| 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 |
We observe that with increasing value of \( j \) the degeneracy increases. For \( j=3/2 \) we end up diagonalizing the same matrix four times. With increasing value of \( j \), it is rather obvious that our insistence on using an uncoupled scheme (or just m-scheme) will lead to unnecessary labor from our side (or more precisely, for the computer). The obvious question we should pose ourselves then is whether we can use the underlying symmetries of the nuclear forces in order to reduce our efforts.