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 He n 7 6 JLAB experiment-E011, Phys. Rev. Lett. 110, 12502 (2013). H n
n t FINUDA collaboration & A. Gal, Phys. Rev. Lett. 108, 042051 (2012). Observation of Neutron-rich -hypernuclei These observations are interesting from the view points of few-body physics as well as physics of unstable nuclei. OUTLINE He 7 n n 1. Introduction
3. 4. He 7 2. 6 4 H H 6 4 He H
n n t 5. Summary n n n p p p 4 H 4
He Section 1 Introduction In neutron-rich and proton-rich nuclei, n Nuclear cluster n Nuclear cluster Nuclear cluster n Nuclear cluster n 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. There are many interesting phenomena in this field as you know. It was considered that nucleus was hardly compressed by the additional nucleons . Nucleus Question:How is the structure modified when a hyperon, particle, is injected into the nucleus? No Pauli principle Between N and particle can reach deep inside, and attract the surrounding nucleons towards the interior of the nucleus. Hypernucleus particle plays a glue like role to produce a dynamical contraction of the core nucleus. Question: How do we observe the dynamical contraction? X
X r N N nucleus r hypernucleus r > r B(E2) value: B(E2) ||2 Propotional to 4th-power of nuclear size Bando, Ikeda, Motoba If nucleus is really contracted by the glue-like role of the particle, then E2 transition in the hypernuclei will become much reduced than the corresponding E2 transition in the core nucleus. 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% + n+p
3 ++n+p + B(E2) 7/2+ 1+ 5/2+ Exp.B(E2)= 9.3 e fm 2 6 4 B(E2) Li 3/2+ p 1/2+ n
7 Li 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. n p By Comparing B(E2) of 6Li and that of 7Li, The shrinkage effect on the nuclear size included by the particle was confirmed for the first time. KEK-E419 The glue like role of particle provides us with another interesting phenomena. particle can reach
Nucleus There is no Pauli Pricliple between N and . deep inside, and attracts the surrounding nucleons towards the interior of the nucleus. hypernucleus Neutron decay threshold nucleus hypernucleus Due to the attraction of N interaction, the resultant hypernucleus will become more stable against the neutron decay. Nuclear chart with strangeness
Multi-strangeness system such as Neutron star Extending drip-line! Interesting phenomena concerning the neutron halo have been observed near the neutron drip line of light nuclei. How does the halo structure change when a particle is injected into an unstable nucleus? Question : How is structure change when a particle is injected into neutron-rich nuclei? n He n 7 6 H n
n t Observed at JLAB, Phys. Rev. Lett. Observed by FINUDA group, 110, 12502 (2013). Phys. Rev. Lett. 108, 042051 (2012). 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 , Light hypernuclei, 3- and 4-nucleon systems, 3-quark systems, multi-cluster structure of light nuclei,
4 He-atom tetramer Section 2 Four-body calculation of n n 7 He n He 6 He n n n
7 6 He : One of the lightest n-rich nuclei He: One of the lightest n-rich hypernuclei 7 Observed at JLAB, Phys. Rev. Lett. 110, 12502 (2013). CAL: E. Hiyama et al., PRC53, 2075 (1996), PRC80, 054321 (2009) 6 He Prom p 2+ 0 MeV t par
7 ticle d eca y He ++n+n 0 MeV +n+n 0+ 5 -1.03 MeV Exp:-0.98 B B
:C AL 5/2+ + 3/2 -4.57 = 5. Jla 36 Ph b e E M 01 ys. xp XP= e eV 25 Re ri 02 v. me 5.6 (2 Lett nt 80 01 .1 .0 3) 10 3 . , 0. He+n+n -6.19
25 1/2+ Halo states CAL: E. Hiyama et al., PRC53, 2075 (1996), PRC80, 054321 (2009) 6 He Prom p 2+ 0 MeV t par 7 ticle d eca y He ++n+n 0 MeV +n+n
0+ 5 -1.03 MeV Exp:-0.98 B B :C AL = He+n+n 5/2+ + 3/2 -4.57 5
Halo states .3 Ob 6 E se X e M x P r + + vthe = 5 of the eV excitation mechanism in neutron-rich 5/2 1/2 Useful Ph pefor ed study r i y . m 01 s. a 6 3/2+ 1/2+ nuclei 1/2+ 25 Re ent t J-L 8 0.
