Can A White Photon Exist?not so easy to Answer

[chemistryknight]

I read this question in a book. but i think a color of a photon is an expression of it's energy . the white light is consist of 7 colours ( 7 frequencies ) . and the photon can exist in one frequency only ( energy = f * h ) .
if we have white light source extremely dim to produce one photon how could we see it ?
can anyone help me

 

[chroot]

"White" light is not monochromatic. White light is composed of at least the three primary colors of light, as defined by the color-sensitive cells in our eyes: red, green, and blue.

You cannot have a single photon of white light. If you have a very dim white light source, which produces only one photon at a time, you will never see a "white" photon. You will, however, see photons of all different colors, as all those colors combined create the color white as seen by the human eye.

- Warren

 

[vanesch]

You cannot have a single photon of white light. If you have a very dim white light source, which produces only one photon at a time, you will never see a "white" photon. You will, however, see photons of all different colors, as all those colors combined create the color white as seen by the human eye.

Well, what about a superposition of 1-photon fock states with different momenta ? This is not the same as an n-photon state.

|psi> = Integral dk g(k) |k>

There is of course a semantic issue if we call "|psi>" a one-photon state...


EDIT: also, there's a difference with "white" light as seen as a thermal _mixture_ of one-photon states |k> of different k, which is not a pure state but an ensemble described by a density matrix.


cheers,
Patrick.

 

[arivero]

The problem is, a single foton should be detected in three different eye detectors to give white. On other hand, a single stimulation of a non-colour detector in the eye should be possible, then producing sort of grey.

 

[Edgardo]

Hi chemistryknight,

I would take the one-photon light source and let the photons go through
a prism, then I would look at how the photons are diffracted.
If the white light consists of different photons, then
you would get dots, that are spread due to dispersion.

-Edgardo

 

[vanesch]

Hi chemistryknight,

I would take the one-photon light source and let the photons go through
a prism, then I would look at how the photons are diffracted.
If the white light consists of different photons, then
you would get dots, that are spread due to dispersion.

-Edgardo

All this doesn't distinguish between a pure superposition of pure-momentum 1-photon states and a mixture, because the measurement apparatus (the eye, or the prism with position detector etc...) measures pure momentum states.

cheers,
Patrick.

 

[Edgardo]

First of all,

yeah, understand what you say vanesch.
I personally thought that a white source consists of photons, whose state is already determined before measurement (more like in classical physics).

In your opinion, a white light source-photon-state is a superposition of all frequency-states,
like you said with the integral? Could also be. But it's quite interesting, if you think
of the (rather naive?) picture of an electron "jumping to another shell" to make the transition from E1 to E2. This jump in the atom is not determined until you measure
the energy of the photon.

Secondly,

I understood chemistryknight's question in another way, namely, if there's
a white photon, that means, if white is a colour. If so, then all photons would
be diffracted to the same spot.

 

[ZapperZ]

Secondly,

I understood chemistryknight's question in another way, namely, if there's
a white photon, that means, if white is a colour. If so, then all photons would
be diffracted to the same spot.

Then the question is moot. If there IS a "white" photon, then it means that there is a SINGLE WAVELENGTH associated with the color "white" that our eyes perceive. This would be no different than a "red" photon, a "blue" photon, etc... i.e. all the colors that can be associated with a single, monochromatic wavelength. If that's a case, a white photon is of no mystery and this question would not have been asked.

But no matter how much we want to force is, there is no "white" photon. We should not let our eyes (which is a very limited and easily unreliable detector) be the primary source of things like this.

Zz.

 

[vanesch]

First of all,

yeah, understand what you say vanesch.
I personally thought that a white source consists of photons, whose state is already determined before measurement (more like in classical physics).

That is a possibility: it is a statistical mixture of "pure color" photons. That is not a pure quantum state, but an ensemble described by a density matrix. It is probably the best description for "sunlight".

In your opinion, a white light source-photon-state is a superposition of all frequency-states,
like you said with the integral?

No, it is probably a state which is much harder to make, but it is what I thought came closest to "a white photon". In fact, the perfect "white photon" would be by a non-destructive POSITION measurement. But I don't know how to do that with photons. It is easier with other particles, such as electrons. Indeed, a position state is a superposition of many momentum states.

cheers,
Patrick.

