structure of few-body hypernuclei

Structure of few-body hypernuclei

Emiko Hiyama 　 (RIKEN)

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.

α Λ

7HeΛ

n

Last year and this year, we had two epoch-making data fromthe view point of few-body problems. Then, in hypernuclear physics, we are so excited.

JLAB experiment-E011,Phys. Rev. Lett. 110,12502 (2013).

FINUDA collaboration & A. Gal,Phys. Rev. Lett. 108, 042051 (2012).

n

t Λ6 HΛ

n n

OUTLINE

1. Introduction

2.

3.

4.

5. Summary

α Λ

7HeΛ

n n7HeΛ

t Λ6 HΛ

n n6 HΛ

n n

Λ

3nΛ

3nΛ

If we take triton away,The system is nnΛ system.This was observed at HyPHIcollaboration as a bound state.

　　　　　　　　　　　　　　 Section 1

Introduction

In neutron-rich and proton-rich nuclei,

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 theadditional nucleons .

Nuclearcluster

Nuclearcluster

Nuclearcluster

Nuclearcluster

n

n

n

n

nn

n n

Nucleus

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

Λ

Hypernucleus

Λ particle can reachdeep inside, and attract the surrounding nucleonstowards the interiorof the nucleus.

No Pauli principleBetween N and Λ

Λ particle plays a ‘glue like role’ to produce a dynamical

contraction of the core nucleus.

Λ

Λ

N

Xr

Λ

nucleus

N

Xr’

Λ hypernucleus

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, thenE2 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

3+

1+

1/2+

3/2+

5/2+

7/2+B(E2)

B(E2)Exp.B(E2)= 9.3 e2fm4

6Li

7LiΛ

α+ n+p α+Λ+n+p KEK-E419Then, Tamura et al.Succeeded inmeasuring thisB(E2) value to be

3.6 ±2.1 e2fm4 from5/2+ to 1/2+ statein 7Li.Λ

Theoretical calculation by Hiyama et al.B(E2: 5/2+ → 1/2+) =2.85 e2fm4

reduced by 22%

The shrinkage effect on the nuclear size included by the Λparticle was confirmed for the first time.

By Comparing B(E2) of 6Li and that of 7Li, Λ α

n pα

n p

Λ

Λ

Nucleus hypernucleus

There is no Pauli Pricliple betweenN and Λ.

Λ particle can reachdeep inside, and attracts thesurrounding nucleonstowards the interiorof the nucleus.

γnucleus

hypernucleus

Due to the attraction ofΛN interaction, theresultant hypernucleuswill become more stable against the neutron decay.

Neutron decay threshold

Λ

Λ

The glue like role of Λ particle provides us with another interesting phenomena.

Nuclear chart with strangeness

Λ

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 Λ particle is injected into an unstable nucleus?

α Λ

7HeΛ

n

Observed at JLAB, Phys. Rev. Lett. 110, 12502 (2013).

Observed by FINUDA group,Phys. Rev. Lett. 108, 042051 (2012).

n

t Λ6 HΛ

n n

Question : How is structure change when a Λ particle is injected into neutron-rich nuclei?

・ A variational method using Gaussian basis functions ・ Take all the sets of Jacobi coordinates

High-precision calculations of various 3- and 4-body systems:

In order to solve few-body problem accurately,

Gaussian Expansion Method (GEM) , since 1987

Review article : E. Hiyama, M. Kamimura and Y. Kino,Prog. Part. Nucl. Phys. 51 (2003), 223.

Developed by Kyushu Univ. Group, Kamimura and his collaborators.

,

Light hypernuclei,

3-quark systems,

Exotic atoms / molecules ,

3- and 4-nucleon systems,

multi-cluster structure of light nuclei, 4He-atom tetramer

　　　　　　　　　　　　　　 Section 2

Four-body calculation of 7HeΛ

α Λ

n n

α Λ

7HeΛ

n

Observed at JLAB, Phys. Rev. Lett. 110, 12502 (2013).

n

α6He

n n 6He : One of the lightest n-rich nuclei

7He: One of the lightest n-rich hypernucleiΛ

0+

2+

γ

α+n+n

-1.03 MeV

0 MeV

BΛ : CAL= 5.36 MeV

γ

1/2+

3/2+

5/2+

5He+n+nΛ

α+Λ+n+n0 MeV

-6.19

-4.57

Halo states

6He7HeΛ

Prompt particle decay

CAL: E. Hiyama et al., PRC53, 2075 (1996), PRC80, 054321 (2009)

BΛ ：

　 EXP= 5.68±0.03±0.25

Jlab experiment

Phys. Rev. Lett.110,

012502 (2013).

