B Photon Continuity in Double-Slit Experiment

[jeremyfiennes]

TL;DR Summary

Is there experimental evidence for photon conitinuity.

In the double-slit experiment, is there experimental evidence that a photon detected passing a slit always results in one and one only screen point?.

 

[vanhees71]

Well, with there's always some probability that a photon is not registered, but if it's registered then at one point and only one point of the screen.

 

[jeremyfiennes]

Thanks. So that a photon behaves as a 'particle' in the classical sense - that if at one point in time it is found at some point, then at a later point in time it will be found at some other (not necessarily predictable) point - is in fact not experiementally demonstrable.

 

[A. Neumaier]

In the double-slit experiment, is there experimental evidence that a photon detected passing a slit always results in one and one only screen point?.

No, because whenever a photon is detected, it ceases to exist.

 

[jeremyfiennes]

Right. So photon continuity is an unverifiable hypothesis. Which doesn't of course mean that it is untrue. But just that in practice it is at present experimentally unverifiable.

 

[A. Neumaier]

So photon continuity is an unverifiable hypothesis.

Nobody claims it anyway.

Photons are just a fancy name for detector clicks in response to faint light. In addition they are a a technical name for ingredients building up the state of electromagnetic radiation.

 

[jeremyfiennes]

But delayed eraser experiments, for instance, DO imply particle-like continuity: that an idler photon arriving in a detector corresponds to a signal photon point on the screen.

 

[A. Neumaier]

But delayed eraser experiments, for instance, DO imply particle-like continuity: that an idler photon arriving in a detector corresponds to a signal photon point on the screen.

This is just a figure of speech for a more complicated situation.

It allows one to reduce the discussion to the essentials but is misleading when taken literally.

 

[jeremyfiennes]

Ok. Thanks.

 

[jeremyfiennes]

In the Mach-Zender 'particle' setup, would it be true to say that a photon is detected either in one detector, or in the other, but never in both simultaneously?

 

[DrChinese]

In the Mach-Zender 'particle' setup, would it be true to say that a photon is detected either in one detector, or in the other, but never in both simultaneously?

Generally, that's the right idea. This experiment spells it out pretty well. 1 photon, 1 click.

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

 

[jeremyfiennes]

Thanks. So photons are at least more trustworthy than atoms.

 

[DarMM]

Note there is no state that will definitely cause $N$ clicks in an appropriate device.

In a sense a "one photon" state just means a state of the electromagnetic field very likely to cause one click in a photon detector, but not certain to.

 

[A. Neumaier]

Generally, that's the right idea. This experiment spells it out pretty well. 1 photon, 1 click.

No. There is always some loss due to detector inefficiency. The 1-1 correspondence is just a convenient simplification.

 

[f95toli]

But delayed eraser experiments, for instance, DO imply particle-like continuity: that an idler photon arriving in a detector corresponds to a signal photon point on the screen.

This is ultimately a philosophical question: what does "exist" mean?
The vast majority of people working with photons do think of them as being as "real" as e.g. atoms, but this does of course not mean that they are in any way classical objects,

 

[DrChinese]

No. ... The 1-1 correspondence is just a convenient simplification.

Well, I used the word "generally" (indicating a "convenient simplification") and cited the specific experiment itself in case anyone wanted to see more depth; I'm not sure how you can say NO to that as an answer to a simple question.

We agree that every experiment has some degree of inefficiency, so I don't see the value of your contrary conclusion. Yes, 1:1 is the relevant takeaway here - and you would get that with perfect efficiency (if there were such a thing). In fact, as mentioned in the reference: "a single photon can only be detected once!*"

IMHO: We do a disservice to some readers when we qualify things so much that the original question gets completely lost. *Originally Grangier, Roger and Aspect, 1986.

 

[A. Neumaier]

We agree that every experiment has some degree of inefficiency [...]

IMHO: We do a disservice to some readers when we qualify things so much that the original question gets completely lost.

