Strange nuclear structures with high density formed by single/double K -- meson

Transcription

1 Strange nuclear structures with high density formed by single/double -- meson 1. Introduction A.Dote (E),Y.Akaishi (E),T.Yamazaki ( 2. ew framework of Antisymmetrized Molecular Dynamics - p- 0 n mixing J & T projections 3. Results ppn -, ppp -, pppn -, 6 Be -, 9 B - and 11 C - 4. Results = extremely proton-rich nuclei = pppp -, ppp - - and pppp Summary & Future plan HYP2003, at Jlab, America

2 Introduction Studies of deeply bound kaonic nuclei by Akaishi & Yamazaki phenomenological potential 1. free scattering data 2. X-ray data of kaonic hydrogen atom 3. binding energy and width of Λ(1405) Analysis of 8 Be - with αα - model Strongly attractive. V I = 0 Binding energy of - = 113 MeV Central density = 5 ρ 0 - deeply bound? p aonic ormal normal nucleus???? n???? Peculiar structure? Highly dense? - What kind of structure does A-nucleons + - system favor?

3 Formalism aonic nuclei unknown structure peculiar structure ex) isovector deformation Method : Antisymmetrized Molecular Dynamics no assumption on clusters, shape, etc. Improving V I = 0 plays an essential role in kaonic nuclei. - 0 p-n mixing - p I=0 int. 0 n Angular momentum & isospin projections

4 Wave function - 0 p/n Essence of mixing ucleon s wave function ϕ Ν n i = C exp υ r α =1 i 2 Z α i ι α σ α σ i α = or υ i 1 ι 1 ι τ α = + γ α p + γ α 2 2 p-n mixing τ i α n = a proton + b neutron = x + y Total wave function Φ = det[ ϕι ] ϕ ± Φ =Φ± PΦ 0 Anti-kaon s wave function ϕ 2 Νk Z α = Cα exp υ r τ α α =1 υ 1 o 1 τα = + γ α + γ α mixing Charge projection ( ˆ ) Z PMΦ = dαexp iα T M Φ ± ± Use P ± MΦ as a trial function

5 J & T projections (VBP) Φ ± After obtaining by Tz-projected AMD cooling, it is projected onto the eigen state of the angular momentum J and the isospin T. J T ± J* P P Φ = dω D ( Ω ) R Ang ( Ω ) M TzTz ' Ang M Ang Ang T * dω D ( Ω ) R iso ( Ω ) Φ iso TzTz ' iso iso ± J projection T projection R Ang ( Ω ) = exp iαj z exp iβj y exp iγ J z R ( ) exp exp exp iso Ω = iαt z iβt y iγt z We calculate various expectation values J T ± with. P M P TzTz ' Φ

6 Formalism 1. Hamiltonian Hˆ = Tˆ+ Vˆ + Vˆ + Vˆ Tˆ Coulomb G { } { } α α α α α α α i i i i 2. Variational parameters X = C, Z, γ, C, Z, γ are determined by Frictional cooling eq. with constraint. 3. G-matrix method Effective interaction bare int = Tamagaki potential (OPEG) bare int = AY potential Vˆ, Vˆ given density and starting energy of G-matrix Repeat until getting consistency AMD calculation density and starting energy of

7 Results ー ppn - ー Model space 2 Gaussian / nucleon 5 Gaussian / kaon Assumption of G-matrix E()=112.8MeV, central density=1.50fm -3 Results Projecting onto J=1/2 and T=0. Parity is positive. E() Width ρ (0) Rrms JT projection simple AMD BHF G-matrix consistency : O! J(J+1) = Quantum numbers After Before J2 (total sys.) J2 ( sys.) L2 ( sys.) S2 ( sys.) JT projection 0.75 : O 0.78! L2 (kaon) T Tz Particle numbers Proton 1.51 eutron bar ppn : pnn =1:1

8 Binding energy of bar and width ucleus threshold ppn ppp pppn 6Be 8Be 9B (simple AMD) E() [ MeV] Σπ threshold Width (Σπ, Λπ) Uncertainty due to the inconsistency between the obtained results and the G-matrix used in the calculation -- is is bound by by ~100 MeV in in these nuclei.

