Can A White Photon Exist?not so easy to Answer

[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.

 

[Edgardo]

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].



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.

I also don't know how to produce this state.

Ask vanesch, , he introduced this state.
But WHY exactly can't this state exist, besides that we both don't know how to produce it.
In the prism example I get the same experimental results, both in the 'statistical mixture' AND in the 'superposition' interpretation.

What I mean is: suppose a theoretical physicist claims that there is such a one-photon-superposition state. He would say something like:
"Hey, I interpret the spectrum behind the prism in another way.
It could also be that this is caused by many photons, each of them in a superposition state."

Then how would you disprove his claim experimentally (e.g. by another experiment)?

 

[ZapperZ]

What I mean is: suppose a theoretical physicist claims that there is such a one-photon-superposition state. He would say something like:
"Hey, I interpret the spectrum behind the prism in another way.
It could also be that this is caused by many photons, each of them in a superposition state."

Then how would you disprove his claim experimentally (e.g. by another experiment)?

You are forgetting the SOURCE of the light that is making this spectrum. Let's look at an incandescant light bulb, for instance. The source of light is the heating of tungsten. As with thermionic electron emission, heating produces a large spectrum of vibrational states. EACH of these states produces a unique transition. So what you get is a lot of photons, each having a unique energy. It is NOT a bunch of photons, each having a superposition of all of these vibrational states. You NEVER hear people analyzing spectra and ever consider such superposition.

As vanesh has mentioned, you cannot deduce superposition this way AFTER an actual measurement of the superposition. The ONLY way to prove such a thing occurs is to make a measurement of a non-commuting observable that does not collapse the superposition. We have seen this in H2 molecules, NH3 molecules, and in SQUID experiments on a superconductor junction. If you think an individual photon can have such superposition, then it is incumbent upon YOU to show evidence of this. A prism and spectral analysis can ONLY tell you that you have a photon with this freq, another photon with another freq., another photon with that freq... Nowhere in the measurement can you ever deduce that it was a photon having a superposition of energy that decided to collapse upon measurement into that particular energy or freq. If you (or that theorist) can make such definitive deduction of superposition based simply on that, then you have drawn conclusions beyond what the results have revealed. As an experimentalist, I am never comfortable with doing such things. Maybe you have a different standard.

Zz.

 

[vanesch]

If you think an individual photon can have such superposition, then it is incumbent upon YOU to show evidence of this.

Well, I reiterate that a femtosecond laser pulse needs such kinds of superpositions. The difficulty is that it is the source, and not the measurement apparatus. I don't know of a femtosecond photomultiplier, so all "position measurement" (with light, this comes down to "time measurement") apparatus is so terribly slow that it is always compatible with almost pure momentum states. But at least we have the *source* (nl. a laser).
If it were a mixture of "pure-energy" photons, you could never obtain a structure in time (or position) which is so lumped.

cheers,
Patrick.

 

[ZapperZ]

Well, I reiterate that a femtosecond laser pulse needs such kinds of superpositions. The difficulty is that it is the source, and not the measurement apparatus. I don't know of a femtosecond photomultiplier, so all "position measurement" (with light, this comes down to "time measurement") apparatus is so terribly slow that it is always compatible with almost pure momentum states. But at least we have the *source* (nl. a laser).
If it were a mixture of "pure-energy" photons, you could never obtain a structure in time (or position) which is so lumped.

cheers,
Patrick.

I don't see how a fs laser "pulse" would produce a superpostion on each individual photon in that pulse. The pulse itself WILL of course consist of possible superposition of various wavelength to make a pulse, but within that pulse, it consists of a gazillion photons, each having a different wavelength.

Again, I deal with laser pulses of about 8 ps in length. We have to produce such pulses not simply by opening and closing shutters (because that will definitely generate a lot of harmonics in the pulse), but using a complicated set of pockel cells, mirrors, etc. in our REGEN simply to do pulse shaping. This is to minimize the presence of photons with energies not within what we want. At no time do we consider the idea of an individual photon within such a pulse having "superposition" of all the various harmonics that together make up the pulse.

