Two-electron states in the first two shells of an isotropic oscillator

The spatial ground orbital of a three-dimensional quantum harmonic oscillator has $l=0$ and energy $3\hslash {\omega }_0/2$; its first excited shell has $l=1$ and energy $5\hslash {\omega }_0/2$. Two identical Electrons both in the ground orbital must form the singlet state, giving one state at total energy $3\hslash {\omega }_0$.

At energy $4\hslash {\omega }_0$, one electron occupies each shell. There are two ground-shell spin orbitals and six first-excited-shell spin orbitals, hence twelve Slater determinants. Equivalently, $0\otimes 1$ gives $L=1$; symmetric spatial combinations with the singlet state give three states, while antisymmetric spatial combinations with the triplet states give nine. Distinct occupied orbitals allow both spatial exchange symmetries, so both spin sectors must be kept.

For the interaction $\lambda \mathbf{S}\cdot \mathbf{L}$ built from the total two-electron spin and orbital angular momentum, set $\mathbf{J}=\mathbf{L}+\mathbf{S}$. The spin-dot-product eigenvalue identity then gives the shift in the title. In the first excited two-electron oscillator manifold, the singlet has $L=1,S=0,J=1$ and zero shift with multiplicity three. The triplet has $L=S=1$, with $J=0,1,2$ shifts $-2\lambda {\hslash }^2,-\lambda {\hslash }^2,+\lambda {\hslash }^2$ and multiplicities $1,3,5$.

The multiplicities sum to twelve, and their weighted shift sums to zero. The specified interaction preserves both $L^2$ and $S^2$; it is distinct from summing separate single-electron spin-orbit interactions. Its shifts are exact within this model's invariant oscillator manifold, with weak coupling ensuring the original level ordering remains useful.