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

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, 042051 (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.

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

　　　　 Section 1 Introduction

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 .

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?

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?

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.

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.

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

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?

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, 12502 (2013). Observed by FINUDA group, Phys. Rev. Lett. 108, 042051 (2012).

・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

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

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, 12502 (2013).

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, 054321 (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, 012502 (2013). 1/2+ -6.19

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, 054321 (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, 012502 (2013). -6.19

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

n n 6H t Λ Λ

n 5 H 6 H n 5H : super heavy hydrogen t t+n+n Γ=1.9±0.4 MeV Phys. Rev. Lett. 108, 042051 (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

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

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) 092501. Γ= 1.9±0.4 MeV 1/2+ 1.7±0.3 MeV Theoretical calculation N. B. Schul’gina et al., PRC62 (2000), 014321. 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

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

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

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

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

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

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, 042051 (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

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) 092501. 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

What is Λ N-ΣN coupling?

Non-strangeness nuclei Σ 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.

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?

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

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, 011301 (R) (2001). H. Nemura et al., Phys. Rev. Lett. 89, 142502 (2002). A. Nogga et al., Phys. Rev. Lett. 88, 172501 (2002).

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

PΣ=1.12 % PΣ=2.21%

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

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, 042051 (2012). 1/2+ FINUDA experiment 1.7±0.3 MeV t+n+n+Λ 0 MeV t+n+n 5 H Cal: -0.87 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

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

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

Preliminary result (ESC08) 3He+Λ 0 MeV 3H+Λ 0 MeV CAL:- 0.45 (Exp: -1.00) CAL: -0.47 (Exp: -1.24) 1+ 1+ (Exp: 0.24 MeV) (cal: 0.01 MeV) CAL: -1.44 (Exp: -2.04) 0+ CAL:-1.46 (Exp. -2.39) 0+ (Exp: 0.35 MeV) (cal. 0.02 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 -0.01 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

　　 Section 5 Summary

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

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

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

Thank you!

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

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).

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

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

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.

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:E01-011 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

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, 054321 (2009) 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?

6Be 6Li 6He 7Be 7Li 7He Without CSB EXP= 5.16 BΛ：CAL= 5.21 EXP= 5.26 JLAB:E01-011 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

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?

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

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

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