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Bose-Einstein Condensation and SuperfluidityPowerPoint Presentation

Bose-Einstein Condensation and Superfluidity

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Bose-Einstein Condensation and Superfluidity

- Lecture 1. T=0
- Motivation.
- Bose Einstein condensation (BEC)
- Implications of BEC for properties of ground state many particle WF.
- Feynman model
- Superfluidity and supersolidity.

- Lecture 2 T=0
- Why BEC implies macroscopic single particle quantum effects
- Derivation of macroscopic single particle Schrödinger equation

- Lecture 3 Finite T
- Basic assumption
- A priori justification.
- Physical consequences
- Two fluid behaviour
- Connection between condensate and superfluid fraction
- Why sharp excitations – why sf flows without viscosity while nf does not.
- Microscopic origin of anomalous thermal expansion as sf is cooled.
- Microscopic origin of anomalous reduction in pair correlations as sf is cooled.

Existing microscopic theory does not explain the

only existing experimental evidence about the

microscopic nature of superfluid helium

Motivation

A vast amount of neutron data has been collected

from superfluid helium in the past 40 years.

This data contains unique features, not observed in any other fluid.

These features are not explained even qualitatively

by existing microscopic theory

J. S. Brooks and R. J. Donnelly, J Phys. Chem. Ref. Data 6 51 (1977).

Normalised condensate fraction

o o T. R. Sosnick,W.M.Snow and P.E. Sokol Europhys Lett 9 707 (1989).

x xH. R. Glyde, R.T. Azuah and W.G. Stirling Phys. Rev. B 62 14337 (2000).

What is connection between condensate fraction and superfluid fraction?

Accepted consensus is that size of condensate

fraction is unrelated to size of superfluid fraction

J. Mayers J. Low. Temp. Phys 109 135 (1997)

109 153 (1997)

J. Mayers Phys. Rev. Lett.80, 750 (1998)

84 314 (2000)

92 135302 (2004)

J. Mayers, Phys. Rev.B64 224521, (2001)

74 014516, (2006)

ħ/L

Bose-Einstein Condensation

T>TB

0<T<TB

T~0

D. S. Durfee and W. Ketterle Optics Express 2, 299-313 (1998).

P.E. Sokol

Phys Rev B 41 11185 (1989)

3.5K

0.35K

Kinetic energy of helium atoms.

J. Mayers, F. Albergamo, D. Timms

Physica B 276 (2000) 811

BEC in Liquid He4

f =0.07 ±0.01

N atoms in volume V

Periodic Boundary conditions

Each momentum state occupies volume ħ3/V

n(p)dp = probability of momentum p →p+dp

BEC

Number of atoms in single momentum state (p=0) is proportional to N.

Probability f that randomly chosen atom occupies p=0 state is independent of system size.

No BEC

Number of atoms in any p state is independent of system size

Probability that randomly chosen atom occupies p=0 state is ~1/N

ħ/L

Quantum mechanical expression for n(p) in ground state

What are implications of BEC for

properties of Ψ?

= overall probability of configuration

s = r2, …rN of N-1 particles

Define

ψS(r) is many particle wave function normalised over r

momentum distribution

for given s

Condensate fraction for given s

|Ψ(r,s)|2 = P(r,s)= probability of configuration r,s of N particles

|ψS(r)|2isconditional probability that particle is at r, given s

Probability of momentum ħp given s

Implications of BEC for ψS(r)

ψS(r) non-zero function of r over length scales ~ L

ψS(r) is not phase incoherent in r – trivially true in ground state

Phase of ψS(r) is the same for all r in the

ground state of any Bose system.

