Structure of four-baryon systems

E. Hiyama (RIKEN)

One of the important and interesting subjects to study Four-baryon systems based on four-body systems

n n

p p

n

p p

Λ n

p

Λ

n p

Λ

n

Λ

4He 4HeΛ

4HΛ

4HΛΛ

(1)Why is it important to study these four-body systems?

(2)What kind of new understandings do we obtain by solving these systems as four-body problems?

・ A variational method using Gaussian basis functions

・ Take all the sets of Jacobi coordinates

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

Our few-body caluclational method

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,

Exsotic atoms / molecules ,

3- and 4-nucleon systems,

multi-cluster structure of light nuclei,

An example of the accuracy of the method:

rR 1

1

rR

r

R22

3

3

C=1 C=2C=3

rR 1

1

rR

r

R22

3

3

C=1 C=2C=3

Basis functions of each

Jacobi coordinate

Determined by diagonalizing H

CNL,lm

An important issue of the variational method is how to selecta good set of basis functions.

What is good set of basis functions?

(1) To describe short-ranged correlation and long-range tail behaviour, highly oscillatory character of few-body wave functions, etc.

(2) Easily to calculate the matrix elements of Hamiltonian

Hin= <Φi | H | Φn >, Nin = <Φi | 1 | Φ ｎ >

For this purpose, we use the following basis function:

νn=(1/rn )2

rn=r1an-1 (n=1-nmax)

The Gaussian basis function is suitable not only for the calculation of the matrix elements but also for describingshort-ranged correlations, long-ranged tail behaviour.

Φlmn(r)

Next, by solving eigenstate problem, we get eigenenergy E and unknown coefficients 　 Cn .

( Hi n) - E ( Ni n ) Cn =0

Hin= <Φi | H | Φn >

Nin = <Φi | 1 | Φ ｎ >

We calculate the energy and overlap matrix elements.

Gauss Expansion Method

Four-body calculation

apply

A good example about the accuracy of the method

n n

p p

n

p p

Λ n

p

Λ

n p

Λ

n

Λ

4He 4HeΛ

4HΛ

4HΛΛ

4He

4-nucleon bound stateNN:AV8

Good agreement among the 7 different method in the binding energy, r.m.s. radius and two-body correlation function

p p

n n

Benchmark-test calculation of the 4-nucleon bound state

1. Faddeev-Yakubovski (Kamada et al.)

2. Gaussian Expansion Method (Kamimura and Hiyama et al.)

3. Stochastic varitional (Varga et al.)

4. Hyperspherical variational (Viviani et al.)

5. Green Function Variational Monte Carlo (Carlson at al.)

6. Non-Core shell model (Navratil et al.)

7. Effective Interaction Hypershperical HarmonicsEIHH (Barnea et al.)

PRC64, 044001(2001)

We have powerful method to solve four-body bound state.

One of the next challenging projects:to calculate properties of the second 0+ state of 4He using realistic force

(1) In investigating the spatial structure of the 0+2 state in 4He,

it is very important to explain the observed (e,e’) transition form factor, to reproduce inelastic electron scattering data.

(2) By the analysis of the low-momentum-transfer part of the form factor Second 0+: 11 % of the energy-weighted E0 sum rule limit

medium-heavy or heavy nuclei: 80-90 % of the limit collective, breathing mode

Where is the major component of the sum rule value Situated?

H. M. Hofmann and G. M. Hale, Nucl. Phys. A613, 69 (1997).

A. Cstoto and G. M. Hale, Phys. Rev. C55, 2366 (1997).

P. Navratil and B. R. Barrett, Phys. Rev. C59, 1906 (1999).

A. C. Fonseca, G. Hale, and J. Haidenbauer, Few-Body Syst.31, 139 (2002).

within the limited model spaceomitted Coulomb force

Use of AV8’ + Coulomb force + phenomenological 3-body force

well reproduces simultaneously

1

B.E.(3H ) : -8.42 MeV (CAL)

-8.48 (EXP)

B.E.(3He) : -7.74 MeV (CAL)

-7.77 (EXP)

B.E.( 4He) : -28.44 MeV (CAL)

-28.30 (EXP)

1.658 fm (CAL)

1.671±0.014 fm (EXP)r.m.s. charge radius of 4He (0+ ) :

How about the second 0+ state?

1

Energy of 0+ of 4He

3He+n20.58

19.82

0.0 MeV(-28.30 MeV) (-28.44)

0.0 MeV

3N+N20.36

0+

0+

0+

0+1 1

2 2

ExEx

3H+p

4He 4He

2

20.2120.31

EXP CAL

Answer to question 1) :

mass density

transition density

The density of the second 0+ state

in the internal region is so much lower than that of theground state.