So again, I still do not see an experiment/measurement that shows the existence of such a superposition on an INDIVIDUAL photon.

Zz.

 

[vanesch]

I don't see how a fs laser "pulse" would produce a superpostion on each individual photon in that pulse. The pulse itself WILL of course consist of possible superposition of various wavelength to make a pulse, but within that pulse, it consists of a gazillion photons, each having a different wavelength.

So you consider such a pulse as a statistical mixture of "single energy photons". But I'm now claiming that a photon of fixed energy cannot be localised in space, so you cannot have such a mixture of which all of them arrive "in time". If it were such a mixture, the probability of detecting them should be independent of the phase with the "lock-in" signal and you'd get a more or less continuous arrival of all the different photons. And IF you filter them in energy, so as to FORCE them into a mixture, then you will get a loss in position (timing) resolution which is corresponding to the energy resolution by which you selected them (and thus truncated the bandwidth over which they could have a superposition).
In fact, it is silly to have a pulsed laser, followed by a very narrow bandwidth filter ! You can then just as well take an incandescent bulb and illuminate the filter You've lost the coherence, and you also loose the time structure. (exactly because you've transformed an interesting superposition into an uninteresting mixture).
But a picosecond isn't yet fast enough: you can easily have a delta lambda / lambda = 10^(-3) and still have picosecond pulses at optical frequencies. So this looks like "reasonably fixed energy" and "pulsed". In order to get dramatic effects of superposition with optical frequency, you need to go to femtosecond lasers. If you then see that they have amazing bandwidths, all of the above takes on much more meaning.

cheers,
patrick.

 

[ZapperZ]

So you consider such a pulse as a statistical mixture of "single energy photons". But I'm now claiming that a photon of fixed energy cannot be localised in space, so you cannot have such a mixture of which all of them arrive "in time".

Not if they're coherent with each other. We know this can happen because the very same thing is going on in a superconductor - the coherent superposition of a number of "plane waves". Each one of them are not localized, which is the very explanation for its zero resistance. yet, due to coherence, they can all "arrive in time", as we can see in any SQUID measurement.

Thus, just because they are extended in space does not negate them forming coherent pulses. There are preceedent for such things already.

Zz.

 

[reilly]

To some degree, the issue of a white photon is a matter of definition. To wit, see the discussion in Mandel and Wolf, Optical Coherence and Quantum Optics (p 480) on localized photons, which they describe by a Gaussian superposition of individual photon momentum/helicity states. Think about one-photon-at-a-time with such a state vector -- over time and many measurements, the location of the photon will be well defined by the mean of the location measurements, and the spread will be proportional to the standard deviation -- pretty much the same idea as a sequence of one photon double slit interference experiments.

So it is perfectly reasonable to define a white light state for a photon as an equally weighted state of optical frequencies. With a very faint source, you will see little if anything. Increase the strength of the source, and you will begin to see white light. What you would see with less than full intensity is very dependent on the optical processing capabilities of your eyes and brain -- perhaps there are conditions in which you would see shifting colors.

If you want to get a sense of this white photon triggering several rods/cones, think of low energy Compton scattering of of three relatively localised electrons, or of multiple photoelectric hits -- with an incident white photon. The probability of anything but the standard processes, that is simultaneous "hits" will be very small-for one incident photon. With many, the situation does indeed change. But not to worry.

To see white light requires quite a few white noise photons.

Mandel and Wolf go over every inch of photon physics -- lasers to thermal sources, coherent fields, light amplifiers, nonlinear effects, and on and on -- in about 1200 pages. So, if you want to understand photon physics, read Mandel and Wolf. Such a read will resolve most of the issues discussed in this thread.

Regards,
Reilly Atkinson

 

[Edgardo]

Hi all,

I thought about a photon that could be in a superposition
of energy-states. And maybe the COMPTON-SCATTERING might
be such a superposition state.

