Pauli principle and Slater determinants

The previous topic settled the case of two electrons. The exchange operator P₁₂ swaps their full coordinates x = (r, σ), position and spin projection together; it squares to the identity, so its only eigenvalues are +1 and −1, and the spin-statistics theorem, imported from relativistic quantum field theory, gives half-integer spin the minus sign. An atom with N electrons owes that minus sign to every pair at once: PᵢⱼΨ = −Ψ for each of the N(N − 1)/2 choices of i and j. Any rearrangement P of the N labels can be assembled from such pairwise swaps. The assembly is not unique, but whether it takes an even or an odd number p of swaps is fixed by P alone, so the separate pair conditions merge into one rule: PΨ = (−1)ᵖΨ, unchanged by even permutations and reversed by odd ones. That is a demanding requirement, one function of N arguments locked to a definite sign across all N! orderings of those arguments. Every Pᵢⱼ commutes with the Hamiltonian of identical particles, so a state built to obey the rule keeps obeying it for all time, and the construction has to be done only once.

Start from the naive independent-particle guess χₐ(x₁)χ_b(x₂), which claims electron 1 is in χₐ. Swapping the labels changes it, so it is illegal. Repair it by subtracting the swapped copy and renormalising: Ψ = [χₐ(x₁)χ_b(x₂) − χ_b(x₁)χₐ(x₂)]/√2, which flips sign on exchange and is normalised whenever the spin-orbitals are orthonormal. Three electrons need all 3! = 6 orderings, each carrying the parity of the permutation that produced it — plus for even, minus for odd. That signed sum over permutations is exactly the Leibniz formula for a determinant, so write it as one: build the N × N matrix whose entry in row i, column j is χⱼ(xᵢ), take the determinant, divide by √(N!). Rows are electrons, columns are spin-orbitals. The compression is free and dramatic — neon's ground state is 10! = 3 628 800 signed products, and the determinant holds them all in one symbol whose algebra then does the physics for you.

Two determinant identities now carry the whole content of the Pauli principle. Interchange two rows and a determinant changes sign: swap two electrons and Ψ flips sign, which is antisymmetry, for free. Give a determinant two identical columns and it is zero: assign two electrons the same spin-orbital and Ψ vanishes at every set of coordinates. Not small, not suppressed — identically zero, so the occupation number of each spin-orbital is 0 or 1, and exclusion is a corollary rather than a postulate. Read the statement carefully: what may not repeat is the whole spin-orbital, space part and spin part together. The 1s orbital holds two electrons because 1s α and 1s β are different columns, and a subshell of orbital quantum number l holds 2(2l + 1) electrons, six for a p subshell. One feature is easy to miss: the determinant assigns nothing to electron 1. Mixing the occupied spin-orbitals among themselves by any unitary transformation leaves Ψ unchanged up to a phase, so which electron occupies which orbital is not a question the state answers.

Set both arguments equal in Ψ(x₁, x₂) = −Ψ(x₂, x₁): you get Ψ(x, x) = −Ψ(x, x), so Ψ(x, x) = 0. Two electrons in the same spin state have zero amplitude to sit at the same point — a hole dug by a sign, with no term in the Hamiltonian doing the digging. Split the two-electron state into space and spin to see the consequence. The antisymmetric spin function is the singlet, one state with S = 0, and it must multiply a symmetric spatial function; the three symmetric spin functions form the triplet, S = 1, and they must multiply an antisymmetric spatial function that vanishes at r₁ = r₂. Parallel spins therefore keep further apart, their expectation of e²/4πε₀r₁₂ is smaller, and the triplet lies lower. Helium's 1s2s configuration shows it directly: 2³S sits 19.82 eV above the ground state while 2¹S sits 20.62 eV above it, a splitting of 0.80 eV = 2K, so the exchange integral for that pair is about 0.40 eV.

Take the expectation of the many-electron Hamiltonian in one determinant and the Slater–Condon rules return E = Σᵢ hᵢᵢ + ½ Σᵢⱼ (Jᵢⱼ − Kᵢⱼ), the sums running over occupied spin-orbitals. The one-electron term hᵢᵢ holds kinetic energy and nuclear attraction. Jᵢⱼ is the ordinary Coulomb repulsion between two charge clouds, exactly what a classical picture would give. Kᵢⱼ has no classical analogue at all: it exists only because of the determinant's minus sign, it enters with a minus, and it vanishes unless χᵢ and χⱼ share a spin projection. Two checks follow. The i = j terms give Jᵢᵢ − Kᵢᵢ = 0 exactly, so no electron repels itself — a result the older Hartree method reaches only by deleting the self term by hand. And the counting is concrete: neon's ten electrons make 45 pairs, of which the 20 same-spin pairs carry an exchange integral and the 25 opposite-spin pairs do not. Minimising E over the orbitals themselves gives the Hartree–Fock equations.

A single determinant is an independent-particle state: each electron feels the average field of the rest, never their instantaneous positions. Same-spin electrons still avoid each other, because antisymmetry forces it, but opposite-spin electrons are left statistically independent, their pair density factorising with no Coulomb hole at all — in helium's ground state the two electrons are permitted to sit on top of one another. The missing energy has a name and a size. A hydrogenic trial with optimised effective charge Z′ = 27/16 gives −77.49 eV; relaxing the orbital shape while keeping one determinant reaches the Hartree–Fock limit, −77.87 eV; the exact non-relativistic value is −79.01 eV. The difference, E_corr = −1.14 eV, is only 1.4% of the total binding, yet it is a quarter of the 4.48 eV bond in H₂, so it decides chemistry. No single determinant can recover it; the cure is a sum of determinants, configuration interaction and its descendants.