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Few-body structure of light hypernuclei

Few-body structure of light hypernuclei. Emiko Hiyama (RIKEN). Last year and this year, we had two 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,

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Few-body structure of light hypernuclei

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  1. Few-body structure of light hypernuclei Emiko Hiyama(RIKEN)

  2. Last year and this year, we had two 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). • Observation of Neutron-rich Λ-hypernuclei • These observations are interesting from the view points of • few-body physics as well as physics of unstable nuclei.

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

  4. Section 1 Introduction

  5. In neutron-rich and proton-rich nuclei, n n n n 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. Thereare many interesting phenomena in this field as you know. It was considered that nucleus was hardly compressed by the additional nucleons .

  6. Λ Nucleus Question:How is the structure modified when a hyperon, Λ particle, is injected into the nucleus?

  7. No Pauli principle Between N and Λ Λ particle can reach deep inside, and attract the surrounding nucleons towards the interior of the nucleus. Λ Hypernucleus Λ particle plays a ‘glue like role’ to produce a dynamical contraction of the core nucleus.

  8. Λ Λ X X r r’ N N Λ hypernucleus nucleus r > r’ B(E2) value: B(E2) ∝|<Ψi |r2 Y2μ(r) |Ψf>|2 Propotional to 4th-power of nuclear size Bando, Ikeda, Motoba If nucleus is really contracted by the glue-like role of the Λ particle, then E2 transition in the hypernuclei will become much reduced than the corresponding E2 transition in the core nucleus.

  9. E. Hiyama et al., Phys. Rev. C59 (1999), 2351 Theoretical calculation by Hiyama et al. B(E2: 5/2+→ 1/2+) =2.85 e2fm4 reduced by 22% α+ n+p α+Λ+n+p KEK-E419 Then, Tamura et al. Succeeded in measuring this B(E2) value to be 3.6 ±2.1 e2fm4 from 5/2+ to 1/2+ state in 7Li. 3+ B(E2) 7/2+ 1+ 5/2+ Exp.B(E2)= 9.3 e2fm4 B(E2) 6Li 3/2+ Λ n p 1/2+ 7Li n p α Λ Λ α By Comparing B(E2) of 6Li and that of 7Li, Λ The shrinkage effect on the nuclear size included by the Λ particle was confirmed for the first time.

  10. The glue like role of Λ particle provides us with another interesting phenomena. Λparticle can reach deep inside, and attracts the surrounding nucleons towards the interior of the nucleus. Λ Λ There is no Pauli Pricliple between N and Λ. Nucleus hypernucleus Λ Due to the attraction of ΛN interaction, the resultant hypernucleus will become more stable against the neutron decay. Neutron decay threshold γ nucleus hypernucleus

  11. 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 does the halo structure change when a Λ particle is injected into an unstable nucleus?

  12. 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).

  13. 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

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

  15. 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).

  16. 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

  17. 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 5/2+→1/2+ 3/2+ →1/2+ ・Useful for the study of the excitation mechanism in neutron-rich nuclei ・Helpful for the study of the excitation mechanism of the halo nucleus γ BΛ：EXP=5.68±0.03±0.25 1/2+ Observed at J-Lab experiment(2012) Phys. Rev. Lett.110, 012502 (2013). -6.19 In n-rich nuclei and n-rich hypernuclei, there will be many examples such as Combination of 6He and 7He. I hope that γ-ray spectroscopy of n-rich hypernuclei will be performed at J-PARC. Λ

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

  19. n n Will be presented by Prof. A. Feliciello tomorrow’s plenary session 6H t Λ Λ

  20. Phys. Rev. Lett. 108, 042051 (2012). Γ=1.9±0.4 MeV 1/2+ FINUDA experiment 1.7±0.3 MeV t+n+n+Λ t+n+n 5H 4H+n+n EXP: BΛ= 4.0±1.1 MeV Λ 0.3 MeV 6H Λ n n 5H : super heavy hydrogen t Λ

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

  22. Framework: To calculate the binding energy of 6H, it is very important to reproduce the observed data 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 Theoretical calculation N. B. Schul’gina et al., PRC62 (2000), 014321. 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. 5H is well described as t+n+n three-body model. t+n+n threshold

  23. Then, I think that t+n+n+Λ4-body model is good model to describe 6H. Then. I take t+Λ+n+n 4-body model. Λ Λ n n n n t t Λ 6H 5H Λ

  24. Before doing the 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 the energies of the 2- and 3-body subsystems. 6H 6H Λ Λ 6H n n Λ n n n n Λ Λ Λ t t t

