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.