The Fourier transformation of the

transition density gives the

form factor of the inelastic electron

scattering.

Hiyama, Gibson, KamimuraPhys. Rev. C70 (2004) 031001(R).

We understand that our method is suitable for describing

both compactly bound states and loosely coupled states

as well as transitions between them.

0+1

0+2

4He

(e,e’)

Answer to question 2) :

Energy-weighted E0 sum rule in

Where is the major component of the E0 sum rule limit situated?

E0 strength of the second 0+ state,

was derived from the data:

4He

0+1

0+2

4He

( This has been a long-standing puzzle of 4He nucleus since 1970’s. )

=1.10 ± 0.16 fm2 (EXP),

But, this contribution from the second 0+ state

exhausts only 11 % of the sum rule limit.

The second 0+ state is not a collective mode.

We solved this puzzle as following:

1) Using the 4-body Gaussian basis functions,

we diagonalized the total Hamiltonian.

2) Since we employed ～ 20,000 basis functions, we obtained the same number of 4-body eigenstates.

Above the second 0+ state, we have presicely-discretized non-resonant continuum states.

3) We calculated the energy-weighted E0 strength to

all the discretized eigenstates, and found the following fact:

precisely-discretized non-resonant continuum states

0+1

0+2

4He

（～ 20,000 )

Energy-weighted E0 sum rule:

　　　　

200 discretized-

continuum states.

Contribution from these continuum states

amounts to 70 % of the sum rule limit.

a major part of the energy-weighted E0

sum rule is exhausted by non-resonant,

low-energy continuum states.

20.2 MeV

60 MeV

0+1

0+2

(17 %)

Energy-weighted E0 sum rule:

4He (CAL)

(70 %)

(13 %)

0 MeV

Answer to question 3) :

Ex

N N

4He

Λ

Replacen

p p

Λ

n

p

Λ

n

4HeΛ

4H

N N

Λ Ｎー Σ Ｎ　 coupling effectin single Λ hypernuclei

In non-strangeness nuclei,NN-NΔcoupling 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.

Interesting Issues for the ΛN-ΣN particle conversionin hypernuclei

(1)How large is the mixing probability of the Σ particle in the

hypernuclei?

(2) How important is the Λ Ｎー Σ Ｎ coupling in the binding energy of the Λ hypernuclei?

(3) How large is the Σ-excitation as effective three-body

ΛNN force?

-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).

ΨJM(A=4)=ΣΦc(rc,Rc,ρc)N=1

8

4He, 4HΛ Λ

VNN : AV8 potential

VYN : Nijmegen soft-core ’97f potential

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.

①N1 Λ N2 N3

Σ

N1 Λ N2 N3

②N1 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).

To summarize the part of A=4 hypernuclei

NN

N Λ+

N N

N Σ

NNNΛ + NNNΣ

(1)NNNΣchannel is essentially important to make A=4 hypernuclei.

(2)ΛN-ΣN coupling as a three-body force is important in 0+ state.

For this purpose, recently, we have updated YN interaction such as ESC08,which 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

NΛ

NN N

N NΣ+

4HeΛ

Furthermore to study ΛN-ΣN coupling, we need updated YN interaction.

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)

NΛ

NN N

N NΣ+

Preliminary result (ESC08)

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

In A=4 hypernuclei, we have one more interesting issue:

YN charge symmetry breaking effect

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

4HΛ

4HeΛ

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).

NΛ

NN N

N NΣ+

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)

NΛ

NN N

N NΣ+

Preliminary result (ESC08)

Binding energies are less bound than theobserved data. How do we consider thisinconsistency?

4HeΛ

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

We get binding energy by emulsion data.

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.

NN

N Λ

replace

Λ

Λ

p

Λ

n

4HΛΛ

Interesting issue: ΛΛ ー ΞN coupling

ΛΛ-ΞN coupling

One of the major goals in hypernuclear physics To study structure of multi-strangeness systems (extreme limit : neutron star)

NN

N N

Λ ΛΛΛΛ

N

N

NMulti-strangeness systems

25MeVΛΛ

ΞN

threshold energy difference is very small! It is considered that ΛΛ→ΞN particle conversionis strong in multi-strangeness system. 　

S1/2

P3/2

n n p p Λ Λ Ξ-

PauliForbidden

　・　 I.R. Afnan and B.F. Gibson, Phys. Rev. C67, 017001 (2003).

・　 Khin Swe Myint, S. Shinmura and Y. Akaishi, nucl-th/029090.

