1. Structure of few-body light Λ hypernuclei
Emiko Hiyama (RIKEN)

2. n n n n 6 H t 7He α 3n Recently, we had three epoch-making data from
the view point of few-body problems. n n n n n n
6 H Λ t Λ Λ
7He Λ α Λ 3n
FINUDA collaboration & A. Gal,
Phys. Rev. Lett. 108,
(2012). Λ
C. Rappold et al.,
HypHI collaboration
Phys. Rev. C 88,
(R) (2013)
JLAB experiment-E011,
Phys. Rev. Lett. 110,
12502 (2013).
E. Hiyama, S. Ohnishi, and M. Kamimura,
Nucl. Phys. A908, 29 (2013).
Observation of Neutron-rich Λ-hypernuclei
These observations are interesting from the view points of
few-body physics as well as unstable nuclear physics.
E. Hiyama, S. Ohanishi. B. F. Gibson,
And Th. A. Rijken, Phys. Rev. C89, (R), (2014).

3. n n
7He Λ α Λ

4. 7He n n α 7He What is interesting to study this hypernucleus?Λ
What is interesting to study
this hypernucleus? n n α
(1) It is important to obtain information about
charge symmetry breaking effect
of n-Λ and p-Λ.
(2) To study structure of hypernuclei due to
the glue-like role of Λ particle. Λ
7He Λ 4 4 4

5. The major goal of hypernuclear physics
1) To understand baryon-baryon interactions
Fundamental and important for the study
of nuclear physics
To understand the baryon-baryon interaction, two-body scattering
experiment is most useful.
Total number of
Nucleon (N) -Nucleon (N) data: 4,000
YN and YY potential models so far proposed
(ex. Nijmegen,
Julich, Kyoto-Niigata) have large ambiguity.
・ Total number of differential cross section
Hyperon (Y) -Nucleon (N) data: 40
・ NO YY scattering data

6. Therefore, as a substitute for the 2-body
limited YN and non-existent YY scattering data,
the systematic investigation of the
structure of light hypernuclei is essential.

7. Hypernuclear g-ray data since 1998 (figure by H.Tamura)
・Millener (p-shell model), 　・ Hiyama (few-body)

8. n n α 7He 4H Charge symmetry breaking (2) ΛN－ΣN coupling
In S= -1 sector,
what are important to study YN interaction?
Charge symmetry breaking
(2) ΛN－ΣN coupling
J-PARC : Day-1 experiment
　　　　　E13
Jlab E05-115, Mainz n n n n Λ p α Λ 4H Λ
7He Λ

9. Exp. 4H 4He In S= -1 sector 3He+Λ 3H+Λ 0 MeV 0 MeV -1.00 -1.24 1+ 1+
-2.04
-2.39 0+
0.35 MeV 0+ n n p n Λ Λ p n Λ p 4H Λ
4He p Λ Λ

10. 4He 4H 4He 4H p n n n Λ Λ p p However, Λ particle has no charge. n p n+ + Λ Λ
So, charge symmetry breaking effect is considered to come from interaction between Sigma and nucleon.Then, in order to CSB effect, these Σ coupling effect should be taken into account. p Σ- p Σ- p n + + + + n p n n n p n n + + Σ0 n Σ0 p Σ+ n Σ+ p

11. Σ In order to explain the energy difference, 0.35 MeV, N N N N + N Λ N
・E. Hiyama, M. Kamimura, T. Motoba, T. Yamada and Y. Yamamoto,
Phys. Rev. C65, (R) (2001).
・A. Nogga, H. Kamada and W. Gloeckle,
Phys. Rev. Lett. 88, (2002)
・H. Nemura. Y. Akaishi and Y. Suzuki,
Phys. Rev. Lett.89, (2002).
Coulomb potentials between charged particles (p, Σ±) are included.

