FOUR-BODY EFIMOV EFFECT

Transcription

1 FOUR-BODY EFIMOV EFFECT Yvan Castin, Christophe Mora LKB and LPA, Ecole normale supérieure (Paris, France) Ludovic Pricoupenko LPTMC, Université Paris 6

2 OUTLINE OF THE TALK Cold atoms in short Introduction and physical motivation for Efimov physics General theory for Efimovian N-mers The fermionic problem

3 COLD ATOMS IN SHORT

4 A brief history: 1980 s: laser cooling, low temperatures but non degenerate regime 1995: first Bose-Einstein condensation (Cornell, Wieman; Ketterle; Hulet) thanks to evaporative cooling 2002: degenerate Fermi gases, strongly interacting regime (Thomas; Jin; Salomon; Ketterle; Grimm; Hulet) thanks to the magnetic Feshbach resonance This young domain of research has received two Nobel prizes. Typical physical parameters for quantum gases: clouds of up to 10 6 atoms in magnetic traps or optical traps very low temperatures < 1µK, down to 0.45 nk (Ketterle, 2003).

5 a gaseous phase: densities < at/m 3 so mean interatomic distance is about 100 times larger than interaction range l VdW 10 9 m. Then interactions mainly characterized by s-wave scattering length a. Tuning the interactions: scattering length may in principle be adjusted at will (in practice, a [10 10 m, 10 5 m] for bosons, ρa 3 < 1 otherwise short lifetime by threebody recombination to two-body bound states (Cornell, Ketterle). Indirect evidence of Efimov trimers from peak of losses (Grimm). Fermi gases are stable over the whole range a R A condensate of dimers for 0 < ρa 3 < 1, a condensate of Cooper-like pairs for 1 < ρa 3 < 0

6 Particularly interesting is the unitary regime a =, which is scale invariant. Exists for arbitrary spin population imbalance: Existence of a polarized normal phase (Ketterle, Hulet, 2006). One can get resonant interactions in p wave, d wave,... channels. Fermions by Salomon (2005), Jin (2007). Not yet p-wave superfluidity. Studying the response to rotations: One can add a stirring laser: A rotating harmonic trap Formation of vortex lattices revealing superfluidiy, for bosons (Dalibard, 2001) and for fermions (Ketterle, 2005)

7 What is a Feshbach resonance? Low energy scattering properties: Low k general scattering theory: 1 f k = a 1 + ik 1 2 k2 r e +... Schematic view of Feshbach resonance: 0.2 V(r 12 ) /k B [10 3 K] 0 E mol Λ V r 12 [nm]

8 How to reach a strongly interacting regime? Try to maximize f k. Step 1: be in the regime r e 1 k typ a, k 1 F. zero range model potential possible and interactions characterized by a only. Step two: increase a to infinity, this is the unitary regime: f k 1 k universal regime, no interaction parameter left

9 A wide range of observable quantities: Static quantities: Density, momentum distribution, correlation functions in position space, in momentum space. Excitations with laser beams, including Raman and Bragg, or with radio-frequency fields Measurement of excitation spectra: Bogoliubov spectrum for bosons (Nir Davidson), collective excitations (sound velocity) and elementary excitations (BCS like gap) for fermions (Grimm, 2005; Jin, 2008) The atom optics toolbox. Dynamics of the Bose condensate phase, by splitting in two pieces and recombining (Cornell). Josephson junction and number squeezing with bosons (Oberthaler; Ketterle, 2007).

10 INTRODUCTION AND PHYSICAL MOTIVATION FOR EFIMOV PHYSICS

11 THE QUANTUM TWO-BODY PROBLEM Two interacting quantum particles in free space in 3D: For a long-range attractive interaction, an infinite number of bound states, with accumulation point at E = 0. Cf. hydrogen atom with V (r) 1/r : E n 1 n 2, n > 0 For short-range attractive interaction, finite number of bound states. Cf. van der Waals interaction O(1/r 6 ) between two atoms. Semi-classical (WKB) argument: If V (r) decays as 1/r 2 or more slowly, the classical action rturning p(r)dr = 2 dr{2µ[e V (r)]} 1/2 + 0 E 0 Can the situation change for more than two particles?

12 THE QUANTUM THREE-BODY PROBLEM From now on, the interaction V (r) is short-range attractive with no two-body bound state with a s-wave resonance: Diverging scattering length a Def. of a from two-body zero-energy scattering state: φ(r) = 1 r + incoming wave + a + O(1/r 2 ) r scattered wave Efimov s breakthrough (1970): Three bosons have an infinite number of l = 0 trimer states, E = 0 accumulation point, geometric spectrum: E (3) n n + E(3) ref e 2πn/ s 3 purely imaginary s 3 = i solves transcendental equation, E (3) ref depends on microscopic details.

