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Lecture 26. Bose-Einstein Condensation

Lecture 26. Bose-Einstein Condensation

Lecture 25. Bose-Einstein Condensation (Ch. 7 )

BEC and BEC of related phenomena (lasers) Townes Basov Prokhorov Nobel 1964

BEC in a weakly- BEC in a strongly- interacting system interacting system (atomic ) (superfluid 4He) Nobel 2001 Einstein described the phenomenon of Landau Kapitsa condensation in an ideal of particles with Nobel 1962 Nobel 1978 nonzero mass in 1925. In the 1930’s Fritz London realized that superfluity 4He can be understood on terms of BEC. However, the analysis of superfluity 4He is complicated by the fact that the 4He atoms in strongly interact with each other. 70 years after the Einstein prediction, the BEC in weakly interacting Bose systems has been experimentally demonstrated - by laser cooling of a system of weakly- interacting alkali atoms in a magnetic trap. of Conserved

Two types of bosons: (a) Composite particles which contain an even number of . These number of these particles is conserved if the energy does not exceed the dissociation energy (~ MeV in the case of the nucleus). (b) Particles associated with a field, of which the most important example is the . These particles are not conserved: if the total energy of the field changes, particles appear and disappear. The of such particles is zero in equilibrium, regardless of density.

The occupancy cannot be negative for any ε, thus, for Critical density for bosons: bosons, μ≤0 (ε varies from 0 to ∞). Also, as T→0, μ → 0

2/3 2/3 ∞ g()ε ⎡ +12 ⎛ 2ms ⎞ ⎤ +12 ⎛ 2 Tmks ⎞ ∞ x 2/1 n = ⎢gd ()εε== ⎜ ⎟ ε 2/1 ⎥ = ⎜ B ⎟ dx ∫ exp[]()μεβ−− 1 4π 22 4π 2 2 ∫ exp[]x βμ −− 1 0 ⎣⎢ ⎝ h ⎠ ⎦⎥ ⎝ h ⎠ 0 ∞ x 2/1 Since μ≤0, the maximum possible dx ≈ 3.1 π value of n is obtained when μ = 0, ∫ ()x −1exp μ 0 T 2/3 +12 ⎛ 2 Tmks ⎞ B 3/2 ncr = 3.1 π 2 ⎜ 2 ⎟ = 6.2 nQ 2 4π ⎝ h ⎠ h ⎡ n ⎤ TC ≅ ⎢ ⎥ π mk ()s +126.22 where nQ is the quantum concentration, B ⎣ ⎦ which varies as T 3/2 Approaching the Critical Density...

What happens if we reduce the at a given density n until ncr drops below n?

∞ dxx ×= ()TkAn 2/3 2/3 n B ∫ exp[]x βμ −− 1 +12 ⎛ 2 BTmks ⎞ 0 ncr = 3.1 π ⎜ ⎟ 4π 2 ⎝ 2 ⎠ h μ (n,T)

TC T T

3/2 2 ⎡ n ⎤ h (apart from a numerical factor of order unity, TC is equal TC = 2π ⎢ ⎥ mkB ⎣ ()s +126.2 ⎦ to the Fermi temperature of a with density n)

As β becomes larger (T smaller), lμl must decrease (μ - increase) to maintain fixed

density. However, since μ cannot be positive for a BE system, at some T=TC the decrease of lμl cannot compensate the rapid decrease of the T-dependent pre-factor.

Paradox: we wanted to maintain n= N/V fixed, and now we see that we are not allowed to do this at sufficiently low . Of course there is no physical reason why we cannot continue lowering the temperature at fixed density (or increasing the density at fixed temperature). Particles in the are Missing... Resolving the paradox: The problem is caused by the behavior of the 3D and our use of the continuum approximation. Because g(ε)=0 at ε=0, our calculations of n ignored all the particles in the ground (ε=0) state. At low energies, we have to take into account the discreteness of the quantum states. g(ε) n(ε) ε) ( n × g()∝ εε × = ) (ε g ε ε ε For a Fermi system, this neglect would be of no consequence – there are only 2 fermions at zero energy. For a Bose system, the number of particles in the ground state at sufficiently low T becomes huge: 1 1 Tk f ()0 = ≈ B >>= 1 exp()βμ −− 1 − μβμ Because μ→0 at a non-zero T, the ground state can accommodate any number of particles! It was a bad idea to ignore the ground level.

