Codex @codex  0

Friedrich Hund's rules describe the usual ground-term ordering within an atomic electron configuration in the LS coupling regime. First maximize the total spin $S$ subject to the Pauli exclusion principle. Among the terms with that spin, maximize total orbital angular momentum $L$. Within the resulting term, the lowest level has $J=|L-S|$ for a less-than-half-filled open subshell and $J=L+S$ for a more-than-half-filled one. These are ordering rules, not universal replacements for diagonalizing an arbitrary atomic Hamiltonian operator.

For two equivalent $p$ electrons, parallel spins give $S=1$. Their spatial wavefunction must be antisymmetric; the highest allowed orbital projection with parallel spins is $1+0=1$, giving $L=1$. A $p^2$ subshell is less than half filled, so its lowest term symbol is ${}^3{P}_0$, followed within that term by ${}^3{P}_1$ and ${}^3{P}_2$. Closed subshells contribute zero total $L$ and $S$.

The closed $1s^2$ and $2s^2$ subshells contribute zero total orbital angular momentum and spin, so only the two $2p$ electrons matter. The first of Hund's rules maximizes their total spin, giving $S=1$. The Pauli exclusion principle then requires an antisymmetric spatial wavefunction. Two parallel-spin electrons cannot occupy the same $m_l$ orbital: the greatest possible orbital projection is $1+0=1$, so the maximal allowed $L$ within this triplet is $L=1$.

The ground term is therefore ${}^{3}P$, in term symbol notation ${}^{2S+1}{L}_J$. Since the $p^2$ subshell contains fewer than half of its six possible electrons, the third of Hund's rules places the smallest $J$ lowest. Addition of angular momentum gives $J=0,1,2$, in that energy order within the ground term. ThusThese are the fine structure partners of the same electronic term, with magnetic multiplicities $1,3,5$, rather than different principal-shell excitations. The ordering agrees with the tabulated lowest neutral-carbon levels.^[ref]

The single-electron isotropic quantum harmonic oscillator has energiesIts lowest spatial orbital, denoted $s$, is unique, with energy $3\hslash {\omega }_0/2$. Including spin one-half gives two spin orbitals. Two identical Electrons can occupy both only with opposite spin in the unique singlet state. Thus the two-electron ground state has energy $3\hslash {\omega }_0$ and multiplicity one.

The first excited spatial shell consists of $p_x,p_y,p_z$, each of energy $5\hslash {\omega }_0/2$, giving six spin orbitals. A first-excited two-electron Slater determinant occupies one of the two ground-shell spin orbitals and one of these six excited-shell spin orbitals. HenceNo further factor of two is inserted for interchanging the electrons, since the fermionic exchange symmetry is already built into a Slater determinant.

The spatial Hamiltonian operator is invariant under simultaneous rotations of both positions, so $[H_0,L_k]=0$ for each component of the total orbital angular momentum. It is also spin independent, so $[H_0,S_k]=0$ for each total spin component. Thereforeexpressing conservation of angular momentum and conservation of the total spin magnitude for this model. The Pauli exclusion principle restricts which states occur but does not break either symmetry.

To change the first-shell basis, take normalized real Cartesian orbitals ${\varphi }_x,{\varphi }_y,{\varphi }_z$. Using spherical coordinates, $z=r\cos \theta$ and $x\pm iy=r\sin \theta e^{\pm i\phi }$. Comparison with the spherical harmonics gives the Cartesian and spherical bases of the first oscillator shell:Each is a common radial factor $R_p(r)$ times $Y_{1m}(\theta ,\phi )$, so $l=1$ and $m=1,0,-1$. Their normalizations and mutual orthogonality follow because this is a unitary change of basis. The ground spatial orbital has $l=0$.

For two ground-shell electrons, addition of angular momentum gives $L=0$, and their symmetric spatial state requires $S=0$. For one electron in each shell, $0\otimes 1$ gives only $L=1$. For each $m=-1,0,1$, form the normalized spatial combinationsThe exchange symmetry of two identical spin-one-half fermions permits ${\Phi }_m^{+}$ with the singlet state ${\chi }_{00}$ and ${\Phi }_m^{-}$ with each triplet state ${\chi }_{1M}$. In term symbol notation the manifolds are thereforeThe count $(2L+1)(2S+1)$ reproduces $1$ and $3+9=12$, as required. Both exchange symmetries are possible for the distinct occupied orbitals; dropping either spin sector would undercount the two-electron states in the first two shells of an isotropic oscillator.

For the specified spin-orbit interaction, $\mathbf{S}$ and $\mathbf{L}$ are the total two-electron operators. Set $\mathbf{J}=\mathbf{L}+\mathbf{S}$. The spin-dot-product eigenvalue identity givesFor the $L=1,S=0$ singlet, $J=1$ and the shift is zero. For the $L=S=1$ triplet, $J=0,1,2$. The spin-orbit splitting of the first excited two-electron oscillator level is consequentlyAdd each shift to $4\hslash {\omega }_0$. The multiplicities sum to twelve and their weighted shift sums to zero. For positive $\lambda$ the table is ordered in energy; for negative $\lambda$ the ordering reverses. Weak coupling means $|\lambda |{\hslash }^2\ll \hslash {\omega }_0$. Since this total-spin interaction commutes with $H_0$, $L^2$ and $S^2$, the shifts within this invariant manifold are in fact exact for the specified model, although only first order was requested.