University Quantum Mechanics II · The Variational Principle · 10.4
Helium's ground state: the repulsion ignored, then averaged
Helium is where hydrogen's exact solution runs out, and where you learn what one order of perturbation theory is worth. Switch the repulsion off and the energy is 38% too low; average it over the product state and it is 5% too high, and provably so, because the first-order energy is an upper bound.
Build the model
Connect the measurement to the mechanism.
With the nucleus fixed and relativity left out, helium is two hydrogenic problems of charge Z = 2 plus one term that couples them, the repulsion e²/(4πε₀|r₁ − r₂|). That term cannot be solved exactly, so the model does the next best thing twice. First it drops the repulsion: the problem separates, both electrons fall into the Z = 2 1s orbital with the spin singlet Unit 7 demands, and the energy is 8E₁ = −108.8 eV, far below the measured −79.0 eV because nothing pushes the electrons apart.
Then it averages the repulsion over that product state, which is all first-order perturbation theory asks. Unit 1's multipole expansion does the angular work in one line, because spherical clouds keep only the l = 0 term, and cutting the radial integral where the electrons pass gives ⟨Vₑₑ⟩ = (5/8)Z e²/(4πε₀a₀) = 34.0 eV. The result, −74.8 eV, is 4.2 eV too high, and it has to be: E⁽⁰⁾ + E⁽¹⁾ is the full Hamiltonian's expectation value in a normalised state, so the variational theorem makes it an upper bound however large the correction.
The method runs out where 1/Z stops being small, not α: the repulsion is 31% of the binding in helium, and the product state knows nothing of the screening and the mutual avoidance that make up the missing 4.2 eV, which later lessons recover.
- Simple definition
- First-order helium replaces the electron–electron repulsion by its average over the state in which both electrons occupy the unscreened hydrogenic 1s orbital of the full nuclear charge, giving E ≈ E⁽⁰⁾ + ⟨Vₑₑ⟩ = −(Z² − 5Z/8) e²/(4πε₀a₀).
- Example
- With e²/(4πε₀a₀) = 27.211 eV and Z = 2: E⁽⁰⁾ = −4 × 27.211 eV = −108.8 eV and ⟨Vₑₑ⟩ = 1.25 × 27.211 eV = 34.0 eV, so E ≈ −74.8 eV, 4.2 eV above the measured −79.005 eV (24.587 eV + 54.418 eV).
Derived: without Vₑₑ the Hamiltonian separates, each electron takes Unit 2's 1s energy Z²E₁, and the symmetric product ψ₁₀₀(r₁)ψ₁₀₀(r₂) pairs with the spin singlet.
E₁ = −13.606 eV, e²/(4πε₀a₀) = 27.211 eV, a₀ = 52.9 pm; Z = 2 gives −108.8 eV
Derived: ∫Pₗ(cos γ) dΩ₂ = 4πδₗ₀, so against a spherical density every l ≥ 1 term vanishes exactly. Nothing is truncated; it is the shell theorem.
Radii in m; from Unit 1's 1/|r₁ − r₂| = Σₗ (r_<ˡ/r_>(l+1)) Pₗ(cos γ), γ the angle between r₁ and r₂
Derived by cutting electron 2's radial integral at r₂ = r₁: charge inside acts from the nucleus, each outer shell adds its own 1/r₂. The figure draws it.
U in J, r₁ in m; tends to Z e²/(4πε₀a₀) at r₁ = 0 and to the point-charge e²/(4πε₀r₁) far outside
Derived: averaging U(r₁) over electron 1 gives ⟨1/r₁₂⟩ = Z/a₀ − 3Z/(8a₀). It grows as Z while E⁽⁰⁾ grows as Z², so E⁽¹⁾/|E⁽⁰⁾| = 5/(8Z).
In J; 1.25 × 27.211 eV = 34.0 eV for helium, against 54.4 eV if the second electron sat on the nucleus
Derived: H₀ + Vₑₑ = H and ψ₀ is normalised and admissible, so the variational theorem applies at any size of E⁽¹⁾. No partial sum past first order carries the guarantee.
In J; −2.75 × 27.211 eV = −74.8 eV for helium, 4.2 eV above the measured −79.005 eV
Rescaling r by a₀/Z makes Vₑₑ (1/Z) times a fixed operator, so 1/Z is the expansion parameter. At Z = 2 the two quoted terms supply 4.17 eV, the gap to −79.0 eV.
