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Classical Electron Oscillator Model, quantum analogy

  1. Oct 11, 2008 #1
    I originally posted this question in the homework section, but I really don't need any help calculating anything, my answers are right. I'm having conceptual trouble, so I figured that this question belongs here.

    So, let's say there is a field driving a single atomic oscillator (hydrogen for arguments sake). As a further simplification, let's say the oscillator has the same frequency as the driving field. I found the time average power absorbed by the oscillator and the time average number of electrons per unit time absorbed. (ie these quantities:)

    [tex] \bigl \langle P \bigr \rangle &=& \bigl\langle\frac{dW}{dt}\bigr\rangle\\ &=& \frac{1}{2}eE_0\omega\nu[/tex]
    (nu is just a constant that is frequency dependent... this equation is correct)
    [tex]\bigl\langle\frac{dN}{dt}\bigr\rangle=\frac{eE_0}{\hbar}\nu[/tex] (this equation should be right but I can't grasp it conceptually.

    What I can't understand is the factor of 2 that is associated with the number of electrons absorbed. Why isn't the number of electrons absorbed per unit time simply the time average power divided by [tex]\hbar\omega[/tex]?

    Does it have something to do with the fact that the oscillator makes two displacements from and back to equilibrium in one period? Or, does it have something to do with emission? Where is the extra energy going, heat? Any help getting my mind around this one would be VERY much appreciated.
  2. jcsd
  3. Oct 11, 2008 #2


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  4. Oct 11, 2008 #3
    Don't think so. I'm actually close to having the result I think. It just hit me that the [tex]\frac{dW}{dt}[/tex] is periodic with freq [tex]\frac{\omega}{2}[/tex], whereas the E-field is periodic with frequency [tex]\omega[/tex] I feel like it comes from there. Gonna work through it now.
  5. Oct 11, 2008 #4


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    I see. I thought you were asking about the factor of 2 in the power equation. For the photons, maybe the two polarizations of an unpolarized electromagnetic wave?
  6. Oct 11, 2008 #5
    Got it. Okay, so basically, how this works is that the work done is periodic with half the period of period of full oscillation. So, the time average of the power leaves us with the average power for 1/2 cycle. So, the power absorbed over a full cycle is twice that. Since the photons are absorbed in a full oscillation, (not the half oscillation of the work function) we get that [tex]\hbar\omega N = 2W[/tex] where N is the number of photons absorbed during the full oscillation. (This is due to the fact that the frequency of oscillation is the same for the atom as it is for the photon. The photons drive the atom through an entire cycle, not just to one side.) There's our factor of 2.
  7. Oct 11, 2008 #6
    Okay, changed my mind, that doesn't make sense, does it? Does this work because we started with a function for a full cycle and it just got reduced when we took the time average?
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