Structure of light Λ hypernuclei by (e,e’K+) reaction

E. Hiyama (RIKEN)

Using these targets, we have these light Λ　 hypernucei by (e, e’K+) reaction at Jlab. Here I will focus on these Λ hypernuclei.

n n

Λ

n n

np

3H 3nΛ

nnp

6 Li

n

6 He

Λ

Λ

４4He ４4He

nnp

7Li

n

7He

Λ

Λ

４4He ４4He

targetResultant hypernucleus

n

n

np

10B

４4He

n

４4He

10Be

４4He

n

４4He

target Resultant hypernucleus

Λ

Λ

Outline

(1) Introduction

(2) n n

Λ

3nΛ

n

6 He

Λ

４4He

n

7He

Λ

Λ

４4He n ４4He

n

４4He

Λ

10BeΛ

These Λ hypernuclei can be produced by (e, e’K+) at Jlab.What is interesting and important to study theseΛ hypernuclei?

Λ

Sec.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

nn

Once the Hamiltonian is determined, it is possible , using few-body calculation method (GaussianExpansion Method), to precisely calculatethe structure of many-body systems consisting of neutrons, protons and hyperons.As a result, we can predict new phenomena such as we have never imagined before.

2) To study the structure of multi-strangeness systems

nn

p

p

Y

Hypernucleus =

Many-body system of

neutrons, protons, hyperons

np

pY

nn

Hypernucleus

No Pauli principleBetween N and Λ

Λ

γnucleus

hypernucleus

Due to the attraction ofΛ N interaction, theresultant hypernucleus willbecome more stable againstthe neutron decay.

neutron decay threshold

Λ

N

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

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

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

α Λ

7HeΛ

n n n n

Λ

3nΛ

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

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

α

Λ6HeΛ

n

　　　　　　　　　　　　　　 Section 2

three-body calculation of 3nΛ

n

　 Λ

n E. Hiyama, S. Ohnishi,B.F. Gibson, and T. A. Rijken,The paper will be publised inPRC as a Rapid communication soon.

What is interesting to study nnΛ system? n

　 Λ

n

S=0

The lightest nucleus to have a bound state is deuteron.

n p

n+p threshold

J=1+

-2.22 MeV

S=-1 (Λ 　 hypernucear sector)

n

　 Λ

3H (hyper-triton)

d

Λ

n+p+Λ

d+Λ

0.13 MeV J=1/2+

Exp.

Exp.

Lightest hypernucleus to have a bound statep

n

　 Λ

n

nnΛ 　 breakup threshold

?They did not report thebinding energy.

Observation of nnΛ system (2013)Lightest hypernucleus to have a bound state

scattering length:-2.68fm

-23.7fm

n

　 Λ

n

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

NN: AV8 potentialWe do not include 3-body force for nuclear sector.

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

nnΛ 　 breakup threshold

?They did not report thebinding energy.

How about YN interaction?

Non-strangeness nucleiN Δ

N N

N

Δ

300MeV

80MeVΛ

Σ

On the other hand, the mass differencebetween Λ and Σ is much smaller, thenΛ 　 can be converted into Σ particle easily.

Nucleon can be converted into Δ.However, since mass difference betweennucleon and Δ is large, then probabilityof Δ in nucleus is not large.

ΛN

N Λ

Σ

n

　 Λ

n n n

　Σ+

YN interaction: Nijmegen soft core ‘97f potential (NSC97f) proposed by Nijmegen group reproduce the observed binding energies of 3H, 4H and 4He ΛΛΛ

To take into account of Λ 　 particle to be converted intoΣ particle, we should perform below calculation usingrealistic hyperon(Y)-nucleon(N) interaction.

n p

Λ

3HΛ

d+Λ

-0.13 ±0.05 MeV

1/2+

Exp.

1/2+

-0.19 MeV

Cal.

0 MeV

-BΛ

-BΛ -BΛ

0 MeV 0 MeV3He+Λ 3H+Λ

1+

0+

-2.39

-1.24

-2.04 0+

1+

-1.00

Exp.Exp.

