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Mar1-12, 12:51 PM
P: 1,863
Quote Quote by M Quack View Post
Note the formula they give,
[itex]\omega_{\mathrm{rec}} = \frac{\hbar k^2}{2m} \ll \Gamma[/itex]

If you multiply both sides (all 3 sides :-) ) by [itex]\hbar[/itex], the formula compares 3 energies.

[itex]\frac{\hbar^2 k^2}{2m}[/itex] is a kinetic energy

[itex]\hbar \omega[/itex] is the energy of an oscillator or wave (written as such to derive a characteristic time scale from the energy)

[itex]\hbar \Gamma[/itex] 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 Quote by M Quack View Post
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 [itex]\Gamma^{-1}[/itex]) are emitted *much* faster than the atom moves, so their effect is zero on average since they are emitted so often?

Thanks in advance.

Best regards,