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Understanding Stimulated Emission: Got the what, how about the why?

  1. Aug 20, 2011 #1
    I've read the literature on the matter, but I'm still not entirely clear what's going on. The general idea I'm getting is: "You have a photon and an excited atom, photon comes near atom, stuff happens, and now you have two (identical) photons, also identical to the first, and traveling in the same direction as the first, and an atom in a less exited state"

    I think I understand the concept of a laser, I know what a gain medium is, and so on, but I really would like to know what constitutes "photon approaches atom" and "stuff happens". Precisely what "stuff" happens? What is the interaction, and how does the incoming photon control the emission of the second photon? Does the first photon have to be absorbed by the atom, or only get "near" it? Can Stimulated Emission only happen when the incoming photon has exactly the same energy as the difference in potential energy between the current and lower electron shell?

    **Edit: not sure if this is a quantum physics question; move if appropriate I guess**
     
  2. jcsd
  3. Aug 20, 2011 #2

    K^2

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    You can use the gain medium with an approaching front, but that's not what is typically done. What usually is done, and what is much easier to think of in therms of "how", is a standing wave between two mirrors. In that case, photons that are being "amplified" are delocolized. So we can forget photon approaching the atom.

    What you are looking at, instead, is the electric and magnetic fields oscillating at the frequency equal to frequency of spontaneous emission from the gain medium. Now, the only question is why all of the atoms in the medium get synchronized.

    While the effect is, indeed, quantum, you can picture this classically. An atom emits radiation because it has an electric dipole. That dipole rotates, and rotating dipole emits energy. Of course, energy in an atom is quantized, so it can only emit energy at the specific frequencies. Now picture a dipole in an electric field. If they are not aligned, there is a torque on the dipole. Now picture a dipole and an electric field rotating at the same frequency. If the dipole leads, the torque is slowing it down. If it lags, the torque makes it catch up. Given some time, all of the rotating dipoles will match phases with the local electric field.

    So now you have a bunch of atoms, ready to emit radiation. All of their dipoles got synchronized by some photons that were emitted earlier, so when they emit, they keep increasing the oscillation of the electric field.

    When you go back to quantum picture, each emission event which adds to the amplitude of the electromagnetic wave that's already there, without changing its phase, just adds another photon to the count of photons in that exact state.

    There is a lot more to it, of course. There is the whole thing with laser light having effectively zero temperature, which is consistent with inverted populations in the gain medium. There is the bit about photons being bosons, which allows you to pump more and more energy into the same state, and so on. But that has more to do with why it's difficult to make a laser work than why it works.
     
  4. Aug 20, 2011 #3

    Claude Bile

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    Think of it as a driven oscillation. The incident EM field causes the electric dipoles to oscillate at the same frequency of the incident field. This, in turn, causes the dipoles to emit at the same frequency, direction, phase and polarisation of the incident field.

    As the photon is already excited, absorption will not occur. The photon needs to get near it in the sense that the atom needs to interact with the EM field that the photon is associated with.

    Yes, after all, it is still emission.

    Claude.
     
  5. Aug 20, 2011 #4
    Thanks, you guys are awesome :). Those are both great explanations and give me a different way to think of it.

    As to what constitutes near, does close enough have to do with the uncertainty of the photon's position? IE: if does the chance of the interaction occurring equal the probability of of the photon passing directly through the excited atom?
     
  6. Aug 20, 2011 #5

    K^2

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    It has to do with intensity of EM field, but since EM field is what gives you the wave function of a single photon, yes, you can think of it as probability of a photon passing through atom.
     
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