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Structure of four-baryon systems

Structure of four-baryon systems. E. Hiyama (RIKEN). One of the important and interesting subjects to study Four-baryon systems based on four-body systems. n. Λ. n. n. n. Λ. Λ. Λ. p. p. n. n. p. p. p. p. 4 He. 4 He. 4 H. 4 H. ΛΛ. Λ. Λ.

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Structure of four-baryon systems

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  1. Structure of four-baryon systems E. Hiyama (RIKEN)

  2. One of the important and interesting subjects to study Four-baryon systems based on four-body systems n Λ n n n Λ Λ Λ p p n n p p p p 4He 4He 4H 4H ΛΛ Λ Λ • Why is it important to study these four-body systems? • What kind of new understandings do we obtain by solving these systems as four-body problems?

  3. Our few-body caluclational method 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: Exsotic atoms / molecules , 3- and 4-nucleon systems, multi-cluster structure of light nuclei, Light hypernuclei, 3-quark systems, An example of the accuracy of the method:

  4. r r 2 R R 3 1 R 2 1 r 3 C=2 C=1 C=3

  5. r r 2 R R 3 1 R 2 1 r 3 C=2 C=1 C=3 Basis functions of each Jacobi coordinate CNL,lm Determined by diagonalizing H

  6. An important issue of the variational method is how to select a good set of basis functions. What is good set of basis functions? • To describe short-ranged correlation and long-range tail behaviour, highly oscillatory character of few-body wave functions, etc. (2) Easily to calculate the matrix elements of Hamiltonian Hin= <Φi | H | Φn >, Nin = <Φi | 1 | Φｎ>

  7. For this purpose, we use the following basis function: Φlmn(r) νn=(1/rn )2 rn=r1an-1 (n=1-nmax) The Gaussian basis function is suitable not only for the calculation of the matrix elements but also for describing short-ranged correlations, long-ranged tail behaviour.

  8. We calculate the energy and overlap matrix elements. Hin= <Φi | H | Φn > Nin = <Φi | 1 | Φｎ> Next, by solving eigenstate problem, we get eigenenergy E and unknown coefficientsCn . ( Hin) - E ( Ni n ) Cn =0

  9. Gauss Expansion Method apply Four-body calculation A good example about the accuracy of the method n Λ n n n Λ Λ Λ p p n n p p p p 4He 4He 4H 4H ΛΛ Λ Λ

  10. Benchmark-test calculation of the 4-nucleon bound state PRC64, 044001(2001) 4He 1. Faddeev-Yakubovski (Kamada et al.) 2. Gaussian Expansion Method (Kamimura and Hiyama et al.) 3. Stochastic varitional (Varga et al.) 4. Hyperspherical variational (Viviani et al.) 5. Green Function Variational Monte Carlo (Carlson at al.) 6. Non-Core shell model (Navratil et al.) 7. Effective Interaction Hypershperical HarmonicsEIHH (Barnea et al.) n n p p 4-nucleon bound state NN:AV8 We have powerful method to solve four-body bound state. Good agreement among the 7 different method in the binding energy, r.m.s. radius and two-body correlation function

  11. One of the next challenging projects: to calculate properties of the second 0+ state of 4He using realistic force (1) In investigating the spatial structure of the 0+2 state in 4He, it is very important to explain the observed (e,e’) transition form factor, to reproduce inelastic electron scattering data . (2) By the analysis of the low-momentum-transfer part of the form factor Second 0+: 11 % of the energy-weighted E0 sum rule limit medium-heavy or heavy nuclei: 80-90 % of the limit collective, breathing mode

  12. Where is the major component of the sum rule value Situated? H. M. Hofmann and G. M. Hale, Nucl. Phys. A613, 69 (1997). A. Cstoto and G. M. Hale, Phys. Rev. C55, 2366 (1997). P. Navratil and B. R. Barrett, Phys. Rev. C59, 1906 (1999). • C. Fonseca, G. Hale, and J. Haidenbauer, Few-Body Syst. 31, 139 (2002). within the limited model space omitted Coulomb force

