Isobaric analog state energy in deformed nuclei: a toy model
IIsobaric analog state energy in deformed nuclei: a toy model
X. Roca-Maza ∗ and G. Col`o † Dipartimento di Fisica “Aldo Pontremoli”, Universit`a degli Studi di Milano, 20133 Milano, Italy andINFN, Sezione di Milano, 20133 Milano, Italy
H. Sagawa ‡ RIKEN Nishina Center, Wako 351-0198, Japan andCenter for Mathematics and Physics, University of Aizu, Aizu-Wakamatsu, Fukushima 965-8560, Japan (Dated: September 21, 2020)A formula to evaluate the effects of a general deformation on the Coulomb direct contributionto the energy of the Isobaric Analog State (IAS) is presented and studied via a simple yet physicalmodel. The toy model gives a reasonable account of microscopic deformed Hartree-Fock-Bogolyubov(HFB) calculations in a test case, and provides a guidance when predicting unknown IAS energies.Thus, deformed HFB calculations, to predict the IAS energies, are performed for several neutron-deficient medium-mass and heavy nuclei which are now planned to be studied experimentally.
PACS numbers: 21.10.Sf, 24.30.Cz
I. INTRODUCTION
Isospin is one of the most important (approximate)symmetries in nuclei. The validity of isospin symmetryhas been established by the experimental observation ofisobaric analogue states (IAS) by charge-exchange reac-tions. Recently, these states have been investigated ex-tensively in connection with the symmetry energy, in par-ticular to determine the so-called slope parameter L [1–3]. The nuclear symmetry energy is one of the fundamen-tal ingredients to describe the nuclear Equation of State(EoS) when dealing with isospin-asymmetric matter. Itsdetermination may entail profound consequences in ourunderstanding of various physical observables; the sym-metry energy governs not only properties of nuclei, butalso numerous facets of astrophysics like neutron starsand supernovae [4–6]. The nuclear EoS and symmetryenergy have been discussed in many different contexts, inwhich both the strong and Coulomb interactions play arole. It should be noticed, however, that our knowledgeof the strong interaction, even in its realistic form em-ployed in ab initio-type calculations, have some room tobe improved for reaching a more robust understanding ofthe nucleus and of the nuclear EoS. On the other hand,the IAS is essentially governed by the well-establishedCoulomb force, which is an advantage when trying toelucidate the EoS in asymmetric nuclear matter.Up to now, theoretical studies of IAS have been mainlyfocused on spherical nuclei such as Ca, Zr and
Pb.The only few exceptions, to the best of our knowledge,are the Skyrme calculations by K. Yoshida [7] and bythe Madrid group [8, 9]. However, a large number ofdeformed nuclei exist in wide regions of the mass table,and play an important role for various nuclear structure ∗ [email protected] † [email protected] ‡ [email protected] problems. While experimental data of IAS exist in somedeformed nuclei, the information on deformation effectson the IAS is still missing in a transparent way besidesonly a few theoretical study The IAS in neutron-rich nu-clei have been considered in most of the previous studiessince the large isospin values of these nuclei make it easierto observe experimentally the IAS as an isolated narrowresonance. On the other hand, neutron-deficient nucleihave some advantages since the differences between theproton and neutron densities (i.e., the size of the neutronskin) is relatively small and, consequently, their absoluteerrors are smaller. This allows evaluating the density dif-ferences ∆ ρ np = ρ n − ρ p with a better systematic accu-racy from charge-exchange reactions. Another interest-ing aspect is that most neutron-deficient isotopes in themedium- to heavy-mass region may have large isospinmixing, which will have a noticiable impact on the IASenergy and its systematics.In this paper, we derive a general formula for thedeformation effects on the Coulomb direct contributionto the energy of the IAS and provide a simple albeitphysical model. In addition, we estimate the deforma-tion