4HeΛ 4H

Λ

-2.28

-0.54

-2.33

-0.57

0+

1+

0+

1+

Cal. Cal.

n

p

p

Λ

4HeΛ4H

Λ

Λ

What is binding energy of nnΛ?

n

p

n

Λ

nnΛ threshold 1/2+

We have no bound state in nnΛ system.This is inconsistent with the data.

Now, we have a question.

Do we have a possibility to have a bound state in nnΛ system tuningstrength of YN potential ?

0 MeVn

　 Λ

n

It should be noted to maintain consistency with the binding energiesof 3H and 4H and 4He.

ΛΛ Λ

-BΛ

VT ΛN-ΣN X1.1, 1.2

n

　 Λ

n

VT ΛN-ΣN VT ΛN-ΣNX1.1 X1.2

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

Λ

n

　 Λ

n

np

Λ

n

p

Λ

n

We have no possibilityto have a bound statein nnΛ system.

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

　 Λ

nT=1, 1S0 stateI multiply component of 1S0 state by 1.13 and 1.35. 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

-1.269 MeV1S0X1.13

1S0X1.35

nnΛ

1/2+

unbound unbound

-1.272 MeV

n

　 Λ

n

N+N+N

1/2+

-13.93 (-13.23)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 n0 MeV

Summary of nnΛ 　 system:Motivated by the reported observation of data suggestinga bound state nnΛ, we have calculated the binding energy of this hyperucleus taking into account ΛN-ΣN explicitly.We did not find any possibility to have a bound state in thissystem. However, the experimentally they reportedevidence for a bound state. As long as we believe the data, we should consider additionalmissing elements in the present calculation. But, I have no idea.Unfortunately, they did not report binding energy.

Then, I would like to suggest experimentalists:To perform search experiment of nnΛ system to concludewhether or not the system exists as bound state experimentally.

n n

Λ

n n

np3H 3nΛ

(e,e’K+)

α

Λ

6HeΛ

n

6HeΛ

α +n

0.89 MeV P3/2

Λ

5He+nΛ -3.12 MeV

1-

0 MeV

5He

6HeΛα Λ

L=05He

Λ

0.17 MeV

α

Λ

n

6HeΛ

Question: Can we reproduce the data ?How do the 3 particle locate to each other?Is there neutron halo in 6He hypernucleus?Λ

Valence neutron is very looselycoupled to the α+Λ system.

α +n+Λ

密度　　 ρn(r) = ∫|Ψ(6He)|2dRdr^

r

Λ

α

Λ

n

r.m.s α-Λ 2.8 fm α-n 5.0 fm

Halo nucleus 6Heα-n 4.5 fm

Larger than 6He

Three-layer structure

7He

Λ

Λ

n n

7He

Λ

α

Λ

7HeΛ

n

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

n

6Hen n 6He : One of the lightest

n-rich nuclei

7He: One of the lightest n-rich hypernucleiΛ

α

α

Λ

0+

2+

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 MeV

-4.57

Neutronhalo states

6He7HeΛ

CAL: E. Hiyama et al., PRC 53, 2075 (1996), PRC 80, 054321 (2009)

BΛ ：

EXP= 5.68±0.03±0.25

Observed at J-Lab experiment Phys. Rev. Lett.110,012502 (2013).

Exp:-0.98Λ

One of the excitedstate was observed at Jlab.

7Li(e,e’K+)7ΛHe

(FWHM = 1.3 MeV)

Fitting results

Third peak?

At present, due to poor statics,It is difficult to have the third peak.Theoretically, is it possible tohave new state?Let’s consider it.

0+

2+

α+n+n0 MeV

6He

-0.98

1.797 MeVΓ=113 ±20 keV

(2+,1-,0+)?

Γ＝ 12.1 ±1.1 MeV

Exp.

-0.97 MeV

1.79 MeV

2.5 MeV

Γ=

Γ=

theoryMyo et al., PRC 84, 064306 (2011).