Coefficients dimensionless; the first two derived here, the rest quoted from tables of high-order perturbation calculations
Switch the repulsion off and helium separates
With the nucleus held fixed and relativity left out, helium's Hamiltonian is H = H₁ + H₂ + Vₑₑ, where Hᵢ = pᵢ²/2mₑ − Ze²/(4πε₀rᵢ) and Vₑₑ = e²/(4πε₀|r₁ − r₂|). Drop Vₑₑ and what remains is two hydrogenic problems in different coordinates, so the eigenstates are products and the energies are sums. The lowest puts both electrons in Unit 2's 1s orbital, ψ₁₀₀ = (Z³/πa₀³)¹⁄² e(−Zr/a₀), whose radius a₀/Z is 26.5 pm. The product ψ₁₀₀(r₁)ψ₁₀₀(r₂) is symmetric under exchange, so Unit 7 pairs it with the spin singlet, and E⁽⁰⁾ = 2Z²E₁ = 8 × (−13.606 eV) = −108.8 eV. The number to compare with comes from a table: removing both electrons costs 24.587 eV + 54.418 eV, so the ground state lies at −79.005 eV. The model is 29.8 eV too low, an error of 38%, and in the direction the missing term predicts: adding a positive operator can only raise the ground-state energy.
The average repulsion, bracketed before it is computed
Unit 6 gives the first correction as the perturbation averaged in the unperturbed state: E⁽¹⁾ = ⟨ψ₀|Vₑₑ|ψ₀⟩. Vₑₑ does not act on spin, so the normalised singlet contributes a factor 1. Both electrons share one orbital, so there is no separate exchange integral: the product is already symmetric, and only the direct integral [e²/(4πε₀)]∫∫ρ(r₁)ρ(r₂)/|r₁ − r₂| d³r₁ d³r₂ appears, with ρ = |ψ₁₀₀|². Fix its form before computing it. The only length in ρ is a₀/Z, so ⟨1/r₁₂⟩ = C Z/a₀ with C a pure number, and E⁽¹⁾ = C Z × 27.211 eV. Bracket C as well. If electron 2 were squeezed onto the nucleus, electron 1 would feel e²/(4πε₀r₁), whose average is set by Unit 2's ⟨1/r⟩ = Z/a₀: C = 1, or 54.4 eV for helium. The real cloud is spread out, and the part lying beyond electron 1 pushes less, so C lies between 0 and 1. The integral is six-dimensional; the next two steps reduce it to one.
The multipole expansion leaves exactly one term
Use Unit 1's series: 1/|r₁ − r₂| = Σₗ (r_<ˡ/r_>(l+1)) Pₗ(cos γ), where r_< and r_> are the smaller and larger of r₁ and r₂ and γ is the angle between the two position vectors. Hold r₁ fixed, point the z axis along it, and integrate over electron 2's directions. Its density does not depend on direction, so each term needs ∫Pₗ(cos θ₂) dΩ₂ = 2π∫₋₁¹ Pₗ(u) du, which is 4π for l = 0 and zero for every l ≥ 1, because Pₗ is orthogonal to P₀ = 1. Every higher multipole vanishes exactly; nothing has been truncated. What survives is 1/r_>, Newton's shell theorem in quantum dress: a thin spherical shell of radius r₂ produces a constant 1/r₂ everywhere inside it and acts from its centre everywhere outside. One spherical density is enough for this. With two electrons in 2p orbitals, as in carbon's outer pair, neither density is spherical, the l = 2 term survives, and it is what splits that configuration into terms.
Cut the radial integral where the electrons pass
Now ⟨1/r₁₂⟩ = ∫∫ P(r₁)P(r₂)/r_> dr₁ dr₂, with the radial probability P(r) = 4(Z/a₀)³r²e(−2Zr/a₀). Fix r₁, write x = Zr₁/a₀, and cut electron 2's integral at r₂ = r₁. The fraction of the cloud inside, 1 − (1 + 2x + 2x²)e(−2x), acts as if it sat at the nucleus and contributes that fraction divided by r₁. The shells outside each contribute their own 1/r₂, which sums to (Z/a₀)(1 + 2x)e(−2x). Together they give the cloud's potential, (1/r₁)[1 − (1 + x)e(−2x)], which tends to Z/a₀ at the centre and to 1/r₁ far outside. Last, average over electron 1. The 1 gives ∫P(r₁)/r₁ dr₁ = Z/a₀; the exponential term gives 4(Z/a₀)³∫r(1 + Zr/a₀)e(−4Zr/a₀) dr = 3Z/(8a₀), from ∫rⁿe(−βr) dr = n!/β(n+1) with β = 4Z/a₀. So ⟨1/r₁₂⟩ = 5Z/(8a₀), in m⁻¹ as it must be, with C = 5/8 inside its bracket, and E⁽¹⁾ = 1.25 × 27.211 eV = 34.0 eV.
