University Physics V · Multi-Electron Atoms · 12.4

Helium: Exchange & the Variational Bound

Helium is the first atom quantum mechanics cannot solve. This lesson shows what to do instead: pick a trial state, prove its energy can only come out too high, then squeeze it. The exchange integral falls out of the same algebra, and with it the reason parallel spins sit lower without any magnetism at all.

Open the lab ↓Practice & worked examples ↓

Build the model

Connect the measurement to the mechanism.

Add a second electron and the Hamiltonian acquires one term, 1/r₁₂, that couples r₁ to r₂ and destroys separability: no product of one-electron functions is an eigenstate, and no closed-form eigenfunction is known. The response is to stop solving and start bounding. The variational theorem says ⟨ψ|H|ψ⟩ ≥ E₀ for any trial ket, because expanding ψ in the exact eigenbasis makes the expectation value a probability-weighted average of eigenvalues, and no average falls below the smallest member.

A guess with an adjustable knob therefore becomes a one-sided measurement, and minimising over the knob tightens it. For helium the knob that pays is the nuclear charge itself: let each 1s orbital see an effective Z′ instead of 2, and the minimum sits at Z′ = 27/16 = 1.6875, giving −77.49 eV against the true −79.01 eV. The missing 5/16 of charge is the other electron's cloud, real physics rather than a fudge.

The same antisymmetry that forces the 1s² ground state into a spin singlet splits every excited configuration in two: the direct integral J shifts both terms and the exchange integral K separates them by 2K, the triplet lower because its antisymmetric spatial function must vanish at r₁ = r₂. What the whole scheme costs is exactness — a single product, however well its orbitals are chosen, cannot let one electron dodge the other instantaneously, and the 1.5 eV it misses is the bill.

Simple definition: The variational bound on helium is the lowest expectation value ⟨ψ|H|ψ⟩ reachable from a chosen family of trial states; it is guaranteed to lie at or above the true ground-state energy, never below it.
Example: The one-parameter family ψ = φ_Z′(r₁)φ_Z′(r₂) gives E(Z′) = Z′² − 4Z′ + (5/8)Z′ hartree, minimised at Z′ = 27/16 with E = −(27/16)² Eh = −77.49 eV — 1.52 eV above the true −79.01 eV, and above it by construction.

The term that blocks separationH = −½∇₁² − ½∇₂² − Z/r₁ − Z/r₂ + 1/r₁₂

Drop the last term and H separates: E = 2(−Z²/2) = −108.85 eV, some 30 eV too deep.

atomic units: ħ = mₑ = e = 4πε₀ = 1, energies in hartree, 1 Eh = 27.211 eV; helium has Z = 2

Variational theoremE[ψ] = ⟨ψ|H|ψ⟩ / ⟨ψ|ψ⟩ ≥ E₀

Turns a guess into a one-sided measurement — between two trial functions, the lower number is the better one, with no further argument.

any ψ in the domain of H, normalised or not; equality only if ψ is the exact ground state

Screened-charge trial energyE(Z′) = Z′² − 2ZZ′ + (5/8)Z′ [hartree]

Three integrals and one knob: the quadratic kinetic term fights the linear attraction and repulsion terms, so a true minimum exists.

⟨T⟩ = Z′², ⟨Vₙₑ⟩ = −2ZZ′, ⟨Vₑₑ⟩ = (5/8)Z′ for the 1s(Z′)² product state

The optimum and the boundZ′* = Z − 5/16, Eₘᵢₙ = −Z′*² [hartree]

Eₘᵢₙ = −⟨T⟩ is the virial theorem showing through: the optimised trial state satisfies ⟨V⟩ = −2⟨T⟩ exactly.

Z = 2 gives Z′* = 27/16 = 1.6875 and Eₘᵢₙ = −729/256 Eh = −77.49 eV

Direct and exchange integralsJ = ⟨ab|1/r₁₂|ab⟩, K = ⟨ab|1/r₁₂|ba⟩

J is cloud-on-cloud Coulomb repulsion; K is built from the overlap density a*b and has no classical reading at all.

orthonormal orbitals a, b; both in hartree, ×27.211 for eV; for the cases here J, K > 0

Singlet–triplet splittingE± = εₐ + ε_b + J ± K, ΔE = E_singlet − Eₜᵣᵢₚₗₑₜ = 2K

He 1s2s measures 2K = 0.796 eV, while magnetic spin–spin coupling at this range is only ~10⁻⁴ eV.

