Quantum gases in the unitary limit and...

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1 Quantum gases in the unitary limit and... Andre LeClair Cornell university Benasque July

2 Outline The unitary limit of quantum gases S-matrix based approach to thermodynamics Application to the unitary limit Hubbard model. (unitary gases work done with Pye-ton How, 2010, JSTAT)

3 Motivations: Intriguing examples of scale invariant theories with z=2 dynamical exponent (Schrodinger symmetry). experimental realizations: cold atoms tuned through a Feshbach resonance. surface of neutron stars. non-relativistic AdS/CFT description? Is there a bound on shear viscosity to entropy density? η/s 4πk B???

4 Unitary limit of quantum gases Model actions for bosons and fermions: S = ( ( ) ) d d xdt iφ t φ φ 2 2m g 4 (φ φ) 2 S = d d xdt ( α=, iψ α tψ α ψ α 2 2m g 2 ψ ψ ψ ψ )

5 Renormalization group: flows to low energy: d<2 0 g g unitary limit d>2 BEC g BCS 0 g d=2 0 g g

6 d=3 case: at the fixed point, a=scattering length diverges. z=2 scale invariant theory. Only energy scales are the chemical potential and temperature. On BEC side, the 2-fermion bound state can condense. BCS side well described by BCS theory at small coupling. (no bound state on this side.) in unitary limit: Very strongly coupled. No small parameter like na 3 new methods are needed.

7 motivate the method with the: d=1 case S-matrix: S = k k ig/4 k k + ig/4 Unitary limit: g S 1 Turns out to be a free fermion. Difficult to see perturbatively, but clear from the TBA.

8 Thermodynamic Bethe Ansatz in 1d free energy: F = 1 β dk log ( 1 + e βε(k)) ε(k) = ω k 1 β dk K(k, k ) log ( ) 1 + e βε(k ) β = 1/T K = i k log S ω k = k 2 /2m = single particle energy In the unitary limit, just a free fermion.

9 The formalism: a TBA-like approach in any dimension density: n = µ F = d d k (2π) d f(k) Making a Legendre transformation in the chemical potential and occupation number f, one can show there exists a functional F where the free energy is given by: variational principle: δϝ δf = 0

10 Starting point: Z = Z π de e βe Tr Im E log Ŝ(E) (Dashen, Ma, Bernstein, 1969) Can derive: Ϝ = Ϝ 0 + Ϝ 1 Ϝ 0 = Ϝ 1 = 1 2 d 2 k (2π) 2 ((ω k µ)f 1β [(f 1) log(1 f) f log f] ) d 2 k (2π) 2 energy d 2 k entropy (2π) 2 f(k ) G(k, k ) f(k) F = E - TS (see Landau-Lifshitz) (keep only 2-body terms) 2πδ (E ω k ω k ) V G(k, k ) = i < k, k log Ŝ(E) k, k >

11 Final result. Variational principle gives: 1 f(k) = Fe βε(k) + 1 ε(k) = ω k µ d d k (2π) G(k, 1 d k ) e βε(k ) + 1 F pseudo-energy integral eqn F = T d 2 k (2π) 2 [ log(1 + e βε ) + β 2 ] 1 e βε (ε ω + µ) + 1 (different signs for bosons) µ = chemical potential

12 Structure of the kernel G G = i 2I log ( ) 1/gR ii/2 1/g R + ii/2 L = i d d p 1 (2π) d E ω p ω K p + 2iɛ S-matrix ( = I + iγ (1-loop integral) renormalized coupling: g R = g 1 gγ/2 K E and are the total energy and momentum of the 2 particles!! Non -perturbative, well-defined expansion in 1/g!!

13 Application to 3d unitary gas ( G(k, k 8π ) = i m k k log ( ) 1/gR ( im k k /16π 1/g R + im k k /16π g R = g ( 1 g/g In the unitary limit, g g g scattering length: a = mg R /2π a ± S-matrix: S 1

14 d>2 bound state BEC g BCS 0 no bound state g a = + (repulsive) a = (attractive) G 8π2 m k k -/+ corresponds to repulsive/attractive (for small g, G = -g )

15 Formally define it as S=-1, i.e. coupling goes to infinity. not an RG fixed point in usual sense, but still scale invariant. Occurs ( at very low energies (infinitely attractive) or very high energy. (infinitely repulsive). G( k ) = 4π m 4 cases: attractive/repulsive bosons/fermions

16 Results Scale invariance implies the scaling form: F = ζ(5/2) ( mt 2π ) 3/2 T c(µ/t ) Critical points must occur at fixed values of These points can be expressed as: c=1 for free boson at zero chem.pot. µ/t. nλ d T c = constant (bosons) λ T = 2π/mT. T c /T F = constant (fermions)

17 Fermions s/n Entropy per particle for fermions critical point T/T F T c /T F 0.1. consistent with lattice Monte Carlo

18 Bosons Evidence for an interacting version of BEC (new) n c λ 3 T = 1.325, (µ/t = x c = ) compare with non-interacting BEC: x c = 0 and n c λ 3 T = ζ(3/2) = 2.61,

19 500 f diverges Occupation number for bosons x c b κ = βk 2 /2m κt (mt ) 3/2 Compressibility for bosons diverges x = µ/t

20 Viscosity to entropy density ratio η/s Viscosity entropy for attractive fermions T/T F η s > πk B In good agreement with experiments

21 0.4 η/s Viscosity to entropy ratio for bosons T/T c η s > πk B a more perfect fluid than fermions

22 High Temperature Superconductivity Start here 4(&)!-5$* 0463!"#$%&'()*&+,-.!),/-+' 0!123 7!'89':;''''''''<26'!"12'''''''''''''''''''''''7"!'=&/$> Not here (doping a Mott insulator)

23 Hubbard Model Gas H = t ( c ri,α c r j,α) t ( c ri,α c r j,α) + U n r n r <i,j>,α=, <i,j>,α=, r diagonalize free part treat as local *free, single particle energies: ω k = 2t (cos(k x a) + cos(k y a)) 4t cos(k x a) cos(k y a) * can treat as a gas with coupling Cuprates: t 0.3, g 13 g = U/t

24 t 0.3, attractive 10 5 G U/t = E/t repulsive 5 10 U/t = 5 U/t = 13.5, 14 U/t = 15 Conclusion: for t =-0.3, g must be greater than 12.8 for an attractive band to exist.

25 g=15, t =0, -0.1, -0.3, -0.4 t = G E/t t = Conclusion: no superconductivity if t =0

26 Fermi surfaces for hole doping h=0,.1,.2,.3, Attractive band in pink

27 ?? Can we see the phase transitions?? T/t UCTION hole doping Dark regions: no solution to pseudogap equation T c /t 0.02

28 !e End