Take the 2-minute tour ×
Physics Stack Exchange is a question and answer site for active researchers, academics and students of physics. It's 100% free, no registration required.

I was discussing laser cooling in class today and I understood that the main principle of the process is to tune a laser to a frequency lower than the absorbtion frequency of the atom and so only the atoms with a certain velocity will absorb the photons. These atoms then re-emit a photon in a random direction thus on average cooling the atom. But I didn't understand the difference between the so called 'doppler cooling limit' and the recoil limit.

I've read on hyperphysics.com that the doppler cooling limit has something to do with random walk motion. Isn't this, though, just the atom recoiling in random directions thus the recoil limit?

Could someone explain the difference?

share|improve this question

1 Answer 1

up vote 3 down vote accepted

The Doppler cooling limit is due to the fact that as the atoms absorb photons and spontaneously emit them in random directions they will not only have momentum no smaller than that of a laser photon, but they must also scatter one photon momentum in a random direction every natural lifetime of the excited state. If this lifetime is short then the random walk in momentum space occurs at a rate too fast to remain near the origin.

On the other hand, if the spontaneous emission rate is not an issue then the atoms really can get to having a single photon momentum and have the associated temperature, $T_\textrm{recoil}=\frac{1}{k_B}\frac{(\hbar k)^2}{2m}$, which is usually quite lower than one can achieve by Doppler cooling. This, however, requires more complex methods like Sisyphus cooling to reach. The difference is that this second class of methods relies on coherent scattering of photons from one beam into a second beam with opposite direction and polarization. Doppler cooling, on the other hand, requires spontaneous emission of photons, and is therefore controlled by the natural lifetime of the excited state - or, equivalently, its inverse, the natural linewidth $\gamma$.

To see how this happens, consider the total heating and cooling power of the beam, at intensities far below saturation. The total heating power is roughly one photon's kinetic energy per natural lifetime, so it's $\frac{(\hbar k)^2}{2m}\gamma$.$^1$ The force on the atoms is roughly the photon momentum times the detuning (why? because $F$ must be proportional to $v$ at low $v$ for the cooling to work, and $v$ can only enter through the detuning $kv$ as this is Doppler cooling. This is a rate and $F$ is a rate of loss of momentum, so one must have $F=\hbar k\cdot kv$). The cooling power is then $Fv=\hbar k^2 v^2$.

If one then equates both these rates of energy flow, as they will be in equilibrium, one gets $\gamma\hbar^2 k^2/2m=\hbar k^2 v^2$, and the mean velocity will be $$ \frac{1}{2}k_B T_\textrm{Doppler}=\frac{1}{2}mv^2=\frac{1}{4}\hbar \gamma. $$

Why then does this scale like that? Think about it in the large-$\gamma$ limit. (How large is large? Naturally, the comparison is $\hbar \gamma \gg \hbar^2 k^2/2m$ so the following effect trumps the recoil limit.) In effect, you not only have to have one-photon kinetic energy at all times, but you also have to add to the atom's velocity a one-photon momentum, in a random direction, every natural lifetime. If $\gamma$ is relatively large then you need to do this quite often! This of course means that you will not get down to single-photon momentum because the random walk insists on carrying you away from that zone around the origin faster than you can cool down the atom. If $\gamma$ increases, then the rate of the random walk increases, and at constant cooling power that will imply a higher mean amplitude for the random walk.


$^1$ Both the cooling and heating power must be multiplied by the intensity, suitably normalized, and near saturation are more complicated functions of it. For the present treatment one can ignore this since both heating and cooling are proportional to the intensity in the far-below-saturation regime.

share|improve this answer

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.