Are clicks proof of single photons?

[A. Neumaier]

Detection probability for first photon arriving at detector will not show any signs of interference.

Neither does the second, third, etc.

Detectors "remembers" phase of arriving photons. Or in other words it undergoes oscillations that affect detection probabilities of photons that arrive later.

How do they do the remembering? How do they get from this the information about the interference pattern that must be realized?

I don't think that you can turn this into a consistent model.

Would you care to explain this in more details.

Corresponding experiments are discussed in the opening chapter of Sakurai's QM book.

Alternatively, you may look at the Bell experiment discussed in slides 46-57 of my lecture http://arnold-neumaier.at/ms/lightslides.pdf

 

[DrChinese]

Alternatively, you may look at the Bell experiment discussed in slides 46-57 of my lecture http://arnold-neumaier.at/ms/lightslides.pdf

Wow, very cool lecture notes covering some fascinating nuances about the subject.

 

[lightarrow]

Photons have straight trajectories except at slit.

It's like how I teach car driving to my grandmother: "you only have to go straight, excepting when you come to a bend"
Photons don't have "Nadelstrahlung" (aciform propagation).
Put a 1 cm object at 10 cm of distance from a 1 mm light source, between the source and your eye: you will still be able to see the light (light has gone around the object). Of course it's called diffraction.

 

[harrylin]

[..]
How do they do the remembering? How do they get from this the information about the interference pattern that must be realized? [..]
I don't think that you can turn this into a consistent model.[..]

I agree with that, and would even go further, but I only realize this from reading this discussion - it's great!

For, it looks to me that even if one could turn it into a consistent model, so that with increasing photon count an interference pattern forms, such a model certainly differs in its prediction for few (up to 10) photon counts (and the experiment can be repeated a number of times). As I recall a few famous pictures of such an experiment, the interference is visible right from the start.

 

[zonde]

Neither does the second, third, etc.

It's not so according to model.
If the first photon is absorbed as thermal energy and the second photon is arriving from other slit then detection probability for second photon is affected by interference effect.

How do they do the remembering? How do they get from this the information about the interference pattern that must be realized?

If we have two photons with zero phase difference arriving from different slits at detector at the same time they undergo constructive interference.
If those photons do not arrive at the same time and first photon is absorbed without causing avalanche it's energy is not instantly converted into scalar energy without any phase information. Instead this energy still keeps phase information only with inverted phase space.
So if the second photon arrives after quarter of period phase of detector will be opposite and photon detection will be affected by destructive interference effect i.e. it will have increased probability of not being detected.
But if the second photon arrives after half a period phase of detector will be the same and photon detection will be affected by constructive interference effect - increased probability of sucesful detection.

Like that:

                      |- first photon arrives at detector and is not detected
1 -/+\-/+\-/+\-/+\-/+\|
D                     |\+/-\+|
2 -/+\-/+\-/+\-/+\-/+\-/+\-/+|
                       black[B]d[/B]- second photon arrives at detector
                       ||-- c stands for constructive interference effect
                       |-- d stands for destructive interference effect

Alternatively, you may look at the Bell experiment discussed in slides 46-57 of my lecture http://arnold-neumaier.at/ms/lightslides.pdf

And in your treatment you assume that all photons after second beam splitter are detected in similar fashion as in Bell theorem.
Therefore to establish correspondence with real experiments you will have to use fair sampling assumption as there are no 100% efficient detectors.
But in my post #114 I said that in my approach you have to drop fair sampling assumption. Meaning that at detector (or at filter if you have one after beam splitter) interference effect takes place between two subensembles from different paths and results in unfair detection (or filtering).

Additional problem with single photon interference is that you can't monitor coincidence count rate versus singlet counts so you have even less idea about detection efficiency for whole setup including any filters after second beam splitter.

So the bottom line of my approach is that as you increase detection efficiency of photons interference visibility tends to zero.

 

[zonde]

It's like how I teach car driving to my grandmother: "you only have to go straight, excepting when you come to a bend"
Photons don't have "Nadelstrahlung" (aciform propagation).
Put a 1 cm object at 10 cm of distance from a 1 mm light source, between the source and your eye: you will still be able to see the light (light has gone around the object). Of course it's called diffraction.

Starting question was about double slit interference and not about single slit diffraction. So I just took single slit diffraction as given.

 

[zonde]

For, it looks to me that even if one could turn it into a consistent model, so that with increasing photon count an interference pattern forms, such a model certainly differs in its prediction for few (up to 10) photon counts (and the experiment can be repeated a number of times). As I recall a few famous pictures of such an experiment, the interference is visible right from the start.

