Are clicks proof of single photons?

[A. Neumaier]

I hope you wouldn't argue that the current in metals is produced by discrete electrons. Continuous charge and current fields are just very rough approximations. The discrete nature of the radiation field was demonstrated by Einstein in his 1905 paper about the photo-electric effect.

In the semiclassical picture known to Einstein 1905, currents are produced by discrete electrons. But now, 100 years later, this picture is known to be approximate only, and that currents in metals are in fact produced by the continuous electron fields of QED.
Discrete semiclassical particles are just very rough approximations.

Moreover, the argument of Einstein put forward for the discrete nature of radiation is spurious. As you can read in the standard reference for quantum optics,
L. Mandel and E. Wolf,
Optical Coherence and Quantum Optics,
Cambridge University Press, 1995.
the clicks in a photon detector are an artifact of photodetection caused by the quantum nature of matter, rather than proof of single photons arriving.

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.

 

[Dr know]

i think a lot of people will probably seem more curious as to how and why they can be proof of single photons this is the first i have heard of this and my curiosity level as just sky rocketed to another level

 

[A. Neumaier]

i think a lot of people will probably seem more curious as to how and why they can be proof of single photons this is the first i have heard of this and my curiosity level as just sky rocketed to another level

The photoelectric effect (http://en.wikipedia.org/wiki/Photoelectric_effect ) is usually explained (following Einstein, who received the Nobel price for this explanation) by saying that a sufficiently energetic photon falling on a photosensitive substance causes the latter to eject a single electron, which is then magnified by a photomultiplier to produce a macroscopic and hence observable effect - the ''click'' of the detector. This is commonly used in discussions of experiments on entangled photons carried out by Alice and Bob, who make statistics on clicks to prove or disprove things, or to communicate secret information.

If the click isn't proof of the arrival of a single photon, one needs to reconsider the traditional explanation of these experiments.

 

[meopemuk]

In the semiclassical picture known to Einstein 1905, currents are produced by discrete electrons. But now, 100 years later, this picture is known to be approximate only, and that currents in metals are in fact produced by the continuous electron fields of QED.
Discrete semiclassical particles are just very rough approximations.

I am very surprised to hear this position. But let me try a different line of argument. I hope you would agree that complete description of any system of electrons, positrons and photons in QED can be achieved if we know the full interacting Hamiltonian H (plus 9 other generators of the Poincare group) in the corresponding Fock space. If we know this Hamiltonian, then we can calculate any dynamical property without additional input.

Now, you are saying that the electron field \Psi(x,t) is the fundamental quantity, which is absolutely necessary for understanding both particle-like and quantum properties of the electrons. On the other hand, I am saying that there is nothing physical about \Psi(x,t). This is just a formal mathematical object, whose only job is to provide a convenient "building block" for the construction of the interacting Hamiltonian H.

Let us see which role is actually played by \Psi(x,t) in QED. First, we use this operator to construct another abstract object called "current density 4-vector"

J^{\mu} (x,t)= \overline{\Psi} (x,t) \gamma^{\mu} \Psi(x,t)

Then we take the product with another abstract "photon field" to form the "interaction Hamiltonian density"

V(x,t) = J^{\mu} (x,t) A_{\mu} (x,t)

Then we integrate the obtained expression by x and set t=0 in order to obtain the interaction Hamiltonian

V = \int dx V(x,0) = \int dx J^{\mu} (x,0) A_{\mu} (x,0)

So, H = H_0 +V is our desired full Hamiltonian, which is a necessary and sufficient tool to do all kinds of calculations. The fields have been totally consumed by integrations. They have played their role as "building blocks". We can forget about them now. They are not needed for further calculations and for physical interpretation of results.

Moreover, parameters x are just integration variables. There is absolutely no reason to identify them with physical positions. Parameter t is set to 0, so it has not direct relevance to measured time. So, it would be less confusing if we used some weird Greek letters for the field arguments: \Psi(\zeta, \chi). Then we would no be tempted to (mistakenly) associate these arguments with physical space and time.