Exp:-0.98

0+

2+

γ

α+n+n

-1.03 MeV

0 MeV

BΛ : CAL= 5.36 MeV

γ

1/2+

3/2+

5/2+

5He+n+nΛ

α+Λ+n+n0 MeV

-6.19

-4.57

Halo states

6He7HeΛ

Prompt particle decay

CAL: E. Hiyama et al., PRC53, 2075 (1996), PRC80, 054321 (2009)

BΛ ：

　 EXP= 5.68±0.03±0.25

Observed at J-Lab

experiment(2012)

Phys. Rev. Lett.110,

012502 (2013).

Exp:-0.98

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 nucleusIn 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

　　　　　　　　　　　　　　 Section 3

Four-body calculation of 6HΛ

E. H, S. Ohnishi, M. Kamimura, Y. Yamamoto, NPA 908 (2013) 29.

t Λ

n n

6HΛt Λ

n n

t+n+n

1/2+

t+n+n+Λ

4H+n+nΛ

EXP: BΛ = 4.0±1.1 M

eV

Phys. Rev. Lett. 108, 042051 (2012).

5H : super heavy hydrogen

1.7±0.3 MeV

Γ=1.9±0.4 MeVFINUDA experiment

0.3 MeV

t Λ

n n

6 HΛ

5 H

Before the experiment, the following authors calculated thebinding 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).

Motivated by the experimental data, I calculated the binding energy of6H and I shall show you my result.Λ

Akaishi et al. pointed out that one of the important subject to study this hypernucleus is to extract information about ΛN-ΣN coupling.

Framework:

To calculate the binding energy of 6H, it is very important to reproduce the observed data of the core nucleus 5H.

Λ

t+n+n threshold

1/2+

1.7±0.3 MeV

Γ= 1.9±0.4 MeVA. Korcheninnikov, et al. Phys. Rev. Lett. 87 (2001) 092501.

Theoretical calculationN. 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+nthree-body model using complex scaling method.

5H is well described as t+n+n three-body model.

transfer reaction p(6He, 2He)5H

6HΛ

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

Λ

t Λ

n n

t

Λ

n n

5H

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 neededto adjust the energies of the 2- and 3-body subsystems.

t

n n

Λ

6HΛ

6HΛ6H

Λ

t

n n

Λt

n n

Λ

t

n n

Λ

6HΛ I take the Λ ｔ　 potential to reproduce

the binding energies of 0+ and 1+ states

of 4H.

In this case, ΛN － ΣN coupling is renormarized into the ΛN interaction.

Λ

6HΛ

t

n n

Λ

I take the Λ Ｎ　 potential to reproduce

the binding energy of 3H.Λ

Λ

t+n+n

1/2+

6HΛ

5H

1.7±0.3 MeV

Γ=1.9±0.4 MeV EXP

1.69 MeV

Γ= 2.44 MeV CAL

0+

1+

Λ

σn ・ σΛ

I focus on the 0+ state.

Λ

t

n n

Then, what is the binding energy of 6H?

t+n+n

0 MeV

½+

1.69 MeV

Γ= 2.44 MeVΓ= 0.91 MeV

1.17 MeV

4H+n+nΛ

E=-0.87 MeV0+

Exp: 1.7 ±0.3 MeV Even if the potential parameters were tunedso as to reproduce the lowest value of theExp. , E=1.4 MeV, Γ=1.5 MeV,we do not obtain any bound state of 6H.

Λ

- 　 2.0 4H+n+nΛ

-2.07 MeV0+

On the contrary, if we tune the potentials to have a bound state

in 6H, then what is the

energy and width of 5H?

t+n+n+ΛΓ=0.23 MeV

Γ=1.9 ±0.4 MeV

Λ

t+n+n

1/2+

t+n+n+Λ

4H+n+nΛ

EXP:BΛ =4.0±1.1 M

eV

Phys. Rev. Lett. 108, 042051 (2012).

5H:super heavy hydrogen

1.7±0.3 MeV

Γ= 1.9±0.4 MeV

FINUDA experiment

0.3 MeV

But, FINUDA group provided the bound state of 6H.

t Λ

6 HΛn n

t

5 H

n n

t+n+n+Λ

Λ

How should I understand the inconsistency between our resultsand the observed data?

