Using the diagonalization method using 3D harmonic oscillator as a basis to expend or approximate the Nilsson orbital, it is not easy to know the good quantum number, i.e. K[Nn_z\Lambda] , here \Lambda = m_l and K is the total spin projected on the body symmetric axis.

One way to find out is using the dominant component |NljK\rangle for small deformation and track the energy curve to the large deformation.

When I listed all the Nilsson orbital in below, it shows a clear pattern.

  1. \Lambda is the projection of orbital angular momentum, thus, l \geq \Lambda
  2. N = n_z gives the lowest energy state
  3. Larger n_z , lower energy
  4. K = 1/2 also gives the lowest energy state
  5. From observation, it is fair to assume that n_z + \Lambda = l
  6. Also, for j = l \pm 1/2, K = \Lambda \pm 1/2

NilssonOrbit to spherical.png

Notice that, in some paper, for example the 2s_{1/2} state was assigned to be 1/2[211]. I think it is no correct, because for s-orbital, it is impossible to have \Lambda = m_l = 1.