Molecular fraction calculations for an atomic-molecular Bose-Einstein condensate model

Transcription

1 Molecular fraction calculations for an atomic-molecular Bose-Einstein condensate model Jon Links Centre for Mathematical Physics, The University of Queensland, Australia. 2nd Annual Meeting of ANZAMP, Mooloolaba 2013

2 Hamiltonian The model consists of three interacting bosonic degrees of freedom: H = U aa N 2 a + U bb N 2 b + U ccn 2 c + U ab N a N b + U ac N a N c + U bc N b N c + µ a N a + µ b N b + µ c N c + Ω(a b c + c ba). It commutes with the total atom number N = N a + N b + 2N c and the atomic imbalance J = N a N b. We introduce k = J/N, k [ 1, 1] as the fractional atomic imbalance. The classical analogue is the non-linear pendulum H = λpφ 2 + 2(α λ)p φ + λ 2α + β ( ) 4φ + 2(1 p φ )(p φ + c + )(p φ + c ) cos N

3 with λ = α = β = ( 2N Uaa Ω 4 + U bb 4 + U cc 4 + U ab 4 U ac 4 U ) bc 4 ( 2N 1 + k U aa + 1 k U bb + 1 Ω U ab 1 + k ) 2N (µ a + µ b µ c ) 2N ( (1 + k) 2 U aa + (1 k) 2 U bb + (1 k 2 )U ab Ω + 2 ) N ((1 + k)µ a + (1 k)µ b ) U ac 1 k U bc 4 c ± = 1 ± 2k.

4 Order parameter - the molecular fraction Define the order parameter O = 2 N c N = 2 N E 0 µ c which measures the average molecular fraction: O = 0 atomic phase, 0 < O < 1 mixed phase, O = 1 molecular phase. The order parameter relates to the momentum of the classical system through O 1 2 (1 p φ). It is not associated with symmetry breaking.

5 Classical fixed points The dynamical evolution is governed by Hamilton s equations: dp φ = H dt φ = 4 ( ) 4φ 2(1 p φ )(p φ + c + )(p φ + c ) sin, N N dφ dt = H = 2λp φ + 2α 2λ p φ + (1 p φ)(2p φ + 2) (p φ + c + )(p φ + c ) 2(1 pφ )(p φ + c + )(p φ + c ) cos The fixed points of the system are determined by the condition H φ = H p φ = 0. Phase boundaries of the parameter space are identified according to fixed point bifurcations. ( ) 4φ. N

6 Bifurcation diagram: k 0 Figure : Regions A, B, C determined by bifurcation analysis.

7 Bifurcation diagram: k = 0 Figure : Regions I, II, III, IV, V determined by bifurcation analysis.

8 Molecular fraction: λ = µ a = µ b = 0 - numerical N=300, J=0 N=301, J=1 N=299, J=1 2N c /N / µ 2N c /N µ Figure : Order parameter. Inset shows the first derivative.

9 Bethe ansatz solution where for J 0 E = AM(M 1) + BM + C Ω Ω(J + 1 v 2 j ) + Bv j v j (Ω + Av j ) = M k j M = (N J)/2, L = (N + J)/2, and A = U aa + U bb + U cc + U ab U ac U bc, M v j, j=1 2 v k v j, j = 1, 2,..., M, B = (1 + 2L 2M)U aa + U bb + (1 2M)U cc + (1 + L M)U ab + (2M L 1)U ac + (M 1)U bc + µ a + µ b µ c, C = (L M) 2 U aa + M(L M)U ac + M 2 U cc + (L M)µ a + Mµ c.

10 Rubeni, Foerster, Mattei, Roditi, Nucl. Phys. B 856 (2012) Ground-state roots: λ = µ a = µ b = 0 Figure : Ground-state root distribution for λ = 0, N = 120.

11 Links and Marquette, work in preparation Large-N limit Ground-state root density: ρ(v) = (b v)(v a) 1 = b Ground-state energy: ( N + 1 E 0 = = µ + 2 Bethe ansatz equation: a dv ρ(v) ) (J + 1)2 2ab ( 1 2πM + J + 1 ) 2πM abv (J + 1)µ 2α 2N ab (ab ) (J + 1)2 ab J + 1 v v µ c Ω = 2M lim ε 0 v ε a dw ρ(w) w v b + 2M lim dw ρ(w) ε 0 v+ε w v

12 Links and Marquette, work in preparation leads to (ab) 2 + 2(1 α 2 )Nab 4(J + 1)α 2N ab 3(J + 1) 2 = 0. For J = O(N 0 ) α > 1 ab 2(α 2 1)N, α = 1 ab 2 5/3 (J + 1) 2/3 N 1/3, ( α < 1 ab (J + 1) 2 2 2α + ) 2 2α (1 α 2 N 1. ) This yields where α 1 O = 1, α 1 O = 1 + f 2 + α f = 2(1 α 2 ) 2 2α + 2α ( 2f + 3 2α f 2 ) df dα

13 Molecular fraction: λ = µ a = µ b = 0 - numerical N=300, J=0 N=301, J=1 N=299, J=1 2N c /N / µ 2N c /N µ Figure : Order parameter. Inset shows the first derivative.

14 Links and Marquette, work in preparation Molecular fraction: λ = µ a = µ b = 0 - analytic curve for all finite J in the N limit Figure : Order parameter. Inset shows the first derivative.

15 Links and Marquette, work in preparation Future work Extend the analysis to cover the entire parameter space.