## Developing a Hartree-Fock program, analyzing the results

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