- #1
savi
- 9
- 0
why can't wave nature of light explain photoelectric effect?
Crosson said: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.
ZapperZ said:Can you show me where such a non-photon scenario is actually used in our modern experiments?
vanesch said:Well, although for more sophisticated photo-electric effects, it is useful to use a quantized EM field, you can look at chapter 9 of Mandel and Wolf (Optical coherence and quantum optics), "Semiclassical theory of photoelectric detection of light" where quite some aspects of the simple photoelectric effect are treated, with a classical EM field and a simple model of an electric dipole interaction with a bound electron.
So, the essential aspects of the photo-electric effect in its most basic form do indeed not need a fully quantized EM field.
ZapperZ said:Well, I did ask for where such a scenario is used to explain our modern experiments. In particular, where does such treatement is used in the physics of ARPES, XPS, RPES, multiphoton photoemission, etc? I've worked in ARPES for many years, and no once was there ever any description of everything that we observed that did not use the photon picture.
Let's get this out of the way once and for all. The "standard photoelectric effect" is a very crude and simplistic experiment. I can show you many experiments which "violates" the standard photoelectric effects "observations" (multiphoton photoemission, for example). The more stringent tests come in through those experiments that I have mentioned. There has been, as far as I know, absolute zero attempt to describe such experiments using anything other than the photon scenario. So this is what I'm asking in case I missed it, since people are already selling that alternative method. Can it do as well as what we already have?
Zz.
Parlyne said:Perhaps a better thing to say would be that the photoelectric effect as described by Einstein can be explained using fully classical light. This is what Mandel and Wolf discuss - that the gross features of the photoelectric effect do not require photons.
That said, there are certainly more detailed experiments that can use the photoelectric effect to demonstrate the need for photons. For example, the semiclassical theory used to derive the gross results without photons predicts that a photocurrent should be shot noise limited. A photon-based treatment allows lower noise, if the photons are created in number states (or, really, and combination with lower noise than a coherent state) and are detected with high efficiency. This effect can actually be seen in a rather simple table-top experiment.
So, photons aren't necessary to get the course photoelectric behavior; but, they are needed to explain the details.
Unfortunately I don't have that book. Can you explain me how that is made? Or point me to some file?Parlyne said:Perhaps a better thing to say would be that the photoelectric effect as described by Einstein can be explained using fully classical light. This is what Mandel and Wolf discuss - that the gross features of the photoelectric effect do not require photons.
Parlyne said:Perhaps a better thing to say would be that the photoelectric effect as described by Einstein can be explained using fully classical light. This is what Mandel and Wolf discuss - that the gross features of the photoelectric effect do not require photons.
That said, there are certainly more detailed experiments that can use the photoelectric effect to demonstrate the need for photons. For example, the semiclassical theory used to derive the gross results without photons predicts that a photocurrent should be shot noise limited. A photon-based treatment allows lower noise, if the photons are created in number states (or, really, and combination with lower noise than a coherent state) and are detected with high efficiency. This effect can actually be seen in a rather simple table-top experiment.
So, photons aren't necessary to get the course photoelectric behavior; but, they are needed to explain the details.
vanesch said:For all clarity, I'm not arguing against photons ! I'm arguing against an erroneous logic that claims the necessity of a certain statement while this doesn't follow from the premisse.
ZapperZ said:But vanesch, if you truly are a stickler for accurate phrasing, why didn't you jump all over crosson when he claimed that the alternative scenario is a "true explanation" for the photoelectric effect? I mean, if claiming the photoelectric effect is a "proof" of photons doesn't sit well with you, then claiming the classsical picuture picture as the valid explanation for that phenomenon should also rub you wrong.
The best that can be said is that for the photoelectric effect experiment alone, it tends to favor the photon explanation, but the classical light picture still cannot be ruled out.
Good enough for what?vanesch said:Eh, yes, I re-read that post and that's true. The semi-classical explanation is not a "true" explanation ; it is just good enough.
tehno said:Good enough for what?
What parts of the photo-electric effect ,in your opinion, "semi-classical" explanation explains and what parts it doesn't explain ?
It turns out, however, that for many purposes the quantization of the electromagnetic field is not necessary at all, and the response of the photdetector can be understood even if we continue to picture the field in terms of classical electromagnetic waves, provided the photoelectrons are treated by quantum mechanics. The field then simply behaves as an external potential that perturbs the bound electrons of the photocathode. Such an approach is sometimes known as semiclassical (...) and it is substantially simpler than the fully quantized treatment. Of course it has certain limitations and if pushed too far the semiclassical treatment will reveal some internal contradictions. However, that does not detract from its usefulness in many circumstances. As we shall see, for those electromagnetic fields for which an adequate classical description exists, the semiclassical and the fully quantized treatments of the photodetection problem yield virtually identical answers.