13 ARE THERE EFIMOVIAN TETRAMERS? E (4) n n + E(4) ref e 2πn/ s 4? Negative results: Amado, Greenwood (1973): There is No Efimov effect for Four or More Particles. Explanation: bosons have Efimov trimers, tetramers decay. Hammer (2007), Greene (2009), Deltuva (2010): The four-boson pb depends only on E (3) ref, no E(4) ref to add. Key point: N = 3 Efimov effect breaks separability in hyperspherical coordinates for N = 4. Idea: Two species: Failure for three bosons + impurity (Adhikari). Here three same-spin-state fermions (mass M) and one impurity (mass m). No three-body Efimov effect if M/m < (Efimov, 1973; Petrov, 2003).

14 GENERAL THEORY FOR EFIMOVIAN N-MERS (N 3)

15 THE ZERO-RANGE MODEL USED BY EFIMOV Interaction replaced by contact conditions. Exact for vanishing binding energies (quantum number n + ). For r ij 0 with fixed ij-centroid C ij = (m i r i +m j r j )/(m i + m j ) different from r k, k i, j (here 1/a = 0): ( 1 ψ( r 1,..., r N ) = 1 ) A ij [ C ij ; ( r k ) k i,j ] + O(r ij ) r ij a Elsewhere, non interacting Schrödinger equation N Eψ = 2 ri ψ 2m i i=1 with correct exchange symmetry Scale invariance: ψ λ ( r 1,..., r N ) ψ( r 1 /λ,..., r N /λ) is another solution with eigenenergy E/λ 2.

16 Exercising with the zero-range model Scattering state of two particles: φ k (r) = e ik r e ikr + f k r For r > 0 this is an eigenstate of the non-interacting problem. Contact condition in r = 0 f k r + (1 + ikf k) + O(r) = A r + O(r) determines scattering amplitude f k : f k = 1 ik

17 Scaling invariance in the non-efimovian case 1 ψ( X) = A ij [ C ij ; ( r k ) k i,j ] + O(r ij ) rij 0 r ij Assume that there is no Efimov effect. Domain of Hamiltonian is scaling invariant: If ψ obeys all the required boundary conditions, so does ψ λ with ψ λ ( X) 1 λ 3N/2 ψ( X/λ) Consequences (also true for the ideal gas): free space box (periodic b.c.) trap N, no bound states ( ) PV = 2E/3 ( ) virial theorem ( ) If ψ of eigenenergy E, ψ λ of eigenenergy E/λ 2. Square integrable eigenfunctions (after center of mass removal) correspond to point-like spectrum, for selfadjoint H. ( ) F(N, V λ 3, T/λ 2 ) = F(N, V, T)/λ 2, derivative in λ = 1 and Gibbs-Duhem F = E TS = µn PV.

18 Virial theorem Particles trapped in general external potential U(r): N H = H Laplace + U(r i ) i=1 Consider eigenstate ψ of energy E. Mean energy of ψ λ : E λ = H Laplace ψ N λ 2 + U(λr i ) ψ i=1 d Eigenstate is stationary point of mean energy: dλ E λ = 0 in λ = 1. Gives energy from density (Thomas, 2008): N E = U(r i ) r i ri U(r i ) ψ = i=1 U harmonic 2E trap For hard walls E = 3 2 PV [ ru = force due to the wall]

19 Useful constraints for Monte Carlo [ξ = µ(t = 0)/EF 0.41 (Carlson, 2009)] Burovski, Prokof ev, Svistunov, Troyer (2006), Tc/TF = 0.152(7); Goulko, Wingate (2010), Tc/TF = 0.173(6) 0.55 s Burovski y gra 0.5 a are te ola vi iti ual es q ne ic i m na ody µ(t)/ef rm the 0.45 Bulgac Goulko E(T)/NEF

20 SEPARABILITY IN HYPERSPHERICAL COORDINATES Werner, Castin (2006) Use Jacobi coordinates to separate center of mass C Hyperspherical coordinates (arbitrary masses m i ): ( r 1,..., r N ) ( C, R, Ω ) with 3N 4 hyperangles Ω and the hyperradius N m u R 2 = m i ( r i C ) 2 i=1 where m u is an arbitrary mass unit. Hamiltonian is clearly separable: H internal = 2 [ R 2 2m + 3N 4 u R R + 1 ] R 2 Ω