- Fundamental result of quantum mechanics
- Ground state wave function of any Bose system has no nodes (Feynman).
- Hence can be chosen as real and positive

Phase of Ψ(r,s,) is independent of r and s

Phase of ψS(r) is independent of r

Not true in Fermi systems

ψS(r) = 0 if |r-rn| < a

ψS(r) = cS otherwise

cS =1/√ΩS

Feynman model for 4He ground state wave function

Ψ(r1,r2, rN) = 0 if |rn-rm| < a a=hard core diameter of He atom

Ψ(r1,r2, rN) = C otherwise

ΩS is total volume within which ψSis non-zero

Calculation of Condensate fraction in Feynman model

Take a=hard core diameter of He atom

N / V = number density of He II as T → 0

Generate random configurations s

(P(s) = constant for non-overlapping

spheres, zero otherwise)

Calculate “free” volume fraction for each randomly generated s with P(s) non-zero

Bin values generated.

J. Mayers PRL 84 314, (2000)

PRB64 224521,(2001)

24

atoms

Δf

192

atoms

Has same value for all

possible s to within terms

~1/√N

Periodic boundary conditions.

Line is Gaussian with same mean and standard deviation as simulation.

f ~ 8%

O. Penrose and L. Onsager

Phys Rev 104 576 (1956)

Gaussian distribution with mean z and variance ~z/√N

N=1022

What does “possible” mean?

Probability of deviation of 10-9 is

~exp(-10-18/10-22)=exp(-10000)!!

Pressure dependence of f in Feynman model

Experimental points

taken from

T. R. Sosnick,W.M.Snow

and P.E. Sokol

Europhys Lett 9 707 (1989).

In generalψS(r) is non –zero within volume >fV.

PRB64 224521

(2001)

For any given fψS(r) non-zero within vol >fV

Feynman model -ψS(r) is non –zero within volume fV.

Assume ψS(r) is non zero within volume Ω

ψS = constant within Ω→ maximum value of f = Ω/V

Any variation in phase or amplitude within Ω

gives smaller condensate fraction.

eg ideal Bose gas → f=1 for ψS(r) =constant

For any possible sψS(r) must connected over macroscopic length scales

1

Loops in ψS(r) over macroscopic

length scales

2

ψS(r) must be non-zero within volume >fV.

In any Bose condensed system

ψS(r) must be phase coherent in r in the ground state

Loops in ψS(r) over

macroscopic (cm) length scales

Rotation of the container creates

a macroscopic velocity field v(r)

Galilean

transformation

if BEC is preserved

Quantisation of circulation

but

Superfluidity

Macrocopic ring of He4 at T=0

At low rotation

velocities v(r)=0

Supersolidity

ψS(r) in solidCan still be connected over macro length

scales if enough vacancies are present

But how can a solid flow?

In frame rotating with ring

Leggett’s argument (PRL 25 1543 1970)

Ω

Ω = angular velocity of ring rotation

R = radius of ring

dR<<R

dR

R

Maintained when

container is slowly rotated

x is distance around the ring.

Simplified model for ψS

F=|ψS|2v(x)

ρ1=|ψ1|2

ρ2=|ψ2|2

Mass density conserved

In ring frame if

ρ1=|ψ1|2

ρ2=|ψ2|2

ρ1=ρ2= ρ→F=ρRΩ

No mass rotates with ring

100% supersolid.

ρ2→0 → F=0

100% of mass rotates

with the ring.

0% supersolid

Superfluid fraction determined by amplitude in connecting regions.

Can have any value between 0 and 1.

Condensate fraction determined by volume in which ψ is non-zero

ψ1→ 0 →50% supersolid fraction in model

connectivity suggests f~10% in hcp lattice.

O single crystal high purity He4

X polycrystal high purity He4

□ 10ppm He3 polycrystal

solid

J. Mayers, F. Albergamo, D. Timms Physica B 276 (2000) 811

M. A. Adams, R. Down ,O. Kirichek,J Mayers

Phys. Rev. Lett. 98 085301 Feb 2007

Supersolidity not due to BEC

in crystalline solid

for all s

Superfluidity and Supersolidity

Summary

BEC in the ground state implies that;

ψS(r) is a delocalised function of r. – non zero over a volume ~V

Mass flow is quantised over macroscopic length scales

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