling me that each photon hitting my cathode is actually composed of not just ONE distinct energy, but can be made up of a range of energy. This renders all the photoemission experiments (and a host of other experiments) to be junk.

Zz.

 

[vanesch]

Because an electron is NOT defined by it's momentum. An individual photon is defined by it's energy (freq. and wavelength) and it's spin. That's about it! We never define an electron with a superposition of charge (let's not get into fractionalize charges here)! It can be in superposition of other observables that are not part of the definition of what an "electron" is.

Well, there is a difference between n-photon states and n-electron states because electrons are of course fermions. But if it is charge that bothers you, switch to neutrons if you like.
However, a photon is not just "a lump of EM energy". It is a genuine particle in QED.

If a photon is a "wave packet state", meaning it has a sum of various other Fourier components (which is the ONLY way to make a wave packet), then it's energy can be any of that Fourier component, and the photoelectric effect is bogus. This is because even when I use a filter to only select a "monochromatic" source, you are telling me that each photon hitting my cathode is actually composed of not just ONE distinct energy, but can be made up of a range of energy.

When it got through the filter, no of course ! The filter is a measurement apparatus, so after that, it is in only those energy states that were allowed by the filter to pass (Born rule). But in doing so, you will also extend the wave packet (render the position less well known).
Only, again, such a filter (which is a measurement in the energy basis) will act upon a "white photon" state (superposition of single-photon momentum states) in exactly the same way as a statistical mixture, with individual colored photons. That's why probably in MOST applications, you can get away with thinking that photons always have to be in pure-momentum states.
But try sending a femto second laser pulse through a monochromatic filter: your bunch length will be extended in space ! Problem is, you will probably not notice this, because I suppose that the best timing (observation) you can do is of the order of 100 ps, which still allows for very tiny bandwidth at optical frequencies. So you'd need a hell of a narrow filter before hitting into measurable times.


cheers,
Patrick.

 

[ZapperZ]

Well, there is a difference between n-photon states and n-electron states because electrons are of course fermions. But if it is charge that bothers you, switch to neutrons if you like.
However, a photon is not just "a lump of EM energy". It is a genuine particle in QED.

But it's a "genuine" QED particle that has a well-defined energy for each individual photon. An individual photon (not a single-photon states) has as one of its definition, an "hf". I haven't seen fundamental characteristics in a superposition for fundamental particles.

When it got through the filter, no of course ! The filter is a measurement apparatus, so after that, it is in only those energy states that were allowed by the filter to pass (Born rule). But in doing so, you will also extend the wave packet (render the position less well known).

Well, I didn't realize that I had a problem with having unknown position. This, I believe, has never been part of my argument anywhere in this thread. Sources that produce "individual particles" are exactly like this. Again, as I've said, such sources does not produce "one-at-a-time" photons via opening-closing of shutters - that will not do because this will impart added Fourier components. Such sources produce photons "naturally" via transition rates that are low enough that at a particular given time and volume, there's only one photon on average that are within such boundary. In essence, the wave function is a plane wave of a particular eigenstate with very low amplitude. This is exactly what you are trying to convince me - that the position is undetermined. I have zero issues with this because this is what I had described. But this automatically implies a very well-defined momentum and energy eigenstate (p and H commute here for a "free particle"). Thus, without having to use any "filters", such sources do produce individual photons with well-defined energy, not a superposition of energies.

It boils down to this:

1. Can a photon have only ONE, well-defined energy?

2. If #1 is YES, then can there be cases where a photon has a superposition of different energies?

Unless I misread what you said, you have indicated that #1 is true, which is what I have been saying. However, we differ in the answer for #2. I don't see how #2 can be true if, by definition, a photon is defined as having a single energy of hf. I've yet to see a measurement of some observables that shows that a photon is in some superposition of a bunch of energy eigenstates the same way that we get the bonding-antibonding states of H2.

Zz.

Only, again, such a filter (which is a measurement in the energy basis) will act upon a "white photon" state (superposition of single-photon momentum states) in exactly the same way as a statistical mixture, with individual colored photons. That's why probably in MOST applications, you can get away with thinking that photons always have to be in pure-momentum states.
But try sending a femto second laser pulse through a monochromatic filter: your bunch length will be extended in space ! Problem is, you will probably not notice this, because I suppose that the best timing (observation) you can do is of the order of 100 ps, which still allows for very tiny bandwidth at optical frequencies. So you'd need a hell of a narrow filter before hitting into measurable times.


cheers,
Patrick.