Right after the scattering of the photon, we don't know it's energy,
so I could describe it by a superposition.

What do you think of it?

-Edgardo

 

[vanesch]

Not if they're coherent with each other. We know this can happen because the very same thing is going on in a superconductor - the coherent superposition of a number of "plane waves". Each one of them are not localized, which is the very explanation for its zero resistance. yet, due to coherence, they can all "arrive in time", as we can see in any SQUID measurement.

But again, what ELSE is this "coherent superposition of plane waves" but a superposition of momentum states ?

Remember that there is NO phase relationship between the individual "quantum waves" (quantum states) in a MIXTURE, and that there IS a definite phase relationship in a SUPERPOSITION. This is in fact the essential difference between a mixture and a superposition. So if you talk about the COHERENT (= phase related) superposition of plane waves, you're talking about a superposition of quantum states, and not about a statistical mixture.

You seem to think that photon number 1 is in a plane wave state with momentum k1 and phase 36 degrees, photon number is in a plane wave state with momentum k2 and phase 42 degrees etc...
But this is not a possible state description in quantum mechanics ! Phase relations only make sense between components of a same state.

You cannot write a mixture of 2 photons as:
state of photon 1 is |k1>
state of photon 2 is exp(i alpha) |k2>
on which you will successively apply a measurement.

Well, you can write it, but it is exactly the same physical situation as:
state of photon 1 is |k1>
state of photon 2 is |k2>

That is because |k2> and exp(i alpha) |k2> describe the same ray in Hilbert space (and each ray corresponds to a unique physical state).

So you cannot COHERENTLY talk about these two states, or "waves" (in the position basis). There is no way to fix any phase relationship between components of a mixture.

The only, very only way to do this is to say that you have prepared the system twice in a single-photon state, described by:

|k1> + exp(i alpha) |k2>

In this case, the exp(i alpha) is NOT arbitrary, and this is the entire meaning of coherence. The "wave" in the position basis is then indeed, the Fourier composition of the two plane waves (with wave vectors k1 and k2) and with a phase relationship alpha.

cheers,
Patrick.

 

[ZapperZ]

But again, what ELSE is this "coherent superposition of plane waves" but a superposition of momentum states ?

But why is this an issue? Have I ever mentioned that such a thing isn't possible? I don't believe this is the source of this discussion.

The issue here has always been the confusion when we use these things to describe (i) an individual photon and (ii) a "pulse" that contains a gazillion photons. I have never (and in fact I have already mentioned it) said that a pulse of light does not contain a superposition of various freq. This is a fact that one can obtain simply via doing a Fourier transform of that pulse. However, this is not a single, individual photon.

Again, taking the lessons from superconductivity, coherency isn't just due to the fact that we describe things via plane-wave states. They become "locked" in step due to some "external" mechanism. For cooper pairs, it's the condensation. For photons, it is due to the source (or the downstream optics, etc). So coherency isn't an automatic property of plane-wave states, but can initiated externally. This is what I was trying to answer to your previous assertion that if plane wave states aren't localized, how can they "in time".

At some point in this thread, there appears to be a switching back and forth between the Fourier superposition of waves, and the quantum superposition of states. I do not see that one automatically implies the other.

But let's go back to the original assertion. If I interpret it correctly, you are saying that if we have a white photon, then I can write this photon as (crudely)

|w> = a1|v> + a2|i> + a3|b> + a4|g> + a5|y> + a6|o> + a7|r>

If I make an observation using my eyes, unless you are claiming that our optical system can view a superposition of all these states (very much similar to the bonding-antibonding states of H2), then even you have stated that you will simply collapse the superposition and view only one of these.

If what I have said is true, then we are left with only 2 issues:

1. There are NO "white photons" within the context that it has never been observed, measured, detected, etc. by our eyes. After all the term "white" is a human optical quality.