 

[masudr]

Even if we did have a superposition of different energy states, the photon would still not be white. What we consider "white" is merely an interpretation of our brain, when neighbouring retinal cells receive photons of varying energies; and hence there is nothing physical about the colour white.

When a photon is detected by the retina, no matter what it's state is, it will end up in an energy eigenstate (as the cell will perform an energy measurement) and hence end up with a definite energy and a definite frequency (assuming nondegenerate eigenstates).

There is no such thing as a white photon. We can have a superposition of different energies, but that doesn't make the photon white.

Indeed, a position state is a superposition of many momentum states.

Yes. Any state is generally a superposition of the eigenstates of some operator (that is what we mean by a complete set of states).

 

[Gonzolo]

It seems to me that the question can mean one of the following two :

1. Can a single photon excite multiple color sensors in our eye such that we see "white"?

or

2. Can the incertainty of a single photon's frequency cover the entire visible spectrum?

 

[ZapperZ]

It seems to me that the question can mean one of the following two :

1. Can a single photon excite multiple color sensors in our eye such that we see "white"?

or

2. Can the incertainty of a single photon's frequency cover the entire visible spectrum?

If it is the 2nd one, then it is a complete misunderstanding of the uncertainty principle. It isn't our inability to accurately measure a single photon's frequency. It is our ability to know what the next one, and the next one, and the next one, etc. is going to be. That is what is contained in the uncertainty principle. The width of a spectrum peak is meaningless when one measures just a single photon. One can obtain this with arbitrary precision not limited by the uncertainty principle. It is the collection of identically-prepared photons that would contribute to the line width of the spectrum. I believe I have tried to explain this in my application of the uncertainty principle in the case of a single-slit measurement.

Zz.

 

[vanesch]

Even if we did have a superposition of different energy states, the photon would still not be white. What we consider "white" is merely an interpretation of our brain, when neighbouring retinal cells receive photons of varying energies; and hence there is nothing physical about the colour white.

Ah, well. I thought as a way of talking, "white" meant "contains all wavelengths with equal intensity/amplitude..."

You have white noise, pink noise and red noise in electronics.

We have beams of white neutrons or X-rays and also monochromatic beams.
...

cheers,
Patrick.

 

[masudr]

I can only echo Zz's message about the HUP. It is not to do with the position (or momentum or energy) of a single particle. It is to do with the variance these things. And that requires repeated measurements of identical systems.

1. Can a single photon excite multiple color sensors in our eye such that we see "white"?

No. I believe the retinal cell absorbs the photon once detected. So a single photon will never be seen as white. Anyway, it requires about ten photons for a cell to register the light.

 

[selfAdjoint]

The retina has several different opsin molecules, three types for the average man, and four for the average woman. Each type responds, by flexing, to a particular narrow band of frequencies. The photon interacts with the molecule, raising it to a higher energy state, which in the case of opsin, causes it to flex. So the number of opsin molecules flexing at a particular time is a (distorted) measure of the photon flux at those energies. The flexing causes a neural pulse to be sent to the visual cortex in the back of the brain. This is the ONLY information the brain receives about the frequency distribution of the ambient light. All our color sensations are constructed in the brain from this information.

The question is then, could a single photon interact with more than one opsin molecule.

 

[masudr]

Thanks selfAdjoint for your knowledge on this. Well it's clearly obvious that once the opsin molecule has been raised to a higher energy state, the original photon is gone from the universe never to come back. Of course when the opsin molecule drops to a lower energy state, it will emit radiation.

In any case, the original question of there being a white photon has been resolved long ago. No photon can be white. A photon could (possibly) excite several different opsin molecules by the process of absorption and emission. To determine these frequencies we will need an approximate solution to the opsin molecule. And so, it may well be possible that a single photon could cause the brain to register the colour white.

So we have the following conclusion: A photon can be in a superposition of different energy states that correspond to "white"; when detected by the human optical system, it will collapse into a definite energy eigenstate. However, by the process of absorption and reabsorption we could register other colours such that we perceive white.

