1. Few-body structure of light hypernuclei
Emiko Hiyama　(RIKEN)
Thank you for chairwomen.

2. Last year and this year, we had two epoch-making data from
the view point of few-body problems.
Then, in hypernuclear physics, we are so excited. n n n n
7He
6 H Λ α Λ t Λ Λ
JLAB experiment-E011,
Phys. Rev. Lett. 110,
12502 (2013).
FINUDA collaboration & A. Gal,
Phys. Rev. Lett. 108,
(2012).
Observation of Neutron-rich Λ-hypernuclei
These observations are interesting from the view points of
few-body physics as well as physics of unstable nuclei.

3. n n n n 6 H 6 H t 7He α 7He 4H 4He 4H 4He OUTLINE Λ Introduction 2. 3.4.
5. Summary
7He Λ n n
6 H
6 H Λ Λ t Λ 4H
4He Λ Λ n n n p Λ Λ p p 4H
4He Λ Λ

4. 　　　　 Section 1
Introduction

5. n n n n n n n n In neutron-rich and proton-rich nuclei,
Nuclear
cluster
Nuclear
cluster
Nuclear
cluster
Nuclear
cluster n n n n
When some neutrons or protons are added to clustering nuclei,
additional neutrons are located outside the clustering nuclei
due to the Pauli blocking effect.
As a result, we have neutron/proton halo structure in these nuclei.
There　are many interesting phenomena in this field as you know.
It was considered that nucleus was hardly compressed by the
additional nucleons .

6. Question:How is the structure modified when a hyperon, Λ particle,
Nucleus
Question:How is the structure modified when a hyperon, Λ particle,
is injected into the nucleus?

7. Λ particle plays a ‘glue like role’ to produce a dynamical
No Pauli principle
Between N and Λ
Λ particle can reach
deep inside, and
attract the
surrounding nucleons
towards the interior
of the nucleus. Λ
Hypernucleus
Λ particle plays a ‘glue like role’ to produce a dynamical
contraction of the core nucleus.
Question: How do we observe the dynamical contraction?

8. r > r’ Λ hypernucleus nucleus B(E2) value:X X r r’ N N
Λ hypernucleus
nucleus
r > r’
B(E2) value:
B(E2) ∝　|<Ψi |r2 Y2μ(r) |Ψf>|2
When a Lambda particle is added to the nucleus, then due to the gluelike role of Lambda particle, size of the corresponding hypernucleus r’ becomes smaller than the size of nucleus. This phenomena appears in B(E2) value. Since B(E2) value is propotinal to 4th-power of nuclear size. So far, Bando et al., pointed out that..
Propotional to 4th-power of nuclear size
Bando, Ikeda, Motoba
If nucleus is really contracted by the glue-like role of the Λ particle, then
E2 transition in the hypernuclei will become much reduced than the
corresponding E2 transition in the core nucleus.

9. α α 6Li 7Li 3.6 ±2.1 e2fm4 from n p n p E. Hiyama et al.,
Phys. Rev. C59 (1999), 2351
Theoretical calculation by Hiyama et al.
B(E2: 5/2+ → 1/2+) =2.85 e2fm4
reduced by 22%
α+ n+p
α+Λ+n+p
KEK-E419
Then, Tamura et al.
Succeeded in
measuring this
B(E2) value to be
3.6 ±2.1 e2fm4 from
5/2+ to 1/2+ state
in 7Li. 3+
B(E2)
7/2+ 1+
5/2+
Exp.B(E2)= 9.3 e2fm4
B(E2)
6Li
3/2+ Λ n p
1/2+
Then, along line, in this paper, I proposed to measure gamma transition from this excites state to the ground state in Li6+Lambda system.
7Li n p α Λ Λ α
By Comparing B(E2) of 6Li and that of 7Li, Λ
The shrinkage effect on the nuclear size included by the Λ
particle was confirmed for the first time.

