thermodynamics of bose-einstein...

Thermodynamics of Bose-Einstein CondensationDavid SchefflerInterface of Quark-Gluon Plasma and Cold Quantum Gases

from NIST/JILA/CU-Boulder

2009/10/29 | TU Darmstadt - IKP | David Scheffler | 1 NHQ

Overview

IntroductionTheory

Partition Function of the Ideal Bose GasMean Occupation NumberBose - Einstein StatisticsTotal Particle NumberChemical PotentialCritical TemperatureMean Occupation of the lowest Energy-StateEnergyHeat Capacity

Comparison with Experimental Resultsλ-Transition of 4HeBEC of 23Na and 87RbBEC of Hydrogen


History

1908 H. K. Onnes liquefies 4He and observes the Onnes effect (T < 2.17 K)1911 H. K. Onnes observes superconductivity in mercury at 4.2 K1924 S. N. Bose sends Einstein a paper where he derives the Planck distribution

law for photons by statistical arguments Bose statistics1924 based upon Bose’s work Einstein publishes “Quantentheorie des

einatomigen idealen Gases” where he predicts a phase transition if totalparticle number is conserved

1938 F. London proposes that superfluidity is related to BEC1941 L. D. Landau publishes a phenomenological theory of superfluidity1957 BCS theory explains superconductivity1972 Lee, Osheroff and Richardson find superfluidity in liquid 3He (T < 3 mK)

which relates superfluidity and superconductivity1995 Cornell and Wieman: first “real” BECs in 87Rb

1995-97 Ketterle at MIT: BEC of 23Na, interference, “atom laser”


Partition Function of the Ideal Bose Gas

microstate with energy Er

all possible distributions of momenta that have a fixed energy Er

r = {np} = (np1 , np2 , np3 , ...)

Z (T , V ,µ) =X

r

exp{−β(Er − µNr )} β =1

kBT

=∞X

np1 =0

exp{−β(εp1 − µ)np1} ·∞X

np2 =0

exp{−β(εp2 − µ)np2} ...

=1

1− exp{−β(εp1 − µ)} ·1

1− exp{−β(εp2 − µ)} · ...

=Y

pi

11− exp{−β(εpi − µ)}


Mean Occupation Number 〈npi 〉

〈npi 〉 =X

r

Pr n(r )pi

=1Z

Xr

exp{−β(Er − µNr )} n(r )pi

=1Z·∞X

np1 =0

exp{−β(εp1 − µ)np1} · ... ·∞X

npi =0

npi exp{−β(εpi − µ)npi } · ...

=∞X

npi =0

npi exp{−β(εpi − µ)npi } ·„

1− exp{−β(εpi − µ)}«

=1β

∂

∂µ

0@ ∞Xnpi =0

exp{−β(εpi − µ)npi }

1A · „1− exp{−β(εpi − µ)}«

=1β

∂

∂µ

„1

1− exp{−β(εpi − µ)}

«·„

1− exp{−β(εpi − µ)}«

=exp{−β(εpi − µ)}

1− exp{−β(εpi − µ)} =1

exp{β(εpi − µ)} − 1


Bose - Einstein Statistics

〈npi 〉 = 1exp{β(εpi−µ)}−1

0pi

0.

〈np i〉

increasing µ

I only µ ≤ 0 is possible otherwise〈npi 〉 has a singularity

I for µ = 0 and εp0 = 0 the meanoccupation of the lowest energydiverges

I how does µ(T , x) look like?I what happens if µ = 0?

next stepderive an expression for µ(T , V , N) from N(T , V ,µ)


Total Particle Number N(T , V ,µ)

N(T , V ,µ) =X

pi

〈npi 〉 =X

pi

exp{−β(εpi − µ)}1− exp{−β(εpi − µ)}

=X

pi

exp{−β(εpi − µ)} ·∞Xj=0

exp{−β(εpi − µ)}j

=X

pi

∞Xj=1

exp{−β(εpi − µ)j} εp =p2

2m

=∞Xj=1

exp{βµj} V(2π~)3

Zd3p exp{−β p2j

2m} p2 =

p′2

j

=∞Xj=1

exp{βµj} V(2π~)3

Zd3p′

j3/2exp{−β p′2

2m} dp =

dp′√j


Total Particle Number N(T , V ,µ)