effects using a microscopic Hartree-Fock-Bogolyubov(HFB) model and test both the general formula and theproposed toy model. Then, we study several neutron-deficient medium-mass and heavy-nuclei, which are nowplanned to be studied experimentally in RCNP, Osakawithin the LUNESTAR project [10]The paper is organized as follows. In Sec. II, the theo-retical model is introduced. In Sec. III, we test the gen-eral formula to account for deformation effects on the IASenergy and the toy model by comparing them with HFBresults. We also provide and discuss deformed HFB pre-dictions for experimentally accessible neutron-decficientnuclei. Conclusions are drawn in Sec. IV. a r X i v : . [ nu c l - t h ] S e p II. MODELA. Definition of the IAS energy
The Isobaric Analog State (IAS) energy E IAS can bedefined as the energy difference between the analog state | A (cid:105) and the parent state | (cid:105) . The parent state is aneigenstate of the Hamiltonian H with N neutrons and Z protons and the analog state can be defined as [11] | A (cid:105) ≡ T − | (cid:105)(cid:104) | T + T − | (cid:105) / , (1)where T + = (cid:80) Ai t + ( i ) and T − = (cid:80) Ai t − ( i ) are the isospinraising and lowering operators, respectively, that followthe usual SU(2) algebra:[ T + , T − ] = 2 T z , [ T z , T ± ] = ± T ± . (2) T z = (cid:80) Ai t z ( i ) and t z has eigenvalues − / / E IAS = (cid:104) A |H| A (cid:105) − (cid:104) |H| (cid:105) = (cid:104) | T + [ H , T − ] | (cid:105)(cid:104) | T + T − | (cid:105) . (3)Assuming good isospin in the parent state T + | (cid:105) = 0, Eq.(3) is rewritten as E IAS = 1 N − Z (cid:104) | [ T + , [ H , T − ]] | (cid:105) . (4)It is important to note that the latter formula can beonly applied to nuclei with N > Z . If isospin mixing isconsidered, namely if T + | (cid:105) (cid:54) = 0, Eq. (4) should be cor-rected as in Eq. (A6) of Ref. [12]. A simple approximateexpression for that correction is [11] (cf. also AppendixA in Ref. [12]) E IMIAS ≈ − ε N − Z + 2 N − Z A − / , (5)where ε is the isospin mixing in the parent state.Specifically, if the parent state wave function is | (cid:105) = (cid:80) n a n | T + n, T (cid:105) , and the states | T + n, T (cid:105) have goodisospin, then ε ≡ a : the admixture of states with totalisospin T > T + 1 is expected to be very small and canbe neglected. Under this assumption, we can define ε ≡ (cid:104) | T − T + | (cid:105) N − Z + 2 (6)where, for our purposes here, it is accurate to evaluate thenumerator on a Bardeen-Cooper-Schrieffer (BCS) groundstate wave function and assuming spherical symmetry . It is important to note here that deformation would be a correc-tion to isospin mixing which is a correction to the IAS energy.The approximate character of Eq. (5) is consistent with the factthat we are dealing with a second-order, small effect.
This will give the simple and closed expression, ε ≡ N − Z + 2 (cid:88) n p ,n n l p = l n j p = j n (2 j p + 1) v p u n O np (7)with the overlap factor between the neutron ( n ) and pro-ton ( p ) radial part R n,l ( r ) of the considered single parti-cle wave function O np ≡ (cid:90) ∞ drr R n p ,l p ( r ) R n n ,l n ( r ) . (8)In these expressions, n is the principal quantum num-ber, and l and j are the orbital and total angular mo-mentum quantum numbers, respectively, while v and u corresponds to the occupation factors. As a test, wehave confirmed that the estimation in Eq. (6) calcu-lated within the Tamm-Dancoff approximation coincideswithin a very good accuracy (1% or below) with the ex-pression in Eq. (7). B. Contributions to the IAS energy