10. Nucleus hypernucleus nucleus hypernucleus
The glue like role of Λ particle provides us with another
interesting phenomena.
Λ particle can reach
deep inside, and
attracts the
surrounding nucleons
towards the interior
of the nucleus. Λ Λ
There is no Pauli
Pricliple between
N and Λ.
Nucleus
hypernucleus Λ
Due to the attraction of
ΛN interaction, the
resultant hypernucleus
will become more stable
against the neutron
decay.
Neutron decay threshold γ
nucleus
hypernucleus

11. Nuclear chart with strangeness
Multi-strangeness system
such as Neutron star
Extending drip-line! Λ
Outline
Interesting phenomena concerning the neutron halo have　been observed near the neutron drip line of light nuclei.
How does the halo structure change when a Λ particle is injected into an unstable nucleus?

12. Question : How is structure change when a Λ particle
is injected into neutron-rich nuclei? n n n n
6 H
7He Λ t Λ Λ α Λ
Observed at JLAB, Phys. Rev. Lett. 110, (2013).
Observed by FINUDA group,
Phys. Rev. Lett. 108, (2012).

13. ・A variational method using Gaussian basis functions
In order to solve few-body problem accurately,
Gaussian Expansion Method (GEM) , since 1987 ,
・A variational method using Gaussian basis functions
・Take all the sets of Jacobi coordinates
Developed by Kyushu Univ. Group,
Kamimura and his collaborators.
Review article :
E. Hiyama, M. Kamimura and Y. Kino,
Prog. Part. Nucl. Phys. 51 (2003), 223.
High-precision calculations of various 3- and 4-body systems:
Exotic atoms / molecules ,
3- and 4-nucleon systems,
multi-cluster structure of light nuclei,
Light hypernuclei,
3-quark systems,
4He-atom tetramer

14. 　　　　 Section 2
Four-body calculation of
7He Λ n n α Λ

15. n n n n 6He α 7He α 6He : One of the lightest n-rich nuclei
n-rich hypernuclei Λ
7He Λ α Λ
Observed at JLAB, Phys. Rev. Lett. 110, (2013).

16. 6He 7He γ α+Λ+n+n 5He+n+n 5/2+ 3/2+ γ 1/2+
CAL: E. Hiyama et al., PRC53, 2075 (1996), PRC80, (2009)
6He
7He Λ
Prompt particle decay 2+ γ
α+Λ+n+n
0 MeV
0 MeV
α+n+n
5He+n+n 0+ Λ
-1.03 MeV
5/2+
Exp:-0.98
Halo states
3/2+
-4.57
BΛ: CAL= 5.36 MeV γ
BΛ： EXP= 5.68±0.03±0.25
Jlab experiment
Phys. Rev. Lett.110,
(2013).
1/2+
-6.19

17. 6He 7He γ α+Λ+n+n 5He+n+n 5/2+ 3/2+ γ 1/2+
CAL: E. Hiyama et al., PRC53, 2075 (1996), PRC80, (2009)
6He
7He Λ
Prompt particle decay 2+ γ
α+Λ+n+n
0 MeV
0 MeV
α+n+n
5He+n+n 0+ Λ
-1.03 MeV
5/2+
Exp:-0.98
Halo states
3/2+
-4.57
BΛ: CAL= 5.36 MeV γ
5/2+ →1/2+
3/2+ →1/2+
・Useful for the study of the excitation mechanism in neutron-rich nuclei
・Helpful for the study of the excitation mechanism of the halo nucleus
In n-rich nuclei and n-rich hypernuclei, there will be many examples such as
Combination of 6He and 7He. I hope that γ-ray spectroscopy of n-rich
hypernuclei will be performed at J-PARC. Λ
BΛ： EXP= 5.68±0.03±0.25
1/2+
Observed at J-Lab experiment(2012)
Phys. Rev. Lett.110,
(2013).
-6.19

18. 6H n n t Four-body calculation of Section 3 Λ
E. H, S. Ohnishi, M. Kamimura, Y. Yamamoto, NPA 908 (2013) 29.

