Are clicks proof of single photons?

[lightarrow]

Let's consider the usual 2-slit setup, but where the detection screen is a wide and very narrow strip, total area being orders of magnitude smaller than say a hemisphere whose radius is that from twin-slit plate to detection strip. This means orders of magnitude smaller cross-section than a single field quanta (as spreading wave) presents to the screen. I share your view there is no possibility of instantaneous physical collapse of such a field quanta - what the screen 'sees' is what the screen 'gets'. OK then - let the light be so attenuated on average only one field quanta passes the slits every minute or so.

But are you really able to say: "it has been sent a photon, then it passes through the slits, let's wait for a detection event...no, it was not detected", or you can only say that a single photon has been sent because you have a detection event?

 

[A. Neumaier]

Not as per my scenario. Recall that the screen area was taken to be several orders of magnitude smaller than needed to fully capture an incident field quanta at some nominal distance from the slits - assuming screen effective cross-section equals it's area. For sake of argument make it a neat factor of 100 (let's not worry about interference fringes for the moment - just take an average screen intensity value). So each of the ~ 1-quanta per minute passing the slits can deposit no more than ~ 1% of their energy to the screen (I think we agree in such a setup the screen can in no way act as a resonant antenna - after all there is no monochromatic stream of radiation). Hence at best the average time between clicks will be around 100 minutes.

OK, this is not the usual scenario when making double-slit experiments. Then wait 100 minutes. You still have full agreement with experiment. The mean waiting time is inversely proportional to the total energy reaching the screen (assuming 100% absorption by electrons).

But that's the real sticking point as I see it. This 'best case scenario' assumes the screen is not only capable of fully absorbing all incident radiation, but losslessly accumulating each hit for perhaps hours until sufficient energy is present to eject one electron. What's more in order to reproduce the interference pattern, a memory of the incident intensity is also dissipationlessly stored. This seems utterly incredible.

No storage is needed. Just a rate of response. That's the nature of probability - even tiny rates accumulate to an almost sure success over a long enough time.

It is like repeatedly throwing 10 dice together. Getting 10 sixes is utterly improbable, but it happens if you wait long enough on the average a number of 1/p throws where p is the tiny probability. The dice don't store their history; indeed, the success might happen the very first throw if the dice are in the right mood.

 

[Q-reeus]

But are you really able to say: "it has been sent a photon, then it passes through the slits, let's wait for a detection event...no, it was not detected", or you can only say that a single photon has been sent because you have a detection event?

Well the latter one would correspond surely to the photon-as-particle picture - deposits all of it's energy in one go - there is a click or scintillation. What I have been trying to distill is the process involved in the continuous field quanta model of QFT. As instantaneous physical collapse of a spherically expanding quanta wave that could be light-years in radius is eschewed, I see no alternative than an accumulation of possibly tiny fractions of a single field quanta per screen 'hit', until an event finally transpires. My view is there are huge problems with that. However the last posting in #107 seems to suggest no accumulation of energy occurs - it's all just a probabilistic thing. So I'm still not clear on how to view it all!

 

[Q-reeus]

OK, this is not the usual scenario when making double-slit experiments. Then wait 100 minutes. You still have full agreement with experiment. The mean waiting time is inversely proportional to the total energy reaching the screen (assuming 100% absorption by electrons).

At first sight this suggests a time extended energy accumulation scenario, but from your later remarks that's evidently not what you are meaning.

No storage is needed. Just a rate of response. That's the nature of probability - even tiny rates accumulate to an almost sure success over a long enough time.

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.

It is like repeatedly throwing 10 dice together. Getting 10 sixes is utterly improbable, but it happens if you wait long enough on the average a number of 1/p throws where p is the tiny probability. The dice don't store their history; indeed, the success might happen the very first throw if the dice are in the right mood.

No problem with dice throwing, but it's the energetics and statistics of interaction between a screen of small cross-section and very infrequent and tenuous field quanta as continuous wave that I still cannot quite picture. When an electron is ejected, can we say anything definite about the transfer mechanism beyond an appeal to stats?

