On Description of the Yrast Lines in IBM-1 V. Garistov Institute of - PDF document

Nuclear Theory’21 ed. V. Nikolaev, Heron Press, Sofia, 2002 On Description of the Yrast Lines in IBM-1 V. Garistov Institute of Nuclear Research and Nuclear Energy, Bulgarian Academy of Sciences, Sofia 1784, Bulgaria Abstract. The geometric representation of IBM-1 is applyed for calculations of the energies of yrast lines in even-even deformed nuclei. The yrast lines can be successfully explained as a crossing of several number of rotational β -bands if the structure of the corresponding bands heads is taken into account. The geometric properties of the interacting boson model are particularly im- portant since they allow one to relate this model to the description of collective states in nuclei by shape variables [1]. There is a large variety of problems that can be attacked with this representation of IBM introduced in nuclear physics by Gilmore and Feng [2], Ginocchio and Kirson [3], Dieperink, Sscolten and Iachello [4], Bohr and Mottelson [5]. In this paper we apply the geometric representation of IBM-1 model follow- ing [6] in description of the yrast lines energies of the even-even deformed nuclei. Recently it was shown that the energies of the yrast lines can be explained with an acceptable accuracy as the crossing of some number of the rotational β − bands even if the energies of the rotational β − bands are calculated within the framework of simple rigid-rotator model [7]. Let us remind that in this pa- per [7]. the distribution of the excited 0 + - states energies as a function of number of bosons p is described with simple formula: E p = Ap − Bp 2 (1) where p is the number of pure monopole bosons ( b + , b ) connected with bosons R + , R − , R 0 through applying the T.Holstein-H.Primakoff [8] transformation: � R + = b † � R 0 = b + b − Ω R − = 2Ω − b † b b ; 2Ω − b † b ; 77

78 On Description of the Yrast Lines in IBM-1 2.5 3.0 194 Pt 2.5 114 Cd 2.0 2.0 1.5 α =0.763518 α =1.19593 1.5 MeV β =0.0719888 1.0 β =0.289235 1.0 0.5 Experiment Experiment 0.5 Calculations Calculations 0.0 0.0 0 1 2 3 4 0 1 2 3 4 5 6 7 8 9 10 11 p p 0 1 2 3 4 2.5 1.5 1.5 168 Yb 172 Yb 2.0 α =1.17889 1.5 α =1.3206 1.0 1.0 MeV β =0.397842 β =0.35529 1.0 0.5 0.5 0.5 Experiment Experiment Calculations Calculations 0.0 0.0 0.0 0 1 2 3 4 0 1 2 3 4 p p 2.0 2.0 156 Gd 178 Hf 1.5 1.5 α =1.21661 α =1.17889 1.0 1.0 β =0.318346 MeV β =0.397842 0.5 0.5 Experiment Experiment Calculations) Calculations 0.0 0.0 0 1 2 3 4 0 1 2 3 4 p p 2.0 3.0 188 Os 2.5 158 Er 1.5 2.0 α =0.762154 α =1.23148 1.0 1.5 MeV β =0.101847 β =0.219335 1.0 0.5 Experiment Experiment Calculations 0.5 Calculations 0.0 0.0 0 1 2 3 4 5 6 7 8 0 1 2 3 4 5 6 p p Figure 1. Comparison of calculated and experimental data of low-lying 0 + excited states.

V. Garistov 79 constructed from pairs of fermions R + = 1 R − = 1 � ( − 1) j − m α † jm α † � ( − 1) j − m α j − m α jm ; j − m ; 2 2 m m R 0 = 1 � ( α † jm α jm − α j − m α † j − m ) , 4 m and commutating as: Ω = 2 j + 1 [ R 0 , R ± ] = ± R ± ; [ R + , R − ] = 2 R 0 ; 2 Figure 1 shows that formula (1) provides perfect description of the experi- mental data for large amount of nuclei. Further we use this classification in our calculations of the energies of the rotational β − bands labeling each 0 + state with its own number of bosons p. We also take the nucleus mean square radius R ms ( p ) to play a role of the carrier of this information about band head collective struc- ture to formulae of the energies of rotational β − bands. From [7,9] we have the expression of the mean square radius of the nucleus in any excited 0 + state with the degree of collectivity determined by number of monopole bosons p : � 3 r 2 0 (15 E 2 0 ( p − 1) p + 80 E 0 pπC 0 + 32 π 2 C 2 0 ) R ms ( r 0 , E 0 , C 0 , p ) = (2) 20 C 0 (3 E 0 p + 8 π 2 C 0 ) 1 3 , C 0 - nuclear surface com- with equilibrium radius of the nucleus r 0 = 1 . 287 A pressibility parameter and E 0 − one phonon excited 0 + state energy. Now for each of three decomposition of U (6) symmetry chain following [6] we write the energies of the bands in terms of geometric representation parame- ters e 0 , ǫ 1 , ǫ 2 , k, k ′ , β,γ and the eigenvalue of the first Casimir N as follow: U (6) ⊃ U (5) ⊃ O (5) ⊃ O (3) ⊃ O (2) : β 2 β 4 1 + β 2 + k ′ N ( N − 1) E 1 = e 0 + ǫ 1 N + ǫ 2 N (5 + N ) + kN (1 + β 2 ) 2 U (6) ⊃ SU (3) ⊃ O (3) ⊃ O (2) : E 2 = e 0 + 6 k ′ Nβ 2 1 + β 2 + ǫ 1 N + ǫ 2 N (5 + N ) � � √ 5 + 11 β 2   N + N ( N − 1) 4 β 2 + β 4 2 β 3 cos(3 γ ) 2 + 2 4 +2 k   1 + β 2 (1 + β 2 ) 2

