# Why can wave nature of light not explain photoelectric effect?

#### tehno

Thank you for the elaboration of these points Quantstart.

#### monish

A reader pointed out to me this discussion and that it was left unresolved.
Interesting. Your discussion took place exactly one year ago, and last week I participated in a very similar discussion which was positively resolved two days ago when ZapperZ used his authority as a monitor to lock down the thread.

I haven't managed to get Mandl and Wolf, but I certainly agree with that....most
of the features of the photoelectric effect can be explained semiclassically.
Yes, this was also brought up in the recent discussion.

Nevertheless, quanta of the electromagnetic field (photons), and not just
quanta of energy, are required to explain it COMPLETELY, including the
simple feature of the "immediate" emission.
Yes indeed! I kept hoping someone would bring up this point because I wanted to respond to it. But we never got past the discussion of the frequency threshold.

Consider the following:

Let electromagnetic radiation start impinging on the area A of a metal of
the work function W.....the amount of energy the classical electromagnetic field has brought on the area A after the time T is given by

A T c epsilon_0 [E^2]

where [E^2] denotes the average value of the electric field E squared,
c is the velocity of light, and epsilon_0 is the vacuum permittivity.
You haven't specified what value you're using for the area.

Therefore, for a very weak light source, i.e., very small [E2], a RELATIVELY
long time (say, T > 10^-8 seconds), given by

T > W /( A c epsilon_0 [E^2] ) ,

is needed for the absorption of the quantum of energy (h nu) which would exceed
the work function W and thus enable the start of the emission of electrons. But,
this is NOT found experimentally...I do not know what the experimental limit is
at present, but the lack of time lags between incident light beam arrival and
emitted photoelectron has long been an established experimental fact.
Therefore, if one insists on the classical EM field....
OK, we know where this argument leads. The spread-out wave energy is much too diffuse to be able to concentrate itself in the tiny cross-sectional area of an electron in such a short time as is observed experimentally. Usually people use the cross-sectional area of an atom to justify this claim.

People who do this kind of thing don't know how to calculate CLASSICAL absorption cross-sections. The absorption cross-section of a classical antenna is sometimes but not always close to the physical size of the antenna. A notable exception is the small lossless dipole. In the limit, the absorption cross-section is on the order of the square of the wavelength, no matter how small the physical antenna.

An isolated hydrogen atom in the superposition of s and p states, according to the Schroedinger equation, is an oscillating dipole antenna. If analyzed classically, its absorption crossection is on the order of 10,000,000 angstroms squared. This is much much greater than the size of the atom. That's why it can absorb enough energy to drive the transition between the two states.

What about the classical photo-electric effect?. I believe that the electron wave functions as calculated by band theory can be as big as the whole piece of metal. So there is no justification for trying to limit the absorption cross-section to a tiny area.
A correct analysis of a classical wave interacting with a real (quantum-mechanical) metal plate should show that the time for photo-emission is limited only be the time it takes for energy to build up over the entire surface of the metal. The cross-sectional area of the atom is irrelevant.

The only reason Einstein and others couldn't see how to explain the PE effect with waves was because they didn't know the Schroedinger equation for the electron.

Marty

#### reilly

If Einstein's photoelectric work was bogus, then so was Bohr's work on hydrogen. And, of course their work was not and is not bogus.Quite the contrary, their work is part of the foundations of today's physics.

Bogus indeed. The plain fact is that both of these brilliant gentlemen gave plausible, intuitive explanations for two of the most vexing problems of the day -- think dark matter, and quantized gravity as equivalent problems. In fact, discrete spectra and the photoelectric effect, as observed, simply could not happen, as in impossible -- by current theory. The explanations made by Einstein and Bohr were among the most gutsy and extraordinary creative leaps of physics, ever. Their impact is, as we say today, huge. Think about it, they started the breakdown of the 19th century view of physics, a breakdown still being fought today.