The Hamiltonian H acts in the Fock space, which is divided into sectors with definite numbers of electrons, positrons and photons. So, the particle interpretation always remains with us. If needed, for each state of the system we can always say what are the probabilities for finding certain numbers of particles in the state. This applies to electrons and photons as well.

Eugene.

 

[A. Neumaier]

Now, you are saying that the electron field \Psi(x,t) is the fundamental quantity, which is absolutely necessary for understanding both particle-like and quantum properties of the electrons.

Yes. Look at its use in the quantum optics book of Mandel and Wolf cited before.

On the other hand, I am saying that there is nothing physical about \Psi(x,t).

This has nothing to do with the topic of the thread, so to keep the discussion focussed, I answered in a new thread: https://www.physicsforums.com/showthread.php?p=3148208

 

[yuiop]

The photoelectric effect (http://en.wikipedia.org/wiki/Photoelectric_effect ) is usually explained (following Einstein, who received the Nobel price for this explanation) by saying that a sufficiently energetic photon falling on a photosensitive substance causes the latter to eject a single electron, which is then magnified by a photomultiplier to produce a macroscopic and hence observable effect - the ''click'' of the detector.

As you say, Einstein demonstrated that the photoelectric effect can only be explained if it assumed that incident light energy is made up of discreet quanta of energy, which later came be called photons. As far as I am aware, the original experiment proposed by Einstein did not require a photo multiplier. The current produced by the photoelectric device was proportional to the incident light energy on the device (if the light frequency is above a certain threshold) and is not multiplied in that experiment.

 

[strangerep]

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.

I presume you're referring to M&W section 12.11 (pp 629-639) and also
ch14? I had not closely studied these sections of M&W previously, so I
thank you mentioning this. [I can now confidently urge other potential
responders in this thread to study these parts before replying. :-) ]

In subsections 12.11.1 to 12.11.4, M&W construct and analyze a
configuration space number operator for quasi-monochromatic photons as
an integral over a finite spatial volume. It seems this can be done
regardless of the nonexistence of a photon position operator, although
some approximations are invoked because optical detectors typically
only respond to a finite range of wavelengths.

As explained at the start of section 12.11, M&W have in mind a cylindrical
volume whose base coincides with the sensitive area of the photodetector.
The cylinder length is c\Delta t, since we are interested in
the number of clicks in a given time interval \Delta t.

At the end of sect 12.11.4, they say:

We see therefore that, provided we do not insist on localizing the
excitation too precisely, we can introduce states of localized
excitations or photons, and we can define a configuration space number
operator that measures the number of photons in a finite volume.
However, it is important to bear in mind that the procedure is
meaningful only because the wavelengths of optical photons are so small
on a laboratory scale.

M&W then move on to the polychromatic case. From subsect 12.11.5
onwards, they construct a similar photon "position" operator and
corresponding wavefunction \Phi(r,t) and show that it
determines the usual probability of locating a particle within a given
spatial volume (eq 12.11.30), which is assumed to be large compared with
optical wavelengths.

It's then curious to find that the energy of the photon and the
probability of photoelectric detection are not co-localized.
(This assumes that the detector interacts with light via the latter's
electric field.) M&W analysis shows that, for a photon strongly localized
near the origin, the photon energy distribution \Psi(r,t)
extends over much larger distances, falling off as r^{-7}.
The electric field distribution follows that of the energy, with the
consequence that it's possible for such a strongly localized photon
field to trigger the detector at a point of distance r from the origin.

IOW, a photon localized "here" can cause a detection event "there".

M&W conclude (end of subsect 12.11.5, top of p639):

From these considerations it is apparent that the concept of the photon as a
localized particle traveling at velocity c can be quite inappropriate and
misleading under some circumstances, even though it works in other cases.

My reading of M&W up to this point thus seemed to partially support, but
also partially contradict, Arnold's position. M&W do indeed construct a
finite-volume wavefunction for photons, and show a relation to the
probability of detection within a finite time interval. Although they
also show that (if it is the electric field which is being detected)
that this field distribution does not precisely coincide with the
photon's position distribution, I do not see a corollary embodying the
stronger proposition that the "discrete clicks" are not evidence of
"discrete photons".