  13. Use of AV8’ + Coulomb force + phenomenological 3-body force well reproduces simultaneously B.E.(3H ) : -8.42 MeV (CAL) -8.48 (EXP) B.E.(3He) : -7.74 MeV (CAL) -7.77 (EXP) B.E.( 4He) :-28.44 MeV (CAL) -28.30 (EXP) r.m.s. charge radius of 4He (0+ ) : 1.658 fm (CAL) 1.671±0.014 fm (EXP) 1 1 How about the second 0+ state?

  14. Answer to question 1): Energy of 0+of 4He 2 3N+N 20.36 3He+n 20.58 20.31 0+ 0+ 20.21 2 2 3H+p 19.82 0+ 0+ 0.0 MeV 0.0 MeV 1 (-28.30 MeV) 1 (-28.44) Ex Ex CAL EXP 4He 4He

  15. mass density The density of the second 0+ state in the internal region is so much lower than that of the ground state. transition density The Fourier transformation of the transition density gives the form factor of the inelastic electron scattering.

  16. Answer to question 2): 0+2 (e,e’) 0+1 4He Hiyama, Gibson, Kamimura Phys. Rev. C70 (2004) 031001(R). We understand that our method is suitable for describing both compactly bound states andloosely coupled states as well as transitions between them.

  17. Energy-weightedE0 sum rule in 4He E0 strength of the second 0+state, was derived from the data: 0+2 =1.10 ± 0.16 fm2 (EXP), 0+1 4He But, this contribution from the second 0+ state exhausts only 11 % of the sum rule limit. The second 0+ state is not a collective mode. Where is the major component of the E0 sum rule limit situated? ( This has been a long-standing puzzle of 4He nucleus since 1970’s. )

  18. Energy-weighted E0 sum rule: precisely-discretized non-resonant continuum states We solved this puzzle as following: （～ 20,000 ) • Using the 4-body Gaussian basis functions, we diagonalized the total Hamiltonian. 2) Since we employed ～20,000 basis functions, we obtained the same number of 4-body eigenstates. Above the second 0+ state, we have presicely-discretized non-resonant continuum states. 3) We calculated the energy-weighted E0 strength to all the discretized eigenstates, and found the following fact: 0+2 0+1 4He

  19. Energy-weighted E0 sum rule: Ex (13 %) 60 MeV Contribution from these continuum states amounts to 70 % of the sum rule limit. 200 discretized- continuum states. (70 %) 20.2 MeV (17 %) 0+2 Answer to question 3): 0 MeV 0+1 a major part of the energy-weighted E0 sum rule is exhausted by non-resonant, low-energy continuum states. 4He (CAL)

  20. Λ n Λ N N Replace N N p p 4He 4He Λ ΛＮーΣＮ　coupling effect in single Λhypernuclei n Λ n p In non-strangeness nuclei, NN-NΔcoupling 4H Λ

  21. 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.

  22. Interesting Issues for the ΛN-ΣN particle conversion in hypernuclei • How large is the mixing probability of the Σ particle in the hypernuclei? (2) How important is the ΛＮーΣＮ coupling in the binding energy of the Λ hypernuclei? (3) How large is the Σ-excitation as effective three-body ΛNN force?

  23. -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

  24. 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).

  25. 8 ΨJM(A=4)=ΣΦc(rc,Rc,ρc) N=1

  26. 4He, 4H Λ Λ VNN : AV8 potential VYN : Nijmegen soft-core ’97f potential

  27. PΣ=1.12 % PΣ=2.21%

  28. 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

  29. 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).

  30. To summarize the part of A=4 hypernuclei Σ N Λ N + N N N N + NNNΛ NNNΣ (1)NNNΣchannel is essentially important to make A=4 hypernuclei. (2)ΛN-ΣN coupling as a three-body force is important in 0+ state.