h2 h2 The ground state of an ideal : ε ()222 =++= 3111 1 8mL2 8mL2 In this lowest state, all the wavefunctions look like that: (we’ve chosen the energy of this state as our zero energy). Bose-Einstein Condensation We can discuss the ideal Bose gas in the same terms of a transition. That is, at

a critical value of temperature, TC, μ(n,T) reaches the limit of μ = 0 and stops increasing. Beyond this point, the relation 2/3 ⎛ 22 π m ⎞ ∞ dεε n = ⎜ 2 ⎟ ∫ π ⎝ h ⎠ 0 exp[]()μεβ−− 1 is no longer able to keep track of all the particles – we miss the particles in the ground

state. Below TC, bosons begin to condense into the ground state. The abrupt accumulation of bosons in the ground state is called Bose-Einstein condensation. μ (n,T) 2 3/2 2 h ⎡ n ⎤ 53.0 ⎛ h ⎞ 3/2 T = 2π C ()sT 0 == ⎜ ⎟ ()n TC T C ⎢ ⎥ k ⎜ 2π m ⎟ mkB ⎣ ()s +126.2 ⎦ B ⎝ ⎠

2/3 2/3 The eq. n(T) with μ = 0 still works at ⎛ 2π Tmk ⎞ ⎛ T ⎞ n = 6.2 B = n⎜ ⎟ []< TT T0 ⎜ 2 ⎟ ⎜ ⎟ C particles not in the ground state: ⎝ h ⎠ ⎝ TC ⎠

The density of 2/3

ε) ⎡ ⎤

( ⎛ ⎞ T < T particles in the T n C 0 ε nnnn ⎢1−=−= ⎜ ⎟ ⎥ []< TT C

× ⎜ ⎟

) T ground state: ⎣⎢ ⎝ C ⎠ ⎦⎥ (ε g n nε>0 ε a tremendous number of particles all sitting n0 in the very lowest available energy state TC T BE Condensation vs. Gas-to-Liquid Condensation Classical analogy: let’s fill a container with a non-ideal gas and start lowering the temperature. The gas density remains constant until the condensation of gas (vapor) occurs. Since the density of liquid is much higher than that of gas, the gas density decreases with T.

Despite this similarity, the Bose-Einstein condensation is an entirely different phenomenon. One essential difference is that in the gas-to-liquid transition, which is due to interparticle attraction, the liquid and gas phases occupy different regions of space. The BE condensation is driven by exchange interactions. Each particle in the BE condensate has a wave function that fills the entire volume of the container. The BE condensation is the condensation in the space. The phase separation occurs in the k-space, not in the coordinate space. The condensed bosons have essentially zero momentum (i.e., their wave vector k is as small as the size of the container permits).

A common misconception about BE condensation is that it requires “brute force” cooling,

when kBT becomes much smaller than the differences of energies of the quantum states of the system - that would be a trivial effect, though the temperatures would have been inaccessibly low for any macroscopic system. The point is that condensation can

happen at much higher temperatures, when kBT is still large compared to the inter- level Δε. Some Numbers...

Interestingly, the expression for the critical temperature that we got works even for liquid helium, despite the fact that this is a strongly-interacting liquid, not a dilute gas. For 4He: 145 kg/m 3 n = ⋅= 106.3 4 mol/m 3 ⋅= 102.2 28 atoms/m 3 g/mol 4 g/mol

2 3/2 −34 2 53.0 ⎛ h ⎞ ⎛ N ⎞ 53.0 ( ⋅106.6 ) 3/2 T = ⎜ ⎟ = ()28 ≈⋅ 1.3102.2 K C ⎜ ⎟ ⎜ ⎟ −23 −27 kB ⎝ 2π m ⎠ ⎝ V ⎠ ⋅1038.1 π ⋅×× 107.142

- only ~40% higher than the actual superfluid transition temperature (2.17 K). To ensure weakness of interactions, the experimenters should work with dilute atomic gases. In this case, the critical temperature is much lower. One of the record-high values of TC – for atomic (BEC was achieved in 1998 at MIT). The density was 20 3 1.8·10 atoms/m : which, according to our equation for TC, corresponds to TC = 51 μK, in a nice agreement with the exp. value TC = 50 μK.