12. 4H 4He 3He+Λ 3H+Λ + 0 MeV 0 MeV -1.00 -1.24 1+ 1+ (Exp: 0.24 MeV)
(cal: MeV(NSC97e))
-2.04 0+
-2.39
(Exp: 0.35 MeV) 0+
(cal MeV(NSC97e)) n n p n Λ p Λ p
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. 4H
・A. Nogga, H. Kamada and W. Gloeckle,
Phys. Rev. Lett. 88, (2002)
4He Λ Λ
・E. Hiyama, M. Kamimura, T. Motoba, T. Yamada and Y. Yamamoto,
Phys. Rev. C65, (R) (2001).
・H. Nemura. Y. Akaishi and Y. Suzuki,
Phys. Rev. Lett.89, (2002). N N N N + Σ N Λ N

13. 4H 4He 3H+Λ 3He+Λ There exist NO YN interaction to reproduce the data.
0 MeV
3H+Λ
0 MeV
3He+Λ
-1.00
-1.24 1+ 1+
(Exp: 0.24 MeV)
(cal: MeV(NSC97e))
-2.04 0+
-2.39
(Exp: 0.35 MeV) 0+
(cal MeV(NSC97e)) n n p n Λ p Λ p 4H
4He Λ Λ
There exist NO YN interaction to reproduce the data.
For the study of CSB interaction, we need more data.

14. 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. p p n n n p α α α
6Be
6Li(T=1)
6He

15. Then, A=7 Λ hypernuclei are also iso-triplet states. 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.

16. 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
JLabE exp.: 5.68±0.03±0.25
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

17. Can we describe the Λ binding energy of 7He observed at JLAB
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 p n n n p p Λ Λ Λ α α α
The detailed study has been done in this paper.
7He
7Be
7Li(T=1) Λ Λ Λ
For this purpose, we study 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)

18. 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?

19. 6Be 6Li 6He 7Be 7Li 7He Without CSB EXP= 5.16 BΛ：CAL= 5.21
JLAB:E experiment
CAL= 5.36
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

20. 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?

21. 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

22. With CSB 5.29 MeV (With CSB) 5.36(without CSB) 5.44(with CSB)
5.28 MeV( withourt CSB)
5.44(with CSB)
5.29 MeV (With CSB)
5.21 (without 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

23. 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.
In order to investigate CSB interaction, it was necessary to perform the energy spectra of
A=4 hypernuclei again at JLab, Mainz and J-PARC.

24. Exp. 4H 4He In S= -1 sector 3He+Λ 3H+Λ 0 MeV 0 MeV -1.00 -1.24 1+ 1+
-2.04
1.406±0.002±0.003
Reported by H. Tamura
-2.39 0+
Esser et al., PRL114, (2015)
Reported by P. Achenbach
0.35 MeV 0+ n n p n Λ Λ p n Λ p 4H Λ
4He p Λ Λ

25. Exp. 4H 4He In S= -1 sector 3He+Λ 3H+Λ Questions to experimentalist:
The ground state of 4ΛHe　is
confirmed?
(2) The excited state of 4ΛH　is
If the answer is no,
・It is important to measure the
excited state of 4ΛH at JLab.
4He(e,e’K+) 4ΛH reaction at JLab is
very powerful way to obtain it.
ground state of 4ΛHe at J-PRAC.
Or please analyze all states of A=4
hypernuclei by emulsion data
by Prof. Nakazawa and
his collaborators.
Exp.
3He+Λ
3H+Λ
0 MeV
0 MeV
-1.00
-0.98 1+ 1+
0.24 MeV
-2.12±0.01±0.09 MeV
1.406±0.002±0.003
Reported by H. Tamura
-2.39 0+
0.35 MeV 0+ n n p n Λ p Λ p 4H
4He Λ Λ

