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Structure of S=-1 systems

Structure of S=-1 systems. Emiko Hiyama (RIKEN). Recently, we had three epoch-making data from the view point of few-body problems. Then, in hypernuclear physics, we are so excited. n. n. n. n. 7 He. 6 H. Λ. α. Λ. t. Λ. Λ. JLAB experiment-E011, Phys. Rev. Lett. 110 ,

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Structure of S=-1 systems

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  1. Structure of S=-1 systems Emiko Hiyama(RIKEN)

  2. Recently, we had three epoch-making data from the view point of few-body problems. Then, in hypernuclear physics, we are so excited. n n n n 7He 6H 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). n n Λ C. Rappold et al., HypHI collaboration Phys. Rev. C 88, 041001 (R) (2013) 3n Λ

  3. OUTLINE 7He n n Λ α Λ • Introduction 2. 3. 4. 5. Summary 7He Λ n n 6H 6H Λ Λ t Λ 3n Λ n n Λ 3n Λ

  4. Section 1 Introduction

  5. 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.

  6. 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

  7. 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 ?

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

  9. 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

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

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

  12. 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+ Halo states Exp:-0.98 3/2+ -4.57 BΛ: CAL= 5.36 MeV γ BΛ：EXP=5.68±0.03±0.25 1/2+ Jlab experiment Phys. Rev. Lett.110, 012502 (2013). -6.19

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

  14. n n 6H Λ Λ t

  15. 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 EXP:BΛ=4.0±1.1 MeV 4H+n+n Λ n n 0.3 MeV t n n 6H Λ Λ t 5H:super heavy hydrogen

  16. Before experiment, the following authors calculated the binding energies by 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. Motivating the experimental data, I calculated the binding energy of 6H and I shall show you my result. Λ

  17. Before doing 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 in the 2- and 3-body subsystems. 6H 6H Λ Λ 6H n n Λ n n n n Λ Λ Λ t t t Among the subsystems, it is extremely important to adjust the energy of 5H core nucleus.

  18. Framework: To calculate the binding energy of 6H, it is very important to reproduce the binding energy 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 To reproduce the data, for example, 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. The calculated binding energy for the ground state of 5H is 1.6 MeV with respect to t+n+n threshold and width has 1.5 MeV. t+n+n

  19. 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 1.69MeV 1.17 MeV 0 MeV t+n+n+Λ Γ=0.23 MeV t+n+n E=-0.87MeV 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

  20. 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 5H EXP:BΛ=4.0±1.1 MeV 4H+n+n Λ t+n+n+Λ n n 0.3 MeV t 6H Λ n n 5H:super heavy hydrogen t Λ But, FINUDA group provided the bound state of 6H. Λ

  21. 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 to measure the energy and width of 5H more preciselyat RCNP by Prof. Tanihatathis year. His proposal was approved recently . Λ 1/2+ 1.7±0.3 MeV t+n+n

  22. We cited this experiment. However, you have many different decay widths. Width is strongly related to the size of wavefunction. Then, I hope that The decay width will be determined by Tanihata san in the future. [3] A.A. Korosheninnikov et al., PRL87 (2001) 092501 [8] S.I. Sidorchuk et al., NPA719 (2003) 13 [4] M.S. Golovkov et al. PRC 72 (2005) 064612 [5] G. M. Ter-Akopian et al., Eur. Phys. J A25 (2005) 315.

  23. In our model, we do not include ΛＮーΣN coupling explicitly. The coupling effect might contribute to the energy of 6H. Λ

  24. Non-strangeness nuclei S=-1 Σ Δ N 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 Δin nuclei is not large. Then, in S=-1 and S=-2 system, ΛN-ΣN and ΛΛ-ΞN couplings might be important.

