Helium: exchange and the variational bound

In atomic units the helium Hamiltonian is H = −½∇₁² − ½∇₂² − 2/r₁ − 2/r₂ + 1/r₁₂. Every term but the last acts on one electron alone, so with 1/r₁₂ dropped the eigenfunctions are products of hydrogenic orbitals and the ground energy is 2 × (−Z²/2) = −4 Eh = −108.85 eV. Measured, helium takes 24.59 eV to lose its first electron and 54.42 eV to lose the second, so the true ground energy is −79.01 eV. The neglected term is therefore worth about 30 eV — more than a third of the total binding — and cannot be called small by anyone. Worse, r₁₂ = |r₁ − r₂| does not factor in any coordinate system, and no closed-form eigenfunction of this H is known. Everything that follows is a way of getting a defensible number without ever writing the eigenfunction down.

Expand a normalised trial ket in the exact eigenstates of H. Helium has an ionisation continuum above −2 Eh as well as bound levels, so completeness needs both parts: |ψ⟩ = Σ cₙ|n⟩ + ∫ c(E)|E⟩ dE, with H|n⟩ = Eₙ|n⟩, H|E⟩ = E|E⟩ (channel labels suppressed) and Σ|cₙ|² + ∫|c(E)|² dE = 1. Then ⟨ψ|H|ψ⟩ = Σ|cₙ|²Eₙ + ∫|c(E)|²E dE ≥ E₀, since replacing every energy by the smallest one, E₀, can only lower the weighted sum; equality demands |c₀| = 1. Three consequences do the work. The statement is rigorous rather than asymptotic, so a computed −77.49 eV is a certificate that the true energy is at most that. It is one-sided, so between two trial functions the lower number wins outright. And it is quadratically forgiving: an error of order ε in the state costs only order ε² in the energy, which is why a crude function still returns a respectable bound — and why a good energy is weak evidence that the wavefunction itself is any good.

Take ψ(r₁, r₂) = φ_Z′(r₁)φ_Z′(r₂) with φ_Z′ the hydrogenic 1s of charge Z′, and treat Z′ as a knob rather than fixing it at 2. Three expectation values are needed, all standard: ⟨T⟩ = Z′², ⟨Vₙₑ⟩ = −2ZZ′ and ⟨Vₑₑ⟩ = (5/8)Z′, in hartree. Notice the competition — kinetic energy grows quadratically in Z′ while both potential terms grow only linearly — so E(Z′) = Z′² − 2ZZ′ + (5/8)Z′ has a genuine minimum. Setting dE/dZ′ = 2Z′ − 2Z + 5/8 = 0 gives Z′ = Z − 5/16, so for helium Z′ = 27/16 = 1.6875: each electron sees the nucleus shielded by 0.3125 of an electronic charge by the other. Back-substituting, Eₘᵢₙ = −Z′² = −2.8477 Eh = −77.49 eV, against −74.83 eV from leaving Z′ at 2. Letting the orbital breathe was worth 2.66 eV, and screening is the leading physical effect a product form can represent.

−77.49 eV sits 1.52 eV, or 1.9%, above the true −79.01 eV, and it is worth knowing where that goes. Push the trial function to the Hartree–Fock limit — the best single determinant built from any orbitals at all, not just hydrogenic ones — and you reach −77.87 eV. Orbital shape is therefore worth a further 0.38 eV only; the remaining 1.14 eV is correlation energy, which no single determinant can hold. The reason is structural: a product state fixes electron 2's distribution regardless of where electron 1 happens to be, so each electron moves in the other's time-averaged cloud and never dodges it. The exact state suppresses |ψ|² as r₁₂ → 0, a Coulomb hole obeying Kato's cusp condition ∂ψ/∂r₁₂ = ½ψ at contact. Capturing it needs explicit r₁₂ dependence in the trial function; Hylleraas built exactly that in, and his 1929 six-term function already came within about 0.01 eV of the exact energy.

The ground configuration 1s² admits only one spatial function, necessarily symmetric, so antisymmetry forces the spin singlet and no 1s² triplet exists. Excited configurations are richer. From a = 1s and b = 2s build the two spatial functions ψ_± = (1/√2)[a(r₁)b(r₂) ± b(r₁)a(r₂)], and pair each with the spin function that makes the whole state antisymmetric: ψ₊ with the singlet, ψ₋ with the triplet. Evaluating ⟨ψ_±|1/r₁₂|ψ_±⟩ produces two distinct integrals. The direct integral J = ∫∫|a(r₁)|²(1/r₁₂)|b(r₂)|² dτ₁dτ₂ is the ordinary electrostatic repulsion of two charge clouds and would survive in a classical model. The exchange integral K = ∫∫a*(r₁)b*(r₂)(1/r₁₂)b(r₁)a(r₂) dτ₁dτ₂ is built from the overlap density a*b and has no classical reading whatever. The result is E_± = εₐ + ε_b + J ± K: J lifts both terms together, K alone separates them, by 2K.

K is positive for orbitals like these, so the singlet lies 2K above the triplet. The mechanism is geometry, not force. Put r₁ = r₂ in ψ₋: the antisymmetric combination vanishes identically, and by continuity stays small nearby. That statistical exclusion zone is the Fermi hole; it holds the two electrons further apart on average, which makes ⟨1/r₁₂⟩ smaller in the triplet. Hund's first rule is this argument generalised. It is tempting to blame two parallel magnetic moments instead, and the numbers close that door: helium's 1s2s levels sit at 20.616 eV (2¹S) and 19.820 eV (2³S), a splitting of 0.796 eV, while the dipole–dipole energy of two Bohr magnetons a Bohr radius apart, μ₀μ_B²/4πa₀³, is about 4 × 10⁻⁴ eV — two thousand times smaller. The spins exert no force on each other; they decide which spatial function is legal, and the spatial function decides the energy.