They were as right with their explanations as anybody could be at that time. And, in many respects their work was the practical beginning of modern quantum theory. The posts in this thread are testament to the importance and durability of their work.
Regards,
Reilly Atkinson

#### Hans de Vries

Gold Member
People who do this kind of thing don't know how to calculate CLASSICAL absorption cross-sections. The absorption cross-section of a classical antenna is sometimes but not always close to the physical size of the antenna. A notable exception is the small lossless dipole. In the limit, the absorption cross-section is on the order of the square of the wavelength, no matter how small the physical antenna.

An isolated hydrogen atom in the superposition of s and p states, according to the Schroedinger equation, is an oscillating dipole antenna. If analyzed classically, its absorption crossection is on the order of 10,000,000 angstroms squared. This is much much greater than the size of the atom. That's why it can absorb enough energy to drive the transition between the two states.
So, the claims of instantaneous emission would be the result of
calculating with the wrong cross-section.... According to this:

http://www.iap.uni-bonn.de/lehre/ss05_laserspectroscopy/Uebungen/Warm%20up.pdf [Broken]

The classical absorption cross-section is

$$\sigma=\frac{3\lambda^2}{2\pi}$$

Which seems to agree with your numbers. Do you have any other links?

Regards, Hans

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#### monish

So, the claims for instantaneous emission would be the result of
calculating with the wrong cross-section.... According to this:

http://www.iap.uni-bonn.de/lehre/ss05_laserspectroscopy/Uebungen/Warm%20up.pdf [Broken]

The classical absorption cross-section is

$$\sigma=\frac{3\lambda^2}{2\pi}$$

Which seems to agree with your numbers. Do you have any other links?

Regards, Hans
I'm not sure I have any more authoritative links than what you've found. Of course there are two formulas, one is for maximum absorption and the other one is for lossless reflection. The second case is four times the area. I'm not sure which of those two formulas you've quoted.

What I can offer is an intuitive explanation. It goes something like this:

1. Any power you receive with an antenna must come from the incident wave.

2. Therefore, the existence of an absorbing antenna can be detected by sensitive field measurements far from the actual antenna.

3. But the far-field radiation pattern of a tiny antenna oscillating at, say, 100 amps, is not very different from a much larger antenna (100x) oscillating at 1 amp.

4. So the interaction of the fields, incident and absorbing, is not strongly dependent on the actual size of the absorbing antenna.

People call this result counter-intuitive, but its all a matter of looking at it the right way.
There is, however, a drastic effect on bandwidth as you get smaller. I don't have a reference handy, but I'm pretty sure if you take the classical bandwidth formulas and apply them to the hydrogen atom at its various frequencies, you get pretty much the correct spectral linewidths. I think those are called the Einstein alpha coeffecients in qm.
But they come right out of the classical antenna formulas.

Marty

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#### Hans de Vries

Gold Member
1. Any power you receive with an antenna must come from the incident wave.

2. Therefore, the existence of an absorbing antenna can be detected by sensitive field measurements far from the actual antenna.

3. But the far-field radiation pattern of a tiny antenna oscillating at, say, 100 amps, is not very different from a much larger antenna (100x) oscillating at 1 amp.

4. So the interaction of the fields, incident and absorbing, is not strongly dependent on the actual size of the absorbing antenna.

People call this result counter-intuitive, but its all a matter of looking at it the right way.
There is, however, a drastic effect on bandwidth as you get smaller. I don't have a reference handy, but I'm pretty sure if you take the classical bandwidth formulas and apply them to the hydrogen atom at its various frequencies, you get pretty much the correct spectral linewidths. I think those are called the Einstein alpha coeffecients in qm.
But they come right out of the classical antenna formulas.

Marty
This goes back to the birth of quantum mechanics.

The cross-section is so large because it is a resonant absorber. Absorption is
An ideal, non-damped, resonant absorber has in fact an infinite cross-section.

