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Dipole Oscillation

  1. Oct 10, 2008 #1
    Let's say that the oscillation of a dipole polarization is given by: [tex]x(t)=u\cos\omega t - \nu\cos\omega [/tex]. The driving field is given by [tex]E(t) = E\cos\omega t[/tex]. (u and nu are the in-phase and out-of-phase components, respectively)

    So, the time average power is [tex]P_{avg} = \frac{1}{2}eE\omega\nu[/tex]
    and the number of photons absorbed per unit time is [tex]N_{avg} = \frac{eE}{\hbar}\nu[/tex]

    I'm just stuck conceptually. If I calculate directly and divide the time averaged power by the energy of a photon, I still have the constant of 1/2.

    Is this dipole absorbing two photons per cycle because each photon is capable of only displacing the dipole to a maximum once, so the subsequent photon displaces it to the opposite maximum? In other words, the work is done to displace the dipole 2 separate times, once from the center to the righthand max and a second time from rest to the lefthand max? I'm getting this because the work completes 2 cycles in the time that it takes the polarization to complete 1.
    Last edited: Oct 10, 2008
  2. jcsd
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