I'm guessing that the human optical system is designed to ignore the emission bands of opsin. Or maybe we have gotten used to seeing colours like this, so we call the excitations caused by a "blue" photon blue only because the colour we see when a photon of such energy enters our eye is a actually a combination of all the secondary effects of opsin emission.

 

[selfAdjoint]

The secondary layer of the visual cortex does subtractions on the electrical impulses coming from the opsin molecules. The source type of the impulses is known (which axon they came in on) and coded differences are generated and passed on to the deeper layers.

BTW, I don't think White exists outside human brains. I don't think any color does. Just bouncing photons of different frequencies.

 

[masudr]

Yes I agree with what you say 100%. White is just merely an interpretation of the frequency/energy of incoming photons.

 

[DB]

When we say that white light (as we see it) is the superposition of many photons, how exactly is this wave and frequency described?

 

[masudr]

White light is not the superposition of many photons. We perceive white light when neighbouring retinal cells register photons of colours from across the spectrum.

When a photon is in a superposition of many states (nothing to do with white photons) it doesn't have a single frequency etc. Upon measurement, it drops into an eigenstate which has a defined energy/frequency.

 

[vanesch]

White light is not the superposition of many photons. We perceive white light when neighbouring retinal cells register photons of colours from across the spectrum.

When a photon is in a superposition of many states (nothing to do with white photons) it doesn't have a single frequency etc. Upon measurement, it drops into an eigenstate which has a defined energy/frequency.

When measuring in a certain basis (here, in the "color" or photon momentum basis) you cannot distinguish between a superposition and a statistical mixture, because Born's rule transforms the former in the latter.
You would have to change measurement basis in order to find out the difference. So as long as you stick to a human eye, which measures "color", you won't see any difference.

But you can easily conceive photons in a superposition of momentum states, and I don't see why we can't call them 'white photons'.
I'm wondering (I'm not sure) if ultra-short pulse lasers don't produce photons in a superposition of momentum states instead of a "mixture". In fact, I think they do, because of their timing: because you have ultra-short pulses in time, you know their position in time (for a given value of t, you know pretty well where they are along the beam) and the only way to do so is to have a *superposition* of momentum states. But ok, they are only 'white' over a very limited bandwidth, probably not the bandwidth of white light.

cheers,
Patrick.

 

[masudr]

Patrick, I completely agree with you. Apart from being easier to say, I don't think there is any reason why we should call a photon white for being in a superposition of states, since we do not perceive that as white. On the other hand a statistical mixture is perceived as white, but that requires several photons to be registered as white so we can't call any single photon white.

Why are you trying to ascribe a colour to a quantum particle?

 

[vanesch]

Why are you trying to ascribe a colour to a quantum particle?

I'm probably a lunatic, but the first reaction I have to "white" is not to associate it with something visual, but to "flat spectrum" and "all momenta".

As I said before, "white" noise, "white" neutron beams, pre-whitening filter,...


Patrick.

 

[Hans de Vries]

When measuring in a certain basis (here, in the "color" or photon momentum basis) you cannot distinguish between a superposition and a statistical mixture, because Born's rule transforms the former in the latter.
You would have to change measurement basis in order to find out the difference. So as long as you stick to a human eye, which measures "color", you won't see any difference.

Interesting related news:

Optics enters the single-cycle regime

http://physicsweb.org/articles/news/9/2/4/1
http://physicsweb.org/articles/news/9/2/4/1/050204

"Since the duration is extremely short, the pulse contained wavelengths
between 410 and 1560 nanometres - a range of 1.9 octaves"


Regards, Hans

 

[vanesch]

"Since the duration is extremely short, the pulse contained wavelengths
between 410 and 1560 nanometres - a range of 1.9 octaves"

Wow !

Patrick

 

[masudr]

I'm probably a lunatic, but the first reaction I have to "white" is not to associate it with something visual, but to "flat spectrum" and "all momenta".

So if i said a white proton, you would think "superposition of various momentum eigenstates"? You are a lunatic! No, to be fair we all are, especially if we study physics.