80 On Description of the Yrast Lines in IBM-1 U (6) ⊃ O (6) ⊃ O (5) ⊃ O (3) ⊃ O (2) : 1 − β 2 � 2 � E 3 = ǫ 0 + kN ( N − 1) − kN ( N + 4) + 4(1 + β 2 ) 2 4 ′ Nβ 2 + Nǫ 1 + N ( N + 5) ǫ 2 + k 1 + β 2 As far as these energies being increasing with the increase of N we put into correspondence to each N the value of the angular momentum L = 2 N that gives us the minimal values of the energies for every chosen N . Further we redefine IBM-1 parameters making them depending on collective structure of correspond- ing β − band head: ε 0 = Ap − Bp 2 , if p = 0 , ε 0 = 0 k g r 0 k = r 0 + ∆ R ms ( r 0 , E 0 /C 0 , p ) , if p = 0 , k = k g k ′ g r 0 k ′ = k ′ = k ′ r 0 + ∆ R ms ( r 0 , E 0 /C 0 , p ) , if p = 0 , g 20 174 Hf 174 Hf 15 SU(3) chain 15 O(6) chain Calculations Calculations k= 0.0012935678 Energy (MeV ) k=-0.009150382723 10 k '= -0.53794476 k'= -3.2103210602 ε 1 = -0.101294519 10 ε 1 = -0.100750 ε 2 = 0.048731415 ε 2 = 0.0487691 E 4 /E 2 = 3.233069683 E 4 /E 2 = 3.23032287 β = 0.2 β = 0.2 γ = π /2 5 E 0 /C 0 = 0.067 5 E 0 /C 0 = 0.06 e 0 = 1.24 e 0 = 1.24 p= 4 p= 4 0 0 0 4 8 12 16 20 24 28 32 0 4 8 12 16 20 24 28 32 36 Angular Momentum Angular Momentum 10 10 236 U 236 U 8 8 O(6) chain SU(3) chain Energy (MeV ) Calculations k=-0.0204038 k'=-1.332301 k= 0.0027752068 6 6 ε 1 = -0.0669377 k'= -0.2238134216 ε 1 = -0.067469256 ε 2 = 0.026137738 ε 2 = 0.02604494449 E 4 /E 2 = 3.269526045 E 4 /E 2 = 3.272793753 β = 0.24 4 4 β = 0.24 E 0 /C 0 =0.06 γ = π /2 e 0 = 0.91 E 0 /C 0 = 0.06 p= 4 e 0 = 0.91 2 2 p= 4 0 0 0 4 8 12 16 20 24 28 32 36 0 5 10 15 20 25 30 35 Angular Momentum Angular Momentum Figure 2. Comparison of the calculated energies with experiment.

V. Garistov 81 ε 1 = ε g 1 ( r 0 + ∆ R ms ( r 0 , E 0 /C 0 , p )) , if p = 0 , ε 1 = ε g 1 r 0 ǫ g 2 r 0 ε 2 = r 0 + ∆ R ms ( r 0 , E 0 /C 0 , p ) , if p = 0 , ε 2 = ε g 2 with � ∆ R ms ( r 0 , E 0 /C 0 , p )= R 2 ms ( r 0 , E 0 /C 0 , p ) − R 2 ms ( r 0 , E 0 /C 0 , 0) � � � � E 0 15 E 0 C 0 ( − 1+ p ) +4 ( − 3 + 20 π ) √ � C 0 p � = 3 r 0 � 60 E 0 C 0 p + 160 π 2 Now we fit all model parameters ǫ 1 , ǫ 2 , k, k ′ , β,γ to the ground β − band part of the yrast line states ( p = 0 ) while the behavior of the rest bands will now de- pend only on the number of bosons p taken from 0 + excited states energies clas- sification (1). So calculated energies of β − bands and comparison of our calcu- 20 20 168 Yb O(5) chain 168 Yb O(6) chain 15 k= -0.0190758646 15 k= -0.0190758646 k'= -2.776302788 Energy ( MeV ) ε 1 = -0.0871292 k'= -2.776302788 Energy ( MeV ) ε 1 = -0.0871292 ε 2 = 0.042995863 ε 2 = 0.042995863 E 4 /E 2 2.9942299 10 β = 0.2 E 4 /E 2 2.9942299 10 β = 0.2 E 0 /C 0 = 0.068 E 0 /C 0 = 0.068 e 0 1.15 e 0 1.15 p= 0 5 p= 3 5 0 0 4 8 12 16 20 24 28 32 36 40 0 20 20 0 4 8 12 16 20 24 28 32 36 40 SU(3) chain SU(3) chain 168 Yb 168 Yb Calculations Calculations 15 15 k=0.00406637 k=-0.0305597784 k '= -0.404296279 k '=-2.40905 ε 1 = -0.093299 Energy ( MeV ) ε 1 =-0.0926557821 Energy ( MeV ) ε 2 =0.04236665 ε 2 = 0.0424456 10 E 4 /E 2 =3.047 10 E 4 /E 2 = 3.0422948 β =0.225 β = 0.225 γ = π /2 γ = 0; π /2 E 0 /C 0 =0.065 E 0 /C 0 = 0.057 e 0 =1.15 5 e 0 =1.15 5 p=3 0 0 0 4 8 12 16 20 24 28 32 36 40 0 4 8 12 16 20 24 28 32 36 40 Angular Momentum Angular Momentum Figure 3. Comparison of calculated ener- Figure 4. Calculations with p = 0 . gies with experiment.