21 Do the contact conditions preserve separability? For free space E = 0, yes, due to scale invariance: ψ 0 ( r 1,..., r N ) = R s N (3N 5)/2 φ( Ω). E = 0 Schrödinger s equation implies [ ( ) ] 3N 5 2 Ω φ( Ω) = s 2 N φ( Ω) 2 with contact conditions. s 2 N discrete real set. For arbitrary E, Ansatz with E = 0 hyperangular part obeys contact conditions [R 2 = R 2 (r ij = 0) + O(rij 2 )]: ψ = F(R)R (3N 5)/2 φ( Ω) Schrödinger s equation for a fictitious particle in 2D: EF(R) = 2 [ F (R) + 1 ] 2m u R F (R) + 2 s 2 N 2m u R 2

22 EF(R) = 2 [ F (R) + 1 ] 2m u R F (R) + 2 s 2 N 2m u R 2 For R 0, F(R) is a linear combination of solutions R ±s N. For s 2 N > 1, only R s N square integrable, no bound state. In a isotropic harmonic trap, spectrum E = E CoM + (1 + s N + 2n) ω, n N. For 0 < s 2 N < 1, at most one bound state. There exist Efimovian N-meres there exists s 2 N < 0 Effective N-body attraction. To have discrete energies, impose N-body contact condition with parameter q: F(R) R 0 (qr) s N + (qr) s N Discrete scale invariance λ 2s N = 1, geometric spectrum.

23 RECAP: CRUCIAL POINTS OF GENERAL THEORY To find N-body Efimov effect, one simply needs to extract the exponents s N from zero-energy solution ψ 0 ( r 1,..., r N ) = R s N (3N 5)/2 φ( Ω). General theory OK if Ω effect for all n < N. self-adjoint: no n-body Efimov

24 THE FERMIONIC PROBLEM

25 INTEGRAL EQUATION Three fermions (mass M, same spin state) and one impurity (mass m) General theory OK for a mass ratio α M m < Calculate E = 0 solution in momentum space. An integral equation for Fourier transform of A ij : [ 1 + 2α 0 = (1 + α) 2(k2 1 + k2 2 ) + 2α ] 1/2 (1 + α) 2 k 1 k 2 D( k 1, k 2 ) d 3 k 3 D( + k 1, k 3 ) + D( k 3, k 2 ) 2π 2 k1 2 + k2 2 + k α 2α ( k 1 k 2 + k 1 k 3 + k 2 k 3 ) D has to obey fermionic symmetry.

26 REDUCTION OF THE INTEGRAL EQUATION Rotational invariance: D is the m l = 0 component of a spinor of spin l: D( k 1, k 2 ) = t ρ D(R k 1, R k 2 ) Clever choice of the rotation matrix R: D( k 1, k 2 ) = t ρ D[k 1 e x, k 2 (cos θ e x + sin θ e y )] }{{} 2l+1 unknown functions f (l) m l (k 1,k 2,θ) Scaling invariance for E = 0: f (l) m l (k 1, k 2, θ) = (k k2 2 ) (s 4+7/2)/2 (cosh x) 3/2 Φ (l) m l (x, θ) with x = ln(k 2 /k 1 ). The integral equation gives M (l) s 4 [ Φ (l) ] = 0. s 4 allowed M (l) s 4 has a zero eigenvalue

27 RESULTS Numerical exploration up to total angular momentum l = 10 Four-body Efimov effect obtained for a single s 4, in channel l = 1 with even parity: D( k 1, k 2 ) = e z in the interval of mass ratio k 1 k 2 k 1 k 2 f(1) 0 (k 1, k 2, θ) < α <

28 s NUMERICAL VALUES OF s 4 ir α=m/m

29 EXPERIMENTAL ASPECTS Few-body problem with large scattering length: use a magnetic Feshbach resonance (Grimm, 2006; Hulet, 2009) Radio-frequency spectroscopy of trimers (Jochim, 2010) Remaining issue: Narrow interval of mass ratio. Solution 1: The right mixture 41 Ca and 3 He have mass ratio α [13.384, ] A priori, s large enough to see two tetramer states 41 Ca has same radioactivity as 239 Pu (half-life 10 5 years) Solution 2: Mass tuning 40 K and 3 He have slightly-off mass ratio α Use optical lattice to tune effective mass (Petrov, Shlyapnikov, 2007)

30 CONCLUSION The four-body Efimov effect has been elusive (even theoretically) for 40 years. We predict that the effect takes place in a 3+1 fermionic system in a narrow interval of mass ratio. Experimental realization with cold atoms plausible, although challenging. Phys. Rev. Lett. 105, (2010)