Structure of neutron-rich Λ hypernuclei E. Hiyama (RIKEN)

Outline n n Λ 3n Λ n Λ ４ n 4He 7He Λ

Major goals of hypernuclear physics 1) To understand baryon-baryon interactions 2) To study the structure of multi-strangeness systems In order to understand the baryon-baryon interaction, two-body scattering experiment is most useful. Total number of Nucleon (N) -Nucleon (N) data: 4,000 Study of NN intereaction has been developed. 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 since it is difficult to perform YN scattering experiment even at J-PARC.

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

Nuclear chart with strangeness Multi-strangeness system such as Neutron star Extending drip-line! Λ Interesting phenomena concerning the neutron halo have　been observed near the neutron drip line of light nuclei. Outline How is structure change when a Λ　particle is injected into neutron-rich nuclei ? 5

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 by FINUDA group, Phys. Rev. Lett. 108, 042051 (2012). Observed at JLAB, Phys. Rev. Lett. 110, 12502 (2013). n n Λ C. Rappold et al., HypHI collaboration Phys. Rev. C 88, 041001 (R) (2013) 3n Λ

3n three-body calculation of E. Hiyama, S. Ohnishi, Λ E. Hiyama, S. Ohnishi, B.F. Gibson, and T. A. Rijken, PRC89, 061302(R) (2014). n n 　Λ

What is interesting to study nnΛ system? 　Λ S=0 The lightest nucleus to have a bound state is deuteron. n+p threshold n p J=1+ d -2.22 MeV Exp. S=-1 (Λ　hypernucear sector) n+p+Λ Lightest hypernucleus to have a bound state n p d+Λ 3H (hyper-triton) 0.13 MeV Λ J=1/2+ 　Λ Exp.

Observation of nnΛ system (2013) -23.7fm nnΛ　breakup threshold n n ? They did not report the binding energy. 　Λ scattering length:-2.68fm Observation of nnΛ system (2013) Lightest hypernucleus to have a bound state Any two-body systems are unbound.=>nnΛ system is bound. Lightest Borromean system.

Theoretical important issue: Do we have bound state for nnΛ system? If we have a bound state for this system, how much is binding energy? nnΛ　breakup threshold n n ? They did not report the binding energy. 　Λ NN interaction : to reproduce the observed binding energies of 3H and 3He NN: AV8 potential We do not include 3-body force for nuclear sector. How about YN interaction?

To take into account of Λ particle to be converted into Σ particle, we should perform below calculation using realistic hyperon(Y)-nucleon(N) interaction. n n n n + 　Λ Σ YN interaction: Nijmegen soft core ‘97f potential (NSC97f) proposed by Nijmegen group reproduce the observed binding energies of 3H, 4H and 4He Λ Λ Λ

-BΛ 0 MeV d+Λ p n 3H Λ Λ 1/2+ 1/2+ -0.19 MeV -0.13 ±0.05 MeV Cal. Exp.

What is binding energy of nnΛ? p n n 4He n Λ Λ 4H Λ -BΛ p Λ p 4He -BΛ Λ 4H Λ Λ 3He+Λ 0 MeV 0 MeV 3H+Λ 1+ 1+ 1+ 1+ -0.57 -0.54 -1.24 -1.00 0+ 0+ 0+ -2.39 -2.04 0+ -2.28 -2.33 Exp. Cal. Cal. Exp. What is binding energy of nnΛ?

Λ VT ΛN-ΣN -BΛ 0 MeV We have no bound state in nnΛ system. 1/2+ nnΛ threshold We have no bound state in nnΛ system. This is inconsistent with the data. n n 0 MeV 　Λ Now, we have a question. Do we have a possibility to have a bound state in nnΛ system tuning strength of YN potential ? It should be noted to maintain consistency with the binding energies of 3H and 4H and 4He. Λ Λ Λ VT ΛN-ΣN X1.1, 1.2

n n 　Λ VT ΛN-ΣN VT ΛN-ΣN X1.1 X1.2 When we have a bound state in nnΛ system, what are binding energies of 3H and A=4 hypernuclei? Λ

Λ n Λ n p We have no possibility to have a bound state in nnΛ system. 　Λ We have no possibility to have a bound state in nnΛ system. n Λ n p

Question: If we tune 1S0 state of nn interaction, Do we have a possibility to have a bound state in nnΛ? In this case, the binding energies of 3H and 3He reproduce the observed data? Some authors pointed out to have dineutron bound state in nn system. Ex. H. Witala and W. Gloeckle, Phys. Rev. C85, 064003 (2012). T=1, 1S0 state I multiply component of 1S0 state by 1.13 and 1.35. What is the binding energies of nnΛ？ n n 　Λ

Λ n n unbound nn 0 MeV -0.066MeV -1.269 MeV 1S0X1.13 1S0X1.35 n n 　Λ 1/2+ -1.272 MeV We do not find any possibility to have a bound state in nnΛ. N+N+N 3H (3He) -7.77(-7.12) -8.48 (-7.72) -9.75 (-9.05) 1/2+ -13.93 (-13.23)MeV Exp. Cal. Cal. 1/2+ Cal.

