physics of radioactive beams1 chapter 4 borromean...

Physics of Radioactive Beams1

Chapter 4Borromean Nuclei

Carlos A. Bertulani, Texas A&M University-Commerce, TX 75429, USA

1These notes consist of a series of lectures presented by the author at the Geselschaft für Schw-erionenforschung, Darmstadt, Germany in the Spring of 1994. GSI-Report 1994-11. This materialwas latter extended and published in the book “Physics of Radioactive Beams”, C.A. Bertulani,M. Hussein and G. Muenzenberg, Nova Science, Hauppage, NY, 2002, ISBN: 1-59033-141-9

0.1. INTRODUCTION

0.1 Introduction

The nucleus 11Li is a prototype of a halo nucleus with fascinating properties. The reason isits unusual large radius and the difficulty in explaining its properties by means of traditionaltheoretical tools. As seen in Table 3.3 the nucleus 10Li is not bound, having a resonantstate at ∼ 500 keV in the continuum. However 11Li is bound by ∼ 350 keV. The valenceneutrons in 11Li have to be treated separately from the core nucleons in order to explainits properties (i.e., radius) with traditional techniques. Thus, it seems that 11Li should betreated theoretically as a three-body problem, 9Li+n+n ; in which to a good approximation9Li is treated as a inert core. This assumption is supported by the fact that the valenceneutrons are distributed in a spatial region which is much larger than the core. As seen inbefore rms(

11Li) ∼= 3.1 fm while rms(9Li) ∼= 2.4 fm. Calling rms(n) the root mean squareradius of the distribution of a valence neutron, one can write

rms(

11Li) ∼= 2

11rms(n) +

9

11

(9Li) ∼= 2

11(6 fm) +

9

11(2.4 fm). (1)

Thus, the rms of the valence distribution is of order of 6 fm, which is much larger thanrms (

9Li) . In other words, the valence neutrons in 11Li “see” the core (9Li) almost like apoint particle.

The properties of the three-body halo nuclei were so intriguing that a new name wascoined for them by J.S. Vaagen and M. Zhukov: Borromean nuclei [1].The name recallsthe Borromean rings, heraldic symbol of the princes of Borromeo, which are carved in thestone of their castle on an island in Lago Maggiore in Northern Italy. The three rings areinterlocked in such a way that if any of them were removed, the other two would also fallapart. The 3-body quantum analog is one where the 3-body system is bound, but wherenone of the two-body subsystems are bound.

Nature has realized Borromean systems in loosely bound halo-like nuclei such as 11Li,which are now produced as secondary beams. To understand these systems one needs to gobeyond the mean field approach, to few-body theoretical procedures such as expansion onhyperspherical harmonics or the coordinate space Fadeev approach. Nuclei like 6He and 11Lican (a realistic starting point!) be modeled as a core surrounded by two loosely bound valenceneutrons. Not only the bound state (6He and 11Li both have only one bound state!), butalso the structure of the continuum and the details of reaction mechanisms can be describedin this way.

Besides the three-body halos (e.g., 6He, 11Li) the two-body halos (e.g., 8Be, 11Be, 19C,etc.) can also be treated in terms of a core and a proton, or neutron, outside the core. Somecharacteristics of these nuclei can be reproduced in this way. However, to obtain the correctassignments of spin and parity and the model has to be improved.

2 Physics of Radioactive Beams - C.A. Bertulani

0.2. THE SINGLET STATE IN THE DEUTERON

Figure 1: The Lithium-11 nucleus consists of a core Lithium-9 nucleus orbited by two neu-trons. If any one of the three bodies is removed the remaining two would be unbound, ratherlike heraldic Borromean rings.

Supplement A

0.2 The singlet state in the deuteron

Before we go ahead and look for the properties of 11Li from the point of view of a three-bodyproblem, let us recall some useful concepts of the two-nucleon system, in particular the deuteron.

The binding energy of the deuteron is ∼ 2.2 MeV which is very small (∼ 1 MeV for eachnucleon) as compared with the binding energy of a nucleon in a large nucleus, which is of orderof 8 MeV. It is also observed experimentally that there is no excited state of the deuteron. Wehave shown that these properties can be obtained by simulating the interaction of the neutronand the proton in the deuteron by a square well potential with depth V0 ∼= 36 MeV and a rangeof r0 ∼ 2 fm. It was also seen that the size of the deuteron is larger than the range of thispotential. Beyond the range of the potential the deuteron wavefunction u(r)/r ∼ e−ηr/r (aYukawa function), with ~2η2/mN = B .

We now look for the continuum states of the n-p system. I.e., for E > 0 . We again consider

Physics of Radioactive Beams - C.A. Bertulani 3


a square-well potential and ` = 0 . Thus

d2uidr2

+ p2ui = 0 , for r ≤ r0,d2u0dr2

+ k2u0 = 0 , for r > r0 (2)

where

p2 =mN(V0 + E)

~2, k2 =

mNE

~2. (3)

Since u→ 0 as r does, the solution of 2, for r 6 r0 is

ui = D sin (pr) . (4)

Similarly, for r > r0 ,u0 = C sin(kr + δ). (5)

Matching the logarithmic derivatives at r = r0 yields

p cot ( p r0) = k cot(kr0 + δ). (6)

For E ∼ 0 we can approximate p cot ( p r0) ∼= −η which is the logarithmic derivative of e−ηr .Thus,

p cot (p r0) ∼= −η = −√mNB

~(7)

andcot(kr0 + δ) ∼= −

η

k. (8)

For E ∼ 0 , kr0 � 1 , δ � 1 and k2 � η2 and the above equation becomes

δ ∼= −k

η− kr0. (9)

The scattering cross section at low energies is

σsc ∼=4π

k2δ2= 4π

(1

η+ r0

)2. (10)

Using 1/η = 4.3 fm, r0 = 2 fm, we find σsc = 5 b . Alternatively, if we neglect r0 ascompared to 1/η we find

σsc ∼=4π

η2= 2.3 b. (11)

Other approximations give values somewhere between these two ranges. The experimental value ishowever σsc = 20.4 b .

This bad agreement can be explained as follows. The n-p system depends very much on its spinproperties. The proton and neutron have spin 1/2 and their spin wave function is to be described



by two states χ(+1/2) and χ(−1/2) , spin-up and spin-down, respectively. The n-p system is acombination of these wave functions with

ξs =1√2

{χ(+)(p)χ(−)(n)− χ(+)(n)χ(−)(p)

}(12)

ξ(+1)t = χ

(+)(p)χ(+)(n) (13)

ξ(0)t =

1√2

[χ(+)(p)χ(−)(n) + χ(+)(n)χ(−)(p)

](14)

ξ(−1)t = χ

(−1)(p)χ(−)(n) (15)

where ξs and ξt are known as singlet and triplet states. They correspond to total spin 0 and 1 ,respectively. Experimentally it is found that the deuteron spin is 1. Therefore, the deuteron is ina triplet state.

When proton and neutron scatter (E > 0) they can be in any of the states described above. Ina general scattering experiment, the spin of the nucleons are randomly oriented and the n-p systemis in the triplet and singlet states in proportion to the corresponding weight factors for these states,which are, respectively, 3/4 and 1/4. The total scattering cross section consists of two parts

σ =3

4σt +

1

4σs (16)

where σt and σs are the cross sections for scattering in the triplet and singlet states respectively.Because of the preference for the triplet state for the deuteron we also assume this preference

for the n-p scattering at E ∼ 0 . Following Eq. 10 we write

σ = 3π

(1

ηt+ rt

)2+ π

(1

ηs+ rs

)2(17)

where

ηt =

√mNB

~2, ηs =

√mN Es

~2. (18)

Es is the energy of a singlet state, which can only exist in the continuum.Assuming rt ∼= rs ∼= 2 fm, using σ = 20.4 b and 1/ηt = 4.3 fm we get∣∣∣∣ 1ηs

∣∣∣∣ = 25 fm (19)which gives |Es| = 66 keV. Eqs. 17 and 18 do not determine the sign of ηs and Es . How-ever, the scattering length as is negative, if one uses the theory to fit the experimental data [2](at = 5.37 fm, as = −23.73 fm) . Hence theEs = 66 keV singlet state is not a bound deuteronstate, but a virtual state.


0.3. TWO LOOSELY-INTERACTING PARTICLES IN A POTENTIAL WELL

0.3 Two loosely-interacting particles in a potential well

The three-body 11Li problem amounts to solving the Hamiltonian

H = H1(r1) +H2(r2) +H ′ (r1, r2) (20)

where r1 and r2 are the coordinates of the two neutrons with respect to9Li, assumed to

be infinitely heavy.The ground state energy of 11Li would be

〈ψ|H|ψ〉 = 2�n + 〈ψ|H ′|ψ〉 (21)

where ψ is the ground-state wavefunction. To find ψ is a big task, involving the solutionof the Fadeev equations (see next Section).

However, a qualitative argument, originally presented by Migdal [3], can be given to showthat, although 10Li is unbound, 11Li can be in a bound state. For this it is necessary that10Li has a resonant state in the continuum.

Then we can assume that the behavior of the neutron wavefunctions is dominated by theresonant state, i.e., ψn ∼ ψn′ ∼ ψres . The energy of this state is �res ∼ 500 keV. If the radialwavefunctions are little affected by H ′ , then

〈ψ|H|ψ〉 ∼= 2�res + 〈H ′ (r1, r2)〉 . (22)

To simplify the calculations, let us assume that H ′(r, r′ ) = −v0(r)δ(r − r′ ) . Then theintegral of the second term reduces to a single integral. That is,

〈H ′〉 ∼=∫ψ4res(r)v0(r)d

3r. (23)

The wavefunction ψres(r) can be approximated by

ψres(r) =A√4π

{(sinkr) /r for r < RΛ for r > R

(24)

when E ∼ 0 . R is the range of the neutron+9Li potential, which we assume R ∼= 3.2 fm.Also, ~k =

√2mN(E + U0) ∼=

√2mNU0 , where U0 ∼ 50 MeV is the depth of the n + 9Li

potential and A is a normalization constant.We now show that the form of the wavefunction for r > R is irrelevant. The neutron-

neutron interaction can be taken as

v0(r) = −v0 e−r2/r20 (25)

where v0 ∼ 50 MeV, r0 ∼ 2 fm.


