6.4 Search for Efimov States in Halo Nuclei like 14 Be, 19 B, 22 C and 20 C
9 1
candidates to investigate the occurrence of Efimov states is the halo
14 Be nucleus.
With this motive in view, we extended [90] the three-body model developed earlier
assuming
14 Be as a three-body system comprising of the two halo neutrons and a
12 Be−core.
For the neutron–neutron pair, we consider the separable potential:
v nn = −
λ n
2μ nn
g( p nn )g
p
nn
, g( p) =
1
p 2 + β 2
(6.49)
taking the strength parameter λ n = 18.6α
3 and the range parameter β = 5.8α already
considered in Sect. 6.1. In the present analysis, we keep these parameters fixed.
However, for the n−
12 Be binary sub-system, it is worth pointing out that while
there is an experimental evidence for a narrow peak in
13 Be corresponding to a
d 5/2 (l = 2) resonance unbound by more than 2 MeV [91], a strong possibility for
the existence of a low-lying s-orbital state which could have been difficult to identify
experimentally cannot be ruled out. It would, therefore, be interesting to explore the
consequences of considering such an intruder state in the context of studying
14 Be
as a three-body system particularly looking for its effect on the possibility of Efimov
states. Thus, for the n−
12 Be potential, we set
v nc = −
λ c
2μ nc
f ( p nc ) f
p
nc
, f ( p) =
1
p 2 + β
2
1
(6.50)
and allow the parameters λ c and β 1 to vary so as to obtain different sets producing
virtual and bound two-body systems near zero energy.
The basic structure of the three-body equation in terms of the spectator functions,
of F( p) and G( p) the
12 Be and of the halo neutrons satisfying the coupled integral
equations is essentially the same as given in Eqs. (4.26)–(4.28). However, for the
purpose of studying the sensitive computational details of the Efimov effect, we here
recast these equations involving only dimensionless quantities. Thus, by defining
τ
−1
n ( p)F( p) ≡ φ( p) and τ
−1
c ( p)G( p) ≡ χ( p),
(6.51)
where
τ
−1
n ( p) = μ
−1
n −
⎡
⎣ β r
β r +
p 2
2a
+ ε 3
2
⎤
⎦
−1
(6.52)
and
τ
−1
c ( p) = μ
−1
c − 2a
1 +
2a
p 2
4c
+ ε 3
−2
,
(6.53)
9 1
candidates to investigate the occurrence of Efimov states is the halo
14 Be nucleus.
With this motive in view, we extended [90] the three-body model developed earlier
assuming
14 Be as a three-body system comprising of the two halo neutrons and a
12 Be−core.
For the neutron–neutron pair, we consider the separable potential:
v nn = −
λ n
2μ nn
g( p nn )g
p
nn
, g( p) =
1
p 2 + β 2
(6.49)
taking the strength parameter λ n = 18.6α
3 and the range parameter β = 5.8α already
considered in Sect. 6.1. In the present analysis, we keep these parameters fixed.
However, for the n−
12 Be binary sub-system, it is worth pointing out that while
there is an experimental evidence for a narrow peak in
13 Be corresponding to a
d 5/2 (l = 2) resonance unbound by more than 2 MeV [91], a strong possibility for
the existence of a low-lying s-orbital state which could have been difficult to identify
experimentally cannot be ruled out. It would, therefore, be interesting to explore the
consequences of considering such an intruder state in the context of studying
14 Be
as a three-body system particularly looking for its effect on the possibility of Efimov
states. Thus, for the n−
12 Be potential, we set
v nc = −
λ c
2μ nc
f ( p nc ) f
p
nc
, f ( p) =
1
p 2 + β
2
1
(6.50)
and allow the parameters λ c and β 1 to vary so as to obtain different sets producing
virtual and bound two-body systems near zero energy.
The basic structure of the three-body equation in terms of the spectator functions,
of F( p) and G( p) the
12 Be and of the halo neutrons satisfying the coupled integral
equations is essentially the same as given in Eqs. (4.26)–(4.28). However, for the
purpose of studying the sensitive computational details of the Efimov effect, we here
recast these equations involving only dimensionless quantities. Thus, by defining
τ
−1
n ( p)F( p) ≡ φ( p) and τ
−1
c ( p)G( p) ≡ χ( p),
(6.51)
where
τ
−1
n ( p) = μ
−1
n −
⎡
⎣ β r
β r +
p 2
2a
+ ε 3
2
⎤
⎦
−1
(6.52)
and
τ
−1
c ( p) = μ
−1
c − 2a
1 +
2a
p 2
4c
+ ε 3
−2
,
(6.53)