Are clicks proof of single photons?

[A. Neumaier]

I agree that this model explains all photo-electric effect observations known to Einstein: there is a light frequency threshold below which the electron emission is not possible; if the frequency is above the threshold then the probability of emission grows with light intensity, so that there are more electrons emitted at places on the screen, where the interference pattern is constructive. So far, so good.

Good. I'd have been very surprised if top quantum optics experts made mistakes in presenting one of the basic results in their field...

However, there is one experimental observation whose explanation I wasn't able to find. It is known that if the photon energy is just above the threshold then only one photo-electron can be emitted.

How is this known experimentally? Experimentally, one commonly _declares_ (in the tradition of Einstein) that each click (or each silver grain) is supposed to be exactly one photon. Thus nothing is to be explained.

In my previous example only one grain of photoemulsion gets blackened. Of course, it may happen that no photo-electrons are emitted at all. This would simply mean that the photon has passed through the material without interaction.

Or that no photon was present - which is much more likely in a faint coherent state, where one can count single detection events. In this case, the vacuum contribution dominates.

Let us ignore such events. The important thing is that one photon has enough energy to kick out only one electron. There is absolutely no chance that two or three electrons are emitted.

I don't see how this fact is explained in the Mandel & Wolf model. They say that the probability of emission is non-zero no matter how weak is the external potential. This means that each electron on the surface has a non-zero chance to be emitted. Since there are billions of electrons on the surface, we should get some non-zero probabilities for n-electron emissions for any n=1,2,3,...

Since a classical field (or what is produced by a laser) corresponds to a coherent state in a quantum mechanical treatment, the source has in such a treatment contributions from N-photon states for arbitrary N - whence the rare coincidences are correctly accounted for. The quantum mechanical treatment of a coherent source in Chapter 14 exactly reproduces the results of Chapter 9; see Section 14.8.2.

 

[A. Neumaier]

Yes and no, there are situations where currents really ARE due to individual electrons. A trivial (classical) example is the presence of shot noise in any circuit, a somewhat less trivial example is single-electron pumps where the current is due to electrons being "pushed" over a potential barrier one after another (meaning the current is given by I=f*e, f begin the frequency you are "pushing" with)

Even a single electron state has its associated current, which is as delocalized as the electron itself. One measures a tiny current peak, not a delta function. The latter is a coarse-grained approximation made for simplicity, since typically the shape of the peak is irrelevant.

Also, the situation with photons is somewhat similar. Photodetectors and other "classical" devices are of course not very good at determining whether or not photons are discrete particles (although I would argue that one can still draw that conclusion by using them to measure distributions and/or coherence functions). However, there are plenty of CQED systems where TLSs of various types (atoms, ions, qubits etc) have been made to generate and then absorb single photons (in the sense that you can use a TLS to create a Fock state with a single photon).
If you can generate "something" using e.g. a single photon maser and that "something" is exciting a resonator one step, I would argue that this "something" is a discrete entity; i.e. a photon ("if it walks like a duck...").

Of course, the classical field only describes coherent states, not the many sorts of nonclassical quantum states on e is interested in quantum optics. Nevertheless, as discussed in detail in slides 14-29 of my lecture http://arnold-neumaier.at/ms/optslides.pdf
there is nothing inherently discrete in the photon states themselves - the single photons on demands are just localized lumps of the quantum field in a state that is a mixture of 0-particle states and 1-particle states. The discreteness comes from the lumping, not from the state information.

 

[A. Neumaier]

I am not sure what you mean by that. In my opinion, the particle picture is still valid no matter how delocalized the particle's wave function may be. After passing the slit(s) the particle (photon, electron, neutron, atom,...) is still a localized particle.

But then why did you complain about a spread out state no longer be reasonably localized? If you are prepared to call a system in an arbitrarily delocalized 1-particle a localized particle, you shouldn't complain that I do this as well.

 

[A. Neumaier]

Which brings up the situation of deep-field astronomy. Nowadays 'photographs' of perhaps several billion ly distant astronomical objects using CCD detectors is routine. Period between individual events (assumed to be single photon absorption) on a CCD array of ~ a few square cm can be several minutes or more in some cases. [...] Isn't the very fact we can see star-light evidence of light's corpuscular nature, given that random phases of an incoherent Maxwellian radiation source should cancel to effectively zero over such vast distances?

I don't think one needs quantum mechanics to explain this, except at the final stage of detection, where the scenario of M&W Chapter 9 suffices.

Astronomical observations are very well explained by geometric optics, together with the Poisson statistics of the recording device, with Poisson parameter proportional to the incident intensity of the radiation.

 

[A. Neumaier]

Isn't the very fact we can see star-light evidence of light's corpuscular nature, given that random phases of an incoherent Maxwellian radiation source should cancel to effectively zero over such vast distances?

I forgot to address the particular reason you gave. Intensity is determined by the mean square deviations, not of the mean field. Cancellations only remove the coherence (i.e., make the mean field term vanish), but don't affect much the intensity integrated over an area large compared to the wavelength squared.

 

[harrylin]

Yes, it looks very different from the imagery Feynman invokes in his lectures. But it is fully based on the most orthodox quantum mechanics.

I was very surprised about that, too, when I started to dig deeper into quantum optics.