However, reading further into ch14, where a more realistic model of a
photodetector is analyzed, we find (in subsect 14.6) that it is necessary
to understand correlations between photoelectric events at different
spacetime points. In general, a detection event at (r1,t1) is not
independent of a detection event at (r2,t2). Some input states exhibit
nontrivial correlations, meaning that the photoelectric pulses produced
by the photodetector are not strictly random. Indeed, such nontrivial
correlations are the basis of the Hanbury-Brown-Twiss effect.

The fact that such correlations between photoelectric events are possible
means that we cannot reliably conclude that detector clicks are evidence
for single independent photons. Rather, the details depend on the precise
nature of the incident field.

 

[A. Neumaier]

I presume you're referring to M&W section 12.11 (pp 629-639) and also ch14? I had not closely studied these sections of M&W previously, so I
thank you mentioning this. [I can now confidently urge other potential
responders in this thread to study these parts before replying. :-) ]

The table of contents of the book is at
http://www.cambridge.org:80/servlet/file/item_9780521417112_frontmatter.pdf?ITEM_ENT_ID=233
I need to get the book to be able to give precise details. But talking from memory, I believe that roughly the following happens:

Sections 9.1-9.5 show that a quantum detector in an external classical e/m field produces discrete Poisson-distributed clicks, although the source is completely continuous, and there are no photons at all in the quantum mechanical model!
This proves that the clicks cannot be taken to be a proof of the existence of photons.

Section 12.11 is about the problems with photon position, and that there is no associated operator (but only a POVM). It is in this section, I believe, that they made the remark I referred to.

Sections 14.1-14.5 show that the semiclassical picture of Chapter 9 holds with small corrections also in the quantum case.

I discussed the situation in some more detail in a public lecture given in 2008, http://arnold-neumaier.at/ms/lightslides.pdf
See Section 3 (pp.35-44); names in smallcaps are accompanied by references, given at the end of the slides.

 

[A. Neumaier]

As you say, Einstein demonstrated that the photoelectric effect can only be explained if it assumed that incident light energy is made up of discreet quanta of energy, which later came be called photons. As far as I am aware, the original experiment proposed by Einstein did not require a photo multiplier.

The photomultiplier is needed to make the current observable. Initially, only a single electron is emitted, which would leave no experimental trace without being magnified.

Einstein's explanation was adequate in 1905, and so important for the development of the subject that he got 1921 the Nobel prize for it, already a few years before modern quantum mechanics was born. The modern concept of a photon was created only much later (1936).

According to today's knowledge, just like Bohr's atomic model, Einstein's explanation of the photoeffect is too simplistic, and is not conclusive.

 

[Cthugha]

The modern concept of a photon was created only much later (1936).

Most people would date the "creation" of the photon concept back to 1926 when the term was coined by Lewis or to 1927 when Dirac published his stuff on the quantum theory of radiation. However, that is nitpicking.

According to today's knowledge, just like Bohr's atomic model, Einstein's explanation of the photoeffect is too simplistic, and is not conclusive.

Of course, but this is well known in the field of QO. The first real evidence for single photon states came already in the seventies with Kimble's and Mandel's antibunching experiments, maybe strongly assisted by the first hints at photon fluctuations measured by Hanbury-Brown and Twiss in 1956. The adequate theory of how to correctly perform optimal measurements to determine the state of a light field has already been given by Glauber in the sixties: A hierarchy of correlation functions up to arbitrary order.

 

[A. Neumaier]

Most people would date the "creation" of the photon concept back to 1926 when the term was coined by Lewis or to 1927 when Dirac published his stuff on the quantum theory of radiation. However, that is nitpicking.

Yes, this was a typo. I meant 1926.

Of course, but this is well known in the field of QO.

Yes. I am not claiming anything new; I only interpret the content of Mandel & Wolf for those who don't have the patience to read through it.

 

[meopemuk]

Perhaps it is easier to think about light shining on a photographic plate or a luminescent screen. We can always adjust the intensity of our light source to be so low that only one photon is present at a time. Then we will clearly see that the image on the plate/screen is made of distinct tiny dots. Each dot corresponds to a grain of photoemulsion that has been hit by a particle - the photon. It might happen that the photoemulsion is not sensitive enough to be triggered by just one photon. Then we can use a higher frequency light (e.g., ultraviolet) to make sure that the "one particle - one dot" rule holds. Isn't it a clear indication that light is a flow of discrete particles rather than a continuous wave?

