University Quantum Mechanics II · The Variational Principle · 10.6
Better helium trial functions, and what an energy certifies
An effective charge left helium 1.5 eV above the truth, and no product of orbitals can recover more than 0.38 eV of it. Here you write the distance between the electrons into the trial, watch one parameter take 70% of the correlation energy, and learn to say exactly what the resulting number proves and what it does not.
Build the model
Connect the measurement to the mechanism.
A product of orbitals, φ(r₁)φ(r₂), lets each electron feel the other's average charge but never lets it see where the other is, and that caps it. Free φ completely and the best product is the Hartree-Fock limit, −77.870 eV from the numerical solution of Unit 7's equations, quoted, still 1.144 eV above the fixed-nucleus energy. That remainder, the correlation energy, is 1.4% of the total but 4.7% of the ionisation energy, and it lives where the electrons meet: the exact state has a kink in r₁₂ = |r₁ − r₂|, derived here from their relative motion, which no finite sum of smooth products can build.
Writing r₁₂ into the trial builds it at once. Hylleraas's factor (1 + cr₁₂) takes 70% of the correlation energy with one parameter, his three-parameter function of 1929 reaches −78.98 eV, and modern expansions fix the fixed-nucleus value, −79.014 eV, to more than twenty figures, both quoted. The last 0.009 eV against the measured −79.005 eV is nuclear motion, relativity and QED, outside this Hamiltonian.
Then the limit: the energy is stationary, so a state error of order ε costs only ε² in E, while every other expectation value moves at order ε. Hylleraas's two-parameter function beats Hartree-Fock by 0.80 eV, yet misses helium's measured diamagnetism, which is set by ⟨r²⟩, by 9% where Hartree-Fock is within 0.2%. A lower energy certifies a better bound, not a better state.
- Simple definition
- The correlation energy is the part of an atom's energy that no single product of orbitals can reach: the exact nonrelativistic, fixed-nucleus energy minus the Hartree-Fock limit, the lowest energy any such product achieves.
- Example
- For helium, Ecorr = −79.014 eV − (−77.870 eV) = −1.144 eV: only 1.4% of the total energy, but 4.7% of the 24.587 eV it takes to remove one electron, which is the number chemistry feels.
The target beyond any product. Zeff = 27/16 sits 0.38 eV above the product floor; one r₁₂ parameter recovers 70% of Ecorr, Hylleraas's three parameters 97%.
Energies in eV; Eexact = −79.014 eV and EHF = −77.870 eV, both fixed-nucleus and nonrelativistic, both quoted
Derived: the kinetic term must cancel e²/(4πε₀r₁₂) as the electrons meet. A smooth product has no term linear in r₁₂, so it cannot make this kink.
ψ̄ is ψ averaged over the direction of r₁ − r₂; a₀ = 52.9 pm, so 1/(2a₀) = 9.45 × 10⁹ m⁻¹; the 2 is the pair's reduced mass mₑ/2
Evaluated from elementary integrals, not printed term by term. Minimum −78.671 eV at γ = 0.198; it crosses the Hartree-Fock line at γ = 0.028.
Trial e(−ζs/a₀)(1 + γζr₁₂/a₀), s = r₁ + r₂; Eₕ = e²/(4πε₀a₀) = 27.211 eV; γ dimensionless; γ = 0 gives −(27/16)²Eₕ
Derived. δ = 5% raises −78.67 eV by only 0.20 eV, while ⟨1/r₁₂⟩ ∝ ζ moves by the full 5% and ⟨r⟩ ∝ 1/ζ by 4.8%.
δ dimensionless; exact for any trial at its optimal scale, since ⟨T⟩ ∝ ζ², ⟨V⟩ ∝ ζ and ⟨V⟩ = 2Eₘᵢₙ there
Derived. For A = H the first-order term is 2εE₀⟨ψ₀|χ⟩ = 0, and that is the only reason the energy converges faster than anything else.
A any observable, in its own unit; ε real and dimensionless; normalising |ψ⟩ changes only the O(ε²) term
Derived in one line from the eigen-expansion. −78.671 eV certifies |⟨ψ₀|ψ⟩|² ≥ 0.983, so an error amplitude up to 0.13 is still allowed.