 

[A. Neumaier]

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.

I don't know how precisely the energy balance can be tested experimentally - it is very difficult to account for efficiency losses. I believe (without having checked that) that the energy fluctuations in a typical solid state device at room temperature is far bigger than the uncertainty in the energy due to a fraction of a photon's energy.

It would perhaps be interesting to know what happens with the experimental photodetection statistics at very low temperature. Maybe Cthugha knows of work in this direction?

 

[Q-reeus]

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.

I don't know how precisely the energy balance can be tested experimentally - it is very difficult to account for efficiency losses. I believe (without having checked that) that the energy fluctuations in a typical solid state device at room temperature is far bigger than the uncertainty in the energy due to a fraction of a photon's energy.

It would perhaps be interesting to know what happens with the experimental photodetection statistics at very low temperature. Maybe Cthugha knows of work in this direction?

Many thanks for clarifying on that - I had been arguing the wrong model all along, but has taken till now to realize that! I imagine this viewpoint will generate some contention! Bed time. :zzz:

 

[zonde]

It would perhaps be interesting to know what happens with the experimental photodetection statistics at very low temperature.

Lowering detector temperature leads to lower rate of dark counts i.e. spontaneous clicks but it does not affect rate of incident photon detection.

I would say that for coincidence counting cases particle approach to photons is way more natural than wave approach.
Say when you increase detection efficiency coincidence rate increases too and it tends toward 100% coincidence rate as you extrapolate detection efficiency towards 100% efficient.
You would not expect such tendency for waves.

 

[A. Neumaier]

Lowering detector temperature leads to lower rate of dark counts i.e. spontaneous clicks but it does not affect rate of incident photon detection.

I would say that for coincidence counting cases particle approach to photons is way more natural than wave approach.
Say when you increase detection efficiency coincidence rate increases too and it tends toward 100% coincidence rate as you extrapolate detection efficiency towards 100% efficient.
You would not expect such tendency for waves.

You would not expect such tendency for classical waves. Indeed, coincidence counts reveal the limitations of the semiclassical picture, but only when fed with nonclassical light. In this case, one needs the corrections from quantum field theory. This together with coherence accounts for nonclassical, nonlocal correlations.

Coincidence counting in case of Bell-type experiment probing the quantum nature on a fundamental level is unnatural in _any_ semiclassical picture, whether in terms of particles or in terms of waves. But interference experiments (which don't need special quantum technology for their performance) remain unnatural for particles, while they are completely natural for light. This let's me prefer the field intuition over the particle intuition (which fails for photons anyway since they are not localizable).

 

[zonde]

Coincidence counting in case of Bell-type experiment probing the quantum nature on a fundamental level is unnatural in _any_ semiclassical picture, whether in terms of particles or in terms of waves.

I do not exactly agree with this.
Photon Bell-type experiments can be analyzed using semiclassical picture in terms of particles.
You have to drop fair sampling assumption in turn making Bell theorem not applicable to photon Bell-type experiments.

But interference experiments (which don't need special quantum technology for their performance) remain unnatural for particles, while they are completely natural for light. This let's me prefer the field intuition over the particle intuition (which fails for photons anyway since they are not localizable).

Resolving photon Bell-type experiments in semiclassical particle picture provides approach for interference experiments too.
It might be less natural than wave approach but that should be Ok if testable predictions specific (or more natural) to particle approach are made.

And I prefer semiclassical particle approach because it conserves energy at the level of single photon.

 

[A. Neumaier]

I do not exactly agree with this.
Photon Bell-type experiments can be analyzed using semiclassical picture in terms of particles.

Of course. I only claimed that it is not natural to think of particles - without any properties between creation and detections.

Resolving photon Bell-type experiments in semiclassical particle picture provides approach for interference experiments too.
It might be less natural than wave approach but that should be Ok if testable predictions specific (or more natural) to particle approach are made.

And I prefer semiclassical particle approach because it conserves energy at the level of single photon.

