binding energies of few-body hypernulear systems and charge symmetry breaking effect
Post on 20-Mar-2016
37 Views
Preview:
DESCRIPTION
TRANSCRIPT
Binding energies of few-body hypernulear systems and charge sy
mmetry breaking effect
E. Hiyama (RIKEN)
・ 7Li(e, e’K+) 7He
・ 10B (e, e’K+) 10Be
Λ
Λ
Recently, at JLAB, the following experiments have been done and analysis is in progress.
To my opinion, these Λ hypernuclei are important forstudy of charge symmetry breaking effect.
Here, I shall talk about the structure of 7He and 10Be.Λ Λ
α
n n
7He
Λ
Λ
α α
n Λ
10BeΛ
Major goals of hypernuclear physics
Fundamental and important for the studyof nuclear physics
1) To understand baryon-baryon interactions
2) To study the structure of multi-strangeness systems
Due to the difficulty of YN and YY 2-body scattering experimentfor the study of baryon-baryon interaction, the systematic investigation of the structure of light hypernuclei is essential.
Hypernuclear -ray data since 1998 by Tamura
・ Millener (p-shell model), ・ Hiyama (few-body)
ΛN interaction (effectively including ΛN -ΣN coupling)
Comparing the theoretical study using few-body technique andshell-model approach (done by Millener, Motoba) with the experimental results,we could succeed in extracting the information about ΛN interaction.
In S= -1 sector, what are the open questions in YN interaction?
(1) Charge symmetry breaking
(2) ΛN - ΣN coupling
・ E13 “γ-ray spectroscopy of light hypernuclei”
by Tamura and his collaborators 11B 4HeΛ Λ
・ E10 “Study on Λ-hypernuclei with the doubleCharge-Exchange reaction”
by Sakaguchi , Fukuda and his collaboratiors
9HeΛ6HΛ
J-PARC : Day-1 experiment
(1) Charge Symmetry breaking
N+N+N
1/2+- 8.48 MeV
0 MeV
- 7.72 MeV1/2+
3H
3He
n np
np
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.
3He+Λ0 MeV
-1.24
-2.39
1+
0+
3H+Λ0 MeV
-1.00
-2.04
1+
0+
0.35 MeV
n n
p Λ
4HΛ
n
p Λ
4HeΛ
p
Exp.
0.24 MeV
Charge Symmetry breaking
n
p Λ
4HeΛ
p n n
p Λ
4HΛ
However, Λ particle has no charge.
n
p Λ
p
p Σ-
pp
p Σ0
p n
p Σ+
n n
+
++
+
4HeΛ
n
p Λ Σ-
pp
Σ0
p n
Σ+
n n
+
++
+
4HΛ
n
n
n n
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).
N
Λ
N
N N
N N
Σ+
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
(Exp: 0.24 MeV)
(Exp: 0.35 MeV)(cal. 0.07 MeV(NSC97e))
(cal: -0.01 MeV(NSC97e))
If we consider that these experimental data are correct, why could notwe reproduce the data?
Talked by Nakamura, Yesterday
According to Nakamura’s talk, the Λ separation energy of 4He is reliable.However, there is energy difference, about 0.2 MeV, between difference 2 modes in 4H.
Λ
Λ
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, somewhere, it is necessary to perform ・・・Furthermore, we need more Λ hypernuclear data to get information on CSB.
For this purpose,It is interesting to investigate the charge symmetry breaking effectin 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.
α
n
α
n pn
α
pp
6He 6Li(T=1) 6Be
α
n
α
n pn
α
pp
7He 7Li(T=1) 7Be
ΛΛ Λ
ΛΛ Λ
Then, A=7 Λ 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.
Exp.1.54
-3.79
BΛ=
5.16
MeV
BΛ=
5.26
MeV
Pre
limin
ary
data
: 5.6
8±0.