Due to the structure of Eq. (3), E IAS depends onlyon isospin symmetry breaking (ISB) parts of H . In nu-clear physics, the main isospin breaking term is known tobe due to the Coulomb interaction. Therefore, the bulkcontribution to Eq. (3) will be due to the difference inthe expectation value of the Coulomb matrix elementsbetween proton and neutron distributions. That is, forthe direct Coulomb (Cd) term, assuming an independentparticle model, one has E CdIAS = 1 N − Z (cid:90) [ ρ n ( r ) − ρ p ( r )] U directC ( r ) d r , (9)where U directC ( r ) is the direct part of the Coulomb poten-tial generated by the electric charge distribution ρ ch ( r ), U directC ( r ) = (cid:90) e | r (cid:48) − r | ρ ch ( r (cid:48) ) d r (cid:48) . (10)Under the same assumption and adopting, in addition,the Local Density Approximation (LDA), the Coulombexchange contribution can be also conveniently written asa function of the neutron and proton density distributionsas E CxIAS = 1 N − Z (cid:90) [ ρ n ( r ) − ρ p ( r )] U exch , LDAC ( r ) d r , (11)where U exch , LDAC ( r ) is the Coulomb exchange part ofthe Coulomb energy potential generated by the electriccharge distribution ρ ch ( r ) within the LDA, U exch , LDAC ( r ) = − e (cid:20) π ρ ch ( r ) (cid:21) / . (12)This contribution will be much smaller than the Coulombdirect part. In what follows we will approximate ρ ch ( r )by ρ p ( r ). This approximation produces a negligible ef-fect for our purposes here, but should be dropped if onewishes a precise value of the IAS energy [2, 3]. QED cor-rections in the fine-structure constant to the Coulombpotential and Coulomb correlation effects will be also ne-glected in the present study.Other ISB effects than Coulomb force come fromthe nuclear strong interaction. Those terms can beparametrized by simple effective interactions solved atthe Hartree-Fock level as it has been recently shown[2, 3, 13, 14]. For example, assuming a Charge Symme-try Breaking (CSB) and Charge Independence Breaking(CIB) effective interaction written as V CSB ( r , r ) = 12 [ t z (1) + t z (2)] s (1 + y P σ ) δ ( r − r ) ,V CIB ( r , r ) = 2 t z (1) t z (2) u (1 + z P σ ) δ ( r − r ) , (13)in analogy to the well known Skyrme interaction andwhere P σ is the exchange operator in spin-space, one findsthe following contributions to Eq. (4): E CSBIAS = − s (1 − y ) N − Z (cid:90) d r (cid:2) ρ n ( r ) − ρ p ( r ) (cid:3) , (14) E CIBIAS = − u (1 − z ) N − Z (cid:90) d r [ ρ n ( r ) − ρ p ( r )] . (15)Hence, the total IAS energy can be accurately estimatedby taking into acount all these contributions [2, 3]. In thepresent study, we will focus only on the terms that arecommonly included in current EDFs, that is, Coulombdirect and Coulomb exchange. In addition, since wewill base our microscopic calculations on a Hartree-Fock-Bogoliubov (HFB) approach, we should correct for thespurious isospin mixing. To do that we will resort to theapproximate formula given in Eq. (5).Finally, the IAS energy will be estimated from Eqs. (5),(9) and (11) as E IAS = E CdIAS + E CxIAS + E IMIAS . (16) C. Deformation effects to the IAS energy
In the following we explicitly evaluate Eqs. (9) and (11)assuming the neutron and proton densities, ρ n ( r ) and ρ p ( r ), can be deformed and relating this result to thespherical case. To this end, we will follow the approachgiven in Ref. [15].We start by introducing some notation and writing thesquare of the distance vector R of a deformed system(ellipsoid) as R = ( x/a ) +( y/b ) +( z/c ) where x , y and z are Cartesian coordinates and a , b and c dimensionlessquantities such that abc = 1 and, thus, the length ofeach semi-axis is aR , bR and cR . The spherical caseis recovered for a = b = c = 1. The relation between the modulus of the deformed distance vector R and themodulus of the spherical distance vector r is R = r (cid:18) sin θ cos φa + sin θ sin φb + cos θc (cid:19) ≡ r S ( θ, φ ) . (17)We are considering a general family of deformations thatconserve the volume. d R = dRd Ω = drd
