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Structure of light hypernuclei

Emiko Hiyama 　 (RIKEN)

α Λ

7HeΛ

Recently, we had three 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).

t Λ6 HΛ

n n

3nΛ

C. Rappold et al., HypHI collaboration Phys. Rev. C 88, 041001 (R) (2013)

OUTLINE

1. Introduction

5. Summary

α Λ

7HeΛ

n n7HeΛ

t Λ6 HΛ

n n6 HΛ

n n

3nΛ

　　　　　　　　　　　　　　 Section 1

Introduction

In order to understand the baryon-baryon interaction, two-body scattering experiment is most useful.

YN and YY potential models so far proposed (ex. Nijmegen, Julich, Kyoto-Niigata) have large ambiguity.

1) To understand baryon-baryon interactions

Total number of

Nucleon (N) -Nucleon (N) data: 4,000

・ NO YY scattering data

・ Total number of differential cross section

Hyperon (Y) -Nucleon (N) data: 40

Study of NN intereactionhas been developed.

since it is difficult to perform YN scattering experiment even at J-PARC.

Major goals of hypernuclear physics

2) To study the structure of multi-strangeness systems

Hypernucleus

No Pauli principleBetween N and Λ

γnucleus

hypernucleus

Due to the attraction of

Λ N interaction, the

resultant hypernucleus will

become more stable against

the neutron decay.

neutron decay threshold

Λ particle can reach deep inside, and attract the surrounding nucleons towards the interiorof the nucleus.

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

α Λ

7HeΛ

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

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

t Λ6 HΛ

n n

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

n n

3nΛ

C. Rappold et al., HypHI collaboration Phys. Rev. C 88, 041001 (R) (2013)

・ 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

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

α6He

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

7He: One of the lightest n-rich hypernucleiΛ

α+n+n

-1.03 MeV

0 MeV

BΛ : CAL= 5.36 MeV

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

　　　　　　　　　　　　　　 Section 3

Four-body calculation of 6HΛ

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

t Λ

n n

n n

Λ6HΛ

t+n+n

1/2+

t+n+n+Λ

4H+n+nΛ

EXP:BΛ =4.0±1.1 M

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

n n

Λ6HΛ

n n

5H:super heavy hydrogen

1.7±0.3 MeV

Γ=1.9±0.4 MeV

FINUDA experiment

0.3 MeV

Before experiment, the following authors calculated thebinding energies by 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).

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

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

Before doing 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 toadjust in the 2- and 3-body subsystems.

6HΛ

nΛ

6HΛ

n nΛ

6HΛ

Among the subsystems, it is extremely important toadjust the energy of 5H core nucleus.

Framework:

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

t+n+n

1/2+

1.7±0.3 MeV

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

To reproduce the data, for example,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.The calculated binding energy for the ground stateof 5H is 1.6 MeV with respect to t+n+n threshold andwidth has 1.5 MeV.

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

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.

- 　 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

t+n+n

1/2+

t+n+n+Λ

4H+n+nΛ

EXP:BΛ =4.0±1.1 M

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

5 H

n n

t+n+n+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 　 this year.His proposal was approved recently .

[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

Non-strangeness nucleiN Δ

N N

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

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.

t+n+n

1/2+

t+n+n+Λ

4H+n+nΛ

EXP:BΛ =4.0±1.1 M

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

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.

How should we understand 6H issue?To investigate it, the study of nnΛ system is suited.

　　　　　　　　　　　　　　 Section 4

three-body calculation of 3nΛ

n E. Hiyama, S. Ohnishi,B.F. Gibson, and T. A. Rijken,The paper will be submitted inPRC this week.

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

nnΛ 　 breakup threshold

?They did not report thebinding energy.

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

n n

The three-body system of nnΛ system isuseful hyperucleus for discussion about whether 6H is bound or unbound.