−74.8 eV is an upper bound, however large the correction
Add the pieces: E⁽⁰⁾ + E⁽¹⁾ = −108.8 eV + 34.0 eV = −74.8 eV, against the measured −79.0 eV. The sum is more than an estimate. Because H₀|ψ₀⟩ = E⁽⁰⁾|ψ₀⟩, it equals ⟨ψ₀|H₀|ψ₀⟩ + ⟨ψ₀|Vₑₑ|ψ₀⟩ = ⟨ψ₀|H|ψ₀⟩, the full Hamiltonian's expectation value in a normalised, admissible state, and this unit's variational theorem puts that at or above the ground-state energy of H. The guarantee does not care that E⁽¹⁾ is 31% of E⁽⁰⁾. Unit 6 agrees: for a ground state every term of E⁽²⁾ = −Σ|⟨m|Vₑₑ|ψ₀⟩|²/(Eₘ − E⁽⁰⁾) is negative, so second order can only move the estimate down. The guarantee belongs to first order alone. The second-order coefficient, quoted from tables, is −0.1577 e²/(4πε₀a₀) = −4.29 eV, which takes helium to −79.12 eV, 0.11 eV below the fixed-nucleus value of −79.014 eV, also quoted. A partial sum past first order is not a bound.
The small parameter is 1/Z, not α
Measure lengths in a₀/Z and energies in Z²e²/(4πε₀a₀), and the Hamiltonian becomes h₀ + (1/Z)(1/ρ₁₂), with h₀ the same for every two-electron ion and ρ₁₂ the scaled separation. The repulsion is a perturbation of strength 1/Z, which is why E⁽¹⁾/|E⁽⁰⁾| = 5/(8Z): 0.31 for helium, 0.16 for Be²⁺, small only for highly charged ions. The fine-structure constant plays no part in that ratio. Since e²/(4πε₀a₀) = α²mₑc², E⁽⁰⁾ and E⁽¹⁾ are both of order α²mₑc² ≈ 27 eV, so binding and repulsion enter at the same order in α. Relativity enters at order α⁴mₑc²: two hydrogenic 1s electrons in Unit 8's fine-structure formula estimate a shift near −6 meV for helium, some 700 times smaller than the 4.2 eV the bound misses. The Z-independent second-order term predicts that gap stays near 4.3 eV along the two-electron ions while the fractional error falls as 1/Z²; the worked examples test that against Be²⁺ and B³⁺.
Change one variable at a time
Make the relationship visible.
At Z = 2 slide r₁ out from 0.1 a₀: near the nucleus the open part, from shells outside r₁, dominates; beyond 1.5 a₀ the bar meets the dashed point-charge curve. Then step Z from 1 to 3: E⁽⁰⁾ grows as Z² while the average repulsion grows only as Z.
E⁽⁰⁾, REPULSION DROPPED-108.8 eV
E⁽¹⁾ = ⟨Vₑₑ⟩34.0 eV
E⁽⁰⁾ + E⁽¹⁾, UPPER BOUND-74.8 eV
1/Z SERIES, QUOTED TERMS-79.0 eV
Live interpretationE⁽⁰⁾, REPULSION DROPPED: −108.8 eV. E⁽¹⁾ = ⟨Vₑₑ⟩: 34.0 eV. E⁽⁰⁾ + E⁽¹⁾, UPPER BOUND: −74.8 eV. 1/Z SERIES, QUOTED TERMS: −79.0 eV
Catch the common trap
Explain before calculating.
First-order perturbation theory puts helium at E⁽⁰⁾ + E⁽¹⁾ = −108.8 eV + 34.0 eV = −74.8 eV. Before any measurement is consulted, what can be said about the sign of the error, the estimate minus the true ground-state energy of the fixed-nucleus Hamiltonian?
Choose an answer to test the model.
Practice & worked examples
Reason from the model, then test the result.
EasyThe ion Be²⁺ has two electrons around a nucleus of charge Z = 4, and removing them costs 153.896 eV and then 217.719 eV. Find E⁽⁰⁾, E⁽¹⁾ and the first-order energy, compare that bound with the measured total, and compare the size of the correction with helium's.
- With the repulsion dropped, both electrons sit in the Z = 4 1s orbital: E⁽⁰⁾ = −Z² e²/(4πε₀a₀) = −16 × 27.211 eV = −435.4 eV.
- Average repulsion: E⁽¹⁾ = (5/8)Z e²/(4πε₀a₀) = 2.5 × 27.211 eV = 68.0 eV. Dimensions: a pure number times an energy.
- First-order energy: −(16 − 2.5) × 27.211 eV = −13.5 × 27.211 eV = −367.35 eV. It equals ⟨ψ₀|H|ψ₀⟩, so it must lie at or above the true ground-state energy.