Ω = d r since the Jacobian of the transformation is 1. Then, (cid:82) d rρ α ( R ) = (cid:82) d Rρ α ( R ) = (cid:82) d rρ α ( r ) or, in otherwords, integrals that depend on an arbitrary power α of the density will not depend on deformation, as longas one does not introduce an explicit dependence on r .Therefore, the result of Eqs. (11) –as well as those inEqs. (14) and (15)– is independent of deformation ef-fects. Only the Coulomb direct contribution shown inEq. (9) will display some dependence on deformation.To deal with the direct Coulomb contribution, we willcompare the result of Eq. (9) between a spherical nucleuswith densities ρ n ( r ) and ρ p ( r ) and a deformed nucleusassuming that the neutron and proton densities satisfy a n ≈ a p , b n ≈ b p and c n ≈ c p and have the same func-tional form of the corresponding spherical nucleus but de-pending on R n ≡ r S n ( θ, φ ) and R p ≡ r S p ( θ, φ ). Thatis, we shall write ρ n ( R n ) and ρ p ( R p ).In order to evaluate Eq. (9) for a deformed nucleuswithin the conditions above described, we will first per-form a Fourier transform E CdIAS = e N − Z π (cid:90) d R n d R (cid:48) p (cid:90) d q [ ρ n ( R n ) − ρ p ( R p )] e ı Q n R n e − ı Q p R (cid:48) p q ρ p ( R (cid:48) p ) . (18)By defining Q ≡ ( aq x , bq y , cq z ) in analogy with R ≡ ( x/a, y/b, z/c ), that is, Q n = q /S n ( θ, φ ) and Q p = q /S p ( θ, φ ) and by taking into account that d R n = d r and d R p = d r , we can write E CdIAS = 8 e N − Z (cid:90) d q q (cid:90) dR (cid:48) p R (cid:48) p ρ p ( R (cid:48) p ) j ( Q p R (cid:48) p ) (cid:110)(cid:90) dR n R n ρ n ( R n ) j ( Q n R n ) − (cid:90) dR p R p ρ p ( R p ) j ( Q p R p ) (cid:111) . (19)Since Q n R n = Q p R p , the last expression can be writtenas E CdIAS = 8 e N − Z (cid:90) d q q (cid:90) dR (cid:48) p R (cid:48) p ρ p ( R (cid:48) p ) j ( Q p R (cid:48) p ) (cid:90) dR p R p (cid:2) λ ρ n ( λR p ) − ρ p ( R p ) (cid:3) j ( Q p R p ) , (20)where λ ≡ Q p Q n = (cid:20) S n ( θ, φ ) S p ( θ, φ ) (cid:21) / (21)depends only on the angles, and λ ρ n ( λR ) corresponds toa volume conserving scaling of the neutron density ρ n ( R ).For λ = 1 we obtain the result for equally deformed neu-tron and proton density distributions. Eq. (20) can befurther developed as E CdIAS = 8 e N − Z (cid:90) d Ω S p ( θ, φ ) (cid:90) dQ p dR (cid:48) p R (cid:48) p ρ p ( R (cid:48) p ) j ( Q p R (cid:48) p ) (cid:90) dR p R p (cid:2) λ ρ n ( λR p ) − ρ p ( R p ) (cid:3) j ( Q p R p ) . (22)We now perform the following manipulations: i) changeof variables ˜ R p = λR p in the integral that goes with ρ n , (cid:90) d ˜ R p ˜ R p ρ n ( ˜ R p ) j ( Q p λ ˜ R p );ii) expand j for λ → λ , j ( Q p λ ˜ R p ) ≈ j ( Q p ˜ R p ) + (cid:104) j ( Q p ˜ R p ) − cos( Q p ˜ R p ) (cid:105) ( λ − ≈ λj ( Q p ˜ R p ) − ( λ −
1) cos( Q p ˜ R p );iii) perform the change of variables R p = ˜ R p , λ (cid:90) dR p R p ρ n ( R p ) j ( Q p R p ) − ( λ − (cid:90) dR p R p ρ n ( R p ) cos( Q p R p ) . By substituting the last expression in Eq. (22), we find E CdIAS = E Cd , sphIAS π (cid:90) d Ω λS p ( θ, φ )+ 8 e N − Z (cid:90) d Ω 1 − λS p ( θ, φ ) (cid:90) dQ p (cid:110)(cid:90) dR (cid:48) p R (cid:48) p ρ p ( R (cid:48) p ) j ( Q p R (cid:48) p ) (cid:90) dR p R p [ ρ n ( R p ) cos( Q p R p ) − ρ p ( R p ) j ( Q p R p )] (cid:111) , (23)where E Cd , sphIAS corresponds to the result assuming spheri-cally symmetric neutron and proton densities. Neglectingthe term in λ − E CdIAS = E Cd , sphIAS π (cid:90) d Ω λS p ( θ, φ ) E CdIAS = E Cd , sphIAS π (cid:90) d Ω [ S n ( θ, φ )] / [ S p ( θ, φ )] / . (24)The last expression differs from the result obtained as-suming spherical symmetry of the neutron and protondensity distributions by a factor14 π (cid:90) d Ω [ S n ( θ, φ )] / [ S p ( θ, φ )] / . (25) Assuming equally deformed neutron and proton distribu-tions ( λ = 1) the factor would be14 π (cid:90) d Ω S ( θ, φ ) , (26)where we have dropped the subindex for obvious reasons.These expressions are valid for any deformation thatpreserves the volume, and where the neutron and protondistributions are deformed in a similar way. To a goodapproximation, this is the case for most deformed nu-clei according to available microscopic calculations [16].Hence, the usefulness of Eq. (24).In order to give some example in terms of the parame-ters commonly used in nuclear physics, one should betterwrite S ( θ, φ ) in terms of spherical harmonics, S ( θ, φ ) − / = (cid:88) lm α lm Y lm ( θ, φ ) + C , (27)taking care of renormalizing the expression so that thevolume is preserved by determining the proper value ofthe constant