N(T , V ,µ) =∞∑j=1

exp{βµj} V(2π~)3 4π

∫dp′

j3/2p′2 exp{−β p′2

2m} �→©

=∞∑j=1

exp{βµj} V(2π~)3

1j3/2

(2mπβ

)3/2

=∞∑j=1

exp{βµj} Vj3/2

(√2mπkBT

2π~

)3

=V

λ(T )3

∞∑j=1

exp{βµj}j3/2

(1)

thermal wavelengthλ(T ) = 2π~√

2πmkBTfrom E = p2

2m = πkBT and p = ~ 2πλ


Intermediate Recapitulation

derived the partition function for the ideal Bose Gas

Z (T , V ,µ) =∏pi

11− exp{−β(εpi − µ)}⇓

calculated the Bose-statistics and found µ ≤ 0

〈npi 〉 =1

exp{β(εpi − µ)} − 1

⇓

asked whether µ(T , V , N) can be zero

N(T , V ,µ) =V

λ(T )3

∞∑j=1

exp{βµj}j3/2


Chemical Potential µ(T , V , N)

fixed number of particlesN(T , V ,µ) = N v = V/N

1 =v

λ(T )3

∞∑j=1

exp{ µjkBT }

j3/2(2)

I T →∞ µ→ −∞I µ = 0 at T = Tc

I no solution for T < Tc

µ(T , V , N)

-1-0.9-0.8-0.7-0.6-0.5-0.4-0.3-0.2-0.1

0 0.1

0 0.5 1 1.5 2 2.5µ

T/Tc

µ can indeed be zero at a critical temperature Tc

particles start to ”condense” into the lowest energy state⇒BEC2009/10/29 | TU Darmstadt - IKP | David Scheffler | 10 NHQ

Critical Temperature Tc of the Ideal Bose Gas

1 =v

λ(T )3

∞∑j=1

exp{βµj}j3/2

µ=0−→ vλ(Tc)3

∞∑j=1

1j3/2

=v

λ(Tc)3 ζ(3/2) (3)

Riemann’s Zeta Function

ζ(x) =∞∑j=1

1jx

ζ(3/2) ≈ 2.612

critical temperature

Tc =2π~2

ζ(3/2)2/3mv2/3kB

constraint for BECparticlesVolume

· λ(T )3 ≥ 2.612


What happens below Tc ? – eq. (2)

for µ = 0 and p2 → 0, 〈np〉 = 1

exp{β( p22m−µ)}−1

∝ 1/p2

N =X

p

〈np〉 = 〈n0〉 +Xp 6=0

〈np〉

N = ∆

Zp〈np〉p2dp = ∆

Z δp

0dp| {z }

=0 δp→0

+ ∆

Z ∞δp〈np〉p2dp

in contradiction to diverging 〈n0〉 for µ = 0 !

treat 〈n0〉 separately below Tc∑p

〈np〉 = 〈n0〉 +∑p 6=0

〈np〉 → 〈n0〉 +4πV

(2π~)3

∫p〈np〉p2dp (4)


Mean Occupation of the lowest Energy-State

N(T , V ,µ) =V

λ(T )3

∞∑j=1

exp{βµj}j3/2

for T > Tc eq. (1)

N(T , V ,µ) = 〈n0〉 +V

λ(T )3 ζ(3/2) for T ≤ Tc eq. (4)

T > Tc

〈n0〉N

= O(1/N) ≈ 0

T ≤ Tc with eq. (3)Vλ3 ζ(3/2) = Vλ3

cvλ3 = N

“TTc

”3/2

〈n0〉N

= 1− VNλ(T )3

ζ(3/2) = 1−„

TTc

«3/2

〈n0(T )〉

0

0.2

0.4

0.6

0.8

1

1.2

0 0.2 0.4 0.6 0.8 1 1.2

<n 0

/N>

T/Tc


E(T , V ,µ) of non-condensed particles

similar to N(T , V ,µ)

E(T , V ,µ) =∑

p

εp〈np〉 = ... =3kBTV2λ(T )3

∞∑j=1

exp{βµj}j5/2

T ≤ Tc, µ = 0 with eq. (3) V = Nλ3c

ζ(3/2)

E(T , V , N) =3kBTV2λ(T )3

∞∑j=1

1j5/2

=3ζ(5/2)2ζ(3/2)

