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The details of the DRHBc theory can be found in Refs. [9, 22, 39, 40]. The DRHBc theory with a point-coupling density functional has been presented in Ref. [24]. For simplicity, we only briefly describe the main formalism here. The relativistic Hartree-Bogoliubov (RHB) equation is
$ \begin{pmatrix} h_{D}-\lambda_{\tau} & \Delta \\ -\Delta^{\ast} & -h_{D}^{\ast}+\lambda_{\tau} \end{pmatrix} \begin{pmatrix} U_{k} \\ V_{k} \end{pmatrix} = E_{k}\begin{pmatrix} U_{k} \\ V_{k} \end{pmatrix},$
(1) which is solved in a Dirac Woods-Saxon basis [23].
$ E_{k} $ is the quasiparticle energy, and$ (U_{k} ,V_{k})^{\mathrm{T}} $ is the quasiparticle wave function.$ \lambda_{\tau} $ ($ \tau = n\ \mathrm{or}\ p $ ) is the Fermi energy. The pairing potential$ \Delta $ is$ \Delta(\boldsymbol{r}_{1},\boldsymbol{r}_{2}) = V^{pp}(\boldsymbol{r}_{1},\boldsymbol{r}_{2})\kappa(\boldsymbol{r}_{1},\boldsymbol{r}_{2}), $
(2) where
$ \kappa(\boldsymbol{r}_{1},\boldsymbol{r}_{2}) $ is the pairing tensor, and$ V^{pp}(\boldsymbol{r}_{1},\boldsymbol{r}_{2}) $ is a density dependent force of zero-range$ V^{pp}(\boldsymbol{r}_{1},\boldsymbol{r}_{2}) = V_{0}\frac{1}{2}(1-P^{\sigma})\delta(\boldsymbol{r}_{1}-\boldsymbol{r}_{2})\left[1-\frac{\rho(\boldsymbol{r}_{1})}{\rho_{\mathrm{sat}}}\right]. $
(3) $ V_0 $ is the pairing strength,$ \rho_{\mathrm{sat}} $ is the saturation density of nuclear matter and$ \dfrac{1}{2}(1-P^{\sigma}) $ is the projector for the spin$ S = 0 $ component. The Dirac Hamiltonian$ h_{D} $ is expressed as$ h_{D} = \boldsymbol{\alpha}\cdot\boldsymbol{p}+V(\boldsymbol{r})+\beta[M+S(\boldsymbol{r})] $
(4) with
$ S(\boldsymbol{r}) $ and$ V(\boldsymbol{r}) $ being the scalar and vector potentials, respectively. For an axially deformed nuclei with the reflection symmetric shape, the potential can be expanded in terms of Legendre polynomials,$ f(\boldsymbol{r}) = \sum\limits_{\lambda}f_{\lambda}(r)P_{\lambda}(\cos\theta),\ \lambda = 0,2,4,6\cdots $
(5) with
$ f_{\lambda}(r) = \frac{2\lambda+1}{4\pi}\int {\rm d}\Omega f(\boldsymbol{r})P_{\lambda}(\cos\theta). $
(6) The relativistic Hartree-Bogoliubov equations are solved in a spherical Dirac Woods-Saxon basis to obtain the wave functions. The total energy of a nucleus can be obtained as [7]
$ \begin{aligned}[b] E_{\mathrm{RHB}} =& \sum\limits_{k>0}(\lambda_{\tau}-E_{k})v^2_k\\& -\int {\rm d}\boldsymbol{r} \Bigg\{\frac{1}{2}\alpha_{S}\rho^2_S+\frac{1}{2}\alpha_{V}\rho^2_V+\frac{1}{2}\alpha_{TV}\rho^2_3\\& +\frac{2}{3}\beta_{S}\rho^3_S+\frac{3}{4}\gamma_{S}\rho^4_S+\frac{3}{4}\gamma_{V}\rho^4_V+\frac{1}{2}\delta_{S}\rho_S \Delta\rho_S\\ & +\frac{1}{2}\delta_{V}\rho_V \Delta\rho_V+\frac{1}{2}\delta_{TV}\rho_3 \Delta\rho_3+\frac{1}{2}eA_{0}\rho_p\Bigg \}\\ &-E_{\mathrm{pair}}+E_{\mathrm{c.m.}}. \end{aligned}$
(7) where
$v^2_k = \displaystyle\int {\rm d}\boldsymbol{r}V^{\dagger}_k(\boldsymbol{r})V_k(\boldsymbol{r})$ , and the densities are$ \begin{aligned}[b] \rho_S(\boldsymbol{r}) =& \sum\limits_{k>0}V^{\dagger}_k(\boldsymbol{r})\gamma_0 V_k(\boldsymbol{r}),\\ \rho_V(\boldsymbol{r}) =& \sum\limits_{k>0}V^{\dagger}_k(\boldsymbol{r})V_k(\boldsymbol{r}),\\ \rho_3(\boldsymbol{r}) =& \sum\limits_{k>0}V^{\dagger}_k(\boldsymbol{r})\tau_3 V_k(\boldsymbol{r}). \end{aligned}$
(8) The pairing energy is calculated as
$ E_{\mathrm{pair}} = -\frac{1}{2}\int {\rm d}\boldsymbol{r} k(\boldsymbol{r})\Delta(\boldsymbol{r}). $
(9) The center-of-mass (c.m.) correction energy is given by
$ E_{\mathrm{c. m.}} = -\frac{1}{2mA} \langle\hat{\boldsymbol P}^2\rangle $
(10) with A the mass number and
$ \hat{\boldsymbol P} $ the total momentum for nucleus in the c.m. frame.The present calculations are carried out with the density functional PC-PK1. The numerical details are the same as those used in Ref. [24]. The Dirac Woods-Saxon basis is chosen with the box size