9 ucleon density distribution ppn - ppp - pppn - proton satellite 3 fm 8.8ρ 0 9.2ρ 0 7.6ρ 0 6 Be - 9 B - 4 fm 5.4ρ 0 4.1ρ 0

10 umber of nucleons near - meson ppn - ucleons one center like two center like ppn ppp pppn 6Be 9B Max dens # of nucleons E() strange structure 1 ( ) 5 ρ MAX Single - meson can interact with limited numbers of nucleons? Saturation of E()

11 Larger nucleus case = 11 C - = ucleons aon 4 fm Results E() = 117 MeV Γ= 37 MeV = 1.49 fm R rms LS force : off 4.8ρ 0 A - meson stays at the center of the system. It attracts three clusters.

12 pppp - ppp - - pppp - - Results of extremely proton-rich kaonic nuclei ppp - total B.E MeV E() 96.7 MeV Proton pppp - total B.E MeV E() 75.1 MeV Γ 12.5 MeV Rrms 0.81 fm ρ max 9.2 ρ 0 ppp - - added 3 fm 3 fm Γ MeV Rrms 0.92 fm ρ max 9.9 ρ 0 These kaonic nuclei have proton-satellite structure. - added - added But not so deeply bound. pppp - - total B.E MeV E() MeV Proton total B.E MeV E() MeV Γ 51.9 MeV Γ 41.2 MeV Rrms 0.65 fm ρ max 21.6 ρ 0 3 fm added 3 fm Rrms 0.73 fm ρ max 14.1 ρ 0 tentative

13 Summary & Future plan - We have improved AMD so that we can treat p/nmixing and perform J & T projections. We have calculated various kaonic nuclei ppn -, ppp -, pppn -, 6 Be -, 9 B - and 11 C -. Our results are E() width ρ (0) Rrms β γ [MeV] [MeV] [fm^- 3] [fm] [deg.] ppn ppp pppn Be B C Saturation of E() is related to the number of nucleons with which a - can interact? Even if the interaction is very attractive, extremely proton-rich nuclei are not always deeply bound. Excited states? interaction? - is very deeply bound and forms very highly dense state. 0

14 Introduction Remarks I ( ) = ( ) I 2 v r vd exp r/ 0.66 fm, I ( ) = ( ) I 2 v r v exp / 0.66 fm,, C r πσ 1 I ( ) = exp ( / 0.66 fm) I v r v, C r πλ 2 2 v = 436 MeV, v = 412 MeV, v = none, I= 0 I= 0 I= 0 D C C 1 2 v = 62 MeV, v = 285 MeV, v = 285 MeV. I= 1 I= 1 I= 1 D C C 1 2 Y. Akaishi and T. Yamazaki, PRC 65 (2002) ot single channel, but coupled channel. Same property as interaction derived from Chiral theory.

15 11 C - Matter Proton E() = 117 MeV Γ= 37 MeV ρ MAX = 0.82 fm -3 R rms = 1.49 fm LS force : off eutron aon size : 4fm x 4fm

16 Introduction Studies of deeply bound kaonic nuclei by Akaishi & Yamazaki phenomenological potential 1. free scattering data 2. X-ray data of kaonic hydrogen atom 3. binding energy and width of Λ(1405) Analysis of 8 Be - with αα - model Strongly attractive. V I = 0 Binding energy of - = 113 MeV Central density = 5 ρ 0 p n???????? normal nucleus - What kind of structure does A-nucleons + - system favor? - deeply bound? highly dense? peculiar structure?

17 Proton satellite in ppp - ppp - proton : Proton P=2.67 P=0.99 = neutron : 0.33 = = Isospin He = p 3 2 H 1,1 = 1 2, ,1 2 T,Tz 2 1 p p pn ,1 1 2, , 0 1 2, Clebsch-Gordan coefficient

18 Introduction According to our study of various kaonic nuclei - is deeply bound in light nuclei. E() 100 MeV. Since the main decay channel, Σπ, is closed, the deeply bound state is a discrete state. Due to the strong attraction from a -, nuclei are drastically shrunk to form dense state. Structures peculiar to kaonic nuclei are found. ρ= 4 ~ 8 ρ 0 Proton satellite 1. Isovector deformation in 8 Be - 2. Proton satellite in ppp - P=0.99 Question The strong attraction from - can make extremely proton-rich nuclei bound? How are their structures?