ψ normalised; E₁ is the lowest level of ψ's symmetry above E₀, helium's 2¹S at 20.62 eV up (tables); energies in eV
One product has a floor, and correlation lies below it
Every helium trial so far has been a product, φ(r₁)φ(r₂) times the spin singlet, and last lesson's 1s orbital with Zeff = 27/16 gave −77.49 eV. Let φ take any radial shape at all and the best product is the Hartree-Fock limit, the self-consistent solution of Unit 7's equations: −77.870 eV, quoted from its numerical solution. Only 0.38 eV was left to gain inside the product family, yet the fixed-nucleus energy is −79.014 eV, so the remaining Ecorr = −1.144 eV is out of reach of every product. The reason is structural. A product's pair density factorises, P(r₁, r₂) = ρ(r₁)ρ(r₂), so where electron 2 is likely to be does not depend on where electron 1 is. Real electrons avoid each other, and the energy that avoidance saves is the correlation energy. At 1.14 eV it is comparable to a chemical bond, which is why no one can ignore it.
The electrons' cusp, and why products converge slowly
Where the electrons meet, e²/(4πε₀r₁₂) diverges, and Hψ = Eψ stays finite only if the kinetic energy diverges to cancel it. Describe the pair by the relative coordinate r = r₁ − r₂, which carries the reduced mass mₑ/2: near r = 0 the s-wave part of the kinetic term is −(ħ²/mₑ)(ψ″ + 2ψ′/r). Matching the 1/r terms gives (2ħ²/mₑ)ψ′ = (e²/(4πε₀))ψ, so ψ rises linearly away from contact with slope ψ(0)/(2a₀), derived, in m⁻¹ as a logarithmic slope must be. The same argument with mass mₑ and charge −Ze² gives the nuclear cusp, slope −Zψ/a₀, hydrogen's 1s slope at Z = 1, which Zeff = 27/16 misses at 1.69/a₀ instead of 2/a₀. The electron cusp is the harder one: a smooth function of r₁ and r₂, averaged over directions, has no term linear in |r₁ − r₂|, so a sum of products can only imitate the kink. By Schwartz's analysis, quoted, the energy gained from partial wave l falls only as (l + ½)⁻⁴, and even infinitely many s-wave products, the s-limit, quoted at −78.342 eV, recover just 41% of Ecorr.
Put r₁₂ in the trial: Hylleraas's functions
For an S state ψ depends only on the triangle's sides r₁, r₂, r₁₂. Hylleraas used s = r₁ + r₂, t = r₁ − r₂ and u = r₁₂; integrating out the three orientation angles turns d³r₁ d³r₂ into π²u(s² − t²) ds dt du over 0 ≤ |t| ≤ u ≤ s, so every matrix element of e(−ζs/a₀) times a polynomial is an elementary integral. The trial e(−ζs/a₀)(1 + cu) adds one parameter to last lesson's. With ζ optimised at each γ = c/ζ, it crosses the Hartree-Fock line at γ = 0.028 and bottoms out at −78.671 eV with ζ = 1.850 and c = 0.366 a₀⁻¹, recovering 70% of Ecorr. Look at c: the cusp asks for 0.5 a₀⁻¹, but a factor linear in u must also serve large u, where it overgrows, and forcing c = 0.5 a₀⁻¹ costs 0.065 eV. Adding c₂t², which lets one electron sit inside while the other sits outside, gives Hylleraas's 1929 three-parameter result, −78.979 eV and 97% of Ecorr, quoted, and reproduced by the same integrals.
Twenty figures, and the 0.009 eV outside the model
Expansions with thousands of Hylleraas-type terms, begun by Pekeris in 1959 and extended since with logarithmic terms, put the fixed-nucleus, nonrelativistic energy at −2.903724377034119598 Eₕ, quoted, which is −79.014 eV with Eₕ = 27.211 eV. The measured value is the sum of the two ionisation energies from tables, 24.587 eV + 54.418 eV, so the ground state lies at −79.005 eV, 0.009 eV higher. That gap is not a failure of the variational method, which has converged; it is physics the Hamiltonian leaves out. Since Eₕ = α²mₑc², the whole Coulomb problem is of order α²mₑc². The nucleus moves: the reduced-mass factor 1 − mₑ/Mα, with mₑ/Mα = 1.37 × 10⁻⁴, raises the energy by about 0.011 eV. Relativity enters at relative order α² ≈ 5 × 10⁻⁵, a few meV, and QED at order α³ ln(1/α), a fraction of a meV; both are named here, not computed. A trial energy between −79.014 eV and −79.005 eV would break no theorem: the bound holds for the model's H, not for nature's.