NkBT(

TTc

)3/2

T > Tc

E(T , V ,µ) =3kBTV2λ(T )3

∞∑j=1

exp{βµj}j5/2

, 1 =V

Nλ(T )3

∞∑j=1

1j3/2

exp{βµj}


Expansion of E(T , V ,µ) for T > Tc

E(T , V ,µ) =3kBTV2λ(T )3

∞Xj=1

exp{βµj}j5/2

, 1 =V

Nλ(T )3

∞Xj=1

1j3/2

exp{βµj}

j = 1 E(T , V ,µ) =3kBTV2λ(T )3

exp{βµ}, 1 =V

Nλ(T )3exp{βµ}

E1(T , V , N) =32

kB N T

j = 2E(T , V ,µ) =

3kBTV2λ(T )3

exp{βµ2}25/2

, (exp{βµ})2 =„

Nλ(T )3

V

«2

E2(T , V , N) =32

kB N Tλ(T )3N25/2V

E(T , V , N) ≈ E1 + E2 =32

kBNT

"1 +

ζ(3/2)25/2

„Tc

T

«3/2#

withVN

=λ3

c

ζ(3/2)eq. (2)


Heat Capacity CV (T )

CV (T ) =(∂Q∂T

)NV

=(∂E∂T

)NV

with dW = 0

T ≤ Tc

CV (T , N) =15ζ(5/2)4ζ(3/2)

NkB

„TTc

«3/2

T > Tc

CV (T , N) ≈ 32

kBN

"1 +

ζ(3/2)27/2

„Tc

T

«3/2#

cV (T )

Tc T

3/2

.

c V[k

B]


Summary

calculated the critical temperature

Tc =2π~2

ζ(3/2)2/3mv2/3kB⇓

observed an inconsistency in our equation for µ for T < Tc

1 =v

λ(T )3

∞∑j=1

exp{ µjkBT }

j3/2

and fixed it

⇓

calculated N(T , V , N) −→ E(T , V , N) −→ cV (T , N)


λ-Transition of 4He

I in 1908 Heike Kamerlingh Onnes liquefiedHelium

I in 1913 Onnes recieved the Nobel prize“for his investigations on the properties of matterat low temperatures which led, inter alia, to theproduction of liquid helium”

I λ-transition at T = 2.17 KI in 1962 Landau recieved the Nobel prize

“for his pioneering theories for condensedmatter, especially liquid helium”

(from http://ltl.tkk.fi/research/theory/helium.html)Ideal Bose Gas

Tc =2π~2

ζ(3/2)2/3mv2/3kBwith v ≈ 46 Å

3, m = 4u ⇒ Tc = 3.11K

well, not bad for an ideal gas model


http://ltl.tkk.fi/research/theory/helium.html

BEC of dilute 23Na and 87Rb gases

I first observation of real BEC in1995 by Cornell and Wieman (Rb,CU-Boulder), Ketterle (Na, MIT)

I Nobel prize in 2001“for the achievement of Bose-Einsteincondensation in dilute gases of alkali atoms, andfor early fundamental studies of the properties ofthe condensates”

I 23Na: T exp ≈ 2µK Tc = 2.2µK

I 87Rb: T exp ≈ 170 nK Tc = 34 nK

upper picture time of flight imaging of a BEC from Ketterle et al., Phys. Rev. Lett. 75, 3969 (1995)lower picture from NIST/JILA/CU-Boulder


BEC of Hydrogen

Ideal Bose gasv =

1n

m = 1u ⇒ Tc = 51µK

v1/3/rB = ∆r/rB ≈ 3300 λ/rB ≈ 4600 rB : Bohr radius

in molecular H2 ∆r/rB ≈ 1.4→ BEC is driven by quantum statistics and not by interactions!


Thanks for your attention!


What is a BEC?


Atom Laser & Interference of BECs

I interference fringes between separatedparts of a BEC cloud ”macroscopicmatter wave” by MIT group

I coherent emission of condensed atoms”pulsed atom laser” by MIT group

more about this inI Fermion-Fermion and Boson-Boson

Interaction at low TI Experimental production of BECs

upper picture from Ketterle et al., Science 275, 637 (1997)lower picture from http://cua.mit.edu/ketterle_group/Nice_pics.htm


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