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Structure of neutron-rich Λ hypernuclei

Structure of neutron-rich Λ hypernuclei. E. Hiyama (RIKEN). Outline. n. n. Λ. 3 n. Λ. Λ. n. ４. n. 4 He. 7 He. Λ. Major goals of hypernuclear physics. 1) To understand baryon-baryon interactions. 2) To study the structure of multi-strangeness systems.

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Structure of neutron-rich Λ hypernuclei

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

  2. Outline n n Λ 3n Λ Λ n ４ n 4He 7He Λ

  3. Major goals of hypernuclear physics 1) To understand baryon-baryon interactions 2) To study the structure of multi-strangeness systems In orderto understand the baryon-baryon interaction,two-body scattering experiment is most useful. Study of NN intereaction has been developed. 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 since it is difficult to perform YN scattering experiment even at J-PARC.

  4. 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

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

  6. Question : How is structure change when a Λ particle is injected into neutron-rich nuclei? n n n n 6H 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

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

  8. What is interesting to study nnΛ system? n n 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 (Λhypernucear sector) n+p+Λ Lightest hypernucleus to have a bound state p n d+Λ 3H (hyper-triton) 0.13 MeV Λ J=1/2+ Λ Exp.

  9. -23.7fm nnΛbreakup 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.

  10. 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? nnΛbreakup 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?

  11. To take into account of Λparticle 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

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

  13. p n n n 4He Λ 4H Λ Λ Λ -BΛ p p 4He -BΛ 4H Λ 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+ 0+ -2.04 -2.39 -2.28 -2.33 Exp. Cal. Cal. Exp. What is binding energy of nnΛ?

  14. -BΛ 1/2+ nnΛ threshold n n 0 MeV Λ We have no bound state in nnΛ system. This is inconsistent with the data. 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 VTΛN-ΣN X1.1, 1.2

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

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

  17. 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

  18. n n unbound nn 0 MeV -0.066MeV -1.269 MeV 1S0X1.13 1S0X1.35 n n unbound unbound 0 MeV nnΛ 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) -13.93 (-13.23)MeV 1/2+ Exp. Cal. 1/2+ Cal. Cal.

  19. Summary of nnΛsystem: 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.

  20. Summary of nnΛsystem: 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 p n (e,e’K+)

  21. 7He submitted in PRC last week Λ n n α Λ 7He Λ

  22. n n 6He : One of the lightest n-rich nuclei 6He α 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).

  23. 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 Neutron halo states 5/2+ 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

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

  25. In S= -1 sector Exp. 3He+Λ 3H+Λ 0 MeV 0 MeV -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 4H 4He Λ Λ

  26. p n n n Λ Λ p p 4He 4H Λ Λ However, Λ particle has no charge. 4He 4H Λ Λ n p n n p p p p + + Λ Λ p p Σ- Σ- p n + + + + n n n n n n p p + + Σ0 n Σ0 p n Σ+ Σ+ p

  27. In order to explain the energy difference, 0.35 MeV, N N N N + N Λ N Σ (3N+Λ） (3N+Σ） ・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.

  28. 3He+Λ 0 MeV 3H+Λ 0 MeV -1.00 -1.24 1+ 1+ (Exp: 0.24 MeV) (cal: -0.01MeV(NSC97e)) -2.04 0+ -2.39 0+ (Exp: 0.35 MeV) (cal. 0.07MeV(NSC97e)) n n p n Λ p Λ p 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

  29. 0 MeV 3H+Λ 0 MeV 3He+Λ -1.00 -1.24 1+ 1+ (Exp: 0.24 MeV) (cal: -0.01MeV(NSC97e)) -2.04 0+ -2.39 0+ (Exp: 0.35 MeV) (cal. 0.07MeV(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.

  30. 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

  31. p n n p p n Λ Λ Λ α α α 7He 7Be 7Li(T=1) Λ Λ Λ Then, A=7Λ 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.

  32. Exp. 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 -3.79 7Be 7Li (T=1) Λ Λ 7He Λ

  33. Important issue: Can we describe the Λ binding energy of 7He observed at JLAB usingΛNinteraction 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 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)

  34. 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?

  35. (Exp: 1.54) Without CSB 6Be (Exp: -0.14) (exp:-0.98) 6Li 6He (T=1) EXP= 5.16 BΛ：CAL= 5.21 EXP= 5.26 BΛ：CAL= 5.28 BΛ：EXP= 5.68±0.03±022 JLAB:E01-011 experiment CAL= 5.36 7Be Λ 7Li (T=1) Λ 7He Λ

  36. 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?

  37. 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

  38. With CSB 5.28 MeV( withourt CSB) 5.21 (without CSB) 5.44(with CSB) 5.29 MeV (With CSB) 5.36(without CSB) 5.16(with CSB) BΛ：EXP= 5.68±0.03±0.22 Inconsistent with the data p n α

  39. 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?

  40. We get binding energy by decay π spectroscopy. decay 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

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

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

  43. CAL: E. Hiyama et al., PRC 53, 2075 (1996), PRC 80, 054321 (2009) 6He Another interestingissue 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 Neutron halo states 5/2+ 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

  44. 7Li(e,e’K+)7ΛHe 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

  45. 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.

  46. 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 1.797 MeV 2+ 1.8 MeV 2+ 1 0 MeV α+n+n 0 MeV α+n+n -0.98 -0.98 0+ 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

  47. 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. 0.8 MeV 2+ 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

  48. Question: What are theoretical results? Γ=1.6 ±0.4 MeV 2.6±0.3 MeV 2+ 2 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 1.8 MeV 2+ 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

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

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

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