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 $j = l \pm 1/2$, $K = \Lambda \pm 1/2$

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 $\Lambda = m_l = 1$.