C . That is, by imposing (cid:90) d r ρ ( r ) = (cid:90) d R ρ ( R )1 = 14 π (cid:90) d Ω S ( θ, φ ) / . (28)As an example, if we assume quadrupole deformationdefining [ S ( θ, φ )] − / = 1 + β Y + C , (29)where α ≡ β , the value of C is now − β / π . Hence, itis easy to evaluate the effect on the energy of the IAS as E CdIAS = E Cd , sphIAS (cid:20) − β n β p π + ( β n − β p )( β n + β p )4 π + ( β n − β p ) π (cid:21) . (30)For the special case β n = β p this reduces to E CdIAS = E Cd , sphIAS (cid:20) − β π (cid:21) , (31)where we have neglected the terms in β . From the re-sult in Eq. (30), one should expect that the larger thequadrupole deformation the smaller the IAS energy . Inorder to have a qualitative idea about the effect of de-formation on the IAS energy, Eq. (31) predicts a realtivedifference that goes as − β / π , which means, for very de-formed nuclei with β ≈ .
8, a relative reduction of E IAS of about 5% with respect to the spherically symmetriccase. So, in general, deformation effects to the IAS en-ergy are expected to be small, about few % at most .Since it will be useful for testing purposes in thenext section, let us use Eq. (31) to define an effectivequadrupole deformation assuming as known the Coulombdirect energies of an axially deformed nucleus and itsspherical counterpart, β eff2 ≡ π (cid:32) − E CdIAS E Cd , sphIAS (cid:33) / . (32) D. Toy model
In order to understand in simple terms the relationbetween the E IAS , the nuclear quadrupole deformationand the neutron skin thickness of a spherical system∆ r sphnp ≡ (cid:104) r n (cid:105) / − (cid:104) r p (cid:105) / , we evaluate E Cd , sphIAS as inRefs. [2, 3] within a simple yet physical model. Withinsuch model that assumes a uniform neutron and protonspherical distributions (sharp sphere approximation) [2],one finds E CdIAS ≈ (cid:18) − β π (cid:19) E Cd , sphIAS ≈ (cid:114) Ze (1 − β / π ) (cid:104) r p (cid:105) / (cid:32) − NN − Z ∆ r sphnp (cid:104) r p (cid:105) / (cid:33) , (33)that is, the IAS energy should decrease with both in-creasing neutron skin thickness and nuclear quadrupoledeformation. For simplicity, within our toy model, wewill assume β n = β p .It is interesting to note that the effect of quadrupoledeformation on the mean square radius is (cid:104) r (cid:105) = (cid:90) d r r ρ ( R )= (cid:90) d r r ρ (cid:32) r (cid:20) β Y − β π (cid:21) − (cid:33) = (1 + 54 π β ) (cid:104) r (cid:105) sph . (34)From the latter result, it is clear that the effect of de-formation on the energy of the IAS cannot be accountedfor only by taking into account the deformation effectson the rms radii of protons and neutrons, although thiswill produde the largest effect. By using the result inEq. (34) we can rewrite Eq. (33), to order β , as follows E CdIAS ≈ (cid:114) Ze (1 + π β ) (cid:104) r p (cid:105) / (cid:32) − NN − Z ∆ r np (cid:104) r p (cid:105) / (cid:33) . (35)This expression allows one to directly use experimentallyknown rms radii and deformations to estimate E CdIAS . Itshould be noticed that the deformation increases the nu-merator, while the charge radius in the denominator is increased even more by the deformation than the numer-ator. The net effect of deformation decreases the IASenergy (35), consistently with Eq. (33).Within this model we can also derive a simple formulafor the Coulomb exchange contribution of Eq. (11). Theresult reads as follows, E CxIAS = − (cid:114) (cid:18) π (cid:19) / e Z / (cid:104) r p (cid:105) / (cid:32) − NN − Z ∆ r np (cid:104) r p (cid:105) / (cid:33) . (36) III. RESULTSA. Deformation effects on a test nucleus ( β e ff ) HFB β -0.304-0.3-0.296-0.292 E I A S C x ( M e V ) Sn FIG. 1. Upper panel: β eff2 in Eq. (32) as a function of themass deformation β m . The gray line corresponds to the limit β eff2 = β m . Lower panel: E CxIAS as a function of β m . Thegray zone is just for a guide to eyes. All calculations havebeen performed with the SAMi functional. In this subsection, we have performed different axiallydeformed constrained HFB calculations for the
Sn nu-cleus, from β = 0 to β = 0 .
