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

  • 3
    $\begingroup$ It would be helpful if you include links to specific derivations with these features, just to set a common reference to talk about. $\endgroup$ Commented Sep 17, 2015 at 12:41

2 Answers 2


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

  • $\begingroup$ 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. $\endgroup$ Commented Sep 18, 2015 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
    Commented Sep 18, 2015 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
    Commented Sep 18, 2015 at 13:51

Suppose we have a two-level atom, at rest, impinged on by an optical field. The steady-state solution for the excited state population of the optical Bloch equations is

$$ \rho = \rho_{ee} = \frac{1}{2}\frac{I/I_{sat}}{1 + I/I_{sat} + 4\Delta^2/\Gamma^2} $$

$\rho=\rho_{ee}$ is the excitated state population, $I$ is the laser intensity at the location of the atom, $I_{sat}$ is that satuation intensity, $\Delta = \omega_L - \omega_A$ is the laser-atom detuning ($\omega_L$ is the laser frequency and $\omega_A$ is the transition frequency) , and $\Gamma$ is the transition linewidth. In atomic/optical steady, the atom will scatter (absorb and emit) photons at a rate $R_{sc} = \rho \Gamma$.

Due to the relativistic Doppler effect, an atom moving with velocity relative to the source of the laser beam will, to first order, see an effective detuning of $\Delta \pm k_L v$ where $k_L = \omega/c$ is the laser wavevector with $c$ the speed of light and $v$ the atomic velocity. Note that in this answer I assume a one-dimensional problem so $v$ is either positive or negative, but does not have a 3D vector character. We select the $+$ sign if the laser is propagating to the left and the $-$ sign if it is propagating to the right.

Each photon carries momentum $p_r=\hbar k_L$ where $p_r$ is called the recoil momentum. When the photon is absorbed this momentum is transmitted to the atom. The repeated absorption of photons results in a force $$ F = \hbar k_L R_{sc} = \hbar k_L \Gamma \rho $$

Each photon that is absorbed is re-emitted in a random direction. This gives the atom a momentum kick of $p_r = \hbar k_L$ in a random direction. If the initial momentum is $p_0$ before the emission then the momentum after the emission is $p_f = p_0 \pm p_r$ and the energy after the emission is $$ \frac{p_f^2}{2m} = \frac{p_0^2}{2m} + \frac{p_r^2}{2m} \pm \frac{p_rp_0}{m} $$ but over many emissions the final terms averages to zero since $p_r$ can be positive or negative (recall we are working in 1D). So the random emissions add an average recoil energy of $E_r = p_r^2/2m$ each time. This gives rise to a heating rate $$ R_H = E_r R_{sc} = \frac{(\hbar k_L)^2}{2m} \Gamma \rho $$

Now suppose we have two counter-propagating laser beams with wavevectors $\pm k_L$. Suppose also that the above formulas hold independently for the two laser beams. This can be arranged, e.g. if the polarizations of the two laser beams are orthogonal. We write $$ \rho_{\pm} = \frac{1}{2}\frac{I/I_{sat}}{1 + I/I_{sat} + 4 \frac{(\Delta \mp k_Lv)^2}{\Gamma^2}} $$ where $\rho_{\pm}$ is the steady state excited state population due to the laser travelling to the right (left). In the limit $k_Lv \ll \Delta$ we can Taylor expand this as \begin{align} \rho_{\pm} \approx& \frac{1}{2}\frac{I/I_{sat}}{1+I/I_{sat} + 4\Delta^2/\Gamma^2} - \frac{1}{2}\frac{I/I_{sat}}{(1+I/I_{sat} + 4\Delta^2/\Gamma^2)^2} \frac{8\Delta(\mp k_L)}{\Gamma^2}v\\ =&\rho_0 \pm \rho_0 \frac{1}{1+I/I_{sat}+4\Delta^2/\Gamma^2} \frac{4\Delta k_L v}{\Gamma^2} \end{align}

The net force on the atom is $$ F = F_+ - F_- = \hbar k_L \Gamma (\rho_+ - \rho_-) $$ From the Taylor expansion above we can see $F \propto \Delta v$. Therefore if $\Delta < 0$, i.e. the laser is red-detuned, then $F\propto - v$ In this case $F$ provides a cooling force on the atom. Specifically, it does work on the atom in an amount $$ R_C = F v = \hbar k_L \Gamma(\rho_+ - \rho_-)v $$ The net heating rate is $$ R_H = \frac{(\hbar k_L)^2}{2m} \Gamma (\rho_+ + \rho_-) $$

