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Why can wave nature of light not explain photoelectric effect?

  1. Nov 22, 2006 #1
    :confused: why can't wave nature of light explain photoelectric effect?
     
  2. jcsd
  3. Nov 22, 2006 #2
    Simple answer: Classically we cannot explain the threshold frequency. Also in classical case the energy of the emitted electrons would depend on the intensity of light, which is not the case in experiments.
     
  4. Nov 23, 2006 #3
    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.
     
  5. Nov 24, 2006 #4

    ZapperZ

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    Really?

    I'm looking at Photoemission texts such as the one by Huffner, and at no part at all in the physics of XPS, ARPES, RPES, etc., is a "non-photon" scenario ever invoked. Can you show me where such a non-photon scenario is actually used in our modern experiments?

    Furthermore, the standard photoelectric effect involves the emission of photoelectrons from solids, typically metals. This means that it is an emission from a many-body system that has formed a collective phenomenon to produce bands, such as the conduction band - the atoms have mostly lost their individual identities. It is different than the phenomenon of photoionization that is from individual atoms and molecules. You do not get "resonances" corresponding to individual atoms. Even in ARPES, you only get the "quasiparticle" peaks corresponding to the band dispersion, not atomic energy level. So I really don't see how one treats the atom "quantum dynamically" and get the same results of a photoemission experiment, when the solid itself defies any description using "the atom".

    Aspect's team has won the nobel prize? What year was this?

    Zz.
     
  6. Nov 24, 2006 #5
    There is a big step not just from Maxwell to Planck in explaining photoeffect,but also between Planck and Einstein.Planck's breakthrough alone wasn't sufficient to explain it.
     
  7. Nov 24, 2006 #6

    vanesch

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    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.
     
  8. Nov 24, 2006 #7

    ZapperZ

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    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.
     
  9. Nov 25, 2006 #8
    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.
     
  10. Nov 25, 2006 #9

    ZapperZ

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    Sure, but that certainly does not make the Mandel-Wolf explanation as the true explanation which is what crosson claimed. In fact, I would say that it is not even close to being the true explanation, especially when it is not even used widely in explaining the various photoemission experiments that I have listed.

    Zz.
     
  11. Nov 25, 2006 #10
    Unfortunately I don't have that book. Can you explain me how that is made? Or point me to some file?
     
  12. Nov 25, 2006 #11

    vanesch

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    Yes, that was the point. The usual textbook arguments of the "need" for photons are wrong (in that you have "immediate" emission, that there is a threshold in frequency and all that, are usually presented as *necessary conditions* to have photons, and these are the things that are also obtained in the semiclassical approximation, as shown by Mandl and Wolf). This is not so widely known. Of course, for more sophisticated setups, the semiclassical approximation doesn't work well anymore, but these are not the arguments usually brought up in elementary textbooks to require the *need* for photons.

    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.

    This is a bit as saying that "the proof of the existance of electrons is electric current creating a magnetic field" or something of the kind. True, electrons exist, but this is not demonstrated by the fact that an electric current generates a magnetic field.
     
  13. Nov 25, 2006 #12

    ZapperZ

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    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.

    In fact, I find such claim to be the "true explanation" to be highly irresponsible. Why? Because there is more of a reason to consider the validity of the photon picture than the classical picture simply from looking at what people use in practice! When it matters the most, it is not the classical picture that is being used. In fact, in none of the experimental descriptions is the Mendel-Wolf scenario is ever, ever used. So how could anyone get away with making such an outrageous claim? There is more validity to claim that the photon picture is the "true explanation" instead simply based on what it can do beyond the naive photoelectric effect experiment.

    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.

    If someone asks "then why do we accept the photon explanation?", we then point out all the phenomena that goes beyond the simple photoelectric effect and show that the photon picture is used everywhere and that the classical light picture is never used nor formulated for such experiments.

    Zz.
     
  14. Nov 25, 2006 #13

    vanesch

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    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.

    Any claim of "true explanation" is of course irresponsible, whether it is classical, semi-classical, quantum mechanical or theological :cool:

    Agree. :approve:

    Or we could say that the photon picture succeeds in explaining the photo-electric effect, amongst other things.
     
  15. Nov 25, 2006 #14
    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 :confused: ?
     
    Last edited: Nov 25, 2006
  16. Nov 25, 2006 #15

    vanesch

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    It doesn't explain those cases where non-classical optical begin and/or end states appear (like those with negative Glauber densities).

    I'll quote Mandl and Wolf (on p 439):
     
  17. Nov 26, 2006 #16
    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!
    As far as concerns me ,that concludes this thread.

    Reference:

    [§]Max Born;"Physics in My Generation",Pergamon Press,London 1956.
     
  18. Nov 26, 2006 #17

    vanesch

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    This is one of the typical misunderstandings propagated by textbooks. The classical EM potential applied in perturbation to a "quantum electron" gives immediately rise to a finite emission probability.
    It is often argued that the "immediateness" of emission (which has indeed been experimentally established) is an argument for the quantization of the EM field, but this is not correct. It is an argument against an *entirely classical* picture (simply because there's not enough energy transmitted at first sight). 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.
    So you need SOME quantum system to obtain this effect. It can be the matter, or it can be the EM field, or of course as is generally accepted, both. But from the moment that one is quantized, one obtains the basic properties of the photo-electric effect.

    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.
     
  19. Nov 26, 2006 #18

    ZapperZ

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    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).

    Zz.
     
  20. Nov 27, 2006 #19

    vanesch

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    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 !

    But there is usually this reference to that old experiment with faint illumination where there isn't enough time for there to be sufficient energy transmitted by the classical beam to get to the energy of one single photo-electron. It is one of the usual textbook arguments to say that the EM field hence must be quantized, but that's not true of course. From the moment that there is the perturbing potential, there is a finite chance of finding the system having undergone a transition.
     
  21. Nov 27, 2006 #20

    ZapperZ

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    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. You can easily no longer get the standard photoelectric effect model. So I am not sure how valid of an explanation that would be using such a form.

    Remember, as soon as you put a higher level of complexity to the standard photoelectric effect, you CAN get differing results.

    Which is why you never see me mention anything about prompt or instantaneous emission. That's why I'm curous what you guys mean by it, regardless of where it came from.

    Zz.
     
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