Description of 178 Hf m2 in the constrained relativistic mean field theory
aa r X i v : . [ nu c l - t h ] O c t Description of Hf m in the constrained relativistic mean fieldtheory ∗ ZHANG Wei † ,
1, 2
PENG Jing, and ZHANG Shuang-Quan School of Physics, and State Key Laboratory of NuclearPhysics and Technology, Peking University, Beijing 100871 School of Electrical Engineering and Automation,He’nan Polytechnic University, Jiaozuo 454003 Department of Physics, Beijing Normal University, Beijing 100875 (Dated: November 5, 2018)
Abstract
The properties of the ground state of
Hf and the isomeric state Hf m are studied withinthe adiabatic and diabatic constrained relativistic mean field (RMF) approaches. The RMFcalculations reproduce well the binding energy and the deformation for the ground state of Hf. Using the ground state single-particle eigenvalues obtained in the present calculation,the lowest excitation configuration with K π = 16 + is found to be ν (7 / − [514]) − (9 / + [624]) π (7 / + [404]) − (9 / − [514]) . Its excitation energy calculated by the RMF theory with time-oddfields taken into account is equal to 2.801 MeV, i.e., close to the Hf m experimental excitationenergy 2.446 MeV. The self-consistent procedure accounting for the time-odd component of themeson fields is the most important aspect of the present calculation. PACS numbers: 21.10.-k, 21.60.-n, 27.70.+q ∗ Supported by the National Natural Science Foundation of China under Grant Nos 10605004 and 10705004,the Natural Science Foundation of He’nan Educational Committee under Grant No 200614003, and theYoung Backbone Teacher Support Program of He’nan Polytechnic University † Email: [email protected] [1, 2, 3]
From the very beginning, it incorporates the important relativisticeffects, and it has achieved success in describing many nuclear phenomena related to stablenuclei [2, 3] , exotic nuclei [4, 5] as well as supernova and neutron stars [6] . The RMF theoryprovides a new explanation for the identical bands in superdeformed nuclei [7] and for theneutron halo in heavy nuclei [4] , it predicts giant neutron halos, a new phenomenon in heavynuclei close to the neutron drip line [5, 8] , it naturally generates the spin-orbit potential,explains the origin of the pseudospin symmetry as a relativistic symmetry [9, 10, 11] , and spinsymmetry in the anti-nucleon spectrum [12] , and also describes well the magnetic rotation [13] ,the collective multipole excitations [14] as well as the properties of hypernuclei [15] , etc. Lately,the ground state properties of about 7000 nuclei have been calculated in the RMF+BCSmodel and good agreements with existing experimental data were obtained [16] . Recent andmore complete reviews of the applications of the RMF model, particularly, those to exoticnuclei, can be found in Refs. [17, 18] .Recently the 31-yr isomer of
Hf (also called Hf m , K π = 16 + , E x =2.446MeV)has attracted extensive attention [19, 20, 21, 22, 23] for its potential to be a good mediumof energy storage [24] . The long half-life of Hf m is connected with the strong in-hibition of spontaneous electromagnetic transitions restricted by the K selection rule,and it supports the point of view that this high- K state has an axially symmetricintrinsic shape [22] . The configuration originally suggested for the isomer Hf m is ν (7 / − [514])(9 / + [624]) π (7 / + [404])(9 / − [514]) [25] , which was further supported by theview of alignment and by the g -factor of the corresponding rotation bands. [26] In this Letter, we will study the properties of the ground state of
Hf and also investi-gate the possible configuration of Hf m within the self-consistent axially symmetric RMFtheory. From the point of views of the adiabatic constrained calculation, nuclear groundstate presents the global minimum of the potential energy surface (PES). To study theexcited states, like isomers, the diabatic (configuration-fixed) constrained approach can beapplied as an effective method. [27] Within the adiabatic constrained approach, nucleons al-ways occupy the lowest levels, while in the diabatic constrained approach, the configurationis kept fixed by the so-called concept of “parallel transport”. In the present paper, both theadiabatic and diabatic constrained RMF approaches are applied in our investigation.The basic ansatz of the RMF theory is a Lagrangian density where nucleons are described2s Dirac particles which interact via the exchange of various mesons and the photon. Themesons considered are the isoscalar-scalar σ , the isoscalar-vector ω and the isovector-vector ρ . The effective Lagrangian density reads [1] L = ¯ ψ (cid:20) iγ µ ∂ µ − M − g σ σ − g ω γ µ ω µ − g ρ γ µ ~τ · ~ρ µ − eγ µ − τ A µ (cid:21) ψ + 12 ∂ µ σ∂ µ σ − m σ σ − g σ − g σ −