 

[vanesch]

I don't see how #2 can be true if, by definition, a photon is defined as having a single energy of hf. I've yet to see a measurement of some observables that shows that a photon is in some superposition of a bunch of energy eigenstates the same way that we get the bonding-antibonding states of H2.

Ok, maybe you don't call it a "photon" then.
If |k1> and |k2> are 2 different 1-photon states in Fock space, how do you call then 0.707 |k1> + 0.707 |k2> ? I would still call that state a "single photon state" but you're free to call it something else, such as a "superposition of photons" or something.
Note that it is a pure state, which is orthogonal to the state |k1,k2> which is a pure 2-photon state.
It is also a different state than the statistical mixture of 50% chance |k1> and 50% chance |k2>.
These are 3 different QM states.

As of experimentally, I don't know. I'm pretty sure that these superpositions of 1-photon states occur in femto-second laser pulses, but they are indeed superposed with n-photon states in general, and with the vacuum state after strong attenuation. *pure* superpositions of single-photon states without vacuum contamination, I don't know if it has been done.
I think I know how to do it in principle (indeed using PDC) but I'm pretty sure that this messes up the wavepacket because there are also spectral conditions, so I will again end up with almost pure momentum states.

cheers,
Patrick.

 

[ZapperZ]

Ok, maybe you don't call it a "photon" then.
If |k1> and |k2> are 2 different 1-photon states in Fock space, how do you call then 0.707 |k1> + 0.707 |k2> ? I would still call that state a "single photon state" but you're free to call it something else, such as a "superposition of photons" or something.

But see, we should not confuse ourselves between the "object" (in this case, the individual photon) and the "system", in this case, the 1-photon STATES that produces such superposition. When we have the 2-slit experiment, the issue here isn't a single photon passing through both slits and interfering with itself, but rather the fact that the system allows two distinct paths, and that the superposition of such paths available to an individual photon is that is more directly responsible for the interference pattern. No where in there do you consider a superposition of an individual photon, but rather the superposition of the paths available to an individual, single-energied photon.

Note that it is a pure state, which is orthogonal to the state |k1,k2> which is a pure 2-photon state.
It is also a different state than the statistical mixture of 50% chance |k1> and 50% chance |k2>.
These are 3 different QM states.

As of experimentally, I don't know. I'm pretty sure that these superpositions of 1-photon states occur in femto-second laser pulses, but they are indeed superposed with n-photon states in general, and with the vacuum state after strong attenuation. *pure* superpositions of single-photon states without vacuum contamination, I don't know if it has been done.
I think I know how to do it in principle (indeed using PDC) but I'm pretty sure that this messes up the wavepacket because there are also spectral conditions, so I will again end up with almost pure momentum states.

cheers,
Patrick.

Again, I haven't and did not intend to discuss single-photon states, which is why I have tried in the last few posts to make sure I say "individual photon" rather than "single photon". I shoot one, just one photon... I want to know how one can tell me that that photon can be defined as a superposition of more than 1 energy states.

Zz.

 

[cartuz]

Excuse me because I have not time now to read the all. But I read the first tread.
The new term WHITE PHOTON can be consider as a short electromagnetic pulse. If we take the uncertainly inequality than we can write
ΔE*Δt=hbar
Δω=(1/(Δt))
or in Latex
$\Delta E\ast \Delta t=\hbar $

$\hbar \Delta \omega \ast \Delta t=\hbar $

$\Delta \omega =\frac{1}{\Delta t}$
From this follow that if the pulse is very shot than we cannot to write the exact wavelength. If you name this pulse as white photon than it is a new term. For this shot pulse we can use the furrier transform and it is mean that
the sum of different photons gives yours white photon.

 

[selfAdjoint]

$\Delta E\ast \Delta t=\hbar $

$\hbar \Delta \omega \ast \Delta t=\hbar $

$\Delta \omega =\frac{1}{\Delta t}$

Cartuz, the $ tags don't work on this site; you have to use tex and /tex with square brackets [ ] around them before and after. So you get
$$\Delta E\ast \Delta t=\hbar$$

$$\hbar \Delta \omega \ast \Delta t=\hbar$$

$$\Delta \omega =\frac{1}{\Delta t}$$

 

[Edgardo]

I agree with vanesch that you can't distinguish between a statistical
mixture and the superposition state. And I also don't see why such
a state shouldn't exist.
Question: How do you produce such a superposition state?

@Zapper: The photon is not defined by it's frequency,
it's rather DESCRIBED by the superposition state according to Vanesch.
(Correct me if I'm wrong Vanesch). Of course we are used to give
the photons a colour, BUT in vaneschs case, the colour would occur AFTER
measurement.


