Motivation

When solving the Hartree-Fock project using a nucleon-nucleon interaction in an uncoupled basis (m-scheme), we found a high level of degeneracy. One sees clear from the table here that we have a degeneracy in the angular momentum \( j \), resulting in \( 2j+1 \) states with the same energy. This reflects the rotational symmetry and spin symmetry of the nuclear forces.

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.