Optical Molasses

An optical molasses is formed by laser cooling of atoms using the spontaneous light force. We study this first by reviewing the spontaneous light force, then investigating a one-dimensional molasses. This provides an excellent scenario to establish an important limit to laser cooling, the Doppler limit, beyond which laser Doppler cooling fails due to the balance established between momentum loss and diffusion of momentum due to the randomness of the classical light field. We then describe the three-dimensional molasses, and conclude with a discussion of laser cooling as an illustration of the fluctuation-dissipation theorem of statistical physics.

The spontaneous light force

Consider a two-level atom with energy spacing $\omega _{0}$, interacting with a single mode laser beam:

Let the laser intensity be $I$, and the interaction matrix element between atom and light be $\hbar \omega _{R}={\vec {E}}\cdot |\langle 1|e{\vec {r}}|2{\rangle }|$, where $\omega _{R}$ is known as the Rabi frequency, ${\vec {E}}$ is the electric field strength, and $e{\vec {r}}$ is the dipole moment of the atom.

It is useful to define a quantity known as the saturation intensity $I_{0}$ as the intensity of light at which the rabi frequency becomes $\omega _{R}=\Gamma /{\sqrt {2}}$, where $\Gamma$ is the spontaneous emission rate (the natural decay rate of the atom from $|2{\rangle }$ to $|1{\rangle }$, excited to ground state). This gives

${\frac {I}{I_{0}}}={\frac {2\omega _{R}^{2}}{\Gamma ^{2}}}\,.$

The rate at which photons are scattered from the atom is known to be

$\gamma _{s}={\frac {\Gamma }{2}}{\frac {I/I_{0}}{1+{\frac {I}{I_{0}}}+(2\delta /\Gamma )^{2}}}\,,$

where $\delta$ is the frequency detuning of the laser from the center of resonance of the atom. Two useful limits of this scattering rate are

${\begin{array}{rcl}\lim _{I\rightarrow \infty }\gamma _{s}&=&{\frac {\Gamma }{2}}={\frac {1}{2\tau }}\\\lim _{I\rightarrow I_{0}}\gamma _{s}&=&{\frac {\Gamma }{4}}\,,\end{array}}$

where we have assumed $\delta =0$ (resonant light). These expressions have a natural physical interpretation: in the limit of infinite intensity, the atomic levels become equally populated between the excited and ground state, and thus only half the atoms (the excited ones) can scatter light. Thus, the scattering rate is $\Gamma /2$ in that limit.

Suppose the force imparted by light on the atom is given by the recoil of photons spontaneously emitted from the atom. This force would then be

$F=\hbar k\gamma _{s}\,,$

where $\hbar k$ is the momentum of each photon. This expression makes several assumptions: that is is the net momentum transfer in absorption, that there is no "stimulated" force, and that $\hbar k/mk\ll \Gamma$, meaning that the jump in the Doppler shift is less than the natural linewidth.

Typically, for alkali atoms, this force is $F\leq 10^{5}g$ times the mass of an atom ($g$ being the acceleration due to gravity). This means that light can stop a sodium atom going at $1000$ m/s in one millisecond, or about half a meter. In comparison to electrostatic forces on ions, this is very small, however: it is comparable to the force exerted by an electric field of 1 millivolt/cm on an ionized sodium atom.

Moving atoms experience a Doppler shift, which we can model as a frequency dependent force, based on Eq.(\ref{eq:ci:lorentzian}), as

$F=\hbar k{\frac {\Gamma }{2}}{\frac {I/I_{0}}{1+{\frac {I}{I_{0}}}+\left[{\frac {2(\delta +kv)}{\Gamma }}\right]^{2}}}\,,$

where $v$ is the velocity of the atom. The effect of a fixed laser frequency on an ensemble of atoms is to modify their Maxwell-Boltzmann thermal velocity distribution:

Note how the initial distribution changes to one with atoms piling up below the velocity group resonant with the laser. The atoms bunch. Historically, this is the first method that was done to cool atoms to Kelvin temperatures.

One-dimensional optical molasses

Let us now turn to a method which allows cooling of atoms to zero velocity. Consider two laser beams incident on an atom from opposite directions. We assume that the total force is the sum of the two forces, ignore standing wave effects, and take the laser intensity to be low compared with the saturation intensity, $I\ll I_{0}$. Taking the force to be the sum of two forces described by Eq.(\ref{eq:ci:vdf}), we find that the two lorentzians sum to give the following force as a function of velocity:

The velocity dependent force is positive from one light beam, and negative from the other. With a detuning chosen such that force is zero at zero velocity, the force around $v=0$ can be expanded linearly, giving

$F(v)=\alpha v\,,$

where $\alpha$ describes the viscosity imparted by the light force to the atom, reflecting the restoring force applied when the atom is not at zero velocity. This configuration is known as an optical molasses, because of this restoring force, which makes the light behave like a thick, viscous medium for the atoms in it. The damping coefficient $\alpha$ can be calculated to be

$\alpha =2\hbar k^{2}{\frac {(2I/I_{0})(2\delta /\Gamma )}{\left(1+\left({\frac {2\delta }{\Gamma }}\right)^{2}\right)^{2}}}\,.$

The Doppler cooling limit

We have seen that the spontaneous light force, characterized by the Lorentzian response of an atom to light, together with the Doppler shift due to movement of the atom, gives a velocity dependent force, which can be zero at zero velocity. Does this mean that the atoms can be cooled to zero temperature?