Eugene.

 

[Cthugha]

We can always adjust the intensity of our light source to be so low that only one photon is present at a time. Then we will clearly see that the image on the plate/screen is made of distinct tiny dots. Each dot corresponds to a grain of photoemulsion that has been hit by a particle - the photon. It might happen that the photoemulsion is not sensitive enough to be triggered by just one photon. Then we can use a higher frequency light (e.g., ultraviolet) to make sure that the "one particle - one dot" rule holds.

I hear this argument quite often, but it is simply wrong. When attenuating a coherent state, it will stay coherent, but just be reduced in intensity. If you split such a beam, and place a detector in each of the split beams and then perform coincidence counting, you will find that the relative rate of simultaneous coincidences normalized to the product of the mean photon count rates in each arm will be the same for the low-intensity and for the high intensity case. You will still have the possibility to have two photons present.
However, this gives you also a good criterion for identifying non-classical light and also single photons: The Cauchy-Schwarz inequality must be violated (the mean number of coincidence counts must be smaller than the products of the mean counting rates). For single photons, the number of coincidence counts must be zero. This is what Kimble and co. showed in resonance fluorescence from a single atom. Although it sounds appealing, lowering the mean intensity does not help at all.

 

[A. Neumaier]

Perhaps it is easier to think about light shining on a photographic plate or a luminescent screen. We can always adjust the intensity of our light source to be so low that only one photon is present at a time. Then we will clearly see that the image on the plate/screen is made of distinct tiny dots. Each dot corresponds to a grain of photoemulsion that has been hit by a particle - the photon. It might happen that the photoemulsion is not sensitive enough to be triggered by just one photon. Then we can use a higher frequency light (e.g., ultraviolet) to make sure that the "one particle - one dot" rule holds. Isn't it a clear indication that light is a flow of discrete particles rather than a continuous wave?

You could as well claim that the fact that a shower emits tiny rays of water is proof that water is composed of discrete rays.

In 1905, when Einstein proposed his explanation, the photoelectric effect was a clear indication of the particle nature of light, since no other model was available that could have explained the process.

However, it is now known (see Chapter 9 of Mandel & Wolf) that a collection of weakly bound electrons responds to a classical external electromagnetic radiation field by emitting electrons according to Poisson-law probabilities, very much like that interpreted by Einstein in terms of light particles. In a photoemulsion, these electrons are magnified via a chemical reaction that produces the tiny dots.

Low intensity coherent or thermal light, which you suggested to employ to make the dots appear one by one, is almost perfectly described by this model. The corrections obtained by using QED instead of the classical external field are very tiny and can be neglected; cf. Chapter 14 in Mandel & Wolf.

Now the state space of this quantum system consists of multi-electron states only. So here the multi-electron system (followed by a macroscopic decoherence process that leads to the multiple dot localization of the emitted electron field) is responsible for the creation of the dot pattern.

Then, how on Earth can the tiny dots be interpreted as an indication that light is a flow of discrete particles rather than a continuous wave? The external field used to generate the pattern _is_ a continuous wave, as one can trivially verify by inspecting the model.

This is enough to make the conclusion invalid that the tiny dots must be regarded as proof of a discrete particle structure of the incident radiation.

On a personal note: 12 years ago, I still had a naive picture of photons, like you. It was a very eye-opening event when I had the occasion to discuss my half-bred views on photons with the quantum optics experimentalists in Zeilinger's group. shortly after he moved to Vienna. These discussions revealed to me that real-life photons are something very different from what superficial discussions seemed to suggest.

 

[phyzguy]

This is all very interesting. I've been reading Mandel and Wolf since you recommended it, but it is quite a tome and will take a while. I have a question for Professor Neumaier. If I understand correctly, you are saying that a continuous electromagnetic field can give rise to quantum effects through its interaction with electrons, because of the quantum nature of the electrons. However, as we see in the analysis of the hydrogen atom, for example, viewing the electron wave function as a continuous field can also give rise to quantum effects because only certain solutions are possible. So is there a possible viewpoint where we view both the EM field and the electron wave function as continuous fields and still get quantum effects? Has such an analysis been done?