What limits the cross-section is the radiation damping $\gamma$ which is not so easy
to specify. Resonant absorber:

$$\ddot{\textbf{x}}+\gamma\dot{\textbf{x}}+\omega_o^2 \textbf{x}\ =\ \frac{e}{m}E_o\ e^{i\omega t}$$

(see for instance (2.178) in Sakurai, advanced QM, or see I, §5 (7) of Heitler's
classic, The Quantum Theory of Radiation)

The radiation damping solves the infinite cross-section in the Kramers-Heisenberg
cross-section when the incoming radiation is equal to the resonance frequency.

$$\sigma_{tot}\ =\ \frac{8\pi_o^2}{3}\frac{\omega^4}{(w_o^2-\omega^2)^2+\gamma^2/4}$$

So the cross-section and thus the response time for photon emission varies over
a wide range depending on the situation as a result of the dependence on the

Regards, Hans

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#### monish

The cross-section is so large because it is a resonant absorber. Absorption is classically the result of emitting radiation opposite to the incoming radiation. An ideal, non-damped, resonant absorber has in fact an infinite cross-section.

What limits the cross-section is the radiation damping which is not so easyto specify. .........So the cross-section and thus the response time for photon emission varies over a wide range depending on the situation as a result of the dependence on the radiation damping.

Regards, Hans
But we should be able to verify this by looking at the data. What I am claiming is that if you apply the classical antenna formulas to the hydrogen atom, you get the correct value for things like the Einstein alpha (and beta?) coefficients. Do you think this is correct?

My main point is that it's wrong to make simple arguments based on the cross-section of the atom to supposedly prove that the wave theory doesn't work.

Marty

#### Hans de Vries

Gold Member
My main point is that it's wrong to make simple arguments based on the cross-section of the atom to supposedly prove that the wave theory doesn't work.

Marty
Yes, but note that the mathematics describes gases, isolated atoms, which is much
simpler as the solid state physics involved in practical photon emission experiments.

Nevertheless it seems indeed a valid argument against "instantaneous emission" which
is by itself important enough.

Regards, Hans.

#### malawi_glenn

Homework Helper
monish: You did not answer Zappers question, nor proving that your view was the correct one.

#### Parlyne

If Einstein's photoelectric work was bogus, then so was Bohr's work on hydrogen. And, of course their work was not and is not bogus.Quite the contrary, their work is part of the foundations of today's physics.

Bogus indeed. The plain fact is that both of these brilliant gentlemen gave plausible, intuitive explanations for two of the most vexing problems of the day -- think dark matter, and quantized gravity as equivalent problems. In fact, discrete spectra and the photoelectric effect, as observed, simply could not happen, as in impossible -- by current theory. The explanations made by Einstein and Bohr were among the most gutsy and extraordinary creative leaps of physics, ever. Their impact is, as we say today, huge. Think about it, they started the breakdown of the 19th century view of physics, a breakdown still being fought today.

They were as right with their explanations as anybody could be at that time. And, in many respects their work was the practical beginning of modern quantum theory. The posts in this thread are testament to the importance and durability of their work.
Regards,
Reilly Atkinson
Reilly, no one here is claiming that Einstein's work (or Bohr's, for that matter) is bogus. In both cases, the work was brilliant and led to huge advances in physics. The question at hand is a subtler one. I doubt that anyone here has a problem with the idea that Bohr model of the atom is taught as leading to fundamentally new ideas about atoms and having new explanatory power but, ultimately, being wrong. So, the fact that we teach that it doesn't correctly explain atomic phenomena is uncontroversial.

In the case of the photoelectric effect, however, Einstein's major idea has turned out to be correct; but, it is not the only structure in currently accepted physics that can correctly explain the effects he was discussing. In other words, the conclusion is correct but the argument is flawed. Of course, Einstein couldn't have known that then; so, this is not an indictment of him or his work. Really, the only important feature of the discussion is the idea that we should stop teaching that the elementary features of the photoelectric effect prove that light is quantized. They don't. The only prove that something is quantized.

#### QuantStart

So, this discussion is revived after a year, by challenging my stressing the role of the absence of the time lag, i.e., the simple feature often (but somewhat imprecisely) called the "instantaneous" emission. Thus, I will refute monish's criticism and elaborate last year's argument, hoping to finish before leaving for New Years Eve...