03±0
.22
7BeΛ(T=1)7LiΛ
7HeΛ
6He
6Li(T=1)
6Be
JLA
B:E
01-0
11 e
xper
imen
t
Emulsion data Emulsion data
Important issue: Can we describe the Λ binding energy of 7He observed at JLAB using ΛN interaction 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
α
n
α
n pn
α
pp
7He 7Li(T=1) 7Be
ΛΛ Λ
ΛΛ 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)
α
Λ
np
Li7Λ
ΛN interaction: Nijmegen ’97fNot original one but simulated one
The ΛN-ΣN coupling interaction can berenomalized into the ΛN-ΛN interaction effectively.
VΛN=V0+σΛ ・ σNVs+(σΛ+σN)/2 ・ VSLS+(σΛ-σN)/2 ・ VALS
Made by Yamamoto so as to reproduce thephase shifts given by the original one
Strengths of Vs,VSLS,VALS are adjusted so as toreproduce of the observed data of 4H, 7Li(T=0), 9Be and 13C.Λ Λ Λ Λ
Now, it is interesting to see as follows:
(1)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?
BΛ
:CA
L=
5.21
BΛ
:CA
L=
5.28
6Be6Li
7BeΛ
(T=1)
(T=1)7LiΛ
6He
7HeΛ
EX
P=
5.26
EX
P=
5.16
CA
L= 5
.36
BΛ
: E
XP
= 5
.68±
0.03
±022
prel
imin
ary
(Exp: 1.54)
(Exp: -0.14)(exp:-0.98)
JLA
B:E
01-0
11 e
xper
imen
t
Without CSB
Next we introduce a phenomenological CSB potential with the central force component only.
3He+Λ0 MeV
-1.24
-2.39
1+
0+
3H+Λ0 MeV
-1.00
-2.04
1+
0+
0.35 MeV
n n
p Λ
4HΛ
n
p Λ
4HeΛ
p
Exp.
0.24 MeV
Strength, rangeare determinedao as to reproduce the data.
BΛ
: E
XP
= 5
.68±
0.03
±0.2
2
α
p n
5.36
(with
out C
SB
)
5.21
(with
out C
SB
)
5.16
(with
CS
B)
5.44
(with
CS
B)
With CSB
Inconsistent with the data5.
28 M
eV( w
ithou
rt C
SB
)
5.29
MeV
(With
CS
B)
Comparing the data of A=4and those of A=7,tendency of BΛ is opposite.
How do we understand these difference?
Why CSB interaction which reproduce the energydifference of A=4 hypernuclei, do not reproduce theenergy difference in p-shell hypernuclei such as A=7 system? (1)In my calculation, ΛN-ΣN coupling effect is not included explicitly and mass difference of Σ.
(2) As talked by Dr. Nakamura, the binding energy of 4H isincorrect. Then, we should measure the binding energy ofthis hypernucleus again. If the experiment will be done successful,It might the energies of 4H and 4He are the same.
Λ
(3) odd-state CSB interaction whose contribution is negligible inA=4 hypernuclei, contribute to p-shell Λ hypernuclei withopposite sign of the even state of CSB interaction.
Λ Λ
α
n
α
n pn
α
pp
7He 7Li(T=1) 7Be
ΛΛ Λ
ΛΛ Λ
A=7 Λ hypernuclei are p-shell nuclei. Then, it is possible that odd stateCSB interaction contribute to those hypernuclei.
Now, let me introduce a phenomenological odd state CSB interaction.
Parameters are adjustedso as to reproduce the observed binding energy of 7He.Λ
In order to check the validity of the odd CSB interaction,It is suited for study the structure of A=10 Λ hypernucleisuch as 10B and 10Be.Λ Λ
α α
n Λ
10BeΛ
α α
p Λ
10BΛ
At JLAB, the analysis is in progress.
Exp. BΛ=9.11±0.22 MeVNumber of event (emulsion data): 3
Exp. BΛ=8.89±0.12 MeVNumber of event (emulsion data): 10
These are p-shell Λ hypernuclei.
α α
n Λ
10BeΛ
α α
p Λ
10BΛ
At JLAB, the analysis is in progress.
Exp. BΛ=9.11±0.22 MeVNumber of event (emulsion data): 3
Exp. BΛ=8.89±0.12 MeVNumber of event (emulsion data): 10
If the odd CSB interaction to reproduce the binding energy of 7Hereproduce the binding energy of 10Be which will be reported soon,we can check the validity of odd state CSB interaction.