 

[A. Neumaier]

This is all very interesting. I've been reading Mandel and Wolf since you recommended it, but it is quite a tome and will take a while.

The book is excellent and pays spending a lot of time on it. It is kind of the bible of quantum optics. Here is a reading guide - first do the quick tour to get an overview, then a second round with more leisure:

At first, you need enough classical background. To update your math, read (or review) Sections 2.1-2.3 and 3.1 and go back to the pieces from Chapter 1 that you need to make sense of these sections. Classical physics (in a simplified setting without polarization) starts in Chapter 4 and 5, where you need at first only 4.1-4.3 and 5.6-5.7 (again, reading omitted stuff you need for understanding that as you go along). Full classical electromagnetism is covered in Chapters 6-8. You need 6.1-6.5.

The quantum part starts in Chapter 9. You'd read 9.1-9.5, 10.1-10.5, 10.9, 10.10, 11.1-8, 11.13, 12.1-12.4, 12.10, 13.1-13.3, 14.1-14.6., 15.1-3, 18.1-4, 20.1-6, 22.4.
Then you have an overview over the central part of quantum optics, and are well prepared to start a second, thorough reading of the whole book (if your time permits).

If I understand correctly, you are saying that a continuous electromagnetic field can give rise to quantum effects through its interaction with electrons, because of the quantum nature of the electrons. However, as we see in the analysis of the hydrogen atom, for example, viewing the electron wave function as a continuous field can also give rise to quantum effects because only certain solutions are possible. So is there a possible viewpoint where we view both the EM field and the electron wave function as continuous fields and still get quantum effects? Has such an analysis been done?

In a sense, yes, though it is not the electron wave function that is treated as a field. Wave functions describe so-called states, whereas fields are so-called observables.

In quantum electrodynamics (QED), both electrons and photons are treated as fields, and the particle view is only a derivative of the latter - covering the special case when the states are sufficiently well localized in phase space, so that a semiclassical picture makes sense. We are currently discussing this in the thread https://www.physicsforums.com/showthread.php?p=3151397
so please contribute there if you have further questions about electron fields.

 

[Cthugha]

This is all very interesting. I've been reading Mandel and Wolf since you recommended it, but it is quite a tome and will take a while.

Just a hint from my personal point of view. When I have to recommend literature about quantum optics to diploma students (or nowadays bachelor and master students), most of them prefer starting with a book providing a low-level intuitive introduction to the topic like the one given in Mark Fox's "quantum optics: an introduction" and then reread the important topics in the Mandel/Wolf where they are treated more rigorously. However, we are experimentalists and might want a different approach to the topic than others do, but I just wanted to share my personal experience of what the students considered the easiest way of getting into the topic.

 

[A. Neumaier]

starting with a book providing a low-level intuitive introduction to the topic

On this, I'd like to recommend the nice booklet by U. Leonhardt, Measuring the Quantum State of Light, Cambridge, 1997.

 

[DrChinese]

Here is a paper with some good references on the subject:

http://people.whitman.edu/~beckmk/QM/grangier/Thorn_ajp.pdf

"While the classical, wavelike behavior of light - interference and diffraction- has been easily observed in undergraduate laboratories for many years, explicit observation of the quantum nature of light i.e., photons is much more difficult. For example, while well-known phenomena such as the photoelectric effect and Compton scattering strongly suggest the existence of photons, they are not definitive proof of their existence. Here we present an experiment, suitable for an undergraduate laboratory, that unequivocally demonstrates the quantum nature of light. Spontaneously downconverted light is incident on a beamsplitter and the outputs are monitored with single-photon counting detectors. We observe a near absence of coincidence counts between the two detectors—a result inconsistent with a classical wave model of light, but consistent with a quantum description in which individual photons are incident on the beamsplitter. More explicitly, we measured the degree of second-order coherence between the outputs to be ..., which violates the classical inequality g(2)(0)>1 by 377 standard deviations."

