Exchange symmetry of two identical spin-one-half fermions

For two identical spin-one-half fermions with no additional antisymmetric internal factor, the antisymmetric spin-one-half singlet state combines with an even-$\ell$ symmetric orbital state, while the symmetric spin-one-half triplet state combines with an odd-$\ell$ antisymmetric orbital state.

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.

For well width $a$, put $\epsilon ={\pi }^2{\hslash }^2/(2m{a}^2)$ and $J=\lambda {\hslash }^2$ when the interaction is $-\lambda {\mathbf{S}}_1\cdot {\mathbf{S}}_2$. The spin-dot-product eigenvalue gives exchange energies $3J/4$ for the singlet state and $-J/4$ for the triplet states. Identical fermions require symmetric spatial orbitals with the singlet and antisymmetric orbitals with the triplet. Two particles in the same orbital can therefore only be singlets.

For $0<J\ll \epsilon$, the three lowest distinct energy levels are the orbital-$11$ singlet at $2\epsilon +3J/4$, orbital-$12$ triplet at $5\epsilon -J/4$, and orbital-$12$ singlet at $5\epsilon +3J/4$, with multiplicities $1,3,1$. Distinguishable particles instead allow an orbital-$11$ triplet, making the lowest three levels $2\epsilon -J/4$, $2\epsilon +3J/4$, and $5\epsilon -J/4$, with multiplicities $3,1,6$. The sixfold level includes both orbital orderings. For negative $J$, the spin-sector orderings reverse within each fixed orbital sum.

For two identical fermions of mass $m$ in the same one-dimensional quantum harmonic oscillator, use center of mass $X=(x_1+x_2)/2$ and separation $r=x_1-x_2$. The corresponding oscillator masses are $2m$ and reduced mass $m/2$, with the same frequency $\omega$. If $N,n$ are their occupation numbers, the unperturbed energy is $\hslash \omega (N+n+1)$. The fermionic exchange symmetry requires even $n$ with a spin-one-half singlet state and odd $n$ with a spin-one-half triplet state.

For a delta potential $\lambda \delta (r)$ the first-order shift is $\lambda |{\phi }_n^{\mathrm{rel}}(0){|}^2$. Every odd relative wavefunction vanishes at contact, so these states remain exact unaffected eigenstates of the spatial problem. For the even relative ground state, the oscillator length is $\sqrt{2}\ell$, where $\ell =\sqrt{\hslash /(m\omega )}$, and the shift is $\lambda /(\sqrt{2\pi }\ell )$, independent of $N$. This separates interaction-sensitive singlets from contact-insensitive triplets.