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

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, 042051 (2012). Λ C. Rappold et al., HypHI collaboration Phys. Rev. C 88, 041001 (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, 061302 (R), (2014).

n n 7He Λ α Λ

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

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

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.

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

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

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

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

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, 011301(R) (2001). ・A. Nogga, H. Kamada and W. Gloeckle, Phys. Rev. Lett. 88, 172501 (2002) ・H. Nemura. Y. Akaishi and Y. Suzuki, Phys. Rev. Lett.89, 142504 (2002). Coulomb potentials between charged particles (p, Σ±) are included.

4H 4He 3He+Λ 3H+Λ + 0 MeV 0 MeV -1.00 -1.24 1+ 1+ (Exp: 0.24 MeV) (cal: -0.01 MeV(NSC97e)) -2.04 0+ -2.39 (Exp: 0.35 MeV) 0+ (cal. 0.07 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 -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. 4H ・A. Nogga, H. Kamada and W. Gloeckle, Phys. Rev. Lett. 88, 172501 (2002) 4He Λ Λ ・E. Hiyama, M. Kamimura, T. Motoba, T. Yamada and Y. Yamamoto, Phys. Rev. C65, 011301(R) (2001). ・H. Nemura. Y. Akaishi and Y. Suzuki, Phys. Rev. Lett.89, 142504 (2002). N N N N + Σ N Λ N

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: -0.01 MeV(NSC97e)) -2.04 0+ -2.39 (Exp: 0.35 MeV) 0+ (cal. 0.07 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.

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

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.

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 JLabE05-115 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

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, 054321 (2009)

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 JLAB:E05-115 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

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 5.29 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

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.

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, 232501 (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

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

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

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 - 7.72 MeV 0.76 MeV 1/2+ Question: Why we have large CSB effect in A=4 hypernuclei? I cannot understand this physics. - 8.48 MeV 3He 1/2+ 3H n p p n n p

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

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

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

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.

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

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, 054321 (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, 012502 (2013). -6.19 MeV

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

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.

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

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, 064306 (2011). Data in 2012 X. Mougeot et al., Phys. Lett. B 718 (2012) 441. p(8He, t)6He

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

My result E=0.96 + 0.14 MeV E=2.81 +4.81 MeV

Exp. 6He 6He Cal. Question: What are theoretical results? Γ=0.12 MeV 2+ 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

6He Cal. How about energy spectra of 7He? Γ=0.14 MeV 2+ 0 MeV α+n+n Λ 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.

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

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 (E05-115 experimental kimenatics)

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, 042051 (2012). C. Rappold et al., HypHI collaboration Phys. Rev. C 88, 041001 (R) (2013)

Thank you!

(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 - 7.72 MeV 0.76 MeV 1/2+ - 8.48 MeV 3He 1/2+ 3H n p p n n p

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