 

[A. Neumaier]

Here is a paper with some good references on the subject:
http://people.whitman.edu/~beckmk/QM/grangier/Thorn_ajp.pdf

A very interesting collection of articles explaining different current views about photons is in
The Nature of Light: What Is a Photon?
Optics and Photonics News, October 2003
http://www.osa-opn.org/Content/ViewFile.aspx?Id=3185
From the discussion, it becomes obvious that the concept of a ''one photon state'' is unambiguous, while the loosely associated concept of ''a photon'' is very fuzzy and ambiguous. It seems that there are as many different definitions of ''photon'' as there are quantum optics experts interested in defining it.

 

[A. Neumaier]

I am moving part of the discussion in another thread here, in order to keep the discussion focussed.

So, photons are "localized lumps of energy". Do they pass through one slit or through both slits at once in the double-slit experiment?

As any wave, localized or not, through both slits.

Only upon recording the photons, they materialize - at a single spot only.

For the double-slit experiment with visible light the distance between two slits can be macroscopic, e.g. 0.1 millimeter, or something like that. This means that the photon "lump" should be no smaller than this size. So, you are saying that the energy lump associated with a single visible-light photon can be as big as 0.1 millimeter or so? And that all this volume is filled with a time-changing electromagnetic field?

Of course. Photons can be very delocalized. This is precisely what happens in a double slit experiment. Each photon in a laser beam has the same shape as the classical field by which this beam is described.

My views about what photons are turned by almost 180 degrees after I had begun to talk to the experimentalists in Zeilinger's group (who moved to Vienna in 1999)...

OK, so we have a macroscopic lump of electromagnetic energy, which falls on the double slit and interferes with itself according to Maxwell equations. Then this macroscopic lump reaches the photographic plate and suddenly collapses to a microscopic point, whose size is comparable to the size of a grain of photo-emulsion. The photon energy, that was previously spread up in a macroscopic lump now gets released within a group of few atoms.

How does Maxwell equation explain this collapse?

The Maxwell equations are valid in vacuum and need not explain their failure in the presence of a detector.

The behavior of the detector in the presence of the incident classical electromagnetic field is fully explained by the detector's quantum structure. See Chapter 9 of the quantum optics book by Mandel & Wolf.

That the photon ''suddenly collapses to a microscopic point, whose size is comparable to the size of a grain of photo-emulsion'' is pure fantasy. The photon is absorbed by the detector, and doesn't survive as a localized photon.

So, is it correct to say that the macroscopically delocalized lump of the photon's EM field is absorbed by the entire photographic plate, and then all this absorbed energy gets channeled somehow to a single grain of photoemulsion?
[...] Do I need to take the quantum structure of the photographic plate into account?

Yes, yes. The e/m field serves as external potential for the entire photographic plate.
The probability density of a particular electron to become free and trigger the subsequent chemical reaction is computed by (a sophisticated version of) Born's rule. It turns out to generate a Poisson process.

 

[meopemuk]

Yes, yes. The e/m field serves as external potential for the entire photographic plate.
The probability density of a particular electron to become free and trigger the subsequent chemical reaction is computed by (a sophisticated version of) Born's rule. It turns out to generate a Poisson process.

So, according to your logic, when we see a blackened grain of photoemulsion we are not allowed to say: "a small particle (photon) hit this grain, passed all its energy to the grain, and thus initiated a chemical decomposition of the grain, which resulted in its blackening". Instead, we should say something like: "a macroscopic lump of EM field and energy (that we call the photon) interacted with a complicated macroscopic quantum system (that we call the photographic plate). The photon's energy got absorbed by the entire plate. Then through some (yet unspecified) chemical process all this energy rushed toward one particular grain of photoemulsion in order to blacken it." Is it what you are saying? I just want to make sure that I understand your position correctly.

Another question. Would you use basically the same picture when electrons are used in the double-slit experiment instead of photons?


Eugene.

 

[A. Neumaier]

So, according to your logic, when we see a blackened grain of photoemulsion we are not allowed to say: "a small particle (photon) hit this grain, passed all its energy to the grain, and thus initiated a chemical decomposition of the grain, which resulted in its blackening".

