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

2. 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 Existing microscopic theory does not explain the only existing experimental evidence about the microscopic nature of superfluid helium

3. What is connection between condensate fraction and superfluid fraction? Accepted consensus is that size of condensate fraction is unrelated to size of superfluid fraction Superfluid fraction 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 x H. R. Glyde, R.T. Azuah and W.G. Stirling Phys. Rev. B 62 14337 (2000).

4. Superfluid helium becomes more ordered as it is heated Why?

5. Line width of excitations in superfluid helium is zero as T → 0. Why?

6. Basis of Lectures 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.B 64 224521, (2001) 74 014516, (2006)

7. Bose-Einstein Condensation T>T B 0<T<T B T~0 D. S. Durfee and W. Ketterle Optics Express 2, 299-313 (1998). ħ/L

8. Kinetic energy of helium atoms. J. Mayers, F. Albergamo, D. Timms Physica B 276 (2000) 811 T.R. Sosnick, W.M Snow P.E. Sokol Phys Rev B 41 11185 (1989) 3.5K 0.35K BEC in Liquid He4 f =0.07 ±0.01

9. N atoms in volume V Periodic Boundary conditions Each momentum state occupies volume ħ 3 /V 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 n(p)dp = probability of momentum p →p+dp Definition of BEC 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.

11. Quantum mechanical expression for n(p) in ground state What are implications of BEC for properties of Ψ? ħ/L

12. |Ψ(r,s)| 2 = P(r,s) = probability of configuration r,s of N particles = overall probability of configuration s = r 2, …r N of N-1 particles |ψ S (r)| 2 is conditional probability that particle is at r, given s Define ψ S (r) is many particle wave function normalised over r momentum distribution for given s Condensate fraction for given s

13. ψ S (r) non-zero function of r over length scales ~ L Implications of BEC for ψ S (r) ψ S (r) is not phase incoherent in r – trivially true in ground state Probability of momentum ħp given s

14. Phase of ψ S (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 ψ S (r) is independent of r Not true in Fermi systems

15. Feynman model for 4 He ground state wave function Ψ(r 1,r 2, r N ) = 0 if |r n -r m | < a a=hard core diameter of He atom Ψ(r 1,r 2, r N ) = C otherwise ψ S (r) = 0 if |r-r n | < a ψ S (r) = c S otherwise Ω S is total volume within which ψ S is non-zero c S =1/√Ω S

16. 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 Bin values generated. Calculate “free” volume fraction for each randomly generated s with P(s) non-zero “free volume” Generate random configurations s (P(s) = constant for non-overlapping spheres, zero otherwise)

17. f ~ 8% O. Penrose and L. Onsager Phys Rev 104 576 (1956) J. Mayers PRL 84 314, (2000) PRB64 224521,(2001) 24 atoms 192 atoms Periodic boundary conditions. Line is Gaussian with same mean and standard deviation as simulation. Has same value for all possible s to within terms ~1/√N ΔfΔf

18. What does “possible” mean? Gaussian distribution with mean z and variance ~z/√NN=10 22 Probability of deviation of 10 -9 is ~exp(-10 -18 /10 -22 )=exp(-10000)!!

19. 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).

20. In general ψ S (r) is non –zero within volume >fV. PRB 64 224521 (2001) Assume ψ S (r) is non zero within volume Ω ψ S = constant within Ω → maximum value of f = Ω/V For any given f ψ S (r) non-zero within vol >fV Any variation in phase or amplitude within Ω gives smaller condensate fraction. eg ideal Bose gas → f=1 for ψ S (r) =constant Feynman model - ψ S (r) is non –zero within volume fV.

21. ψ S (r) must be non-zero within volume >fV. ψ S (r) must be phase coherent in r in the ground state In any Bose condensed system Loops in ψ S (r) over macroscopic (cm) length scales For any possible s ψ S (r) must connected over macroscopic length scales 2 1 Loops in ψ S (r) over macroscopic length scales

22. Superfluidity Macrocopic ring of He4 at T=0 In ground state Rotation of the container creates a macroscopic velocity field v(r) Galilean transformation At low rotation velocities v(r)=0 if BEC is preserved but Quantisation of circulation

23. ψ S (r) in solid Can still be connected over macro length scales if enough vacancies are present But how can a solid flow? BEC Supersolidity

24. Ω = angular velocity of ring rotation R = radius of ring dR<<R R dR Ω Leggett’s argument (PRL 25 1543 1970) Maintained when container is slowly rotated In frame rotating with ring In ground state

25. Mass density conserved In ring frame if x is distance around the ring. F=|ψ S | 2 v(x) ρ 1 =|ψ 1 | 2 ρ 2 =|ψ 2 | 2 Simplified model for ψ S

26. ρ 1 = ρ 2 = ρ → F=ρRΩ No mass rotates with ring 100% supersolid. ρ 2 → 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 ψ 1 → 0 → 50% supersolid fraction in model ρ 1 =|ψ 1 | 2 ρ 2 =|ψ 2 | 2 connectivity suggests f~10% in hcp lattice.

27. O single crystal high purity He4 X polycrystal high purity He4 □ 10ppm He3 polycrystal solid liquid 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

28. Summary ψ S (r) is a delocalised function of r. – non zero over a volume ~V for all s Mass flow is quantised over macroscopic length scales BEC in the ground state implies that; Superfluidity and Supersolidity