26. Exp. 4H 4He In S= -1 sector 3He+Λ 3H+Λ Questions to experimentalist:
The ground state of 4ΛHe　is
confirmed?
(2) The excited state of 4ΛH　is
If the answer is yes,
From now, this is homework to
theory side.
・In A. Gal, PLB 744, 352 (2015),
Using purely central-type ΛN-
ΣN coupling reproduce the current
energy-splittings of A=4 hypernuclei,
and p-shell nuclei of A=7, 8,10.
However, I do not think that
ΛN-ΣN coupling should be purely
central force.
In S= -1 sector
Exp.
3He+Λ
3H+Λ
0 MeV
0 MeV
-1.00
-0.98 1+ 1+
0.24 MeV
-2.12±0.01±0.09 MeV
1.406±0.002±0.003
Reported by H. Tamura
-2.39 0+
0.35 MeV 0+ n n p n Λ p Λ p 4H
4He Λ Λ

27. Charge Symmetry breaking
Energy difference comes from
dominantly Coulomb force
between 2 protons.
Charge symmetry breaking
effect is small.
In S=0 sector
Exp.
N+N+N
0 MeV
MeV
0.76 MeV
1/2+
Question:
Why we have large CSB effect in
A=4 hypernuclei?
I cannot understand this physics.
MeV
3He
1/2+ 3H n p p n n p

28. Σ Now, as a few-body physicist, we
should perform both of four-body NNNΛ
+NNNΣ coupled channel calculation
And (α+Λ+N+N)+(α+Σ+N+N)coupled channel
calculation with updated YN realistic interaction. N N N N + N Λ N Σ
(3N+Λ）
(3N+Σ）
Now, we have next stage to discuss
on which part,
theory side or experiment side should
have homework. n n n n ＋ Λ Σ α α
A=7 hypernuclei

29. 7He n n α 7He What is interesting to study this hypernucleus?Λ
What is interesting to study
this hypernucleus? n n α
(1) It is important to obtain information about
charge symmetry breaking effect
of n-Λ and p-Λ.
(2) To study structure of hypernuclei due to
the glue-like role of Λ particle. Λ
7He Λ 29 29 29

30. Second of major goals in hypernuclear physics
Why is it interesting to study neutron-rich Λ hypernucleus
such as 7ΛHe?
Second of major goals in hypernuclear physics
To study the structure of multi-strange systems

31. 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.

32. No Pauli principle
Between N and Λ
Λ particle can reach deep inside, and attract the surrounding nucleons towards the interior
of the nucleus. N Λ
Due to the attraction of
Λ N interaction, the
resultant hypernucleus will
become more stable against
the neutron decay.
Hypernucleus Λ
neutron decay threshold γ
nucleus
hypernucleus

33. 6He 7He Λ -4.57 α+Λ+n+n 5He+n+n 5/2+ 3/2+ 1/2+
CAL: E. Hiyama et al., PRC 53, 2075 (1996), PRC 80, (2009)
6He
Another interesting　issue is to study the excited states of 7He. Λ
7He Λ 2+
α+Λ+n+n
0 MeV
0 MeV
α+n+n
One of the excited
state was observed at JLab.
5He+n+n 0+ Λ Λ
Exp:-0.98
-1.03 MeV
5/2+
Neutron
halo states
3/2+
-4.57
BΛ: CAL= 5.36 MeV
BΛ： EXP= 5.68±0.03±0.25
1/2+
Observed at J-Lab experiment
Phys. Rev. Lett.110,
(2013).
-6.19 MeV

34. 7Li(e,e’K+)7ΛHe (FWHM = 1.3 MeV) Fitting results Cal: Ex=1.70MeV
Good agreement with my prediction

35. Question: In 7He, do we have any other new states?
If so, what is spin and parity? Λ
First, let us discuss about energy spectra of 6He core nucleus.