I never claimed that ["collapse"] happens instantaneously. Chemical reactions take
time. Instantaneous collapse is one of the fairy tales of the kindergarden textbooks.
[..]
In any given time interval Delta t, each of the zillions of potential photoelectrons on the photographic plate has a certain probability of firing. This probability is calculated in the standard, orthodox way using a quantum mechanical model of an array of independent electrons (shape doesn't matter) interacting with the classical electromagnetic field - a simple N-electron model with a time-dependent interaction. Note that no photons are present in the model!

The probability turns out to be proportional to the incident field strength and to the Delta t. Therefore, eventually, one of these zillions of potential photoelectrons fires - most likely one in the part of interference region where the intensity is high, least likely one where the intensity is low. This accounts for the interference pattern if you observe enough electrons by making Delta t large enough. The fired electron starts a chemical reaction, which, when completed, makes the effect irreversible and causes the collapse of the whole system. For classical coherent light as produced by a faint laser, this happens randomly according to a Poisson distribution.


In the model, there is no particle causing an electron to fire. There is only a time-dependent external potential, incident with different intensities at _every_ electron on the plate. But the model explains all the stuff Einstein knew when he proposed his particle interpretation.
[..]
the particle picture simply stops to be applicable when the field is no longer reasonably localized (according to whatever definition of ''reasonable'' you'd like to propose). One could equivalently say that the photon loses its particle character after passing the slit. This doesn't change the situation. In quantum field theory, you always have the field, and in some cases you are entitled to talk about the particle. Since the latter is only a semiclassical notion, you have the choice of thinking of a thinly spread out particle or no particle at all.

Thank you, your explanations help me to make sense of QM.

Therefore, I would appreciate your comment on atom interferometers (see post #25)...
Would you say that atoms are simply much more localised? But then, how can they still significantly interfere, and would not in this case, clicks really correspond to atoms?

 

[Q-reeus]

I forgot to address the particular reason you gave. Intensity is determined by the mean square deviations, not of the mean field. Cancellations only remove the coherence (i.e., make the mean field term vanish), but don't affect much the intensity integrated over an area large compared to the wavelength squared.

Agreed. But let's consider the sun as model of a typical star. From http://en.wikipedia.org/wiki/Sun and http://en.wikipedia.org/wiki/File:EffectiveTemperature_300dpi_e.png we can gather a few facts:
Sun's output power W = 3.846×10²⁶ Watt
Surface temp = 5,778 K
Mean wavelength ~ 700 nm -> mean frequency f ~ 3*10⁸/(7*10^-7) = 4.3*10¹⁴ Hz
Then using E = hf = 6.62606896*10⁻³⁴*4.3*10¹⁴ = 28.5*10^-20 joule (energy per 'average' solar photon)
Total solar output of average photons/sec = W/E = 3.846*10²⁶/(2.85*10^-19) = 1.35*10⁴⁵ photons/sec
Total # photons/average wavelength = (W/E)*7*10^-7/(3*10⁸) = 1.35*10⁴⁵*2.333*10^-15 = 3.15*10³⁰

This last figure is of course just a very rough guide, but indicates that assuming classical EM model of photon propagating as spherical wavefront, there is out to infinity a fixed and enormous average density of random phase photons crammed into any given very small wavefront spatial interval. All that happens at very large distances in this picture is a dilution of energy density, but NOT photon number density, in stark contrast to the corpuscular photon picture.
And how much mean square deviation would there be, even allowing that the real situation is one of a continuous near black body distribution (with however nearly half the total energy confined to the visible range of frequencies)? Very very little I would suggest. Which, by the conservation of energy, further suggests that stars in general aught by this 'semi-classical' reasoning be extremely poor radiators, with the (obviously incorrect) expectation of the need to radiate at probably x-ray wavelengths. No such dilemma for the corpuscular photon model.

Do you really believe that at astronomical distances some infinitesimal fraction of each of ~ 3*10^30 random phase photons/wavelength can excite anything - CCD plate, photographic emulsion, or whatever? In #29 I wrote "The 'photon energy ratio' A_ccd/(4*pi*r²) (A_ccd being the CCD device effective area, r being the distance between source and detector) here is utterly minute. The source is completely incoherent radiation. How can this possibly be consistent with photons propagating as spherical waves in vacuo a la Maxwell's Equations? The only remote possibility might be to assume the CCD plate acts as a completely lossless accumulator of energy for minutes of more, until some threshold energy initiates a quantum 'spike'." That part was excised from your reply in #34, but I believe is relevant. Can you explain in 'plain English' how this would be feasible given the extremely compressed and totally random character of the extremely dilute incident radiation? Please don't quote chapters from M&W as I don't have access to that tome.

 

[A. Neumaier]

Therefore, I would appreciate your comment on atom interferometers (see post #25)...
Would you say that atoms are simply much more localised? But then, how can they still significantly interfere,

They are not much more localized but they are much heavier. This makes interference more difficult, since it is far more difficult to keep them coherent.

They also follow a Klein-Gordon or Dirac equation rather than a Maxwell equation.
But these are stiil wave equations, so the qualitiative behavior - a superposition of two spherical waves after passing the slits, and the resulting form of the interference pattern - is the same.

would not in this case, clicks really correspond to atoms?

Since Newton's time, there have been two competing pictures for the behavior of light and matter - particles and waves (or rather fields). Some phenomena can be reasonably interpreted in both pictures. But on the submicroscopic level, where experience can only be indirect, both pictures are somewhat limited because a quantum field - the reality underlying it all - is too abstract to be easily visualized.