Nevertheless, the only correct B-level answer to the original question

In the double-slit experiment, is there experimental evidence that a photon detected passing a slit always results in one and one only screen point?

is no.

The experimental evidence is overwhelming that a photon passing undetected a slit always results in zero or one screen point, never more. A zero count is due to detector inefficiencies.

On the other hand, if the photon is detected passing a slit it disappears and always results in no point on the screen.

 

[jeremyfiennes]

"The experimental evidence is overwhelming that a photon passing undetected a slit always results in zero or one screen point, never more."
In SPDC terms: "there is overwhelming experimental evidence that if an idler photon is detected, then in principle one and only signal photon will necessarily be detected. And if it isn't, this will be attributed to detector inefficiency" (correct?).

"No. ... The 1-1 correspondence is just a convenient simplification."
This is the essence of my query. Is the near 1:1 ratio due to a given wave amplitude generally resulting in a photon detection at both points (slit and screen)? Or is is a necessary condition due to the particle properties of photons. And that if a photon is detected at one point and not the other, then this must be due to detector inefficiency at the no-show point.

 

[A. Neumaier]

"The experimental evidence is overwhelming that a photon passing undetected a slit always results in zero or one screen point, never more."
In SPDC terms: "there is overwhelming experimental evidence that if an idler photon is detected, then in principle one and only signal photon will necessarily be detected. And if it isn't, this will be attributed to detector inefficiency" (correct?).

If only one photon is around at the time of measurement, there can be only one detection event.

(But you seem to be discussing here the case of an entangled pair of photons, an idler photon and a signal photon. For this, there are two possible detection events.)

"No. ... The 1-1 correspondence is just a convenient simplification."
This is the essence of my query. Is the near 1:1 ratio due to a given wave amplitude generally resulting in a photon detection at both points (slit and screen)? Or is is a necessary condition due to the particle properties of photons. And that if a photon is detected at one point and not the other, then this must be due to detector inefficiency at the no-show point.

No. One can detect a given photon only once, as it is destroyed by the detection process. Hence only at the slit or at the screen, not twice. But because of detector inefficiency, both detectors can possibly miss it.

 

[jeremyfiennes]

But you seem to be discussing here the case of an entangled pair of photons, an idler photon and a signal photon. For this, there are two possible detection events.)

But do these in principle (ie. ignoring detection inefficiency) always correlate: if one is detected, then the other necessarily will be; and if not then not?

 

[A. Neumaier]

But do these in principle (ie. ignoring detection inefficiency) always correlate: if one is detected, then the other necessarily will be; and if not then not?

The detection failures are independent. But assuming 100%efficiency, two photons imply two detection events. This has nothing to do with correlations.

 

[DrChinese]

But do these in principle (ie. ignoring detection inefficiency) always correlate: if one is detected, then the other necessarily will be; and if not then not?

You are talking about PDC pairs. So the answer is YES, despite everything being said that might indicate otherwise. When one of the pair goes through a double slit, and since it is a photon: it only has the opportunity to be detected once. That is what A. Neumaier is pointing out, and there is no argument about that. The key takeaways are:

a. There is 1 click at both detectors, or no clicks at either*. That is what you refer to as "correlated"; which is often used with PDC to mean something a bit different (again why A. Neumaier's answer is different than mine). But you are correct generally, that is the entire point of the reference I provided.

b. There are never cases of 1 click at one detector*, and 2 (or more) clicks at the other. That is because a PDC pair - sometimes referred to as a biphoton - is a special state with a known photon number (2 in this case). Such a state is called a Fock state. Most light does not appear in this state, as photon number is not usually a conserved quantity. In fact, PDC photon pairs are created from a single input photon (in those cases where a biphoton results).

c. Although often ignored with a double slit setup: it IS possible to determine which slit a photon goes through without destroying it. That's another subject though.

d. You should never make assumptions about what quantum particles are doing - at least not rigorous assumptions - when not being observed. They have a nasty habit of doing the impossible when you make an assumption which is reasonable yet wrong.

