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))

N=1-6.png