・ T. Yamada and C. Nakamoto, Phys. Rev.C62, 034319 (2000).

p

6HeΛΛ

V ΛΛ--ΞN

(3 protons in S1/2)

Effect of ΛΛ － ΞN coupling is small in 　　 6He which was observed as NAGARA event.

ΛΛ

4H 5H ( 5He)ΛΛ

・ I.N. Filikhin and A. Gal, Phys. Rev. Lett. 89, 172502 (2002)

・ Khin Swe Myint, S. Shinmura and Y. Akaishi, Eur. Phys. J. A16, 21 (2003).

・ D. E. Lanscoy and Y. Yamamoto, Phys. Rev. C69, 014303 (2004).

・ H. Nemura, S. Shinmura, Y. Akaishi and Khin Swe Myint, 　 Phys. Rev. Lett. 94, 202502 (2005).

Λ Λ

nn

ΛΛ

Λ Λ

pnΛΛ

For the study of ΛΛ － ΞN coupling interaction,

s-shell double Λ hypernuclei such as

4H and 5H ( 5He) are very suitable.ΛΛ ΛΛ ΛΛ

p

S1/2

P3/2

4HΛΛ

n p Λ Λ Ξ- p

4HΛΛ

V ΛΛ － ΞN

(2 protons in S1/2)

Due to NO Pauli plocking, the ΛΛ － ΞN coupling can be large in 4H

ΛΛ

B.F. Gibson, I.R. Afnan, J.A.Carlson and D.R.Lehman,Prog. Theor. Phys. Suppl. 117, 339 (1994).

Λ Λ

n p

(No data)

n p

Λ Λ

4HΛΛ

The important issue: Does the YY interaction which designed to reproduce the binding energy of 6He make 4H bound?And how does the effect of ΛΛ ー ΞN coupling play important role in the binding energy of 6He and 4H?

ΛΛ ΛΛ

ΛΛ ΛΛ

1)I.N. Filikhin and A. Gal, Phys. Rev. Lett. 89, 172502(2002)2)H. Nemura, Y. Akaishi et al., Phys. Rev. C67, 051001(2002)

n

p

Λ

Λ

VΛΛ

α

Λ Λ

NOT BOUND !

4HΛΛ

6HeΛΛ

NAGARA event

α+Λ+Λ

7.25±0.1 MeV

0+

Did not include ΛΛ-ΞN coupling

ΛΛ-ΞN coupling => ・ significant in 4HΛΛ

n p

Λ Λ

4HΛΛ

・ Not so important in 　 6HeΛΛ

α

Λ Λ 6HeΛΛ

n p

Λ Λ

N

N

N

Ξ

+

4HΛΛ

One of the most numerically difficult 4-body problem

E. HiyamaDr. Nemuran n

Λ Σ

n n

Σ Σ

E(MeV)

0

-3.12

-7.25 0+

Exp.(KEK-E373)

CAL.

5He+ΛΛ

α+Λ+Λ6He

ΛΛwith Pauli blocking

E(MeV)

n+p+Λ+Λ

3H+ΛΛ

NoExp.

unbound

NNΛΛchannel only

4H with no Pauli blockingΛΛ

α

0MeV0.03 MeV

YY: Extended soft core 04VΛΛ-ΞNx1.1

NNΛΛ+ NNNΞIf the bound state of 4H is observed in the future, we can obtain useful

information about ΛΛ-ΞN coupling mechasim.ΛΛ

NNΛΛ+NNNΞ

n+p+Λ+Λ

3H+ΛΛ

0.03MeV

0

28MeV

NNNΞ ？？

4HΛΛ

E(MeV)

For the study of ΞN interaction, it is important to study

the structure of Ξ hypernuclei.

However, so far there was no observed Ξ hypernuclei.

Then, it is important to predict theoretically what kinds of

Ξ hypernuclei will exist as bound states.

nucleus

Ξ Ξ hypernucleus ΞN -ΞN interaction

11B

pK-

Ξ-

11B

12C Ξ hypernucleus

・ E05 “Spectroscopic study of Ξ-Hypernucleus, 12Be, via the 12C(K-,K+)

Reaction” by Nagae and his collaboratorsDay-1 experiment

First observation of Ξ hypernucleus

K+

Approved proposal at J-PARC

This observation will give information about ΞN interaction.

Results ESC04

(3N)+Ξ0 MeV

0+

1+

-2.3

-0.86N N

N Ξ

1+: [12V(1,1)+V(1,0)+10V(0,1)+3V(0,0)]/260+:[V(1,0)+V(0,1)]/2

T,S

= =

repulsive strongly attractive

repulsive

weakly repulsive strongly attractive

Multi-strangeness systemsuch as Neutron star

J-PARC

Concluding remark

GSIJLABMainzJ-PARC

Thank you!