0.4. VARIATIONAL METHODS

The integral 23 becomes

〈H ′〉 = −A4

4πv0

{∫ R0

sin4 kr

r2e−r

2/r20dr + Λ4∫ ∞R

e−r2/r20

r2dr

}

= −A4

4πv0{

0.887 + 3.20× 10−3}, (26)

using the values given above.It is also natural to normalize ψres(r) inside the range r < R since in a resonant state

the particle has a larger amplitude to be found within the range of the potential. Neglectingthe second term in 26 we obtain

〈H ′〉 ∼ −2.5× 10−2v0 ∼ −1.25 MeV. (27)

Using this value in 22 we find that

E11Liground ∼ −0.25 MeV, (28)

which is close to the experimental value.Of course, this result is based on the assumption that the n − n interaction can be

considered as a perturbation, but it comes out larger than 2 �res . Thus, the conclusion isthat H ′ is essential to produce the binding in 11Li. Although the dineutron system is notbound, it can became bound in the presence of a core nucleus [3].

Based on this idea Hansen and Jonson [10] proposed that much of the properties of 11Lican be obtained by assuming an inert 9Li-core bound to a particle called by di-neutron.As in the case of the deuteron the wavefunction for 11Li at large distances is given byψ2n−9Li(r) ∼ e−ηr/r , with η =

√2µB/~ ∼= 6 fm−1.

Supplement B

0.4 Variational methods

The Schrödinger equationH|ψ〉 = E|ψ〉 (29)

is equivalent to the variational equation

δE(ψ) = 0 (30)


0.4. VARIATIONAL METHODS

where

E(ψ) =〈ψ|H|ψ〉〈ψ|ψ〉

. (31)

The variation is obtained by modifying the initial (normalized) wavefunction by

ψ′ = ψ + δψ. (32)

Replacing 32 in 31, we obtain

E(ψ′) =〈ψ′|H|ψ′〉〈ψ′|ψ′〉

=〈ψ + δψ|H|ψ + δψ〉

〈ψ′|ψ′〉

= E +〈ψ|[H − E]|ψ〉〈ψ|ψ〉

(33)

where in the last step we assumed δψ small and 〈δψ|ψ〉 = 0 .Hence δE = E(ψ′) − E = 0 implies 29. Also, δE is quadratic in δψ and if E0 is the

lowest eigenvalue of H , then (H −E0 ) is a positive operator and E(ψ′) ≥ E0 . This observationleads to the conclusion that E(ψ) must be minimized with respect to the parameters on which ψdepends in order to approximate the true lowest eigenvalue of H . That is, for ψ = ψ(αi)

∂E

∂αi=

1

(〈ψ|ψ〉)2

{〈ψ|ψ〉

[〈 ∂ψ∂αi|H|ψ〉+ 〈ψ|H| ∂ψ

∂αi〉]−

− 〈ψ|H|ψ〉[〈 ∂ψ∂αi|ψ〉+ 〈ψ| ∂ψ

∂αi〉]}

=1

〈ψ|ψ〉

{〈 ∂ψ∂αi|H − E(ψ)|ψ〉+ 〈ψ|H − E(ψ)| ∂ψ

∂αi〉}

= 0 (34)

where we have used Eq. 33.The important step in this method is to choose a parametrized trial wavefunction ψ that

reflects the main features of the interaction.For the three-body system a trial function can be a simple product

ψ = φ(r1)φ(r2)φ(r3). (35)

We would then be looking for the best φ that would lead to a stationary E .As an illustration consider a simple attractive two-body potential

V (r) = −V0e−r/r0

(r/r0). (36)


0.5. A VARIATIONAL CALCULATION FOR 11LI

A possible trial wavefunction for a system of 3 identical particles is

ψ1 = e−α(r1+r2+r3) (37)

with the non-linear parameter α that is varied to yield the lowest ground state energy. A secondtrial wavefunction is

ψ2 =N∑i=1

ai Ŝ e−α(jr1+kr2+`r3) (38)

for which the parameters ai, i = 1, · · · , N are varied while α is kept fixed. Ŝ is the symmetriza-tion operator and j, k, ` are integers depending on the value of i .

In general, as the two-body interaction becomes more complicated by the addition of suchfeatures as the repulsive core, the tensor component, etc., the search for a satisfactory trial wave-function becomes more involved, too.

Completely equivalent to a variational procedure is a diagonalization calculation. In this caseψ is expanded in terms of a complete set of wavefunctions ψν

ψ =∑ν

aν ψν . (39)

The exact wavefunction for a given E is determined by diagonalizing the matrix Hµν =〈ψν |H|ψν〉 . For a complete set of antisymmetric ψν (with ν → ∞ ), the diagonalization of thematrix 〈ψµ|H|ψν〉 would result in a series expansion of Ψ . But taking a large set of ψν wouldinvolve a prohibitive amount of work, and generally only a few terms on the right side of Eq. 39 areretained. Actually, a few values of ν are sufficient to determine a ψ which is a good approximationto the exact wavefunction, provided the set of ψν

′s chosen is such that some of them are fairly

close to the actual wavefunction.

0.5 A variational calculation for 11Li

The Hamiltonian for 11Li is given by

H =p21 + p

22

2mN+

p232m9

− (p1 + p2 + p3)2

2 (2mN +m9)

+Vnn (|r1 − r2|) + Vn−9 (|r1 − r3|) + Vn−9 (|r2 − r3|) (40)

where the two neutrons and 9Li are labeled by 1,2 and 3, respectively. The center of masskinetic energy is subtracted from Eq. 40 (3rd term on the rhs). Transforming to Jacobi



coordinates the corresponding momenta are

pρ =1

2(p1 − p2) (41)

pλ =m9 (p1 + p2)− 2mNp3

m9 + 2mN(42)

R =mN (r1 + r2) +m9 r3

m9 + 2mN, pR = p1 + p2 + p3 (43)

so that

H =p2ρmN

+p2λ2µ

+ Vnn(ρ)

+Vn9

(|λ− 1

2ρ |)

+ Vn9

(|λ+ 1

2ρ |)

(44)

where µ = 2mNm9/m11 is the reduced mass of the9Li-dineutron system. The interactions

depend on the absolute values of λ and ρ and the angle θ between their directions. Thethree remaining degrees of freedom correspond to relations of the entire three particle system.For the calculation of the ground-state (S = 0) one may neglect the rotational degrees offreedom.

Johannsen, Jensen, and Hansen [5] used the following potentials

Vnn(r) =∑k

Skρ exp[−(r/bkρ)2

]Vn9(r) =

∑k

Skλ exp[−(r/bkλ)2

]. (45)

The ground state wavefunctions can be expanded in terms of harmonic oscillator wave-functions,

ψ(ρ, λ, θ) = (αβ)3/4∑nρ`

a`nρRn`(αρ2)Rρ`(βλ

2)

(`+

1

2

)1/2P`(cos θ) (46)

where P` is the `th Legendre Polynomial and Rn` is the properly normalized radial wave-function. Only even `-values need to be included as the Hamiltonian is even under thetransformation θ → π− θ and the odd values of ` correspond to excited states. For a givensize of the basis the precision is improved by treating also the scale parameters α and βas variational parameters, in addition to the expansion coefficients a`nρ . The Schrödingerequation can now be solved by diagonalization to obtain the energy E11 .



The neutron-neutron potential is chosen so as to reproduce the low-energy scatteringdata, which through the known ` = 0 scattering length and effective range parameterdetermines the two parameters of a single Gaussian uniquely: Sρ = −31 MeV, bρ = 1.8 fm.This is a good approximation and no additional terms in Eq. 45 is needed.

No available data is known of scattering of neutrons on 9Li. To construct the potentialone can fold a Gaussian (or harmonic oscillator) density distribution of 9Li with an averagerange of the nucleon-nucleon interaction. One obtains bλ ∼= 2.4 fm. With this value, apotential depth Sλ ∼= −10 MeV corresponds to the limit of no binding of 10Li. The Gaussianpotentials fall off more rapidly than the correct long-range weakly attractive potential. Thus,various types of Vng potentials can be tested [5].

For each set of potentials parameters a variational calculation yields

Eρ = 〈ψ| −~2

mN∇2ρ + Vnn(ρ)|ψ〉

Eλ = 〈ψ| −~2

2µ∇2λ + Vn9

(|λ− 1

2ρ|)

+ Vn9

(|λ+ 1

2ρ|)|ψ〉

δ〈r2〉 = 111

{18

11〈ψ|λ2|ψ〉+ 1

2〈ψ|ρ2|ψ〉

}〈p2n〉 = 〈ψ|p2ρ|ψ〉+

1

4〈ψ|p2λ|ψ〉 (47)

where

δ〈r2〉 ≡ 〈r2〉11 −9

11〈r2〉9 (48)

represents (2/11)th of the mean-square radius of a neutron in the halo.

A set of parameters (b(1)λ , S

(1)λ = 2.0 fm, 15 MeV and b

(2)λ , S

(2)λ = 2.5 fm, −17 MeV) was

found [5] that reproduce the experimental value of δ〈r2〉 ∼= 6 fm and E11 ∼= −0.3 MeV. Oneobserves that the first parameters correspond to a repulsive Gaussian potential.

Contrary to the simple estimates, it was found [5] that Eρ is always positive, for all setof parameters tested. This shows that a three body system has subtle intrinsic structure,which are in most times not obtainable from two-body arguments only.

In this model, the intrinsic (ρ) and relative (λ) coordinates have a most probable valueof 3.3 fm and 3.1 fm, respectively. A sphere with radius of 2.2 fm (the core of 11Li) has aprobability of 0.8 for having both neutrons outside the sphere. For comparison, a dineutronmodel with (normalized) wavefunction

ψ(r) =η

2πe−ηr (49)

yields a value of P (r > r0) = e−2ηr0 . Thus, the probability that both neutrons are outside

the core is ∼ 0.5 for the dineutron model. This value can increase somewhat if one corrects



Eq. 49 for the distortion caused by a finite range potential. For a square-well potential withradius R the wavefunctions for r > R can be approximated by [10]

ψ(r) =

√η

2π

eηR√1 + ηR

e−ηr

r(50)

With this wavefunction the probability that both neutrons are at r > R = 2.2 fm is 0.6.If one switches off the n − n interaction (i.e., Sρ = 0) it is not possible to get 11Li

bound and 10Li unbound at the same time. Also, in this case, the most probable separation1

between the neutrons is now 5.6 fm and the dineutron-core separation in 2.9 fm. Thus then− n interaction is very important in determining the properties of 11Li. It also brings theneutrons closer together in the halo.

1In case one takes 10Li as bound.


0.6. TWO-BODY SCATTERING AT LOW ENERGIES

Supplement C

0.6 Two-body scattering at low energies

0.6.1 Partial wave decomposition

If E > 0 , the energy of a two-body system can attain any value. For a spherically symmetricpotential and for each value of E the wave equation reduces to

d2u`dr2

+

{2µ

~2[E − V (r)]− `(`+ 1)

r2

}u` = 0 (51)

where

ψ(r ) =∞∑`=0

a`(k)u`(r)

rP`( cos θ) , with k =

√2µE

~2. (52)

For r →∞, V (r)→ 0 and `(`+ 1)/r2 → 0 . The solution of Eq. 51 is

u`(r) ∼ sin(kr − `π

2+ δ`

)(53)

where δ` is the scattering phase shift.The total wavefunction is composed of as incident plane wave and a scattered wave

ψ ∼= eik·r + f(θ)eikr

r.