Among other things, monish said:
... ....
OK, we know where this argument leads. The spread-out wave energy is much too diffuse to be able to concentrate itself in the tiny cross-sectional area of an electron in such a short time as is observed experimentally. Usually people use the cross-sectional area of an atom to justify this claim.
...
... ... If analyzed classically, its absorption crossection is on the order of 10,000,000 angstroms squared. This is much much greater than the size of the atom. That's why it can absorb enough energy to drive the transition between the two states.

What about the classical photo-electric effect?. I believe that the electron wave functions as calculated by band theory can be as big as the whole piece of metal. So there is no justification for trying to limit the absorption cross-section to a tiny area. ...
No. In fact, I did not rely on assuming a tiny area scale anywhere in
my argument; you ascribe me this assumption groundlessly. I did not specify
a value for the area A since it was not essential for the argument. (In a
quantitative illustration I give below, I will even use a macroscopic value for A.)

However, you seem to have missed my point that weakening of the CLASSICAL
electric field squared is bound to produce an experimentally unacceptable
time lag if the conservation of energy holds. I mean the experiments which,
in the post #19, vanesch refers to as "that old experiment with faint
illumination ..." where even for very low intensity of radiation no time
delays "were ever observed, at least none longer than 10^-9 sec" (last quote is
from Gasiorowicz's "Quantum Physics", 2nd edition, Wiley 1996, which may be
a typical example of textbooks criticized here because of photo-effect). While
vanesch argued that the absence of such time lags does not provide anything
against classical EM field (i.e., in favor of photons) because the probability
of transition is immediately non-zero (which is correct), his argument neglects
the issue of the energy conservation.

These old experiments and analyses showed already at the order of 10^-9 seconds
that the classical EM field has the trouble with energy conservation, and the term
"instantaneous emission" seemed appropriate then.
Now one should speak of time delays (lags), since nowadays, as ZapperZ
pointed out in post #18, experiments can measure the finite response time,
ranging from fs to ps time scale. So no wonder that these old experiments
seem somewhat forgotten, since fs to ps time scale is 1000 to 10^6 times
SHORTER than 10^-9 s, the time scale which was however already sufficient
to demonstrate (by these old experiments) the trouble with energy conservation
for the classical field in photo-effect.

Namely, as I pointed out in my first post (post #25 of this thread), the
energy exceeding the work function W can be supplied by the CLASSICAL
EM field on the surface A only in time exceeding

W /( A c epsilon_0 [E^2] )

where [E^2] denotes the average value of the electric field E squared,
c is the velocity of light, and epsilon_0 is the vacuum permittivity.

Let us take the case of a gold (Au) laboratory sample, where W = 5.1 eV.

Also, since several posts after monish's post #27 deal with the cross-section of
a classical antenna and the like, claiming that the argument of "instantaneous
emission/no time delay" is the result of calculating with the wrong cross-section,
i.e., single-atom cross-section, let us assume a MACROSCOPIC value for the
surface, one millimeter squared,
A = 1 mm^2

which is a conceivable surface area of our laboratory sample.

Then, for the electric field E = 1 V/m, the time exceeds 0.3 nanoseconds,
as W /( A c epsilon_0 [E^2] ) = 0.307 x 10^-9 s,

and for the electric field E = 0.1 V/m, the time lag exceeds 3x10^-8 s,
as W /( A c epsilon_0 [E^2] ) = 3.07 x 10^-8 s,
etc., etc.

To summarize:
The classical EM field has a continuous Poynting vector, continuous energy
and momentum density, and delivers energy in a continuous manner, so that
the issue of the energy conservation (stressed in my first post) cannot be
neglected; that is, it is among the simple properties of photo-effect, and
through the absence of time lags for weak EM fields shows the need for the
EM field quanta - photons, although other simple features of the photoelectric
effect can be explained semiclassically.

#### monish

So, this discussion is revived after a year, by challenging my stressing the role of the absence of the time lag, i.e., the simple feature often (but somewhat imprecisely) called the "instantaneous" emission. Thus, I will refute monish's criticism and elaborate last year's argument, hoping to finish before leaving for New Years Eve...

Among other things, monish said: (here QuantStart quotes from my message of Dec. 22: "OK, we know where this argument leads. The spread-out wave energy is much too diffuse to be able to concentrate itself in the tiny cross-sectional area of an electron in such a short time as is observed experimentally. Usually people use the cross-sectional area of an atom to justify this claim."