19 aonic nuclei に期待できること 1 ハドロン物理の観点 - が核子を引き寄せ 非常に高密度な状態が核内に 高密度だが 割と冷たい 高密度核物質 カイラル対称性の回復 高密度下での / 相互作用 凝縮 strange matter 2 核構造の観点 - からのアイソスピンに依存した強い引力 従来のクラスター構造が変化 原子核が shrink isovector deformation - は陽子をより強くひきつける extreme proton-rich nuclei pp ppp pppp pp p pppp など 今までの原子核にない新しい構造様式? 多彩な構造?

20 Results // 8 Be and 8 Be - // Binding energy of - = 104 MeV 7 fm Density (/fm^3) Density (/fm^3) Be 8 Be - rmsr = 2.46 fm β = 0.63 Central density = 0.10 /fm^3 force: MV1 case 3 rmsr = 1.42 fm β = 0.55 Central density = 0.76 /fm^3 4.5 normal density

21 Isovector deformation in 8 Be - density distributions along Z-axis V I = 0 >> VI = 1 V p >> V n ew type of neutron skin eutrons are peripherally distributed, yet with quite high density.

22 Double kaonic nuclei Motivation We have a simple question: How do kaonic nuclei behave, if extra-one - is added? Multi - system is related to - condensation, strange quark matter, and so on. Double kaonic nuclei, which contain two - s, are the simplest case of Multi - systems. Formalism Wave func.: [ ] Φ = det S[ ] Symmetrized calculate in quite the same way as single kaonic nuclei. V = 0 since we have no information in the present stage.

23 Double kaonic nucleus // ppn - - // 4 fm 4 fm 4 fm E() = 110 MeV E(2) = 213 MeV Density [fm -3 ] Density [fm -3 ] Density [fm -3 ] ppn ppn - total B.E. = 6.0 MeV central density = 0.14 fm -3 R rms = 1.59 fm total B.E. = 118 MeV central density = 1.50 fm -3 R rms = 0.72 fm ppn - - total B.E. = 221 MeV central density = 3.01 fm -3 R rms = 0.69 fm

24 Double kaonic nucleus // pppn - - // 3 fm 3 fm pppn - E() = 105 MeV central density = 1.29 fm -3 R rms = 0.97 fm pppn - - E(2) = 222 MeV central density = 2.33 fm -3 R rms = 0.73 fm

25 11 C - Matter Proton E() = 137 MeV Γ= 33 MeV ρ MAX = 0.80 fm -3 R rms = 1.43 fm LS force : on eutron aon size : 4fm x 4fm

26 H ppn 密度の解釈 = P Φ Φ 3 T = 0 Z 古いモデルと新しいモデル + 3/2 = PT Z = 0 PT ( P P 1/2 1/2 ) Z = m Φ + Φ TZ =+ TZ = = m= 3/2 = P Φ P Φ + P Φ P Φ T =+ 1/2 T = 1/2 T = 1/2 T =+ 1/2 Z Z Z Z 0 - ppn 前回 pnn p,n, - P T =+ Φ 1/2 P T = Φ 1/2 Z Z 今回 p,n, -,0bar P T = 0 Φ Φ Z Proton eutron

27 3 H = P T = 0 ΦΦ Z - = + ppn pnn ; AMD wave function 0 Here, In addition, ppn H = H = ppn + pnn pnn H = normalized wave function pnn e iπ T y ppn = pnn 0 = e π i T y ppn - Therefore, 1 3 i T - H e θ y ppn θ= 0, π 2 = Intrinsic state in in isospin space 3 holds good in AMD calculation of H( T=0 ).