5. We have slightly modifiedthe code HFBTHO [17] to use the SAMi interaction andcalculate the displacement energies shown in this paper.The number of oscillator shells of the basis is 20, and wecalculate pairing correlations with particle number pro-jection after variation, with a density dependent pairinginteraction of the type V ( r , r ) = V (cid:20) x (cid:18) ρ ( r + r ) ρ (cid:19)(cid:21) δ ( r − r ) , (37)with x = 0 . V is fixed as V = − . in order to reproduce the neutronpairing gap in Sn when the pairing cut-off energy is60 MeV.In Fig. 1, the upper panel, we show ( β eff2 ) from Eq.(32) as a function of β m , where β q = (cid:114) π Q q (cid:104) r q (cid:105) N q , (38)with q = n, p, m , and Q m = Q n + Q p (39)from the HFB calculations. Note that β m is obtainedfrom the HFB densities and could be different from theparameter defining the deformation of the HFB potential.This is a consistency test of Eq. (31), or a consistency testbetween densities and Coulomb potentials. Results in theupper panel of Fig. 1 do not deviate substantially fromEq. (31), i.e. from the grey solid line. This result sug-gests that our model evaluation of deformation effects onthe IAS is quite acceptable. Note that we have assumed β n = β p to define β eff2 while HFB calculations give somedifference for yhe axial quadrupole deformat of neutronsand protons.In the lower panel of the same figure, we check nu-merically the contribution to the IAS energy from theCoulomb exchange term in LDA approximation. Thegrey area delimits the numerical variation of the Coulombexchange (within Slater approximation) with deforma-tion. As we discussed in the previous subsection, in prin-ciple this variation should be zero. Numerically we findan error of few keV which is also quite acceptable. B. Toy model test: the Sn isotopic chain
In Fig. 2, we compare the HFB results for the Coulombdirect and exchange contributions in Eqs. (9) and (11),respectively, with their corresponding toy model counter-parts, that is, the direct and exchange ones in Eqs. (35)and (36), respectively.In Table I we first show the isospin mixing probabili-ties, both without and with pairing correlations. Thosehave been calculated as in Eq. (7), from the overlap ofthe ground state single-particle neutron and proton wavefunctions. The laters have been evaluated within theHF-BCS approach, using the same type of pairing inter-action and model space as in the HFB calculations sothat to reproduce the neutron pairing gap in
Sn. Thisapproximation is not expected to significatively changethe numerical value of ε and, thus, it is enough for ourpurposes here to estimate the isospin mixing in the wavefunction. In connection to that, it is also important toremind the approximate character of our estimate of theisospin mixing contribution to the energy of the IAS bymeans of Eq. (5). Hence, the values in Table I should betaken as semi-quantitative results. More precise results
100 104 108 112 116 120 124 128 132 A E I A S ( M e V ) ExpHFBTOY δ E d C d ( M e V )
100 110 120 130 A δ E d C x ( M e V ) Sn SAMi
FIG. 2. Comparison of the HFB results (empty squares) withthe toy model results (plus symbols) as predicted by the SAMiinteraction along the Sn isotopic chain. Experimental dataare also shown when available [18]. In the insets, the absolutedeviation of the Coulomb direct term between HFB (9) andthe toy model (35) are shown in the upper panel, and thosebetween the Coulomb exchange of HFB (11) and the toymodel (36) are shown in the lower panel. removing spurious mixings by using the Quasi-particleRandom Phase Approximation –that exactly restore spu-rious contributions– will be discussed in a future publi-cation. As it is expected, the isospin mixing probabili-ties are larger in neutron-deficient nuclei and smaller inneutron-rich nuclei. As previously investigated [8, 19],pairing may enhance the effect of isospin impurites as itis actually found in our case as well. The effect of theisospin mixing on the IAS energy amounts to − Sn at the largest, and −