Probably what you remember is simulation of double slit experiment. In real experiments nobody installs huge detector array as a whole screen but instead scans the area line by line.

 

[harrylin]

Probably what you remember is simulation of double slit experiment. In real experiments nobody installs huge detector array as a whole screen but instead scans the area line by line.

I remember a textbook reproduction of a photographic plate, but I don't remember which one.
However, with Google I now found the following image series:

http://www.tnw.tudelft.nl/live/pagina.jsp?id=f1a85c5d-ed42-4f63-b70b-e682b39735c4&lang=en

Putting one of the later images next to the first one shows that the interference effect is about as strong at only 8 spots as it is for a large number of spots.

Cheers,
Harald

PS note the remark "Thus, the purely wave interpretation of light is not valid", which is based on "very localized impacts" - which is itself unfounded interpretation.

 

[lightarrow]

It's not so according to model.
If the first photon is absorbed as thermal energy and the second photon is arriving from other slit then detection probability for second photon is affected by interference effect.

Then it's not two distinct photons.

If we have two photons with zero phase difference

Phase difference of photons? So you are talking of an EM wavepacket model for a photon? I think you are moving in the wrong direction...

 

[zonde]

I remember a textbook reproduction of a photographic plate, but I don't remember which one.
However, with Google I now found the following image series:

http://www.tnw.tudelft.nl/live/pagina.jsp?id=f1a85c5d-ed42-4f63-b70b-e682b39735c4&lang=en

Putting one of the later images next to the first one shows that the interference effect is about as strong at only 8 spots as it is for a large number of spots.

From the text it is not clear whether upper images are real photographs. It's quite clear that pictures near the end of article are real photographs (with first image having around 25 dots). But I believe you are speaking about first image from the upper set.

But nonethless:
1. from documentation about used image intensifier (it is actual measurement device in this setup that converts single photon input into macroscopic signal) I found that it's quantum efficiency is around 4% for particular wavelength. So for 8 dots you have ~200 photons arriving at image intensifier.

2. and even more serious - from description it seems like shutter is part of photocamera so that image intensifier receives input continuously and the shutter operates at the level where single photons are already converted into classical signal.

So it does not seems that this text can back your argument.

 

[zonde]

Then it's not two distinct photons.

How did you arrived at that?
Are you implying something like - "click" in detector=photon ?

Phase difference of photons? So you are talking of an EM wavepacket model for a photon? I think you are moving in the wrong direction...

Wavefunction describes phase in complex plane, right?
So when you find interference term you convert phase difference in complex plane for two components of wavefunction into scalar term that either increases or decreases final probability.

 

[lightarrow]

How did you arrived at that?
Are you implying something like - "click" in detector=photon ?

No. How can there be interference between 2 photons if we are talking of photons sent one at a time? Then it must be at least two photons sent with a very little time interval, that is, two at a time.

Wavefunction describes phase in complex plane, right?

Which is exactly the wavefunction you are describing? Can you write it?

 

[A. Neumaier]

It's not so according to model.
If the first photon is absorbed as thermal energy and the second photon is arriving from other slit then detection probability for second photon is affected by interference effect.


If we have two photons with zero phase difference arriving from different slits at detector at the same time they undergo constructive interference.

But to do so, your photons must be wave packets rather than semiclassical particles.

 

[harrylin]

From the text it is not clear whether upper images are real photographs. It's quite clear that pictures near the end of article are real photographs (with first image having around 25 dots). But I believe you are speaking about first image from the upper set.

But nonethless:
1. from documentation about used image intensifier (it is actual measurement device in this setup that converts single photon input into macroscopic signal) I found that it's quantum efficiency is around 4% for particular wavelength. So for 8 dots you have ~200 photons arriving at image intensifier.

2. and even more serious - from description it seems like shutter is part of photocamera so that image intensifier receives input continuously and the shutter operates at the level where single photons are already converted into classical signal.

So it does not seems that this text can back your argument.

Good points! I do notice that your arguments strongly lean on limited sampling efficiency. Thus you seem to agree with me that your hypothesis is in principle open to experimental testing, as its predictions deviate from QM.

Harald

 

[StevieTNZ]

Mandel and Wolf write (in the context of localizing photons), about the temptation to associate with the clicks of a photodetector a concept of photon particles. [If there is interest, I can try to recover the details.] The wording suggests that one should resist the temptation, although this advice is usually not heeded. However, the advice is sound since a photodetector clicks even when it detects only classical light! This follows from the standard analysis of a photodetector, which treats the light classically and only quantizes the detector.

Thus the discreteness of the clicks must be caused by the quantum nature of matter, since there is nothing discrete in an incident classical external radiation field.