28 Treatment of - 0 p/n mixing Usual practice = Coupled channel ex) ppn - 0 Φ = C ppn + D pnn Multi-Slater determinants But some problems { } Z i ACZ ACZ 1 1. common for all slater det., { i } a + slater determinants! ( ) 2. different Z a for each slater det., how many slater determinants? 3. How is the effective frictional cooling for multi-slater det.? 4. Calculation of non-diagonal matrix elements is somewhat tedious. a a b ppn b 0 V pnn b Single Singleslater determinat with with charge-mixed s.p. s.p. wave wave func. func., Φ = In the single-particle state, p and n are mixed, and - and 0bar are mixed. contains ppn and pnn = a p + b n = x + y 0 0 But total Tz is restored with the Tz-projection.

29 Treatment of - 0 p/n mixing Φ = C ppn + D pnn Usual practice = Coupled channel ex) ppn - 0 But in AMD treatment non-diagonal matrix element can t be calculated. Calculation of ppn 0 V pnn needs the inverse matrix of the overlap matrix B ij = ϕ ϕ ι j B p p n α β 0 = 0 0 γ 0 0 δ p n n 1 B does not exist! In the Slater determinant Φ =, ppn and pnn 0 are mixed. In the single-particle state, p and n are mixed, and - and 0bar are mixed. = a p + b n = x + y 0 But total Tz is restored with the Tz-projection.

30 Introduction If interaction is very strongly attractive We can obtain a highly dense state by implanting - into normal nuclei. low temperature, high density high temperature, high density Change of nuclear structure : well-developed clustering structure vanishes?? Extending the drip line of unstable nuclei???

31 Stability for strong interaction pppn -- 6 Be p+p+p+n+ - ppn+p+ - ppp - +n pppn - ppn - +p pppn - is unstable for strong interaction p+p+ p+p+n+n+ - ppn+ p+p+n+ - ppnn+ p+p+ - ppp - + p+n+n ppp - + pnn pppn - +p+n pppn - +pn ppn - + p+p+n ppn - + ppn 6 Be Be - is stable for strong interaction.

32 Formalism Hamiltonian Hˆ Tˆ ˆ ˆ ˆ Tˆ ˆ = T + VC + VCoul. + Vˆ TG ˆ In kinetic energy part, taking the mass difference between a nucleon and - into consideration. Frictional cooling eq. with constraint X i α = ( λ + iµ ) negative 1 i H X G η and c.c. i* i* α X α W + G = W G 2 R + P G 2 Constraint condition The center of mass of the system is fixed to the origin. { i } { i i X = C Z, C, Z } α α, α α α complex numbers A m r + m r R P p p i A i= 1 G =, G = i + Am + m i= 1

33 G-matrix calculation around normal density existing effective interaction This time, extremely high density???? G-matrix method bare interaction and model space int.: Tamagaki potential (OPEG) int.: Akaishi-Yamazaki potential model space : 2 Gaussian wave packets for one nucleon, 5 Gaussian wave packets for one anti-kaon. procedure given density and starting energy of G-matrix Repeat until getting a consistent result. AMD calculation density and starting energy of

34 Introduction Changing the matter density of 8Be Y.Akaishi solved αα - system with ATMS method. interaction = Hasegawa-agata o.1 which has 1.6 GeV repulsive core. interaction = Y.A. & T.Y. force. Binding energy = 113 MeV Γ Λ+π = 38 MeV Central density = 5 ρ 0 u αα ( r) 2 The distribution of the αα relative motion in αα system. - causes 8Be to be shrunk. 8Be implanted - loses two α structure.

35 Akaishi-Yamazaki interaction 1, free scattering data A.D.Martin, ucl.phys.b179(1981)33 a I=0 = i 0.46 fm, a I=1 = i 0.60 fm cf) A.D.Martin : a I=0 = i 0.68 fm, a I=1 = i 0.60 fm 2, X-ray data of kaonic hydrogen atom a -p = ( a I=0 +a I=1 ) /2 = i 0.53 fm M.Iwasaki et al. Phys.Rev.Lett.78(1997)3067 T.M.Ito et al. Phys.Rev.C58(1998)2366 cf) T.M.Ito et al. : a -p = (-0.78 ± 0.15 ± 0.03) + i (0.49 ± 0.25 ± 0.12) fm 3, Binding energy and decay width of Λ(1405) B.E.= MeV, Γ= 40 MeV cf) Λ(1405) exists 27 MeV below - +p threshold regarding Λ(1405) as I=0 bound state of V = is much more attractive than!! I 0 V I = 1

