Slater determinants and occupation numbers

Take N orthonormal spin-orbitals φ₁ … φ_N, each a vector in L²(R³) ⊗ C², and form the Hartree product φ₁(x₁)φ₂(x₂)…φ_N(x_N) with x = (r, σ). That product is not a fermionic state: swap x₁ and x₂ and it turns into a different function rather than into minus itself. The repair is a projector. Define A = (1/N!) Σ_P sgn(P) P̂, summing over all N! permutations of the coordinate labels with sgn(P) = +1 for even and −1 for odd. A is Hermitian and idempotent, A² = A, so it projects the N-particle space onto its totally antisymmetric subspace. Apply it and restore the norm: Ψ = √(N!) A [φ₁(x₁)…φ_N(x_N)] = (1/√N!) det[φᵢ(xⱼ)]. The prefactor 1/√N! is exactly right only because the orbitals are orthonormal; if they are not, the overlaps survive and ⟨Ψ|Ψ⟩ = det S.

Write the determinant with orbitals down the rows and electron coordinates across the columns. Exchanging two electrons exchanges two columns, and a determinant changes sign under a column swap — antisymmetry delivered by the notation rather than imposed on it. Put the same spin-orbital in two rows and the determinant is identically zero: not a suppressed state, not a rare one, but the zero vector, which cannot be normalised and so is not a state at all. That is the whole content of the exclusion principle. One further property matters and is usually skipped. Replace the occupied orbitals by any unitary mixture φ′ᵢ = Σⱼ Uⱼᵢ φⱼ and the determinant becomes Ψ′ = det(U) Ψ, a phase. The individual orbitals are therefore not observable; the occupied subspace is, carried by the projector ρ̂₁ = Σᵢ |φᵢ⟩⟨φᵢ|. Canonical Hartree–Fock orbitals and localised bonding orbitals are the same determinant in two coordinate systems.

Once an ordered one-particle basis is fixed, a determinant is specified entirely by which orbitals are filled: no assignment of electrons to orbitals is left to record. So write |n₁ n₂ n₃ …⟩ with nᵢ ∈ {0, 1}, defined as (a₁†)^(n₁)(a₂†)^(n₂)… acting on the vacuum, the creation operators standing in ascending index order from the left. The ordering convention is not cosmetic. Because aᵢ†aⱼ† = −aⱼ†aᵢ†, moving a creation operator into its slot costs one sign for every occupied mode it has to cross, so aᵢ†|…⟩ = (−1)^{(Σ_(j⟨i) nⱼ)}|… nᵢ = 1 …⟩ — the Jordan–Wigner string. What the string counts is a parity, not a distance. With four modes, a₄† acting on |1, 1, 0, 0⟩ crosses two filled modes and returns +|1, 1, 0, 1⟩, while a₂† acting on |1, 0, 0, 0⟩ crosses one and returns −|1, 1, 0, 0⟩; the operator that travels further is the one that costs nothing. Drop the string and every matrix element you compute afterwards carries an arbitrary sign, and terms that ought to cancel do not.

The algebra is (aᵢ, aⱼ†) = δᵢⱼ, {aᵢ, aⱼ} = 0 and (aᵢ†, aⱼ†) = 0. Set i = j in the last relation: 2(aᵢ†)² = 0, so aᵢ† applied twice to anything gives the zero vector. That is exclusion in one line, with no appeal to quantum numbers. The occupation operator sharpens it. With n̂ᵢ = aᵢ†aᵢ, use aᵢ aᵢ† = 1 − aᵢ†aᵢ to get n̂ᵢ² = aᵢ†(aᵢ aᵢ†)aᵢ = aᵢ†aᵢ − (aᵢ†)²(aᵢ)² = n̂ᵢ. An operator obeying n̂² = n̂ has eigenvalues satisfying λ² = λ, so λ = 0 or 1 and nothing else. Now change one sign. Bosons obey [bᵢ, bⱼ†] = δᵢⱼ, so bᵢ bᵢ† = 1 + bᵢ†bᵢ and n̂ᵢ² = n̂ᵢ + (bᵢ†)²(bᵢ)², which is not idempotent; the eigenvalues run 0, 1, 2, … with no ceiling. The Pauli limit and Bose–Einstein condensation are one equation with a commutator in place of an anticommutator.

Because the state is a single determinant, every expectation value collapses to sums over occupied orbitals. For Ĥ = Σᵢ ĥ(i) + ½ Σ_(i≠j) 1/rᵢⱼ the diagonal rule reads ⟨Ψ|Ĥ|Ψ⟩ = Σᵢ hᵢᵢ + ½ Σ_{(i, j)} (Jᵢⱼ − Kᵢⱼ), with direct integral Jᵢⱼ = ⟨ij|ij⟩ and exchange integral Kᵢⱼ = ⟨ij|ji⟩. Kᵢⱼ carries a spin overlap ⟨σᵢ|σⱼ⟩², so it vanishes unless the two spin-orbitals share a spin. Geometrically this is the Fermi hole: the pair density of a determinant is P(x₁, x₂) = ρ(x₁)ρ(x₂) − |ρ₁(x₁, x₂)|², and at coincident space-and-spin points ρ₁(x, x) = ρ(x), so P(x, x) = 0 exactly. Parallel-spin electrons never meet. Nothing here referred to a magnetic force between spins — the hole is a consequence of the determinant, and it is why the triplet of a configuration lies below the singlet.

The algebra that bounds n̂ᵢ also exposes the limitation. For a single determinant the one-body density matrix is idempotent, ρ̂₁² = ρ̂₁, so its eigenvalues — the natural occupation numbers — are exactly 0 or 1. No exact interacting ground state passes that test: helium's leading natural orbital holds about 1.98 electrons of the 2, not 2.00. Equivalently P(x₁, x₂) factorises into ρρ up to one exchange term, so antiparallel electrons in the same determinant stay statistically independent exactly where 1/r₁₂ diverges — a Fermi hole but no Coulomb hole. The deficit is defined as E_corr = E_exact − E_HF. For helium E_HF = −2.8617 Eₕ against the exact non-relativistic −2.9037 Eₕ, giving E_corr = −0.0420 Eₕ = −1.14 eV: only 1.4% of the total energy, but a quarter of a typical covalent bond. Recovering it means adding determinants to the wavefunction, and there is no shortage of them to add: 10 electrons in 20 spin-orbitals already furnish C(20,10) = 184 756.