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### Structure of neutron-rich Λ hypernuclei

E. Hiyama (RIKEN)

Major goals of hypernuclear physics

1) To understand baryon-baryon interactions

2) To study the structure of multi-strangeness systems

In orderto understand the baryon-baryon interaction,two-body scattering

experiment is most useful.

Study of NN intereaction

has been developed.

Total number of

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

YN and YY potential models so far proposed

(ex. Nijmegen,

Julich, Kyoto-Niigata) have large ambiguity.

・ Total number of differential cross section

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

・ NO YY scattering data

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

Between N and Λ

Λ particle can reach deep inside, and attract the surrounding nucleons towards the interior

of the nucleus.

N

Λ

Due to the attraction of

Λ N interaction, the

resultant hypernucleus will

become more stable against

the neutron decay.

Hypernucleus

Λ

neutron decay threshold

γ

nucleus

hypernucleus

Nuclear chart with strangeness

Multi-strangeness system such as Neutron star

Extending drip-line!

Λ

Interesting phenomena concerning

the neutron halo havebeen

observed near the neutron drip line

of light nuclei.

How is structure change when a Λparticle is injected

into neutron-rich nuclei ?

Question : How is structure change when a Λ particle

is injected into neutron-rich nuclei?

n

n

n

n

6H

7He

Λ

t

Λ

Λ

α

Λ

Observed by FINUDA group,

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

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

n

n

Λ

C. Rappold et al., HypHI collaboration

Phys. Rev. C 88, 041001 (R) (2013)

3n

Λ

3n

Λ

E. Hiyama, S. Ohnishi,

B.F. Gibson, and T. A. Rijken,

PRC89, 061302(R) (2014).

n

n

Λ

What is interesting to study nnΛ system?

n

n

Λ

S=0

The lightest nucleus to have a bound state is deuteron.

n+p threshold

n

p

J=1+

d

-2.22 MeV

Exp.

S=-1 (Λhypernucear sector)

n+p+Λ

Lightest hypernucleus to have a bound state

p

n

d+Λ

3H (hyper-triton)

0.13 MeV

Λ

J=1/2+

Λ

Exp.

nnΛbreakup threshold

n

n

?

They did not report the

binding energy.

Λ

scattering length:-2.68fm

Observation of nnΛ system (2013)

Lightest hypernucleus to have a bound state

Any two-body systems are unbound.=>nnΛ system is bound.

Lightest Borromean system.

Do we have bound state for nnΛ system?

If we have a bound state for this system, how much is

binding energy?

nnΛbreakup threshold

n

n

?

They did not report the

binding energy.

Λ

NN interaction : to reproduce the observed binding energies

of 3H and 3He

NN: AV8 potential

We do not include 3-body force for nuclear sector.

How about YN interaction?

To take into account of Λparticle to be converted into

Σ particle, we should perform below calculation using

realistic hyperon(Y)-nucleon(N) interaction.

n

n

n

n

+

Λ

Σ

YN interaction: Nijmegen soft core ‘97f potential (NSC97f)

proposed by Nijmegen group

reproduce the observed binding energies of 3H, 4H and 4He

Λ

Λ

Λ

n

n

n

4He

Λ

4H

Λ

Λ

Λ

-BΛ

p

p

4He

-BΛ

4H

Λ

Λ

Λ

3He+Λ

0 MeV

0 MeV

3H+Λ

1+

1+

1+

1+

-0.57

-0.54

-1.24

-1.00

0+

0+

0+

0+

-2.04

-2.39

-2.28

-2.33

Exp.

Cal.

Cal.

Exp.

What is binding energy of nnΛ?

1/2+

nnΛ threshold

n

n

0 MeV

Λ

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 tuning

strength of YN potential ?

It should be noted to maintain consistency with the binding energies

of 3H and 4H and 4He.

Λ

Λ

Λ

VTΛN-ΣN

X1.1, 1.2

n

Λ

VTΛN-ΣN

VTΛN-ΣN

X1.2

X1.1

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

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 reproduce

the observed data?