The two lasers will extract energy from the system, slowing the atom down, until the heating and cooling rates sum to zero $$ R_C + R_H = \hbar k_L \Gamma(\rho_+ - \rho_-)v + \frac{(\hbar k_L)^2}{2m} \Gamma (\rho_+ + \rho_-) = 0 $$ We begin some algebra \begin{align} \frac{\hbar k_L}{2mv} =& \frac{\rho_- - \rho_+}{\rho_+ + \rho_-}\\ =& -\frac{1}{2} \frac{\Delta/\Gamma}{1 + I/I_{sat} + 4\Delta^2/\Gamma^2} 8 \frac{k_Lv}{\Gamma}\\ =& -4 \frac{\Delta/\Gamma}{1 + I/I_{sat} + 4\Delta^2/\Gamma^2}\frac{k_Lv}{\Gamma} \end{align} Re-arranging $$ \frac{1}{2}mv^2 = -\hbar \Gamma \frac{1}{16} \frac{1 + I/I_{sat} + 4\Delta^2/\Gamma^2}{\Delta/\Gamma} $$ So we have solved for the steady-state kinetic energy when the velocity is such that the cooling rate equals the heating rate. Since the problem is one-dimensional the equipartition theorem tells us $$ \frac{1}{2}m v^2 = \frac{1}{2}k_BT $$ So we can solve $$ k_B T = -\hbar \Gamma \frac{1}{8} \frac{1 + I/I_{sat} + 4\Delta^2/\Gamma^2}{\Delta/\Gamma} $$

If we have $I\ll I_{sat}$ then we see that this is minimized for $\Delta = -\Gamma/2$ and we get $$ k_BT_D = \frac{1}{2} \hbar \Gamma $$

For reference we express the total cooling force $$ F = -8\hbar k_L \frac{I/I_{sat}}{(1+I/I_{sat} + 4\Delta^2/\Gamma^2)^2} \frac{\Delta}{\Gamma} k_L v $$ We can write this as $F = -\beta v$ with damping coefficient $$ \beta = 8\hbar k_L^2 \frac{I/I_{sat}}{(1+I/I_{sat} + 4\Delta^2/\Gamma^2)^2} \frac{\Delta}{\Gamma} $$

We can plot the equilibrium temperature $T_D$ and damping coefficient $\beta$ as functions of $I/I_{sat}$ and $\Delta/\Gamma$

enter image description here

(Just made these plots now, no guarantee of correctness..)

We can see that for $\Delta = -\Gamma/2$ as $I/I_{sat}$ decreases we approach the Doppler temperature $\hbar \Gamma/2$. However, as $I/I_{sat}$ decreases the damping rate becomes arbitrarily small. The damping rate is maximized around $\Delta = -\Gamma/2$ but at $I/I_{sat}\approx 2$. So we see that there is a tradeoff between the minimum achievable temperature and the maximum cooling rate.

Frankly, I haven't been able to come up with a more intuitive or simple explanation for the Doppler temperature than this. Often people say something like "The Dopler temperature must be Gamma because it is limited by spontaneous emission". But polarization gradient cooling is also limited by spontaneous emission, but it reaches the recoil temperature.

I think an answer might be realized with slightly less algebra by approximation $I\ll I_{sat}$ earlier to simplify equations and then recognizing $$ \frac{\rho_- - \rho_+}{\rho_+ + \rho_-} $$ as being related to the derivative of $\rho$ divided by $\rho$, noting that $\rho$ is simply a Lorentzian centered at $\Delta$. Then all of the physics comes from the recoil momentum, the recoil energy, the scattering rate, and a factor of order unity related to the ratio of a Lorentzian lineshape to its derivative. But still, no answer that can be given with words alone.

  • $\begingroup$ There's a factor of 2 issue in this answer. An algebra error was made in copying over the Taylor expansion of $\rho_{\pm}$ when calculating $(\rho_- - \rho_+)/(\rho_+ + \rho_-)$. Had it been copied correctly it would have given an incorrect Doppler temperature of $\hbar \Gamma/4$. I think the problem is I'm not including the heating due to absorption of photons, only emission. I'm editing this answer to correct this and make everything more clear. $\endgroup$
    – Jagerber48
    Commented Apr 22 at 16:03

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

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