+ with the symmetric spatial function (S = 0), − with the antisymmetric one (S = 1)

Where separability dies

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.

The variational theorem is an inequality, not an estimate

Expand a normalised trial ket in the exact eigenbasis, |ψ⟩ = Σ cₙ|n⟩ with H|n⟩ = Eₙ|n⟩ and Σ|cₙ|² = 1. Then ⟨ψ|H|ψ⟩ = Σ |cₙ|²Eₙ ≥ E₀ Σ|cₙ|² = E₀, since replacing every eigenvalue by the smallest one can only lower the 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.

Screening as the variational parameter

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.

What the bound still misses

−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 matched the exact energy to six figures with a handful of terms.

Two ways to build 1s2s, and the integrals that separate them

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.

Why the triplet lies lower, and why it is not magnetism

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.

Change one variable at a time

Make the relationship visible.

Interactive model

Trial nuclear charge Z′2.0000

Repulsion switched on, λ1.00

Start at Z′ = 2 with λ = 1 — the unscreened answer, −74.83 eV — and drag Z′ down: the curve bottoms at 27/16 and −77.49 eV, still 1.52 eV above the dashed line. Then pull λ to 0 and watch the minimum walk back to Z′ = 2 and dive to −108.85 eV.

TRIAL ENERGY E(Z′)-74.83 eV

GAP ABOVE EXACT4.18 eV

OPTIMUM Z′ = Z − 5λ/161.6875

REPULSION (5/8)λZ′34.01 eV

Live interpretationTRIAL ENERGY E(Z′): −74.83 eV. GAP ABOVE EXACT: 4.18 eV. OPTIMUM Z′ = Z − 5λ/16: 1.6875. REPULSION (5/8)λZ′: 34.01 eV

Catch the common trap

Explain before calculating.

Misconception check

The commonest error reads the singlet–triplet splitting as a magnetic interaction between the two electron spins, parallel moments repelling and antiparallel attracting. Nothing in the Hamiltonian here is magnetic: H holds kinetic terms and Coulomb potentials only, and no spin operator appears in it. Spin enters as a constraint. The total state must be antisymmetric, so a triplet spin function compels an antisymmetric spatial function, which vanishes at r₁ = r₂. The difference is ⟨1/r₁₂⟩ evaluated on two different spatial distributions. A size check settles it: magnetic dipole coupling between two electron moments at atomic separation is of order 10⁻⁴ eV, against helium's 0.796 eV 1s2s splitting — a factor of two thousand. A second version of the slip hunts for a 1s² triplet; there is only one 1s² spatial function, it is symmetric, and antisymmetry leaves the singlet alone.

A variational calculation on helium with ψ = φ_Z′(r₁)φ_Z′(r₂) returns −77.49 eV at Z′ = 27/16. A colleague reports that a more flexible trial function, carrying explicit r₁₂ dependence, returns −79.42 eV. The exact non-relativistic ground-state energy of helium is −79.01 eV. What should you conclude?

Choose an answer to test the model.

Practice & worked examples

Reason from the model, then test the result.

EasyTreat helium's 1s² ground state first with the electron–electron repulsion switched off, then switch it on to first order using ⟨1s²|1/r₁₂|1s²⟩ = (5/8)Z hartree for the unscreened hydrogenic product. Give both energies in eV and compare with the measured −79.01 eV. Take 1 hartree = 27.211 eV.

With 1/r₁₂ dropped, H separates into two independent hydrogenic problems of charge Z = 2, each with ground energy −Z²/2 = −2 Eh. So E⁽⁰⁾ = −4 Eh = −4 × 27.211 = −108.85 eV.
That lies 29.84 eV below the measured −79.01 eV. A term worth almost 30 eV is not a small perturbation by any standard.
First-order correction: E⁽¹⁾ = ⟨1s²|1/r₁₂|1s²⟩ = (5/8)Z = (5/8)(2) = 1.25 Eh = 34.01 eV.
Sum: E⁽⁰⁾ + E⁽¹⁾ = −4 + 1.25 = −2.75 Eh = −74.83 eV, now 4.18 eV above the measured value.
It lands above, not below, exactly as required: the first-order sum is ⟨ψ⁽⁰⁾|H|ψ⁽⁰⁾⟩ for a normalised state, so it is already a legitimate variational upper bound — the Z′ = 2 member of the family used next.

