- #1

Niles

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Hi

When talking about atoms absorbing a photon, a classical picture is utilized. The intuitive picture often presented is that when an atom absorbs a photon, a recoil of [itex]\hbar k[/itex] is obtained by the atom. After a time [itex]\tau=1/\Gamma[/itex] the atom emits the photon again. So the force in this picture is (under the condition that the atom is in resonance with the light so that the probability of absorbing a photon is maximal)

[tex]

F = \frac{\Delta p}{\Delta t} = \frac{\hbar k }{\tau} = \hbar k \Gamma

[/tex]

However utilizing another picture, the force will be given by the scattering rate multiplied by the momentum recoil, i.e.

[tex]

F = \hbar k \Gamma \rho_e

[/tex]

where ρ

Why does these two expressions differ? What is it that my first scenario has not taken into account?

Best,

Niles.

When talking about atoms absorbing a photon, a classical picture is utilized. The intuitive picture often presented is that when an atom absorbs a photon, a recoil of [itex]\hbar k[/itex] is obtained by the atom. After a time [itex]\tau=1/\Gamma[/itex] the atom emits the photon again. So the force in this picture is (under the condition that the atom is in resonance with the light so that the probability of absorbing a photon is maximal)

[tex]

F = \frac{\Delta p}{\Delta t} = \frac{\hbar k }{\tau} = \hbar k \Gamma

[/tex]

However utilizing another picture, the force will be given by the scattering rate multiplied by the momentum recoil, i.e.

[tex]

F = \hbar k \Gamma \rho_e

[/tex]

where ρ

_{e}is the probability to be in the excited state, which saturates to the maximum value 1/2.Why does these two expressions differ? What is it that my first scenario has not taken into account?

Best,

Niles.

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