Running the program, one finds that the lowest-lying states for a nucleus like \( {}^{16}\mbox{O} \), we see that the nucleon-nucleon force brings a natural spin-orbit splitting for the \( 0p \) states (or other states except the \( s \)-states). Since we are using the \( m \)-scheme for our calculations, we observe that there are several states with the same eigenvalues. The number of eigenvalues corresponds to the degeneracy \( 2j+1 \) and is well respected in our calculations, as see from the table here.
The values of the lowest-lying states are (\( \pi \) for protons and \( \nu \) for neutrons)
| Quantum numbers | Energy [MeV] |
| \( 0s_{1/2}^{\pi} \) | -40.4602 |
| \( 0s_{1/2}^{\pi} \) | -40.4602 |
| \( 0s_{1/2}^{\nu} \) | -40.6426 |
| \( 0s_{1/2}^{\nu} \) | -40.6426 |
| \( 0p_{1/2}^{\pi} \) | -6.7133 |
| \( 0p_{1/2}^{\pi} \) | -6.7133 |
| \( 0p_{1/2}^{\nu} \) | -6.8403 |
| \( 0p_{1/2}^{\nu} \) | -6.8403 |
| \( 0p_{3/2}^{\pi} \) | -11.5886 |
| \( 0p_{3/2}^{\pi} \) | -11.5886 |
| \( 0p_{3/2}^{\pi} \) | -11.5886 |
| \( 0p_{3/2}^{\pi} \) | -11.5886 |
| \( 0p_{3/2}^{\nu} \) | -11.7201 |
| \( 0p_{3/2}^{\nu} \) | -11.7201 |
| \( 0p_{3/2}^{\nu} \) | -11.7201 |
| \( 0p_{3/2}^{\nu} \) | -11.7201 |
| \( 0d_{5/2}^{\pi} \) | 18.7589 |
| \( 0d_{5/2}^{\nu} \) | 18.8082 |