People (including quantum optics experts) say it all the time, but it must be understood in a figurative sense, on the same level as when we talk about that an atom consists of electrons moving around a nucleus. It invokes pictures of Bohr's atomic model which contain some truth and intuition, but are very misleading if taken literally. In the same way, Einstein's picture should not be taken literally.

Instead, we should say something like: "a macroscopic lump of EM field and energy (that we call the photon) interacted with a complicated macroscopic quantum system (that we call the photographic plate). The photon's energy got absorbed by the entire plate.

So far, yes. This is a much more accurate description than Einstein's pre-QM picture, conforming to quantum field theory.

Then through some (yet unspecified) chemical process all this energy rushed toward one particular grain of photoemulsion in order to blacken it."

That process is not unspecified. This energy causes some photoelectron to be emitted, according to the rules of quantum field theory. This is a probabilistic process predicted in M&W Chapter 9 by a model of an array of weakly bound electrons in an external potential (with quantum corrections in case of nonclassical light, derived in M&W Chapter 14). The photoelectron then starts a chemical reaction in the metastable photoemulsion, blackening a grain according to well-understood chemistry.

Another question. Would you use basically the same picture when electrons are used in the double-slit experiment instead of photons?

Yes. Photons are lumps of energy of the electromagnetic field, reasonably localized in phase space. Electrons are lumps of energy of the electron field, reasonably localized in phase space.

 

[meopemuk]

That process is not unspecified. This energy causes some photoelectron to be emitted, according to the rules of quantum field theory. This is a probabilistic process predicted in M&W Chapter 9 by a model of an array of weakly bound electrons in an external potential (with quantum corrections in case of nonclassical light, derived in M&W Chapter 14). The photoelectron then starts a chemical reaction in the metastable photoemulsion, blackening a grain according to well-understood chemistry.

This is so much different from quantum mechanics that I've learned from Feynman Lectures and other places.

Let us assume for definiteness that we have a single photon with energy of ~1 eV. The photographic plate has the surface of 10cm x 10cm. It is not difficult to make our double-slit such that the interference picture covers entire surface. Of course, in order to see the entire picture we'll need to shoot many photons one-by-one to accumulate statistics. So far we are talking about a single photon. There are billions of photoemulsion grains (or, as you call them, weakly bound electrons) exposed to the photon's field in this case. The photon field's energy per each grain is less that one billionth's of eV. Experiment tells us that only one electron (grain) gets excited in the whole array, so that the entire amount of photon's energy (~1 eV) is concentrated on this one electron. If I understand correctly, you are saying that there is a physical process, by means of which the widely delocalized energy density is concentrated on one electron only. This "energy localization" process happens instantaneously. It is quite remarkable that there is only one concentration point. It never happens that the photon's energy gets shared between two or three emulsion grains.

Also, your explanation presumes that the energy concentration (or collapse) should somehow depend on the physical nature of the observation screen. So, it would be reasonable to assume that the shape of the interference picture could be different depending on whether we use a photographic plate or a luminescent screen or a light-sensitive array of a digital camera. In fact, experimentally, there is absolutely no difference. The shape of the interference picture depends *only* on the geometry of the two slits and on the momentum (energy) of photons. It will be the same even if we project our interference picture on a brick wall. The shape of the picture will stay the same if instead of photons we use any other particle with the same momentum (electron, neutron, atom, etc.). So, your explanation of the collapse by some physical process inside the photographic plate does not look plausible to me.

Yes. Photons are lumps of energy of the electromagnetic field, reasonably localized in phase space. Electrons are lumps of energy of the electron field, reasonably localized in phase space.

I don't think that 10 cm x 10 cm can be called "reasonably localized".

Eugene.

 

[meopemuk]

Arnold,

here is a reference to the double-slit experiment with atoms:

O. Carnal, J. Mlynek, Young’s double-slit experiment with atoms: A simple atom interferometer. Phys. Rev. Lett. 66, 2689–2692 (1991). http://prl.aps.org/pdf/PRL/v66/i21/p2689_1

It follows the same pattern as other types of double-slit experiments. However, I don't think one can claim that some matter or energy redistribution processes happen in the projection screen. Atoms simply fall wherever they may and this does not depend on the physical nature of the substrate.