-6.19 02 v. (2 ab 0 0 Helpful for the study of the excitation mechanism of the halo L 3 (2 ett 12 01 hypernuclei, 0. will be many examples such as . ) In n-rich nuclei and n-rich there nucleus 25 3) 110 6 7 . , Combination of He and He. I hope that -ray spectroscopy of n-rich hypernuclei will be performed at J-PARC. Section 3 Four-body calculation of
n n t 6 H E. H, S. Ohnishi, M. Kamimura, Y. Yamamoto, NPA 908 (2013) 29. n n 6 t H 1/2 Phys. Rev. Lett. 108, 042051 (2012). =1.90.4 MeV
+ FINUDA experiment 1.70.3 MeV t+n+n+ H 1 .0 =4 B 5 P: EX t+n+n 4 .1 eV M 0.3 MeV 6
5 H : super heavy hydrogen H+n+n H n n t 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 N-N coupling.N coupling. Motivated by the experimental data, I calculated the binding energy of 6 H and I shall show you my result. Framework: To calculate the binding energy of 6H, it is very important to reproduce the observed data of the core nucleus 5H. transfer reaction p(6He, 2He)5H
A. Korcheninnikov, et al. Phys. Rev. Lett. 87 (2001) 092501. = 1.90.4 MeV 1/2+ 1.70.3 MeV t+n+n threshold Theoretical calculation N. B. Schulgina 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. 5 H is well described as t+n+n three-body model. Then, I think that t+n+n+ 4-body model is good model to describe 6H. Then. I take t++n+n 4-body model. n n t 5 H n
n t 6 H 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. Namely, all the potential parameters are needed to adjust the energies of the 2- and 3-body subsystems. 6 6 H H H 6
n n t n n n t n t H 6 n n
I take the potential to reproduce the binding energies of 0+ and 1+ states of 4H. t In this case, N NN coupling is renormarized into the N interaction. H 6 n n I take the potential to reproduce t the binding energy of 3 H. =1.90.4 MeV EXP = 2.44 MeV CAL
1/2+ 1.69 MeV 1.70.3 MeV I focus on the 0+ state. t+n+n n 1+ n t n 0+ H 6 5 H Then, what is the binding energy of 6H? Even if the potential parameters were tuned so as to reproduce the lowest value of the
Exp. , E=1.4 MeV, =1.5 MeV, we do not obtain any bound state of 6H. Exp: 1.7 0.3 MeV =1.9 0.4 MeV + = 2.44 MeV = 0.91 MeV 1.69 MeV 1.17 MeV 0 MeV t+n+n+ 4 H+n+n =0.23 MeV t+n+n 0+ E=-0.87 MeV - 2.0 H+n+n
4 -2.07 MeV 0+ On the contrary, if we tune the potentials to have a bound state in 6 H, then what is the energy and width of 5H? = 1.90.4 MeV 1/2 Phys. Rev. Lett. 108, 042051 (2012). + FINUDA experiment 1.70.3 MeV 5 H
t 5 0.3 MeV eV n n H+n+n 4 M t+n+n+ t+n+n+ .1 1 .0 =4 B P: EX t+n+n 6
H H:super heavy hydrogen But, FINUDA group provided the bound state of 6H. n n t How should I understand the inconsistency between our results and the observed data? (1) We need more precise data of 5H. A. A. Korcheninnikov, et al. Phys. Rev. Lett. 87 (2001) 092501. =1.90.4 MeV 1/2+ 1.70.3 MeV To get bound state of 6H, the energy should be lower than the present data. It is planned that experiment to measure the energy and width of 5H more precisely at RCNP next year. t+n+n (2) In our model, we do not include NN coupling explicitly.
The coupling effect might contribute to the energy of 6 H. What is N-N coupling?N coupling? Non-strangeness nuclei N S=-1 N coupling. 80 MeV S=-2 N N 25MeV 300MeV N Probability of not negligible. N in nuclei is
In hypernuclear physics, the mass difference is very small in comparison with the case of S=0 field. Then, in S=-1 and S=-2 system, N-N coupling.N and -N couplings might be important. Interesting Issues for the N-N coupling?N particle conversion in hypernuclei (1)How large is the mixing probability of the N coupling? particle in the hypernuclei? (2) How important is the N coupling? coupling in the binding energy of the hypernuclei? -B -B He 4 H 4 0 MeV
3 He+ -1.24 -2.39 1 0 MeV + 3 -1.00 0+ -2.04 Exp. Exp. N N N 4
He H 4 H+ 1+ 0+ N N N N N coupling? + NNN N +
N NNNN coupling? 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). 4 He, H 4 VNN : AV8 potential VYN : Nijmegen soft-core 97f potential PN coupling?=1.12 % PN coupling?=2.21% A. Gal and D. J. Millener, arXiv:1305.6716v3 (PLB725 445(2013)). They pointed out that N-N coupling?N coupling is important for 6H. It might be important to perform the following calculation: H 6 n
t n n n + 3N N =1.90.4 MeV Phys. Rev. Lett. 108, 042051 (2012). 1/2+ 1.70.3 MeV FINUDA experiment t+n+n+ 5 H .1 1 .0 =4 B P:
EX t+n+n 0 MeV M eV Cal: -0.87 MeV 4 H+n+n N-NN 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 N-NN coupling. N In hypernuclear physics, currently, it is extremely important to get information about N-N coupling.N coupling. Question: Except for 6H, what kinds of hypernuclei are suited for extracting information on the N-N coupling.N coupling? Answer: H, 4 He
4 n n n p p p 4 H 4 He Recently, we have updated YN interaction such as ESC08 which has been proposed by Nijmegen group.
