Structure of four-baryon systems
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Transcript of Structure of four-baryon systems
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!