2. Detection measurements we have of photons (be it energy, momentum, wavelength, freq., etc) have not detected the presence of |w> states superposition. All experiements using photons have yielded them to be in a definite state and not a superposition of them. Only a repeated measurement of an identical system will yield the HUP relationship between two non-commuting observable.

So which part of any of the above do you disagree?

 

[vanesch]

At some point in this thread, there appears to be a switching back and forth between the Fourier superposition of waves, and the quantum superposition of states. I do not see that one automatically implies the other.

??

Then what is the quantum-mechanical description, according to you, of "a Fourier superposition of waves" ? No classical-QM switching back and forth allowed, only Fock states (which, according to QED, completely describe all possible pure states of the EM field).

But let's go back to the original assertion. If I interpret it correctly, you are saying that if we have a white photon, then I can write this photon as (crudely)

|w> = a1|v> + a2|i> + a3|b> + a4|g> + a5|y> + a6|o> + a7|r>

If I make an observation using my eyes, unless you are claiming that our optical system can view a superposition of all these states (very much similar to the bonding-antibonding states of H2), then even you have stated that you will simply collapse the superposition and view only one of these.

Of course. That's what I said: if you do an energy measurement, of course you will only find one definite result for one photon. But that doesn't mean that the photon was in such a state BEFORE the measurement. So an energy measurement is the worst kind of measurement one can think off to try to show the existence of the |w> state!
But what you are saying amounts to the following: "Because I only have a vertical and horizontal polarizer, light can only be vertically or horizontally polarized. It cannot be 45 degree polarized ; that just consists of a mixture of purely vertically and purely horizontally polarized photons."
If I see half of my light intensity get through my vertical polarizer, and half of my intensity get through half of my horizontal polarizer, that means that 50% of those photons are vertically polarized, and 50% are horizontally polarized.
Then I come in and I say: no, you have two possibilities.
The case you cite (a statistical mixture) IS a possibility. But your light could also be 45 degree polarized. And you come back and tell me that light, by definition, can only be vertically or horizontally polarized, look, when I use vertical and horizontal polarizers I always get ONE PRECISE ANSWER for each of them. So what's that silly business of *superpositions of polarization* ?? Only horizontal or vertical, look at my results !
But then I say: well, I can make the difference between the 50-50 HV mixture, and the 45 degree case, by having a 45 degree and a 135 degree polarizer: your mixture gets through those just as well 50/50, while the 45 degree polarization gets through the first 100% and is fully blocked by the second. So the worst possible measurement to distinguish between a mixture HV and 45 degree pure are HV polarizers of course !

In my example, there is an analogy between HV polarization and momentum states, and 45 / 135 degree polarization and position states.

If what I have said is true, then we are left with only 2 issues:

1. There are NO "white photons" within the context that it has never been observed, measured, detected, etc. by our eyes. After all the term "white" is a human optical quality.

2. Detection measurements we have of photons (be it energy, momentum, wavelength, freq., etc) have not detected the presence of |w> states superposition. All experiements using photons have yielded them to be in a definite state and not a superposition of them. Only a repeated measurement of an identical system will yield the HUP relationship between two non-commuting observable.

So which part of any of the above do you disagree?

Obviously I disagree with 1, and 2 is trivial because it is not the appropriate measurement to detect such a superposition.

cheers,
patrick.

 

[ZapperZ]

?? Of course. That's what I said: if you do an energy measurement, of course you will only find one definite result for one photon. But that doesn't mean that the photon was in such a state BEFORE the measurement. So an energy measurement is the worst kind of measurement one can think off to try to show the existence of the |w> state!
But what you are saying amounts to the following: "Because I only have a vertical and horizontal polarizer, light can only be vertically or horizontally polarized. It cannot be 45 degree polarized ; that just consists of a mixture of purely vertically and purely horizontally polarized photons."
If I see half of my light intensity get through my vertical polarizer, and half of my intensity get through half of my horizontal polarizer, that means that 50% of those photons are vertically polarized, and 50% are horizontally polarized.
Then I come in and I say: no, you have two possibilities.
The case you cite (a statistical mixture) IS a possibility. But your light could also be 45 degree polarized. And you come back and tell me that light, by definition, can only be vertically or horizontally polarized, look, when I use vertical and horizontal polarizers I always get ONE PRECISE ANSWER for each of them. So what's that silly business of *superpositions of polarization* ?? Only horizontal or vertical, look at my results !
But then I say: well, I can make the difference between the 50-50 HV mixture, and the 45 degree case, by having a 45 degree and a 135 degree polarizer: your mixture gets through those just as well 50/50, while the 45 degree polarization gets through the first 100% and is fully blocked by the second. So the worst possible measurement to distinguish between a mixture HV and 45 degree pure are HV polarizers of course !