Now obviously, and as A. Neumaier correctly points out, there are many details being glossed over in our discussion. You can read more at the reference I provided. It is very good as it is intended for an undergrad lab. Note that this experiment soundly refutes one of the classical views of light. *Of course many photons that are part of PDC pairs never make it through the double slit in the first place (in your setup); so we are ignoring that for the purposes of our discussion.

 

[vanhees71]

Nevertheless, the only correct B-level answer to the original question

is no.

The experimental evidence is overwhelming that a photon passing undetected a slit always results in zero or one screen point, never more. A zero count is due to detector inefficiencies.

On the other hand, if the photon is detected passing a slit it disappears and always results in no point on the screen.

This is a bit strange formulation for a beginner.

Fact is that there is some probability for a single photon, which state can be prepared today easily via parametric downconversion, to be detected by a detector or not.

If photon is detected and if the detector resolves the position of the detection event a single photon state is registered at one "point" (a point has of course necessarily a finite size) and not as a smeared distribution. That's one of the most simple indications for the necessity of field quantization. It cannot be explained within classical electrodynamics.

 

[A. Neumaier]

This is a bit strange formulation for a beginner.

What is strange?? What you wrote does not contradict what I wrote.

 

[vanhees71]

No, it doesn't contradict anything, but I hope it makes it more clear to a beginner in QED.

 

[jeremyfiennes]

Dr Chinese: thanks. Ok, for simplicity let's restrict discussion to PDC pairs, and assume hypothetical 100% detection efficiency.

"There is 1 click at both detectors, or no clicks at either."
That is what I have been calling "correlated". Sorry, I didn't realize it could have another meaning.

So when Einstein said that
"A particle can only appear as a limited region in space where the field strength, or the energy density, is particularly high. We can consider as matter those regions of space when the field is extremely intense."
this doesn't exclude a particle (e.g. a photon) being detected at one point but not at a subsequent one, and so doesn't fit the present model. (Right ?)

"It IS possible to determine which slit a photon goes through without destroying it. That's another subject though."
Could you give a reference? Does it confirm photon continuity: that one detected passing a slit always corresponds to one and one only screen point?

"You can read more at the reference I provided. The experiment soundly refutes one of the classical views of light."
I read and liked it. Which classical view does it refute?

 

[vanhees71]

I don't know, what precisely you mean by "photon continuity". As I said before, if you have prepared a single photon and it hits a position resolving detector it leaves only one spot though the expectation value of the em-field energy density, which is by definition the intensity of the em. field, can be a broad distribution. That distinguishes a single-photon state from a classical em.-wave state (QFT-wise something like a coherent state or some other state, e.g., thermal radiation like a Planckian black body).

One very elegant way to gain which-way information in the double-slit experiment with photons is to start with linearly polarized photons (say horizontally polarized wrt. the $x$ direction) and put quarter-wave plates into the slits one ("slit 1") oriented with $\pi/4$ relative to the $x$ direction and the other (slit 2) with $-\pi/4$. Then a photon running through slit 1 gets left-circularly polarized and a photon running through slit 2 gets right-circularly polarized, and thus you can (in principle or really) determine behind the slits through which each photon came. Since the corresponding states are orthogonal to each other there's no double-slit interference pattern (but only an incoherent addition of the two single-slit interference patterns). As I said before, to get a distribution you need many photons. Each single photon leaves one point-like spot on the screen.

 

[DrChinese]

Dr Chinese: thanks. Ok, for simplicity let's restrict discussion to PDC pairs, and assume hypothetical 100% detection efficiency.

"There is 1 click at both detectors, or no clicks at either."
That is what I have been calling "correlated". Sorry, I didn't realize it could have another meaning.

So when Einstein said that
"A particle can only appear as a limited region in space where the field strength, or the energy density, is particularly high. We can consider as matter those regions of space when the field is extremely intense."
this doesn't exclude a particle (e.g. a photon) being detected at one point but not at a subsequent one, and so doesn't fit the present model. (Right ?)