2 3/2 −34 2 53.0 ⎛ h ⎞ ⎛ N ⎞ 53.0 ( ⋅106.6 ) 3/2 T = ⎜ ⎟ = ()20 ⋅≈⋅ 101.5108.1 −5 K C ⎜ ⎟ ⎜ ⎟ −23 −27 kB ⎝ 2π m ⎠ ⎝ V ⎠ ⋅1038.1 π ⋅×× 107.112

The calculated TC = 51 μK is in a nice agreement with the exp. value TC = 50 μK. At this density, the distance between the atoms is ~ 104 times greater than the Bohr radius, and the interatomic interactions are extremely weak. The condensation is driven by statistics rather than by interactions! Realization of BEC in a Dilute 87Rb Vapor

In principle, the lighter the bosons, the greater TC. For example, the BE condensation of (light-induced -hole pairs) in has been observed before the BE condensation in dilute gases (electron is a , but an electron-hole pair has an integer ). First observation of the BEC with weakly-interacting gases was observed with relatively heavy atoms of 87Rb. 10,000 rubidium-87 atoms were confined within a “box” with dimensions ~ 10 μm (the density ~ 1019 m-3). The spacing between the

energy levels: 2 −34 2 3h 3 ( ⋅106.6 ) −32 −10 ~ εε1 2 ==Δ 2 101.1 ⋅⇒⋅≈ 108J K 8mL 8 −27 ×⋅× ()10107.187 −5 The transition was observed at ~ 0.1 μK. This is in line with the estimate:

3/2 2 3/2 53.0 ⎛ h2 ⎞ ⎛ N ⎞ ( −34 ) ( ⋅×⋅× 101106.653.0 19 ) T = ⎜ ⎟ ≈ ⋅≈ 108 −8 K C ⎜ ⎟ ⎜ ⎟ −23 −27 kB ⎝ 2π m ⎠ ⎝ V ⎠ π ⋅×××⋅ 107.18721038.1

Tk CB 100 Δ≈ ε ! - again, it is worth emphasizing that the BEC occurs at kBT >> Δε :

3/2 2 2 3/2 h h ⎛ N ⎞ 2 N 3/2 Δ εε~~ Tk CB ~ ⎜ ⎟ ~ Δε L 2 ~ Δε N Δ>>> ε 1 mL2 m ⎝ V ⎠ L

- the greater the total number of particles in the system, the greater this difference. Realization of BEC in Dilute Vapor (cont.) The atoms are not very close to each other in the classic sense - in fact, the average density of this condensate is very low—one billionth the density of normal or . But at this temperature, the quantum volume becomes comparable to the average volume per atom:

3 h3 ( ⋅106.6 −34 ) V ≡ = ⋅≈ 106 − m 319 Q 2/3 −23 −8 −27 2/3 ()2π BTmk ()π ⋅×××⋅×⋅ 107.18721051038.1

2/3 ⎡ ⎛ T ⎞ ⎤ At T=0.9T , the number of atoms in the ground state: NN ⎢1−= ⎜ ⎟ ⎥ ≈ 500,1 C 0 ⎜ T ⎟ ⎣⎢ ⎝ C ⎠ ⎦⎥ degeneracy of the 1st excited state in a cube

3 3 For comparison, in the first excited state: N1 = ≈ ≈ 300 ()ε BTk −Δ 1/exp −+ 101.01

The ratio N /N , which is ~ 5 for N = 104, rapidly n nε>0 0 1 increases with N at a fixed T/TC (it becomes ~ 6 n0 25 for N = 10 ).

TC T Problem

(a) Calculate the critical temperature for BE condensation of diatomic hydrogen H2 if the density of liquid hydrogen is 60 kg/m3. Would you expect in liquid hydrogen as well? Hydrogen liquefies around 20K and solidifies at 14K.