  31. Furthermore to study ΛN-ΣN coupling, we need updated YN interaction. For this purpose, 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

  32. 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

  33. In A=4 hypernuclei, we have one more interesting issue: YN charge symmetry breaking effect

  34. 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

  35. 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

  36. 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

  37. 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.

  38. 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

  39. 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. How do we consider this inconsistency? N N N N + Σ N Λ N

  40. We get binding energy by emulsion data. 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

  41. Λ Λ Λ N n Λ p N N 4H replace ΛΛ Interesting issue: ΛΛーΞN coupling

  42. ΛΛ-ΞN coupling One of the major goals in hypernuclear physics To study structure of multi-strangeness systems (extreme limit : neutron star) N N N Multi-strangeness systems N N N N Λ Λ Λ Λ Λ ΞN 25MeV ΛΛ threshold energy difference is very small! It is considered that ΛΛ→ΞN particle conversion is strong in multi-strangeness system.

  43. Effect of ΛΛ－ΞN coupling is small in 6He which was observed as NAGARAevent. ΛΛ Pauli Forbidden V ΛΛ--ΞN P3/2 Ξ- Λ Λ n n p p p S1/2 (3 protons inS1/2) 6He ΛΛ ・I.R. Afnan and B.F. Gibson, Phys. Rev. C67, 017001 (2003). ・Khin Swe Myint, S. Shinmura and Y. Akaishi, nucl-th/029090. ・T. Yamada and C. Nakamoto, Phys. Rev.C62, 034319 (2000).

  44. For the study of ΛΛ－ΞN coupling interaction, s-shell double Λ hypernuclei such as 4H and 5H ( 5He) are very suitable. ΛΛ ΛΛ ΛΛ Λ Λ Λ Λ 4H 5H ( 5He) ΛΛ n p n p ΛΛ ΛΛ n ・I.N. Filikhin and A. Gal, Phys. Rev. Lett. 89, 172502(2002) ・Khin Swe Myint, S. Shinmura and Y. Akaishi, Eur. Phys. J.A16, 21(2003). ・D. E. Lanscoy and Y. Yamamoto, Phys. Rev. C69, 014303(2004). ・H. Nemura, S. Shinmura, Y. Akaishi and Khin Swe Myint, Phys. Rev. Lett. 94, 202502 (2005).

  45. 4H Λ Λ ΛΛ n p V ΛΛ－ΞN P3/2 Ξ- Λ Λ n p p S1/2 4H (2 protons inS1/2) (No data) ΛΛ Due to NO Pauli plocking, the ΛΛ－ΞN coupling can be large in 4H ΛΛ B.F. Gibson, I.R. Afnan, J.A.Carlson and D.R.Lehman, Prog. Theor. Phys. Suppl. 117, 339 (1994).

  46. p n 4H ΛΛ Λ Λ The important issue: Does the YY interaction which designed to reproduce the binding energy of 6He make 4H bound? And how does the effect of ΛΛーΞN coupling play important role in the binding energy of 6He and 4H? ΛΛ ΛΛ ΛΛ ΛΛ

  47. 1)I.N. Filikhin and A. Gal, Phys. Rev. Lett. 89, 172502 (2002) 2)H. Nemura, Y. Akaishi et al., Phys. Rev. C67, 051001 (2002) 6He Λ n Λ Λ ΛΛ VΛΛ Λ p α NAGARA event NOT BOUND ! α+Λ+Λ 4H ΛΛ 7.25±0.1 MeV 0+

  48. Did not include ΛΛ-ΞN coupling ΛΛ-ΞNcoupling => ・ significant in 4H ΛΛ n p 4H ΛΛ Λ Λ ・Not so important in6He ΛΛ Λ Λ 6He ΛΛ α

  49. N N n p + Ξ Λ Λ N 4H ΛΛ One of the most numerically difficult 4-body problem E. Hiyama Dr. Nemura n n n n Σ Σ Λ Σ

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