Why can wave nature of light not explain photoelectric effect?

[Hans de Vries]

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
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 \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
radiation damping. Regards, Hans

 

[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]

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]

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]

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]

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]

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]

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]

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]

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]

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

 

[lightarrow]

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

How do you deduce that? I don't think QM does it at all: how could QM describe diffraction then? Light can "circle" around obstacles. There is not "Nadelstrahlung" of photons.

 

[lightarrow]

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.

Agreed.

 

[monish]

How do you deduce that? I don't think QM does it at all: how could QM describe diffraction then? Light can "circle" around obstacles. There is not "Nadelstrahlung" of photons.

I don't think I'm that far off the mark here. All I said was that directional radiation is one of the features that distinguishes photons from classical em. An atom that radiates a single quantum of energy semi-classically would be expected to give off a wave in all directions. According to photons it goes in one direction, and the atom recoils.

If there is existing ambient radiation, the distinction becomes blurred; even the classical case tends to radiate into the existing stream. But now there are claims (in the paper referenced earlier in this thread) that people can manufacture photons at will, one photon at a time. This raises some questions.

You obviously can't get one photon at a time with stimulated emission. So they must be using spontaneous emission. In fact, I found one reference on the internet as to how this was done. Yes the energy was directional and not spherical. But the way they did this was to put the atom between two mirrors, like a little laser cavity. So it really works like stimulated emission where the "stimulating field" is bootstrapped from the atom itself.

It's not exactly like the case of an isolated atom radiating all by itself. So I still have to wonder: where does the directionality come from with an isolated atom?

 

[Mentz114]

Monish:

So I still have to wonder: where does the directionality come from with an isolated atom?

I'm not sure what you mean by this, because all the theory I've seen assumes the direction of emission will be random. Like in cooling atoms by resonant lasers. The cooling is effected because the absorption is always in one direction, but subsequent emissions are random, resulting in a net loss of momentum.

Einstein established (1916) that emission must carry momentum randomly in order to get a black-body spectrum for matter interacting with an EM field.

 

[monish]

Monish:

I'm not sure what you mean by this, because all the theory I've seen assumes the direction of emission will be random. Like in cooling atoms by resonant lasers. The cooling is effected because the absorption is always in one direction, but subsequent emissions are random, resulting in a net loss of momentum.

Einstein established (1916) that emission must carry momentum randomly in order to get a black-body spectrum for matter interacting with an EM field.

I'm not questioning the randomness of the emission. I'm asking why it has any direction at all. The obvious semi-classical emission mode is spherically symmetric. (Not quite; a small dipole puts out a wave which has a directional pattern symmetric about the z axis; it's as "spherically symmetric" as things get in em theory.)

This situation changes if there is ambient radiation ("stimulated emission"). Now the emission tends to line up with the ambient field, but this is clearly understandable from classical em theory.

What I don't understand is how a single isolated atom manages to give off a wave that is not spherically symmetric. I can't see any mechanism that generates recoil, even in a random direction. I thought we might gain some insight by looking at the experiments which claim to produce single photons, one at a time. It turns out that in the case I was able to look at, they put the atom in a classical resonant cavity and that's how they get the direction. I still don't see how a single isolated atom can emit anything other than a spherically symmetric em wave.

 

[Mentz114]

Hi Monish:

I still don't see how a single isolated atom can emit anything other than a spherically symmetric em wave

There's no such thing as a spherically symmetric solution to Maxwells equations. An early attempt (1924) to account for emissions with spherical wave (Bohr, Kramer, Slater) had to invoke non-conservation of energy and was soon discounted.

 

[monish]

Hi Monish:



There's no such thing as a spherically symmetric solution to Maxwells equations. An early attempt (1924) to account for emissions with spherical wave (Bohr, Kramer, Slater) had to invoke non-conservation of energy and was soon discounted.

Oh come on. I already said I wasn't talking about a COMPLETELY symmetrical wave:

(quoting my own previous post:)

"...a small dipole puts out a wave which has a directional pattern symmetric about the z axis; it's as "spherically symmetric" as things get in em theory.)"

 

[Mentz114]

Oh, apologies for not paying attention. In that case, I can't see your problem.

[edit]

What I don't understand is how a single isolated atom manages to give off a wave that is not spherically symmetric. I can't see any mechanism that generates recoil, even in a random direction.

Found your problem. It's possible for a bunch of waves to interfere so that only a uni-directional component survives. If the wave carries momentum, you have a recoil.

It would be good to understand the mechanism, but it's pure speculation until some experimental data is available. Photonic band gap crystals are interesting.

 

[monish]

Oh, apologies for not paying attention. In that case, I can't see your problem.

Right. Sorry for snapping at you.

Found your problem. It's possible for a bunch of waves to interfere so that only a uni-directional component survives. If the wave carries momentum, you have a recoil.

It would be good to understand the mechanism, but it's pure speculation until some experimental data is available. Photonic band gap crystals are interesting.

Yes, something like this explains the directionality in the case of stimulated emission. But not for spontaneous emission. That's the problem.

Yes, I think it is nice when there are mechanisms for these things. There are a lot of cases in qm when you can find a convincing mechanism by looking at the details of the wave functions. It's funny that when they artificially make single-photon sources, they build little mechanisms like classical resonators to control the directionality. That makes sense. But there is no way an isolated atom can classically give off any kind of em wave that will cause it to recoil. So what is the mechanism?

 

[Mentz114]

But there is no way an isolated atom can classically give off any kind of em wave that will cause it to recoil. So what is the mechanism?

I'll risk infraction points by speculating. Classically there is no mechanism, but we can invoke the vacuum in a quantum theory. So maybe it's a vacuum fluctuation that stimulates the 'spontaneous' emission.

I can't resist quoting from the 1916 paper "... the molecule suffers a recoil of hv/c in a direction which is only determined by 'chance', according to the current state of the theory". The great man was baffled.