 

[vanesch]

So if i said a white proton, you would think "superposition of various momentum eigenstates"? You are a lunatic! No, to be fair we all are, especially if we study physics.

If you said a white proton, I would be in doubt between:

- indeed a superposition of momentum eigenstates
- or the QCD meaning as a singlet state of the color charge

Whoops ! The men in white coats are coming


Patrick.

 

[masudr]

or the QCD meaning as a singlet state of the color charge

I should have expected that one...

 

[selfAdjoint]

When measuring in a certain basis (here, in the "color" or photon momentum basis) you cannot distinguish between a superposition and a statistical mixture, because Born's rule transforms the former in the latter.
You would have to change measurement basis in order to find out the difference. So as long as you stick to a human eye, which measures "color", you won't see any difference.

But you can easily conceive photons in a superposition of momentum states, and I don't see why we can't call them 'white photons'.
I'm wondering (I'm not sure) if ultra-short pulse lasers don't produce photons in a superposition of momentum states instead of a "mixture". In fact, I think they do, because of their timing: because you have ultra-short pulses in time, you know their position in time (for a given value of t, you know pretty well where they are along the beam) and the only way to do so is to have a *superposition* of momentum states. But ok, they are only 'white' over a very limited bandwidth, probably not the bandwidth of white light.

cheers,
Patrick.

But can the photon, however configured, interact with more than one retinal cone? The opsin molecule in that cone is only capable of telling the brain "I was excited by a wavelength somewhere in a distribution about X", where X = 556 nm, 535 nm, or 44 nm, depending on the molecule. It's like putting a complex frequency pattern through a band-pass filter. See http://users.rcn.com/jkimball.ma.ultranet/BiologyPages/V/Vision.html , for example. This signal is then input to the visual cortex which does a lot of data processing on it befor the color sensation is generated. So it appears to me that one cole can't generate the sensation white.

As for thinking of a photon in a superposition of frequencies as white, surely Patrick, that is an abus de langage caused by ignoring the complex relation between the physical observable frequency (or wavelength) and the living sensation color.

 

[masudr]

Yes, sA, that's exactly what I was trying to say. We ruled out secondary re-emissioin effects before, and so it's pretty clear that one photon can't be seen as white.

And again, I agree, that otherwise calling a photon white is just a case of language; the term could be accepted, if by common usage, as long as people knew that we didn't mean the photon would look white one absorbed by the retinal cones. But such an interpretation will ulikely become common usage, and hence we can't call objects white simply because they exist in a superposition of various momentum states.

 

[vanesch]

But can the photon, however configured, interact with more than one retinal cone? The opsin molecule in that cone is only capable of telling the brain "I was excited by a wavelength somewhere in a distribution about X", where X = 556 nm, 535 nm, or 44 nm, depending on the molecule.

There are two possible responses to this, according to one's view on QM.

The first one ("traditional") is: if you consider that this interaction is a *measurement* then the answer is of course no. You apply the Born rule to your photon superposition in the momentum basis, and thus you pick out a probability for each of those components to be chosen.

My pet interpretation, however, which is more MWI-like is:
This interaction of the EM quantum field and those opsine molecules is just a unitary process described in QED language. This then means that you simply entangle the different states of the different opsine molecules with the photon superposition. This is what a quantum chemist should tell you, after he has written out his hamiltonian of the interaction of photons with his molecules.
So yes, this then entangles with different nerve states (K/Na balances in superposition) etc... until your consciousness has to make the ultimate choice, using the Born rule, whatever that may mean. Upon making this choice, you can then track back that this came down to exactly one excitation of one molecule (in the chosen branch).

In practice however, there is of course no observational difference because all these entanglements are so hopelessly complicated that we will never be able to use them to perform interference experiments, discriminating between superpositions and stochastical mixtures.

So, if our eye were sensitive to a single-photon impact, you would get A SINGLE color sensation, red, green or blue, upon the impact of a white photon.

But as our real eye needs several photons in order to do so, and as each one will give rise to a different sensation, this will probably never be experienced.

A single photon: you don't see anything (because we need more)
Several photons: we will get stimuli from different types of molecules and experience "white".

cheers,
Patrick.