If one chooses the particle picture, one earns paradoxes about being nowhere unless measured, passing to through both slits or only one, explaining why an interference pattern exists at all, etc.. There is no way to get a particle picture show wave effects - all that becomes very counterintuitive. (But the Copenhagen brain washing still shows its power - people became used to the fact that these alleged ''particles'' are very strange objects.) On the other hand, waves can simulate particle properties by regarding the latter as wave packets - localized excitations of waves.

Now light showed obvious interference effects, difficult to interpret with particles, and (polarization effects, impossible to explain by even the most contrived particle picture. As a result, since around 1850 if not earlier (the double slit experiment dates from 1801), physicists universally adopted the wave view of light.

For matter, the development was much slower since the wave properties of matter such as electron diffraction http://en.wikipedia.org/wiki/Electron_diffraction were only discovered after the advent of quantum mechanics. But here also the field view proved to be more versatile and won the competition. Modern QED (the most accurate of all physical theories) and all high energy physics (based on the standard model and derived field theories) are framed in the language of quantum field theory, where quantum fields are the primary objects and particles are regarded as localized excitations of these fields.

So. if one knows that one produced single photons or atoms at a time, as localized wave packets, it is appropriate to talk about particles. In that case, the wave arriving and the particle arriving are synonymous.

But their fate as particles after passing a slit more narrow than their wavelength (or a beam splitter and similar devices) becomes dubious since the slit delocalizes the field to an expanding spherical wave. Now this wave arrives at a far away detector at all places simultaneously, while the detector responds randomly at one place. Thus the particle view and the wave view diverge, and only the latter is a reasonable (i.e., semiclassical) explanation of what happens.

... except if one adheres to a strict Copenhagen view, according to which one cannot assert anything about a quantum system when it is not observed. This view has severe difficulties, though, once the quantum system becomes big enough to behave classically.

 

[A. Neumaier]

Total # photons/average wavelength = (W/E)*7*10^-7/(3*10⁸) = 1.35*10⁴⁵*2.333*10^-15 = 3.15*10³⁰
This last figure is of course just a very rough guide, but indicates that assuming classical EM model of photon propagating as spherical wavefront, there is out to infinity a fixed and enormous average density of random phase photons crammed into any given very small wavefront spatial interval. All that happens at very large distances in this picture is a dilution of energy density, but NOT photon number density, in stark contrast to the corpuscular photon picture.

In the quantum version, this is not different. In QED, there is no concept of a photon number density, whereas the local mean energy density is still well-defined.

And how much mean square deviation would there be, even allowing that the real situation is one of a continuous near black body distribution (with however nearly half the total energy confined to the visible range of frequencies)? Very very little I would suggest.

It is very little if and only if very little energy arrives. For the energy density _is_ the mean squared amplitude of the electromagnetic field.

The only remote possibility might be to assume the CCD plate acts as a completely lossless accumulator of energy for minutes of more, until some threshold energy initiates a quantum 'spike'."

No. It is enough that the detector elements on the plate respond locally according to a Poisson process with probability rate determined by the incident intensity. This means it fires randomly at the rate determined at each moment from the incident faint field. No memory is needed, and energy loss is irrelevant (except for the efficiency of the process). The local detector elements will respond independently and rarely but occasionally, and waiting long enough will precisely reproduce the averaged intensity profile - the goal of the imaging exercise.

Can you explain in 'plain English' how this would be feasible given the extremely compressed and totally random character of the extremely dilute incident radiation?

The response has nothing to do with the randomness in the incident field. The phase is random but the energy density arriving is not (though it could be - in any case, one would see only the average effect). The latter is small but - as the experimental results prove - noticeable.

 

[A. Neumaier]

there is out to infinity a fixed and enormous average density of random phase photons crammed into any given very small wavefront spatial interval. All that happens at very large distances in this picture is a dilution of energy density, but NOT photon number density, in stark contrast to the corpuscular photon picture.

To be very explicit: It doesn't make sense to count photons classically and pretend that each one when created in a distant star is a spherical wave spreading out through space to be ''collapsed'' when entering the CCD detector. The detector doesn't see myriads of these extremely faint spherical waves and decides to collapse just one of them. Instead, it ''sees'' the energy density, and feels (according to the value) more or less ''motivated'' to respond. The reason is that in QED, the local mean energy density is an observable field, whereas the concept of a photon number density cannot even be defined.

 

[PhilDSP]

Since Newton's time, there have been two competing pictures for the behavior of light and matter - particles and waves (or rather fields). Some phenomena can be reasonably interpreted in both pictures. But on the submicroscopic level, where experience can only be indirect, both pictures are somewhat limited because a quantum field - the reality underlying it all - is too abstract to be easily visualized.

Aren't there really three competing pictures? (The third is particle plus wave along the lines of De Broglie Theory)

After reading the lecture material you linked to last week I thought "certainly he sees the necessity of wave plus particle". At least what I thought I understood about what you called "Photon Field State" was either the wave or the possible wave channel which a separate particle could pass through.

 

[A. Neumaier]

Aren't there really three competing pictures? (The third is particle plus wave along the lines of De Broglie Theory)

This picture doesn't exist since Newton's time. Though it enjoys some popularity with a minority of quantum physicists (including some who are active on PF), I don't consider it to be a serious alternative. For the reasons, see http://de.arxiv.org/abs/quant-ph/0001011

After reading the lecture material you linked to last week I thought "certainly he sees the necessity of wave plus particle".

