The Helium Atom and Electron Repulsion

Introduction

Hydrogen can be solved exactly because it has one electron. Add a second electron and everything changes. Helium, with a nucleus of charge +2e and two electrons, is the simplest many-electron atom, yet its Schrödinger equation has no exact analytical solution. The culprit is a single term: the repulsion between the two electrons. Studying helium shows why approximation methods are unavoidable in chemistry and sets up the variational treatment that follows.

Core explanation

The Hamiltonian. Treat the nucleus as fixed (it is about 7300 times heavier than an electron). Label the electrons 1 and 2, with distances r₁ and r₂ from the nucleus and r₁₂ from each other. In atomic units the Hamiltonian is

Ĥ = −½∇₁² − ½∇₂² − 2/r₁ − 2/r₂ + 1/r₁₂

The five terms are: the kinetic energy of electron 1, the kinetic energy of electron 2, the attraction of each electron to the nucleus (charge Z = 2), and the repulsion between the electrons.

Why it cannot be solved exactly. Without the last term, Ĥ splits into two separate hydrogen-like Hamiltonians, ĥ(1) + ĥ(2), each depending on the coordinates of only one electron. The Schrödinger equation would then separate, just as a two-dimensional box separates into two one-dimensional problems. The term 1/r₁₂ depends on the positions of both electrons at once, so the motion of one electron is correlated with the motion of the other. The equation no longer separates, and no closed-form solution is known. The problem is not a lack of cleverness: the three-body problem has no general analytical solution in classical mechanics either.

The independent-electron guess. The simplest approximation drops 1/r₁₂. Each electron then occupies a hydrogen-like 1s orbital for Z = 2, with energy −Z²/2 = −2 Eₕ. The two-electron wavefunction is the product

ψ(1,2) = 1s(1) 1s(2), with 1s(r) = (Z³/π)^(1/2) e^(−Zr)

and the energy is the sum, E = −4 Eₕ ≈ −108.8 eV.

Comparison with experiment. The experimental ground-state energy is the energy needed to remove both electrons: the first ionisation energy (24.6 eV) plus the second (54.4 eV), giving −79.0 eV, or −2.904 Eₕ. The independent-electron energy is far too low — by about 30 eV, or roughly 38%. Neglecting repulsion lets the electrons crowd close to the nucleus with nothing to push them apart.

Putting repulsion back on average. A better estimate keeps the product function but evaluates the full Hamiltonian with it. The average repulsion ⟨1/r₁₂⟩ for two 1s electrons is 5Z/8 Eₕ, which for Z = 2 is 1.25 Eₕ (34.0 eV). Adding this gives E ≈ −2.75 Eₕ ≈ −74.8 eV, now only about 5% above experiment. This number is a genuine variational estimate, so it must lie above the true energy — and it does.

Spin. The spatial function 1s(1)1s(2) is symmetric under exchange of the electrons, so the Pauli principle requires an antisymmetric spin part: the singlet combination with paired spins.

Step-by-step reasoning

1. List the particles and their pairwise interactions. 2. Write a kinetic energy term for each electron. 3. Write an attraction term −Z/rᵢ for each electron. 4. Add +1/r₁₂ for the electron pair. 5. Note that dropping 1/r₁₂ gives separable one-electron problems. 6. Sum the one-electron energies, then estimate the repulsion as a correction.

Visual explanation

Imagine two small clouds of electron density centred on the helium nucleus. In the independent-electron picture each cloud is shrunk tightly by the full +2 charge and the clouds pass through each other freely. In reality each electron feels the other pushing outwards, so the true clouds are more expanded, and the electrons tend to be found on opposite sides of the nucleus.

Real-world analogy

Two dancers on a small stage both drawn towards a spotlight in the centre. If they ignored each other they would both stand in the brightest spot. In reality they avoid collisions, so each must move in a way that depends on where the other is — their motions are correlated.

Real-world example

The ionisation energies of helium, measured by photoelectron spectroscopy and atomic spectra, are among the most precisely known numbers in chemistry. They serve as a benchmark: any new quantum-chemical method is tested against the helium ground-state energy of −2.9037 Eₕ.

Why?

Why is the independent-electron energy too low rather than too high? The repulsion term is always positive because like charges repel. Removing a positive term from the Hamiltonian can only lower the energy, so the neglect of 1/r₁₂ gives an energy below the true value.

Common misconception

"Helium's energy is just twice hydrogen's energy." This ignores both the larger nuclear charge, which scales each orbital energy by Z², and the electron repulsion. Twice −13.6 eV is −27.2 eV, nowhere near the actual −79.0 eV.

Worked example

Question: Estimate the first ionisation energy of helium using the first-order energy of −2.75 Eₕ.

Reasoning: He⁺ is hydrogen-like with Z = 2, so its energy is −Z²/2 = −2.00 Eₕ exactly. Ionisation energy = E(He⁺) − E(He) = −2.00 − (−2.75) = 0.75 Eₕ. Converting, 0.75 × 27.21 ≈ 20.4 eV.

Answer: About 20.4 eV, compared with the experimental 24.6 eV — the right size but noticeably too small.

Quick check

1. Which term in the helium Hamiltonian prevents the Schrödinger equation from separating into one-electron equations? Answer: The electron–electron repulsion term 1/r₁₂, because it depends on the coordinates of both electrons at once.

Exam focus

Write the helium Hamiltonian with every term labelled, and state clearly whether you are using atomic units or SI units (in SI each Coulomb term carries e²/4πε₀). Examiners reward a comparison of −4 Eₕ (no repulsion), −2.75 Eₕ (repulsion averaged) and −2.904 Eₕ (experiment).

Advanced insight

The remaining gap between the best single-orbital-product energy and the exact energy is called the correlation energy; for helium it is about 0.042 Eₕ (roughly 1.1 eV). It arises because the true wavefunction contains an explicit dependence on r₁₂. Hylleraas included r₁₂ directly in trial functions in the late 1920s and obtained helium energies of remarkable accuracy.

Summary

The helium Hamiltonian contains two kinetic terms, two nuclear attractions and the electron repulsion 1/r₁₂. The repulsion couples the electrons and prevents an exact solution. Ignoring it gives −4 Eₕ (−108.8 eV), far below the experimental −2.904 Eₕ (−79.0 eV). Averaging the repulsion over the product function, 5Z/8 = 1.25 Eₕ, gives −2.75 Eₕ, much closer to experiment.

Practice questions

1. Write the helium Hamiltonian in atomic units and name each term. Answer: Ĥ = −½∇₁² − ½∇₂² − 2/r₁ − 2/r₂ + 1/r₁₂: two kinetic energy terms, two electron–nucleus attractions and the electron–electron repulsion. 2. Calculate the independent-electron energy of Li⁺, which also has two electrons, in hartrees. Answer: Each electron has −Z²/2 = −9/2 = −4.5 Eₕ, so the total is −9.0 Eₕ. 3. Explain why the independent-electron energy of helium lies below the experimental value. Answer: The neglected repulsion term is positive, so leaving it out removes a positive contribution and lowers the energy below the true value. 4. Use ⟨1/r₁₂⟩ = 5Z/8 to find the first-order energy of Li⁺. Answer: −9.0 + 5(3)/8 = −9.0 + 1.875 = −7.125 Eₕ.