Thread: Time scale of the atom View Single Post
P: 1,863
 Quote by M Quack Note the formula they give, $\omega_{\mathrm{rec}} = \frac{\hbar k^2}{2m} \ll \Gamma$ If you multiply both sides (all 3 sides :-) ) by $\hbar$, the formula compares 3 energies. $\frac{\hbar^2 k^2}{2m}$ is a kinetic energy $\hbar \omega$ is the energy of an oscillator or wave (written as such to derive a characteristic time scale from the energy) $\hbar \Gamma$ is the width of an emission line (for example), which is finite because of the finite life time of the initial state.
Ah, thanks for making that clear.

 Quote by M Quack adiabatic in this context means without transfer of energy.
I have to admit I still don't fully understand "adiabatic" in this context. So the internal dynamics follows the external COM-motion without any energy transfer(?). I'm not even sure I know what that means.

Does it refer to the fact that the spontaneously emitted photons (characterized by $\Gamma^{-1}$) are emitted *much* faster than the atom moves, so their effect is zero on average since they are emitted so often?