 

[ZapperZ]

There are two possible responses to this, according to one's view on QM.

The first one ("traditional") is: if you consider that this interaction is a *measurement* then the answer is of course no. You apply the Born rule to your photon superposition in the momentum basis, and thus you pick out a probability for each of those components to be chosen.

My pet interpretation, however, which is more MWI-like is:
This interaction of the EM quantum field and those opsine molecules is just a unitary process described in QED language. This then means that you simply entangle the different states of the different opsine molecules with the photon superposition. This is what a quantum chemist should tell you, after he has written out his hamiltonian of the interaction of photons with his molecules.
So yes, this then entangles with different nerve states (K/Na balances in superposition) etc... until your consciousness has to make the ultimate choice, using the Born rule, whatever that may mean. Upon making this choice, you can then track back that this came down to exactly one excitation of one molecule (in the chosen branch).

In practice however, there is of course no observational difference because all these entanglements are so hopelessly complicated that we will never be able to use them to perform interference experiments, discriminating between superpositions and stochastical mixtures.

So, if our eye were sensitive to a single-photon impact, you would get A SINGLE color sensation, red, green or blue, upon the impact of a white photon.

But as our real eye needs several photons in order to do so, and as each one will give rise to a different sensation, this will probably never be experienced.

A single photon: you don't see anything (because we need more)
Several photons: we will get stimuli from different types of molecules and experience "white".

cheers,
Patrick.

The issue here is whether a photon can be in a superposition of several different wavelengths, at least within the setting of this problem. I haven't seen any convincing argument that it can!

Thus, if we go by this, a "white light" isn't a superposition of a photon having different wavelengths, but rather a composition of photons, each having a spectrum of wavelengths. What's my justification for this? Look at this light using a spectrometer. You'll see a band of wavelengths, not one or the other, or the other, or the other... So the question of whether a single photon is responsible for the excitation of all our optic nerves is moot.

Zz.

 

[vanesch]

Thus, if we go by this, a "white light" isn't a superposition of a photon having different wavelengths, but rather a composition of photons, each having a spectrum of wavelengths. What's my justification for this? Look at this light using a spectrometer. You'll see a band of wavelengths, not one or the other, or the other, or the other... So the question of whether a single photon is responsible for the excitation of all our optic nerves is moot.

You're right (I already said this before) that something like _sunlight_ is just a statistical mixture. But have a look at ultra-short laser pulses. I do not see how you can get around having them in a superposition of momentum states in order to make these short wave packets.
Such light, when NOT analysed in the time domain, but by spectral analysis, should give you identical results as with "thermal" white light.

cheers,
Patrick.

 

[ZapperZ]

You're right (I already said this before) that something like _sunlight_ is just a statistical mixture. But have a look at ultra-short laser pulses. I do not see how you can get around having them in a superposition of momentum states in order to make these short wave packets.
Such light, when NOT analysed in the time domain, but by spectral analysis, should give you identical results as with "thermal" white light.

cheers,
Patrick.

But to be able to do a "spectral analysis", you need more than just ONE photon. In fact, you need a conglomorate of photons in a single pulse. How you truncate the pulse in the time domain dictates the Fourier components within that pulse. This is something that I've had to deal with since I work with 8 ps laser pulses that impinges on a photocathode.

But it still doesn't mean that a single photon has that superposition. Single photon sources do not operate via such truncation in time.

Zz.

 

[vanesch]

But to be able to do a "spectral analysis", you need more than just ONE photon. In fact, you need a conglomorate of photons in a single pulse.

We both agree that using spectral analysis on a population cannot distinguish between a statistical mixture of "single color" photons and pure state "superpositions of different wavelength photons". They give you both a broad spectrum.

How you truncate the pulse in the time domain dictates the Fourier components within that pulse. This is something that I've had to deal with since I work with 8 ps laser pulses that impinges on a photocathode.

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) ?

If you had an incoherent mixture of individual photons of different wavelengths, and not a single coherent superposition, you wouldn't have the correct phase relationships in order to get a sharp pulse in time, but you'd just have white noise (white thermal light).

cheers,
Patrick.

 

[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.