Part II: Applications - Bose-Einstein Condensation SDSMT, Physics 204 Fall
Introduction Historic Remarks 2 Bose-Einstein Condensation Bose-Einstein Condensation The Condensation Temperature 3 The observation of BEC How to realize the BEC The observation of BEC
From prediction to observation BEC and related phenomena BEC of photons (lasers) Townes Basov Prokhorov Nobel 964 BEC in a stronglyinteracting system (superfluid 4 He) BEC in a weaklyinteracting system (atomic gases) Nobel 200 Landau Nobel 962 Kapitsa Nobel 978 925: Einstein described the phenomenon of condensation in an ideal gas of particles with nonzero mass. 930 s Fritz London realized that superfluity 4 He can be understood on terms of BEC. However, the analysis of superfluity 4 He is complicated by the fact that the 4 He atoms in liquid 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.
The B.E. distribution. For Bosons that have zero chemical potential, like photons, the total number of them is determined by the condition of thermal equilibrium - which is basically determined by the temperature of the radiator. 2. For bosons that have non-zero chemical potential, the B.E. distribution: n BE = e (ɛ µ)/kt. One example is shown in this diagram: µ = 5.0 kt = 0, 5,
Why it is complicated for non-zero mass Bosons? From this plot, one can see when T 0, more and more bosons will pile up on the state that has energy ɛ µ. Looking at the B.E. distribution n BE =, one can find: e (ɛ µ)/kt if ɛ µ = 0 + δ, n BE + if ɛ µ = 0 δ, n BE Since n BE 0, so, the area of interest is: 0 < (ɛ µ)/kt <<. For a given µ, the lower the temperature, the faster Bosons can pile up on the ground state. 2. At a given T, how many Bosons are on the ground state depends on how close the ground state energy ɛ 0 is to µ. So, it depends on µ. But it is more complicated because of µ is a nontrivial function of the density and temperature: µ = µ( n BE, T ). Here, n BE corresponding to density : high density more particles on a state.
When T is small B.E. distribution: n BE =. For condensation to happen on the e (ɛ µ)/kt ground state, we need 0 < (ɛ 0 µ)/kt <<. This leads to the following relations: e (ɛ0 µ)/kt + ɛ 0 µ + ( ɛ0 ) µ 2 kt 2! kt + n BE (ɛ 0) + ɛ 0 µ kt + kt n BE (ɛ 0) ɛ 0 µ( n BE, T ) So, at low T, in order to have n BE (ɛ 0), ɛ 0 µ( n BE, T ) has to be very small. µ( n BE, T ) = ɛ 0 kt n BE (ɛ 0) This tells us: () Condensation happens when µ approaches ɛ 0 from below. (2) When the number of particles n is large, µ depends only weakly on T. () (2)
When T is small - cnt. The question is, how low must be the temperature be in order for n BE (ɛ 0) to be very large so that condensation happens. µ is a characteristic of a system. Particles move from system with bigger µ to a system with smaller µ. The general condition that determines µ is that the sum of the B.E. distribution over all states must add up to the total number of atoms, N. g(ɛ) is the density of states. See the Figure on next slide. N = (3) e (ɛs µ)/kt all s Or, converting the sum to integration: N = g(ɛ) dɛ (4) e (ɛ µ)/kt 0
... g(ε) g( ε ) ε n(ε) = g(ε) n(ε) ε ε ε g(ε) g( ε ) ε n(ε) = g(ε) n(ε) ε ε ε Density of states B.E. distribu3on Par3cle distribu3on
When T is small - cnt. N = g(ɛ) dɛ 0 e (ɛ µ)/kt The problem is not solved - This equation cannot be solved for µ analytically! Can we make some rough estimations by assuming something reasonable? - Yes. The logic is: When condensation happens, there are a large number of particles on the ground state. A large number of particles piling up on the ground state (N 0 ) happens only when the increase in µ is very very small: N 0µ 0, which means µ 0. Now we can use this approximation and see how the number of particles changes with the temperature.
When T is small - cnt. N = 0 µ 0 g(ɛ) dɛn e (ɛ µ)/kt 0 g(ɛ) dɛ (5) e ɛ/kt Let s take spin-zero bosons confined in a volume V as an example: For electron, spin=/2: g(ɛ) = π(8m)3/2 V ɛ 2h 3 For spin=0 boson: g(ɛ) = π(8m) 3/2 V ɛ = 2 2 2h 3 So, ) N 2 3/2 V 0 N x=ɛ/kt N 2 ( 2πm π h 2 ( 2 2πm π h 2 ( 2πmkT π h 2 ( 2πm π h 2 ) 3/2 V 0 kt x e x ) 3/2 V 0 ) 3/2 V ɛ ɛ dɛ (6) e ɛ/kt ktdx (7) x dx (8) e x N 2 π ( 2πmkT h 2 ) 3/2 V (2.3) (9)
The Condensation Temperature We arrive at [textbook Eq. (7.25)]: ( ) 3/2 2πmkT N 2.62 V (0) This says, the total number of particles in the system is a function of temperature. Given a N, you have T : h 2 ( ) ( ) h 2 2/3 N T = 0.527 () 2πmk V We call this T the condensation temperature, noted as T c. This should be a spacial solution to Eq.(4).