No. In fact, I did not rely on assuming a tiny area scale anywhere in
my argument; you ascribe me this assumption groundlessly. I did not specify
a value for the area A since it was not essential for the argument.
I can't believe you claim that I ascribe such an assumption to you. I specifically noted in my post that you refrained from using a specific area in your calculation. Therefore, in order to attempt to refute your argument, it was necessary for me to fill in the blanks.
I certainly did not ascribe any assumption to you; I distinctly stated that this the assumption commonly made by others who wish to promote the photon theory.

However, you seem to have missed my point that weakening of the CLASSICAL
electric field squared is bound to produce an experimentally unacceptable
time lag if the conservation of energy holds.
No, I didn't miss your point: you chose not to make your point. You chose to remain silent on the question of what would be the relevant cross-sectional area for the calculation you presented. Possibly you were being clever; but in these circumstances you can't blame me for supposedly missing your point.

...since several posts after monish's post #27 deal with the cross-section of
a classical antenna and the like, claiming that the argument of "instantaneous
emission/no time delay" is the result of calculating with the wrong cross-section,
i.e., single-atom cross-section, let us assume a MACROSCOPIC value for the
surface, one millimeter squared,
A = 1 mm^2

which is a conceivable surface area of our laboratory sample.

Then, for the electric field E = 1 V/m, the time exceeds 0.3 nanoseconds,
as W /( A c epsilon_0 [E^2] ) = 0.307 x 10^-9 s,

and for the electric field E = 0.1 V/m, the time lag exceeds 3x10^-8 s,
as W /( A c epsilon_0 [E^2] ) = 3.07 x 10^-8 s,
etc., etc.

To summarize:
The classical EM field has a continuous Poynting vector, continuous energy
and momentum density, and delivers energy in a continuous manner, so that
the issue of the energy conservation (stressed in my first post) cannot be
neglected; that is, it is among the simple properties of photo-effect, and
through the absence of time lags for weak EM fields shows the need for the
EM field quanta - photons, although other simple features of the photoelectric
effect can be explained semiclassically.
This is definitely not the "prompt emission" argument that is usually made in physics textbooks. I have a paper in front of me from by Muthukrishnan, Scully, and Zubairy ("The concept of the photon - revisited", OPN Trends, Oct 2003) in which they make a similar argument. Understand that these are people who are working at the leading edge of semi-classical interpretations...and even THEY use the atomic cross-section in their calculation, which appears to me to be obviously incorrect:

"....if we persist in thinking about the field classically, energy is not conserved. Over a time interfal t, a classical field E brings ain a flux of energy epsilon-E-squared-At to bear on the atom, where A is the atomic cross-section. For short enough time intervals...."

Yes, you can modify this calculation, as you have, by putting in a macroscopic area, but are you quite sure the experiments have been done to back this up? It's not obvious to me that this is so easy to do. How do you turn a light source on and off with that kind of precision? And if you really could do the experiment, and you found that energy wasn't conserved, well...wouldn't that be a problem for the photon theory as well? It's not so obvious to me that you get around the conservation of energy by just by saying that light is made of particles.

But the real problem with all arguments of this kind is that they fail to come to grips with the question of why we NEED photons in the first place. Historically, photons were brought in because people couldn't understand some basic physical phenomena involving interaction of radiation and matter. It wasn't a question of picosecond time delays and tiny discrepancies....it was a case of all kinds of things that just "shouldn't have happened AT ALL" if light was a wave. But once the true nature of the electron was understood in 1926, many of these puzzles were cleared up. It turned out that you could explain most or all of these mysteries with the wave theory of light. So what was left? You go down to the very fringes of measurement, where you're able to supposedly isolate "one photon at a time". And there you find them. Supposedly.