No. I see the sufficiency of standard quantum mechanics. Bohmian mechanics adds nothing to understand the photoelectric effect:

There is no Bohmian mechanics version of QED using electron, positron and photon particles.

There is no Bohmian mechanics explanation of the photoelectric effect by means of photon particles absorbed by an array of electrons.

Photon absorption in Bohmian mechanics requires giving up the deterministic nature of the standard Bohmian approach, even in a simplified toy model:
Detlef Dürr et al 2003 J. Phys. A: Math. Gen. 36 4143
http://arxiv.org/pdf/quant-ph/0208072

There is no Bohmian mechanics version of photon absorption consistent with Maxwell's equation, though models for Bohmian photons with a Maxwell pilot wave exist: http://arxiv.org/pdf/0907.2667

But even if this were improved, I don't see any reason for postulating unobservable Bohmian ghost particles with nonlocal instantaneous influences over galactic spacetime regions, just in order to explain seeing the light of stars.

 

[meopemuk]

How is this known experimentally? Experimentally, one commonly _declares_ (in the tradition of Einstein) that each click (or each silver grain) is supposed to be exactly one photon. Thus nothing is to be explained.


Since a classical field (or what is produced by a laser) corresponds to a coherent state in a quantum mechanical treatment, the source has in such a treatment contributions from N-photon states for arbitrary N - whence the rare coincidences are correctly accounted for. The quantum mechanical treatment of a coherent source in Chapter 14 exactly reproduces the results of Chapter 9; see Section 14.8.2.

Let us not discuss coherent states with undetermined numbers of photons. Such states introduce unnecessary complications. Let us discuss sources, which emit one and only one photon at a time. Such sources are not difficult to prepare, at least, theoretically. We can take one atom with a radioactive nucleus (possibly embedded in a piece of inert material). This nucleus decays and emits exactly one gamma photon. If the energy of the photon is too high to be used in our double-slit experiment, then we can always make our source moving, so that the energy (=frequency) is Doppler-shifted to any desired value.

Let us also assume that each emitted photon is captured by the photographic plate or the "charge coupled device" or whatever detector we have chosen. Then the rule "1 photon = 1 click" must be strictly obeyed. I don't see how Maxwell's continuous field representation of the photon is going to enforce this rule.

Eugene.

 

[A. Neumaier]

Let us not discuss coherent states with undetermined numbers of photons. Such states introduce unnecessary complications.

Of course a classical external field can fully capture only the properties of classical light, which is modeled on the quantum level by coherent states. But this is enough to show that the interpretation of clicks as photons is a pure convention, not a consequence of easily available experimental evidence.

Let us discuss sources, which emit one and only one photon at a time. Such sources are not difficult to prepare, at least, theoretically.

Theoretically, perhaps. Practically, it is quite difficult to create approximate 1-particle photon states. See the entry ''What is a photon?'' in Chapter B2 of my theoretical physics FAQ at http://arnold-neumaier.at/physfaq/physics-faq.html#photon

We can take one atom with a radioactive nucleus (possibly embedded in a piece of inert material). This nucleus decays and emits exactly one gamma photon.

First you need to bring the nucleus into an excited state. How do you do that in a way that you can predict or check when this single nucleus decayed?

If the energy of the photon is too high to be used in our double-slit experiment, then we can always make our source moving, so that the energy (=frequency) is Doppler-shifted to any desired value.

Then you need to make sure that the decay happens just at the right time that the photon jumps from the fast moving source precisely into the detector! But radioactive decay is a random process!

Let us also assume that each emitted photon is captured by the photographic plate or the "charge coupled device" or whatever detector we have chosen.

This is again very difficult to achieve experimentally. You are heavily idealizing the situation, just in order to have a situation that is favorable to your particle-only ideas!

But ok, let us assume this.

Then the rule "1 photon = 1 click" must be strictly obeyed. I don't see how Maxwell's continuous field representation of the photon is going to enforce this rule.

In this case, I can't tell what happens with a photographic plate. But I can tell you what happens with an array of photomultipliers.

Because you send one photon at a time, and because you make sure that it is captured (which you can be sure only by observing a click), energy conservation (together with the information about the momentum of the emitted electron - which can be predicted by a quantum calculation as in Chapter 9 of M&W and can be measured in principle) tells that exactly one electron must have fired.

 

[meopemuk]

But then why did you complain about a spread out state no longer be reasonably localized? If you are prepared to call a system in an arbitrarily delocalized 1-particle a localized particle, you shouldn't complain that I do this as well.

The difference is that in my case the delocalized "thing" is the wave function, which is not a material substance, but sort of unvisible imaginary "probability amplitude density". The wave function simply tells us where the particle can be found and with what probability. When we actually find the particle (the photon or whatever) we always find it as a small indivisible lump of concentrated energy.

In your approach (if I understand it correctly) you insist that one photon itself is a kind of spread-out material substance. The energy of one photon can be distributed evenly across hundreds of light years. This raises the question of how this thinly spread distribution of energy collapses to one point, when we observe it by means of photographic plates or CCD devices?