19. n n 6H t Λ Λ

20. n 5 H 6 H n 5H : super heavy hydrogen t t+n+n Γ=1.9±0.4 MeV
Phys. Rev. Lett. 108, (2012).
Γ=1.9±0.4 MeV
1/2+
FINUDA experiment
1.7±0.3 MeV
t+n+n+Λ
t+n+n
5 H
4H+n+n
EXP: BΛ= 4.0±1.1 MeV Λ
0.3 MeV
6 H Λ n n
5H : super heavy hydrogen t Λ

21. Before the experiment, the following authors calculated the
binding energy using shell model picture and G-matrix theory.
(1) R. H. Dalitz and R. Kevi-Setti, Nuovo Cimento 30, 489 (1963).
(2) L. Majling, Nucl. Phys. A585, 211c (1995).
(3) Y. Akaishi and T. Yamazaki, Frascati Physics Series Vol. 16 (1999).
Akaishi et al. pointed out that one of the important subject to study
this hypernucleus is to extract information about ΛN-ΣN coupling.
Motivated by the experimental data, I calculated the binding energy of
6H and I shall show you my result. Λ

22. To calculate the binding energy of 6H, it is very important
Framework:
To calculate the binding energy of 6H, it is very important
to reproduce the observed data of the core nucleus 5H. Λ
transfer reaction p(6He, 2He)5H
A. Korcheninnikov, et al. Phys. Rev. Lett.
87 (2001)
Γ= 1.9±0.4 MeV
1/2+
1.7±0.3 MeV
Theoretical calculation
N. B. Schul’gina et al., PRC62 (2000),
R. De Diego et al, Nucl. Phys. A786 (2007), 71.
calculated the energy and width of 5H with t+n+n
three-body model using complex scaling method.
5H is well described as t+n+n three-body model.
t+n+n threshold

23. n n n n t t 6H 5H Then, I think that t+n+n+Λ 4-body model
is good model to describe 6H.
Then. I take t+Λ+n+n 4-body model. Λ Λ n n n n t t Λ 6H 5H Λ

24. Before doing the full 4-body calculation,
it is important and necessary to reproduce the observed binding energies of all the sets of subsystems in 6H. Λ
Namely, all the potential parameters are needed
to adjust the energies of the 2- and 3-body subsystems. 6H 6H Λ Λ 6H n n Λ n n n n Λ Λ Λ t t t

25. t t n n n n 6H Λ 6H Λ I take the Λｔ potential to reproduce
the binding energies of 0+ and 1+ states
of 4H.
In this case, ΛN－ΣN coupling is
renormarized into the ΛN interaction. n Λ n Λ Λ t 6H Λ n n
I take the ΛＮ potential to reproduce
the binding energy of 3H. Λ Λ t Λ

26. t n n Then, what is the binding energy of 6H? 0+ 6H 5H
Γ=1.9±0.4 MeV  EXP
Γ= 2.44 MeV  CAL
1/2+
1.69 MeV
1.7±0.3 MeV
I focus on the 0+ state.
t+n+n 1+ n n
σn・σΛ 0+ Λ t
When a Λ particle is added to this nucleus, due to spin-spin interaction between Λ and nucleon, we have 0+ and 1+ state.
Now, I focus on 0+ state. 6H Λ 5H
Then, what is the binding energy of 6H? Λ

27. On the contrary, if we tune the potentials to have a bound state
Exp: 1.7 ±0.3 MeV
Even if the potential parameters were tuned
so as to reproduce the lowest value of the
Exp. , E=1.4 MeV, Γ=1.5 MeV,
we do not obtain any bound state of 6H.
Γ=1.9 ±0.4 MeV Λ
Γ= 2.44 MeV ½+
Γ= 0.91 MeV
1.69 MeV
1.17 MeV
0 MeV
Γ=0.23 MeV
t+n+n+Λ
t+n+n
E=-0.87 MeV 0+
4H+n+n Λ
-　2.0
4H+n+n 0+
-2.07 MeV Λ
On the contrary, if we tune the
potentials to have a bound state
in 6H, then what is the
energy and width of 5H? Λ