$ R_{ \rm{box}} = 20 $ fm and the mesh size$ \Delta r = 0.1 $ fm. The numbers of states in the Dirac sea and Fermi sea are set to be the same. The energy cutoff for the Woods-Saxon basis is taken as$ E^{+}_{ \rm{cut}} = 300 $ MeV, and the angular momentum cutoff is$ J_{ \rm{max}} = 23/2\hbar $ . The density-dependent zero-range pairing force is adopted, and the pairing strength is determined by the experimental odd-even differences in binding energies. For the particle-particle channel, with a pairing window of$ 100 $ MeV, the saturation density$ \rho_{ \rm{sat}} = 0.152 $ fm$ ^{-3} $ , and pairing strength$ V_0 = -325.0 $ MeV fm$ ^3 $ . The convergence check of the Legendre expansion has been performed [41], and$ \lambda_{ \rm{max}} = 10 $ is used in the present work. -
We take Ds isotopes near the neutron drip line as an example to reveal the microscopic mechanism of reentrant binding beyond the drip line. Table 1 shows the detailed results from
$ ^{362} {\rm{Ds}}$ to$ ^{404} {\rm{Ds}}$ . The single neutron energy levels of the nuclei around$ ^{368} {\rm{Ds}}$ are shown in Fig. 2. One can see that for the spherical nuclei$ ^{368} {\rm{Ds}}$ ,$ ^{370} {\rm{Ds}}$ , and$ ^{372} {\rm{Ds}}$ , a large shell gap appears for these three nuclei at$ N = 258 $ , which indicates the nature of spherical shell closure. According to the two-neutron separation energies,$ ^{368} {\rm{Ds}}$ is a bound nucleus, whereas$ ^{370} {\rm{Ds}}$ and$ ^{372} {\rm{Ds}}$ are not. This is because the neutron Fermi surface of$ ^{368} {\rm{Ds}}$ is just at the$ N = 258 $ spherical shell. For the ground state, all the 258 neutrons occupy the single-particle levels below the continuum threshold (shown by the dot-dashed line in Fig. 2). For$ ^{370} {\rm{Ds}}$ ($ ^{372} {\rm{Ds}}$ ), there are two (four) neutrons mainly occupying the$ 1k_{15/2} $ orbital, which are just above the continuum threshold. Therefore, the Fermi energies of$ ^{370,372} {\rm{Ds}}$ are positive, and their two-neutron separation energies are negative, and thus,$ ^{370,372} {\rm{Ds}}$ are unbound against two-neutron emission. For$ ^{374} {\rm{Ds}}$ , six more neutrons, in comparison with$ ^{368} {\rm{Ds}}$ , lead to deformation of$ \beta_{2} = 0.064 $ . Three deformed single-particle levels stemming from the spherical orbitals$ 2i_{13/2} $ and$ 1k_{15/2} $ drop, and their energies become negative. The increased deformation leads to a much smaller$ N = 258 $ shell gap as the neutron number N increases. For the ground state, the six neutrons mainly occupy the three levels just below the continuum threshold, and the Fermi energy becomes negative with a very small absolute value. However,$ ^{374} {\rm{Ds}}$ is still unbound against two-neutron emission due to the negative two-neutron separation energy, which can be explained by a loss of pairing energy. For$ ^{376} {\rm{Ds}}$ , the deformation increases, five deformed single-particle levels drop below the continuum threshold, and there is no energy gap around the Fermi surface. The two-neutron separation energy is positive, and the Fermi energy is negative.$ ^{376} {\rm{Ds}}$ is bound against two-neutron emission but unbound against multi-neutron emission.$ ^{378} {\rm{Ds}}$ is very similar to$ ^{376} {\rm{Ds}}$ . For the heavier isotopes up to$ ^{400} {\rm{Ds}}$ , the deformation$ \beta_{2} $ keeps increasing, and more deformed single-particle levels stemming from the$ 2i_{13/2} $ and$ 1k_{15/2} $ orbitals drop below the continuum threshold.