Expanding in Legendre polynomials

ψ(r, θ) ∼=∞∑`=0

[(2`+ 1) i`j`(kr) + f`

eikr

r

]P`(cos θ) (54a)

wheref(θ) =

∑`

f` P`(cos θ) (55)

and j`(kr) is the spherical Bessel function. At r →∞

j`(kr) −→1

krsin(kr − `π/2). (56)



Using Eq. 53, and comparing Eq. 52 with Eq. 54a one finds

f(θ) =1

k

∞∑`=0

(2`+ 1)eiδ` sin δ` P`(cos θ). (57)

The probability current is given by

S =~µ

Im [ψ∗∇ψ] . (58)

For the incident wave

Si =~µ

Im

[e−ikz

d

dzeikz]

=~kµ

= v. (59)

For the scattered wave, the radial current is

Sr =~m

Im

{f ∗(θ)

e−ikr

r

∂

∂r

[f(θ)

eikr

r

]}=

v

r2|f(θ)|2. (60)

The differential scattering cross section is defined as

dσ

dΩ=r2SrSi

= |f(θ)|2 = 1k2

∣∣∣∣∣∞∑`=o

(2`+ 1) eiδ` sin δ` P`(cos θ)

∣∣∣∣∣2

. (61)

The total cross section is

σ =

∫dσ

dΩdΩ =

4π

k2

∞∑`=0

(2`+ 1) sin2 δ`. (62)

It is maximum whenever

δ` =

(n+

1

2

)π , n = 0,±1,±2, · · · (63)

which is a resonance condition.If E → 0 (k → 0) and V is very small, f` → 0 , and

ψ(r ) '∞∑`=0

B`(kr)P`(cos θ)

whereB`(kr) = (2`+ 1)i

` j`(kr) ,



i.e.,

B0(kr) =sin(kr)

kr∼= 1−

(kr)2

6+ · · ·

B1(kr) = i3

[sin(kr)

(kr)2− cos(kr)

kr

]∼= i

[kr − (kr)

3

10+ · · ·

](64)

Thus, at r ∼ r0 , ∣∣∣∣B1B0∣∣∣∣2 ∼= (kr0)2 (65)

For a neutron-proton system scattering at 1 MeV in the laboratory system, and using r0 ∼ 2 fm,only 9% of the scattering is due to ` = 1 . In fact, for laboratory energies up to 10 MeV, s-scattering(` = 0 ) is predominant.

In the absence of a potential, and ` = 0

ψ`=0 ∼sin(kr)

kr=eikr − e−ikr

2ikr. (66)

In the presence of a potential, Eq. 53 yields

ψ(r)`=0 =

u`=0(r)

r−→r→∞

sin(kr + δ0)

r∼ e

i(kr+2δ0) − e−ikr

2ikr, (67)

since this is a mere proportionality.The difference between Eq. 67 and Eq. 66 is the scattered wave

ψsc ∼eikr

2ikr

(e2iδ0 − 1

). (68)

For elastic scattering, δ0 is real, and the probability currents are

|ingoing wave | = |outgoing wave |.

Also, for ` = 0 , the total scattering cross section is

σsc =4π

k2sin2 δ0 (69)

which is angle-independent.



Figure 2: A scattering state with (a) positive and (b) negative scattering length.

0.6.2 Scattering length

For E → 0 , and ` = 0 (unnormalized!),

u = rψ = eiδsin(kr + δ)

k(70)

where δ ≡ δ0 . If u is finite, δ must approach zero as k does. Defining

a = limk→0

(−sin δ

k

)(71)

we get from Eq. 69

σsc = 4πa2. (72)

Thus a has the significance of the radius of a hard sphere from which the scattering occurs(note that classically σsc = πa2 ).

Eq. 71 also means δ/k = −a , and we can rewrite 70 as

limk→0

u ∼ krk

+δ

k= r − a. (73)

Thus, the scattering length a is the intercept of u on the r-axis (see Fig. 2.



0.6.3 Effective range theory

The energy dependence of σsc can be expressed in terms of the scattering length and anotherparameter which is called effective range. We will denote it by r0 .

For E = ~2k2/2µ and E = 0 ,

u′′

+

(k2 − 2µV

~2

)u = 0 , and u

′′

0 −2µV

~2u0 = 0. (74)

Multiplying by u0 and u , respectively, and subtracting, we obtain

d

dr(uu′0 − u0 u′) = k2uu0. (75)

The solutions for V = 0 , are denoted by v and v0 , obeying the equations

v′′

+ k2v = 0 , and v′′

0 = 0. (76)

As in 75,d

dr(vv′0 − v0v′) = k2v0v (77)

Subtracting 75 from 76 and integrating from r = 0 to ∞ we get

(uu′0 − u0u′ − v v′0 + v0v′)|r=∞r=0 = k

2

∫ ∞0

(uu0 − v v0)dr. (78)

Since (u, u0)→ (v, v0) at r →∞ ,

(v v′0 − v0v′)|r=0 = k2

∫ ∞0

(uu0 − v v0)dr. (79)

We can choose the asymptotic form of v as

v = C sin(kr + δ). (80)

We also choose the normalization so that v = 1 at r = 0 Thus C = 1/ sin δ and

u(r) −→r→∞

v(r) ≡ sin(kr + δ)sin δ

. (81)

The solution for v0 is a straight-line and

u0(r) −→r→∞

v0(r) ≡ D(r − a).

With v0(0) = 1,

v0(r) ≡ 1−r

a. (82)


0.7. SHELL MODEL + SINGLE-PARTICLE CORRELATIONS

Now, Eq. 79 becomes

−1a− k cot δ = k2

∫ ∞0

(uu0 − vv0)dr (83)

For r > a, u → v and v → v0 slowly. For r < a u ≈ u0, v ≈ v0 if E is small. Then wedefine the effective range r0 that

r0 = 2

∫ ∞0

(v20 − u20

)dr. (84)

In terms of the effective range, Eq. 83 becomes approximately,

k cot δ = −1a

+ r0k2

2. (85)

Using this in 69, we find

σ =4π

k2 + (1/a− r0k2/2)2. (86)

In a general scattering experiment of neutrons on protons, or vice-versa, the neutron beamhas its spin randomly oriented and the n-p system is in a triplet (spin parallel) and singlet (spinantiparallel) states in proportion to the corresponding weight factors for these states, which are,respectively, 3/4 and 1/4. The total scattering cross section consists of two parts

σsc =3

4σt +

1

4σs. (87)

For σt and σs we can write solutions of the form 86. We thus have to consider four parameters,at, σ0t, as and σ0s to describe n-p scattering at low energies.

0.7 Shell model + single-particle correlations

Three-body calculations become increasingly more complicated as the details of the inter-action are considered. It is thus useful to study a simpler problem based on single particlemodel with correlations. This has been done by Bertsch and Esbensen [6]. In a single-particle approach including correlations one solves the same Hamiltonian as in Eq. 40, butthe basis of the calculations are single-particle wavefunctions for the nucleons, obtained bysolving the single-particle Hamiltonian (the recoil of the core is neglected here)[

− ~2

2m∇2 + V (r)

]ψi(r ) = �i ψi(r ). (88)



The single-particle potential, including spin-orbit, is parametrized in the usual way [7]

V (r) = −V0(

1− 0.44 r20 (l.s)1

r

d

dr

) [1 + exp

(r −Ra

)]−1(89)

with a = 0.67 fm, R = r0(A− 2)1/3 and r0 = 1.27 fm, appropriate for 11Li. Note that theradius R is calculated as a function of the core mass, A− 2 . For the protons this potentialis supplemented with the Coulomb interaction with the other (Z − 1) protons, representedby a uniformly charged sphere.

The strength of the potential for core nucleons is adjusted to give a reasonable estimateof the binding of the 1p3/2 proton and neutron states in p-shell nuclei which contain N = 8neutrons. This leads to

V0 =

(52− 33 N − Z

Aτ3

)MeV (90)

where τ3 = 1/2 (−1/2) for neutrons (protons). This is very close to the value commonlyused [7].

For the valence neutrons the strength of the single-particle potential is adjusted so thatthe resonant p1/2 state appears at the energy Er = 800 keV, which is achieved for V0 =30.2 MeV.

The effective interaction (medium modified) between the valence neutrons is taken as adensity dependent delta interaction.

veff (r1, r2) = δ (r1, r2)

{v0 + vρ

[ρc

(r1 + r2

2

)/ρ0

]P}. (91)

The delta interaction form simplifies the calculation greatly. v0 is chosen to simulate theinteraction between free nucleons. The density dependent term contains the ratio of the coredensity, ρc(r) , and the nuclear matter density, ρ0 = 0.16 fm

−3 . vρ and P are adjusted sothat the pairing energies in p-shell nuclei are reasonably well reproduced.

A zero-range interaction lead to divergencies. I.e., an infinite number of momenta arepresent in the Fourier transform of δ(r ) . A zero-range interaction is only meaningful aseffective interaction within a truncated space of states.

To understand how to handle with a truncated space let us neglect the density dependentterm in 91 for the moment. The Schrödinger equation for relative motion of the valenceneutrons, with reduced mass m/2 , is(

−~2

m∇+ v0 δ(r )

)ψ(r ) = Eψ(r ). (92)

The formal solution is

ψ(r ) = ψ(0)(r )−∫

d3r′G0 (r, r′, E) v0δ(r

′)ψ(r ′) (93)



Figure 3: Radial wavefunction for the relative motion of two neutrons. The fully drawncurve is the sine-integral function Si(k, r), obtained from the local pairing interaction eq.97, which reproduces a bound state at zero energy. The dashed curve is the result obtainedfrom the Reid soft-core potential.

G0 is the Green’s function in the truncated space, i.e., �k ≤ Ec or k ≤ kc ,

G0 (r, r′, E) =

1

(2π)3

∫�k≤Ec

d3keik·(r−r

′)

�k − E. (94)

To simulate the fact that the scattering length for free neutrons is very large we requirethat 93 has a bound state solution at zero energy. This implies that ψ(0) = 0 (ψ(0) = thesolution of the equation without the interaction). At r = 0

1 = − v0(2π)3

∫�k≤Ec

d3k1

�k(95)

which implies that

v0 = −2π2(

~2

m

)3/2E−1/2c . (96)

Choosing Ec = 40 MeV (∼ Fermi energy of a usual nucleus) one gets v0 = −831 MeV fm3 .