The Condensation Temperature - cnt. We obtained one special solution T c under assumption of µ = 0. N = g(ɛ) 0 eɛ/ktc dɛ Let s see what happens when T T c. We have to take a look at the constraint Eq.(4) again: N = g(ɛ) 0 e (ɛ µ)/kt dɛ T > T c: To keep Eq.(4) valid for a given N, µ should become smaller than zero, µ < 0!. Since generally speaking, increasing the density of particles in a system always increases the chemical potential, this means particles will not be gathering together when T becomes bigger than T c. T < T c: To keep Eq.(4) valid for a given N, µ should become positive, µ > 0. ( This is equally odd because we know µ = S ) 0. T N U,V What is wrong? The cause of this is the transition from to.
The Condensation Temperature - cnt. What is WRONG in the transition from to? Let s look at the original (always correct) formula again: N = all s e (ɛs µ)/kt When T is small and µ 0, the number of particles in the ground state will become enormous. This is NOT reflected in the integration: N = g(ɛ) 0 e (ɛ µ)/kt dɛ On the other hand, this to transition may still be valid if we only count the particles on the excited states: N excited = g(ɛ) 0 e (ɛ µ)/kt dɛ So, at T T c, we can write: ( N excited 2.62 2 2πmkT ) 3/2 π V, when T < Tc. ( Using T c = 0.527 h 2 h 2 2πmk ) ( N ) 2/3, Nexcited can be written as V ( ) 3/2 T N (T < T c) (2) N excited T c
BEC - cnt. In this case, the number of particles on the ground state becomes: [ ( ) ] 3/2 T N 0 = N N excited N (T < T c). (3) T c N N 0 N excited T C T
BEC - cnt. The chemical potential as a function of temperature looks like this. µ (n,t)/kt c 0-0.4-0.8 T C T
The observation of BEC Each frame corresponds to the distance the atoms have moved in about /20 s after turning off the trap.
The observation of BEC For T > T c, 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 of the velocity distribution. It corresponds to atoms which are hardly moving at all. Atoms are first trapped and cooled down in the system, using magnetic trap. To observe the distribution of velocities of atoms in the system, the magnetic trap is turned off. Because they have some residual velocity, they just fly apart. Use a laser beam to take a snapshot of atoms after they have flown apart for some time: photons scattered by the atom cloud. How to cool the gas of atoms down to 0.µK? Laser Cooling + Evaporative Cooling.
Cooling - Laser Cooling If the laser frequency is tuned slightly below E 2 E, 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. Works for a dilute gas of neutral atoms (cannot be applied to cool solids) E 2 E For photon absorption or emission, the photon energy hν must be equal to E 2 -E The 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 E k. The single-photon recoil temperature limit (for Na) ~ T!! 0 "0 ev #0 4 K / ev =µk "=kt, k=8.67#0 "5 ev / K ~ 0 "4 ev / K. (for Na)
Cooling - Evaporative Cooling Radio-frequency forced evaporative cooling: The resonance excitation flips the spins and those atoms with higher energies are ejected (evaporated). In other words, the magnetic trapping potential can be modified by a radio frequency (RF) signal.
Experiment Aparatus The picture to the right shows the velocity distribution of atoms in the cloud at the time of its release, instead of the spatial distribution. A view of the BEC setup For T < T c, 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).
Classical Physics Analogy BE Condensation vs. Gas-to-Liquid Condensation: The Bose-Einstein condensation is an entirely different phenomenon. Phase transition: A container filled with a non-ideal gas 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 the temperature. One essential difference: In the gas-to-liquid transition, which is due to inter-particle attraction, the liquid and gas phases occupy different regions of space. The BE condensation is driven by exchange interactions, a specific mutual effect of identical particles that is manifested in a certain special type of interaction. The exchange interaction is a purely quantum- mechanical effect that has no analogue in classical physics. Each particle in the BE condensate has a wave function that fills the entire volume of the container.
Two examples 7-0. Bose-Einstein Condensation In the first achievement of Bose- Einstein condensation with atomic hydrogen [Fried, PRL 8(998)38], a gas of 2 0 0 atoms was trapped and cooled until its peak density was.8 0 4 atoms/cm 3. Calculate the condensation temperature for this system and compare to the measured value of 50 µk. Note: To achieve such low temperature, people use helium dilution refrigerator + laser cooling techniques.
Two examples 7-. Thermal Quantities of Boson Gas: The figure below shows the heat capacity of a Bose gas as a function of temperature. Explain the features of curve: The peak, the limit at high temperature. C V /Nk B 2.5 0 T/T C Q: Write down the expression for the total energy of a gas of N bosons confined to a volume V, in terms of an integral. Q2: For T < T c, set µ = 0, calculate the heat capacity C V = ( ) U. T V Compare the result to the figure. Q3: What the heat capacity should be at high-temperature limit?