Identical particles and exchange symmetry

A two-particle wavefunction Ψ(r₁, r₂) carries labels 1 and 2 that no apparatus can read. Define the exchange operator by P₁₂Ψ(1,2) = Ψ(2,1). For identical particles the Hamiltonian is symmetric in those labels — same mass, same charge, same interaction — so [H, P₁₂] = 0 and the exchange symmetry is a constant of the motion: decided once, kept for all time. Because swapping twice restores the original arrangement, P₁₂² = 1, so the eigenvalues are +1 and −1 and nothing else. That is why 'somewhere in between' is not on offer. A state that was neither symmetric nor antisymmetric would make |Ψ|² change when you relabelled, which is the same as saying the labels are measurable. Particles taking +1 are bosons and those taking −1 are fermions; that the choice follows the spin — integer or half-integer — is the spin-statistics theorem, proved inside relativistic quantum field theory and imported here without proof.

Take two orthonormal one-particle orbitals ψₐ and ψ_b. The plain product ψₐ(r₁)ψ_b(r₂) is not an eigenstate of P₁₂, but the two combinations ψ±(r₁, r₂) = [ψₐ(r₁)ψ_b(r₂) ± ψ_b(r₁)ψₐ(r₂)]/√2 are, with eigenvalues ±1. Now set b = a. The symmetric state survives as ψₐ(r₁)ψₐ(r₂), while the antisymmetric one collapses to ψₐ(r₁)ψₐ(r₂) − ψₐ(r₁)ψₐ(r₂) = 0 at every point. There is no state at all — not a forbidden one, an absent one. That is the Pauli exclusion principle, and notice where it came from: not from a rule about how many electrons a level holds, but from antisymmetry, which came from indistinguishability. Written for N particles the same object is a Slater determinant, and a determinant with two equal rows vanishes for exactly this reason. Bosons, taking the plus sign, carry no such restriction, and pile into a single mode happily.

The antisymmetry demanded of electrons applies to the whole state, spin included. Two spin-half particles combine into one antisymmetric singlet, χ = (↑↓ − ↓↑)/√2 with S = 0, and three symmetric triplet states, ↑↑, (↑↓ + ↓↑)/√2 and ↓↓, with S = 1. If the state factorises as Ψ = ψ(space)⋅χ(spin), the two factors must carry opposite exchange signs, because their product has to be antisymmetric. So a spin singlet is locked to the symmetric spatial function and a spin triplet to the antisymmetric one — para and ortho helium in the older spectroscopic language, two families whose separate line series once looked like two different gases. Nothing here says that spin interacts with anything. Spin is the bookkeeping partner: fixing S fixes the spatial symmetry, and it is the spatial symmetry that the Coulomb energy actually sees.

Work out the mean-square separation of the two symmetrised states in one dimension and the algebra returns ⟨(x₁−x₂)²⟩± = ⟨x²⟩ₐ + ⟨x²⟩_b − 2⟨x⟩ₐ⟨x⟩_b ∓ 2|⟨x⟩_ab|², where ⟨x⟩_ab = ∫ψₐ*(x) x ψ_b(x) dx. The first three terms are what two distinguishable particles would give. The last term is new, and its sign is set by the exchange symmetry: symmetric states sit closer together, antisymmetric states further apart. Two consequences follow. First, no force has been introduced — the Hamiltonian is untouched, and what changed is which configurations the wavefunction permits. Second, the effect requires overlap, because ⟨x⟩_ab integrates over the region where both orbitals are appreciable. Two electrons in orbitals that never overlap — one here, one on the Moon — feel nothing from each other's statistics, and symmetrising them is a formality with no measurable consequence.

Take helium's 1s2s configuration and treat the electron-electron repulsion V = e²/4πε₀r₁₂ as a first-order perturbation on ψ±. The expectation value splits into two integrals. The direct integral J = ⟨ab|V|ab⟩ is the electrostatic repulsion of the two charge clouds and is identical for both states. The exchange integral K = ⟨ab|V|ba⟩ multiplies the cross term and so enters with the exchange sign, giving E± = Eₐ + E_b + J ± K. For the Coulomb repulsion K is positive — a property of the 1/r₁₂ kernel itself, not merely of V being positive — so the triplet, carrying the minus sign, lies 2K below the singlet. Experiment agrees: the 2³S level sits 19.820 eV above helium's ground state and the 2¹S sits 20.616 eV, so 2K = 0.796 eV and K = 0.398 eV. Evaluating K with unscreened hydrogenic Z = 2 orbitals gives (16/729)Z hartree = 1.19 eV — right sign, right order, three times too big, because the real 2s electron is screened toward Z_eff ≈ 1, so its orbital is about twice as wide and overlaps the 1s far less.

Two debts stay outstanding. The first is spin-statistics itself: nothing in non-relativistic quantum mechanics forces half-integer spin to take the minus sign. It follows from demanding a relativistic field theory with an energy bounded below whose observables commute at spacelike separation. In two spatial dimensions, where exchanging twice need not restore the state, the choice is no longer limited to two signs, which is how anyons become possible. The second is correlation. Even after symmetrising, ψ± is assembled from single-particle orbitals, so each electron moves in an average field. Real electrons dodge one another instant by instant, digging a Coulomb hole around each that no single determinant can reproduce. The exchange hole is real but partial — it separates parallel-spin electrons and does nothing for antiparallel ones. Helium's exact non-relativistic ground state is −79.01 eV against −77.87 eV at the Hartree-Fock limit, and that 1.14 eV shortfall is precisely what the term 'correlation energy' names.