  25. Gal and D. J. Millener, arXiv:1305.6716v3 (To be published in PLB. 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

  26. 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 0 MeV t+n+n 5H 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

  27. t+n+n+Λ No peak?! 4H+n+n Λ 0.3 MeV FINUDA data 6H Λ Theoretically, we might understand by the following reason. If the state is resonant state, the reaction cross section would be much smaller than that we expect. => I should calculate reaction cross section 6Li (π,K) 6H with Harada’s help. Λ

  28. How should we understand 6H issue? To investigate it, the study of nnΛ system is suited. Λ Section 4 three-body calculation of 3n Λ E. Hiyama, S. Ohnishi, B.F. Gibson, and T. A. Rijken, The paper will be submitted in PRC soon. n n Λ

  29. nnΛbreakup threshold ? They did not report the binding energy. HypHI collaboration observed bound state of nnΛ system. It is considered that ΛN-ΣN coupling play an important role to make bound state. n n Λ

  30. t If we inject triton cluster into nnΛsystem, we have 6H. Λ n n Λ n n Λ t The three-body system of nnΛ system is useful hyperucleus for discussion about whether 6H is bound or unbound. Namely, if we have a bound state in nnΛ system, there is possibility to have a bound state in 6H. Λ Λ

  31. Important issue: Do we have bound state for nnΛ system? If we have a bound state for this system, how much is binding energy? n n n n + Λ Σ NN interaction : to reproduce the observed binding energies of triton and 3He NN: AV8 potential

  32. In S=0 sector N+N+N 0 MeV 1/2+ - 7.72 MeV -7.12 MeV 1/2+ 1/2+ -7.77 MeV Cal. 3He - 8.48 MeV 1/2+ Cal. 3H n p p n n p Exp. Exp.

  33. Important issue: Do we have bound state for nnΛ system? If we have a bound state for this system, how much is binding energy? n n n n + Λ Σ YN interaction: to reproduce the observed binding energies of 3H, 4H and 4He NSC97f potential Λ Λ Λ

  34. 4He N N Λ 4H What is binding energy for nnΛ? Λ -BΛ 4He Λ -BΛ N 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. p N d+Λ 3H Λ Λ 1/2+ 1/2+ -0.19 MeV -0.13 ±0.05 MeV Cal. Exp.

  35. 1/2+ nnΛ We have no bound state in nnΛ system. Now, we have a question. Do we have a possibility to have a bound state in nnΛ system tuning strength of YN potential ? When we have a bound state in nnΛ system, what are binding energies of 3H and A=4 hypernuclei? Λ VTΛN-ΣN X1.1, 1.2

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

  37. We have no possibility to have a bound state in nnΛ system. And then, it would be difficult to say to have a bound state in 6H. Λ

  38. 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). n n T=1, 1S0 state I multiply component od 1S0 state by 1.13 and 1.15. What is the binding energies of nnΛ？ Λ

  39. n n unbound 0 MeV nn -0.066MeV -0.118 MeV 1S0X1.13 1S0X1.15 n n unbound unbound nnΛ Λ 1/2+ -0.043MeV 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) -10.09 (-9.38)MeV 1/2+ Exp. Cal. 1/2+ Cal. Cal.

  40. Section 5 Summary

  41. Summary 1) Motivated by observation of neutron-richΛhypernuclei, 7He,6H , nnΛ, 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

  42. 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 reported to have no peak. Λ Λ To study 6H issue, I calculated nnΛ system taking ΛN-ΣN coupling explicitly. However, I could not find any bound state keeping consistency with observed data of 3H, A=4 hypernuclei, 3H and 3He. This fact would suggest to have no bound state in 6H. Λ n n Λ Λ Λ

  43. To conclude whether or not 6H has a bound state, it think that it would be better to perform search experiment at Mainz. I hope to have conclusion for this issue in the future. Λ

  44. Thank you!

  45. 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 the ΛN-ΣN coupling? Λ Answer: 4H, 4He Λ Λ n n n p Λ Λ p p 4H 4He Λ Λ

  46. -BΛ -BΛ 4He 4H Λ Λ 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

  47. 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).

  48. PΣ=1.12 % PΣ=2.21%

  49. Another interesting role of Σ-particle in hypernuciei, namely effective ΛNN 3-body force generated by the Σ-particle mixing. N1 N2 N3 Λ ② N1 N2 N3 Λ ① Σ Σ N1 N2 N3 Λ N1 N2 N3 Λ 3N+Λ space N1 N2 N3 N1 N2 N3 Λ Λ Effective 3-body ΛNN force Effective 2-body ΛN force How large is the 3-body effect? N1 N2 N3 N1 N2 N3 Λ Λ

  50. 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).

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