28. n n 5 H n t 6 H n t 5H:super heavy hydrogen FINUDA experiment t+n+n+Λ
Γ= 1.9±0.4 MeV
Phys. Rev. Lett. 108, (2012).
1/2+
FINUDA experiment
1.7±0.3 MeV
t+n+n+Λ
t+n+n
5 H
EXP:BΛ=4.0±1.1 MeV
4H+n+n Λ
t+n+n+Λ n n
0.3 MeV t
6 H Λ n n
5H:super heavy hydrogen t Λ
But, FINUDA group provided the bound state of 6H. Λ

29. How should I understand the inconsistency between our results
and the observed data?
We need more precise data of 5H.
A. Korcheninnikov, et al. Phys. Rev. Lett.
87 (2001)
Γ=1.9±0.4 MeV
To get bound state of 6H, the energy should
be lower than the present data.
It is planned that experiment to measure
the energy and width of 5H more precisely　
at RCNP next year. Λ
1/2+
1.7±0.3 MeV
t+n+n
In our model, we do not include ΛＮーΣN coupling explicitly.
The coupling effect might contribute to the energy of 6H. Λ

30. What is Λ N-ΣN coupling?

31. Non-strangeness nucleiΣ N Δ
80 MeV Λ
S=-2 ΞN N N
25MeV ΛΛ Δ
In hypernuclear physics,
the mass difference is very small
in comparison with the case of S=0
field.
300MeV N
Probability of Δ　in nuclei is
not negligible.
Then, in S=-1 and S=-2 system, ΛN-ΣN and ΛΛ-ΞN
couplings might be important.

32. Interesting Issues for the ΛN-ΣN particle conversion in hypernuclei
How large is the mixing probability of the Σ particle in the
hypernuclei?
(2) How important is the ΛＮーΣＮ coupling in the binding
energy of the Λ hypernuclei?

33. Exp. Exp. -BΛ -BΛ 0 MeV 3He+Λ 0 MeV 3H+Λ 1+ 1+ -1.00 -1.24 0+ 0+ -2.04
-2.39
Exp.
Exp. N N
4He Λ Λ 4H N Λ

34. E. Hiyama et al., Phys. Rev. C65, 011301 (R) (2001). Σ N Λ N + N N N N +
NNNΛ
NNNΣ
E. Hiyama et al., Phys. Rev. C65, (R) (2001).
H. Nemura et al., Phys. Rev. Lett. 89, (2002).
A. Nogga et al., Phys. Rev. Lett. 88, (2002).

35. 4He, 4H VNN : AV8 potential VYN : Nijmegen soft-core ’97f potential Λ

36. PΣ=1.12 %
PΣ=2.21%

37. Gal and D. J. Millener, arXiv:1305.6716v3 (PLB725 445(2013)).
They pointed out that Λ N-ΣN coupling is important for 6H. Λ
It might be important to perform the following calculation: n n n n 6H + Λ t Λ 3N Σ

38. 5 H FINUDA experiment t+n+n+Λ t+n+n EXP:BΛ=4.0±1.1 MeV Cal: -0.87 MeV
Phys. Rev. Lett. 108, (2012).
1/2+
FINUDA experiment
1.7±0.3 MeV
t+n+n+Λ
0 MeV
t+n+n
5 H
Cal: MeV
EXP:BΛ=4.0±1.1 MeV
4H+n+n Λ
ΛN-ΣN coupling ？
Exp: -2.3 MeV
This year, at J-PARC, they performed
a search experiment of (E10 experiment) of 6H.
If E10 experiment reports more accurate energy,
we can get information about ΛN-ΣN coupling. Λ Σ