(1) We need more precise data of 5H.

t+n+n

1/2+

1.7±0.3 MeV

Γ=1.9±0.4 MeV

A. Korcheninnikov, et al. Phys. Rev. Lett. 87 (2001) 092501.

To get bound state of 6H, the energy shouldbe lower than the present data.

It is planned to measure the energy and widthof 5H more precisely 　 at RCNP by Prof. Tanihata.He submitted his experimental proposalto RCNP and PAC meeting will be heldin 11th March.

Λ

[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.

We cited this experiment.However, you have manydifferent decay widths.Width is strongly related tothe size of wavefunction.Then, I hope thatThe decay width will bedetermined by Tanihata sanin the future.

(2) In our model, we do not include Λ Ｎー ΣN coupling explicitly.

The coupling effect might contribute to the energy of 6H.Λ

How should I understand the inconsistency between our resultsand the observed data?

(1) We need more precise data of 5H.

t+n+n

1/2+

1.7±0.3 MeV

Γ=1.9±0.4 MeV

A. A. Korcheninnikov, et al. Phys. Rev. Lett. 87 (2001) 092501.

To get bound state of 6H, the energy shouldbe lower than the present data.

Question:Is it possible to measure the energy and widthof 5H more precisely 　 somewhere 　 again?

Λ

(2) In our model, we do not include Λ Ｎー ΣN coupling explicitly.

The coupling effect might contribute to the energy of 6H.Λ

Non-strangeness nucleiN Δ

N N

N

Δ

300MeV

Probability of Δ 　 in nuclei is not large.

25MeVΛΛ

ΞN

In hypernuclear physics,the mass difference is very smallin comparison with the case of S=0field.

80 MeVΛ

Σ

S=-1

S=-2

Then, in S=-1 and S=-2 system, ΛN-ΣN and ΛΛ-ΞNcouplings might be important.

It might be important to perform the following calculation:

+6HΛ

t Λ

n n

3N Σ

n n

A. 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

t+n+n

1/2+

t+n+n+Λ

4H+n+nΛ

EXP:BΛ =4.0±1.1 M

eV

Phys. Rev. Lett. 108, 042051 (2012).

1.7±0.3 MeV

Γ=1.9±0.4 MeV

FINUDA experiment

0 MeV

Cal: -0.87 MeV

Exp: -2.3 MeVΛN-ΣN coupling ？

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.

5 H

Σ

Λ

4H+n+nΛ

0.3 MeV

6 HΛ

t+n+n+Λ

FINUDA data

No peak?!

Theoretically, we might understand by the following reason.If the state is resonant state, the reaction cross section would be much smaller thanthat we expect. => I should calculate reaction cross section 6Li (π,K) 6H with Harada’s help.

Λ

Now, we have open question whether or not6H is bound state.

Λ

We should answer to this issue.

I think that there is one possibility to perform experiment at Mainz.

Concept of Decay pion spectroscopy

Mass resolution : 100 keVMass accuracy : 30 keV

Core-to-Core Workshop at Vila Lanna in Czech, 5-7/Dec/2013

Hypernuclei from 9Be targetbreak-up mode Q value (MeV) - decay p (MeV/c)

9Li - 9Be + - 121.18

p + 8He -13.817 8Li + - 116.40

n + 8Li -3.756 8Be + - 124.12

2p + 7H -40.328 (B=6.1) 7He + - 135.17

d + 7He -12.568 7Li + - 114.61

2n + 7Li -12.218 7Be + - 108.02

3He + 6H -29.608 (B=5.1) 6He + - 133.47

3H + 6He -9.745 6Li + - 108.39

3n + 6Li -18.957 6Be + - 100.58

+ 5H -11.749 (B=4.1) 5He + - 133.42

n + + 4H -12.005 4He + - 132.95

6He + 3H -18.183 3He + - 114.29

Other targets, 7Li or 12C or so, should be measured in future.

Core-to-Core Workshop at Vila Lanna in Czech, 5-7/Dec/2013

I hope that at Mainz we can have conclusion about 6H.Λ

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, 4HeΛ Λ

n n

p Λ

4HΛ

n

p Λ

4HeΛ

p

-BΛ

-BΛ

0 MeV 0 MeV3He+Λ 3H+Λ

1+

0+

-2.39

-1.24

-2.040+

1+

-1.00

Exp. Exp.

4HeΛ

4HΛ

N

N

N

Λ

4HeΛ

4HΛ

NN

N Λ+

N N

N Σ

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).

PΣ=2.21%

PΣ=1.12 %

Another interesting role of Σ-particle in hypernuciei,namely effective ΛNN 3-body force generated by the Σ-particle mixing.