Using 93, 95 and 96 the bound state wavefunction at zero energy is

ψE=0(r ) = ψ(0)Si(kcr)

kcr(97)

where Si(x) =

∫ x0

dt sin(t)/t . This wave function, Si(kcr) , (solid line) is shown in Fig. 3,

together with the radial wavefunction obtained from a commonly used finite range inter-action, namely the Reid soft core potential [8] (dashed curve). The agreement is quitereasonable.

The wavefunctions of the core nucleons are needed only to construct the core-density, tobe used is 93. They are constructed by using the single-particle wavefunction obtained from88 and 89,

ψn`jm(r ) = φn`j(r)∑m`,ms

(`m`1/2ms|jm)Y`m(r̂)χms. (98)

The neutron-neutron wave function is obtained by coupling these solutions to the totalangular momentum J . What is needed are the two-particle wavefunction at r1 = r2 ≡ r tocompute the effect of the zero-range residual interaction, Eq. 91.

For example, coupling to J = 0 yields the two-particle wavefunction

ψ(2)nn′`j(r ) =

φn`j(r)φn′`j(r)√4π

(−1)`√

2`+ 1

8π

× 1√2

(χ

(1)1/2χ

(2)−1/2 − χ

(1)−1/2χ

(2)1/2

). (99)

The spin part is seen to be an S = 0 state, which is typical when r1 = r2 ≡ r , since thespatial part is automatically symmetric and the whole wavefunction has to be antisymmetric.The wavefunction for J 6= 0 is a bit more complicated [6].

The single-particle wavefunctions φn`j(r) are bound states as well as continuum states,but bound states occupied by the core nucleons are explicitly excluded from the space. Thisprocedure is equivalent to solve the Bethe-Goldstone equation [7] for the two valence neutronsin the presence of the core.

One can find the eigenfunctions and energies of the two-particle Hamiltonian, Eq. 20,by expanding in terms of the basis ψ

(2)J followed by diagonalization. However, a more

numerically interesting approach reported in Ref. [6].The two-particle Green’s function is defined as

G(0)2 (E) =

1

H(0)2 − E − iη

=∑2p

|2p〉〈2p|�

(0)2p − E − iη

(100)

in the space of two-particle states |2p〉 with energies �(0)2p .



Figure 4: Contour plots of the two-particle density of the valence neutrons in 11Li. Thedensity has been weighted by 4πr214πr

22, and it is illustrated (in units of fm

−2) for threedifferent positions r1 of one the neutrons. The contours are shown as functions of thecoordinates r2 cos (θ12), r2 sin (θ12) of the second neutron.

The correlated two-particle Green’s function is (with the residual interaction)

G2(E) = (H2 − E − iη)−1 =∑

f2p|2̃p〉〈2̃p|

�2p − E − iη(101)

where | 2̃p〉 are the correlated two-particle states, with eigenvalues �2p . H2 is Eq. 20.We note that

1 +G(0)2 (E)veff = G

(0)2

(H

(0)2 + veff − E − iη

)= G

(0)2 (E)G

−12 (E) (102)

or

G2(E) =(

1 +G(0)2 (E)veff

)−1G

(0)2 (E). (103)



Thus

〈2p0|δ (r1 − r2)G2(E)|2p〉 =∑

f2p〈2p0|δ (r− r2) |2̃p〉〈2̃p|2p〉

�2p − E − iη(104)

=〈2p0|δ (r1 − r2) (1 +G(0)2 (E)veff)−1|2p〉

�(0)2p − E − iη

(105)

where we used Eqs. 100, 101, and 103.Eqs. 104 and 105 solve the problem. 104 diverges when E is identical to one of the

correlated energy eigenvalues. Keeping a finite (but small) η , the imaginary part of thisexpression will have a local maximum when E coincides with an eigenvalue. A simplenumerical search can be made to locate the ground state energy of the pair. The amplitudesof the ground-state wavefunction on two-particle shell model states can also be extracted.Eqs. 104 and 105 show that

〈2p̃|2p〉 = N Im〈2p0|δ (r1 − r2)

[1 +G

(0)2 (E)veff

]−1|2p〉

�(0)2p − E − iη

(106)

when E is set equal to the ground state energy of the pair. N is a constant that can beeliminated by the normalization of the ground state wavefunction. The state |2p0〉 is fixedin this process but it can otherwise be any of the two-particle states.

The local independent two-particle Green’s function can be constructed from the two-particle states, e.g., Eq. 99. For a pair coupled to J = 0 one gets [6]

G(0)2 (r, r

′E) =∑nn′`j

φn`j(r)φn′`j(r)φn`j(r′)φn′`j(r

′)

�n`j + �n′`j − E − iη

× 2j + 18π

Y00(r̂)Y00(r̂′). (107)

The main complication to calculate 107 is the inversion of the operator 1 +G(0)2 (E)veff .

This inversion is similar to the one used in ordinary RPA theory [9], and it can be performedon a finite radial grid.

The numerical results depend on the choice of the parameters vρ and P in Eq. 91. Theseparameters were chosen to fit the two-neutron binding energies of 11C and 12Be, which like11Li have eight valence neutrons. The best fit [6] is obtained for P = 1.2 and vp = 930 MeVfm3.



Figure 5: Single particle density of the valence neutrons in 11Li, weighted by 4πr2 . Theradial cutoff is 20 and 30 fm for the two dashed curves, and it is 40 fm for the fully drawncurve. The dots represent the best p3/2 shell model fit.

The two-particle wavefunctions |2̃p〉 ≡ ψ(2) (r1, r2) are obtained by solving Eqs. 104 and105 for the eigenvalues and correlated amplitudes, respectively. That is, using Eq. 99 thenon-correlated two particle wavefunctions |2p〉 ≡ ψ(2)nn′j(r, r ) are constructed (see AppendixA of [6]). Then the correlated two-particle wavefunction

ψg.s (r1, r2) =∑nn′`j

αnn′`j Ψ(2)nn′j (r1, r2) (108)

is constructed. The αnn′`j are then found by the Green’s function method, as describedabove.

In Figure 4 we see the two-particle density ρ(2) (r1, r2, θ21) obtained from the correlatedtwo-particle wavefunction. r1 and r2 are the radial coordinates of the neutrons with respectto the core and θ21 is the angle between them. The density has been weighted by 4πr

214πr

22 .

When one of the particles is required to be close to the core (upper part of Fig. 4) thereare two peaks in ρ(2) , one is close to the selected particle and one is at θ21 = 90

o . The laterpeak is associated with the S = 1 component of the two-particle wavefunction whereas thepeak near the particle is dominated by the S = 0 component. When one of the particles isfar away from the core (lower part of the Figure) the peak at θ21 = 90

o almost disappears.


0.8. SCATTERING STATES AND RESONANCES

The single particle density is obtained from ρ(2) by integration over one of the variables,e.g., r2 . In figure 5 the single particle density ρ

(1)(r) weighted by 4πr2 is plotted againstr . The long dashed, short-dashed, and the solid curve are obtained by increasing the radialcutoff of the numerical integration for 20, 30 and 40 fm, respectively. The dots are obtainedfrom the density of a p1/2 bound state in a Woods-Saxon potential of form 89, with r0 =1.52 fm and V0 = 26.53 MeV. This yields a single-particle energy of −0.296 MeV, not farfrom the two-particle binding energy.

One can compare the results obtained via the Green’s function method [6] with thevariational three-body calculations presented in the last Section [5]. The variables ρ and λ

are here equivalent to r1 − r2 and1

2(r1 + r2) (see Fig. 6(b)). One finds

〈r2〉 = 39 fm2, 〈ρ2〉 = 43− 54 fm2 (109)

1

4〈|r1 + r2|2〉 = 24.4 fm2, 〈ψ2〉 = 21− 28 fm2 (110)

where the range of 〈ρ2〉 and λ2 from Ref. [5] depend on their choices for the potential param-eters. The agreement is quite reasonable. Based on the results of Bertsch and Esbensen [6]one concludes that the di-neutron model for 11Li has at least qualitative validity.

The Green’s function method [6] is a very handy, technically simple, method to quicklyassess the characteristics of a three-body system. However, the model ignores finite-range(and hence p-wave) effects in pairing interactions. The simplicity of the model is lost if adelta interaction between the valence neutrons is not used. The recoil of the core is also nottreated properly. Thus, while reasonable for 9Li+n+n (11Li) it is a bad approximation for4He+n+ n (6He) .

Supplement D

0.8 Scattering states and resonances

The total cross section for a scattering process is given by

σt = σsc + σr (111)

where σsc is the elastic scattering cross section treated in the Supplement C. To derive an expression



for the reaction cross section we use

eikz =∞∑`=0

i`(2`+ 1)j`(kr)P`(cos θ)

∼∑`

i`+1

2kr(2`+ 1)

{exp

[−i(kr − `π

2

)]− exp

[i

(kr − `π

2

)]}P`(cos θ) (112)

for r →∞ In a reaction the scattering amplitude of the outgoing spherical wave part of the planewave is modified. The wavefunction ψ(r) describing the outgoing wave after interaction is givenby

ψ(r) ∼∑`

i`+1

2kr(2`+ 1)

{exp

[−i(kr − `π

2

)]−η` exp

[i

(kr − `π

2

)]}P`(cos θ) (113)

for r →∞.We can write η` = |η`|e2iδ` . If |η`| = 1 the scattering is elastic. If |η`| < 1 , then both elastic

scattering and reaction will occur. The elastic scattered wavefunction is the difference between112 and 113

ψsc =∑`

i`+1

2kr(2`+ 1)(1− η`) exp

[i

(kr − `π

2

)]P`(cos θ). (114)

Using the same steps as in 59 – 61 we find

dσsc(θ)

dΩ=

π

k2

∣∣∣∣∣∞∑`=0

√2`+ 1(1− η`)Y`0(θ)

∣∣∣∣∣2

and

σsc =

∫dσscdΩ

dΩ =∞∑`=0

σsc,` (115)

whereσsc,` =

π

k2(2`+ 1) |1− η`|2 . (116)

To calculate the total reactions cross section we use Eq. 58 to obtain the flux of particles towardsa sphere of radius r . The next flux will be the number of particles removed from the beam. Thuswe have to use ψ = eikz + ψsc in 60. It is left to the reader to show that

Sr =~πµk

∑`

(2`+ 1)(1− |η`|2

).