14 Ω µν Ω µν + 12 m ω ω µ ω µ + 14 c ( ω µ ω µ ) − ~R µν · ~R µν + 12 m ρ ~ρ µ · ~ρ µ − F µν F µν (1)in which the field tensors for the vector mesons and the photon are, respectively, defined as Ω µν = ∂ µ ω ν − ∂ ν ω µ ,~R µν = ∂ µ ~ρ ν − ∂ ν ~ρ µ ,F µν = ∂ µ A ν − ∂ ν A µ . (2)From the Lagrangian, the equation of motion for the nucleon is { α · [ − i ∇ − V ( r )] + V ( r ) + β [ M + S ( r )] } ψ i = ε i ψ i , (3)with the attractive scalar potential S ( r ) = g ω ω ( r ), the usual repulsive vector poten-tial V ( r ) = g ω ω ( r ) + g ρ τ ρ ( r ) + e − τ A ( r ) , and the nuclear magnetic potential V ( r ) = g ω ω ( r ) + g ρ τ ρ ( r ) + e − τ A ( r ). The Klein-Gordon equations for the mesonsand electromagnetic fields are ( −∇ + m ζ ) ζ ( r ) = S ζ ( r ) , (4)where S ζ ( r ) is the source term and all other notations are the same as in Ref. [18].In the RMF approaches which are widely used, only the time-even fields are essential forthe physical observables, since the time-odd components of vector fields do not exist becauseof the time reversal symmetry for the ground state of an even-even nucleus. For an odd- A or odd-odd nucleus, the unpaired valence nucleon will give non-vanishing contribution tothe nuclear current which provides the time-odd component of vector fields, i.e., the nuclear3agnetic potential. It is found that the nuclear magnetic potential has small influence onthe root-mean-square radii and quadrupole moments while it plays an important role inthe single-particle properties and magnetic moments in odd- A or odd-odd nuclei. [28, 29] Oneshould keep in mind that for the excited states in the even-even nuclei, there may also existthe unpaired nucleons, which result in non-vanishing time-odd field. Therefore the time-oddfields should also be treated carefully for some isomeric states in the even-even nuclei, asin the odd- A or odd-odd nuclei. In the calculation without current, the nuclear magneticpotential V ( r ) will be neglected.For the adiabatic constrained approach, the binding energy at a certain deformation isobtained by constraining the mass quadruple moment h ˆ Q i to a given value µ , i.e. h H ′ i = h H i + 12 C ( h ˆ Q i − µ ) . (5)where C is the curvature constant parameter, and µ is the given quadrupole moment. Theexpectation value of ˆ Q is h ˆ Q i = h ˆ Q i n + h ˆ Q i p , where h ˆ Q i n,p = h r P (cos θ ) i n,p . Thedeformation parameter β is related to h ˆ Q i by h ˆ Q i = 3 √ π Ar β , r = R A / ( R = 1 . A is the mass number. By varying µ , the binding energy at different deformationscan be obtained [30] .For the adiabatic constrained approach, the occupied levels are determined by the so-called “parallel transport” [27] , i.e., h ψ i ( q ) | ψ j ( q + ∆ q ) i| ∆ q → ≈ δ ij , (6)where i and j enumerate all the single-particle levels of two adjacent configurations. Insuch a way, the original configuration at q can be traced and the corresponding PES canbe obtained as a function of the deformation [27] . In principle, if ∆ q is small enough, theconfigurations at q and at q + ∆ q should be the same. In the calculation, the two-stepprocedure is adopted: first, the wave functions and the configuration at the initial q arerecorded. Second, the wave functions | ψ i ( q ) i are mapped to | ψ j ( q + ∆ q ) i one by one bysearching the largest overlap in | ψ j ( q + ∆ q ) i with the same quantum number Ω π . Theconfiguration is transferred by copying the occupation number from | ψ i ( q ) i to the mapped | ψ j ( q + ∆ q ) i . The wave functions and the configuration at this q + ∆ q are also recorded.The second step is repeated until enough points on the diabatic PES are obtained.The constrained RMF calculations are carried out with parameter set PK1 [31] . The full N = 20 deformed harmonic-oscillator shells for fermions and bosons are taken into account4s the basis. This basis is large enough to produce a converged binding energy at certaindeformation.The PES of Hf obtained in adiabatic (open circles) and diabatic (lines) constrainedRMF calculations are plotted in Fig. 1. The calculated energy E =-1434.0 MeV and thedeformation β =0.283 of the ground state (denoted as a black asterisk in Fig. 1) are in goodagreement with the experimental energy -1432.8 MeV [32] and deformation 0.280 [33] . In thisfigure, the adiabatic PES can be decomposed into three regions by the discontinuity, i.e., β =0.22 ∼ β = 0.35 ∼ β = 0.43 ∼ π (7 / + [404]) − (1 / − [541]) ,where microscopically the proton level 7 / + [404] below the Fermi surface in region 1 be-comes unoccupied while the proton level 1 / − [541] above the Fermi surface in region 1becomes occupied. Since a pair of protons change the levels at the same time, the K -valuesremain zero for region 2. Similarly, from region 2 to region 3, the configuration changesfrom π (7 / + [404]) − (1 / − [541]) to ν (5 / − [512]) − (1 / + [660]) π (7 / + [404]) − (1 / − [541]) and the K -values also remain zero for region 3.Based on the single-particle spectra of the ground state, one can construct excited stateswith high K -values. In a deformed, axially symmetric nucleus, a high- K state is made bysumming the contributions from several unpaired quasiparticles. To form low-lying high- K states, several high-Ω single-particle (both neutron and proton) levels lying close to the Fermisurface are necessary. The well-deformed nuclei with A ≈ Hf, satisfy thisrequirement very well. With the restriction of the total Ω and parity as 16 + , the candidateconfigurations of Hf m will be constructed.For the ground state of Hf, the neutron (proton) single-particle levels close to theFermi surface are shown in the first column of the left (right) panel in Fig. 2. Each level islabeled by the Nilsson notation Ω π [ N n z m l ] of its main component. It can be seen that inFig. 2, the energy required to excite one neutron or one proton is not less than 0.63 or 1.845eV. As the experimental excitation energy of Hf m equals 2.446 MeV, it is sufficient toconsider one- or two-neutron excitations, together with one- or two-proton excitations. Fortwo-neutron (two-proton) excitations, the following cases are considered: in the first case, apair of particles below the Fermi surface are excited to two different levels above the Fermisurface; in the other case, two particles occupying different levels below the Fermi surfaceare excited to form a new pair. Those cases which involve four or more single-particle levelsare not included for simplicity. All possible configurations are constructed by restricting the K (total Ω) value to 16 and the nuclear parity to +. Thus the configurations with the lowestexcitation energies are obtained and labeled by 16 +1 ,16 +2 , 16 +3 , etc.The detailed configurations of the five lowest K π = 16 + states of Hf are listed in column2 of Table I. In this table, the first 4 states are one-neutron plus one-proton excitation, whilethe last one is a pair of neutrons excited to high-Ω levels combined with the one-proton exci-tation. The configuration of 16 +1 is ν (7 / − [514]) − (9 / + [624]) π (7 / + [404]) − (9 / − [514]) with respect to the ground state, i.e., one formerly paired neutron in the level 7 / − [514]becomes unpaired and excited to the level 9 / + [624], and another formerly paired protonin level 7 / + [404] is excited to the level 9 / − [514]. This configuration is consistent withthe former assignment of Hf m . Note that the excitation energy of 16 +1 given bythe sum of single particle excitations is 3.954 MeV which is about 1.5 MeV higher than theexperimental value 2.446 MeV.In order to obtain the self-consistent excitation energy for state 16 +1 microscopically, thediabatic constrained RMF calculations are carried out with the respective configurationinformation. As discussed before, the time-odd fields [29] caused by the unpaired nucleonsshould be taken into account carefully for the high- K state. The time-even calculation isalso done for comparison. The PES of state 16 +1 with (without) current is plotted as a solid(dashed) red curve in Fig. 1, together with the PES of state 16 +2 . The local minima of 16 +1 (16 +2 ) PES are denoted as up (down) triangles in Fig. 1. For state 16 +1 , the excitation energyaccording to the calculations without and with current is, respectively 3.579 MeV and 2.801MeV, which clearly shows that both the self-consistent calculation and the considerationof the time-odd fields are crucial effects to obtain the reasonable excitation energy. Theexcitation energy of 16 +1 ( E x = 2 .