But...but... don't you see? THIS is the VERY type of measurement that I've been asking for to show the existence of superposition of energy states in an individual photon! I am CONVINCED that a photon can exist in a superposition of what you call "polarization". You got no arguments from me there! The experiment you described is VERY convincing in showing that. What I asked is something in the same, analogous form to show the same effect for a photon! We can't just say "well, since superpostion can occur for that observable, then it makes sense that it should be possible for OTHER observables". This is simply not true since this is very system-dependent. Just becaue an electron can be in superpostion of spin, doesn't mean its charge must also be in such a state. So far, the arguments that superpostion of energy/wavelength/freq/ for an individual photons have been built via that line.

Zz.

 

[vanesch]

We can't just say "well, since superpostion can occur for that observable, then it makes sense that it should be possible for OTHER observables". This is simply not true since this is very system-dependent. Just becaue an electron can be in superpostion of spin, doesn't mean its charge must also be in such a state. So far, the arguments that superpostion of energy/wavelength/freq/ for an individual photons have been built via that line.

Concerning the electron's charge, there are different answers possible, but the simplest is that EVERY electron has one and the same charge. So this is a Hilbert space with only one dimension, which makes superposition trivial.
The superposition principle states that ALL POSSIBLE STATES can occur in a superposition. Some people have introduced a few superselection rules, such as boson/fermion interdiction, or charge interdiction. But it is a result of decoherence that you do not need those superselection rules: they are dynamically generated. So the superposition principle is universal.

The example I showed you was a femtosecond laser. Of course it are gazilion photons that come out, but I wanted to argue that they can ONLY have a time relationship if they are all in the same, pure state which is a superposition of momentum states.
Unfortunately, we don't have femtosecond photomultipliers. So it is only a preparation, and not a measurement. Nevertheless, I hope you believe that out of a femtosecond laser, come pulses which have a very short duration of which the photons are synchronized in time (or bunched in space, for a given value of time). Otherwise you've been had by the salesman of the laser!
So in the same way as out of a polarizer come only 45 degree polarised photons, even if there are a lot of them (in identical states), I am trying to convince you that out of such a laser come photons which are (at a given t) bunched in position, all of them, in identical "position" states. And the only way to have that is by having them in a superposition of momentum states.
But it is true that I don't have a SINGLE PHOTON source. It is a MANY IDENTICAL PHOTON source. And in that respect it is different from an incandescent bulb, which sends out a statistical mixture of different photons, in different states.
But any momentum filter, or spectral analysis, will screw up your nice difference between the pulsed source and the light bulb, like your HV filter screwed up your nice 45 degree polarized light as compared to unpolarized light from a light bulb. The only measurement proving this would be a femtosecond photomultiplier. But I don't know of any.

cheers,
patrick.

 

[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

i think ,
to solve this problem we must not cocentrate on photons produced from light source , or eye ( sensor ) respectors. But the answer is near if we take into account the way that the white light source produce this photons.
I read all what you wrote , and this guide me to this idea .
The answer of my question - i think - is " white one photon cannot exist ". The light source excited and in the delay to ground state it emit 3 photons at the same time , the 3 photon product of 3 (e.g.) eleectrons exictation.
and the rest of my idea is now clear 2 you

if i on the right way?

