Ultracold atoms and optical lattices

At room temperature, a rubidium atom zips around at about 300 m/s — faster than a high-speed train. In today’s cold-atom laboratories, the same atom is cooled to a few tens of nanokelvin: its speed drops to a few millimetres per second, a hundred million times colder than the cosmic microwave background (2.7 K). Nothing this cold has ever appeared in the observable history of the universe — it exists only inside vacuum chambers on Earth.

Why go this cold? Recall the de Broglie wavelength $\lambda=h/p$: the slower the atom, the longer its matter wave. Around microkelvin temperatures $\lambda$ reaches the scale of an optical wavelength and the atoms’ wave natures begin to “overlap”; colder still, bosons pile en masse into a single quantum state — Bose–Einstein condensation (BEC, achieved in 1995), a million atoms sharing one macroscopic wavefunction. From that moment on, an atomic gas is no longer a bit-player in the thermodynamic background: it becomes a piece of macroscopic quantum matter you can see, photograph, and tune.

And the first step of it all sounds the most implausible: cooling atoms with light.

How a laser can possibly be a refrigerator

Light heats what it shines on — that is everyday experience. The trick of laser cooling is to make only the atoms coming towards you absorb the light.

The mechanism is Doppler cooling: tune the laser frequency slightly below the atomic transition (red detuning).

• An atom running into the laser sees the light blue-shifted by the Doppler effect in its own frame — right into resonance, so it absorbs photons copiously. Every absorbed photon delivers a head-on momentum kick of $\hbar k$.
• An atom running with the laser sees the light shifted further to the red, further from resonance, and absorbs almost nothing.

After each absorption the atom spontaneously re-emits the photon, but in a random direction, so on average the emission carries away no momentum. Net effect: whoever runs into the beam gets kicked — a damping force proportional to velocity and opposed to it. With six beams in three counter-propagating pairs covering all directions, the atoms swim as if through syrup, which is why the arrangement is called “optical molasses”.

Order of magnitude: how many photons to stop an atom? How cold does it get?basic~8 minexpand

How hard is one kick? A rubidium atom (mass $m\approx87\,u$) absorbing a 780 nm photon recoils at

$$v_r=\frac{\hbar k}{m}=\frac{h}{m\lambda} =\frac{6.63\times10^{-34}}{87\times1.66\times10^{-27}\times780\times10^{-9}} \approx 6\ \text{mm/s}\tag{13.2.1}$$

How many kicks are needed? To stop from a room-temperature speed of 300 m/s takes

$$N\approx\frac{300}{6\times10^{-3}}\approx5\times10^4\ \text{photons}\tag{13.2.2}$$

How long does the kicking take? The absorption–emission cycle rate is capped by the excited-state lifetime; rubidium’s spontaneous emission rate is $\Gamma\approx2\pi\times6\ \text{MHz}$, giving a saturated scattering rate of about $\Gamma/2\sim2\times10^7$ per second. So the stopping time is $\sim5\times10^4/2\times10^7\approx3\ \text{ms}$, over a stopping distance of about 1 metre — which is exactly the length of the “Zeeman slower” tube in the lab. A tabletop-scale apparatus, just right.

How cold can it get? The damping cannot go on forever: each spontaneous emission delivers a random recoil, a continuous “heating noise”. Balancing cooling against heating leaves the atom with a residual energy of order the excited state’s energy width $\hbar\Gamma$ (velocity differences within the linewidth are invisible to the Doppler mechanism). The careful calculation gives the Doppler limit

$$T_D=\frac{\hbar\Gamma}{2k_B} =\frac{1.05\times10^{-34}\times2\pi\times6.07\times10^6}{2\times1.38\times10^{-23}} \approx1.5\times10^{-4}\ \text{K}\approx146\ \mu\text{K}\tag{13.2.3}$$

From 300 K to 150 μK — six orders of magnitude — in a few milliseconds with a few laser beams. Getting lower needs new tricks: polarisation-gradient cooling reaches a few microkelvin, and the final step, evaporative cooling (repeatedly skimming off the hottest atoms and letting the rest rethermalise — the same principle as blowing on a bowl of soup), delivers the gas into the nanokelvin regime and BEC.

Optical lattices: building a crystal out of standing waves

Two counter-propagating laser beams form a standing wave with intensity profile $\cos^2(kx)$. A detuned light field shifts the atomic energy levels by an amount proportional to the intensity (the AC Stark shift — you met its mechanism in the second-order perturbation theory of section 7.5), so the atom feels a spatially periodic potential

$$V(x)=V_0\cos^2(kx)\tag{13.2.4}$$

Three beam pairs make a three-dimensional simple-cubic “lattice” with spacing half an optical wavelength. Cold atoms loaded into these wells are in exactly the situation of electrons in the ionic lattice of a solid — except that this “crystal”:

• has no impurities, no defects, no lattice vibrations — the potential is printed by laser light, perfectly periodic;
• has adjustable depth — turning the laser-power knob changes the barrier height;
• runs ten orders of magnitude slower — electrons hop between sites in femtoseconds, atoms in milliseconds, slow enough to photograph the atoms site by site through a microscope (the quantum gas microscope, mature since 2009).