My conclusion from this is that double-slit interference is the manifestation of the fundamental single-particle quantum effect often called "wavefunction collapse". This effect is basically the same for photons, electrons, atoms, buckyballs, etc. Fields (either quantum or classical) play no role there.

Eugene.

 

[A. Neumaier]

This is so much different from quantum mechanics that I've learned from Feynman Lectures and other places.

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.

The photon field's energy per each grain is less that one billionth's of eV. Experiment tells us that only one electron (grain) gets excited in the whole array, so that the entire amount of photon's energy (~1 eV) is concentrated on this one electron. If I understand correctly, you are saying that there is a physical process, by means of which the widely delocalized energy density is concentrated on one electron only. This "energy localization" process happens instantaneously.

I never claimed that this process happens instantaneously. Chemical reactions take
time. Instantaneous collapse is one of the fairy tales of the kindergarden textbooks.

It is quite remarkable that there is only one concentration point. It never happens that the photon's energy gets shared between two or three emulsion grains.

Yes. Remarkable, and predicted by orthodox quantum mechanics without photons. Quantum mechanics is a very remarkable theory!

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.

My conclusion from this is that double-slit interference is the manifestation of the fundamental single-particle quantum effect often called "wavefunction collapse".

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.

your explanation of the collapse by some physical process inside the photographic plate does not look plausible to me.

It is not my explanation. It is only my putting in intelligible words the quantitative quantum mechanical derivation in Sections 9.2-9.5 of Mandel & Wolf. So you must find fault with their derivation, not with my perhaps imperfect words.

The derivation in M&W takes only 7 pages (pp.439-446) of elementary mathematics, the first two of which consist of a review of what is needed from quantum mechanics - so it should be worth your time studying it in detail.

I am looking forward to your diagnosis of the error in their formal reasoning.

I don't think that 10 cm x 10 cm can be called "reasonably localized".

What is reasonable depends on the context. In astronomy even a huge quantum object like a giant star is reasonably localized.

On the other hand, 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.

 

[f95toli]

But now, 100 years later, this picture is known to be approximate only, and that currents in metals are in fact produced by the continuous electron fields of QED.

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)

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...").

 

[meopemuk]

On the other hand, 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.

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 stil a localized particle. So, it hits only one emulsion grain or whatever small detecting spot there is on the screen. It is a different matter that we cannot predict where exactly this hit will occur. There is an inherent quantum uncertainty in the particle's behavior. This uncertainty is expressed in the probabilistic language of (often delocalized) wave functions.

This is how I understand quantum mechanics. Something tells me that your understanding is different. I'll go and find Mandel & Wolf to see what they have to say on this matter.

Eugene.

 

[Q-reeus]

..What is reasonable depends on the context. In astronomy even a huge quantum object like a giant star is reasonably localized...

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. 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'. But is that at at all realistic? 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?

 

[meopemuk]

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.

It is not my explanation. It is only my putting in intelligible words the quantitative quantum mechanical derivation in Sections 9.2-9.5 of Mandel & Wolf. So you must find fault with their derivation, not with my perhaps imperfect words.

The derivation in M&W takes only 7 pages (pp.439-446) of elementary mathematics, the first two of which consist of a review of what is needed from quantum mechanics - so it should be worth your time studying it in detail.

I am looking forward to your diagnosis of the error in their formal reasoning.

OK, I got a copy of Mandel & Wolf, but this is a Russian translation, so page numbers are not the same. They have a model in which each electron on the surface is in a bound state with binding energy -E_0. The surface is subjected to a periodic time-dependent potential with frequency \omega. A simple quantum mechanical calculation shows that if the frequency is sufficiently high \hbar \omega > E_0, then there is a non-zero probability for each electron to jump to a continuous spectrum state with a positive energy. This means that electrons can be ejected from the surface.

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.

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. 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. 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,...

The rule "one incident photon results in one (or zero) emitted electron" is an important component in the corpuscular interpretation of light. Perhaps this rule is explained in later chapters of the M&W book, but I haven't found this explanation yet.

Eugene.

 

[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.