Why the energy converges before anything else
Write a trial as |ψ₀⟩ + ε|χ⟩ with ⟨ψ₀|χ⟩ = 0, whose norm differs from 1 only at order ε². This unit's theorem puts E − E₀ at order ε², but any other observable picks up 2ε Re⟨ψ₀|A|χ⟩, first order, and that term vanishes for H only because H|ψ₀⟩ = E₀|ψ₀⟩. Scaling makes the contrast exact. Stretch a scale-optimised trial by ζ → (1 + δ)ζ: ⟨T⟩ ∝ ζ² and ⟨V⟩ ∝ ζ with ⟨V⟩ = 2Eₘᵢₙ at the optimum, so E = Eₘᵢₙ(1 − δ²), while ⟨1/r₁₂⟩ ∝ 1 + δ and ⟨r⟩ ∝ 1/(1 + δ). Take the two-parameter optimum, whose ⟨1/r₁₂⟩ is 3.0% high and ⟨r⟩ 3.5% low against exact values from tables, and set δ = −3%: the energy rises by 78.67 eV × 0.0009 = 0.07 eV, and the errors shrink to 0.1% and 0.5%. The energy-optimal state is not the property-optimal one. Nor does the virial ratio −⟨V⟩/⟨T⟩ = 2 help: every scale-optimised trial satisfies it exactly, so it certifies the scale and nothing else.
What an energy certifies, and what it does not
Expand the trial on the exact levels: E − E₀ = Σ|cₙ|²(Eₙ − E₀) ≥ (E₁ − E₀)(1 − |c₀|²), with E₁ the lowest level of the trial's symmetry, helium's 2¹S, 20.62 eV above the ground state in tables. That is Eckart's criterion, and it is all the energy says about the state. Hartree-Fock, 1.144 eV above E₀, guarantees |c₀|² ≥ 0.945; the two-parameter trial, 0.343 eV above, guarantees 0.983. Now compare predictions, Hartree-Fock values from its numerical solution and exact ones from tables. ⟨r²⟩ is 1.1848 a₀² from Hartree-Fock and 1.0768 a₀² from the two-parameter trial, against 1.1935 a₀²: 0.7% low against 9.8% low. The density at the nucleus goes the same way, 0.7% against 9.4%, and even the three-parameter trial, 1.11 eV below Hartree-Fock, is worse on both. An allowed error amplitude of 0.13 can point anywhere, and the energy never says where. To certify a property, converge that property as the trial grows.
Change one variable at a time
Make the relationship visible.
Slide γ from 0: the curve drops below the Hartree-Fock line by 0.03 and bottoms out near 0.20. Push on to 0.30: the energy rises 0.14 eV while the ⟨1/r₁₂⟩ error passes through zero. Back at 0.20, set δ to −3%: the energy climbs only 0.07 eV and both property errors fall below 0.6%.
⟨H⟩ OF THIS TRIAL-78.671 eV
ENERGY ERROR0.43 %
⟨1/r₁₂⟩ ERROR2.9 %
⟨r⟩ ERROR-3.5 %
Live interpretation⟨H⟩ OF THIS TRIAL: −78.671 eV. ENERGY ERROR: 0.43 %. ⟨1/r₁₂⟩ ERROR: 2.9 %. ⟨r⟩ ERROR: −3.5 %
Catch the common trap
Explain before calculating.
Two trial functions for helium give ⟨H⟩ = −78.67 eV (trial 1) and −77.87 eV (trial 2), and the exact fixed-nucleus energy is −79.01 eV. Which statement is justified?
Choose an answer to test the model.
Practice & worked examples
Reason from the model, then test the result.
EasyHold the nucleus fixed, so He⁺ has the exact hydrogenic energy −Z²Eₕ/2 = −54.423 eV. Predict helium's first ionisation energy from the Hartree-Fock limit, −77.870 eV, and from Hylleraas's three-parameter trial, −78.979 eV, compare both with the measured 24.587 eV, and say on which side of the fixed-nucleus value each must fall.
- The first ionisation energy is the cost of He → He⁺ + e⁻ with the electron left at rest far away: IE = E(He⁺) − E(He).
- Hartree-Fock: IE = −54.423 eV + 77.870 eV = 23.447 eV, which is 4.6% below the measured 24.587 eV.
- Three-parameter trial: IE = −54.423 eV + 78.979 eV = 24.556 eV, only 0.13% below.
- Direction: E(He⁺) is exact and E(He) is an upper bound, so each IE is a lower bound on the fixed-nucleus value, 79.014 eV − 54.423 eV = 24.591 eV. Both land below it, as they must.
- Hartree-Fock's shortfall from 24.591 eV is 1.144 eV, the whole correlation energy. It is 1.4% of the total energy but 4.7% of the ionisation energy, because an ionisation energy is a small difference of two large numbers.
AnswerIE = 23.447 eV from Hartree-Fock and 24.556 eV from the three-parameter trial, 4.6% and 0.13% below the measured 24.587 eV; both are lower bounds on the fixed-nucleus 24.591 eV.