$ ^{380-400} {\rm{Ds}}$ are all bound nuclei. Therefore, we see that the nuclear deformation can strongly influence the single-particle levels and the shell structures. It plays a vital role in reentrant binding beyond the drip lines in the presence of shell effects at neutron closure$ (N = 258) $ .A N B/MeV $ S_{2n} $ /MeV$ \lambda_{n} $ /MeV$ \beta_{2} $ $ Z=110 $ (Ds)362 252 2284.175 3.311 −1.468 −0.084 364 254 2287.042 3.311 −1.570 −0.051 366 256 2290.220 3.178 −1.421 −0.034 368 258 2293.352 3.132 −0.834 0.000 370 260 2292.776 −0.576 0.255 0.000 372 262 2292.224 −0.552 0.237 0.000 374 264 2291.931 −0.293 −0.113 0.064 376 266 2292.448 0.517 −0.325 0.097 378 268 2293.115 0.667 −0.337 0.118 380 270 2293.724 0.609 −0.317 0.136 382 272 2294.359 0.635 −0.358 0.155 384 274 2295.126 0.767 −0.454 0.177 386 276 2296.139 1.013 −0.555 0.201 388 278 2297.258 1.119 −0.555 0.218 390 280 2298.299 1.041 −0.481 0.232 392 282 2299.146 0.847 −0.405 0.243 394 284 2299.886 0.741 −0.370 0.253 396 286 2300.566 0.679 −0.330 0.260 398 288 2301.156 0.590 −0.275 0.265 400 290 2301.593 0.437 −0.141 0.269 402 292 2301.674 0.081 0.008 0.273 404 294 2301.555 −0.119 0.071 0.277 Table 1. Ground-state properties of Ds isotopes with
$ 252\leqslant N\leqslant294 $ calculated using the DRHBc theory. The two-neutron unbound nuclei are underlined.Figure 2. (color online) Single-neutron energy levels in the canonical basis near the Fermi surface of Ds isotopes with
$ 256\leqslant N\leqslant270 $ . The neutron Fermi energy$ \lambda_{n} $ of each isotope is denoted by hollow stars.As the existence of the bound nuclei beyond the drip line is affected strongly by the nuclear deformation, it is essential to investigate how strong the influence of the different orders of deformation is on this phenomenon. We perform the DRHBc constrained calculations at
$ \beta_{2} = 0 $ and unconstrained calculations with the Legendre expansion truncation$ \lambda_{\mathrm{max}} = 2 $ (only including quadrupole deformation$ \beta_{2} $ ),$ \lambda_{\mathrm{max}} = 4 $ (including quadrupole and hexadecapole deformations$ \beta_{2} $ and$ \beta_{4} $ ), and$ \lambda_{\mathrm{max}} = 6 $ (including$ \beta_2 $ ,$ \beta_4 $ , and$ \beta_6 $ ), respectively. The calculated total energies relative to that of$ ^{368} {\rm{Ds}}$ $ (N = 258) $ as a function of the neutron number for$ ^{364-402} {\rm{Ds}}$ are presented in Fig. 3. One can see that for the calculations with$ \beta_{2} = 0 $ ,$ E-E_{N = 258} $ increases at$ N>258 $ . No bound nuclei would exist beyond the drip line. For the calculation of$ \lambda_{\mathrm{max}} = 2 $ , there are 4 unbound, 3 two-neutron emission bound while multi-neutron emission unbound, and 12 bound nuclei beyond the neutron drip line. Comparing with the result of$ \lambda_{\mathrm{max}} = 10 $ , there is 1 additional two-neutron emission bound but multi-neutron emission unbound nucleus beyond the drip line. For the calculation of$ \lambda_{\mathrm{max}} = 4 $ , except for the quantitive differences, the obtained bound nuclei are the same as those of the$ \lambda_{\mathrm{max}} = 10 $ calculation. For the calculations of$ \lambda_{\mathrm{max}} = 6 $ , the results are very similar to those of$ \lambda_{\mathrm{max}} = 10 $ , except that there is 1 additional bound nucleus obtained in the calculations with$ \lambda_{\mathrm{max}} = 6 $ . It demonstrates clearly that the nuclear deformation is vital to the existence of the bound nuclei beyond the drip lines.Figure 3. (color online) DRHBc calculated total energies relative to that of
$ ^{358} {\rm{Ds}}$ (N = 258) under different expansion orders of deformation. The bound and unbound nuclei are denoted by the solid and open symbols, respectively. The nuclei that are bound against two-neutron emission while unbound against multi-neutron emission are denoted by the crossed open symbols.
Possible existence of bound nuclei beyond neutron drip lines driven by deformation
- Received Date: 2021-06-10
- Available Online: 2021-10-15
Abstract: Based on the relativistic calculations of the nuclear masses in the transfermium region from No