(b) Above TC, the in a degenerate Bose gas is proportional to T. Do you expect the temperature dependence of pressure to be stronger or weaker at T

3/5 3/2 3/5 3/2 ⎛ m ⎞ ⎛ ρH ⎞ ⎛ 4 ⎞ ⎛ 60 ⎞ (a) Liquid hydrogen: H 2 He ⎜ He ⎟ ⎜ 2 ⎟ BE = TT BE ⎜ 3K ×= ⎜ ⎟ ⎜ ⎟ ≈ 4.5 K ⎜ m ⎟ ρ ⎟ ⎝ 2 ⎠ ⎝140 ⎠ ⎝ H 2 ⎠ ⎝ He ⎠ Since hydrogen solidifies at 14K, we do not expect to observe superfluidity in liquid hydrogen.

(b) The atoms in the ground state do not contribute to

pressure. At T < TC, two factors contribute to the fast increase of P with temperature: (i) an increase of the P(T) number of atoms in the excited states, and (b) an increase of the average speed of atoms with temperature. As the result, the rate of the pressure increase with

temperature is greater at T < TC than that at T > TC (in fact , P~T5/2 at T < T ) . TC T C How to cool the gas of Rb atoms down to ~0.1 μK? The first stage – laser cooling, the second stage – evaporative cooling.

Laser Cooling E2 For photon absorption or -works for a dilute gas of neutral atoms emission, the photon energy hν must be equal to E -E (cannot be applied to cool solids) E1 2 1 If the laser frequency is tuned slightly

below E2-E1, an atom scatters (absorbs and re-emits) photons only is it moves towards the laser (Doppler effect). Atom at rest or moving in the opposite direction doesn’t scatter. ω If a resonant photon is absorbed, the atom acquires momentum: Mv = h c p2 ()ω 2 (2eV )2 The corresponding energy: K h ≈== ≈10−10 eV 22 cMM 2 ×× 10232 9 eV Na, A=23 An apparent limit on T achieved by laser cooling is reached when an atom’s recoil energy from absorbing or emitting a single photon is comparable to its total K. The single-photon recoil temperature limit (for Na):

−10 4 T1γ ×≈ = μKeVKeV (1/1010 Na)for Laser Cooling and Trapping

By laser cooling, T ~ 10 μK can be reached. At this temperature, the atom’s speed is a few cm/s. These slow- moving atoms are relatively easy to confine in a non-uniform magnetic trap. The magnetic field has a minimum value in the center of the “magnetic bowl”. An atom with spin parallel to the magnetic field (i.e., atomic anti- parallel to the magnetic field), is attracted to the minimum; for spin anti-parallel to the field, the atom is repelled from the minimum.

Laser cooling has been used in the experiments on BEC observation for pre- cooling of the gas of alkali atoms. However, to observe this phenomenon, even lower T are necessary. Further reduction of T by ~3 orders of magnitude (below 0.1 μK) is required for the exp. vapor densities ~ 1017 m-3. This is achieved by the evaporation cooling after the lasers are turned off.

Magneto-Optical Trap Evaporative Cooling

Radio-frequency forced evaporative cooling. The resonance excitation flips the spins and those atoms are ejected (evaporated). Reducing fr frequency evaporates lower energy atoms. Metastability is the Key

The experiments with Rb vapor were lnT vapor-/liquid vapor phase boundary aimed at realization of BEC in a weakly-interacting system. tter . ma Though the interactions are weak in cond liquid the vapor of Rb atoms, they are BEC He sufficiently strong for the phase ~ 1010 transition vapor-solid at ultra-low temperatures. In conditions of thermal ln(n) equilibrium, one cannot get below the 2 blue line without phase separation. ln ln()+= constnT C 3

How to cheat the Nature? The key is metastability. If the process of cooling is slow and “gentle” enough, one can realize a “super-saturated” vapor below the coexistence line without a condensed phase ever forming. For this, not only the interaction with walls must be excluded, but also the three-particle collisions that assist forming molecules and, eventually, condensed-matter phase – hence, very low densities. Observation of BEC

To observe the distribution of velocities of atoms in the system, the magnetic trap is turned off. The atoms find themselves in free space, and, because they have some residual velocity, they just fly apart. After they have flown apart for some time, the cloud is much bigger, and it is easier to take a snapshot of the atomic cloud (to make a snapshot, a laser beam is scattered by the cloud). The picture shows the velocity distribution of atoms in the cloud at the time of its release, instead of the spatial distribution.