Core nucleus

1/2+

3/2+5/2+

5He+n+nΛ

α+Λ+n+n0 MeV

-6.19 MeV

-4.57

Neutronhalo states

7HeΛ

Λ7HeΛ

n n

α

-3.12

-1.88 Γ=0.7 MeV5/2+

3/2+ -2.04, Γ=0.5 MeVNew prediction

0+

2+

α+n+n0 MeV

6He

-0.98

1.797 MeV

Γ=113 ±20 keV

(2+,1-,0+)?

Γ＝ 12.1 ±1.1 MeV

Exp.1/2+

3/2+5/2+

5He+n+nΛ

α+Λ+n+n0 MeV

-6.19 MeV

-4.57

Neutronhalo states

-3.12

-1.88 Γ=0.7 MeV5/2+

3/2+ -2.04, Γ=0.5 MeV

7HeΛ

2+

If we find these two excited states at Jlab,in 6He, existence of the second 2+ stateis promising. Please search the second 2+

state in 7He at Jlab. Λ

If we find these two states at Jlab, these existencecontribute to unstable nuclear physics.

Response of super-deformed states by addition of Λ particle

Section 3

10BeΛ

spherical Normal deformed (ND)

Super-deformed(SD)

We have various kinds of nuclear shape, such as spherical, normal deformed and super-deformed shapes. Especially,

it is known that super-deformed shape is related to the appearance of a large shell gap at a 2:1 axis ratio of the nuclear deformation for specific particle numbers.

What happens when a Λ particle is added to such ND and SD states?

Λ Λ

0+1

0+2

NormalDeformed (ND) states

Super-deformed (SD) states

core nucleus

2+1

2+2

Does energy gain goin parallel way for both typesof states?

Λ

Λ hypernucleus

No !

What happens when a Λ particle is added to such ND and SD states?

0+2

2+2

0+1

2+1

Energy gain by Λ-particle addition

ΔE (ND state) > ΔE (SD state)

ND state

SD state

core nucleus

Λ hypernucleus

Level crossing

ND state

One example in10BeΛ

10BeΛ

Y. Zang, E. H., and Y. Yamamoto, Nucl. Phys. A 881, 288 (2012):

E. H. and Y. Yamamoto, Prog. Theor. Phys. 128, 105, (2012).

Λ

9Be10BeΛ

α+α+n

3/2-

1/2+5/2-

Interesting phenomena such as level crossing is seen in 9Be and 10Be combination.Λ

It is well known that , in the 1/2+ state,a valence neutron is located in 1S 1/2 orbit

in the simple shell model configuration.It is a kind of SD state.

When a Λ 　 particle is injected into each state,is there any change of energy spectra?

10BeΛ

9BeΛ

ND

SD

ND

5/2

1/2+

3/2-

α ＋ α ＋ｎ α ＋ α ＋ｎ＋ Λ

-10.52 MeV

-8.14

-7.35

-1.58

+0.1

+0.6 MeV

MeV

BΛ = 8.94 M

eV

CAL

(　B

Λ = 9.11 0.22 MeV)

EXP

10BeΛ

9Be

Level reversion is occurred by addition of a Λ particle.

10B(e,e’K+)10Be (Jlab E01-11)

ND

ND

SD

1-, 2-

2-, 3- 0+ , 1+

Small spin-splitting is

neglected to show.

Λ

If we observepositive states andnegative states,we find thatΛ-separationenergies are dependenton the degree ofdeformation.Please observe the positive parity states at Jlab.

n n

Λ

n n

np

3H 3nΛ

nnp

6 Li

n

6 He

Λ

Λ

４4He ４4He

nnp

7Li

n

7He

Λ

Λ

４4He ４4He

n

n

np

10B

４4He

n

４4He

10Be

４4He

n

４4He

Λ

Λ

Summary

Please confirm whether ornot we have bound state,

It is interesting to have ahuge halo structure.

Please observesecond 3/2+ and 5/2+ state.This is important for the study of neutron-rich nuclei.

Please observethe positive parity state.

In this way, the experiment at Jlab isvery important for theorists to study the structure of multi-strangeness systems.

I hope we have many data at Jlab in the future.

Thank you!