In other words, there still exists the problem known as the "wave function collapse". Quantum approach to photons has its own way of solving this problem, which we are not discussing here. But if you insist on the classical description of the "photon's electromagnetic field" you must explain how *you* describe the collapse phenomenon. It is definitely not in Maxwell equations, and M&W example doesn't do a good job either, in my opinion.

Eugene.

 

[meopemuk]

Since Newton's time, there have been two competing pictures for the behavior of light and matter - particles and waves (or rather fields).

And it appears that this epic battle still rages on. Fascinating!

If one chooses the particle picture, one earns paradoxes about being nowhere unless measured, passing to through both slits or only one, explaining why an interference pattern exists at all, etc.. There is no way to get a particle picture show wave effects - all that becomes very counterintuitive. (But the Copenhagen brain washing still shows its power - people became used to the fact that these alleged ''particles'' are very strange objects.)

All these paradoxes are explainable in the particle picture.

... except if one adheres to a strict Copenhagen view, according to which one cannot assert anything about a quantum system when it is not observed. This view has severe difficulties, though, once the quantum system becomes big enough to behave classically.

Yes, this is the only way to do physics without paradoxes, in my opinion. The rule is that one cannot assert anything about something which has not been observed. Only results of observations can be discussed scientifically. This refers to all kinds of systems - small or big.

Eugene.

 

[meopemuk]

Of course a classical external field can fully capture only the properties of classical light, which is modeled on the quantum level by coherent states. But this is enough to show that the interpretation of clicks as photons is a pure convention, not a consequence of easily available experimental evidence.

Do I understand you correctly? This means that M&W arguments cannot explain detection of single-photon states (which can be called "quantum light" as opposed to multi-photon "classical light")? Then we are in agreement.

First you need to bring the nucleus into an excited state. How do you do that in a way that you can predict or check when this single nucleus decayed?

For example, I can shoot a single neutron at a piece of proper material (I hope you wouldn't deny that a single-neutron state can be predictably prepared.) The neutron gets absorbed by only one nucleus. This nucleus goes to an excited state, which decays later by emitting just one gamma quantum. I don't know nuclear physics so well, but I guess that such a sequence of events can be arranged.

Then you need to make sure that the decay happens just at the right time that the photon jumps from the fast moving source precisely into the detector! But radioactive decay is a random process!

If the lifetime of the nucleus is 10 hours I can wait for 10 hours, no problem. If you don't like the idea of the source receding at high velocity, then I can place my radioactive nucleus on a rotating platform. Or I can place it in a strong gravitational potential, so that the frequency (energy) of the emitted photon is red-shifted. There are various realistic ways to achieve the goal.

Because you send one photon at a time, and because you make sure that it is captured (which you can be sure only by observing a click), energy conservation (together with the information about the momentum of the emitted electron - which can be predicted by a quantum calculation as in Chapter 9 of M&W and can be measured in principle) tells that exactly one electron must have fired.

But M&W Chapter 9 does not guarantee the emission of only one electron. I don't see such a guarantee there. Besides, the energy conservation argument doesn't work. In their model the oscillating classical potential serves basically as an unlimited reservoir of energy. It doesn't respond (e.g., by changing the amplitude of oscillations) when an electron is emitted and a quantum of energy is absorbed by the electron subsystem.

Eugene.

 

[A. Neumaier]

The difference is that in my case the delocalized "thing" is the wave function, which is not a material substance, but sort of unvisible imaginary "probability amplitude density". The wave function simply tells us where the particle can be found and with what probability. When we actually find the particle (the photon or whatever) we always find it as a small indivisible lump of concentrated energy.

No. Even according to a hypothetical Hamiltonian multiparticle theory accounting for photodetection, one could ''find'' a photon in this way only when it no longer exists! The very act of attempting to observe a photon kills it!

Therefore, one cannot use observations to prove the existence of photons! They are ghosts conjured by Einstein and Dirac! Whereas in quantum field theory, one just has states of the electromagnetic field, and the number of photons in a given region is a
non-observable.

In your approach (if I understand it correctly) you insist that one photon itself is a kind of spread-out material substance.

No. The electromagnetic field is a kind of spread-out material substance. A photon is just a semiclassical epiphenomenon of the field.

The energy of one photon can be distributed evenly across hundreds of light years.

Formally, yes, but photon number is meaningless in QED. Trying to insist on its meaning creates the infrared problems, since single photons spread out across hundreds of light years or further exist in infinite amounts. What is meaningful is only the electromagnetic field they belong to.

This raises the question of how this thinly spread distribution of energy collapses to one point, when we observe it by means of photographic plates or CCD devices?

There is no such collapse. The fields simply ends at its boundary, and its energy/momentum distribution creates a propensity for the valence electrons to move. Sometimes one of them actually moves is detected, and we say (by convention) that a full, discrete photon arrived. Whereas we only have proof that an electron arrived.

But if you insist on the classical description of the "photon's electromagnetic field"

No. I don't insist on a classical description of the electromagnetic field. The correct description is by the quantum field of QED. The classical description is well-known to be only an approximation, valid for coherent radiation fields.

I insist, however, on the observation that the fact that a quantum detector responds to a classical radiation field (whose modeling does not involve the concept of a photon) with clicks that reproduce the photoelectric effect is a convincing demonstration that clicks cannot be proof of single photons arriving.

Instead, the clicks are measurements of the intensity of a continuous field by an array of binary detectors. It is obvious that this can only give a random and inaccurate measurement of the field.