 

[kaonyx]

The original question asks if the intensity of white light is reduced to individual photon detections, what will they represent? Will there be individual photons that represent a range of energy (ie frequencies) as we understand white light to be? That is a good question. If so then we would be quite justified in calling that a "white photon" regardless of the make up of our eye, or issues of physiological perception etc. Just remember that photons are the quanta of the field, and E=hv was discovered by Einstein himself. Remember the photoelectric effect, he increased the voltage and just got more photons, not more energy per photon.
In the case of white light, just follow Newton. The separation of frequencies is dead easy, just use diffraction. Glass prisms will do. Ultimately you will find that extra prisms don't make a lot of difference, you can pretty well decide that you have a spectral distribution of frequency, which according to quantum theory is simply a spectral distribution of energies. And you will not be able to split your light any more than that.
That should convince you you don't have white photons. You might be tempted to think that the glass prisms "caused" the "photons to split". But then you would be inventing new physics about diffraction that you can't justify.

 

[kaonyx]

Notice that I said nothing about atoms and energy transitions. There is nothing to stop the energy distribution in the light beam from being as a continuum, it doesn't just have to be generated or for that matter absorbed, from the process of atomic transitions. So what does that do to the possibility of having white photons? Sorry but it is the same deal. You see, instead of thinking about particles having energy, it is more useful in the case of the light field to think of energy having particles. Again this was realized when Einstein sponsored Bose to publish what became known as the Bose-Einstein statistics. You see photons are not conserved like electrons. They invent themselves all the time, according to the energy requirements. So you see your continuum beam would happily go through the glass prism and produce a population of photons, with a range of energies, about which we could not say much apart from the overall probabilities. If you still not happy about the glass, then imagine your beam is so weak that you are only detecting one photon per day. It would be detected at one position after refraction. Thats it.

 

[Cthugha]

The original question asks if the intensity of white light is reduced to individual photon detections, what will they represent? Will there be individual photons that represent a range of energy (ie frequencies) as we understand white light to be?
[...]
If you still not happy about the glass, then imagine your beam is so weak that you are only detecting one photon per day. It would be detected at one position after refraction.

It is a common misconception that reducing the intensity of some arbitrary beam can give you single photons. This is not the case. Single detection events do not mean that you have single photons present. If you reduce the intensity of some coherent beam, the photon number distribution will stay Poissonian and even if you have on detection per day on average, you will still get some two-photon detection events as predicted by Poissonian statistics. The only way to show that you have true single photon states lies in measuring the variance of the photon number distribution or equivalently second-order correlations.

 

[ZapperZ]

Wow! Not sure how one digs out something from way back in 2005 to necropost.

Zz.

 

[Cthugha]

Oh, ouch. I did not even notice that only the last two posts before mine are new and the others are ancient. Sorry for taking part in resurrecting the dead.

 

[kaonyx]

It is not a popular misconception, it is a fact. The Poisson distribution simply means that photons arrive at purely random times. They are detected singly because they are just that, single photons. The randomness is in the arrival times between successive photons. The physical meaning of this randomness is that the electric field always contains some uncertainty in its value. You can modify this distribution of arrival times by quantum mechanical technology, such as non-linear optical systems that produce phase effects such as quadrature squeezing. What that achieves is that either the times between photon arrivals are a bit closer together than you expect classically, or conversely they can be a bit less closer together - commonly known as bunching or anti-bunching and is quantum mechanical, not classical. Photons tend to bunch slightly anyway, and electrons tend to anti-bunch, and you can demonstrate both effects with either particle. Photon number is simply a conjugate property of the phase, these things are how we describe the wave as a quantum mechanical wave function, so none of this changes the fact that these are photons, pure and simple, quanta of energy with a characteristic frequency, or if you prefer - colour.
So yes you can reduce the intensity down to a single photon (a gedankenexperiment) and you will detect a red photon and then a blue photon etc coming out of your prism with plenty of random time lag between them to convince yourself that there are no white photons. Case closed.