Dipole Oscillation: Calculation of Time Avg. Power & Photons Absorbed

[buttersrocks]

Let's say that the oscillation of a dipole polarization is given by: $$x(t)=u\cos\omega t - \nu\cos\omega$$. The driving field is given by $$E(t) = E\cos\omega t$$. (u and nu are the in-phase and out-of-phase components, respectively)

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

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.

 

[BigJDubs]

The answer is not exactly two photons per cycle because the dipole is only absorbing an average of 1/2 photon per cycle. This is because the dipole is only oscillating between the maximum and minimum values of polarization, so the work done by each photon is used to displace the dipole from one extreme to the other. This means that in each cycle, the dipole absorbs an average of 1/2 photon.