Why can wave nature of light not explain photoelectric effect?

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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.
 
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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.
 
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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

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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.
 
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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

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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.
 
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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

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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.
 
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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

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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.
 

Cthugha

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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.
Why maybe? Spontaneous emission happens of course due to coupling to the quantized EM-field.

The Purcell-effect shows nicely that the spontaneous emission rate can even be enhanced by altering the mode structure of the field.
 
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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.
If that's the mechanism, then it supports the wave picture as well as the photon picture. If the quantum field is allowed to have vaccum fluctations, then the classical field is allowed to have bits of stray energy that are always present because they're just below the quantum absorption threshold of ordinary matter.
 

Mentz114

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Why maybe? Spontaneous emission happens of course due to coupling to the quantized EM-field. The Purcell-effect shows nicely that the spontaneous emission rate can even be enhanced by altering the mode structure of the field.
Thanks for the information, Cthugha.

Monish:
If that's the mechanism, then it supports the wave picture as well as the photon picture. If the quantum field is allowed to have vaccum fluctations, then the classical field is allowed to have bits of stray energy that are always present because they're just below the quantum absorption threshold of ordinary matter.
Classical fields don't have fluctuations or 'stray' bits of energy.
 
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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.

[...]

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?
Now it's clear what you intended and your question is indeed interesting. The only spatial asymmetry that comes to my mind is electron's (or nucleus's) spin, and the fact a photon has spin too (s = 1) but I don't know how much that could have to do with it.
 
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Actually, the true explanation of the photoelectric effect does not involve "photons", in fact the quantum calculations show that light is behaving as a classical wave during this effect! Treating the atom quantum dynamically, we derive the existence of resonant frequencies which are responsible for the "thresholds" which characterize this effect.

Ironically Einstein's nobel prize winning work was bogus, the photoelectric effect does not require photons to be explained.

Beginning in the 1980's experiments done by Aspect et al detected photons for the first (non-bogus) time, for which the team won a nobel prize.
That's why Einstein was a genius. He could come to a right conclusion even from wrong assumption
 
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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.





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.




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?
Well, since to the argument I presented you reacted (in post #27) by saying this:
"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.", you ascribed such an assumption also to me at least implicitly.

Also, I obviously belong among those "who wish to promote the photon theory", and that is why you wanted to refute the argument presented in my post, and not because I "chose not to make" my point. Obviously, my point in my first post (#25) was that the COMPLETE 'prompt emission argument' includes the conservation of energy setting the bound that time "...

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, the photoelectric effect would imply the non-conservation of energy. On the other hand, the quantized EM field, i.e., the photon concept, does not have the above problem, as the absorption of the quantum of energy happens "at once",
when the electron and the EM field quantum interact."

(end of my quote from #25)

For this argument it is essential that for the classical, continuous EM field, [E^2] can, at least in principle, take arbitrarily small values. Since I did not do any computations with concrete values, there was no need to fix A. But, you obviously thought ("OK, we know where this argument leads..." - in #27) that this argument, i.e., the bound on the time lags based on W/(A c epsilon_0 [E^2]) can in practice work only with a "tiny cross-sectional area" A, and in your opinion there was the mistake because you advocated much larger A, from "CLASSICAL absorption cross-sections" of some "10,000,000 angstroms squared" (= 10^-13 m^2) to hints that area "as big as the whole piece of metal" may be relevant.

Therefore, in my second post (#36) I did concrete calculations which showed that even for this latter, extreme choice of A being the surface of "the whole piece of metal", W/(A c epsilon_0 [E^2]) can set the limit T > 10^-9 seconds.
To that you answer "but are you quite sure the experiments have been done to back this up?" and "How do you turn a light source on and off with that kind of precision?" Since my interest in this is only pedagogical, I do not want to search for references, BUT it is enough to recall that in post #18, ZapperZ pointed out that nowadays some experiments (on metals) are so precise that they measure the finite response time to be on a femtosecond (10^-15 s) scale!
That is the factor 10^-6 times shorter than what I got for W/(A c epsilon_0 [E^2]) in my second post (#36) with A = 1 mm^2 and E = 1 to 0.1 V/m, which means that again with the macroscopic, extremely high area A = 1 mm^2, one can get W/(A c epsilon_0 [E^2]) above the established femtosecond scale already with [E^2] not much below 10^6 V^2/m^2 ... and mind you, assuming A = 1 mm^2 , which is extremely large. This assumption is useful to show that the conservation of energy implies, through W/(A c epsilon_0 [E^2]), the quantization of EM field, even if one adopts your most EXTREME viewpoint on A.

On the other hand, the viewpoints on A of Muthukrishnan, Scully, Zubairy may well be correct. You correctly point out that they are "at the leading edge of semi-classical interpretations". Well, as experts in using semi-classical interpretations, they most probably have strong arguments
for using the atomic cross-section. I do not know what is in that paper you quote, but even as a non-expert I can think of of situations where any macroscopic area A cannot be conceivably justified: see, it is not obligatory to study photelectric effect on metal plates/samples. If one tries to eject electrons from some rarefied noble gas, you cannot argue that electron wave functions can extend beyond microscopic sizes. Although I showed above that the conservation-of-energy argument works even for a macroscopic A and thus a microscopic A is not absolutely necessary, it is clear that a microscopic A (i.e., from several to many orders of magnitude smaller than A = 1 mm^2) enables

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

to exceed nanosecond, and especially nowadays relevant femtosecond time-scale, for much larger intensities [E^2] than considered above in the A = 1 mm^2 case.
 
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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...."
Let me just add to my reply above that it is very helpful that among the authoritative experts on semi-classical interpretations you mentioned Scully, because exactly his work is sometimes credited with making photons unnecessary. For example, Greenstein and Zajonc state in their book "Quantum Challenge" (page 23, first edition, 1997, Jones & Bartlett Publishers) that "... in 1969, Jaynes and Lamb and Scully showed that one can account for the photoelectric effect without recurse to the concept of the photon at all."

So, it is very instructive to note that Scully, in spite of being one of not only experts but even AUTHORS of the semi-classical interpretation, actually advocates the view that in the case of photoelectric effect, the classical EM field cannot quite consistently handle the conservation of energy, in spite of all others successes (which he himself partially discovered or clarified).
 
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So, it is very instructive to note that Scully, in spite of being one of not only experts but even AUTHORS of the semi-classical interpretation, actually advocates the view that in the case of photoelectric effect, the classical EM field cannot quite consistently handle the conservation of energy, in spite of all others successes (which he himself partially discovered or clarified).
That's what makes it all the more baffling to me that he would use the cross-section of the atom in his calculation, which is clearly wrong.
 

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