I don't quite understand - aren't photons discrete quantum matter? So photons are setting it off?
Classical light = incident classical external radiation field?

 

[zonde]

No. How can there be interference between 2 photons if we are talking of photons sent one at a time?

Well, not directly bit with measurement equipment playing the role of mediator.

Which is exactly the wavefunction you are describing? Can you write it?

Let's take as a simple example wavefunction for Mach–Zehnder interferometer experiment:
\psi =\frac{1}{\sqrt{2}}(e^{i(kL_{1}+\phi_{1})}+e^{i(kL_{2}+\phi_{2})})

 

[zonde]

But to do so, your photons must be wave packets rather than semiclassical particles.

You mean - photons must be wave packets to speak about phase difference?
Well, but if we speak about soliton waves rather than wave packets? They are rather very similar.

 

[zonde]

I do notice that your arguments strongly lean on limited sampling efficiency. Thus you seem to agree with me that your hypothesis is in principle open to experimental testing, as its predictions deviate from QM.

Certainly

 

[StevieTNZ]

Are you implying something like - "click" in detector=photon ?

You'd think so, if its a photon dectector.

First post:

Thus the discreteness of the clicks must be caused by the quantum nature of matter, since there is nothing discrete in an incident classical external radiation field.

and I say:

I don't quite understand - aren't photons discrete quantum matter? So photons are setting it off?

*utterly confused*

 

[lightarrow]

Well, not directly bit with measurement equipment playing the role of mediator.

Very obscure.

Let's take as a simple example wavefunction for Mach–Zehnder interferometer experiment:

\psi =\frac{1}{\sqrt{2}}(e^{i(kL_{1}+\phi_{1})}+e^{i(kL_{2}+\phi_{2})})

What are k and \phi ?

 

[lightarrow]

Are you implying something like - "click" in detector=photon ?

You'd think so, if its a photon dectector.

Because it's written in the kit label? Or because it's written in some books? The fact it's a "photon detector" say nothing. Firstly you should explain what exactly is a photon for you, then by which mechanism your photon interacts with a photon detector expelling an electron.
Then you will notice that with a classical EM field is much simpler...

 

[zonde]

Very obscure.

But does it matter if based on this we can make testable predictions?
I would say that you don't have to look far for an example.

What are k and \phi ?

k is wavenumber and \phi is phase shift along particular path.

 

[lightarrow]

k is wavenumber and \phi is phase shift along particular path.

So you are describing nothing else than the classical EM field. No need for photons at all. We have come to an agreement

 

[A. Neumaier]

I don't quite understand - aren't photons discrete quantum matter? So photons are setting it off?
Classical light = incident classical external radiation field?

Classical light = light in a mixture of coherent quantum states.

 

[A. Neumaier]

You mean - photons must be wave packets to speak about phase difference?
Well, but if we speak about soliton waves rather than wave packets? They are rather very similar.

Photons must satisfy the Maxwell equations. These don't have soliton solutions.

 

[Q-reeus]

From #110:
Originally Posted by Q-reeus:
'Well if there is no energy storage between 'hits', the implication seems to be either an actual collapse of all the energy into the screen of what could be a very extended single field quanta, or a purely probabilistic event where energy conservation is just a time-averaged thing. Please clarify.'

"QM claims energy conservation only in the ensemble average, which is usually equated (with practical success but no fully convincing demonstration) with the time average..."

Context for this exchange was in #105.
Disappointed no-one took issue with the claim conservation of energy is merely statistical at the quantum level. Motivated by an entry in another thread (#14 in https://www.physicsforums.com/showthread.php?t=250361) claiming it always exactly holds (which is my understanding), let's look at this again. Suppose as per #105 the screen is highly efficient but presents a very small cross-section relative to an approximately hemispherical surface that has as radius r the distance from the slit or slits to the screen. Let the mean rate of photons passing through the slits be one per minute. Let the ratio of screen area A_s to 2*pi*r² be say 0.001. Then on a photon-as-particle picture an event rate of roughly once every 1000 minutes on average applies. What about the situation with photon-as-continuous-spherical-wave? Realistically if for each 'hit' some tiny fraction .001 of a single photon energy can actually be absorbed by the screen, we might expect exponential decay-rate dissipation as heat with a time constant in the nano-second range. For all practical purposes there is simply no memory between hits, and given the vast insufficiency in available energy for an electron ejection event ('click') by a single photon-as-spherical-wave, how could we ever expect one - ever?
Further, even taking the position that energy can in fact be 'borrowed' on time scales of ~ 1000 minutes in order to statistically match the photon-as-particle count rate, how would the screen 'know' how to get the stats right - when to eject an electron, given the total obliteration of evidence between hits? Is there some quantum ghost that hovers about, doing a count? In solving one 'quantum mystery' (interference), two 'mysteries' at least as serious seem to have taken it's place. Does M & W cover this situation?

