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Im doing an essay on laser cooling, and I've got two questions about the derivation of the two limits. Firstly, in the derivation of the Doppler limit, it is stated in many places that the scattering rate is dependent on kv (the wavenumber of the light, k, multiplied by the velocity of the atom, v,). Now I sort of get how the rate at which photons arrive at the atom is dependent on those variables but I'm not sure exactly why there aren't other variables involved and would appreciate it if someone could explain this to me.

Secondly, the recoil limit includes a factor of a half, however this seems odd to me as the Energy gained by the recoiling atom has a factor of a half in it, and since we equate it with kT/2 surely the factors of a half cancel out? Many thanks for any help

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    $\begingroup$ It would be helpful if you include links to specific derivations with these features, just to set a common reference to talk about. $\endgroup$ – Emilio Pisanty Sep 17 '15 at 12:41
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The Doppler cooling limit should only depend on the natural linewidth of the transition, $\gamma$, since it is a process related to spontaneous emmision. The Doppler cooling limit is given is given by

$$ T_D = \hbar\gamma/2k_B $$

Even though the spontaneous emmision provides momentum kicks that average out to zero, the mean squared velocity will still be non-zero, resulting in some heating of the system. This is because the momentum kicks that you get from a broad transition upon spontaneous emission will naturally have a statistical distribution related to the linewidth, resulting in a range of final velocities for the atom (i.e. temperature).

The $kv$ term in the scattering rate equation is actually related to how the Doppler shift changes the scattering rate. You'll usually see this in the laser frequency detuning term of the scattering rate equation. There are other variables involved in the scattering rate as well, notably the laser intensity, transition linewidth and the saturation intensity of the atomic transition. The steady state excited state fraction for a two-level system is given by:

$$ \sigma = \frac{I/2I_{sat}}{1+I/I_{sat}+4\delta^2/\gamma^2} $$

where $\delta$ is the laser detuning, $\gamma$ is the linewidth of the transition, $I_{sat}$ is the saturation intensity of the atomic transition, and $I$ is the laser intensity. You can see that as the intensity goes to infinity, the excited state fraction approaches 1/2, so as to not create a population inversion. The scattering rate is the excited state fraction, $\sigma$, times the natural linewidth, $\gamma$ of the transition for low intensities (which are usually the case for laser cooling), since you can only scatter the laser light as fast as atoms are repopulating the ground state when in a steady-state configuration. A plot of the scattering rate as a function of laser detuning is seen below:

A plot of the scattering rate vs detuning

The black line represents the laser detuning in the lab frame, which is to the red of the atomic resonance. The blue line represents the effective detuning for an atom that is moving opposite to the direction of laser propagation, resulting in a Doppler shift to higher frequency (blue-shift), and a higher scattering rate (the momentum kicks from this will slow the atom). The red line represents the effective detuning for an atom moving along the direction of laser propagation, resulting in a Doppler shift to lower frequency (red-shift), and a lower-scattering rate (the momentum kicks from this will increase the velocity of the atom, resulting in heating for a cloud of atoms).

To ensure that on average, the momentum kicks provided will lower atoms' velocities, resulting in cooling, one can counter-propagate two red-detuned lasers (in 1-D), such that an atom moving toward either laser will experience a higher scattering rate from the laser that propagates opposite to its motion compared to the laser propagating along its direction of motion. The force on such an atom is given by the scattering rate times the momentum kick, $\hbar k$.

The Doppler cooling force in 1-D given by the following (for a two-level system):

$$ F = \frac{\hbar k\gamma}{2} \left(\frac{I/I_{sat}}{1+I/I_{sat}+4(\frac{\delta-kv}{\gamma})^2}-\frac{I/I_{sat}}{1+I/I_{sat}+4(\frac{\delta+kv}{\gamma})^2}\right) $$

You can see that the $kv$ term is added and subtracted to the laser detuning term, $\delta$. This comes from the Doppler shift as the atom moves in the light field. A plot of the cooling force as a function of velocity is seen below:

Doppler cooling force as a function of velocity, plotted with different laser detunings

In this plot, you can see that for low velocities, the Force on an atom varies linearly with velocity at low velocities. At high velocities, the force begins to decrease, and atoms with velocities outside the linear region are said to be outside the "capture range". As you change the detuning of the laser, you can see that the slope of the Force as a function of velocity changes. The linear restoring force as a function of velocity is necessary for cooling to work, since the goal of this system is to drive the atoms toward low velocity.

For a good explanation about the recoil limit, see this answer: Doppler cooling limit vs recoil limit

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  • $\begingroup$ I've actually been using that explanation in the link provided, and I suppose my uncertainty stems from one of the points made in it. In that explanation it is said that "v can only enter through the detuning kv" and I don't understand what is meant by this. $\endgroup$ – Lewis Keeble Sep 18 '15 at 10:44
  • $\begingroup$ Okay, I understand what the confusion is. In Doppler cooling, for the appropriate setup, the cooling Force is proportional to velocity for low velocity atoms. You'll see this if you plot the the equation for F above for the right parameters. If you change the detuning, the slope of the force changes. I'll work on an edit to my answer to explain this further an provide a plot like a I described for some different parameters later this morning. $\endgroup$ – tmwilson26 Sep 18 '15 at 11:45
  • $\begingroup$ I edited my answer above to add in the plots I discussed earlier. Let me know if this clears things up for you, and if you have any other questions, let me know and I'll try to address them. $\endgroup$ – tmwilson26 Sep 18 '15 at 13:51

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