MediumHylleraas's two-parameter optimum has ζ = 1.8497, γ = 0.1978, ⟨H⟩ = −78.671 eV, ⟨1/r₁₂⟩ = 0.9743 a₀⁻¹ and ⟨r⟩ = 0.8968 a₀. A student uses ζ = 1.80 but keeps γ = c/ζ, so every length in the trial is stretched uniformly. Find the new ⟨H⟩, ⟨Vₑₑ⟩ and ⟨r⟩, and compare them with the exact 25.737 eV and 0.9295 a₀.
- Keeping γ makes the change a pure rescaling: δ = 1.80/1.8497 − 1 = −0.0269, so δ² = 7.2 × 10⁻⁴.
- Energy: E = Eₘᵢₙ(1 − δ²) = −78.671 eV × (1 − 0.00072) = −78.614 eV, only 0.057 eV higher and still 0.74 eV below the Hartree-Fock limit.
- ⟨1/r₁₂⟩ scales as ζ: 0.9743 a₀⁻¹ × 0.9731 = 0.9481 a₀⁻¹, so ⟨Vₑₑ⟩ = 27.211 eV × 0.9481 = 25.80 eV, 0.24% above the exact 25.737 eV. The optimum's 26.51 eV was 3.0% above.
- ⟨r⟩ scales as 1/ζ: 0.8968 a₀ / 0.9731 = 0.9216 a₀, 0.8% below the exact 0.9295 a₀, where the optimum was 3.5% below.
- A trial 0.057 eV worse in energy predicts ⟨Vₑₑ⟩ twelve times and ⟨r⟩ four times more accurately: the energy is flat to first order about its minimum, the properties are not.
Answer⟨H⟩ = −78.614 eV, 0.057 eV higher; ⟨Vₑₑ⟩ = 25.80 eV, 0.24% high instead of 3.0%; ⟨r⟩ = 0.922 a₀, 0.8% low instead of 3.5%.
HardHartree-Fock gives ⟨H⟩ = −77.870 eV and ⟨r²⟩ = 1.1848 a₀² per electron; Hylleraas's two-parameter trial gives −78.671 eV and 1.0768 a₀². With E₀ = −79.014 eV and the 2¹S level 20.62 eV higher, bound each trial's overlap with the exact state, then predict helium's molar diamagnetic susceptibility from each and compare with the measured −1.88 × 10⁻⁶ cm³ mol⁻¹, which is −2.36 × 10⁻¹¹ m³ mol⁻¹.
- Eckart: 1 − |c₀|² ≤ (E − E₀)/(E₁ − E₀). Hartree-Fock: 1.144 eV/20.62 eV = 0.0555, so |c₀|² ≥ 0.945. Two-parameter: 0.343 eV/20.62 eV = 0.0166, so |c₀|² ≥ 0.983. The second trial earns the higher floor, but a floor is not an overlap: these bounds alone cannot say which trial is actually closer to the exact state.
- Unit 3's A² term with A = ½B × r adds e²B²(x² + y²)/(8mₑ) per electron, and a spherical state has ⟨x² + y²⟩ = (2/3)⟨r²⟩, so ΔE = e²B²Σ⟨r²⟩/(12mₑ). The moment −∂ΔE/∂B gives χₘ = −μ₀NA e²Σ⟨r²⟩/(6mₑ); with L = S = 0 there is no paramagnetic term.
- Numerically μ₀NA e²a₀²/(6mₑ) = 9.953 × 10⁻¹² m³ mol⁻¹ per a₀² of Σ⟨r²⟩. Hartree-Fock: Σ⟨r²⟩ = 2.3697 a₀², χₘ = −2.36 × 10⁻¹¹ m³ mol⁻¹. Two-parameter: 2.1536 a₀², χₘ = −2.14 × 10⁻¹¹ m³ mol⁻¹.
- Against the measured −2.36 × 10⁻¹¹ m³ mol⁻¹, Hartree-Fock is within 0.2% and the two-parameter trial falls 9% short, although its energy is 0.80 eV lower and its guaranteed overlap floor higher.
- Why: r² weights the tail. One exponent, ζ = 1.85, serves both electrons, so when one strays outward its amplitude dies as e(−1.85r/a₀), while the exact state dies as e(−κr) with κa₀ = √(2I/Eₕ) = √(2 × 0.9037) = 1.344, fixed by the 24.59 eV ionisation energy. Hartree-Fock's orbital, with energy −0.918 Eₕ, decays as e(−1.355r/a₀), nearly right; the energy barely notices the tail.
AnswerOverlap floors 0.945 (Hartree-Fock) and 0.983 (two-parameter); χₘ = −2.36 × 10⁻¹¹ and −2.14 × 10⁻¹¹ m³ mol⁻¹ against the measured −2.36 × 10⁻¹¹ m³ mol⁻¹, so the higher-energy trial predicts it within 0.2% and the lower misses by 9%.