"It IS possible to determine which slit a photon goes through without destroying it. That's another subject though."
Could you give a reference? Does it confirm photon continuity: that one detected passing a slit always corresponds to one and one only screen point?

"You can read more at the reference I provided. The experiment soundly refutes one of the classical views of light."
I read and liked it. Which classical view does it refute?

Correlated... I understood what you meant.

Einstein's comment is a little vague to be useful here. There was a lot of quantum optics theory/experiment developed after his death: Fock states and PDC for example. One good thing about PDC is that you can essentially get 1 photon to herald (announce) the impending presence/detection of another.

Most of the time, particles (photons) can be thought to move in a classical path. But you want to be careful about assumptions about any quantum particle's trajectory between detection events. The assumption goes only so far, and breaks down completely where there are interference effects, multiple indistinguishable paths, etc. So the specific setup is important.

Here is a reference to an experiment where which-slit can be determined using polarization information. Polarizers are placed by the slits and varied, causing interference to appear or disappear. This occurs because it is merely the possibility of obtaining the which-slit information which controls the interference - you do not need to actually learn it.

http://sciencedemonstrations.fas.ha...-demonstrations/files/single_photon_paper.pdf

 

[vanhees71]

It doesn't make sense to say a quantum particle moves along a classical path. For photons you don't even have a position observable to begin with!

 

[DrChinese]

It doesn't make sense to say a quantum particle moves along a classical path. For photons you don't even have a position observable to begin with!

Just wondering if you read entire sentences. Yes, MOST OF THE TIME, you can think of a photon as moving in a straight line at c (that would be a classical path, my friend). You can also predict its arrival time at a specified position within very narrow limits. You can also manipulate its path. OTHER TIMES: this is not possible with quantum particles, such as photons.

Really, why make every answer so complex that no one can discuss anything without referencing the entire body of physics? Any scientist doing an experiment on photons operates with basic assumptions that are useful for the job at hand. That including how a photon travels. They don't set up fiber, mirrors, optics and detectors at random places. They pair PDC entangled photons using precise estimates of path and travel time - and that in fact assumes a specific velocity. So I don't see the point of stating that there are exceptions to general rules when someone states to begin with (as I did)... it's a general rule. It is a general rule, and it does apply MOST OF THE TIME exactly as I said.

The idea that a photon lacks a position observable is laughable, as they are regularly observed at specific points in hundreds of experiments (as both detector clicks and in other manners) exactly as predicted. What you mean to say is that field theory does not allow for a photon position observable in the same sense as other quantum particle observables.

 

[jeremyfiennes]

It doesn't make sense to say a quantum particle moves along a classical path. For photons you don't even have a position observable to begin with!

I'm not talking of paths, but of detection at specific points.

 

[A. Neumaier]

I'm not talking of paths, but of detection at specific points.

You cannot detect the same photon twice, at the slit and at the screen!

 

[jeremyfiennes]

You cannot detect the same photon twice, at the slit and at the screen!

This is a thought exercise: whether in principle the 'particle' model holds, whether a single slit photon always corresponds to a single screen photon and vice versa. In the analogous SPDC case of whether signal and idler photons are always either both found or not, this can be verified experimentally. But again "in principle", due to detection inefficiency.

 

[vanhees71]

The idea that a photon lacks a position observable is laughable, as they are regularly observed at specific points in hundreds of experiments (as both detector clicks and in other manners) exactly as predicted. What you mean to say is that field theory does not allow for a photon position observable in the same sense as other quantum particle observables.

Well, we discuss science here, not wishful thinking.

The PDC photons used in Bell-test experiments are rather send through filters that they have a rather well-defined momentum to ensure a sufficient degree of coherence to enable these very experiments, and indeed you describe it right but draw the wrong conclusions: There's a position of detection events, determined by the position of massive detectors. What's measured by photon detectors are the appropriate one- or two-photon correlation functions.

