few-body structure of light hypernuclei

Few-body structure of light 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

p Λ

4HΛ

n

p Λ

4HeΛ

p

4HΛ4HeΛ

　　　　　　　　　　　　　　 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

nn

nn

nnn 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).

nt Λ

6 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 nWill be presented by Prof. A. Feliciellotomorrow’s plenary session

t+n+n

1/2+

t+n+n+Λ

4H+n+nΛ

EXP: BΛ = 4.0±1.1 MeV

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 n6 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 nΛ t

n nΛ

t

n n

Λ

6HΛ I take the Λ ｔ　 potential to reproduce

the binding energies of 0+ and 1+ statesof 4H.

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

Λ

6HΛ

t

n nΛ

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 MeVΓ= 2.44 MeV CAL

0+

1+

Λ

σn ・ σΛ

I focus on the 0+ state.

Λ

tn 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 MeV 0+

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 MeV

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 HΛ

n nt

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. 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

ΣS=-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 MeV

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. (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.

5 H

Σ

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

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.

n n

p Λ

4HΛ

n

p Λ

4HeΛ

p

Charge Symmetry breaking

N+N+N

1/2+- 8.48 MeV

0 MeV

- 7.72 MeV1/2+

3H

3He

n np

n p

0.76 M

eV

Energy difference comes fromdominantly Coulomb force between 2 protons.

Charge symmetry breaking effect is small.

In S=0 sector Exp.

p

3He+Λ0 MeV

-1.24

-2.39

1+

0+

3H+Λ0 MeV

-1.00

-2.04

1+

0+

0.35 MeV

n

p Λ

n

p Λ

Exp.

0.24 MeV

In S=-1 sector

4HΛ4HeΛ

np

n

p Λ

4HeΛ

pn n

p Λ

4HΛ

However, Λ particle has no charge.

n

p Λ

p

p

pp

p Σ0

p n

p Σ+

n n

+

++

+

4HeΛ

n

p Λ

pp

Σ0

p n

Σ+

n n

+

++

+

4HΛ

n

n

n n

Σ-Σ-

In order to explain the energy difference, 0.35 MeV,

N N

N Λ+

N N

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.

3He+Λ0 MeV

-1.24

-2.39

1+

0+

3H+Λ0 MeV

-1.00

-2.04

1+

0+

n n

p Λ

4HΛ

n

p Λ

4HeΛ

p

・ A. Nogga, H. Kamada and W. Gloeckle, Phys. Rev. Lett. 88, 172501 (2002)

(Exp: 0.24 MeV)

(Exp: 0.35 MeV)(cal. 0.07 MeV(NSC97e))

(cal: -0.01 MeV(NSC97e))

・ 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

Λ

N

N N

N N

Σ+

3He+Λ0 MeV

-1.24

-2.39

1+

0+

3H+Λ0 MeV

-1.00

-2.04

1+

0+

n n

p Λ

4HΛ

n

p Λ

4HeΛ

p

(Exp: 0.24 MeV)

(Exp: 0.35 MeV)(cal. 0.07 MeV(NSC97e))

(cal: -0.01 MeV(NSC97e))

How do we consider this situation?

There exist NO YN interaction to reproduce the data.

4HeΛ

π-+1H+3He →2.42 ±0.05 MeV π-+1H+1H+2H → 2.44 ±0.09 MeV

We get binding energy by decay π spectroscopy.

decay

Total: 2.42 ±0.04 MeV

Then, binding energy of 4He is reliable.Λ

4HΛ

π-+1H+3H →2.14 ±0.07 MeV π-+2H+2H → 1.92 ±0.12 MeV

decay

Total: 2.08 ±0.06 MeV

Two different modesgive 0.22 MeV

This value is so large to discuss CSB effect.

Then, for the detailed CSB study, we should perform experiment toconfirm the Λ separation energy of 4H.

Λ

　 4He (e, e’K+) 4HΛ

For this purpose, at JLAB and Mainz, it is planned to perform ・・・Key experiment to get information about CSB.

　 4He (e, e’K+) 4HΛ JLAB and Mainz experiment

Possible new understanding

(1)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 thatCSB 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 usingupdated realistic YN interaction.

Λ Λ

Λ Λ

Recently, we have updated YN interaction such as ESC08which 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. ΛΛ

At present, I use AV8 NN interaction.Soon, I will use new type of Nijmegen NN potential.

4HΛ

N

Λ

N

N N

N N

Σ+

4HeΛ

3He+Λ0 MeV

CAL: -0.47 (Exp: -1.24)

CAL:-1.46 (Exp. -2.39)

1+

0+

3H+Λ0 MeV

CAL:- 0.45 (Exp: -1.00)

CAL: -1.44 (Exp: -2.04)

1+

0+

n n

p Λ

4HΛ

n

p Λ

4HeΛ

p

(Exp: 0.24 MeV)

(Exp: 0.35 MeV)(cal. 0.02 MeV

(cal: 0.01 MeV)

N

Λ

N

N N

N N

Σ+

Preliminary result (ESC08)

Binding energies are less bound than theobserved data. We need more updatedYN interactions.

　　　　　　　　　　　　 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 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

Λ Ｎ－ 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 hypernucleiusing 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 forlight 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

Thank you!

You are welcome to join us to research hypernuclear physics!

　　　　　　　　　　　　　　 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 data5.

28 M

eV( w

ithou

rt C

SB)

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?