If the rate of energy loss due to cooling is

${\dot {E}}_{\rm {cool}}=Fv=-\alpha v^{2}=-{\frac {2\alpha }{M}}E\,,$

then we should reach zero velocity, and zero temperature. Indeed, the kinetic energy decays exponentially. However, the spontaneous force has a random character, and thus has fluctuations which limit the minimum temperature achievable.

Momentum diffusion limit

This limit is determined by momentum diffusion. The force imparted can be described by a random walk. The final momentum is

$p_{\rm {final}}^{RMS}=\hbar k{\sqrt {N}}\,,$

on average, due to the random walk. Note that the momentum spread is

${\begin{array}{rcl}\langle p^{2}\rangle &=&(\hbar k)^{2}N\\{\frac {d\langle p^{2}{\rangle }}{dt}}&=&(\hbar k)^{2}\gamma _{s}\,.\end{array}}$

This describes heating which arises due to photons randomly scattering in all directions, such that the net momentum almost adds up to zero, but not quite. There is also a similar term due to absorption: some atoms will absorb more or fewer photons, due to the Poissonian statistics of absorption.

Time variation of kinetic energy

The time variation of the kinetic energy due to the spontaneous emission is forces is

${\dot {E}}_{\rm {heat}}^{\rm {em}}={\frac {\hbar ^{2}k^{2}\gamma _{s}}{2M}}={\frac {D^{\rm {em}}}{M}}\,,$

where $D$ is the momentum diffusion coefficient

$D={\frac {1}{2}}{\frac {\mathrm {d} }{\mathrm {d} t}}\langle p^{2}\rangle \,,$

And $D^{\rm {em}}$ is $D$ in the special case that we only consider the heating due to spontaneous emission. Later we'll see that $D$ is a correlation function of the force fluctuations.

Accounting for the Poissonian variance in number of absorbed photons adds a factor of two to the heating rate, which yields

${\dot {E}}_{\rm {heat}}={\frac {2\hbar ^{2}k^{2}\gamma _{s}}{2M}}={\frac {D}{M}}\,,$

Balance of heating and cooling

Let us now derive the Doppler limit for cooling. In equilibrium, ${\dot {E}}_{heat}={\dot {E}}_{cool}$. This means

${\frac {D}{M}}={\frac {2\alpha }{M}}E\,.$

The heating rate is independent of kinetic energy, whereas the cooling rate is a function of kinetic energy. So as the atoms cool down, the cooling rate slows down, resulting in a final temperature equilibrium being reached:

$2E_{final}=kT_{\rm {doppler}}={\frac {D}{\alpha }}$

$\alpha$ is a viscosity parameter: it reflects transport. $D$ reflects mobility. Thus, this is an Einstein relation, a universal expression in statistical mechanics resulting from the fundamental theorem which relates dissipation to fluctuations.

The Doppler limit temperature

We've now obtained an expression for the Doppler limit temperature, a limit on the temperature an ideal two-level atom can be cooled to by laser beams,

$kT_{\rm {doppler}}={\frac {\hbar \Gamma }{2}}\,.$

This optimal temperature is achieved for $I\ll I_{0}$, and detuning of $\delta =-\Gamma /2$ (half a linewidth). Physically, at low temperatures, the atom cannot determine whether the photon comes from left or right; at higher temperatures, the atom can discriminate whether photons come from left or right, thus cooling. For sodium, this temperature is $240$ $\mu$K, corresponding to a velocity of $30$ cm/s.

3D molasses, high intensities

To cool atoms along not just one axis, but along three axes, use six counter-propagating laser beams. This configuration is called a 3D molasses. Everything we've discussed in one dimension can be applied; just sum up the forces. Some care must be taken, however, if interference patterns are created between the beams. As long as the atoms move a distance greater than the wavelength, interference may be neglected. But large field gradients can add extra forces and heating.

One can also alternate between the six beams, but having simultaneous beams actually turns out to be better, due to the interference between the beams. In particular, it gives polarization gradients and other subtle effects which provide extra cooling. This wasn't initially forseen, but when implemented it was rapidly recognized that 3D cooling with six simultaneous beams was much more powerful than originally thought. A significant landmark was achieved when, in 1985, Steve Chu used chirped slowing and a 3D molasses configuration to obtain atoms colder than $1$ mK, for the first time (Original paper on optical molasses, Chu et al.).

Cooling at high intensities

Let us consider an example, of laser cooling at high laser intensities. Keep in mind that laser cooling works because $F=-\alpha v$ and $\alpha >0$. Assume we have a detuning of about one linewith, $\delta =-\Gamma$. Now plot the friction coefficient $\alpha$ as a function of intensity:

Initially, at small intensities, $\alpha$ increases as a function of intensity. Don't be confused by the fact that the Doppler limit is achieved at low intensities. The diffusion coefficient is also linear in intensity at low intensity. $\alpha$ increases with $I/I_{0}$ at first, and peaks around $0.5$, but above the saturation limit $\alpha$ actually changes sign and starts heating. When $\alpha <0$ then, counter-intuitively, blue detuned light can be used to cool atoms.