 

[jeremyfiennes]

A different slant. The Schroedinger equation gives the probability of a measurement detecting a particle (photon or electron ) at a point. These probabilities always add up to unity. In other words, one cannot predict with certainty where the particle will be found. But one can predict with certainly that it will be found. Correct?

 

[DrChinese]

... indeed you describe it right but draw the wrong conclusions: There's a position of detection events, determined by the position of massive detectors. ...

It "right" because I make reference to representative experiments where this description is "useful". Those detectors are measuring arrival times of (bi)photons traveling on precise classical paths once they are corralled. I am not drawing any conclusion past what the experimenters are doing. You should be able to see the irony of asserting photons don't have position (operator), when you can measure that position to a precision limited only by experimental setup and the usual constraints of the Heisenberg Uncertainty Principle. That it is done with a detector is irrelevant. ALL particle detection events EVERYWHERE are brought to us by some detector or film or similar - you can't just deny the obvious position for photons and accept the same evidence for anything else just because it doesn't fit with your theory. The theory is not reality, but it can be a useful representation of reality.

Generally accepted science is: Photons exist, and their future position can be accurately predicted - within obvious constraints of the actual setup. This does not mean that photons are classical particles, they aren't - and I repeatedly say that it is a mistake to think of them in that manner. However, there are certainly many times when a photon exhibits the behavior of a classical particle - and such description can be useful. Certainly it is useful when the experimenter decides where to place the apparatus.

 

[Nugatory]

A different slant. The Schroedinger equation gives the probability of a measurement detecting a particle (photon or electron ) at a point. These probabilities always add up to unity. In other words, one cannot predict with certainty where the particle will be found. But one can predict with certainly that it will be found. Correct?

With the appropriate disclaimers about detector efficiency, yes. But this thread is about photons, so the non-relativistic Schrodinger's equation in the position basis isn't so relevant. Instead you have to look at how the single-photon state evolves over time.

 

[Nugatory]

You should be able to see the irony of asserting photons don't have position (operator), when you can measure that position to a precision limited only by experimental setup and the usual constraints of the Heisenberg Uncertainty Principle.

As a moderately interested bystander... it sure looks to me as if you and @vanhees71 are just rerunning the old discussion about whether it is more correct to say "it has position x" or "a detector at point x will trigger at time t". The first includes a few micrograms of unnecessary interpretational content, is more often useful, and is (IMO) more likely to reinforce B-level misconceptions.

 

[DrChinese]

As a moderately interested bystander... it sure looks to me as if you and @vanhees71 are just rerunning the old discussion about whether it is more correct to say "it has position x" or "a detector at point x will trigger at time t". The first includes a few micrograms of unnecessary interpretational content, is more often useful, and is (IMO) more likely to reinforce B-level misconceptions.

Maybe. This is the quantum forum, so it is always useful to add a caveat at the beginning of a discussion precisely to insure that such misconceptions are not propagated. I try to do that (and I think most are pretty good about that too, including @vanhees71). And once the OP acknowledges that, we should be able to describe the general case, and occasionally throw in an alternative scenario. On the other hand: we can also take things so far that there is no statement that can be made that is completely true in all cases. Hard to get a useful message across when every answer is "that's wrong" - which I sometimes see more often than "that's correct". I don't think every caveat is needed in a B level discussion as long as the key points are being made.

In scientific papers: I see "it has position x" statements about 10 times as often as a statement with caveat such as "a detector at point x will trigger at time t". Ditto in textbooks and the like. The exceptions and caveats should be brought in as the general rule gets into better focus, and there is relevance. That's my take, anyway.

 

[DarMM]

As a moderately interested bystander... it sure looks to me as if you and @vanhees71 are just rerunning the old discussion about whether it is more correct to say "it has position x" or "a detector at point x will trigger at time t". The first includes a few micrograms of unnecessary interpretational content, is more often useful, and is (IMO) more likely to reinforce B-level misconceptions.