Summary of nnΛ　system: Motivated by the reported observation of data suggesting a bound state nnΛ, we have calculated the binding energy of this hyperucleus taking into account ΛN-ΣN explicitly. We did not find any possibility to have a bound state in this system. ‘Nonexistence of a Λnn bound state’ H. Garcilazo and A. Valcarce, Phys. Rev. C 89, 057001 (2014). They did not find any bound state in nnΛ system. However, the experimentally they reported evidence for a bound state. As long as we believe the data, we should consider additional missing elements in the present calculation. But, I have no idea. Experimentally, they did not report any binding energy.

3n 3H Summary of nnΛ system: It might be good idea to perform search experiment of nnΛ system at Jlab to conclude whether or not the system exists as bound state experimentally. Or I hope to perform search experiment at GSI again in the Future. n n n n 3H 3n Λ Λ n p (e,e’K+)

7He submitted in PRC last week Λ n n α Λ 7He Λ 22 22 22

n n n n 6He α 7He α 6He : One of the lightest n-rich nuclei Λ n n 7He: One of the lightest n-rich hypernuclei Λ 7He Λ α Λ Observed at JLAB, Phys. Rev. Lett. 110, 12502 (2013).

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 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

The ground state of 7He is important to study of CSB interaction between Λn and Λp. 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 p Λ p 4H 4He Λ Λ

4He 4H 4He 4H p n n n Λ Λ p p However, Λ particle has no charge. n p n + + 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. Λ Λ 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+ -2.04 0+ -2.39 0+ n (Exp: 0.24 MeV) (cal: -0.01 MeV(NSC97e)) -2.04 0+ -2.39 0+ (Exp: 0.35 MeV) (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 0 MeV 0 MeV -1.00 -1.24 1+ 1+ -2.04 0+ -2.39 0+ n n (Exp: 0.24 MeV) (cal: -0.01 MeV(NSC97e)) -2.04 0+ -2.39 0+ (Exp: 0.35 MeV) (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 JLAB:E01-011 experiment Preliminary data: 5.68±0.03±0.22 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 EXP= 5.26 JLAB:E01-011 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 5.21 (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

How do we understand these difference? Let me compare with the experimental data of A=4 hypernuclei and data of A=7 hypernuclei. Comparing the data of A=4, and those of A=7 and A=10, tendency of BΛ is opposite. How do we understand these difference?

We get binding energy by decay π spectroscopy. π-+1H+3He →2.42 ±0.05 MeV 4He Λ π-+1H+1H+2H → 2.44 ±0.09 MeV Total: 2.42 ±0.04 MeV Then, binding energy of 4He is reliable. Λ decay π-+1H+3H →2.14 ±0.07 MeV Two different modes give 0.22 MeV 4H Λ π-+2H+2H → 1.92 ±0.12 MeV Total: 2.08 ±0.06 MeV This value is so large to discuss CSB effect. Then, for the detailed CSB study, we should perform experiment to confirm the Λ separation energy of 4H. Λ For this purpose, at Mainz, they performed ・・・ 　4He (e, e’K+) 4H Λ

Preliminary data 4ΛH Preliminary 0+ state Data was taken by Tohoku Univ. Group (S. Nagao et atl.) at Mainz.

A=4 H He p n Λ p n Λ 4 Λ 4 Λ 3H + Λ -BΛ (MeV) -1.00 0+ 1+ -2.04 -2.39 -1.24 3He + Λ JLab (e,e’K+) J-PARC E13 ΔBΛ=0.28 MeV! still we have CSB effect. Preliminary BΛ = 2.11±0.03(stat.)±0.09(syst.)

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 At present, due to poor statics, It is difficult to have the third peak. Theoretically, is it possible to have new state? Let’s consider it. (FWHM = 1.3 MeV) Fitting results 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+ 0 MeV How about theoretical result? Γ=1.6 ±0.4 MeV 1.6±0.3 MeV 2+ 2 2.5 MeV Γ= Γ=0.12 MeV Decay with Is smaller than Calculated with. 2+ 0.8 MeV Γ= 1 0 MeV α+n+n 0.79 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

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

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+ 2 2 Γ=0.12 MeV Γ=0.14 MeV 2+ 0.8 MeV 2+ 0.8 MeV 1 1 0 MeV α+n+n 0 MeV α+n+n -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. (preliminary) Motoba san recently estimated differential cross sections for each state. At Elab=1.5 GeV　and θ=7 deg (E05-115 experimental kimenatics)

7Li(e,e’K+)7ΛHe (FWHM = 1.3 MeV) Fitting results At present, due to poor statics, It is difficult to have the third peak. But, I hope that next experiment at Jlab will observe the third peak. (FWHM = 1.3 MeV) Fitting results Good agreement with my prediction Third peak???

Neutron-rich Λ hypernuclei Summary n n Λ n Λ ４ n 4He 3n 7He Λ Λ Neutron-rich Λ　hypernuclei

Thank you!