Therefore, the total reaction cross section is

σr =Srv

=∞∑`=0

σr,` (117)

whereσr,` =

π

k2(2`+ 1)

{1− |η`|2

}. (118)

The total cross section is

σ = σsc + σr =π

k2(2`+ 1)

{|1− η`|2 + 1− |η`|2

}=

2π

k2(2`+ 1) {1− Re η`} . (119)

When there is no reaction, |η`|2 = 1 . Since η` = |η`|e2iδ` , this means purely real δ` . In thecase of any reaction, at least one of the δ` is complex, so |η`| < 1 . This is because the outgoingparticles cannot exceed the number of incident particles.

Therefore, the maximum value of σsc,` is obtained for

η` = −1 : σsc,`,max =4π

k2(2`+ 1) (120)

and σr,` = 0 .The maximum value of σr,` is for η` = 0 , in which case σsc,` 6= 0 . This means that there is

no reaction without scattering.For scattering on an object with a sharp edge with radius R we can say that all particles with

` <pR

~= kR (121)

will react with the target. Thus, η` ∼= 0 for ` < kR and η` = 1 for ` > kR . We obtain

σr ∼=π

k2

kR∑`=0

(2`+ 1) = π(R + 1/k)2 ∼= πR2

for kR� 1 (wavelength much smaller than the radius). Similarly, for the scattering cross section

σsc ∼= πR2. (122)

The total cross section isσT ∼= 2πR2. (123)

And is twice the classical cross section, πR2 . This is due to the extra-piece arising from thescattering cross section 122. The wave is diffracted around the target and this takes out flux fromthe incident wave, resulting in 122. This extreme limit, valid for kR� 1 (high energies) is calledby shadow scattering.



0.8.1 Breit-Wigner formula for ` = 0

In Supplement C we found that

limk→0

δ

k= −ak (124)

k cot δ = − 1ak

+r0k

2

2. (125)

We now define

k cot δ = − 1a(k)

(126)

for all values of k . If there is no reaction, a(k) is real. For σr 6= 0 a(k) is in general complex.Denoting now a(k) = a ,

η = ei2δ = (cos δ + i sin δ)2

=

(1− ika√1 + k2a2

)=

1− ika1 + ika

. (127)

Thus, for ` = 0 ,

σsc =π

k2|1− η| = 4π

|1/a+ ik|2(128)

σr =π

k2(1− |η|2) = Im(1/a)

|1/a+ ik|2. (129)

For real a , σr is zero. In this situation, and resonance energy E0 , the phase shift δ ∼ π/2 ,so that

1

a(E0)= 0. (130)

Expanding 1/a(E) about E0 , we get

1

a(E)= 0 + (E − E0)

[d

dE

(1

a

)]E0

+ · · · (131)

Defining, [d

dE

(1

a

)]E0

=2k

Γs(132)

we find

σsc =4π

|(E − E0)22k/Γs + ik|2=

(π/k2)Γ2s|(E − E0) + iΓs/2|2

=(π/k2)Γ2s

(E − E0)2 + Γ2s/4. (133)



This is known as a Breit-Wigner formula.To derive an expression for the reaction cross section we keep in mind that a(E) is a complex

function and assuming that 1/a(E′0) = 0 for E

′0 = E0 − iΓR/2, and defining

Re

[d

dE

(1

a

)]E

′0

=2k

ΓS; Im

[d

dE

(1

a

)]E

′0

= kα

we find that

1

a(E)=

(E − E0 +

iΓR2

)(2k

ΓS+ ikα

)

=2k

ΓS(E − ER) + ik

[ΓRΓS

+ α(E − E0)]

(134)

where

ER = E0 +αΓRΓS

4. (135)

The scattering cross section is now

σsc =4π∣∣ 1

a+ ik

∣∣2 = (π/k2)Γ2S(E − ER)2 + [ΓS + ΓR + αΓS(E − E0)]2 /4 . (136)For a sharp resonance α is small, and 136 becomes

σsc =(π/k2)ΓRΓS

(E − ER)2 + (ΓS + ΓR)2 /4. (137)

Using 134 in 129, the reaction cross section is

σr =(π/k2) [ΓRΓS + αΓ

2S(E − E0)]

(E − ER)2 + [ΓS + ΓR + αΓS(E − E0)]2 /4. (138)

Again neglecting α(E − E0) we obtain

σr =(π/k2)ΓRΓS

(E − ER)2 + (ΓS + ΓR)2 /4. (139)

To get an insight of these formulas, let us assume a scattering by a potential such that V (r) = 0for r > R . The radial wavefunction can be written as

rψ(r) = u(r) = uout(r) + uin(r) (140)

whereuout(r) = u

(+) exp(ikr) , and uin(r) = u(−) exp(−ikr). (141)



For r ≥ R we know that

ψ(r ≥ R) =√π

kr

∑`

i`+1√

2`+ 1

{exp

[−i(kr − `π

2

)]

− η` exp[i

(kr − `π

2

)]}Y`0(θ). (142)

The radial wavefunction for ` = 0 is (r ≥ R)

(rψ)`=0 = u(r) =

√π

ki {exp(−ikr)− η exp(ikr)} . (143)

Comparing 140 with 143, we find

u(+) = −√π

kiη , and u(−) =

√π

ki. (144)

Now

η =−u(+)(∞)u(−)(∞)

=−uout(R)e−ikR

uin(R)eikR≡ η(R)e−2ikR (145)

where

η(R) = −uout(R)uin(R)

. (146)

From 127, η = (1− ika) / (1 + ika). If we now define a(R) so that

η(R) =1− ika(R)1 + ika(R)

(147)

we get

a(R) =1

ik

1− η(R)1 + η(R)

. (148)

We note that [ru′(r)

u(r)

]r=R

= R−ik e−ikR − ikη eikR

e−ikR − η eikR= − R

a(R). (149)

As in 129 we get, using 145 and 148,

σr =π

k2(1− |η|2) = 4π

k

Im(

1a(R)

)∣∣∣ 1a(R) + ik∣∣∣2 . (150)

Thus, the reaction cross section is a function of a(R) which depends on the wavefunction atr = R through Eq. 149. Well below the resonance and far from the zero of 1/a(R) , we findσr ∼ 1/k ∼ 1/v . This is the well known “1/v law”.



In terms of a(R) the scattering cross section is

σsc =π

k2|1− η|2 = 4π

∣∣∣∣1− η2ik∣∣∣∣2

= 4π

∣∣∣∣∣e2ikR − 12ik + 11a(R)

+ ik

∣∣∣∣∣2

= 4π

∣∣∣∣∣eikR sin kRk + 11a(R)

+ ik

∣∣∣∣∣2

. (151)

The first term is called by potential scattering amplitude whereas the second term is the res-onance scattering amplitude. We recall that in terms of a, rather than a(R) , the scatteringamplitude is given by Eq. 128.

We can again define behavior of 151 near a resonance by expanding a(R) as in 131. Since[d

dE

(1

a

)]E0

=2k

ΓS> 0 (152)

since ΓS > 0 , we have that, below the resonance 1/a < 0 . For low energies

eikr sin kR

k∼= R > 0.

Therefore, for low-energy scattering, we have destructive interference with the potential ampli-tude below the resonance and constructive interference above.

Away from the resonance, a(R) → 0 and the second term in Eq. 151 ' 0 , leaving only thefirst term to contribute to the scattering cross section. In this case (a(R) ∼ 0) , Eq. 149 givesu(r = R) ∼= 0 , the boundary condition for an almost impenetrable sphere. Thus for low energies,kR� 1 , Eq. 151 gives σsc ∼= (4π/k2) sin2 kR ∼= 4πR2 , the potential scattering for a rigid sphere.

Near a resonance a(R)→∞ , which means that u′(R) = 0 . This indicates that the amplitudeof the incident wave is a maximum at the surface, and the continuity conditions require the samefor the internal wave function.

0.8.2 Breit Wigner formula for all `

If V (r) ∼= 0 for r > R we define

f` = R

(1

u`

du`dr

)r=R

(153)



where u` satisfies the equation

d2u`dr2

+

(k2 − `(`+ 1)

r2

)u`(r) = 0. (154)

The solution can be written as

u`(r) = a+u(+)` (r) + a−u

(−)` (r) (155)

with

u(±)` (r) = ± ikr [j`(kr)± in`(kr)]

= ikr

{h

(1)` (kr)

−h(2)` (kr)

∼ exp[±i(kr − `π

2

)]for r →∞. (156)

[For ` = 0, u0(r) = a+eikr + a−e−ikr ].Comparing 113 with 155, using 156, we find

η` = −a+a−. (157)

Now,

f` = Ra+u

′(+)` (R) + a−u

′(−)` (R)

a+u(+)` (R) + a−u

(−)` (R)

. (158)

Using the definition[R

1

u(±)`

d u(±)`

dr

]r=R

= S` ± iK` ; [S0 = 0 , K0 = kr]

and 157 we get

η` =f` − S` + iK`f` − S` − iK`

=

(1 +

2iK`f` − S` − iK`

)e2iξ` (159)

whereu

(−)` (R)

u(+)` (R)

= e2iξ` ; [ξ0 = −kR]. (160)

Using 116 and 119 we find

σsc =π

k2

∑`

|A`(pot) + A`(res)|2 (161)



where

A`(pot) = e−2iξ` − 1 (162)

A`(res) =−2iK`

f` − S` − iK`(163)

Also,

σr =π

k2

∑`

(2`+ 1) · 2K` (f` − f ∗` )[12

(f` + f ∗` )− S`]2

+[

12

(f` − f ∗` )−K`]2 . (164)

For an impenetrable sphere, u`(R) = 0 . Then from Eq. 158, f` → ∞ and thereforeA`(res)→ 0. In this case σr → 0 .

To be more explicit let us assume that V (r) = −V0 for r < R , then

uin(r) = Ain sin

[Pr − `π

2+ α(E)

], r < R (165)

uout(r) = Aout sin

[kr − `π

2+ β(E)

], r > R (166)

where α and β are complex functions of the energy, and

P =

√2m(E + V0)

~, k =

√2mE

~. (167)

From 158 and 165 we find

f` = PR cot

[PR− `π

2+ α(E)

]

= kR cot

[kR− `π

2+ β(E)

]. (168)

From

uin(R) = uout(R) ,AinAout

=sin [kR− `π/2 + β(E)]sin [PR− `π/2 + α(E)]

=kR√

f 2` + k2R2

√f 2` + P

2R2

PR(169)

Treating f` as real, f` & PR → Ain/Aout ∼ k/P � 1 , if E � V0 . But for f` =0 , Ain/Aout = 1 , and the particle penetrates the region of r < R with the maximum amount.Thus f` = 0 corresponds to a resonance.



