Laser cooling can be achieved by exciting an atomic system, in order to "pump" atoms in the excited state. This gives a recoil momentum $\hbar k$ the the atom, which then emits spontaneously. As spontaneous emission is an isotropic process, this does not give (statistically speaking) e recoil momentum to the atom, which is in this way "slowed down" by the laser. What is was wondering is: in which way stimulated emission can be ignored in this process ? Maybe both spontaneous and stimulated emission happen, but as the latter does not give momentun change (considering both absorption and emission), is considered as irrelevant? Or maybe that's linked to the choice of alkali, which have some sort of particular property?
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2 Answers
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You've hit the nail on the head.
Maybe both spontaneous and stimulated emission happen, but as the latter does not give momentum change (considering both absorption and emission), is considered as irrelevant?
This is exactly what happens. If the absorbed photon is emitted via stimulated emission, there is no net change in the momentum and the atom is neither cooled nor heated.
answered Oct 7, 2015 at 15:04
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$\begingroup$ So stimulated emission is not treated simply basing the study on the statistical point of view? Every brief I've read does not say anything about this, while I think that a word should be spent on this point. $\endgroup$– KlopmintCommented Oct 7, 2015 at 15:22
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Cycles of absorption / stimulated emission plays a crucial role in laser cooling. It is true the cycle does not change the momentum of the atom if the absorption and the emission take place in the same photon mode. But photon can be absorbed in one mode, and emitted in a second one.
This results in the so called optical dipole force, which is related to the spatial variations of the light intensity $$ F(r)=-\frac{3\pi c^2 \Gamma}{2\omega \Delta} \nabla I(r), $$ where $\Delta$ is the detuning of the light frequency $\omega$ with respect to the atomic transition and $\Gamma$ is the transition linewidth. Note that the involvement of different modes is included in the presence of the gradient - for a single mode (plane wave), the intensity is homogeneous.
This is how we trap optically atoms, before we perform optical evaporation - this is how the lowest temperatures are reached.
To get an alternative picture, the dipole force is related to the reactive part of the response function $\chi$ of the atom to the light. The radiative pressure, which is due to cycles of absorption / spontaneous emission, is related to the dissipative part of $\chi$.