39. 4H 4He In hypernuclear physics, currently, it is extremely important
to get information about ΛN-ΣN coupling.
Question: Except for 　6H, what kinds of Λ hypernuclei are
suited for extracting information on
the ΛN-ΣN coupling? Λ
Answer: 4H, 4He Λ Λ n n n p Λ Λ p p 4H
4He Λ Λ

40. 4H 4He Recently, we have updated YN interaction such as ESC08
which has been proposed by Nijmegen group.
Then, using this interaction directly,
I performed four-body calculation of 4H and 4He. Λ Λ
At present, I use AV8 NN interaction.
Soon, I will use new type of Nijmegen NN potential. 4H
4He Λ Λ N N N N + Σ N Λ N

41. Preliminary result (ESC08)
3He+Λ
0 MeV
3H+Λ
0 MeV
CAL: (Exp: -1.00)
CAL: (Exp: -1.24) 1+ 1+
(Exp: 0.24 MeV)
(cal: 0.01 MeV)
CAL: (Exp: -2.04) 0+
CAL:-1.46 (Exp ) 0+
(Exp: 0.35 MeV)
(cal MeV n n p n Λ
Among these calculation, Nogga and his collaborators investigated the charge symmetry breaking effect by sophisticated 4-body calculation using modern realistic YN and NN interactions. These are results using Nijmegen soft core ’97e model. The calculated energy difference in the ground state is 0.07 MeV. And this value in the excited state is MeV. Both of energy difference in the ground state and the excited state are inconsistent with the data. At the present, there exist no YN interaction to reproduce the charge symmetry breaking effect. p Λ p 4H
4He Λ Λ
Binding energies are less bound than the
observed data. We need more updated
YN interactions. N N N N + Σ N Λ N 44

42. 　　 Section 5
Summary

43. Summary 1) Motivated by observation of neutron-rich Λhypernuclei,
7He and 6H, I performed four-body calculation of them. Λ Λ
In 7He, due to the glue-like role of Λ particle,
We may have halo states, the 3/2+ and 5/2+ excited state.
We wait for further analysis of the JLAB experiment. Λ

44. I performed a four-body calculation of 6H.
But, I could not reproduce the FINUDA data of 6H .
The error bar of data is large. Analysis of E10 experiment at J-PARC is in progress.
If they provide with more accurate data in the future and confirm to have a bound state, then we could obtain information about ΛN-ΣN coupling. Λ Λ

45. To obtain this information, four-body calculations of
4H and 4He is important.
Then , four-body calculations of A=4 hypernuclei
using realistic YN and NN interaction were performed.
However, the results are not in good agreement with the data.
we need more sophisticated YN interaction.
And in the future, we have many experimental data for
light Λ hypernuclei at J-PARC, Jlab and Mainz.
By comparison with future experiment and theoretical
effort, we expect to have more reliable YN interaction in the
future. Λ Λ

46. Thank you!

47. Another interesting role of Σ-particle in hypernuciei,
namely effective ΛNN 3-body force generated by the
Σ-particle mixing. N1 Λ N2 N3 ② N1 Λ N2 N3 ①
In the calculation of
6 H, this effect is
included in ΛN
interaction. Σ Λ Σ N1 Λ N2 N3 N1 Λ N2 N3
3N+Λ space N1 Λ N2 N3 N1 Λ N2 N3
Effective 3-body
ΛNN force
Effective
2-body ΛN
force
How large is the
3-body effect? N1 Λ N2 N3 N1 Λ N2 N3

48. Y. Akaishi, T. Harada, S. Shinmura and Khin Swe Myint,
Phys. Rev. Lett. 84, 3539 (2000).
3He
3He Σ + + Λ +
They already pointed out that three-body force effect is
Important within the framework of (3He+Λ)+(3He+Σ).