So here are the two different interpretations:
1) statistical mixture:
$\hat{\rho} = \int p(k) |k \rangle \langle k| dk$

2) superposition state:
$| \Psi \rangle = \int c(k) |k \rangle dk$

(Correct if that's wrong)

or sum (instead of the integral)
$\hat{\rho} = \sum_k p_{k} |k \rangle \langle k|$

$| \Psi \rangle = \sum_{k} c_{k} |k \rangle$

 

[ZapperZ]

I agree with vanesch that you can't distinguish between a statistical
mixture and the superposition state. And I also don't see why such
a state shouldn't exist.
Question: How do you produce such a superposition state?

@Zapper: The photon is not defined by it's frequency,
it's rather DESCRIBED by the superposition state according to Vanesch.
(Correct me if I'm wrong Vanesch). Of course we are used to give
the photons a colour, BUT in vaneschs case, the colour would occur AFTER
measurement.

So then, according to you, there's no point in creating a monochromatic light source, because you just never know what energy/wavelength/freq/ you might get. If that's the case, please explain why we should buy Einstein's photoelectric effect explanation. Even if its energy is defined only after measurement, if I do this enough number of times, since your "photon" is in a superposition of different energies, I should get several peaks in the spectrum corresponding to each energy/freq. So show me where we have seen this.

Zz.

 

[Edgardo]

So then, according to you, there's no point in creating a monochromatic light source, because you just never know what energy/wavelength/freq/ you might get. If that's the case, please explain why we should buy Einstein's photoelectric effect explanation.

Hmm...I didn't say that every photon is a superposition of every possibly frequency state. I refer to the WHITE photon being the superposition state.

And I absolutely agree with you that a monochchromatic light has
an energy before measurement. It's simply because the light is in
a pure state, namely it's already in an energy eigenstate. And
a measurement doesn't of course change this state anymore.

For monochromatic light the state reads:

$\hat{ \rho} = |k \rangle \langle k|$

$| \Psi \rangle = |k \rangle$

Even if its energy is defined only after measurement, if I do this enough number of times, since your "photon" is in a superposition of different energies, I should get several peaks in the spectrum corresponding to each energy/freq. So show me where we have seen this.

Zz.

Hmm...white (?!) light incident on a prism. You see different colours on your screen corresponding to different energies/frequencies.

 

[dextercioby]

Sorry to interviene,but,for the record,the quantum state of a photon (uniparticle state) is described not only by the (3) momentum $\vec{k}$,but also by the eigenvalue of the helicity operator (which in this case is +/- 1).

Just for rigurosity.Someone else reading the thread might think otherwise.

Daniel.

 

[ZapperZ]

Hmm...I didn't say that every photon is a superposition of every possibly frequency state. I refer to the WHITE photon being the superposition state.

And I absolutely agree with you that a monochchromatic light has
an energy before measurement. It's simply because the light is in
a pure state, namely it's already in an energy eigenstate. And
a measurement doesn't of course change this state anymore.

For monochromatic light the state reads:

$\hat{ \rho} = |k \rangle \langle k|$

$| \Psi \rangle = |k \rangle$





Hmm...white (?!) light incident on a prism. You see different colours on your screen corresponding to different energies/frequencies.

But then you've missed practically ALL of the arguments put forth here on why the term "white" light is misleading. Show me where you have observed "white" light that is due to only ONE photon! I will bet you my year's salary that you have NEVER observe, with your eye, a white light other than due to a stream of light source that is a composition of several different wavelengths that hits your eye! You are confusing your perception, and how your brain interprets signals coming from your nerves, with what is really going on.

I think that at this stage, this thread should be left alone to its natural death.

Zz.

 

[vanesch]

But then you've missed practically ALL of the arguments put forth here on why the term "white" light is misleading. Show me where you have observed "white" light that is due to only ONE photon! I will bet you my year's salary that you have NEVER observe, with your eye, a white light other than due to a stream of light source that is a composition of several different wavelengths that hits your eye!

You're right, but that's a cheap weasel-out

If I convert that to, say, electrons, you're saying that you've never observed several momentum states at once with a momentum analyser for a single particle, and that, in order to have a whole spectrum of answers from a momentum analyser, you need many particles, which then turn out, after measurement, to be in different momentum states. Yeah, that's true.
Given that a momentum analyser gives only one answer per particle, you can hardly have several momenta for one single particle, can you
Like you cannot measure several positions of one single electron.

cheers,
Patrick.

