Universal angular probability distribution of three particles near zero-energy threshold

Abstract

We study bound states of a 3-particle system in R 3 described by the Hamiltonian H(λ n ) = H 0 + v 12 + λ n (v 13 + v 23 ), where the particle pair {1, 2} has a zero energy resonance and no bound states, while other particle pairs have neither bound states nor zero energy resonances. It is assumed that for a converging sequence of coupling constants λ n → λ cr the Hamiltonian H(λ n ) has a sequence of levels with negative energies E n and wave functions ψ n , where the sequence ψ n totally spreads in the sense that lim n→∞ |ζ|≤R |ψ n (ζ)| 2 dζ = 0 for all R > 0. We prove that for large n the angular probability distribution of three particles determined by ψ n approaches the universal analytical expression, which does not depend on pair-interactions. The result has applications in Efimov physics and in the physics of halo nuclei.

Universal low-energy behavior in three-body systems

“…For an operator A acting on a Hilbert space D(A), σ(A) and σ ess (A) denote the domain, the spectrum, and the essential spectrum of A respectively [19]. Recently a new type of universality in 3-body systems has been discovered in [5]. Consider the Hamiltonian of the 3-particle system in R 3…”

“…The pair-interactions v ik are operators of multiplication by real V ik (r i − r k ) ≤ 0 and r i ∈ R 3 are particle position vectors. For pair potentials we require like in [5] that…”

“…We need only to prove that the limit on the lhs of (29) exists and is positive. Note that all requirements R1-R3 in [5] are satisfied and the sequence ψ n totally spreads (see Sec. Theorems 1,3 in [17] for the proof).…”

“…(60)Then using Eq. (98) in[5] we obtain the inequalityM −(R,∞) (|η|)φ 0 |v 12 | (R,∞) (|η|)φ 0 |v 12 | (R,∞) (|η|)φ 0 |v 12 | |v 13 | 1/2B 13 (k n )Ψ (i) n := |v 13 | 1/2B−1 13 (k n ) H 0 + k |v 12 ||ψ n |,(63)Ψ (2) n := λ n |v 13 | 1/2B−1 13 (k n ) H 0 + k |v 23 ||ψ n |,(64)Ψ (3) n := λ n |v 13 | 1/2 H 0 + k |v 13 | 1/2 1 − λ n |v 13 | 1/2 H 0 + k |v 13 | 1/2 −1 ×B −1 13 (k n )|v 13 | 1/2 H 0 + k |v 12 | + λ n |v 23 | |ψ n |.(65)Eqs. (62) and (63)-((65)) come from Eqs.…”

“…By Eq. (103) in[5] F (i) n (η, ζ) ≤1 2 7/2 π 5/2 d 3 η ′ d 3 p ζ V 13 (α ′ η ′ ) 1/2 e − √ p 2 ζ +k 2 n |η−η ′ | |η − η ′ |t n (p ζ ) Ψ (i) n (η ′ , p ζ ) , (66) whereΨ (i) n = F 13 Ψ (i)n . Using the exponential falloff of V 12 (x) from(27) one can easily derive the inequalityφ 0 (x) V 12 (αx) ≤b 1 e −b 2 |x| ,(67)whereb 1,2 are constants.…”