AnswerE⁽⁰⁾ = −108.85 eV; E⁽⁰⁾ + E⁽¹⁾ = −74.83 eV, an upper bound sitting 4.18 eV above the measured −79.01 eV.

MediumFor the screened trial function the expectation values are ⟨T⟩ = Z′², ⟨Vₙₑ⟩ = −2ZZ′ and ⟨Vₑₑ⟩ = (5/8)Z′, all in hartree. Minimise E(Z′) for Z = 2, quote the bound in eV, verify the virial ratio ⟨V⟩/⟨T⟩ at the optimum, and say what the screening was worth.

E(Z′) = Z′² − 2ZZ′ + (5/8)Z′ = Z′² − 4Z′ + 0.625Z′ = Z′² − 3.375Z′ for Z = 2.
dE/dZ′ = 2Z′ − 3.375 = 0 gives Z′ = 1.6875 = 27/16, and d²E/dZ′² = 2 > 0 confirms a minimum. The screening is Z − Z′ = 5/16 = 0.3125 of an electronic charge.
Eₘᵢₙ = (1.6875)² − 3.375(1.6875) = 2.84766 − 5.69531 = −2.84766 Eh, which is exactly −Z′². In eV: −2.84766 × 27.211 = −77.49 eV.
Virial check: ⟨T⟩ = Z′² = 2.84766 Eh, and ⟨V⟩ = −2(2)(1.6875) + 0.625(1.6875) = −6.75 + 1.05469 = −5.69531 Eh. So ⟨V⟩/⟨T⟩ = −2.000, as a Coulombic bound state requires.
Worth of the screening: at Z′ = 2, E = 4 − 6.75 = −2.75 Eh = −74.83 eV. Relaxing to 27/16 gained 2.66 eV, and the bound now sits 1.52 eV above the exact −79.01 eV.

AnswerZ′ = 27/16 = 1.6875, Eₘᵢₙ = −(27/16)² Eh = −77.49 eV, with ⟨V⟩/⟨T⟩ = −2. Screening was worth 2.66 eV, leaving a 1.52 eV gap to −79.01 eV.

HardFor helium's 1s2s configuration with unscreened hydrogenic orbitals (Z = 2) the standard integrals are J = (17/81)Z Eh and K = (16/729)Z Eh. Predict the singlet–triplet splitting in eV, compare it with the measured levels 2¹S at 20.616 eV and 2³S at 19.820 eV above the ground state, and account for the discrepancy.

J = (17/81)(2) = 34/81 = 0.41975 Eh = 11.42 eV, and K = (16/729)(2) = 32/729 = 0.043896 Eh = 1.194 eV.
E_± = ε₁s + ε₂s + J ± K, with + for the singlet (symmetric spatial function) and − for the triplet, so the predicted splitting is ΔE = 2K = 2.389 eV, singlet above triplet.
Measured: 20.616 − 19.820 = 0.796 eV, also singlet above triplet. The ordering is right; the magnitude is too large by a factor of 2.389/0.796 = 3.0.
The fault is the orbitals, not the formula. The 2s electron sits outside a nearly complete screen of the 1s charge, so it sees Z_eff close to 1, not 2, and its true orbital is roughly twice as extended as the Z = 2 function assumes.
K is built entirely from the overlap density a*(r)b(r). Spreading the 2s function out cuts that overlap everywhere, so K falls faster than J does — which is the observed pattern: J-type shifts survive while the exchange splitting shrinks threefold.
The sign is what the physics guarantees. K > 0 for real orbitals of this type, so the triplet is always the lower term of a configuration — Hund's first rule — however badly a hydrogenic estimate misses the size.

AnswerPredicted 2K = 2.39 eV against a measured 0.796 eV: right sign and ordering, three times too large, because unscreened Z = 2 orbitals make the 2s far too compact.