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Standing Waves and Radiation Emission

  1. Dec 1, 2011 #1
    In Lewis Epstein’s book Thinking Physics, he says that, since an electron behaves as a standing wave around the nucleus (exhibiting no known position changes, and thus no acceleration and no emission of radiation), something has to push part of the wave into a lower orbit to get the radiation process started. Since the electron would then have probability in both upper and lower orbits, there are essentially two standing waves now, which interfere with each other, constructively at some points and destructively at others, to make a more localized wave packet. This packet behaves like a particle orbiting the nucleus, and so emits a photon.

    Epstein further says that in empty space, an atom with enough energy can emit a photon because virtual photons, allowed by the uncertainty principle, are the initiators of the radiation process by colliding with the electron wave.

    I had never heard of this way of thinking in school, and was wondering if what Epstein says is true. To me, it would make more sense if a photon hitting an electron wave simply interferes with it, making a localized wave packet without the “two standing waves” middle step that Epstein describes. Anyway, I’m skeptical because the need for the virtual photon to explain spontaneous radiation seems too important and interesting to not be in textbooks. But is this really what goes on?
  2. jcsd
  3. Dec 2, 2011 #2
    David Griffiths mentions something like this in his QM textbook. After working out stimulated emission of radiation, he mentions the idea that we can think of the electromagnetic field as fluctuating slightly even in the vacuum, and these fluctuations can be thought of as stimulating what we normally call spontaneous emission. However I think this is a very heuristic and non-rigorous way of viewing things. It may be possible to talk about spontaneous emission as being the result of "virtual photons fluctuating in the vacuum" or some such, but I think it is not standard.

    The reasoning in your book sounds a little fishy. The idea that only accelerating charges radiate electromagnetic waves is a very classical idea. In quantum mechanics there is no well-defined concept of "acceleration." In fact electrons in the "standing wave" (atomic orbital) states can emit radiation and I would argue there is no need to invoke virtual photons to explain this.

    If one could somehow turn off the electromagnetic field (radiation), but keep the Coulomb attraction that binds electrons to nuclei, then an electron in any atomic orbital would stay there forever, and not drop down to the lowest. This is because the atomic orbitals are essentially *defined* as the unchanging, static, standing wave states of electrons in the absence of electromagnetic field interactions.

    I would make the following analogy to describe the effect of including the possibility of electromagnetic radiation. If I have a single pendulum and set it oscillating under ideal conditions it will oscillate forever. This is like an electron in some atomic orbital in the absence of electromagnetic field interactions. The oscillating pendulum is a simplification of the electron standing wave. Now let me introduce a second, initially unmoving pendulum representing the electromagnetic field and connect the weight of this pendulum to the weight of the electron pendulum by a spring. The spring is not very stiff, so it only offers a small resistance to the electron pendulum's oscillation. However over a period of time the electron pendulum will push on the EM field pendulum through the spring and set the EM field pendulum oscillating. As a result the electron pendulum transfers energy to the EM field pendulum and the electron pendulum's oscillations die down.

    This is an analogy for how the electron standing wave can excite a wave in the EM field (a photon), and as a result die down, dropping into a lower energy state of its atom.
  4. Dec 2, 2011 #3

    Ken G

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    Gold Member

    It's pretty much correct, but very simplified. What he seems to basically be talking about is something called a Rabi oscillation. A Rabi oscillation is, if you have a superposition between two energy eigenstates, the different modes acquire an oscillating phase difference. That phase difference oscillates at the frequency of the transition, so if you also have a photon passing through at that frequency, it gives it something to resonate with, and this can "stimulate" the downward transition. If the state is fully in the excited state to begin with, you might think it wouldn't Rabi oscillate because it is an energy eigenstate all by itself, but it's not an energy eigenstate any more once you include that passing photon. So the excited state is not an eigenstate of the atom + photon, and hence Rabi oscillates, which again gives that passing photon something to resonate with (if it's at the Rabi frequency, which is also the transition frequency).

    If there is no passing photon, then a "virtual photon" from the vacuum can indeed do the trick. That's called spontaneous emission. The virtual photon picture is just that, a useful picture, but what makes it seem like a nice picture indeed is the fact that the stimulated plus spontaneous rate is proportional to 1+n, where n is the photon occupation number in the mode that resonates with the transition. Given that n is thus the "number of photons passing by" at any given time, the "1" sure looks like the virtual photon that is becoming promoted to a real photon (by the energy available in the transition). Quantum field theory makes this correspondence more precise, but simple Einstein coefficients suffice to get this "1+n" effect.
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