36. Exp. Exp. Core nucleus 6He 6He Data in 2002 (2+,1-,0+)?
Γ＝12.1 ±1.1 MeV
(2+,1-,0+)?
Γ=1.6 ±0.4 MeV
2.6±0.3 MeV 2+ 2
Γ=0.11 ±0.020 MeV
Γ=0.12 MeV 2+
1.797 MeV 2+
1.8 MeV 1
0 MeV
α+n+n
0 MeV
α+n+n
-0.98 0+
-0.98 0+
6He
6He
Exp.
Exp.
Data in 2002
Data in 2012
X. Mougeot et al., Phys. Lett. B
718 (2012) 441. p(8He, t)6He
Core nucleus

37. theory Exp. 6He Myo et al., PRC 84, 064306 (2011). Γ=0.12 MeV 2+ α+n+n
How about theoretical result?
Γ=1.6 ±0.4 MeV
2.5 MeV
1.6±0.3 MeV Γ= 2+ 2
Decay with
Is smaller than
Calculated with.
Γ=0.12 MeV Γ= 2+
0.8 MeV 1
α+n+n
0.79 MeV
0 MeV
-0.98 0+
What is my result?
-0.97 MeV
6He
theory
Exp.
Myo et al., PRC 84, (2011).
Data in 2012
X. Mougeot et al., Phys. Lett. B
718 (2012) 441. p(8He, t)6He

38. Exp. 6He Question: What are theoretical results? Γ=0.12 MeV 2+ 0 MeV
These are resonant states.
I should obtain energy position
and decay width.
To do so, I use complex scaling
method which is one of powerful
method to get resonant states.
Γ=0.12 MeV 2+
1.8 MeV 1
0 MeV
α+n+n
-0.98 0+
6He
Exp.
Data in 2012
X. Mougeot et al., Phys. Lett. B
718 (2012) 441. p(8He, t)6He

39. My result
E= MeV
E= MeV

40. Exp. 6He 6He Cal. Question: What are theoretical results? Γ=0.12 MeV2+
2.81 MeV 2
Γ=1.6 ±0.4 MeV
1.6±0.3 MeV 2+ 2
Γ=0.12 MeV
Γ=0.14 MeV 2+
0.8 MeV 2+
0.8 MeV 1 1
α+n+n
0 MeV
α+n+n
0 MeV
-0.98
-0.98 0+ 0+
6He
6He
Exp.
Cal.
Data in 2012
X. Mougeot et al., Phys. Lett. B
718 (2012) 441. p(8He, t)6He

41. 6He Cal. How about energy spectra of 7He? Γ=0.14 MeV 2+ 0 MeV α+n+nΛ Λ
Γ=4.81 MeV
2.81 MeV 2+ 2
3/2+,5/2+
Γ=0.14 MeV 2+
0.8 MeV 1
0 MeV
α+n+n
-0.98 0+
6He
Cal.

42. E=0.07 MeV+1.13 MeV
The energy is measured
with respect to
α+Λ+n+n threshold.

43. 4.3 nb/sr
I propose to
experimentalists to
observe these states.
3.4 nb/sr
40% reduction
11.6 nb/sr
10.0 nb/sr
49.0 nb/sr
I think that
It is necessary to estimate
reaction cross section
7Li (e,e’K+).
Motoba’s Cal.
Motoba san recently
estimated differential cross
sections for each state.
At Elab=1.5 GeV　and θ=7 deg (E experimental kimenatics)

44. n n n n 6 H t Conclusion 7He α 3n Λ Λ n n Λ JLAB experiment-E011,
Phys. Rev. Lett. 110,
12502 (2013).
FINUDA collaboration & A. Gal,
Phys. Rev. Lett. 108,
(2012).
C. Rappold et al.,
HypHI collaboration
Phys. Rev. C 88,
(R) (2013)

45. Thank you!

46. (1) Charge Symmetry breaking
Energy difference comes from
dominantly Coulomb force
between 2 protons.
Charge symmetry breaking
effect is small.
In S=0 sector
Exp.
N+N+N
0 MeV
MeV
0.76 MeV
1/2+
MeV
3He
1/2+ 3H n p p n n p

47. Complex scaling is defined by the following transformation.
As a result, I should solve this Schroediner equation.
E=Er + iΓ/2