Namely, if we have a bound state in nnΛ system,there is possibility to have a bound state in 6H.

n n n

　Σ+

NN interaction : to reproduce the observed binding energies of triton and 3He NN: AV8 potential

Important issue: Do we have bound state for nnΛ system?If we have a bound state for this system, how much is binding energy?

N+N+N

1/2+- 8.48 MeV

0 MeV

- 7.72 MeV1/2+

3He

n np

n p

In S=0 sector

Exp.

1/2+

-7.77 MeV

Cal.

Exp.

1/2+

-7.12 MeV

Cal.

n n n

　Σ+

YN interaction: to reproduce the observed binding energies of 3H, 4H and 4He NSC97f potential

ΛΛΛ

Important issue: Do we have bound state for nnΛ system?If we have a bound state for this system, how much is binding energy?

-BΛ -BΛ

0 MeV 0 MeV3He+Λ 3H+Λ

-2.39

-1.24

-2.040+

-1.00

Exp.Exp.

4HeΛ 4H

N p

3HΛ

-2.28

-0.54

-2.33

-0.57

Cal. Cal.

4HeΛ4H

d+Λ

-0.13 ±0.05 MeV

1/2+

Exp.

1/2+

-0.19 MeV

Cal.

What is binding energy for nnΛ?

nnΛ1/2+

We have no bound state in nnΛ system.

Now, we have a question.

Do we have a possibility to have a bound state in nnΛ system tuningstrength of YN potential ? When we have a bound state in nnΛ system, what are binding energies of 3H and A=4 hypernuclei?

VT ΛN-ΣN X1.1, 1.2

VT ΛN-ΣN VT ΛN-ΣNX1.1

When we have a bound state in nnΛ system, what are binding energies of 3H and A=4 hypernuclei?

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

Question: If we tune 1S0 state of nn interaction,Do we have a possibility to have a bound state in nnΛ?In this case, the binding energies of 3H and 3He reproducethe observed data?

n T=1, 1S0 stateI multiply component od 1S0 state by 1.13 and 1.15. What is the binding energies of nnΛ ？

Some authors pointed out to have dineutron bound state innn system. Ex. H. Witala and W. Gloeckle, Phys. Rev. C85, 064003 (2012).

nnunbound

-0.066MeV

0 MeV

-0.118 MeV1S0X1.13

1S0X1.15

nnΛ

1/2+

unbound unbound

-0.043MeV

N+N+N

1/2+

-10.09 (-9.38)MeV-8.48 (-7.72)

Exp.

1/2+

3H (3He)

Cal.

-7.77(-7.12) -9.75 (-9.05)

We do not find any possibility to have a bound statein nnΛ.

Cal.Cal.

n n

　　　　　　　　　　　　 Section 5

Summary

Summary

1) Motivated by observation of neutron-rich Λhypernuclei, 7He, 6H , nnΛ, 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.

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 study 6H issue, I calculated nnΛ system taking ΛN-ΣN coupling explicitly.However, I could not find any bound statekeeping consistency with observed data of 3H, A=4 hypernuclei, 3H and 3He.This fact would suggest to have no bound statein 6H.

To conclude whether or not 6H has a bound state,

it think that it would be better to perform search experiment

at Mainz.

I hope to have conclusion for this issue in the future.

Thank you!

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

p Λ

4HeΛ

-BΛ

0 MeV 0 MeV3He+Λ 3H+Λ

-2.39

-1.24

-2.040+

-1.00

Exp. Exp.

4HeΛ

4HΛ

4HeΛ

4HΛ

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

② 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).

n n

6HΛ

Then, I use t+n+n+Λ 　 4-body model for 6Hand take the same t-n potential employedby N. B. Schul’gina et al., Nucl. Phys. A597, 197 (1996 ） .

t n

3H-n scattering

n n

It was difficult to reproduce theexperimental energy and width of 5H,then, they introduced phenomenologicalattractive t-n-n three-body force.

Using complex scaling method which is one of the powerful method to obtainenergy and width of resonant state, I tune the strength , S3b, tnn three-body force.

S3b=-57 MeV