It is like measuring the amount of water flowing over a water mill by the number of times you hear a spike pass some obstacle. When the creek is fast-flowing, this gives a reliable measurement, but if the creek trickles only, the spike makes its sound at odd times.
Fortunately, nobody takes this as proof that the water consists of bucket-sized particles!

This classical parable has its limits, but according to this parable, a photodetector is sort of a quantum mill for electromagnetic radiation.

 

[meopemuk]

I insist, however, on the observation that the fact that a quantum detector responds to a classical radiation field (whose modeling does not involve the concept of a photon) with clicks that reproduce the photoelectric effect is a convincing demonstration that clicks cannot be proof of single photons arriving.

Instead, the clicks are measurements of the intensity of a continuous field by an array of binary detectors. It is obvious that this can only give a random and inaccurate measurement of the field.

It is like measuring the amount of water flowing over a water mill by the number of times you hear a spike pass some obstacle. When the creek is fast-flowing, this gives a reliable measurement, but if the creek trickles only, the spike makes its sound at odd times.
Fortunately, nobody takes this as proof that the water consists of bucket-sized particles!

This classical parable has its limits, but according to this parable, a photodetector is sort of a quantum mill for electromagnetic radiation.

I think I understand what you are saying. And I agree that in the M&W model the detector will produce distinct random clicks even if the incident radiation is modeled by a continuous classical oscillating potential.

My only disagreement is that this model cannot explain the "1 photon = 1 click" rule. I insist that 1-photon states can be, in principle, prepared, and that this rule can be verified experimentally. So, the M&W model cannot explain all experimental data.

Eugene.

 

[meopemuk]

photon number is meaningless in QED.

I disagree. The most basic calculation in QED is the calculation of the Compton scattering cross-section. It assumes explicitly the initial state consisting of *exactly* one photon and *exactly* one electron. So, the photon number has a very precise meaning there.

Well, you can argue that this is true only for asymptotic states where the electron and the photon are not interacting. But in this thread we are discussing non-interacting free-propagating photons as well. So, even within QED formalism the photon number is well-defined for our purposes.

Eugene.

 

[harrylin]

[..]
So. if one knows that one produced single photons or atoms at a time, as localized wave packets, it is appropriate to talk about particles. In that case, the wave arriving and the particle arriving are synonymous.

But their fate as particles after passing a slit more narrow than their wavelength (or a beam splitter and similar devices) becomes dubious since the slit delocalizes the field to an expanding spherical wave. Now this wave arrives at a far away detector at all places simultaneously, while the detector responds randomly at one place. Thus the particle view and the wave view diverge, and only the latter is a reasonable (i.e., semiclassical) explanation of what happens. [..]

Now here you lump photons and atoms together in a single description, and this is what I have an issue with (due to my intuition, which may be right, or wrong).
For I have no problem to imagine that a photon isn't really a particle, so that it doesn't need to be localised; that fits rather well with "classical" concepts and anyway, nobody has ever really seen a photon and you can't hold a bunch of photons in your hand.

But it's different with atoms: you can hold them in your hand and count them with a microscope. Surely you do not suggest that atoms can be split or destroyed by slits, or do you? It would for me be a revolutionary way of splitting the atom!
Please clarify.

 

[A. Neumaier]

And it appears that this epic battle still rages on. Fascinating!

Yes. And the future might perhaps view our discussion here as one of the last few rearguard actions of the particle faction.

Do I understand you correctly? This means that M&W arguments cannot explain detection of single-photon states (which can be called "quantum light" as opposed to multi-photon "classical light")? Then we are in agreement.

The M&W arguments in Chapter 9 do not try to explain detection of single-photon states. It is not even possible to state in the model used in Chapter 9 what a single-photon state should be (except in the wave packet approximation described below).

M&W treat the full QED case in Chapter 14, where they show that for coherent states, one gets exactly the results from Chapter 9. For other states, they find deviations from the Poisson distribution and differences in the distribution of the time between two clicks.

The result is very far from your postulated idealized 1-1 correspondence between photons and clicks (which is wishful thinking without an experimental basis). They get (after some approximations - the formulas are not valid for sufficiently long times) the formula (14.8-16) for the probability of getting n clicks in a time interval of length T, given an m-photon Fock state in which only a single momentum mode is occupied. For n>m, this is zero.

For example, I can shoot a single neutron at a piece of proper material (I hope you wouldn't deny that a single-neutron state can be predictably prepared.)

Well, how would you do it reliably?

The neutron gets absorbed by only one nucleus.

The problem is that it gets absorbed by at most one nucleus. And you don't know whether or not it does.

This nucleus goes to an excited state, which decays later by emitting just one gamma quantum. I don't know nuclear physics so well, but I guess that such a sequence of events can be arranged.

According to conventional wisdom, the resulting gamma quant is emitted in a random direction. And you don't know whether it is in the direction of your detector unless you record it there!

If the lifetime of the nucleus is 10 hours I can wait for 10 hours, no problem.

The decay happens at an unpredictable random time. After 10 hours, the probability of emission is only a factor of 1/e.

If you don't like the idea of the source receding at high velocity, then I can place my radioactive nucleus on a rotating platform.

Well, with the rotating platform, it won't be easier to hit the target detector!

Or I can place it in a strong gravitational potential, so that the frequency (energy) of the emitted photon is red-shifted.