At the cost of not being able to say anything about the field between the source and the target. In view of the fact that as the number of photons get larger, suddenly an approximate classical picture of the field appears out of nowhere, this seems to me very strange.

I prefer to have a reasonably realistic view of what happens at every point in space and time, which turns naturally into the classical picture as the intensity get large. This is possible with a field picture but not with a particle picture. It removes most of the weirdness of quantum mechanics with one single stroke.

 

[zonde]

Of course. I only claimed that it is not natural to think of particles - without any properties between creation and detections.

At the cost of not being able to say anything about the field between the source and the target. In view of the fact that as the number of photons get larger, suddenly an approximate classical picture of the field appears out of nowhere, this seems to me very strange.

I prefer to have a reasonably realistic view of what happens at every point in space and time, which turns naturally into the classical picture as the intensity get large. This is possible with a field picture but not with a particle picture. It removes most of the weirdness of quantum mechanics with one single stroke.

Hmm, no.
When I talk about semiclassical picture I mean that we have trajectories at macro level.
And we have more or less sharp volume that defines photon at micro level, but not point particles.

Maybe I should have said classical picture but that people tend to associate with some billiard ball like model.

 

[A. Neumaier]

Hmm, no.
When I talk about semiclassical picture I mean that we have trajectories at macro level.
And we have more or less sharp volume that defines photon at micro level, but not point particles.

How then does such a photon go through a double slit and create the interference pattern? Could you please describe the trajectories and explain why the resulting patterns are different with one slit closed and with both slits open?

And what happens when such a photon passes a beam splitter?

 

[zonde]

How then does such a photon go through a double slit and create the interference pattern? Could you please describe the trajectories and explain why the resulting patterns are different with one slit closed and with both slits open?

Photons have straight trajectories except at slit. But when photons reach detector (or filter if you have such present) they interfere with state of measurement device (or filter). That state is created by photon ensemble (previous photons) including those photons coming from other slit. Result of that interference in detector is that photon is either absorbed as thermal energy or produce avalanche.
In case of destructive interference on average more photons are absorbed as thermal energy but in case of constructive interference on average more photons produce avalanche.
When one slit is closed state of measurement device is different because photons coming from other slit are absent.

And what happens when such a photon passes a beam splitter?

Nothing. It either goes one way or another with equal probabilities for 50/50 beam splitter (with change in phase by half of wavelength when it undergoes front side reflection).
It's a bit more difficult with polarization beam splitter. Say when photon with polarization angle 30° encounters polarization beam splitter with outputs of 0° and 90° polarizations it undergoes -30° phase shift if it appears in 0° output but 60° phase shift if it appears in 90° output.

 

[A. Neumaier]

Photons have straight trajectories except at slit.

So which trajectory does the photon have after having passed the slit? Straight in a random direction?

But when photons reach detector (or filter if you have such present) they interfere with state of measurement device (or filter). That state is created by photon ensemble (previous photons) including those photons coming from other slit.

This is not enough to guarantee that the first few photons will not place themselves at a position of destructive interference. In a particle picture, the detector can know the inference pattern at best when enough photons had already arrived to display it.

Result of that interference in detector is that photon is either absorbed as thermal energy or produce avalanche.
In case of destructive interference on average more photons are absorbed as thermal energy but in case of constructive interference on average more photons produce avalanche.

When one slit is closed state of measurement device is different because photons coming from other slit are absent.[/QUOTE]
How could the detector know about the difference?

Nothing. It either goes one way or another with equal probabilities for 50/50 beam splitter

This is inconsistent if you apply different polarization filters to the two resulting beams and then recombine the beams.

 

[zonde]

So which trajectory does the photon have after having passed the slit? Straight in a random direction?

Straight according to probabilities of single slit diffraction.

This is not enough to guarantee that the first few photons will not place themselves at a position of destructive interference. In a particle picture, the detector can know the inference pattern at best when enough photons had already arrived to display it.

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

How could the detector know about the difference?

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

This is inconsistent if you apply different polarization filters to the two resulting beams and then recombine the beams.

Would you care to explain this in more details.

 

[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 $$\phi$$ ?