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research34 min studyLesson 481 in Physics

Doppler and sub-Doppler laser cooling

Balance velocity-selective radiation pressure against recoil diffusion, then identify why multilevel polarization-gradient cooling can beat the two-level Doppler limit.

Before you begin
On this page

The question for today

What you’ll understand

  • Determine the damping sign for counterpropagating red-detuned light.
  • Use angular linewidth consistently in the Doppler temperature.
  • Separate two-level Doppler cooling from multilevel optical-pumping mechanisms.

Laser cooling reduces the random motion of atoms by transferring energy and entropy to scattered light. A moving atom absorbs preferentially from a beam opposing its motion when the light is tuned below resonance. Spontaneous emission does not cancel this mean force, but its random recoil heats the velocity distribution. Cooling is therefore a competition between a velocity-dependent drift and momentum diffusion. The familiar Doppler limit belongs to a specified two-level model rather than to every possible laser-cooling mechanism.

Velocity changes which beam is closer to resonance

Let two weak opposing beams have the same laboratory angular detuning, defined as laser frequency minus atomic resonance frequency. An atom moving along the positive axis sees the opposing beam shifted toward a red-detuned resonance and absorbs more photons from it. Each absorption imparts directed momentum, while spontaneous photons average to zero directed momentum over many cycles. Near zero velocity the difference of scattering forces is proportional to velocity with a negative coefficient for red detuning.

The low-intensity force formula treats each beam independently and ignores saturation cross-coupling, multilevel optical pumping, and magnetic shifts. Its linear expansion also requires Doppler shift small compared with the frequency scale over which the scattering rate changes. At larger speeds the force is nonlinear and the capture range matters. Blue detuning reverses the small-velocity damping sign in this elementary model, so increasing scattering alone does not guarantee cooling.

F(v)=ℏkΓs2[11+4(Δ−kv)2/Γ2−11+4(Δ+kv)2/Γ2]≃−αvF(v)=\frac{\hbar k\Gamma s}{2}\left[\frac1{1+4(\Delta-kv)^2/\Gamma^2}-\frac1{1+4(\Delta+kv)^2/\Gamma^2}\right]\simeq-\alpha v
k > 0 is optical wavenumber in m⁻¹, Γ is excited-state decay rate in s⁻¹, Δ is angular detuning in rad/s, and s ≪ 1 is the on-resonance saturation parameter of each beam. For Δ < 0, α > 0 has units kg/s.↔ Scroll wide equations horizontally.

A temperature follows from damping and diffusion

Absorption timing and spontaneous emission produce fluctuations around the mean force. Recoil diffusion prevents indefinite narrowing of the velocity distribution even while the mean velocity is damped. In the ideal low-intensity two-level three-dimensional Doppler model, balancing those effects gives a detuning-dependent equilibrium temperature. Its minimum occurs at red detuning equal to minus half the natural angular linewidth. The derivation assumes a semiclassical velocity distribution broad enough that recoil discreteness is not the dominant scale.

A linewidth quoted in megahertz often means the ordinary-frequency width, whereas the decay rate in the formula is that width multiplied by two pi. Confusing them makes the predicted temperature too small by that factor. Define the recoil temperature separately from the Doppler temperature; conventions sometimes differ by two in the recoil energy definition. Neither temperature is a direct measure of the mean drift of a cloud, which must be removed before assigning a random-motion temperature.

TD=ℏΓ2kB,Δopt=−Γ2,ER=(ℏk)22m,TR=ERkBT_D=\frac{\hbar\Gamma}{2k_B},\qquad \Delta_{\mathrm{opt}}=-\frac\Gamma2,\qquad E_R=\frac{(\hbar k)^2}{2m},\quad T_R=\frac{E_R}{k_B}
TD is the minimum of the ideal low-intensity two-level Doppler model, with Γ an angular linewidth. m is atomic mass in kg, kB is in J/K, ER in joules, and TR uses the explicitly declared single-recoil-energy convention.↔ Scroll wide equations horizontally.

Internal structure opens another cooling channel

Real alkali atoms have multiple magnetic sublevels. A spatially changing light polarization produces sublevel-dependent light shifts and position-dependent optical-pumping probabilities. In a Sisyphus cycle, an atom tends to climb a light-shift hill and then be pumped into a lower-energy internal state. The emitted photon can carry away that energy. This mechanism can cool below the two-level Doppler temperature because it uses internal-state structure omitted from that model.

Sub-Doppler cooling is sensitive to polarization, magnetic fields, laser intensity, and the optical-pumping timescale. It is not available for every transition or geometry, and at sufficiently low energy quantum recoil and coherent dark-state mechanisms require other descriptions. An observed narrow cloud is also not enough to establish a temperature: release-and-recapture, time-of-flight, or spectroscopy needs a calibrated dynamical model. Agreement between independent thermometers was central to identifying cooling below the original Doppler prediction.

