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
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
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 2 2 2 U ab 1 + k 4 + 1 ) 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.
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.
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
Bifurcation diagram: k 0 Figure : Regions A, B, C determined by bifurcation analysis.
Bifurcation diagram: k = 0 Figure : Regions I, II, III, IV, V determined by bifurcation analysis.
Molecular fraction: λ = µ a = µ b = 0 - numerical 1 0.95 N=300, J=0 N=301, J=1 N=299, J=1 2N c /N 0.9 0.85 0.8 0.75 0.7 0.65 0.6 / µ 2N c /N 0 0.01 0.02 0.03 0.04 0.05 30 25 20 30 28 26 24 22 20 18 16 14 12 µ Figure : Order parameter. Inset shows the first derivative.
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.
Rubeni, Foerster, Mattei, Roditi, Nucl. Phys. B 856 (2012) 698-715 Ground-state roots: λ = µ a = µ b = 0 Figure : Ground-state root distribution for λ = 0, N = 120.
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
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 2 + 6 2(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 2 + 6. ( 2f + 3 2α f 2 ) df dα
Molecular fraction: λ = µ a = µ b = 0 - numerical 1 0.95 N=300, J=0 N=301, J=1 N=299, J=1 2N c /N 0.9 0.85 0.8 0.75 0.7 0.65 0.6 / µ 2N c /N 0 0.01 0.02 0.03 0.04 0.05 30 25 20 30 28 26 24 22 20 18 16 14 12 µ Figure : Order parameter. Inset shows the first derivative.
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.
Links and Marquette, work in preparation Future work Extend the analysis to cover the entire parameter space.