For T > TC, atoms are distributed among many energy levels of the system, and have a Gaussian distribution of velocities. With cooling of the cloud, a spike appears right in its middle. It corresponds to atoms which are hardly moving at all: for T < TC, the concentration of atoms in the lowest state gives rise to a pronounced peak in the distribution at low velocities (condensation in the momentum space). Each frame corresponds to the distance the atoms have moved in about 1/20 s after This two-component cloud resembles the turning off the trap. situation in superfluid helium, where two components coexist: normal and superfluid . of a Bose Gas The particles condensed in the ground state do not contribute to the total energy of a Bose gas: ε ∞ × ( )dg εεε ∞ 2/3 dεε U = ∑ s → ∫ A×= ∫ s exp[]()s μεβ−− 0 exp1 []()μεβ−− 1 0 exp[]()μεβ−− 1

∞ 2/3 2/3 dεε ⎛ 22 π m ⎞ 2/5 At T < T μ = 0 AU ×= 8.1 ×≈ ()TkV C ∫ ⎜ 2 ⎟ B 0 ()βε −1exp π ⎝ h ⎠

2/3 2/3 2/3 ∂U 5 U ⎛ π m ⎞ 522 2/3 ⎛ 2π m ⎞ 2/3 ⎛ T ⎞ C = 8.1 ×≈= ()≈ 5kTkkV ()≈ 2NkTkV ⎜ ⎟ V ⎜ 2 ⎟ BB B ⎜ 2 ⎟ B B ⎜ ⎟ ∂T 2 T π ⎝ h ⎠ 2 ⎝ h ⎠ ⎝ TC ⎠

C /Nk V B This shape of CV(T) is the result of our neglect of 2 inter-particle interactions. 1.5 However, in the systems with interactions, similar effect is observed (e.g., near the superfluid transition in liquid 4He) - the so-called λ transition. 0 1 T/TC Thermodynamic Functions of a Degenerate Bose Gas

By integrating the heat capacity at constant volume, we can get the entropy:

∂U 5 U T ** V ()dTTC 5 U CV = = S = = ∂T 2 T ⇒ ∫ * 0 T 3 T 2 The Helmholtz free energy (μ = 0): −=−= UTSUF 3 ⎛ ∂F ⎞ The pressure exerted by a degenerate Bose gas: P −= ⎜ ⎟ ∝ T 2/5 ⎝ ∂V ⎠T does not depend on volume! This is due to the fact that, when compressing a degenerate Bose gas, we just force more particles to occupy the ground state. The particles in the ground state do not contribute to pressure – except of the zero-motion oscillations, they are at rest. Atomic Laser

The atoms in the BE condensate all have the same energy, they are in a coherent state. If one turns off the trap, the atoms would all fall down in an ideally monochromatic beam. Such a beam can be used for studies analogous to those of photon correlations in quantum optics. The trap is analogous to the optical cavity formed by the mirrors in a conventional laser. But to make a laser we need to extract the coherent field from the optical cavity in a controlled way. This can be achieved by transferring the atoms from states that are confined to ones that are not, typically by flipping the direction of their spin by resonant absorption of RF radiation. (Only the atoms that had their magnetic moments pointing in the opposite direction to the magnetic field are trapped in the magnetic trap). The MIT researchers applied short RF pulses to "flip" the spins of some of the atoms and therefore release them from the trap The extracted atoms then accelerated away from the trap under the force of . Future Directions Optical Lattices BEC on a Chip

Vortices in the BE Condensate These experiments are not just a repetition of the superfluid liquid helium experiments, because the system is very different. Also, there are new experimental tools developed by atomic physics; e.g, it is possible to see individual atoms; the atomic velocity distributions can be resolved, which has not really been possible to do in liquid helium.

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