It's a big difference from what I was told in high school: you shine ultraviolet light on a piece of metal, and an electric current flows in the circuit. "Shouldn't happen if light is a wave." Well, now it turns out, yes, it should too. So you redesign the experiment, bring in a lot of expensive and complicated instrumentation, and you claim that you can isolate it down to individual photons. (Which I don't think you really can, because everything you measure now has to be interpreted within your theoretical framework. It's not just a deflection on an ammeter anymore.) But even assuming you do find your single photons at the fringes of measurement...what did you really need them for? Everything you originally said you couldn't do without photons...the photo-electric effect, the Compton effect, the laser, you name it...it turns out you can get it from ordinary e-m radiation. Give or take a few picoseconds. So what do we really need photons for?

#### ZapperZ

Staff Emeritus
2018 Award
It's a big difference from what I was told in high school: you shine ultraviolet light on a piece of metal, and an electric current flows in the circuit. "Shouldn't happen if light is a wave." Well, now it turns out, yes, it should too. So you redesign the experiment, bring in a lot of expensive and complicated instrumentation, and you claim that you can isolate it down to individual photons. (Which I don't think you really can, because everything you measure now has to be interpreted within your theoretical framework. It's not just a deflection on an ammeter anymore.) But even assuming you do find your single photons at the fringes of measurement...what did you really need them for? Everything you originally said you couldn't do without photons...the photo-electric effect, the Compton effect, the laser, you name it...it turns out you can get it from ordinary e-m radiation. Give or take a few picoseconds. So what do we really need photons for?
Are you still claiming that there are no phenomena that cannot be described using the classical wave theory, or are you still sore that you've been "lied" to about the photoelectric effect? Because if you are claiming the former, I've listed tons of phenomena in an earlier locked thread. The photon antibunching phenomena and the which-way experiments are two prime and easy examples. If you are only focusing on the photoelectric effect, take note that none of the published papers on using classical wave theory description of the photoelectric effect have any resemblance to your simplistic model, meaning that what you are claiming you can do have not been verified to be valid.

One could also argue using your question "So what do we really need light wave for?", especially when ALL light phenomena can be described via such photon model while the wave model is strangely deficient in describing all of those phenomena that I've mentioned. One would think that based on this, the photon model has a wider range of applicability, and so we should simply teach the kids in school this model.

I can actually answer that. The same way we still teach diffraction and interference phenomena using the wave model instead of the full blown quantum mechanical approach, using the wave picture can actually make things simply to describe whenever they are equally valid. The QM approach under such condition can be excruciatingly difficult. Thus, rather than delay the introduction of such phenomena until students have a good grasp of QM, we show them how such things can be described simply using the wave picture.

However, the reverse now is true for the photoelectric effect. Try looking at all the published papers that claim that one can actually use classical wave picture of light to describe this. The wave picture is now the more complicated aspect when compared to the photon picture. You can check this yourself if you don't believe me (and you might want to do that to compare why I did not buy your model). So now, it is the photon picture that has the pedagogical advantage!

Now, considering that the photon picture can in fact describe ALL the observed phenomena of light, and that the classical wave picture still don't, maybe it is you who need to explain why we need the classical wave picture.

Zz.

#### monish

monish: You did not answer Zappers question, nor proving that your view was the correct one.
Last time I tried to reply to one of Zappers questions he used his power as a system monitor to lock down the discussion. So I prefer not to get involved in a discussion with him.

Marty

#### ZapperZ

Staff Emeritus
2018 Award
Last time I tried to reply to one of Zappers questions he used his power as a system monitor to lock down the discussion. So I prefer not to get involved in a discussion with him.

Marty
That was because the last time you replied to my questions, you were pulling things out of thin air without any valid citations, of which you were warned a couple of times based on our guidelines. It had nothing to do with me. It had everything to do with you.

Zz.

#### lightarrow

One could also argue using your question "So what do we really need light wave for?", especially when ALL light phenomena can be described via such photon model while the wave model is strangely deficient in describing all of those phenomena that I've mentioned. One would think that based on this, the photon model has a wider range of applicability, and so we should simply teach the kids in school this model.
What exactly do you mean with "photon model"? Surely you don't mean an object spatially localized moving from source to detector.

#### ZapperZ

Staff Emeritus
2018 Award
What exactly do you mean with "photon model"? Surely you don't mean an object spatially localized moving from source to detector.
What does "photon model" have anything to do with "object spatially localized"?

