Axially Deformed Nuclei

For spherical nuclei, Skyrme HFB equations are best solved in the coordinate space, because Eq. (16) reduces in this case to a set of radial differential equations [29]. In the case of deformed nuclei, however, the solution of a deformed HFB equation in coordinate space is a difficult and time-consuming task. For this reason, here we use the method proposed by Vautherin [30], which combines two different representations. The solution of the deformed HFB equation is carried out by diagonalizing the HFB hamiltonian in the configurational space of wave-functions with appropriate symmetry, while evaluation of the potentials and densities is performed in the coordinate space. Such a method is applicable to nonaxial deformations [16], but typical computation time for large-scale mass-table calculations is prohibitively large. In the present implementation, we make the restriction to axially-symmetric and reflection-symmetric shapes in order to obtain HFB solutions within a much shorter CPU time.

In the case of axial symmetry, the third component $J_z$ of the total angular momentum is conserved and provides a good quantum number $\Omega_k$. Therefore, quasiparticle HFB states can be written in the following form:

$$\left( \begin{array}{c} U_k ({\bf r},\sigma,\tau) \\ V_k ({... ...ght) e^{\imath \Lambda^+ \varphi} \chi_{-1/2}(\sigma) \right],$$ (24)

where $\Lambda^\pm=\Omega_k \pm 1/2$ and $r$, $z$, and $\varphi$ are the standard cylindrical coordinates defining the three-dimensional position vector as ${\bf r}=(r\cos\varphi,r\sin\varphi,z)$, while $z$ is the chosen symmetry axis. The quasiparticle states (24) are also assumed to be eigenstates of the third component of the isospin operator with eigenvalues $q_k=+1/2$ for protons and $q_k=-1/2$ for neutrons.

By substituting ansatz (24) into Eq. (16), the HFB equation reduces to a system of equations involving the cylindrical variables $r$ and $z$ only. The same is also true for the local densities, i.e.,

$$\begin{array}{rcll} \rho (r,z) & = & \displaystyle\sum_{k} &... ...k}^{+}}(r,z)+V_{k}^{-}(r,z){U_{k}^{-}}(r,z)\right), \end{array}$$ (25)

where $\nabla_r=\partial/\partial r$ and $\nabla_z=\partial/\partial z$. In addition, when tensor forces are considered, the following additional densities have to be calculated:

$$\begin{array}{rcll} J_{r \varphi} (r,z) & = & \displaystyle\... ...da^{+}}{r} V_{k}^{-}(r,z)V_{k}^{-}(r,z) \right),\\ \end{array}$$ (26)

where indices denote the cylindrical components of the tensor ${\bf J}_{ij}$, while all remaining components vanish due to the cylindrical symmetry, i.e., $J_{r r}(r,z) = J_{z z}(r,z) = J_{\varphi\varphi }(r,z) = J_{r z}(r,z) = J_{z r}(r,z)$=0

Due to the time-reversal symmetry, if the $k$th state, defined by the set $\{U_k^+,U_k^-,V_k^+,V_k^-,\Omega_k\}$, satisfies the HFB equation (16), then the $\bar{k}$th state, corresponding to the set defined by $\{U_k^+,-U_k^-,V_k^+,-V_k^-,-\Omega_k\}$, also satisfies the HFB equation for the same quasiparticle energy $E_k$. Moreover, all wave functions in cylindrical coordinates are real. Contributions of time-reversal states $k$ and $\bar{k}$ are identical (we assume that the set of occupied states is invariant with respect to the time-reversal), and we can restrict all summations to positive values of $\Omega_k$ while multiplying total results by a factor two. In a similar way, one can see that due to the assumed reflection symmetry, only positive values of $z$ need to be considered.