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Binding energies of few-body hypernulear systems and charge symmetry breaking effect

Binding energies of few-body hypernulear systems and charge symmetry breaking effect. E. Hiyama (RIKEN). Recently, at JLAB, the following experiments have been done and analysis is in progress. 　・　 7 Li(e, e’K + ) 7 He. Λ. 　・ 10 B (e, e’K + ) 10 Be. Λ.

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Binding energies of few-body hypernulear systems and charge symmetry breaking effect

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  1. Binding energies of few-body hypernulear systems and charge symmetry breaking effect E. Hiyama (RIKEN)

  2. Recently, at JLAB, the following experiments have been done and analysis is in progress. 　・　7Li(e, e’K+) 7He Λ 　・10B (e, e’K+) 10Be Λ To my opinion, these Λ hypernuclei are important for study of charge symmetry breaking effect. Here, I shall talk about the structure of 7He and 10Be. Λ Λ n n Λ n Λ α α α 10Be 7He Λ Λ

  3. Major goals of hypernuclear physics 1) To understand baryon-baryon interactions Fundamental and important for the study of nuclear physics 2) To study the structure of multi-strangeness systems Due to the difficulty of YN and YY 2-body scattering experiment for the study of baryon-baryon interaction, the systematic investigation of the structure oflight hypernuclei is essential.

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

  5. ΛN interaction(effectively including ΛN -ΣN coupling) Comparing the theoretical study using few-body technique and shell-model approach (done by Millener, Motoba) with the experimental results, we could succeed in extracting the information about ΛN interaction.

  6. In S= -1 sector, what are the open questions inYN interaction? • Charge symmetry breaking • (2) ΛN－ΣN coupling J-PARC : Day-1 experiment ・E13 “γ-ray spectroscopy of light hypernuclei” by Tamura and his collaborators 11B 4He Λ Λ ・E10 “Study on Λ-hypernuclei with the doubleCharge-Exchange reaction” by Sakaguchi , Fukuda and his collaboratiors 9He 6H Λ Λ

  7. (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

  8. Charge Symmetry breaking Exp. 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 4H 4He Λ Λ

  9. 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

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

  11. 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

  12. 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 4He Λ Λ If we consider that these experimental data are correct, why could not we reproduce the data?

  13. Talked by Nakamura, Yesterday According to Nakamura’s talk, the Λ separation energy of 4He is reliable. However, there is energy difference, about 0.2 MeV, between difference 2 modes in 4H. Λ Λ Then, for the detailed CSB study, we should perform experiment to confirm the Λ separation energy of 4H. Λ For this purpose, somewhere, it is necessary to perform ・・・ 4He (e, e’K+) 4H Λ Furthermore, we need more Λ hypernuclear data to get information on CSB.

  14. For this purpose, 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

  15. 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.

  16. 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 Λ 7Li (T=1) Λ 7He Λ

  17. 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)

  18. 7 Li Λ ΛN interaction: Nijmegen ’97f n Not original one but simulated one p Λ The ΛN-ΣN coupling interaction can be renomalized into the ΛN-ΛN interaction effectively. α VΛN=V0+σΛ・σNVs+(σΛ+σN)/2・VSLS+(σΛ-σN)/2・VALS Made by Yamamoto so as to reproduce the phase shifts given by the original one Strengths of Vs,VSLS,VALS are adjusted so as to reproduce of the observed data of 4H, 7Li(T=0), 9Be and 13C. Λ Λ Λ Λ

  19. 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?

  20. (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 preliminary 7Be Λ 7Li (T=1) Λ 7He Λ

  21. Next we introduce a phenomenological CSB potential with the central force component only. Strength, range are determined ao 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

  22. 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

  23. Comparing the data of A=4 and those of A=7, tendency of BΛ is opposite. How do we understand these difference?

  24. Why CSB interaction which reproduce the energy difference of A=4 hypernuclei, do not reproduce the energy difference in p-shell hypernuclei such as A=7 system? • In my calculation, ΛN-ΣN coupling effect is not included • explicitly and mass difference of Σ. (2) As talked by Dr. Nakamura, the binding energy of 4H is incorrect. Then, we should measure the binding energy of this hypernucleus again. If the experiment will be done successful, It might the energies of 4H and 4He are the same. Λ Λ Λ (3) odd-state CSB interaction whose contribution is negligible in A=4 hypernuclei, contribute to p-shell Λ hypernuclei with opposite sign of the even state of CSB interaction.