Zz.

#### lightarrow

What does "photon model" have anything to do with "object spatially localized"?

Zz.
Ok.
What is exactly the "photon model"?

#### ZapperZ

Staff Emeritus
2018 Award
Ok.
What is exactly the "photon model"?
Quantum electrodynamics.

Zz.

#### lightarrow

Quantum electrodynamics.

Zz.
Ok, and what exactly would you teach to kids in the school about it? You can't teach them Quantum electrodynamics.

One would think that based on this, the photon model has a wider range of applicability, and so we should simply teach the kids in school this model.

#### ZapperZ

Staff Emeritus
2018 Award
Ok, and what exactly would you teach to kids in the school about it? You can't teach them Quantum electrodynamics.
I won't.

I know this is hard to remember, but I've mentioned earlier on why we still teach kids the wave picture - because it is easier to handle and describe when one deals with diffraction and interference. After all, would you want to be shown what Marcella has done in his Eur. J. Phys. paper in intro physics? However, when we deal with Compton effect and photoelectric effect, the photon picture of quantized clumps of energy is way simpler and more transparent than invoking Stochastic Electrodynamics, for example.

In other words, I would continue to do what is typically done in the regular class.

Zz.

#### lightarrow

I won't.

I know this is hard to remember, but I've mentioned earlier on why we still teach kids the wave picture - because it is easier to handle and describe when one deals with diffraction and interference. After all, would you want to be shown what Marcella has done in his Eur. J. Phys. paper in intro physics? However, when we deal with Compton effect and photoelectric effect, the photon picture of quantized clumps of energy is way simpler and more transparent than invoking Stochastic Electrodynamics, for example.

In other words, I would continue to do what is typically done in the regular class.

Zz.
I agree completely with the idea of "quantized clumps of energy"; I don't think that is even questionable in any way; the problem is to avoid making kids in the school believe these "clumps of energy" are "corpuscles moving from source to detector", do you agree?

#### ZapperZ

Staff Emeritus
2018 Award
I agree completely with the idea of "quantized clumps of energy"; I don't think that is even questionable in any way; the problem is to avoid making kids in the school believe these "clumps of energy" are "corpuscles moving from source to detector", do you agree?
Well, they are moving from source to detector. We just should not give the impression that these are "ping pong-like balls", i.e. the common concept of "particles" that we are familiar with, which they are not. To me, that is the most common misconception of what people get when we tell them about "particles".

I know for a fact that, in other threads on the "size" of a photon, I've mentioned many times that the concept of a photon never defined it as having a definite size and shape in real space. This is contrary to what we normally call as particles, which occupies a definite boundary in real space and thus, can be defined with a volume/size.

Zz.

#### f95toli

Gold Member
It is actually quite remarkable that ordinary EM theory works so well even for single photons, despite the fact that the "wave nature" of light is implicit in Maxwell's equations.
A striking example is recent experiments with single microwave photon sources where the components (resonators etc) are very large (many mm) and are designed using ordinary microwave theory.
In some way it is of course only natural that there is a "smooth" transition from "classical" EM to QFT since we know that the classical model work well in most situations; but I still find it quite faccinating.

#### monish

It is actually quite remarkable that ordinary EM theory works so well even for single photons, despite the fact that the "wave nature" of light is implicit in Maxwell's equations.
A striking example is recent experiments with single microwave photon sources where the components (resonators etc) are very large (many mm) and are designed using ordinary microwave theory.
In some way it is of course only natural that there is a "smooth" transition from "classical" EM to QFT since we know that the classical model work well in most situations; but I still find it quite faccinating.
Can you tell us more about how they make single microwave photons? I'm interested in how they become directional. One of the biggest differences between classical em and quantum theory is that quantum theory demands the photons be directional while classically the radiation is spherically distributed. Actually, when there is an ambient field ("stimulated emission") the classical behavior becomes the same as the quantum behavior; but when you talk about photons one at a time, they do not seem to be reconcilable with the classical picture. Maybe the experiments you refer to can give us some insight on this.

Marty

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