In a deep lattice each site is approximately a harmonic-oscillator well (section 2.9), the atoms occupy only the lowest orbital, and the physics of the whole system reduces to two ingredients: tunnelling to the neighbouring site (amplitude $t$ — the barrier penetration of section 2.11) and the repulsion energy $U$ of two atoms on the same site. Written in the language of section 8.5, this is the celebrated Bose–Hubbard model:

$$H=-t\sum_{\langle ij\rangle}\hat a_i^\dagger \hat a_j +\frac{U}{2}\sum_i \hat n_i(\hat n_i-1)\tag{13.2.5}$$

The first term lets atoms delocalise and lower their kinetic energy; the second penalises two atoms crowding onto one site.

The superfluid–Mott transition: kinetic energy versus interactions, in a tug-of-war

Imagine exactly one atom per site on average, and tune $t/U$:

• Shallow lattice ($t\gg U$): tunnelling wins. Each atom spreads its wavefunction over the whole lattice, and the whole gas condenses into a single phase-coherent entity — the superfluid phase. The price: large fluctuations in the atom number on each site.
• Deep lattice ($U\gg t$): repulsion wins. Squeezing an extra atom onto a site costs $U$, so the system simply locks exactly one atom onto every site, and nobody moves — the Mott insulator phase. With the number pinned down, the phase becomes completely uncertain (number and phase are a conjugate pair, bound by the uncertainty principle just like $x$ and $p$).

In 2002, Greiner and colleagues filmed this tug-of-war: switch the lattice off suddenly and let the atoms fly free, and the interference pattern their wavefunctions form reveals the coherence — the superfluid phase gives sharp interference peaks; ramp the lattice deeper and the peaks vanish (into the Mott phase); ramp it back down and the peaks come back to life. A quantum phase transition, demonstrated reversibly, back and forth.

∎Key formulas

• Photon recoil velocity

  $$v_r=\frac{\hbar k}{m}=\frac{h}{m\lambda}\approx 6\ \text{mm/s (Rb, 780 nm)}$$

  Stopping a room-temperature atom takes ~5×10⁴ photons and a few milliseconds

• Doppler limit

  $$T_D=\frac{\hbar\Gamma}{2k_B}\approx 146\ \mu\text{K (Rb)}$$

  Balance point between cooling and random spontaneous-emission recoils

• Optical lattice potential

  $$V(x)=V_0\cos^2(kx)$$

  AC Stark shift; the depth V₀ is set directly by laser power

• Bose–Hubbard model

  $$H=-t\sum_{\langle ij\rangle}\hat a_i^\dagger \hat a_j+\frac{U}{2}\sum_i \hat n_i(\hat n_i-1)$$

  Large t/U: superfluid; small t/U (integer filling): Mott insulator

?Self-check3 questions

1. 1.

   In Doppler cooling, why is the laser tuned slightly below the atomic resonance (red detuning)?

   • So that only atoms moving against the beam are Doppler-shifted into resonance and preferentially absorb, feeling a head-on damping force
   • Red photons carry less energy and therefore do not heat the atoms
   • Red-detuned light is not scattered by atoms
   • To prevent the atoms from being ionised

   Submit
2. 2.

   The excited-state spontaneous emission rate of rubidium is Γ = 2π × 6.07 MHz. Find the Doppler limit temperature T_D = ħΓ/(2k_B) in μK. (ħ = 1.055×10⁻³⁴ J·s, k_B = 1.381×10⁻²³ J/K)

   μK1000% relative tolerance

   Submit
3. 3.

   Which of the following statements about the superfluid–Mott insulator transition are correct? (Select all that apply.)

   Select all that apply

   • Deepening the lattice to increase U/t can drive the superfluid into the Mott insulator phase
   • In the Mott phase the atom number per site is definite and phase coherence is lost
   • The transition is an ordinary thermal transition driven by raising the temperature
   • The interference pattern of released atoms serves as an experimental probe of coherence

   Submit

What comes next

Cold atoms scale “artificial quantum matter” to a million atoms, but each atom’s energy levels come pre-set by nature. The next section goes the opposite way: build a single “atom” out of a circuit board — energy levels, transition frequency, and coupling to photons all decided by the lithography mask.