The calculation use 84 Nilsson basis from 0s1/2, up to 6i13/2. Although some lines are broken, it is kind of nice. And we notices that, when the same j orbitals approach each other, they repulse. The straight line states are the (almost) pure state, which only consist with 1 spherical orbital.

Int the calculation, $\kappa = 0.05$ and

$\mu(N) = \begin{pmatrix} 0 & N=1 \\ 0 & N=2 \\ 0.35 & N=3 \\ 0.625 & N = 4 \\ 0.625 & N=5 \\ 0.63 & N=6 \\ 0.63 & N=7 \end{pmatrix}$

The parameter $\mu$ is for adjusting the energy to match with experimental data for spherical nuclei.

The spherical energy, which is the diagonal element of the Hamiltonian of spherical basis, is

$\displaystyle E_0(n,l,j,k) = n + \frac{3}{2} - \kappa(2 l\cdot s + \mu l(l+1))$

$\displaystyle l \cdot s = \frac{1}{2}(j(j+1) - l(l+1) - s(s+1))$