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Wave-particle duality and the photo-electric effect

  1. May 4, 2009 #1
    After completing a course in Quantum Mechanics and feeling rather dissatisfied with my physical understanding of the theory, I turned to Introductory Quantum Mechanics by Liboff, as I was advised that it was a much more clear treatment. I have barely started reading it and I am already immensely grateful for this book as my understanding is already deepened by the introductory chapters, but questions are also already forming. I am hoping someone will be able to provide me with some answers.


    I have two questions pertaining to the wave-particle duality in nature. Liboff explains it in terms of the Heisenberg uncertainty principle for position and momentum. The more accurately you know the position, the more localized your wave function is and therefore the more like a particle your system is. The more accurately you know the momentum, the less localized your wave function and the more wavy your system is.

    This makes a great deal of sense to me and it works wonders in explaining a double slit experiment. If you manage to know position accurately enough to say you are launching a single photon, you won't get any interference. If you don't know position very accurately, you have a wave and you'll get interference.

    I am having trouble, however, applying this explanation to, for example, the photo-electric effect. In this experiment, your photons are apparently always particles. However, the experiment merely requires you to send light onto a metal, it does not seem to require that you know the positions of your photons. Why could you not have a situation in this experiment where your light is a wave, rather than a stream of particles? Is there something inherently different?


    My second question pertains to photons and the probability density interpretation. If matter waves give you the probability of finding your particle somewhere, I'm assuming electromagnetic waves give you the probability of finding photons somewhere. Electromagnetic waves do not, however, ever seem to be treated as such, say in Electrodynamics. I think of electromagnetic waves of waving vector fields that permeate space. I am not sure what they are supposed to tell me about photons.

    Let's say we have an electromagnetic plane wave. Plane waves are momentum eigenstates, describing free particles. Presumably, an electromagnetic plane wave describes a completely delocalized photon. What strikes me as odd, perhaps unjustly, is that an electromagnetic plane wave can be very extensive. Should I think of it, then, as describing a single photon? Let's say we have a source emitting such a wave (let's say it has been emitting since t = -infty and we have a real plane wave, not a wave packet), or perhaps a spherical wave which we view at a large distance from the source. Is that source really putting out a single photon? That seems odd to me, especially since this wave can be detected in any direction (the electromagnetic wave hardly collapses into a localized photon when we detect it somewhere, I would think) at any time, since the fields can exert forces. Perhaps there is not such a clear link between electromagnetic waves and wave functions as I am proposing?


    I hope my train of thought in this is clear enough and that someone will be able to answer some of my questions.
     
  2. jcsd
  3. May 4, 2009 #2
    You can consider the photoelectric effect as an electromagnetic device that "measures" the energy of a photon; it requires the wave to have a specific energy near a specific location, but the electromagnetic wave is not restricted to a specific energy, time, or location at any earlier time. Using the uncertainty principle delta E delta t = hbar, if the photoelectron energy is a measurement of the photon energy, there has to be a corresponding uncertainty in the timing of photoelectron emission.

    H-bar is 6.65 x 10^-16 eV-sec, so "measuring" photon energy to 1 eV by creating a photoelectron then corresponds to a time uncertainty of nearly 10^-15 sec. This time uncertainty in turn is the transit time of a 1 eV wavelength/2 pi (1.2 microns/2 pi).
     
  4. May 4, 2009 #3

    ZapperZ

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    The photoelectric effect cannot determine the "localization" of a photon. It isn't meant to do that. All it does is produce a strong argument for the discreteness of energy of photons.

    Other experiments, such as the which-way experiments, are more designed to show such a thing. See, for example, http://people.whitman.edu/~beckmk/QM/grangier/Thorn_ajp.pdf" [Broken].

    Zz.
     
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