36 いろいろな ポテンシャルの作り方 W.Weise Julich グループ (. Holinde ) 赤石 山崎 基礎理論 Chiral SU(3) effective Lagrangian meson exchange 目的 系全体を合わす,Y,, 全てを合わす の bound state にのみ着目 Channel ポテンシャルの導入 p, n, Λπ, Σ π, Σ π, Σ π threshold 下にΛ(1405) というresonance カイラル摂動論で処理しようとすると ある種のダイヤグラムを無限オーダーとらねばならない ポテンシャルで処理 q2 まで取ったカイラルラグラジアンから Born 近似で求めた s 波散乱長と等しい散乱長を持つようなポテンシャル 但しポテンシャルの形は仮定 * * ( 同左 ) +,, 3 相互作用ラグラジアンW = d x L int から W:2 次 one boson exchange, resonance diagram W:4 次 box diagram のポテンシャル 相互作用から G-parity 変換で決まるものはそれを用いる その他は free parameter vertex の form factor は仮定 p, n, Λπ, Σ π, Σ π, Σ π とにかくポテンシャルの形は Gauss 型としてしまえ! 全チャネルで v = v r I x I 2 cxexp[ ( / 0.66) ] local : separable : Yukawa 型 α α 2 2 αi j i + k i α j + kj 再現 Λ(1405) の位置と幅 低エネルギー散乱の全断面積 ( : 60~300 MeV/c in LAB ) Branching ratio Λ(1405) の位置と幅 低エネルギー散乱の全断面積 ( : 60~300 MeV/c in LAB ) Λ(1405) の位置と幅 低エネルギー散乱の散乱長 kaonic hydrogen atom のスペクト 核物質中での散乱振幅の振る舞いは Weise と consistent 論文 ucl.phys.a594(1995)325 ucl.phys.a513(1990)557 Phys.Rev.C65(2002)044005

37 Julich Quasi-potential p LAB = MeV / c Dominant Minor π, ρ ρ ω, ρ σ Λ,Σ Δ π, ρ * Δ ρ Λ π Λ π Λ π Σ π の場合 全てのメソンが引力的に働く * Σ ρ Λ(1405) の形成 Λ π Weise とも consistent local potential を用いた場合 レンジパラメータはこれらのメソン質量に対応 Λ(1405) を作る Σ π Σ π Σ * π Λ,Σ Σ Σ ρ π π

38 potential (G-matrix) for 6 Be - r^2 V [M ev fm ^2] I=1 2 r V() r I= 0 I= 1 1E E I=0 r [fm ]

39 ppn - Density distributions of PTzΦ in ZX plane. Matter Proton eutron 3 fm A=3 P=1.51 =1.49 0bar - =1 0=0.49 -=0.51 ρ = - 0 ppn : pnn = 1:1-3 ( 0) 1.50fm central density

40 ー model space 2 Gaussian / nucleon 5 Gaussian / kaon force E()=94.5MeV, central density=1.49fm -3 Results Projecting onto J=3/2 and T=1. Parity is negative. B.E. width dens0 rmsr beta gamma JT projection G-matrix consistency is a little violated. Quantum numbers After Before J2 (total sys.) J2 ( sys.) L2 ( sys.) S2 ( sys.) L2 (kaon) T Tz Particle numbers Proton 2.67 eutron bar 0.33 Even if we start AMD cooling from various initial value γ which is concerned to the isospin, most of all solutions are converged to this result. JT projection is correctly working.