Cooling with blue detuned light

This is a non-trivial result (where does the energy go?), which can be undersood in the context of the optical Bloch equation and the dressed atom model (see Gordon and Ashkin). Specifically, the underlying physical reason which allows blue detuned light to cool is that at low intensities, the force seen by an atom comes from adding two Lorentzians, as we have seen; this fails at higher intensities, however. In particular, the optical Bloch equation component $u$ combines with the usual steady state term a new a velocity dependent term:

$u=u_{st}+v()\,.$

One can approximate that $u({\vec {r}})\approx u_{st}({\vec {r}}-{\vec {v}}\Delta t)$ When you average the light force in the standing wave over an optical wavelength, then you find an average force which is a friction force,

${\vec {F}}=-\alpha {\vec {v}}$

At weak intensities, $\alpha _{sw}=2\alpha _{tw}$, but at high intensities $\alpha$ changes sign.

Momentum and spatial diffusion

Let us return to the physics of the randomness of spontaneous light force induced cooling, and revisit the behavior of the diffusion of the cooled atom's momentum and spatial position. We shall see that the physical balance involved is an excellent example of the important fluctuation-dissipation theorem of statistical mechanics.

Momentum diffusion

First, consider diffusion of the momentum of an atom being cooled. The momentum diffusion coefficient is defined as

${\begin{array}{rcl}2D^{p}={\frac {d}{dt}}\langle (p-\langle p{\rangle })^{2}{\rangle }\,.\end{array}}$

This can be directly calculated if we have a fluctuating force, using the fact that $d{\vec {p}}/dt={\vec {f}}$ is a force:

${\begin{array}{rcl}2D^{p}&=&{\frac {d}{dt}}\left[{\langle }{\vec {p}}\cdot {\vec {p}}{\rangle }-{\langle }{\vec {p}}{\rangle }{\langle }{\vec {p}}\rangle \right]\\&=&2\left[{\langle }{\vec {p}}\cdot {\vec {f}}{\rangle }-{\langle }{\vec {p}}{\rangle }{\langle }{\vec {f}}\rangle \right]\\&=&2\int _{-\infty }^{0}{\langle }{\vec {f}}(0)\cdot {\vec {f}}(t)\rangle -\langle {\vec {f}}(0){\rangle }{\langle }{\vec {f}}(t)\rangle \,dt\,,\end{array}}$

showing that the diffusion is given by the integral of the force-force correlation function. Essentially:

${\begin{array}{rcl}2D^{p}&=&2\int _{-\infty }^{0}\langle {\vec {f}}(t)\cdot {\vec {f}}(0)\rangle \,dt\,.\end{array}}$

This results due to the fluctuation-dissipation theorem.

Spatial diffusion

Spatial diffusion is less frequently discussed in the literature compared with momentum diffusion, but it is of practical importance in experiments. Suppose the atoms start in a single point, embedded in a 3D optical molasses. How does the point distribution expand? On the time scale determined by $\alpha$, the atoms loose their memory of their original velocities. The molasses has a nearly perfect thermal distribution, despite atoms in the cloud never interacting with each other, because they thermalize to the laser beam.

The damping time is

${\frac {1}{\gamma }}={\frac {M}{\alpha }}\,.$

Spatial diffusion can be described by a random walk (in space), with a step size $\ell$ given by the RMS velocity of the atoms and the damping time,

$\ell ={\frac {v_{rms}}{\gamma }}\,,$

Thus, starting from a point distribution, by the standard random walk result, after time $t_{d}$, we obtain

$\langle r^{2}\rangle =2\ell ^{2}t_{d}\gamma \,,$

where the number of steps is $t_{d}\gamma$ and the extra factor of $2$ comes from a more rigorous treatment.. This is

${\begin{array}{rcl}\langle r^{2}\rangle &=&2{\frac {v_{rms}^{2}}{\gamma }}t_{d}\\=2{\frac {D^{p}t_{d}}{\alpha ^{2}}}\,.\end{array}}$

Now recall the definition

$\langle r^{2}\rangle =2D^{x}t_{d}\,,$

where $D^{x}$ is the spatial diffusion coefficient. This gives a relation between the spatial and momentum diffusion coefficients,

$D^{x}={\frac {D^{p}}{\alpha ^{2}}}={\frac {kT}{\alpha }}\,.$

Note the similarity of this expression with the Einstein relation for carriers in semiconductors, $D/\mu =kT/q$.

Example: laser cooling of sodium atoms

These expressions are useful in the laboratory context, as an example illustrates. How long does a typical trapped alkali atom (eg cesium or sodium) take to diffuse out by $0.5$ cm at the Doppler temperature? Using the formulas above, we get $t_{d}=1$ second. This is very accessible in the laboratory, and is one of the reasons why optical molasses are so useful in practice.