We define a resonance energy formally as in

Re[f`(E

`r)]

= 0. (170)

Thus

f`(E) = i Im f`(E`r) + (E − E`r)

(∂f`∂E

)E=E`r

+ · · · (171)

and Eq. 162 yields

A`(res) =iΓα

E − E`res + iΓ/2(172)

where

Γα =−2K`

(∂f`/∂E)E=E`r; Γr =

2Imf`(E`r)

(∂f`/∂E)E=E`r(173)

Γ = Γα + Γr (174)

(175)

E`res = E`r +

S`(∂f`/∂E)E+E`r

(176)

Γ is the total width and Γr is the reaction width. E`res is the actual resonance energy whichdiffers from Er by a small amount S`/(∂f`/∂E)E=E`r (equal to zero for ` = 0 ).

The expression for σ(`)sc and σ(`)r are now given by

σ(0)sc =π

k2

∣∣∣∣[e−2ikR − 1]+ iΓαE − E0res + iΓ/2∣∣∣∣2 for ` = 0 (177)

σ(`)sc =π

k2(2`+ 1)

Γ2α(E − E`res)2 + (Γ/2)

2 for ` =≥ 1 (178)

σ(`)r =π

k2(2`+ 1)

ΓαΓr

(E − E`res)2 + (Γ/2)2 . (179)

where in σ(`)sc the potential scattering term is dropped and in σ(0)sc we used ξ0 = −kR .

Γr maybe interpreted as the sum of all widths except the incident channel width Γα

Γr =∑β 6=α

Γβ , Γ =∑

Γβ = Γα + Γr (180)

The scattering cross section σ(`)β←α from channel α into channel β is given by 173

σ(`)β←α =

π

k2(2`+ 1)

Γα Γβ

(E − E`res)2 + (Γ/2)2 . (181)


0.9. THE EFIMOV STATES

Figure 6: (a) Behavior of states of a three-body system as a function of the strength of theinteraction. (b) Jacobi coordinates.

0.9 The Efimov states

11Li is an example of a general class of three-body systems with a common characteristic:the three subsystem (i.e., 1 + 2, 1 + 3 and 2 + 3) are “resonant”. By resonant we meanthat they have resonant, almost bound states. As we have seen in Supplement A the n-nsystem has a resonant state (the singlet state of the deuteron) close to threshold.2 The 10Li(9Li− n) system has also a resonant state at about 500 keV.

Efimov [11, 12] has shown that resonant two-body forces give rise to a series of levelsin three-particle systems. For resonant systems, the range of the force r0 is considerablysmaller than the scattering length, a (see Supplement B). A barely bound (unbound) statehas a very large scattering length.

For simplicity consider three spinless neutral particles of equal mass, interacting througha potential gV (r) , with V (r) > 0. At g = g0 two particles get bound in their first s-state.For values of g close to g0, the two-particle scattering length a is large, and this is the regionof interest. In the Figure 6 the series of levels for the three-body particle is shown. Thehatched boundary is the continuum boundary .

As g grows, approaching g0 , three-particle bound states emerge one after the other. Atg = g0 (infinite scattering length) their number is infinite. As g grows beyond g0 , levelsleave into the continuum one after the other.

2Due to the charge-independence of the nuclear interaction, the n-n system and the n-p system areconsidered to be identical.


0.10. THE FADEEV EQUATIONS

The number of levels is given by the equation [11]

N ∼=1

πln(|a|/r0). (182)

The physical cause for this effect is in the emergence of effective attractive long-rangeforces of radius a in three-body systems. One can show that they are of the 1/R2 kind;R2 = r212 + r

223 + r

231 . With a → ∞ the number of levels becomes infinite as in the case of

two particles interacting with attractive 1/r2 potential.These are the basic results obtained by Efimov. They are of general validity and 11Li is

a good example. If the di-neutron model is considered, rnn � rn−9Li and the 1/R2 attractionterm deduced by Efimov means that the 9Li + 2n-system behaves as an atom, possessing“Rydberg” states, with a di-neutron-core potential of the form 1/R22n− 9Li . As far as weknow, only one of these states is known: the ground state. But other states are not ruledout. In fact, it has been shown [13] that depending on the pairing energy of the two valenceneutrons, several Rydberg states may exist in 11Li.

Supplement E

0.10 The Fadeev equations

The He atom, the 3H and 3He nuclei are examples of systems for which a three-body solution ofthe Schrödinger equation is necessary. Numerical solution is now done in a very precise manner[28]. Not only the bound states, but also scattering observable and responses to external probesare now being investigated.

There are many techniques to solve the three-body problem variational calculations [4, 30], theuse of hyperspherical harmonics [29] and Green’s function Monte Carlo [33]. The most commonlyused method is solving the Fadeev equations [31, 32].

Let us consider three particles interacting via two-body interactions V (rij) . The Hamiltonianis

H =3∑i=1

ti +∑i

0.10. THE FADEEV EQUATIONS

It is convenient to employ the Jacobi coordinates

ρi = rj − rk , and λi =1

2(rj + rk)− ri (185)

where ri are the position of the three nucleons, and i, j, k = 1, 2, 3 are taken cyclically.It is also convenient to use the notation

Vi ≡ V (ρi). (186)

The Fadeev equations are obtained by writing the Schrödinger wave function ψ as

ψ = ψ1 (ρ1, λ1) + ψ2 (ρ2, λ2) + ψ3 (ρ3, λ3)

= ψ1 + ψ2 + ψ3. (187)

Substitution in 184 yields (Fadeev equations)

(T + Vi − E)ψi = −Vi (ψj + ψk) ; i, j, k = 1, 2, 3 (188)

with i, j, k permuted cyclically and T =∑

i ti .For identical particles, like the 3N -system (i.e., 3H, 3He, etc), the ψi(i = 1, 2, 3) amplitudes

all have the same functional form so that once ψ1 is found, the other two can be constructed bya rotational transformation. I.e., counterclockwise rotations of the particle coordinates of 185 (seealso Fig. 6(b)) generate the sets

ρ3 = −1

2ρ1 − λ1 , and λ3 =

3

4ρ1 −

1

2λ1 (189)

and

ρ2 = −1

2ρ1 − λ1 , and λ2 = −

3

4ρ1 −

1

2λ1. (190)

Hence, for these systems one may focus on the solution of one of the Fadeev amplitudes, sayψ (ρ1, λ1) = ψ (ρ, λ ).

For a asymmetrical three-body system, the equivalent to the Jacobi coordinates can be labelledas x and y, so that

xi =

√1

m

mjmk(mj +mk)

(rj − rk) , yi =

√1

m

(mj +mk)mi(mj +mk) +mi

(mjrj +mkrkmj +mk

− ri), (191)

where i, j, k is a cyclic permutation of 1,2,3; mi and ri are the masses and coordinates of the threeparticles.

Summing up the three Fadeev equations gives the Schrödinger equation. The solutions of theFadeev equations satisfy therefore also the Schrödinger equation. The opposite statement does nothold, however, since Ψ can vanish identically even when the three components Ψi individually arenon-zero functions. Concerning the total function, Ψ, the Schrödinger and the Fadeev equationsprovide identical solutions.


0.11. STRICT THREE-BODY CALCULATIONS

0.11 Strict three-body calculations

In the last two Sections we have presented the simplest and most transparent calculations fora three-body system, e.g., 11Li or 6He. However, as we have seen, the models rely on strongassumptions. In the first the spin variables of the neutrons were neglected. In the second adelta interaction was used. These assumptions are needed for the sake of simplicity. Pauliblocking was also not included properly in the variational calculations discussed previously.

The Efimov effect was discovered theoretically as a peculiar behavior of a three-bodysystem with short-range two-body potentials [11]. The effect appears when at least two ofthe binary subsystems have extremely large scattering lengths or bound states at nearly zeroenergy. Then a number of three-body bound states arises with enormous spatial extensionand exceedingly small binding energies. Even when none of the binary subsystems is bound,the three-body system may have a large number of bound states. One surprising consequenceis that more three-body bound states can appear by weakening the potentials such that thescattering lengths increase.

These states have not been seen experimentally yet, but their possible existence have beeninvestigated theoretically in a few systems like the three 4He-atoms (4He-trimer) [14]. Suchinvestigations have been rather demanding in terms of computer capacity as the extremelysmall binding energies and the large spatial extensions of the systems put strict requirementson the numerical accuracies. However, an efficient method was suggested for solving theFadeev equations in coordinate space [15]. The method is particularly suitable for studyingthe Efimov effect.

The characteristic properties of halo nuclei off hand match the description of Efimovstates rather well. The fact that they often are describable as highly clusterized two- orthree-body systems furthermore improves the resemblance.

In this Section we first describe the hyperangular function method for the Fadeev equa-tions in some detail and discuss the asymptotic properties of the equations. Using thismethod we then investigate the conditions for possible existence of the Efimov effect in thetwo-neutron halo nuclei. We summarize the results obtained by M. Zhukov, D. Fedorov, J.Vaagen, and collaborators [1].

0.11.1 Hyperangular functions

The ordinary set of Jacobi coordinates {x,y} does not facilitate the analysis of the asymp-totic behavior of a three-body system since the boundary conditions in this set may changesignificantly within the two-dimensional {x, y} space [17]. It is more convenient to use a set ofcoordinates with one (hyper-) radial variable and five dimensionless angles. The asymptoticplay then runs on a graspable one-dimensional stage yet allowing mathematically correctthree-body boundary conditions [18]. This set of so called hyperangular coordinates [19] isgiven as {ρ, αi, xixi ,

yiyi} ≡ {ρ,Ωi}, where the hyperradius ρ and the hyperangles αi are defined



as

ρ =√x2i + y

2i , αi = arctan(

xiyi

), i = 1, 2, 3. (192)

We shall from now on omit the index when any one of the three sets of coordinates is implied.The kinetic energy operator in these variables is

T =~2

2m(−ρ5/2 d

2

dρ2ρ5/2 +

15/4

ρ2) +

~2

2mρ2Λ̂2 , (193)

Λ̂2 = − 1sin(2α)

d2

dα2sin(2α)− 4 + l̂

2x

sin2 α+

l̂2ycos2 α

. (194)

The components of the total wavefunctions are now expanded in terms of hyperangularfunctions Φn(ρ,Ω) =

∑3i=1 Φ

(i)n (ρ,Ωi)

Ψ(ρ,Ω) = ρ−5/2∑n

fn(ρ)Φn(ρ,Ω). (195)

The hyperangular functions for each ρ are chosen to be the eigenfunctions of the angularpart of the Fadeev equations

Λ̂2Φ(i)n +2mρ2

~2ViΦn = λn(ρ)Φ

(i)n . (196)

The radial functions fn(ρ) satisfy the coupled set of radial equations(− d

2

dρ2+λn(ρ)

ρ2+

15/4

ρ2− 2mE

~2

)fn =

∑n′

(2Pnn′

d

dρ+Qnn′

)fn′ ,

Pnn′(ρ) =

∫dΩ Φn(ρ,Ω)

d

dρΦn′(ρ,Ω), Qnn′ =

∫dΩ Φn(ρ,Ω)

d2

dρ2Φn′(ρ,Ω).