50. 　　　　 Section 4.2
Λ-Σ coupling and CSB in
7He Λ n n α Λ

51. It is interesting to investigate the charge symmetry breaking effect
in p-shell Λ hypernuclei as well as s-shell Λ hypernuclei.
For this purpose, to study structure of A=7 Λ hypernuclei is suited,
because, core nuclei with A=6 are iso-triplet states. n n n p p p α α α
6Li (T=1)
6Be
6He

52. n n n p p p α α α 7He 7Be 7Li(T=1) Λ Λ Λ
Then, A=7　Λ hypernuclei are also iso-triplet states.
It is possible that CSB interaction between Λ and valence nucleons
contribute to the Λ-binding energies in these hypernuclei.

53. Exp. 6He 6Be 6Li 7Be 7Li 7He Emulsion data Emulsion data BΛ=5.16 MeV
1.54
Emulsion data
Emulsion data
6He
6Be
6Li
(T=1)
BΛ=5.16 MeV
BΛ=5.26 MeV
Preliminary data: 5.68±0.03±0.22
JLAB:E experiment
These are experimental data of A=7 hypernuclei. These two data of Be7L and Li7L are taken by emulsion data. As mentioned by the previous speaker, Hashimoto san, we got new data of He7L.
The value is like this. We see that as the number of neutron increase, Lambda separation energy increase.
-3.79
7Be
7Li
(T=1) Λ Λ
7He Λ 56

54. Important issue:
Can we describe the Λ binding energy of 7He observed at JLAB
using　ΛN　interaction to reproduce the Λ binding energies of
7Li (T=1) and 7Be ?
To study the effect of CSB in iso-triplet A=7 hypernuclei. Λ Λ Λ
The detailed study has been done in this paper.
For this purpose, we studied structure of A=7 hypernuclei within
the framework of α+Λ+N+N 4-body model.
E. Hiyama, Y. Yamamoto, T. Motoba and M. Kamimura,PRC80, (2009) 57

55. Now, it is interesting to see as follows:
What is the level structure of A=7 hypernuclei
without CSB interaction?
(2) What is the level structure of A=7 hypernuclei
with CSB interaction?

56. 6Be 6Li 6He 7Be 7Li 7He Without CSB EXP= 5.16 BΛ：CAL= 5.21 EXP= 5.26
JLAB:E experiment
CAL= 5.36
preliminary
These are results of A=7 hypernuclei without CSB interaction. We see that our Λ binding energies of 7BeL and Li7L are in good agreement with the data.
Let see the Λ separation energy of He7L. Our result is not inconsistent with data.
7Be Λ
7Li
(T=1) Λ
7He Λ 59

57. Now, it is interesting to see as follows:
What is the level structure of A=7 hypernuclei
without CSB interaction?
(2) What is the level structure of A=7 hypernuclei
with CSB interaction?

58. Exp. 4H 4He Next we introduce a phenomenological CSB potential
with the central force component only.
Strength, range
are determined
so as to reproduce
the data.
0 MeV
3He+Λ
0 MeV
3H+Λ
-1.00
-1.24 1+ 1+
0.24 MeV
-2.04
-2.39 0+
0.35 MeV 0+ n n p n Λ p Λ p
Exp. 4H
4He Λ Λ

59. With CSB 5.21 (without CSB) 5.29 MeV (With CSB) 5.44(with CSB)
5.28 MeV( withourt CSB)
5.44(with CSB)
5.21 (without CSB)
5.29 MeV (With CSB)
5.36(without CSB)
5.16(with CSB)
BΛ： EXP= 5.68±0.03±0.22
This is our results with a phenomenological CSB interaction. The binding of Li7 with and without CSB is almost the same. Because, there is cancellation between ｎΛ and pΛ CSB interaction. On the other hand, in the case of Be7L, the energy of Be7L with CSB make deeper bound by 0.2 MeV comparing with this value. And in the case of He7L, the energy with CSB interaction make less bound by 0.2 MeV comparing with this. Then, we found that binding energies with CSB interaction of He7L and Be7L became inconsistent with the data.
Inconsistent with the data p n α 62

60. Comparing the data of A=4 and those of A=7,
tendency of BΛ is opposite.
Let me compare with the experimental data of A=4 hypernuclei and data of A=7 hypernuclei.
How do we understand these difference? 63