Some authors pointed out to have dineutron bound state in

nn system. Ex. H. Witala and W. Gloeckle, Phys. Rev. C85,

064003 (2012).

T=1, 1S0 state

I multiply component of 1S0 state by 1.13 and

1.35. What is the binding energies of nnΛ？

n

n

Λ

n

unbound

nn

0 MeV

-0.066MeV

-1.269 MeV

1S0X1.13

1S0X1.35

n

n

unbound

unbound

0 MeV

nnΛ

Λ

1/2+

-1.272 MeV

We do not find any possibility to have a bound state

in nnΛ.

N+N+N

3H (3He)

-7.77(-7.12)

-8.48 (-7.72)

-9.75 (-9.05)

-13.93 (-13.23)MeV

1/2+

Exp.

Cal.

1/2+

Cal.

Cal.

Motivated by the reported observation of data suggesting

a 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 this

system.

‘Nonexistence of a Λnn bound state’

H. Garcilazo and A. Valcarce,

Phys. Rev. C 89, 057001 (2014).

They did not find any bound state in nnΛ system.

However, the experimentally they reported

evidence for a bound state. As long as we believe the data,

we should consider additional

missing elements in the present calculation. But, I have no idea.

Experimentally, they did not report any binding energy.

It might be good idea to

perform search experiment of nnΛ system at Jlab to conclude

whether or not the system exists as bound state experimentally.

Or I hope to perform search experiment at GSI again in the

Future.

n

n

n

n

3H

3n

Λ

Λ

p

n

(e,e’K+)

n

6He : One of the lightest

n-rich nuclei

6He

α

Λ

n

n

7He: One of the lightest

n-rich hypernuclei

Λ

7He

Λ

α

Λ

Observed at JLAB,

Phys. Rev. Lett. 110, 12502 (2013).

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

6He

7He

Λ

2+

α+Λ+n+n

0 MeV

0 MeV

α+n+n

One of the excited

state was observed at Jlab.

5He+n+n

0+

Λ

Λ

Exp:-0.98

-1.03 MeV

Neutron

halo states

5/2+

3/2+

-4.57

BΛ: CAL= 5.36 MeV

BΛ：EXP=5.68±0.03±0.25

1/2+

Observed at J-Lab experiment

Phys. Rev. Lett.110,

012502 (2013).

-6.19 MeV

n

n

n

Λ

Λ

p

p

4He

4H

Λ

Λ

However, Λ particle has no charge.

4He

4H

Λ

Λ

n

p

n

n

p

p

p

p

+

+

Λ

Λ

p

p

Σ-

Σ-

p

n

+

+

+

+

n

n

n

n

n

n

p

p

+

+

Σ0

n

Σ0

p

n

Σ+

Σ+

p

In order to explain the energy difference, 0.35 MeV,

N

N

N

N

+

N

Λ

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.

0 MeV

3H+Λ

0 MeV

-1.00

-1.24

1+

1+

(Exp: 0.24 MeV)

(cal: -0.01MeV(NSC97e))

-2.04

0+

-2.39

0+

(Exp: 0.35 MeV)

(cal. 0.07MeV(NSC97e))

n

n

p

n

Λ

p

Λ

p

4H

・A. Nogga, H. Kamada and W. Gloeckle,

Phys. Rev. Lett. 88, 172501 (2002)

4He

Λ

Λ

・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

N

N

N

+

Σ

N

Λ

N

3H+Λ

0 MeV

3He+Λ

-1.00

-1.24

1+

1+

(Exp: 0.24 MeV)

(cal: -0.01MeV(NSC97e))

-2.04

0+

-2.39

0+

(Exp: 0.35 MeV)

(cal. 0.07MeV(NSC97e))

n

n

p

n

Λ

p

Λ

p

4H

4He

Λ

Λ

There exist NO YN interaction to reproduce the data.

