Optical Molasses: Is Polarization Important?

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Hi

When introducing optical molasses, the polarization of the counter-propagating beams are never taken into account. Is it really correct that the polarization of the beams are irrelevant, when trying to reach the Doppler-limited temperature?


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Yes. The Doppler limit can be derived by considering the interplay between the optical molasses cooling force and the heating effect of random spontaneous emission. The spontaneous emission can be modeled as a random walk in momentum space with steps of \hbar k. Nowhere in this derivation is the polarization of the beams relevant.

However, the counter-propagating beams do have a polarization, and this polarization can be used in conjunction with the atoms' magnetic structure to realize sub-Doppler temperatures. In this case, the polarization of the beams is highly relevant.
 
I see, thanks.

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