Last time, we studied the finite spherical square well and calculated the energy levels for difference angular momentum. In that case, the good quantum numbers (quantities that is commute with the Hamiltonian, i.e. conserved, invariance with time) are the principle quantum number , which dictated the number of node and the angular momentum , and the spin state . A eigen state can be denoted as . The degeneracy for each state is .

With the spin-orbital potential

The angular momentum and spin are not good quantum numbers. The sum, total angular momentum and it z-component are. The new eigen state is .

The Hamiltonian inside the well is

The energy for dose not change, as the is still a good quantum number. The spin-orbital coupling is evaluated like

then,

Since the spin has only 2 spin states, thus,

Note that, the mean energy of the L-states are unchanged. The degeneracy for is , for is . Also, the s-states are not affected, because .

Thus, the wave vector inside the well becomes

the wave vector outside the well

And the rest is matching the boundary condition.

We can also add one more term, an additional term so that the centroid of the L-state is shifted. The energy for this potential is .

Here is the energy levels for and .

We can see, we reproduced the shell ordering 0s1/2, 0p3/2, 0p1/2, 0d5/2, 1s1/2, 0d3/2, 0f7/2, 1p3/2, 1p1/2, 0f5/2….

The magic number 2, 8, 20, and 40 are reproduced. May be, if I adjust the strength of the spin-orbit coupling, my get the magic number 28 and 50. Since the energy level of the s-states are fixed, I need to adjust the shift of the centroid of the others, and the splitting.

Here is the result of

It seems that the magic number of 28, which is from the isolation of 0f7/2, is recreated. And the isolation of 0g9/2, created the magic number of 50.