12m⟨(vx−⟨vx⟩)2⟩=12kBTx\frac12m\langle(v_x-\langle v_x\rangle)^2\rangle=\frac12k_BT_x
Tx is a directional temperature in kelvins inferred from a velocity variance when a thermal description is appropriate. A nonthermal, anisotropic, or coherent momentum distribution need not admit one common temperature; subtracting the mean separates drift from random motion.↔ Scroll wide equations horizontally.

Work it through

Convert a natural width into a cooling scale

A transition has Γ/(2π) = 6.0 MHz. Use ℏ = 1.05 × 10⁻³⁴ J s and kB = 1.40 × 10⁻²³ J/K to estimate the ideal Doppler temperature and optimal angular detuning.

  1. The angular decay rate is Γ=2π(6.0×106)≃3.77×107 s−1\Gamma=2\pi(6.0\times10^6)\simeq3.77\times10^7\,\mathrm{s^{-1}}, rather than 6.0 × 10⁶ s⁻¹.

  2. The temperature is TD=(1.05×10−34)(3.77×107)/(2.80×10−23)≃1.41×10−4 KT_D=(1.05\times10^{-34})(3.77\times10^7)/(2.80\times10^{-23})\simeq1.41\times10^{-4}\,\mathrm K.

  3. The optimal angular detuning is −Γ/2, approximately −1.88 × 10⁷ rad/s. In ordinary-frequency units it is −3.0 MHz, half the quoted linewidth to the red.

The result

The model gives TD approximately 141 μK and optimal detuning −1.88 × 10⁷ rad/s, equivalent to −3.0 MHz.

The living notebook

Explore the model

Interactive lab

Interactive lab · oscillator

Damping reduces motion in a local illustration

Reset and set BOTH Damping ratio ζ to 0.1 and Natural frequency ω₀ to 1; predict slowly damped oscillations. Then Reset and set BOTH ζ to 0.5 and ω₀ to 1; predict faster envelope decay. The lab has positive restoring stiffness and starts at x = 1 with zero velocity. It illustrates damping only; it has no recoil diffusion, optical transition, Doppler temperature, or sublevel pumping.

Damping reduces motion in a local illustrationUnit mass, x(0) = 1 and v(0) = 0. The analytic solution includes underdamped, critically damped and overdamped cases; time is measured in seconds. Current readings: Displacement x(t): 1; Velocity v(t): 0; Regime: underdamped. All parameters are adjustable with the labeled controls below.02.557.510−1.15−0.5700.571.15Time · sDisplacementunderdamped
Displacement x(t)Current time
Displacement x(t)
1
Velocity v(t)
0
Regime
underdamped
t = 0 s

Unit mass, x(0) = 1 and v(0) = 0. The analytic solution includes underdamped, critically damped and overdamped cases; time is measured in seconds. Parameter changes pause playback and recompute the model at the current elapsed time. Reset restores both parameters and returns time to zero. Playback starts paused. Adjust either parameter, then choose Play or Step.

Look forDamping of a mean trajectory is only one part of a cooling model; the final distribution also depends on fluctuations.

Pause & reason

Check your understanding

Why does observation below TD not contradict the ideal two-level Doppler calculation?

Your turn

Try it yourself

A narrow momentum packet absorbs N = 100 photons all propagating along +x. Each photon has momentum ℏk = 8 × 10⁻²⁸ kg m/s, and atomic mass is 10⁻²⁵ kg. Find the mean velocity change from absorption. Explain why randomly directed spontaneous emission still affects temperature.

Take a moment to make a prediction or work it out on paper. Open a hint when you need one.

Give me a hint

Sum directed absorbed momentum for the mean; distinguish a zero emission mean from its nonzero variance.

Compare with the solution

The directed momentum change is 100(8×10−28)=8×10−26 kg m/s100(8\times10^{-28})=8\times10^{-26}\,\mathrm{kg\,m/s}, so mean velocity increases by 0.80 m/s. An isotropic emission distribution has zero mean x recoil but a positive mean squared recoil. Repeated emissions therefore broaden momentum even when their average directed force vanishes. The calculation determines mean drift from absorption, not a final cooling temperature.

Keep the idea

Take it with you

  • Red detuning gives a velocity-restoring force near zero speed in the weak two-beam model.
  • Doppler and recoil temperatures use distinct energy scales and declared frequency conventions.
  • Sub-Doppler optical pumping requires internal structure and suitable light polarization.

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