Two spin-one-half fermions on a ring ($E={\hslash }^2(l_1^2+l_2^2)/(2m{R}^2)$)

For free fermions on a ring, each one-particle orbital has integer angular momentum label $l$ and two spin states. Two particles occupy the $l=0$ orbital in a unique singlet state at zero kinetic energy. At the next distinct energy ${\hslash }^2/(2m{R}^2)$, orbitals $0,+1$ or $0,-1$ can be combined with one singlet state or three triplet state spin states, giving total quantum degeneracy eight. The lowest triplet state has six states and its antisymmetric spatial factor vanishes when the particles coincide.