The potential would have to be so strong that it swallows all your recording equipment!

There are various realistic ways to achieve the goal.

So far, your proposals didn't sound even remotely realistic.

But M&W Chapter 9 does not guarantee the emission of only one electron. I don't see such a guarantee there.

They don't since under their assumptions (coherent input light) there is a small but nonzero probability for emitting more than one electron in an arbitrarily short time interval. And their results are very well in agreement with experiment! Thus the requested guarantee would not conform to experimental reality.

Besides, the energy conservation argument doesn't work. In their model the oscillating classical potential serves basically as an unlimited reservoir of energy.

Yes, but they treat a coherent state, not a single photon. A coherent state produces an unlimited number of photons if you wait long enough. (This is why you can't represent it in Fock space.)

But if only a single photon arrives (as you tried to arrange), there is a definite total energy. In the classical setting the single photon appears as a modulated wave packet, with nonzero energy density only for the time it took your nucleus to decay, and an integral that matches the 1-photon energy hbar times omega. Then the energy conservation argument works. (Their analysis generalizes since they split the time
anyway into many small time intervals, where the intensity can be considered constant. only the integration would be different, and hence the probability distribution.)

 

[A. Neumaier]

My only disagreement is that this model cannot explain the "1 photon = 1 click" rule. I insist that 1-photon states can be, in principle, prepared, and that this rule can be verified experimentally.

Please cite a paper that verified this rule.

As I mentioned in my just finished post, the QED result (which agrees with experiment) is different from what your rule claims.

 

[A. Neumaier]

I disagree. The most basic calculation in QED is the calculation of the Compton scattering cross-section. It assumes explicitly the initial state consisting of *exactly* one photon and *exactly* one electron. So, the photon number has a very precise meaning there.

Only in the tree approximation. If you compute radiative corrections you get infrared divergences due to an unlimited number of soft photons. Phys. Rev. 85, 231–244 (1952).

And if you look at how Compton scattering is experimentally verified, you see that they use not 1-photon states but ordinary laser light, which form coherent states
(or even ordinary light, which is a mixture of coherent states).

 

[meopemuk]

But it's different with atoms: you can hold them in your hand and count them with a microscope. Surely you do not suggest that atoms can be split or destroyed by slits, or do you? It would for me be a revolutionary way of splitting the atom!
Please clarify.

Surely, atoms cannot be split by the slits. Each atom exists as one localized entity each time we look at it. Then you may say that in order to interfere on the double-slit setup the atom must go through both slits simultaneously. How these two views can co-exist?

My answer is that when the atom is "passing through slits" we are not actually looking at the atom. The conclusion about "passing simultaneously through both slits" is not a result of direct measurement but a result of some logical reasoning. But logical speculations have a lesser value than direct experimental observations. So, it would be a safer bet to stick to atoms as single indivisible localized entities, which are seen in experiments.

If you ask me: "but what about atoms passing through slits"? My answer would be "I don't know". In science I am not obliged to answer questions, whose answers cannot be directly verified by experiment. "But what about atoms passing through slits"? is exactly this type of question. There is no shame if I answer simply "I don't know". Yes, I can introduce atom wave function and calculate the probability density and predict the shape of the interference picture (i.e., the final outcome of the experiment). But this does not mean that I *know* what atoms are doing while passing through the slits.

By the way, the same reasoning should be applied to photons. Each time we actually observe individual photons with sufficiently high resolution (using photographic plates, photo-multipliers, CCD devices, etc) we see them as separate countable undivisible particles. So, it is a safer bet to assume that this is exactly what the photons are. The wave properties appear in situations when no actual observations are made (e.g., when a photon passes through two slits). So, just as in the case of atoms above, we can describe these wave properties by an abstract wave function, which interferes with itself, collapses, and does various nasty (but not troubling) things while we are not looking.

Eugene.

 

[A. Neumaier]

Now here you lump photons and atoms together in a single description, and this is what I have an issue with (due to my intuition, which may be right, or wrong).

They are all described by wave equations, though different ones due to their different masses and spins. But they behave quite similarly.

For I have no problem to imagine that a photon isn't really a particle, so that it doesn't need to be localised; that fits rather well with "classical" concepts and anyway, nobody has ever really seen a photon and you can't hold a bunch of photons in your hand.

In quantum field theory, a photon is as much or as little a particle as an electron, depending on your precise definition of a particle.

You can see a photon as a flash in your eyes, whereas you cannot see an atom.

But it's different with atoms: you can hold them in your hand and count them with a microscope.

You can hold a crystal in your hands but not an atom.

If you look at an atom under an field emission microscope or a similar high resolution device, you see a continuous surface whose bulges are interpreted as atoms. This is just what one would expect from a field! A macroscopic body with many macroscopic bulges is indeed not viewed as being composed of macroscopic balls (''atoms'') but as a matter field described by elasticity theory.

Surely you do not suggest that atoms can be split or destroyed by slits, or do you?

No. The atom field splits at the slits and unites afterwards again, just like a water wave or any other wave.

In an atom interferometer, you have a diluted atom field in the form of a high-frequency beam, which moves through the splits like any other field, and expands behind the slits as a superposition of two spherical waves. Then it reaches the screen and causes some detectors to fire, with probability rates proportional to the incident field energy. This is precisely the same as what happens with the photon field; except that the internal degrees of freedom of atoms and photons are different. (But these are not relevant in a standard double slit experiment)

 

[meopemuk]

Well, how would you do it reliably?