150 keV in
Sn at thesmallest.The differences between the microscopic HFB calcula-tions and the macroscopic toy model are shown in theinsets of Fig. 2. Our toy model is quite reasonable inthe description of the Coulomb direct part (with an er-ror of 1% at most), and it reproduces the correct trendof the IAS energy, while the estimate of the Coulomb ex-change term is satisfactory only for stable and neutron-rich nuclei. The reason might be twofold: sharp sphereapproximation in the toy model; and the larger relevanceof surface effects in the Coulomb exchange term whencompared to the Coulomb direct one. The latter can beseen from Eqs. (35) and (36) where the surface correc-tion predicted by the Toy model –term that goes with theneutron skin– is 6 times larger for the Coulomb exchange.In summary, the toy model deviation from the self-consistent HFB calculations is about 5% for
Sn anddecreases smoothly up to 1% for
Sn as it an be seenfrom Fig. 2.
TABLE I. Isospin mixing probability ε in the Sn isotopes.The energy E IMIAS is calculated by using (5), with the mixingprobability ε in the case with pairing included. Energiesare in MeV. Some calculations without pairing did not reachconvergence, and correspond to the entries in the Table thatare left blank A ε (%) ε (%) E IMIAS (MeV)(w/o pairing) (with pairing)102 2.161 2.363 − − − − − − − − − − − − − − − C. Neutron-deficient medium- and heavy-nucleiand IAS energies
In Table II, we show the results of deformed HFBcalculations as predicted by SAMi for several neutron-deficient nuclei which are now planned to be studied ex-perimentally by LUNESTAR project [10]. E HFBIAS is thesum of the direct Coulomb, exchange Coulomb and theisospin mixing contributions in Eqs. (5), (9) and (11).The deformation effect on the IAS energy has been es-timated in different ways: from the HFB calculations∆ E HFBIAS ≡ E HFBIAS ( β n , β p ) − E HFBIAS ( β n = 0 , β p = 0); byusing HFB calculations to input the quantities appearingin the r.h.s. of Eq. (30); and by using HFB calculationsto input the quantities appearing in the r.h.s. of Eq. (31).The deformation effect varies from −
212 keV in
Yb atthe maximum to zero when the nucleus is not deformedlike, for example, in the case of
Nd. In general, thedeformation effect is small, nevertheless it might be im-portant –together with other neglected contributions notobject of this study– for a precise prediction of the IASenergy.The IAS of these nuclei are planned to be measured bycharge-exchange ( He , t ) reaction at E lab = 420 MeV atRCNP, Osaka University, under the LUNESTAR project[10]. This projectile energy is very selective to exciteisospin states with respect to spin-isospin states. Theexperimental campaign will give us new information ofIAS states outside of the valley of stability. Specifically,it is planned to investigate the available most neutrondeficient stable nuclei, and the ( He, t ) reaction will leadto unstable daughter nuclei. Using ( He, t ) ensures thebest spectral resolution and the most precise determina-tion of the IAS excitation energy in the daughter nucleus. The mass of the daughter nucleus will be measured by aPenning trap experiment. The two experiments, ( He, t )and the mass measurement, will be combined to extractthe excitation energy of IAS with high accuracy. The re-sults shown in Table II reproduce the few experimentallyknown IAS energies within an error of about 100 keVand, thus, may give a good guide for experimental searchof new IAS states of these neutron-deficient nuclei.The nuclei listed in Table II have relatively small neu-tron skin size, so that accurate experimental cross sectionmeasurements will not only provide the neutron skin size(small) but also the radial dependence of skin density∆ ρ np = ρ n − ρ p . Then, the deformation effect may playan important role to establish the link between ∆ ρ np andthe symmetry energy in these unstable nuclei, while thecontributions to the absolute IAS energy are rather small. IV. SUMMARY