- Measured: E = −(153.896 + 217.719) eV = −371.615 eV. The bound holds, 4.27 eV above, a 1.15% error against helium's 4.17 eV and 5.3%.
- The correction relative to E⁽⁰⁾ is 5/(8Z) = 0.156, half of helium's 0.3125. The energies grew fourfold while the absolute gap barely moved, as a Z-independent second-order term predicts.
AnswerE⁽⁰⁾ = −435.4 eV, E⁽¹⁾ = +68.0 eV, E⁽⁰⁾ + E⁽¹⁾ = −367.35 eV, 4.27 eV (1.15%) above the measured −371.615 eV; the correction is 16% of E⁽⁰⁾, against 31% for helium.
MediumIn helium's unperturbed state, electron 1 is at r₁ = a₀/2, its most probable radius. Find the fraction of electron 2's cloud inside r₁, the repulsion electron 1 feels from that inner part and from the outer shells, and the total. Repeat at r₁ = 2a₀, then explain why the average over electron 1, 34.0 eV, lies below the value at the most probable radius.
- With Z = 2, x = Zr₁/a₀ = 1. Fraction inside: 1 − (1 + 2x + 2x²)e(−2x) = 1 − 5e⁻² = 0.323.
- The inner part acts from the nucleus: 27.211 eV × 0.3233 × (a₀/r₁) = 27.211 eV × 0.6467 = 17.6 eV.
- Each outer shell contributes its own 1/r₂: 27.211 eV × Z(1 + 2x)e(−2x) = 27.211 eV × 6e⁻² = 22.1 eV. The total, 39.7 eV, matches the closed form 27.211 eV × (a₀/r₁)[1 − (1 + x)e(−2x)] = 27.211 eV × 2(1 − 2e⁻²) = 39.7 eV.
- At r₁ = 2a₀, x = 4: the fraction inside is 1 − 41e⁻⁸ = 0.986, giving 13.42 eV, and the outer shells add 27.211 eV × 18e⁻⁸ = 0.16 eV. The total, 13.58 eV, is within 0.2% of the point-charge value 27.211 eV × (a₀/r₁) = 13.61 eV: far outside, the cloud acts as if it sat on the nucleus.
- The repulsion falls steadily as r₁ grows, and electron 1 lies beyond a₀/2 with probability 5e⁻² = 0.68, where it feels less than 39.7 eV, down to about 13.6 eV at 2a₀. Weighting by P(r₁) therefore pulls the average down to 34.0 eV.
AnswerAt r₁ = a₀/2: 32% of the cloud lies inside, giving 17.6 eV, and the outer shells give 22.1 eV, 39.7 eV in all. At r₁ = 2a₀: 13.58 eV, within 0.2% of a point charge. The average, 34.0 eV, is lower because electron 1 lies beyond a₀/2 with probability 0.68.
HardA two-electron ion's first ionisation energy is I = E(one electron) − E(two electrons). Using only derived results, turn the first-order bound into a bound on I for nuclear charge Z, evaluate it for He and for B³⁺ (Z = 5, measured I = 259.37 eV), and say what it proves at Z = 1.
- The one-electron ion is hydrogenic, so within the model its energy is Z²E₁ exactly. The two-electron energy obeys E ≤ −(Z² − 5Z/8) × 27.211 eV = −(2Z² − 5Z/4) × 13.606 eV.
- Subtracting an upper bound reverses the inequality: I = Z²E₁ − E ≥ −13.606 eV × Z² + 13.606 eV × (2Z² − 5Z/4) = 13.606 eV × Z(Z − 5/4). An upper bound on the energy is a lower bound on the ionisation energy, never an upper one. Dimensions: an energy times a pure number.
- Helium, Z = 2: I ≥ 13.606 eV × 2 × 0.75 = 20.41 eV. Measured 24.587 eV, so the bound holds with 4.18 eV to spare, the same 4.2 eV gap as in the total energy, because E(He⁺) = −54.42 eV carries no error in the model.
- B³⁺, Z = 5: I ≥ 13.606 eV × 5 × 3.75 = 255.11 eV, against the measured 259.37 eV: 4.26 eV to spare, 1.6% low. The gap has hardly moved while I grew more than tenfold.
- Z = 1, H⁻: 13.606 eV × 1 × (−0.25) = −3.40 eV. A negative lower bound on I is no information: the first-order state does not prove H⁻ bound, and it does not prove it unbound either.
AnswerI ≥ 13.606 eV × Z(Z − 5/4): at least 20.4 eV for helium (measured 24.59 eV) and 255.1 eV for B³⁺ (measured 259.37 eV). At Z = 1 the bound is −3.4 eV and proves nothing.