Crosson said: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.
tehno said:Just on the contrary:
The lack of time lags between incident light beam arrival and emmited photoelectron was the key point.
The electron is emmited at instant,that is and was measured experimental fact."Semi-classical" explanation of this aspect of the photoelectric effect,taking only Planck's quantization in the account,turns out to be insufficient there.
This is where Einstein steps in:EM field itself must be quantizied too.
Not having said that photon hypotesis explained more elegantly other aspects of the phenomenon,Einstein deserved Nobel solely on the basis of this insight.
Moreover,it was so revolutionary that even Planck or Bohr couldn't believe it at first.
Well,I'm in the process of reading [§],and it is my revelation to be informed that Bohr was,at one stage,more willing to accept breaking the law of energy conservation than Einstein's photons!
vanesch said:Now, when Einstein studied the photo-electrical effect, there was not yet a quantum theory of matter.
But if you apply simply Fermi's golden rule to a classical harmonic potential (as is done in every quantum text that respects itself) you will find that the probability of transition is immediately non-zero.
ZapperZ said:How does this fit in into the quantum description of a metal? We don't use a SHO potential for the conduction electrons.
As a side note, what do people mean by the emission of photoelectrons being either immediate or instantaneous? I mean, there is a finite, non-zero response time in such a process, which gets progressively longer as one goes from a metal to a semiconductor (fs to ps time scale).
vanesch said:I meant: the potential of the perturbation, due to a harmonic EM wave. Not the binding potential of the quantum system without it.
If the field is of the form E = E0 exp(i w t - i k x) + cc and we assume only dipolar electric perturbation (W = p.E) then you get a perturbation term which is time-dependent in an exp(iwt) way. I was simply pointing this out as the elementary treatment to reach Fermi's golden rule.
You are the expert on that, not me !
ZapperZ said:That's a dubious model to use. It means that you require a plane-polarized light source. The standard photoelectric effect requries a non-polarized source. Once you polarize it, a number of things can occur, especially on metal where the crystal orientation makes a difference.
vanesch said:If you know the responses to polarized light, then the unpolarized case should be a statistical mixture, no ?
Now, I would like to lift a misunderstanding which is still propagating here. I take it on your word (because you are the expert, not me, on this), that detailled calculations and angular distributions, and time profiles and anything detailled for specific photo-electric effects on specific materials might need the quantization of the EM field. If you say so, I can accept that.
What I thought what was under discussion here (and a detailled re-reading of the post by crosson made me doubt after the fact that this was what was actually being said, but I initially thought it was), was the following:
- the elementary properties of the photo-electric effect (which are: frequency threshold, "immediate" response and a few others ; the very first empirically known effects, which Einstein was aware of) do not strictly speaking require the quantization of the EM field.
You can, with a simple toy model, also crudely obtain those properties with the EM field acting as a classical perturbing potential on a quantized set of bound electrons. As I said, for more detailled aspects, I take your word for it that the EM quantization makes a difference, but we were talking here about the bare-bones properties of the photo-electric effect (those things that are done in a first or second year course on general physics).
Nevertheless, many textbooks claim the opposite (namely that those bare-bones properties require photons), without them arguing that it are in the detailled and sophisticated aspects of photo-electric effect that one has to look for indications of the quantization of the EM field.
ZapperZ said:I don't know. If we are using the classical picture of light, then an unpolarized light should have... what... the superposition of E field in all directions?
If you use that, then you really don't have a simple, 1D SHO potential anymore. Then what do you put the "bound electrons" in?
A spherically symmetry 3D SHO? Would this be something Mendel and Wolf used? Would such a thing produce the same effect? Or does one consider only a 2D SHO considering that only the in-plane E-field would be the one most responsible for the energy transfer? I don't know and that's why I'm asking.
I mean, even the old caloric theory of heat had SOME agreement with observations, but you don't see anyone still insisting that we must mention such a thing in thermodynamics textbooks just because it had some successes. This is why I am not that concerned when textbooks cite the photoelectric effect as "evidence" for the validity of the photon picture. While it isn't 100% correct, it is accurate in the sense that it is the only description that fits everything we have verified, not just the most simplistic scenario.
Yep,I'm positive (and I was also surprised to get known of it).vanesch said:Are you sure about that ? I thought is was in beta-emission that Bohr was ready to give up on energy conservation (because of the undetected neutrino). I didn't know he also was ready to do so for the photo-electric effect.
QuantStart said:A reader pointed out to me this discussion and that it was left unresolved.
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.
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.
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.
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...
monish said: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.
Hans de Vries said: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
The classical absorption cross-section is
[tex]\sigma=\frac{3\lambda^2}{2\pi}[/tex]
Which seems to agree with your numbers. Do you have any other links?
Regards, Hans
monish said: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
Hans de Vries said: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
monish said: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
reilly said: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