(197)

For illustration we consider the case of three identical spin zero particles and s-waves only.The extension to higher orbital momenta, non-zero spins and different masses is straightfor-ward but somewhat tedious.

We then write the hyperangular function as

Φn(ρ,Ω) =3∑i=1

φn(ρ, αi)

sin(2αi). (198)

The three Fadeev equations are identical in this case. To write the equation explicitly inone of the Jacobi sets we have to “rotate” two other components into the chosen set. This



rotation, which expresses one set of coordinates by another and subsequently integrates outthe four angular variables of x and y is performed by an operator R̂ given for s-waves andidentical particles can be easily found to be

R̂φ =2√3

∫ π/2−|π/6−α||π/3−α|

dα′φ(ρ, α′). (199)

The angular part of the Fadeev equation is then(− d

2

dα2− λ̃)φ(ρ, α) + ρ2v(ρ sinα)

(φ(ρ, α) +

4√3

∫ π/2−|π/6−α||π/3−α|

dα′φ(ρ, α′)

)= 0 (200)

where v(z) ≡ 2mV (√

2z)/~2 and λ̃ ≡ λ+ 4.

0.11.2 Asymptotic behavior of the angular eigenvalues

Let us now consider the asymptotic behavior of λ̃ for small and large ρ. For ρ = 0 thesolution of Equation 200, which obeys the boundary conditions of being zero for both α = 0and π/2, is sin(2nα) with an eigenvalue λ̃n = 4n

2. This is in fact the hyperharmonicalspectrum for the s-waves [19]. The eigenvalue λ̃ for small ρ is now obtained in first orderperturbation theory as

λ̃n = 4n2 + 3v(0)ρ2, n = 1, 2, ... . (201)

For large ρ we consider separately the cases corresponding to the bound and continuumstates of the potential v.

For a bound state and for ρ much larger than the spatial extension of this state wecan neglect the exponentially small integral term in the Equation 200. Introducing a newvariable z = ρ sinα one rewrites the equation as(

− d2

dz2− λ̃ρ2

+ v(z)

)φ = 0. (202)

With the boundary condition φ(z = 0) = φ(z = ρπ/2→∞) this equation is equivalent to thetwo-body problem with the potential v. The two-body bound state with the binding energy� > 0 and the radial function u(z) gives the following angular eigenvalue and eigenfunction

λ̃1 = −2m�

~2ρ2 +

∫dz u(z) (u′(z)z + u′′(z)z2),

φ1(ρ, α) =√ρ u(ρ sinα),

Q11 = −1

4

1

ρ2+

1

ρ2

∫dz u(z) (u′(z)z + u′′(z)z2). (203)



The eigenvalue corresponding to a two-body bound state therefore diverges parabolicallywith ρ with a factor proportional to the two-body binding energy. Note that the secondterms in the expressions for λ̃ and Q cancel when put in the radial equation. Thus theproper two-body radial asymptotics is restored which describes the motion of one of theparticles against the bound system of the other two particles.

Now, we shall consider the case of a short range potential, that is the potential whichcan be neglected outside a finite range r0. The results, however, are valid for any potentialwhich falls off faster than 1/r2 at large distances [18]. For ρ � r0 Equation 200 reduces tofirst order in r0/ρ to(

− d2

dα2− λ̃(ρ)

)φ(ρ, α) + ρ2v(ρα)

(φ(ρ, α) + α

8√3φ(ρ, π/3)

)= 0.

To solve this equation one divides the α-space into two regions I and II, where ρ sinα isrespectively more and less than r0. In region I, where α > r0/ρ, the potential is negligibleand this equation becomes (

− d2

dα2− λ̃)φ(I)(ρ, α) = 0 (204)

with the solution sin(√λ̃(α − π/2)) vanishing at π/2 independent of λ. In region II, where

α < r0/ρ, one has instead the equation(− d

2

d(ρα)2+ v(ρα)

)φ(II)(ρ, α) = −v(ρα)α 8√

3φ(I)(ρ, π/3), (205)

where the small term λ/ρ2 has been neglected. The total solution of Eq. 205, the homoge-neous plus the inhomogeneous one, is now

φ(II)(ρ, α) ∝ u0(ρα)− α8√3φ(I)(ρ, π/3), (206)

where the homogeneous solution u0(ρα) is the wave function describing the state of zeroenergy in the potential v, which outside the range of the potential (ρα > r0) has the formρα + a, where a is the scattering length.

The eigenvalue equation for λ̃ now arises by matching of the derivative of the logarithmof the two solutions at the point α = r0/ρ. To first order in r0/ρ one then immediatelyobtains

−√λ cos(

√λπ

2) +

8√3

sin(√λπ

6) =

ρ

asin(√λπ

2)(1 + λ

r0ρ

r0 + a

ρ). (207)

Neglecting the first order term in r0/ρ one gets the zero order equation of Efimov [11]. Forfinite a� r0, the first two terms of λ in powers of a/ρ are



λ̃n = 4n2(1− 4

π

3a

ρ), φn =

sin(√λ̃(α− π/2))√

π/4− sin(√λ̃π)/(4

√λ̃)

, n = 1, 2, ...

P12 = −P21 =a3

ρ454

π3(8π2 − 3), Q11 = −

a2

ρ46(4 +

3

π2) ... . (208)

The mixing terms P and Q fall of as ρ−4 and therefore the radial equations are asymptoticallydecoupled thus allowing to impose mathematically correct boundary conditions in each ofthe coupled radial equations independently.

For finite scattering length λ̃n therefore return back to the hyperharmonical spectrumwith the scattering length as a governing parameter.

For infinite scattering length (a = ∞) Eq. 207 has a solution λ̃∞ = −1.01251, whichsubstituted into the radial Eq. 197 implies that the equation asymptotically has an effectiveattractive potential given by −1.26251/ρ2. Such a potential is known to produce the “fallingtowards the center” phenomenon [20] but when corrected at small ρ according to 201 insteadresults in an infinite number of bound states [15]. This anomaly is called the Efimov effect.

If only two of the three scattering lengths are infinitely large the effect still takes place.One infinite scattering lengths does not produce the effect as this gives too small asymptoticvalue for λ̃.

Let us consider now the Coulomb repulsion vc which enters the angular equation as

ρ2vc(ρ sinα) =ηρ

sinα≡ ηρ+ ηρ1− sinα

sinα, (209)

where η = 2mZ2e2/√

2~2. The constant term can be included exactly while for the secondwe shall use the first order perturbation theory. The applicability of the perturbation theoryis justified when the correction to the eigenvalue is small. The relative contribution of thesecond term is in our case 0.7 for the first continuum eigenfunction. This value although beingnot much less than unity still allows to trace the general trend. For the lowest continuumstates we then have

λ1(ρ) = 4 +16

3πηρ, λ2(ρ) = 16 +

704

105πηρ, ... . (210)

The Coulomb repulsion therefore leads to the eigenvalues which grow linearly with ρ. Thisdependence restores the 1/ρ repulsion in the radial equations. The Efimov effect can nottake place in this case.

The results 203 for the bound state are unchanged since the only assumption madeconcerned the exponential fall off of the two-body radial function which is still valid in thepresence of the Coulomb field.



Figure 7: The six lowest lying angular eigenvalues λ as function of ρ for a neutron-coregaussian potential, strength Scn =10.7 MeV and range parameter 2.55 fm, correspondingto one slightly bound state of binding energy Bcn = 40 keV. The core mass is 9 times thenucleon mass.

0.11.3 Application to two-neutron halo

The neutron-neutron interaction is rather well known [21]. Since for loosely bound systemsonly the low energy properties of the potentials are important [18] and since the s-waves arethe ones responsible for the Efimov effect, we shall use a simple attractive central potential ofGaussian form, which reproduces the measured s-state scattering length and effective range,i.e. −31 MeV exp(−(r/1.8fm)2) [5]. The neutron-core potential can also be assumed to begaussian, i.e. −Scn exp(−(r/2.55fm)2) [5].

The neutron-core two-body system is first investigated as function of the strength Scn.The scattering length is computed for each potential and the angular part of the three-bodyproblem is solved. Eigenvalues λn(ρ) and wavefunctions Φn are obtained for ρ-values up to20 times the largest scattering length. The resulting lowest lying eigenvalues are shown inFig. 7 as function of ρ for a neutron-core potential with one slightly bound state with thebinding energy Bcn =40 keV. We recognize the hyperharmonic spectrum at ρ = 0, namelyλ = K(K + 4) where K is a non-negative even integer [19]. The odd integers are connectedwith odd parity. The convergence towards the same spectrum at ρ = ∞ is clearly seen



Figure 8: Binding energies of the two-body ground state (Bcn solid curve) and the firstexcited state (B1 dashed curve) of the three-body system as functions of potential strengthScn. The insert also includes the three-body ground state (B0 long-dashed curve). The coremass is 9 times the nucleon mass.

for the curves number 3,4 and 6 (starting from bottom) , whereas number 1, 2 and 5 haveonly just started to bend over, respectively towards the parabolic divergence, 0 and 32.The slow divergence of the lowest eigenvalue is the signature of a barely bound state in theneutron-core potential.

The radial Eqs. 197 are now solved and the resulting energies for the ground state and thefirst excited state are shown in Fig. 8 as function of Scn. The two- and three-body systemsare bound respectively for Scn > 9.510 MeV or Scn > 6.6 MeV. Extremely close to thetwo-body threshold, but still below (Scn ≈ 9.505 MeV), appears the first excited three-bodystate, where the neutron-core scattering length is about 18000 fm. Infinitely many boundstates appear, one after the other, when Scn is increased from 9.505 MeV to 9.510 MeV.The closer we are to the two-body threshold the larger is the number of bound states. Thisnarrow region could be called the Efimov region.

The convergence of the λ-expansion in Eq. 195 is very fast for both ground and excitedstates. The lowest λ accounts for more than 99.9% of the norm of both wavefunctions andprovides binding energies with an accuracy of about 2%. The binding energy of the firstexcited state increases until the state disappears into the two-body continuum at Scn ≈



Figure 9: The mean square hyperradius < ρ2 > of the three-body system as function of anenergy defined as the one- or two-neutron separation energy above and below the two-bodythreshold, respectively. The arrow indicates the direction of increasing potential strength.The core mass is 9 times the nucleon mass.