  25. p n n p p n Λ Λ Λ α α α 7He 7Be 7Li(T=1) Λ Λ Λ A=7Λ hypernuclei are p-shell nuclei. Then, it is possible that odd state CSB interaction contribute to those hypernuclei. Now, let me introduce a phenomenological odd state CSB interaction. Parameters are adjusted so as to reproduce the observed binding energy of 7He. Λ

  26. In order to check the validity of the odd CSB interaction, It is suited for study the structure of A=10 Λ hypernuclei such as 10B and 10Be. Λ Λ These are p-shell Λ hypernuclei. Λ Λ n p α α α α 10B 10Be Λ Λ Exp. BΛ=9.11±0.22 MeV Number of event (emulsion data): 3 Exp. BΛ=8.89±0.12 MeV Number of event (emulsion data): 10 At JLAB, the analysis is in progress.

  27. Λ Λ n p α α α α 10B 10Be Λ Λ Exp. BΛ=9.11±0.22 MeV Number of event (emulsion data): 3 Exp. BΛ=8.89±0.12 MeV Number of event (emulsion data): 10 At JLAB, the analysis is in progress. If the odd CSB interaction to reproduce the binding energy of 7He reproduce the binding energy of 10Be which will be reported soon, we can check the validity of odd state CSB interaction. Λ Λ

  28. (Exp: 1.54) 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= 5.18 →5.66 preliminary 7Be Λ 7Li (T=1) Λ 7He Λ

  29. Now, I shall show you the Λ separation energy of 10B and 10Be. Λ Λ • Results without CSB interaction • Results with CSB interaction Λ Λ n p α α α α 10B 10Be Λ Λ Exp. BΛ=9.11±0.22 MeV Number of event (emulsion data): 3 Exp. BΛ=8.89±0.12 MeV Number of event (emulsion data): 10 At JLAB, the analysis is in progress.

  30. Without CSB interaction Λ Λ p n α α α α 10B 10Be Λ Λ CAL:BΛ=8.76 MeV CAL:BΛ=8.56 MeV Exp. BΛ=9.11±0.22 MeV Number of event (emulsion data): 3 Exp. BΛ=8.89±0.12 MeV Number of event (emulsion data): 10 At JLAB, the analysis is in progress.

  31. With CSB interaction which reproduce the observed binding energy of 7He Λ Λ Λ p n α α α α 10B 10Be Λ Λ BΛ=8.56 MeV (without CSB) BΛ=8.76 MeV(without CSB) BΛ=8.35 MeV (with CSB) BΛ= 8.97 MeV (with CSB) Exp. BΛ=8.89±0.12 MeV Number of event (emulsion data): 10 Exp. BΛ=9.11±0.22 MeV Number of event (emulsion data): 3 If the calculated results are consistent with the observed data at JLAB in the future, we can extract information on the odd state of CSB interaction.

  32. Concluding remark For the study of CSB interaction, we need BΛ within 100keV accuracy. For this purpose, (e,e’K+) reaction might be powerful tool. And I calculated Λ separation binding energies of 7He, 7Li(T=1), 7Be, and 10Be and 10B. Λ Λ Λ Λ Λ For the study of CSB interaction, Maintz? JLAB? 4He (e,e’K+) 4H Λ I need the following data. Where? 10B(π+,K+) 10B Λ I hope to measure this hypernucleus somewhere.

  33. For detailed study of CSB effect in A=4 hypernuclei 4H Λ n Σ+ n p n Σ- n Σ0 + + + n Λ n n n p p p 4He realistic NN and YN interaction such as ESC08a Λ n Σ+ n p p Σ- n Σ0 + + + p p Λ n p p p p

  34. Thank you!

  35. Λ p We have one more interesting issue in 10B. α α Λ 10B Λ 3/2- α+α+p+Λ α+α+p p 2- ΔE<100keV? or 2- is ground state? α α γ 9B 1- We could not obtain observed γ-ray. What value in my calculation?

  36. ΛN interaction: Nijmegen ’97f Λ N Not original one but simulated one α α The ΛN-ΣN coupling interaction can be renomalized into the ΛN-ΛN interaction effectively. 10Be Λ VΛN=V0+σΛ・σNVs Strengths of even state and odd state spin-spin interactions are adjusted so as to reproduce of the observed data of 4H , 4He and 7Li(T=0). Λ Λ Λ Firstly, I shall show you the results without CSB interaction.

  37. Λ p 10B Λ α α 0.275 3/2- α+α+p+Λ 0 MeV α+α+p BΛ=8.825 MeV p 1- -8.25 α α 2- -8.33 2- 9B -8.28 BΛ=8.56 MeV 1- -8.55 even state odd state The same tendency in 10Be Λ

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