Then, using this interaction directly, I performed four-body calculation of 4H and 4He. At present, I use AV8 NN interaction. Soon, I will use new type of Nijmegen NN potential. 4 H 4 He N N N + N N N N coupling. Preliminary result (ESC08) 3
3 n p 4 4 He Binding energies are less bound than the observed data. We need more updated YN interactions. H N N N + N N
N N coupling. Section 5 Summary Summary 1) Motivated by observation of neutron-rich hypernuclei, He and 6H, I performed four-body calculation of them. 7 In 7He, due to the glue-like role of particle, We may have halo states, the 3/2+ and 5/2+ excited state. We wait for further analysis of the JLAB experiment. 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 N-N coupling.N coupling. To obtain this information, four-body calculations of 4 4 H and He 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 hypernuclei at J-PARC, Jlab and Mainz. By comparison with future experiment and theoretical effort, we expect to have more reliable YN interaction in the future. Thank you! Another interesting role of N coupling?-particle in hypernuciei, namely effective NN 3-body force generated by the N coupling?-particle mixing. N1 N2 N3 In the calculation of 6 H, this effect is included in N interaction. N coupling? N1
N1 N2 N3 N2 N3 N coupling? N1 N2 N3 N1 N2 N3 3N+ space N1 N2 N3 Effective 3-body NN force Effective 2-body N force N1 N 2 N3 How large is the 3-body effect?
N1 N2 N3 Y. Akaishi, T. Harada, S. Shinmura and Khin Swe Myint, Phys. Rev. Lett. 84, 3539 (2000). 3 He + + 3 He + N coupling? They already pointed out that three-body force effect is Important within the framework of (3He+)+(3He+N coupling?). Section 4.2 -N coupling. coupling and CSB in n n
7 He It is interesting to investigate the charge symmetry breaking effect in p-shell hypernuclei as well as s-shell hypernuclei. For this purpose, to study structure of A=7 hypernuclei is suited, because, core nuclei with A=6 are iso-triplet states. n n n 6 He p p p
6 Li (T=1) 6 Be n n 7 He n p p p
7 Li(T=1) 7 Be Then, A=7 hypernuclei are also iso-triplet states. It is possible that CSB interaction between and valence nucleons contribute to the -binding energies in these hypernuclei. 7He eV B =5.16 M He eV B =5.26 M riment e p x e 1
1 -0 1 0 :E JLAB .22 6 .030 0 8 6 . 5 : a t da Preliminary Exp. 1.54 Emulsion data 6 Li (T=1) Li 7
(T=1) Emulsion data 6 Be -3.79 7 Be Important issue: observed at JLAB Can we describe the binding energy of 7He using 7 Li N interaction to reproduce the binding energies of ? (T=1) and 7Be To study the effect of CSB in iso-triplet A=7 hypernuclei. For this purpose, we studied structure of A=7 hypernuclei within
the framework of ++N+N 4-body model. E. Hiyama, Y. Yamamoto, T. Motoba and M. Kamimura,PRC80, 054321 (2009) Now, it is interesting to see as follows: (1)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? 6 He 7 He 6 Li (T=1) 6 EXP= 5.1 5.21 = L A C B (exp:-0.98)
26 EXP= 5. 5.28 B CAL= riment e p x e 1 1 -0 1 0 :E JLAB preliminary 0.03022 8 .6 5 = P X B E CAL= 5.36 (Exp: 1.54) Without CSB
(Exp: -0.14) 6 7 Li (T=1) Be 7 Be Now, it is interesting to see as follows: (1)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? 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
3 -1.24 He+ 0 MeV 0.24 MeV -2.39 p Exp. 4 He 1+ -2.04 0+ 0.35 MeV p n H+
-1.00 1+ 0+ 3 n n p 4 H SB ) p n SB ) B) S C t
u o h t 5.21 (wi C 5.44(with t CSB) r u o h it w ( 5.28 MeV ith C W ( V e M 5.29 ) 5.16(with CSB CS B ) t u o h
it w ( 6 3 5. 30.22 0 . 0 8 .6 5 B EXP= With CSB Inconsistent with the data Comparing the data of A=4 and those of A=7, tendency of B is opposite. How do we understand these difference?
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