Λ
Λ
BΛ
:CA
L=
5.21
BΛ
:CA
L=
5.28
6Be6Li
7BeΛ
(T=1)
(T=1)7LiΛ
6He
7HeΛ
EX
P=
5.26
EX
P=
5.16
CA
L= 5
.18
→5.
66B
Λ:
EX
P=
5.6
8±0.
03±0
22pr
elim
inar
y
(Exp: 1.54)
(Exp: -0.14)(exp:-0.98)
JLA
B:E
01-0
11 e
xper
imen
t
Now, I shall show you the Λ separation energy of 10B and 10Be.Λ Λ
α α
n Λ
10BeΛ
α α
p Λ
10BΛ
At JLAB, the analysis is in progress.
Exp. BΛ=9.11±0.22 MeVNumber of event (emulsion data): 3
Exp. BΛ=8.89±0.12 MeVNumber of event (emulsion data): 10
(1)Results without CSB interaction(2)Results with CSB interaction
α α
n Λ
10BeΛ
At JLAB, the analysis is in progress.
Exp. BΛ=9.11±0.22 MeVNumber of event (emulsion data): 3
Without CSB interaction
CAL:BΛ=8.76 MeV
α α
p Λ
10BΛ
Exp. BΛ=8.89±0.12 MeVNumber of event (emulsion data): 10
CAL:BΛ=8.56 MeV
With CSB interaction which reproduce the observed binding energy of 7HeΛ
α α
n Λ
10BeΛ
Exp. BΛ=9.11±0.22 MeVNumber of event (emulsion data): 3
BΛ=8.76 MeV(without CSB)
α α
p Λ
10BΛ
Exp. BΛ=8.89±0.12 MeVNumber of event (emulsion data): 10
BΛ=8.56 MeV (without CSB)
BΛ= 8.97 MeV (with CSB) BΛ=8.35 MeV (with CSB)
If the calculated results are consistent with the observed data at JLAB in thefuture, we can extract information on the odd state of CSB interaction.
Concluding remark
For the study of CSB interaction, we need BΛ within 100keVaccuracy.For this purpose, (e,e’K+) reaction might be powerful tool.
For the study of CSB interaction,
4He (e,e’K+) 4HΛ Maintz? JLAB?
And I calculated Λ separation binding energies of 7He, 7Li(T=1), 7Be,and 10Be and 10B. Λ Λ Λ
Λ Λ
10B(π+,K+) 10BΛ Where?I need the following data.
I hope to measure this hypernucleus somewhere.
For detailed study of CSB effect in A=4 hypernuclei
4HΛ
n p
n Λ
n Σ-
p p
Σ0
p
n Σ+n
n nn+ + +
4HeΛ
n p
Λ
Σ-
p p
Σ0
p
n Σ+n
n+ + +
p
p
p p
realistic NN and YN interaction such as ESC08a
Thank you!
α α
p Λ
10BΛ
We have one more interesting issue in10B.Λ
α+α+p
3/2-
1-
2-
α α
p
9B
α+α+p+Λ
γ
We could not obtain observed γ-ray.
ΔE<100keV?or 2- is ground state?
What value in my calculation?
ΛN interaction: Nijmegen ’97fNot original one but simulated one
The ΛN-ΣN coupling interaction can berenomalized into the ΛN-ΛN interaction effectively.
VΛN=V0+σΛ ・ σNVs
Strengths of even state and odd state spin-spin interactionsare adjusted so as to reproduce of the observed data of4H , 4He and 7Li(T=0).Λ Λ Λ
α α
N Λ
10BeΛ
Firstly, I shall show you the results without CSB interaction.
3/2-
α+α+p
α+α+p+Λ
0.275
-8.55
BΛ =8.825 M
eV
1-
2- -8.33
even state
2-
1-
-8.28
-8.25
odd state
BΛ=8.56 MeV
α α
p
9B
α α
p Λ10BΛ
The same tendency in 10BeΛ
0 MeV
top related