For the study of CSB interaction, we need more data.

It is interesting to investigate the charge symmetry breaking effect

in 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.

p

p

n

n

n

p

α

α

α

6Be

6Li(T=1)

6He

n

n

p

p

n

Λ

Λ

Λ

α

α

α

7He

7Be

7Li(T=1)

Λ

Λ

Λ

Then, A=7Λ 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.

1.54

Emulsion data

Emulsion data

6He

6Be

6Li

(T=1)

BΛ=5.16 MeV

BΛ=5.26 MeV

JLAB:E01-011 experiment

Preliminary data: 5.68±0.03±0.22

-3.79

7Be

7Li

(T=1)

Λ

Λ

7He

Λ

Can we describe the Λ binding energy of 7He observed at JLAB

usingΛNinteraction to reproduce the Λ binding energies of

7Li (T=1) and 7Be ?

To study the effect of CSB in iso-triplet A=7 hypernuclei.

Λ

Λ

Λ

p

n

n

n

p

p

Λ

Λ

Λ

α

α

α

7He

7Be

7Li(T=1)

Λ

Λ

Λ

For this purpose, we study 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)

Now, it is interesting to see as follows:

- 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?

Without CSB

6Be

(Exp: -0.14)

(exp:-0.98)

6Li

6He

(T=1)

EXP= 5.16

BΛ：CAL= 5.21

EXP= 5.26

BΛ：CAL= 5.28

BΛ：EXP= 5.68±0.03±022

JLAB:E01-011 experiment

CAL= 5.36

7Be

Λ

7Li

(T=1)

Λ

7He

Λ

Now, it is interesting to see as follows:

- 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

3He+Λ

0 MeV

3H+Λ

-1.00

-1.24

1+

1+

0.24 MeV

-2.04

-2.39

0+

0.35 MeV

0+

n

n

p

n

Λ

p

Λ

p

Exp.

4H

4He

Λ

Λ

5.28 MeV( withourt CSB)

5.21 (without CSB)

5.44(with CSB)

5.29 MeV (With CSB)

5.36(without CSB)

5.16(with CSB)

BΛ：EXP= 5.68±0.03±0.22

Inconsistent with the data

p

n

α

and those of A=7 and A=10,

tendency of BΛ is opposite.

How do we understand these difference?

We get binding energy by decay π spectroscopy.

decay

π-+1H+3He →2.42 ±0.05 MeV

4He

Λ

π-+1H+1H+2H → 2.44 ±0.09 MeV

Total: 2.42 ±0.04 MeV

Then, binding energy of 4He is reliable.

Λ

decay

π-+1H+3H →2.14 ±0.07 MeV

Two different modes

give 0.22 MeV

4H

Λ

π-+2H+2H → 1.92 ±0.12 MeV

Total: 2.08 ±0.06 MeV

This value is so large

to discuss CSB effect.

Then, for the detailed CSB study, we should perform experiment to

confirm the Λ separation energy of 4H.

Λ

For this purpose, at Mainz, they performed ・・・

4He (e, e’K+) 4H

Λ

Preliminary data

4ΛH

Preliminary

0+ state

Data was taken by Tohoku Univ. Group (S. Nagao et atl.) at Mainz.

A=4

H

He

4

4

Λ

Λ

JLab (e,e’K+)

ΔBΛ=0.28 MeV!

still we have CSB effect.

Preliminary

BΛ = 2.11±0.03(stat.)±0.09(syst.)

p

p

p

n

n

-BΛ (MeV)

n

Λ

Λ

3H + Λ

3He + Λ

-1.00

-1.24

1+

1+

J-PARC E13

-2.04

0+

-2.39

0+

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

6He

Another interestingissue is to study the excited states of 7He.

Λ

7He

Λ

2+

α+Λ+n+n

0 MeV

0 MeV

α+n+n

One of the excited

state was observed at JLab.