Maybe. This is the quantum forum, so it is always useful to add a caveat at the beginning of a discussion precisely to insure that such misconceptions are not propagated. I try to do that (and I think most are pretty good about that too, including @vanhees71). And once the OP acknowledges that, we should be able to describe the general case, and occasionally throw in an alternative scenario. On the other hand: we can also take things so far that there is no statement that can be made that is completely true in all cases. Hard to get a useful message across when every answer is "that's wrong" - which I sometimes see more often than "that's correct". I don't think every caveat is needed in a B level discussion as long as the key points are being made.

I think both sides here are very understandable. On the one hand one wishes to be correct, on the other completely correct QM almost nullifies conversation.

What's a particle? Well according to fully general Quantum Field Theory a "particle" is a quantum state that at very late or very early times will "probably" make a certain type of experimental probe click once.

Why probably? Well no state will definitely make a detector click due to the Reeh-Schleider theorem.

Well at least we can always speak of properties like angular momentum etc right? Well no, most POVMs cannot be read as quantizations of classical properties. Thus the observable some pieces of equipment are measuring when they "click" are simply the observable representing that piece of equipment clicking! https://arxiv.org/pdf/quant-ph/0207020.pdf
As Peres says in that paper most quantum observables don't even have names in our vocabulary.

There's just preparations and clicks! No picture like classical mechanics. However people are used to talking about objects like photons etc

Hard to know what to do.

 

[A. Neumaier]

The Schroedinger equation gives the probability of a measurement detecting a particle (photon or electron ) at a point.

No. Phoons and electrons are not described by a Schrödinger equation but by the free Maxwell and Drac equation, respectively. Solutions of the Maxwell equations don't hve a probabilistic interpretation as probability density.

 

[vanhees71]

It "right" because I make reference to representative experiments where this description is "useful". Those detectors are measuring arrival times of (bi)photons traveling on precise classical paths once they are corralled. I am not drawing any conclusion past what the experimenters are doing. You should be able to see the irony of asserting photons don't have position (operator), when you can measure that position to a precision limited only by experimental setup and the usual constraints of the Heisenberg Uncertainty Principle. That it is done with a detector is irrelevant. ALL particle detection events EVERYWHERE are brought to us by some detector or film or similar - you can't just deny the obvious position for photons and accept the same evidence for anything else just because it doesn't fit with your theory. The theory is not reality, but it can be a useful representation of reality.

Generally accepted science is: Photons exist, and their future position can be accurately predicted - within obvious constraints of the actual setup. This does not mean that photons are classical particles, they aren't - and I repeatedly say that it is a mistake to think of them in that manner. However, there are certainly many times when a photon exhibits the behavior of a classical particle - and such description can be useful. Certainly it is useful when the experimenter decides where to place the apparatus.

A WRONG description is never useful. Physics is an exact science, and we should do our best to describe it correctly. Again: You cannot define the position of a photon (to some extent maybe you can define "transverse position" somehow). All you can define are probability distributions for a photo-detector, placed at some position (which is well-defined, because as a massive object you can always define a position observable for it). As it turns out from a quite straight-forward analysis of the detection process in terms of, e.g., the photoeffect (i.e., an electromagnetic wave kicks out an electron from a bound state to the continuum) the detection probability distributions of various kinds (usually 1- or 2-photon detection probabilities as function of time(s) and position(s)) can be calculated from autocorrelation functions of the electric-field components.

Of course, photons "exist". They don't have a position in the strict sense, and one should avoid to use undefinable ideas. To define what a photon is and how to measure relevant observables of it, you need a theory, and the best theory we have about them is QED. Of course, theory is not reality, but you cannot describe and investigate reality without it.

 

[DarMM]

Of course, photons "exist". They don't have a position in the strict sense, and one should avoid to use undefinable ideas. To define what a photon is and how to measure relevant observables of it, you need a theory, and the best theory we have about them is QED

I'm not going to get into "exist", but certainly there are photons in QED. The only difficulty is that in QED they are a type of excitation distribution in idealized probes/detectors placed at spatial and temporal asymptotic infinity. Strictly speaking they aren't well defined at finite times or outside detectors nor are they associated with a fixed number of clicks (though this last point can usually be ignored).
That's quite far from what people might normally have in their heads.