801 MeV) is close to the experimental excitation energy( E x = 2 .
446 MeV) of Hf m . The deformation ( β = 0 .
30) obtained for this isomer is similarto that of the ground state. In Table I, all the excitation energies from calculations without6nd with current as well as the deformation calculated for the states 16 +1 to 16 +5 are alsogiven. For all five states, the excitation energies calculated with current are 0.6 ∼ Hf m (experimentally K π =8 − , E x =1.147MeV). The configuration obtainedfor 8 − is π (7 / + [404]) − (9 / − [514]) , and the excitation energy without and with current is,respectively 1.552 MeV and 1.315 MeV.In Fig. 2, the neutron and proton single-particle levels of state 16 +1 from the RMF withoutand with current have been plotted in columns 2, 3 in both panels. In the time-odd calcula-tion each single-particle level splits into two levels due to the breaking of the time-reversalsymmetry, the level with positive Ω being energetically favored. Such splittings will changestraightforward the energy gap between two single-particle levels. In particular, for one-neutron and one-proton excitation of the state 16 +1 discussed here, the neutron energy gapbetween ν (9 / + [624]) and ν (7 / − [514]) in the calculation without current decreases from2.00 MeV to 1.25 MeV , which is the gap between ν (9 / + [624] , +Ω) and ν (7 / − [514] , − Ω)in the calculation with current. Correspondingly the energy gap for the proton excitationconcerned decreases from 1.34 MeV to 0.54 MeV. As a result, the calculation with currentdecreases the excitation energy considerably.In summary, the properties of the ground state of
Hf and the isomeric state Hf m are investigated by the adiabatic and diabatic constrained RMF approaches. The con-strained RMF theory reproduces well the binding energy and deformation for the groundstate of Hf. With the single-particle levels of the ground state obtained in the adiabaticconstrained RMF theory, by restricting K π = 16 + , the configuration with the lowest exci-tation energy is found to be ν (7 / − [514]) − (9 / + [624]) π (7 / + [404]) − (9 / − [514]) , whichis consistent with the former configuration assignment. The excitation energy based on thesingle-particle spectra is 3.954 MeV, which is much higher than the experimental excitationenergy 2.446 MeV of Hf m . By applying the self-consistent time-even and time-odd RMFcalculation, the excitation energy of this configuration is decreased to 3.579 MeV and 2.801MeV, respectively. Therefore both the self-consistency and the consideration of current areimportant factors for studying nuclear isomers.The authors gratefully acknowledge Professor Jie Meng, Professor L. N. Savushkin, and7r. Jiangming Yao for their helpful suggestions and discussions. [1] Serot B D and Walecka J D 1986 Adv. Nucl. Phys. Rep. Prog. Phys. Prog. Part. Nucl. Phys. Phys. Rev. Lett. Phys. Rev. Lett. Compact Stars (Springer-Verlag, New York)[7] K¨onig J and Ring P 1993
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Phys. Rev. C Nucl. Phys. A At. Data Nucl. Data Tables .20 0.25 0.30 0.35 0.40 0.45 0.50-1436-1434-1432-1430-1428-1426-1424-1422-1420-1418 PK1 / (adiabatic/diabatic) g.s. PES/ 16 +1 PES (no / with current)/ 16 +2 PES (no / with current) E ( M e V ) Hf FIG. 1: (color online) Potential energy surfaces (PES) of
Hf from adiabatic (open circles) anddiabatic (solid curves) RMF calculations, as well as the PES of the first two 16 + states 16 +1 (red)and 16 +2 (blue) from the adiabatic constrained RMF calculations with (solid curves) and without(dashed curves) current. The RMF parameter set PK1 is adopted. The ground state as well asthe local minima 16 +1 (16 +2 ) are denoted as a black asterisk and red (blue) triangles, respectively. With currentWith current
No current
Neutron g.s.11/2+[615]9/2-[505]1/2-[510]3/2-[512]9/2+[624]5/2-[512]7/2+[633]7/2-[514]1/2-[521]5/2+[642]3/2+[651] S i ng l e p a r t i c l e e n e r g y ( M e V ) +1 -9-8-7-6-5-4-3 Proton S i ng l e p a r t i c l e e n e r g y ( M e V ) +1 No current
FIG. 2: (color online) Neutron (left panel) and proton (right panel) single-particle levels of theground states (1st column), state 16 +1 of Hf obtained from the constrained RMF approachwithout (2nd column) and with (3rd column) current. Each level is labeled by the Nilsson notationΩ π [ N n z m l ] of its main component. The Fermi surfaces are marked by neutron and proton numbersin the grey circles. The red (blue) lines denote the parity + ( − ), and the dashed (solid) lines denotethe positive (negative) signs of Ω, respectively. The configuration difference is illustrated by filledcircles, open circles and arrows. ABLE I: The configurations, excitation energies as well as the deformations obtained from theadiabatic and diabatic constrained RMF approaches for the first five K π = 16 + states of Hf.The excitation energies (in MeV) include the sum of single-particle excitations X ( ε j − ε i ), andthe excitation energies E x calculated without (with) current.State Configuration X ( ε j − ε i ) E x E x β (no current) (with current) (with current)16 +1 ν (7 / − [514]) − (9 / + [624]) π (7 / + [404]) − (9 / − [514]) +2 ν (7 / + [633]) − (9 / + [624]) π (7 / − [523]) − (9 / − [514]) +3 ν (5 / − [512]) − (11 / + [615]) π (7 / + [404]) − (9 / − [514]) +4 ν (7 / + [633]) − (9 / − [505]) π (7 / + [404]) − (9 / − [514]) +5 ν (5 / − [512]) − (9 / + [624]) (7 / − [503]) π (7 / + [404]) − (9 / − [514])1.835