5He+n+n

0+

Λ

Λ

Exp:-0.98

-1.03 MeV

Neutron

halo states

5/2+

3/2+

-4.57

BΛ: CAL= 5.36 MeV

BΛ：EXP=5.68±0.03±0.25

1/2+

Observed at J-Lab experiment

Phys. Rev. Lett.110,

012502 (2013).

-6.19 MeV

7Li(e,e’K+)7ΛHe

At present, due to poor statics,

It is difficult to have the third peak.

Theoretically, is it possible to

have new state?

Let’s consider it.

(FWHM = 1.3 MeV)

Fitting results

Good agreement with my prediction

Question: In 7He, do we have any other new states?

If so, what is spin and parity?

Λ

First, let us discuss about energy spectra of 6He core nucleus.

(2+,1-,0+)?

Γ=1.6 ±0.4 MeV

2.6±0.3 MeV

2+

2

Γ=0.11 ±0.020 MeV

Γ=0.12 MeV

1.797 MeV

2+

1.8 MeV

2+

1

0 MeV

α+n+n

0 MeV

α+n+n

-0.98

-0.98

0+

0+

6He

6He

Exp.

Exp.

Data in 2002

Data in 2012

X. Mougeot et al., Phys. Lett. B

718 (2012) 441. p(8He, t)6He

Core nucleus

Γ=1.6 ±0.4 MeV

1.6±0.3 MeV

2+

2

2.5 MeV

Γ=

Γ=0.12 MeV

Decay with

Is smaller than

Calculated with.

0.8 MeV

2+

Γ=

1

0 MeV

α+n+n

0.79 MeV

-0.98

0+

What is my result?

-0.97 MeV

6He

theory

Exp.

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

Data in 2012

X. Mougeot et al., Phys. Lett. B

718 (2012) 441. p(8He, t)6He

Question: What are theoretical results?

Γ=1.6 ±0.4 MeV

2.6±0.3 MeV

2+

2

These are resonant states.

I should obtain energy position

and decay width.

To do so, I use complex scaling

method which is one of powerful

method to get resonant states.

Γ=0.12 MeV

1.8 MeV

2+

1

0 MeV

α+n+n

-0.98

0+

6He

Exp.

Data in 2012

X. Mougeot et al., Phys. Lett. B

718 (2012) 441. p(8He, t)6He

Complex scaling is defined by the following transformation.

As a result, I should solve this Schroediner equation.

E=Er+ iΓ/2

Question: What are theoretical results?

Γ=1.6 ±0.4 MeV

Γ=4.81 MeV

1.6±0.3 MeV

2+

2.81 MeV

2+

2

2

Γ=0.12 MeV

Γ=0.14 MeV

0.8 MeV

2+

0.8 MeV

2+

1

1

0 MeV

α+n+n

0 MeV

α+n+n

-0.98

-0.98

0+

0+

6He

6He

Exp.

Cal.

Data in 2012

X. Mougeot et al., Phys. Lett. B

718 (2012) 441. p(8He, t)6He

How about energy spectra of 7He?

Λ

Λ

Γ=4.81 MeV

2.81 MeV

2+

2

3/2+,5/2+

Γ=0.14 MeV

0.8 MeV

2+

1

0 MeV

α+n+n

-0.98

0+

6He

Cal.

I propose to

experimentalists to

observe these states.

3.4 nb/sr

40% reduction

11.6 nb/sr

10.0 nb/sr

49.0 nb/sr

I think that

It is necessary to estimate

reaction cross section

7Li (e,e’K+).

Motoba’s Cal.

(preliminary)

Motoba san recently

estimated differential cross

sections for each state.

At Elab=1.5 GeVand θ=7 deg (E05-115 experimental kimenatics)

7Li(e,e’K+)7ΛHe

At present, due to poor statics,

It is difficult to have the third peak.

But, I hope that next experiment

at Jlab will observe the third peak.

(FWHM = 1.3 MeV)

Fitting results

Good agreement with my prediction

Third peak???

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