 

[ZapperZ]

You're right, but that's a cheap weasel-out

If I convert that to, say, electrons, you're saying that you've never observed several momentum states at once with a momentum analyser for a single particle, and that, in order to have a whole spectrum of answers from a momentum analyser, you need many particles, which then turn out, after measurement, to be in different momentum states. Yeah, that's true.
Given that a momentum analyser gives only one answer per particle, you can hardly have several momenta for one single particle, can you
Like you cannot measure several positions of one single electron.

But you CAN measure the CONSEQUENCES of the superposition of position of a single electron. I believe I do not have to repeat my often-used explanation of the H2 molecule, do I? Find me the corresponding non-commutating observable, and measure that! That measurement will not cause the collapse of the superposition the same way the measurement of the energy gap between the bonding and antibonding state in an H2 molecule did not cause a collapse of the electron position. This, I haven't seen done for a single photon, nor has anyone even proposed such a thing.

Again, I hate to repeat this, but a "position" characteristics, or a "momentum" characteristics, is NOT part of the fundamental identity of what an electron is. You NEVER define an electron based on where it is, or how fast it is moving. It's position, momentum, energy, etc. etc. are all observables that can be in as many superposition as necessary based on the geometry of the system (it has nothing to do with the electron).

NOw this is different for the photon. Just from the Einstein's model itself, a photon IS defined as a quanta of energy. It isn't defined with a definite size, or where it is, or what hair-piece it has - those can be in as many superpositions as our heart's content. All the examples of superpostions that you have mentioned is due to the originating system/source that created the photon, not a superposition of an individual photon itself.

Look, there's nothing to prevent me from obtaining a photon from an H atom making the 2p --> 1s transition, is there? Is that individual photon, even BEFORE measurement, a superposition of a number of energy states?

Zz.

 

[Edgardo]

But then you've missed practically ALL of the arguments put forth here on why the term "white" light is misleading. Show me where you have observed "white" light that is due to only ONE photon! I will bet you my year's salary that you have NEVER observe, with your eye, a white light other than due to a stream of light source that is a composition of several different wavelengths that hits your eye! You are confusing your perception, and how your brain interprets signals coming from your nerves, with what is really going on.
Zz.

It's clear that one single photon can never cause the
white colour (cause the impression for our brain) since it is, like you say,
a consequence of many photons with different colours.
In my first post to this thread I said that such a photon, that will
cause the white colour impression doesn't exist (I used the prism to show that).

But still, I don't see why such a superposition photon state shall
not exist. Of course one such photon will always have ONE colour
after measurement and can never be white.

 

[ZapperZ]

But still, I don't see why such a superposition photon state shall
not exist. Of course one such photon will always have ONE colour
after measurement and can never be white.

Then show me the situation, system, setup, etc. that produces an individual photon in such a superposition. I have already given you an example where a system produces an individual photon that is NOT in such a superposition.

Zz.

 

[Edgardo]

Then show me the situation, system, setup, etc. that produces an individual photon in such a superposition.

Say I have such a photon with
$| \Psi \rangle = \sum_{k} c_{k} |k \rangle$

Then I let it pass a prism which means I measure its energy.
What happens? The photon will collaps into one of the energy-eigenstates
and appears for example red (a red dot on the screen).

$| \Psi \rangle \rightarrow | \alpha \rangle$

The probability for a certain energy $E_{\alpha}$ would be:

$P = |c_{\alpha}|^{2} = | \langle \alpha | \Psi \rangle |^{2}$

There's nothing weird with the superposition state. (Except of course the question is how to produce it)

I have already given you an example where a system produces an individual photon that is NOT in such a superposition.

I don't exactly know to which example you refer to. But even if you have an example of a photon that is NOT in a superposition, it does not disprove
my claim. If you want to disprove it you have to find an example (experiment) where my claim leads to a contradiction, that is my superposition state leads to wrong physical results.

Of course there are photons in a pure state (monochromatic light beam).

 

[ZapperZ]

Say I have such a photon with
$| \Psi \rangle = \sum_{k} c_{k} |k \rangle$

WHOA! No! THIS is exactly what I want you to show! What system produces ONE photon with such superposition! You cannot "say" this if you can't demonstrate that this thing actually exist.

[the rest of your posting is moot if this doesn't exist].

I don't exactly know to which example you refer to.

You mean the H atom transition from 2p to 1s isn't good enough? Or do you think this isn't possible or isn't a realistic example?

Zz.