  25. 6H I take the Λｔpotential to reproduce the binding energies of 0+ and 1+ states of 4H. In this case, ΛN－ΣN coupling is renormarized into the ΛN interaction. n Λ n Λ Λ t 6H Λ n n I take the ΛＮpotential to reproduce the binding energy of 3H. Λ Λ t Λ

  26. Γ=1.9±0.4 MeV  EXP Γ= 2.44MeV  CAL 1/2+ 1.69 MeV 1.7±0.3 MeV I focus on the 0+ state. t+n+n 1+ n n σn・σΛ 0+ Λ t 6H Λ 5H Then, what is the binding energy of 6H? Λ

  27. 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

  28. 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

  29. 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. Question: Is it possible to measure the energy and width of 5H more preciselysomewhereagain? Λ 1/2+ 1.7±0.3 MeV t+n+n • In our model, we do not include ΛＮーΣN coupling explicitly. • The coupling effect might contribute to the energy of 6H. Λ

  30. 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.

  31. 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

  32. 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. (to be reported by Prof. T. Fukuda in Parallel session C1, today) If E10 experiment reports more accurate energy, we can get information about ΛN-ΣN coupling. Λ Σ

  33. 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

  34. n n n p Λ Λ p p 4H 4He Λ Λ ΛN-ΣN coupling is related to YN charge symmetry breaking interaction. Then, if we can obtain information on ΛN-ΣN coupling, we can get information on YN CSB interaction simultaneously. Study of YN CSB interaction is interesting in hypernuclear physics. I will explain it.

  35. Charge Symmetry breaking In S=0 sector Energy difference comes from dominantly Coulomb force between 2 protons. Charge symmetry breaking effect is small. 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

  36. 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+ p n n n Λ p Λ p 4He 4H Λ Λ

  37. p n n n Λ p Λ 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 p n n n n n p + + Σ0 Σ0 p n Σ+ Σ+ n p

  38. 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.

  39. 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

  40. 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. How do we consider this situation?

  41. 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 JLAB and Mainz, it is planned to perform ・・・ Key experiment to get information about CSB. 4He (e, e’K+) 4H Λ

  42. If the binding energy of 4H is the same as that of 4He, • we find that we have no charge symmetry breaking effect • in S=-1 sector. Λ Λ (2) If the binding energy of 4H is quite different from that of 4He, we find that we have charge symmetry breaking (CSB) effect between Λn and Λp interaction. (It should be noted that CSB interaction in S=0 is very small.) We need the JLAB and Mainz experiment. And as a few-body physicist, we should calculate 4H and 4He using updated realistic YN interaction. Λ Λ JLAB and Mainz experiment 4He (e, e’K+) 4H Λ Possible new understanding Λ Λ

  43. Recently, we have updated YN interaction such as ESC08 which has been proposed by Nijmegen group. CSB interaction is included in this potential. Then, using this interaction directly, I performed four-body calculation of 4H and 4He. Λ Λ At present, I use AV8 NN interaction. Soon, I will use new type of Nijmegen NN potential. 4H 4He Λ Λ N N N N + Σ N Λ N

  44. Preliminary result (ESC08) 3He+Λ 0 MeV 3H+Λ 0 MeV CAL:- 0.45 (Exp: -1.00) CAL: -0.47 (Exp: -1.24) 1+ 1+ (Exp: 0.24 MeV) (cal: 0.01MeV) CAL: -1.44 (Exp: -2.04) 0+ CAL:-1.46 (Exp. -2.39) 0+ (Exp: 0.35 MeV) (cal. 0.02MeV n n p n Λ p Λ p 4H 4He Λ Λ Binding energies are less bound than the observed data. We need more updated YN interactions. N N N N + Σ N Λ N

  45. Section 5 Summary

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

  47. 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 is in progress. If they provide with more accurate data in the future and confirm to have a bound state, then we could obtain information about ΛN-ΣN coupling. Λ Λ

  48. ΛＮ－ΣN coupling is related to YN charge symmetry breaking effect. To obtain this information, four-body calculations of 4H and 4He is important. Then , four-body calculations of A=4 hypernuclei using realistic YN and NN interaction were performed. However, the results are not in good agreement with the data. we need more sophisticated YN interaction. And in the future, we have many experimental data for light Λ hypernuclei at J-PARC, Jlab and Mainz. So, now, it is good timing to study light Λ hypernuclei from view points of few-body problem. Λ Λ

  49. You are welcome to join us to research hypernuclear physics! Thank you!

  50. Section 4.2 Λ-Σ coupling and CSB in 7He Λ n n α Λ

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