11.9 MeV, where the neutron-core scattering length is about −15 fm. Therefore the one-neutron separation energy for this state (B1−Bcn) will first increase and then decrease as thepotential strength is increased. Note that the states preferentially occur for bound neutron-core systems. The ground state wavefunction is concentrated in the pocket region near 4 fm.The excited state has unlike the ground state a node in the radial wavefunction, which ispeaked at ρ-values much larger than the ranges of the nuclear potentials. Consequently theground state and the first excited state are orthogonal although they both have the sameangular momentum quantum numbers.

The mean square hyperradius of the first excited state is shown in Fig. 9 as function ofan energy, which is defined as the one- or two-body separation energy above and below thetwo-body threshold, respectively. Close to the three-body threshold the mean square radiusshould obey the three-body asymptotics, i.e. go as a logarithm of the three-body bindingenergy [22]. This occurs outside the range shown in the Fig. 9. When the binding energyincreases towards about 2 keV, the mean square hyperradius decreases to about 104 fm2.Then both turn around until the excited three-body state disappears into the two-bodycontinuum. The mean square radius has here reached the asymptotic region and scales



inversely with energy [22].We thus conclude that the two-neutron halo nuclei is a type of system where the Efimov

effect might take place. The most promising candidate would be a system where a neutronis added to a weakly bound one neutron halo. For the most distinct case the scatteringlength in the neutron-core system should be about −20 fm corresponding to the neutron-core binding energy of ≈ 100 keV. The spatial extension of the first three-body excited stateat this point is estimated to be about 100 fm with the binding energy about 2 keV.

More elaborate three-body calculations with spin-spin, spin-orbit, Pauli blocking andfinite range interactions have also been performed. For a review, see Ref. [1]. A variationalapproach using an expansion on hyperspherical harmonics is one of the techniques used. Interms of the Jacobian coordinates and in a LS representation the total wavefunction of thethree-body system is (for simplicity, we use again the notation of Fig. 6(b))

ΨTJm =1

R5/2

∑χLSk`ρ`λ(R)

{Y`ρ`λKL (Ω5)⊗XS

}JM·XTMT (211)

where the “hyper radius” R is defined as

R2 = ρ2 + λ2.

It is a collective translationally, rotationally and permutation invariant variable. The “hy-perharmonic” basis functions Y`ρ`λKL (Ω5) are functions of 5 spherical angles Ω5 = (α, ρ̂, λ̂)containing the hyperangle α = arctan (ρ/λ) which extracts part of the radial correlations.XS(1, 2) is the coupled two-nucleon spin function and likewise for isospin. The “hypermo-ment” K is connected with the principal quantum number, K = `ρ+`λ+2n (n = 0, 1, 2, · · · ).

The hyperspherical harmonics have the explicit form

Y`ρ`λKLML(Ω5) = φ`ρ`λK (α)

{Y`ρ(ρ̂)⊗ Y`λ(λ̂)

}LML

(212)

φ`ρ`λk (α) = N

`ρ`λK (sinα)

`ρ(cosα)`λ · P `ρ+1/2, `λ+1/2n (cos 2α) (213)

where Pα,βn is the Jacobi polynomial, N`ρ`λK a normalization coefficient and n = (K − `ρ −

`λ)/2 .When the expansion 211 is inserted in Eqs. 183, 184, with the definitions 185, a set of

one-dimensional coupled differential equation in the variable R is obtained. These are solvednumerically as explained in detail in Refs. [23, 24]. These calculations converge rapidly forthe wave function but slowly for the eigenenergies, as we have seen in the last Section.

Another strict three-body approach is to solve the Fadeev equations. In coordinate-spacethe i−th wave function is given a bipolar angular momentum decomposition

ψTJM(i) =1

ρλ

∑ψSjΣ`ρ`λ

(ρ, λ){Y`λ(λ̂)⊗

(χk ⊗

[Y`ρ(ρ̂)⊗XSi

]j

)Σ

}JM·XTMT . (214)



Figure 10: Density contour plots for the valence neutrons in 8He in shell model and clustermodel.

When these expansion are inserted into the Fadeev equations a set of two-dimensionalcoupled differential equations are obtained and solved numerically [25].

The three-body calculations are very sensitive to the choice of the neutron-neutron andneutron-core interactions. Several authors have performed these calculations for 11Li and 6He(see Ref. [1], and references therein). With a consistent choice of potentials which reproduceknown data for neighboring nuclei, the binding energies, resonant states, and rms radii arewell reproduced within most approaches (even the simpler ones of the last two Sections).

Results concerning details which are not experimentally obtainable are sometimes con-troversial and not a good basis to distinguish if a model is better than another. As anexample we cite the search for di-neutrons and cigar structures in three-body systems. Thiscan be done, in principle, by looking at the probability of finding the neutrons in a positionof interest. For convenience, let us introduce [34] the function

Ω(R, r) = 8πR2 r2∫ π

0

|ψ|2 sin θdθ (215)

where R = λ, r = ρ , in terms of the Jacobian coordinates for a very heavy (recoil neglected)core. ψ is the wavefunction of the system and θ in the angle between R and r . Integrationover the spin coordinates are implicit. In Refs. [26] and [27] two peaks in the contour plotof Ω(R, r) were obtained. One could argue if they represent two geometrical conformations,the cluster 4He + 2n dineutron structure and a cigar-shaped structure with two neutronson the opposite side of 4He. However, as claimed by T. Suzuki [34] these two peaks donot necessarily mean the existence of such configurations, especially the cigar-structure. Forexample, if the wavefunction of the two-neutrons is taken as a product of two p3/2-harmonic



oscillator wavefunction of a shell-model with size parameter a:

Ω(R, r) =16

9πa2

(8

5

)5 (rR

2a2

)2 [(R

a

)4+( r

2a

)4− 4

3

(rR

2a2

)2]

× exp

[−8

5

{(R

a

)2+( r

2a

)2}]. (216)

This function has a reflection symmetry Ω(R, r) = Ω (r/2, 2R) so that two peaks natu-rally appear as displayed in left-side of Fig. 10. If the wavefunction ψ is given by a clusterconfiguration, one obtains a single peak in the plot of Ω(R, r) as schematically shown inright-side of Fig. 10. Thus, while a dineutron configuration can be easily identified, otherkind of structures (like cigars) may be not.

Questions like the existence of a dineutron structure in 11Li or 6He are not easily solvedby three-body calculations. They depend on the technique, choice of potentials, etc. usedin these models. To be more specific about the structure of such nuclei these models haveto be tested by experiments looking for additional variables other than the binding energiesand rms radii, only.


Bibliography

[1] M.V. Zhukov, B.V. Danilin, D.V. Fedorov, J.M. Bang, I.J. Thompson and J.S. Vaagen,Phys. Rep. 231 (1993) 231.

[2] G.L. Squires and A.T. Stewart, Proc. Roy. Soc. (London) A230 (1955) 19.

[3] A.B. Migdal, Sov. J. Nucl. Phys. 16 (1973) 238.

[4] L.M. Delves, Adv. Nucl. Phys. 5 (1972) 1.

[5] L. Johannsen, A.S. Jensen and P.G. Hansen, Phys. Lett. B244 (1990) 357.

[6] G.F. Bertsch and H. Esbensen, Ann. of Phys. 209 (1991) 327.

[7] A.Bohr and B.R. Mottelson, “Nuclear Structure”, Vol. I, Benjamim, New York, 1969.

[8] R. Reid, Ann. Phys. (N.Y.) 50 (1968) 411.

[9] G. Bertsch and S.F. Tsai, Phys. Rep. 18 (1975) 127.

[10] P.G. Hansen and B. Jonson, Europhys. Lett. 4 (1987) 409.

[11] V. Efimov, Phys. Lett., B33 (1970) 563; Sov. Jour. Nucl. Phys. 12 (1971) 589.

[12] V. Efimov, Nucl. Phys. A210 (1973) 157.

[13] C.A. Bertulani, L.F. Canto and M.S. Hussein, Phys. Rep. 226 (1993) 281.

[14] V.N. Efimov, Comm. Nucl. Part. Phys. 19 (1990) 271.

[15] D.V. Fedorov and A.S. Jensen, Phys. Rev. Lett. 71 (1993) 4103.

[16] P.G. Hansen, Nucl. Phys. A553 (1993) 89c.

[17] S.P. Merkuriev, Sov.J.Nucl.Phys. 19 (1974) 222.

49

BIBLIOGRAPHY BIBLIOGRAPHY

[18] D.V. Fedorov, A.S. Jensen and K. Riisager, Phys. Rev. C49 (1994) 201; C50 (1994)2372.

[19] R.I. Jibuti and N.B. Krupennikova, The Hyperspherical Functions Method in few-bodyquantum mechanics, Mecniereba, Tbilisi, 1984 (in Russian).

[20] L.D. Landau and E.M. Lifshitz, “Quantum Mechanics – Non-relativistic Theory”, Perg-amon Press, 1975.

[21] O. Dumbrajs et al., Nucl. Phys. B216 (1983) 277.

[22] D.V. Fedorov, A.S. Jensen and K. Riisager, Phys. Lett. B312 (1993) 1; Phys. Rev. C49(1994) 201.

[23] B.V. Danilin, M.V. Zhukov, A.A. Korsheninnikov, L.V.Chulkov and V.D. ’Efros, Sov.Jour. Nucl. Phys. 49 (1989) 351.

[24] B.V. Danilin, M.V. Zhukov, A.A. Korsheninnikov and L.V. Chulkov, Sov. Jour. Nucl.Phys. 53 (1991) 71.

[25] J.M. Bang anf I.J. Thompson, Phys. Lett. B279 (1992) 201.

[26] V.I. Kukulin, V.M. Krasnopol’sky, V.T. Voronchev and P.B. Suzanov, Nucl. Phys. A453(1986) 365.

[27] B.V. Danilin, M.V. Zhukov, A.A. Korsheninnikov V.D.’Efros and L.V. Chulkov, Sov.Jour. Nucl. Phys. 48 (1988) 766.

[28] W. Glöckle, in “Computational Nuclear Physics I”, K. Langanke, J.A. Maruhn and S.E.Koonin, eds., Springer, Berlin,1991.

[29] M. Fabre de la Ripelle, Lecture Notes in Physics 273 (1987) 283.

[30] R.B. Wiringa, Few-Body Systems Suppl. 1 (1986) 130.

[31] L.D. Fadeev, “Mathematical aspects of the three-body problem in the quantum scat-tering theory ” (Steklov Mathematical Institute, Leningrad, 1963) [English translation:Israel Program for Scientific Translation, Jerusalem, 1965].

[32] W. Glöckle, “The quantum mechanical few-body problem”, Springer, Berlin, 1983.

[33] K.E. Schmidt, Lectures Notes in Physics 273 (1987) 363.

[34] Y. Suzuki, Nucl. Phys. A528 (1991) 395.


physics of radioactive beams1 chapter 4 borromean...

Documents