The problem is that it gets absorbed by at most one nucleus. And you don't know whether or not it does.

According to conventional wisdom, the resulting gamma quant is emitted in a random direction. And you don't know whether it is in the direction of your detector unless you record it there!

The decay happens at an unpredictable random time. After 10 hours, the probability of emission is only a factor of 1/e.

Well, with the rotating platform, it won't be easier to hit the target detector!

The potential would have to be so strong that it swallows all your recording equipment!

So far, your proposals didn't sound even remotely realistic.

These are all technical issues. Let's leave them to experimentalists.

But if only a single photon arrives (as you tried to arrange), there is a definite total energy. In the classical setting the single photon appears as a modulated wave packet, with nonzero energy density only for the time it took your nucleus to decay, and an integral that matches the 1-photon energy hbar times omega. Then the energy conservation argument works. (Their analysis generalizes since they split the time
anyway into many small time intervals, where the intensity can be considered constant. only the integration would be different, and hence the probability distribution.)

Can we then agree that single photon states can be prepared, at least in principle? I don't see how you can disagree, because a 1-photon sector is present even in QED. Since we are talking about non-interacting freely propagating light here, you cannot use arguments about non-Fock spaces.

So, we take this one-photon state and let it pass through a double slit setup and let it fall on a CCD detector. According to you, after passing the slits the photon's electromagnetic wave diverges, so that it covers the whole detector area. You also say that there is a non-zero probability that more than one pixel will catch a signal. Is this what you're saying?

On the other hand, my claim is that no more than one pixel will "click". Photon is a particle that cannot be split in smaller pieces.

I don't know experiments performed exactly in this way, but I'll try to look for them. Only one of us can be correct. Hopefully, we can resolve this contradiction peacefully, without engaging in rearguard skirmishes.

Eugene.

 

[meopemuk]

In an atom interferometer, you have a diluted atom field in the form of a high-frequency beam, which moves through the splits like any other field, and expands behind the slits as a superposition of two spherical waves. Then it reaches the screen and causes some detectors to fire, with probability rates proportional to the incident field energy. This is precisely the same as what happens with the photon field; except that the internal degrees of freedom of atoms and photons are different. (But these are not relevant in a standard double slit experiment)

Instead of using detectors, which respond to the "field energy", I can simply allow the atoms to be deposited on the screen/substrate. So that over time thicker atomic layers grow in regions of constructive interference and there is less stuff deposited in the regions of descructive interference. How would you then deny that each atom passed through the slits landed at a very specific point on the substrate? So, atoms behave exactly as individual particles. The only thing that behaves continuously in this case is the probability density for each atom to land at a specific point. This probability density is obtained from the regular one-particle quantum mechanical wave function. There is no need to invoke fields, either classical or quantum.

Eugene.

 

[A. Neumaier]

Can we then agree that single photon states can be prepared, at least in principle?

The standard field interpretation of QED doesn't depend on this. The states commonly prepared by Nature or by physicists are far from single photon states. Usually they are coherent states or squeezed states, or entangled tensor products of such states.

The best I know is that it is possible to create mixtures of superpositions of the vacuum state and 1-photon states in the sense of my lecture http://arnold-neumaier.at/ms/lightslides.pdf

So, we take this one-photon state and let it pass through a double slit setup and let it fall on a CCD detector. According to you, after passing the slits the photon's electromagnetic wave diverges, so that it covers the whole detector area. You also say that there is a non-zero probability that more than one pixel will catch a signal. Is this what you're saying?

No. Given a mixture of superpositions of the vacuum state and 1-photon states, there is a nonzero probability for recording some event, while the probability for more than one event is exactly zero. This can be deduced from the estimate in M&W (14.8-16).

On the other hand, my claim is that no more than one pixel will "click".

Here we agree.

Photon is a particle that cannot be split in smaller pieces.

This is irrelevant. It is neither needed to derive the probability statement, nor does it follow from it.

 

[A. Neumaier]

Instead of using detectors, which respond to the "field energy", I can simply allow the atoms to be deposited on the screen/substrate. So that over time thicker atomic layers grow in regions of constructive interference and there is less stuff deposited in the regions of destructive interference. How would you then deny that each atom passed through the slits landed at a very specific point on the substrate?

I deny this on the basis of several facts:
1. The interpretation rules are inconsistent with the atomic nature of the atom:
An atom is indivisible (at the energies of the experiment) and can therefore go only through one of the slits. Assuming it went through the left or the right slit, the atoms' arrival pattern would (according to the same arguments you give for your interference pattern) have to be different from what is observed. Contradiction.
2. One cannot say how each atom arrived at a constructive interference spot rather than at a destructive interference spot. The atom comes from the source, is nonexistent (or in a state of limbo where it has no physical property at all) during a period of time corresponding to the flight of time to the screen, and then miraculously is called into existence by the detector, consistent with an interference pattern of which neither the atom nor the detector knows the slightest thing. This is far too miraculous a story to believe, though this fairy tale has been told umpteen times since the old Copenhagen times. The emperor wears no clothes, but nobody dares to say so.

The quantum field view has none of these problems.

So, atoms behave exactly as individual particles.

They would need a Bohmian guiding field in order to behave like that. I reject this for the reasons given in post #42.