We have studied the isobaric analog state in spheri-cal and deformed nuclei in the medium- and heavy-massregion. We propose a general formula [Eq. (24)] to ac-count in a simple way for the effects of deformation onthe energy of the IAS and a toy model based on suchformula to explore both, deformation effects on IAS en-ergy and its dependence with the neutron skin thickness.We examine the validity of the presented model by com-paring it with deformed HFB calculations in Sec. III.We have found that the expression (24) describe well de-formation effects and that the toy model works well todescribe the Coulomb direct contribution –that is, themain contribution– to the IAS energy, within an accu-racy at the 1% level. The toy model expression for theCoulomb exchange within the Slater approximation givesalso a good account in the stable Sn isotopes, but it showsnon-negligible differences for Sn isotopes with N ∼ Z .We have also performed deformed HFB calculations ofmany neutron-deficient nuclei outside the valley of sta-bility. Our model reproduces well the empirical IAS en-ergies, for those nuclei whose IAS energies have been al-ready measured. Thus, our HFB results may provide areasonable guide for future experiments in the neutron-deficient nuclei proposed in the LUNESTAR project atRCNP, Osaka University [10]. ACKNOWLEDGMENTS
We would like to thank A. Tamii for encouraging usto study the deformed IAS energies. We thank also K.Yoshida for illustrative discussions on charge exchangeQPA for IAS states. Thanks are for Dieter Frekers, andTommi Eronen for informing us details of experimentalproject LUNESTAR. This work was supported in part byJSPS KAKENHI Grant Numbers JP19K03858. Fundingfrom the European Union’s Horizon 2020 research and
TABLE II. Results of deformed HFB calculations with the SAMi EDF. All energies are in MeV and proton rms radii andneutron skin thickness in fm. E HFBIAS is the sum of the direct Coulomb, exchange Coulomb and the isospin mixing contributionsin Eqs. (5), (9) and (11). The deformation effect on the IAS energy has been estimated in different ways: from the HFBcalculations [∆ E HFBIAS ≡ E HFBIAS ( β n , β p ) − E HFBIAS ( β n = 0 , β p = 0)]; by using HFB calculations to input the quantities appearingin the r.h.s. of Eq. (30); and by using HFB calculations to input the quantities appearing in the r.h.s. of Eq. (31) (for theCoulomb direct term in both cases).Nucl. β n β p β m (cid:104) r p (cid:105) / ∆ r np E HFBIAS E expIAS ∆ E HFBIAS ∆ E IAS ∆ E IAS
Ref. [18] [Eq.(30)] [Eq.(31)]
Pd 0.186 0.174 0.180 4.421 0.042 13.061 − − − Cd 0.255 0.256 0.256 a − − − Sn 0.192 0.199 0.195 b − − − Te 0.039 0.045 0.042 4.620 0.074 14.237 − − − Xe 0.000 0.000 0.000 c Ba 0.000 0.000 0.000 4.744 0.069 14.954 0.000 0.000 0.000
Ce 0.126 0.158 0.140 d − − − Ce 0.040 0.047 0.043 4.820 0.087 15.067 − − − Nd 0.000 0.000 0.000 4.864 0.079 15.540 0.000 0.000 0.000
Sm 0.000 0.000 0.000 4.893 0.061 16.008 16.075(15) 0.000 0.000 0.000
Dy 0.201 0.225 0.211 5.066 0.072 16.499 − − − Dy 0.197 0.217 0.205 5.076 0.087 16.444 − − − Er 0.348 0.378 0.360 e − − − Er 0.346 0.377 0.359 f − − − Yb 0.385 0.413 0.396 5.293 0.086 16.860 − − − Hf 0.304 0.319 0.310 5.289 0.091 17.426 − − − Hf 0.267 0.276 0.271 5.281 0.104 17.341 17.388(7) − − − W 0.278 0.299 0.286 5.334 0.093 17.718 − − − Os 0.335 0.355 0.343 5.418 0.085 18.035 − − − Pt -0.147 -0.141 -0.145 5.359 0.106 18.468 − − − Hg -0.180 -0.187 -0.183 5.437 0.105 18.736 − − − a Deformed secondary energy minimum for β ∼ .
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