①

N1 Λ N2 N3

Σ

N1 Λ N2 N3

② N1 Λ N2 N3

Σ

N1 Λ N2 N3

N1 Λ N2 N3

N1 Λ N2 N3

N1 Λ N2 N3

N1 Λ N2 N3

3N+Λ space

Effective2-body ΛNforce

Effective 3-bodyΛNN force

How large is the3-body effect?

Y. Akaishi, T. Harada, S. Shinmura and Khin Swe Myint,Phys. Rev. Lett. 84, 3539 (2000).

3He Λ3He Σ+ ++

They already pointed out that three-body force effect isimportant within the framework of (3He+Λ)+(3He+Σ).

This effect (ΛN-ΣN coupling) might contribute toAnother s-shell Λ hypernuclei.

n p

　 Λ

n+p+Λ

d+Λ

½+Exp. 0.13 MeV bound

n

　 Λ

n Take Saito observed bound stateof nnΛ system. It is considered that ΛN-ΣN couplingplay an important role to make boundstate.

n

　 Λ

n

t If we inject triton cluster intonnΛ 　 system, we have 6H.

Λ

t

n n

Λ

Then, it is expected that ΛN-ΣN couplingmight contribute to 6H.

Λ

At GSI, they observed a bound state of nnΛ system.

nnΛ

½+

They did not report binding energy.

n

　 Λ

n

n

　 Λ

n n n

　Σ+

NN interaction : to reproduce the observed binding energies of triton and 3He

YN interaction: to reproduce the observed binding energies of 3H, 4H and 4HeΛΛΛ

What is the binding energy for nnΛ?

We have no possibilityto have a bound statein nnΛ system.And then, it would bedifficult to say to have abound state in 6H.Λ

　　　　　　　　　　　　 Section 5

Summary

Summary

1) Motivated by observation of neutron-rich Λhypernuclei, 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

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.To confirm to have bound state of this hypernucleus or not, it think that it would be better to perform searchExperiment at Mainz.I hope to have conclusion fro this issue in the future.

Λ

Λ

Thank you!

　　　　　　　　　　　　　　 Section 4.2

Λ-Σ coupling and CSB in

Λ

n n

7HeΛ

α

It is interesting to investigate the charge symmetry breaking effectin 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.

α

n

6He 6Li (T=1) 6Be

n

α

n p

α

pp

7He 7Li(T=1) 7BeΛ Λ Λ

Then, A=7 　 Λ 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.

α

n n

Λ α

n

Λ

p

αΛ

pp

Exp.1.54

-3.79

B Λ=5

.16

MeV

B Λ=5

.26

MeV

Prel

imin

ary

data

: 5.6

8±0.

03±0

.22

7BeΛ(T=1)7LiΛ

7HeΛ

6He

6Li(T=1)

6Be

JLAB

:E01

-011

exp

erim

ent

Emulsion data Emulsion data

Important issue:

Can we describe the Λ binding energy of 7He observed at JLAB using 　 ΛN 　 interaction to reproduce the Λ binding energies of

7Li (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 α+Λ+N+N 4-body model.E. Hiyama, Y. Yamamoto, T. Motoba and M. Kamimura,PRC80, 054321 (2009)

Λ

Λ Λ

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?

B Λ：

CAL=

5.2

1

B Λ：

CAL=

5.2

8

6Be6Li

7BeΛ

(T=1)

(T=1)7LiΛ

6He

7HeΛ

EXP

= 5.

26

EXP

= 5.

16

CAL=

5.3

6 B Λ

：　E

XP=

5.6

8±0.

03±0

22pr

elim

inar

y

(Exp: 1.54)

(Exp: -0.14)(exp:-0.98)

JLAB

:E01

-011

exp

erim

ent

Without CSB

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.

3He+Λ0 MeV

-1.24

-2.39

1+

0+

3H+Λ0 MeV

-1.00

-2.04

1+

0+

0.35 MeV

n n

p Λ

4HΛ

n

p Λ

4HeΛ

p

Exp.

0.24 MeV

Strength, rangeare determinedso as to reproduce the data.

B Λ：　E

XP=

5.6

8±0.

03±0

.22

α

p n

5.36

(with

out C

SB)

5.21

(with

out C

SB)

5.16

(with

CSB

)

5.44

(with

CSB

)

With CSB

Inconsistent with the data

5.28

MeV

( with

ourt

CSB

)

5.29

MeV

(With

CSB

)

Comparing the data of A=4and those of A=7,tendency of BΛ is opposite.

How do we understand these difference?