 

[vanhees71]

Of course, we can philosophize endless about the "ontology of photons/elementary particles". There's enough paper wasted on this fruitless subject ;-)). For me the particles of the standard model simply "exist" in the very specific sense you describe it: There are predictions of the theory (QED) which are confirmed to amazing precision by all measurements.

It's also clear that QED tells us that a particle (photon) description is only possible in the sense of asymptotic free states and strictly speaking it also tells us that this is a far from trivial concept, because of the long-range nature of the em. interaction (aka the "masslessness" of the photon field), but that's another story.

 

[DarMM]

For me the particles of the standard model simply "exist" in the very specific sense you describe it: There are predictions of the theory (QED) which are confirmed to amazing precision by all measurements.
...
It's also clear that QED tells us that a particle (photon) description is only possible in the sense of asymptotic free states

Yeah no disagreement of course. Just the chasm between that what people normally think and QFT says is quite a hard one to cross and it's difficult to know what to say in a "B" thread.

I think the common notion is that a photon is a thing flying around out there with maybe some "quantum weirdness" associated with the uncertainty principle. This is the sort of picture you might get from non-relativistic QM where particles might have uncertainties and so on associated with their observables, but at least you clearly have the particle itself had all times.

In QFT though we lose even that, having a particle only as a late time excitation in a specific detector type as you said.

 

[vanhees71]

The problem is that even in otherwise good textbooks this wrong picture of a "photon" as some localized lump of matter (like a miniature "billard ball") persists. It's just laziness of textbook writers to introduce QM in an "as simple as possible but not simpler" way. In the case of popular-science writing it's not shear laziness of the authors but the impossibility to really describe photons without the adequate math.

But still there, I think you can do a better job:

I think it's sufficient to first tell the readers about the classical concept of light as an electromagnetic wave. It's not so hard to make the classical-field concept understandable without any math. Then you can refer to what we perceive as light (intensities, colors etc.) from a physical point of view. When this is made clear you can argue with Planck's finding that the energy exchange of electromagnetic fields with a typical frequency $\omega$ happens only in "discrete packets" of $E=\hbar \omega$. You can also tell that the electromagnetic field carries momentum, and that it also comes in "packets" of the size $p=2 \pi \hbar/\lambda$. I think to specify photons as these "energy-momentum packets" is a better picture than the "bullet picture". One can then also stress the "indivisibility" of these packets, i.e., that whenever they are detected on a photo plate or with a cell-phone camera (CCD) they leave a "single spot", and that is the only way you can associate a (transverse) position of these packets, and that's the only "particle-like feature" such a photon has and that in general you cannot say in advance, where a specific "packet" will be registered on the photo plate but that one can with the comoplicated math of QED only predict the probability distribution that it hits the plate at a given place.

 

[DarMM]

that whenever they are detected on a photo plate or with a cell-phone camera (CCD) they leave a "single spot"

Yes I think this gets to the heart of the matter. People are lead to think photons are something that hits the camera and causes the excitation of a pixel, where as under QED it is more the case that a photon is the excitation of a pixel in a camera and QED gives rules for the probability of a given pixel being excited a given amount.

 

[vanhees71]

The irony is that this view was already provided by Planck, but Einstein's view of 1905 prevailed though Einstein himself said that he was still puzzled by what "radiation really is" till his death. It's also ironic that Einstein got his Nobel prize for the only theory of his that didn't stand the later development of physics and that his Nobel certificate is the only (?) one that cites for which achievement he definitely did not get the prize, i.e., relativity ;-))).

 

[PeterDonis]

A WRONG description is never useful. Physics is an exact science

I think you're going to have a very hard time justifying those statements in the face of the obvious facts that physicists use approximations all the time and that all measurements have some finite error.

 

[vanhees71]

Of course you have to use approximations and any measurement has a finite error, but you shouldn't refer to wrong qualitative descriptions to begin with, at least if you know better!