41 ppp - Density distributions of PTzΦ in ZX plane. Matter Proton eutron 3 fm P=0.99 A=3 P=2.67 =0.33 0bar - =1 0=0.33 -=0.67 He = p 3 2 H ρ = 2 1 = pp pn ( 0) 1.57fm central density - 0

42 ー model space 2 Gaussian / nucleon 5 Gaussian / kaon force E()=99.6MeV, central density=1.31fm -3 Results Projecting onto J=1 and T=1/2. Parity is negative. B.E. width dens0 rmsr beta gamma JT projection G-matrix consistency is a little violated. Quantum numbers Particle numbers After Before J2 (total sys.) J2 ( sys.) L2 ( sys.) S2 ( sys.) L2 (kaon) T Proton 2.60 eutron bar 0.40 Tz JT projection is correctly working.

43 pppn - Density distributions of PTzΦ in ZX plane. Matter Proton eutron 3 fm A=4 P=2.60 =1.40 0bar - =1 0=0.40 -=0.60 ρ = -3 ( 0) 1.29fm central density

44 Results ー 6Be - ー model space force 2 Gaussian / nucleon 5 Gaussian / kaon E()=106MeV, central density=0.96fm -3 Results Projecting onto J=0 and T=1/2. Parity is negative. B.E. width dens0 rmsr beta gamma JT projection G-matrix consistency is accomplished. Quantum numbers Particle numbers After Before J2 (total sys.) J2 ( sys.) L2 ( sys.) S2 ( sys.) L2 (kaon) T Proton 3.79 eutron bar 0.21 Tz JT projection is correctly working.

45 6 Be - Density distributions of PTzΦ in ZX plane. Matter Proton eutron 4 fm A=6 P=3.79 =2.21 0bar - =1 0=0.21 -=0.79 ρ = -3 ( 0) 0.91fm central density

46 9 B - model space 2 Gaussian / nucleon 5 Gaussian / kaon force E()=106MeV, central density=0.96fm -3 for 6 Be - Results Projecting onto J=3/2 and T=0. Parity is negative. E() width dens0 rmsr beta gamma JT projection G-matrix consistency is violated. Quantum numbers Particle numbers After Before J2 (total sys.) J2 ( sys.) L2 ( sys.) S2 ( sys.) L2 (kaon) T Proton 4.69 eutron bar 0.31 Tz JT projection is almost correctly working.

47 interaction Is Is interaction very very strongly attractive? Energy -14 ev??? -27 MeV 1s 2p Λ(1405) int. r Coulomb kaonic hydrogen puzzle repulsive attractive??? atomic 1s shift caused by interaction atomic 2p 1s X-ray measurement : attractiv analysis of low energy data : repulsive Prof. Iwasaki investigated very precise repulsive!! *: M.Iwasaki et.al. Phys.Rev.Lett.78(1997)3067 We recognize Λ(1405) is - p bound state. Adding int., a new bound state (= Λ(1405) ) is generated below the atomic 1s state. Then it must be orthogonal to the new bound state. The original atomic 1s state has a node and its kinetic energy increases. As a result it shifts upward. = repulsive-like Λ(1405) = 3 quark state??

48 lowered 2s = 1s (1sもどき!!) Seeing with nuclear scale, it has a node. So it is 2s state. Seeing with atomic scale, the shape of its wave function is almost that of original 1s state. kaonic hydrogen atom about atomic 1s state shifted by interaction Coulomb only Coulomb + interaction 2s r r 1s r r 50fm 1fm ~ range of interaction 2s state lowered by - interaction is very similar to the original atomic 1s state. If ignoring a node. this lowered 2s state seems to be the solution obtained by changing the boundary condition a little --- ψ=0 at r=0 fm ψ=0 at r=1 fm --- So lowered 2s state appears energetically near the original 1s state.

49 ppn- // dependency on the number of wave packets // Density distribution in YZ plane Matter eutron Proton - 2 Gaussian y β= 0.19 γ= 0 3 Gaussian β= 0.13 γ= 55 y prolate oblate As expected, the neutron distribution is strongly dependent on the number of wave packets. As a result, the shape of the system is largely changed. z z

50 9 B - (tentative) Density distributions of PTzΦ in ZX plane. Matter Proton eutron 4 fm A=9 P=4.74 =4.26 0bar - =1 0=0.26 -=0.74 Similar structure to 8 Be -