The Nilsson orbital can be decomposed into series of orbitals of  3D-harmonic oscillator, such that

$\displaystyle [Nn_z\lambda]K = \sum_{N'lj} C_{N'lj}^{N n_z \lambda} |N'ljK\rangle$

with eigen energy $\epsilon_{Nn_z\lambda K} (\beta)$ and

$n_z + \lambda = l$

$n_z + K = j$

Since the Nilsson orbital is normalized

$\displaystyle \sum_{N'lj} \left(C_{N'lj}^{N n_z \lambda}\right)^2 = 1$

Since the number of orbital for fixed $l,j$ is $2j+1$, thus using an inverse transformation from spherical orbital to Nilsson orbital, we have,

$\displaystyle \sum_{N n_z \lambda} \left(C_{N'lj}^{N n_z \lambda}\right)^2 = 2j+1$

I cannot prove it, but

$\displaystyle \sum_{N n_z \lambda} \epsilon_{N n_z \lambda K}(\beta) \left(C_{N'lj}^{N n_z \lambda}\right)^2 = \epsilon_{N n_z \lambda K} (0)$

Thus, the single-particle energy fro Nilsson orbital is as same as the spherical orbital !!!