 

[A. Neumaier]

From #110:
Originally Posted by Q-reeus:
'Well if there is no energy storage between 'hits', the implication seems to be either an actual collapse of all the energy into the screen of what could be a very extended single field quanta, or a purely probabilistic event where energy conservation is just a time-averaged thing. Please clarify.'

"QM claims energy conservation only in the ensemble average, which is usually equated (with practical success but no fully convincing demonstration) with the time average..."

Context for this exchange was in #105.
Disappointed no-one took issue with the claim conservation of energy is merely statistical at the quantum level. Motivated by an entry in another thread (#14 in https://www.physicsforums.com/showthread.php?t=250361) claiming it always exactly holds (which is my understanding),

Conservation of energy holds on the Feynman diagram level. But a Feynman diagram doesn't describe a real process in time - only a contribution to a scattering amplitude between t=-inf and t=+inf. Thus this is irrelevant for the interpretation of processes that happen in time.

Given the Schroedinger equation i hbar d/dt psi(t) = H psi(t) , one can easily prove that d/dt psi(t)^*A psi(t) = 0 whenever A commutes with H. This is the case for any function of H. Therefore, <f(H)> is time-independent for any function of H. This is the strongest conservation statement concerning energy that can be proved. It is only a statistical conservation law.

Further, even taking the position that energy can in fact be 'borrowed' on time scales of ~ 1000 minutes in order to statistically match the photon-as-particle count rate, how would the screen 'know' how to get the stats right

Firing at random with the correct rate gives the correct statistics.

 

[Q-reeus]

It is only a statistical conservation law...

Alright let's take that as so. There still seems to be a monumental gulf in terms of what Heisenberg's Uncertainty Principle can allow here - ie delta E * delta t >= h. Taking the delta E to be hf (photo-electron ejection energy), and delta t as the time between 'clicks' ~ 1000 minutes, this comes to hf*1000*60 ~ h*6*10¹⁴*6*10⁴ = h*3.6*10¹⁹ , (using f~ 6*10¹⁴ Hz for typical visible light). Many orders of magnitude greater than h! And is this a totally inapplicable use of the HUP? I find it unreasonable to imagine quantum energy-time fluctuations at this level. It also seems hard to see how one does not in fact have a net gain in energy here, as there is no apparent mechanism to cancel the energy of the incident photons (unlike say for monochromatic radiation interacting with classical oscillators).

..Firing at random with the correct rate gives the correct statistics.

But how does that statement physically explain a situation where as I have claimed there is a screen total state reset between photon hits? Each hit might as well be the first. If 0.001 of the energy needed to generate a click is available that time, that's all that is available every other time, unless what physical process intervenes exactly?

 

[A. Neumaier]

Alright let's take that as so. There still seems to be a monumental gulf in terms of what Heisenberg's Uncertainty Principle can allow here - ie delta E * delta t >= h.

This is not Heisenberg's uncertainty relation, since t is not an observable in the usual QM sense.

Taking the delta E to be hf (photo-electron ejection energy), and delta t as the time between 'clicks' ~ 1000 minutes,

You cannot choose arbitrary energies and times in such a relation. delta E is the _uncertainty_ in measuring the energy E = h*nu, and delta t is the uncertainty in the timing of a measurement (i.e., of the order of the duration of a click).

Each hit might as well be the first.

Just like each throw of a die may be the first. This doesn't invalidate correctly getting the statistics of dice or clicks.

 

[Q-reeus]

..You cannot choose arbitrary energies and times in such a relation. delta E is the _uncertainty_ in measuring the energy E = h*nu, and delta t is the uncertainty in the timing of a measurement (i.e., of the order of the duration of a click)...

But the relationship is more than just a measurement phenomenon - does it not lie at the heart of say the ZPF energy spectrum?
I'd still like an explanation of how photo-ejection does not simply add to net energy with photons-as-spherical-waves picture. Where does cancellation of these incident photon energies ever occur?

Just like each throw of a die may be the first. This doesn't invalidate correctly getting the statistics of dice or clicks.

Sure, but in this case the energy deficit is enormous - where does the 0.999 fraction needed to make up the deficit come from? When the incident energy is relatively high, 'normal' statistics are OK, but here? The implied linearity of count rate vs photon incident rate seems way suss here for reasons already given (exponential seems more likely), but is perfectly natural in the particle picture.