Perturbation Theory versus the Variation Method

Introduction

Chemistry relies on two great strategies for approximating the Schrödinger equation. The variation method guesses a trial function and minimises its energy. Perturbation theory starts from a problem we can solve exactly and adds corrections for whatever was left out. Both appear throughout quantum chemistry — often in the same calculation — and knowing their strengths and weaknesses is essential for judging any computed result.

Core explanation

The perturbation idea. Write the Hamiltonian as

Ĥ = Ĥ⁽⁰⁾ + Ĥ′

where Ĥ⁽⁰⁾ has known eigenfunctions ψₙ⁽⁰⁾ and energies Eₙ⁽⁰⁾, and Ĥ′ is a small extra term. The exact energy is expanded as a series:

E = E⁽⁰⁾ + E⁽¹⁾ + E⁽²⁾ + …

First order. The first correction is the average of the perturbation over the unperturbed state:

E⁽¹⁾ = ⟨ψ⁽⁰⁾ Ĥ′ ψ⁽⁰⁾⟩

No new wavefunction is needed. Notice that E⁽⁰⁾ + E⁽¹⁾ = ⟨ψ⁽⁰⁾ Ĥ ψ⁽⁰⁾⟩, the expectation value of the full Hamiltonian over ψ⁽⁰⁾. That is precisely a variational energy with ψ⁽⁰⁾ as trial function, so the first-order result is itself an upper bound.

Second order. The second correction mixes in other unperturbed states:

E₀⁽²⁾ = Σₙ≠₀ H′ₙ₀ ² / (E₀⁽⁰⁾ − Eₙ⁽⁰⁾), with H′ₙ₀ = ⟨ψₙ⁽⁰⁾ Ĥ′ ψ₀⁽⁰⁾⟩

For the ground state every denominator is negative, so E⁽²⁾ is always negative. States close in energy and strongly coupled contribute most. Beyond first order there is no upper-bound guarantee: the sum through second order can fall below the exact energy.

Helium as a test case. Take Ĥ⁽⁰⁾ as two hydrogen-like Hamiltonians with Z = 2 and Ĥ′ = 1/r₁₂.

Method Energy / Eₕ --- --- Zeroth order (no repulsion) −4.000 Zeroth + first order −2.750 Variation, optimised ζ = 1.6875 −2.848 Zeroth + first + second order −2.908 Exact (non-relativistic) −2.904

The second-order result is closer to the exact value in magnitude but lies slightly below it — an illustration that perturbation theory offers no bound. The variational energy is less accurate here but is guaranteed to be above the truth.

Comparing the strategies.

- Variation method: gives an upper bound; flexible in the choice of trial function; the quality depends on a good guess; naturally handles large effects; the lowest root is reliable, excited states need care. - Perturbation theory: systematic and needs no guessing of functional form; works best when Ĥ′ is genuinely small compared with the level spacing; gives explicit formulae for how properties respond to fields; no upper bound beyond first order, and the series can converge slowly or diverge.

In practice. Modern methods combine both. Hartree–Fock is variational; Møller–Plesset perturbation theory (MP2) then adds the correlation energy as a second-order correction. Spectroscopists use perturbation theory for the effects of electric and magnetic fields (Stark and Zeeman effects) and for anharmonic corrections to vibrations.

Step-by-step reasoning

1. Identify the part of Ĥ you can solve exactly; call it Ĥ⁽⁰⁾. 2. Define Ĥ′ = Ĥ − Ĥ⁽⁰⁾. 3. Use ψ⁽⁰⁾ to find E⁽¹⁾ = ⟨ψ⁽⁰⁾ Ĥ′ ψ⁽⁰⁾⟩. 4. If higher accuracy is needed, estimate E⁽²⁾ from couplings to other states. 5. Compare with a variational estimate to judge the reliability.

Visual explanation

Plot energy on a vertical axis with the exact energy as a horizontal line. Variational estimates approach the line from above, stepping downwards as the trial function improves. Perturbation estimates jump from −4.0 up to −2.75 and then down to −2.908, crossing just past the line — the partial sums can land on either side.

Real-world analogy

Tuning a guitar string two ways: you can try different tensions until the note sounds best (variation), or you can start from a string you know is exactly in tune and calculate how much a slight change in thickness shifts the pitch (perturbation). The second works beautifully for small changes and poorly for large ones.

Real-world example

MP2 calculations, a second-order perturbation correction on top of Hartree–Fock, are widely used to describe dispersion interactions that hold DNA base-pair stacks and protein–ligand complexes together — effects that Hartree–Fock alone misses entirely.

Why?

Why is the second-order correction to a ground state always negative? Mixing higher-energy states into the ground state lets the wavefunction adapt to the perturbation. The ground state is already the lowest; any variationally allowed adaptation can only lower its energy, and every term in the sum has a positive numerator over a negative denominator.

Common misconception

"Perturbation theory is just a less accurate version of the variation method." Neither is universally better. For helium, second-order perturbation theory is closer to the exact energy than the one-parameter variational result, although it gives no bound.

Worked example

Question: A ground state with E₀⁽⁰⁾ = −10.0 eV couples to one excited state at E₁⁽⁰⁾ = −6.0 eV with H′₁₀ = 0.40 eV. Estimate E₀⁽²⁾.

Reasoning: E₀⁽²⁾ = 0.40 ² / (−10.0 − (−6.0)) = 0.16 / (−4.0) = −0.040 eV.

Answer: The ground state is lowered by about 0.040 eV by the coupling.

Quick check

1. Why is the zeroth-plus-first-order energy of helium guaranteed to lie above the exact energy? Answer: It equals the expectation value of the full Hamiltonian over the unperturbed function, which is a variational energy and therefore an upper bound.

Exam focus

State E⁽¹⁾ and E⁽²⁾ formulae with correct notation, and apply first order to helium using ⟨1/r₁₂⟩ = 5Z/8. Examiners often ask which method provides an upper bound and why second-order corrections to a ground state are negative.

Advanced insight

For helium, one can scale the Hamiltonian as Ĥ⁽⁰⁾ + λĤ′ and expand the energy as a power series in 1/Z. The series converges for helium and gives the exact energy to many digits, but for systems with near-degenerate states perturbation series can diverge, a real difficulty in computational chemistry.

Summary

Perturbation theory splits Ĥ into a solvable part and a perturbation, giving E⁽¹⁾ as the average of Ĥ′ and a negative E⁽²⁾ for the ground state. First-order energies are also variational upper bounds, but higher orders carry no bound. For helium, perturbation theory gives −2.750 then −2.908 Eₕ, while the variational ζ method gives −2.848 Eₕ, against the exact −2.904 Eₕ.

Practice questions

1. Write the expression for the first-order energy correction. Answer: E⁽¹⁾ = ⟨ψ⁽⁰⁾ Ĥ′ ψ⁽⁰⁾⟩, the average of the perturbation over the unperturbed wavefunction. 2. Using ⟨1/r₁₂⟩ = 5Z/8, give the zeroth-plus-first-order energy of He in hartrees. Answer: −4.000 + 1.250 = −2.750 Eₕ. 3. State one advantage of the variation method over perturbation theory. Answer: It always provides an upper bound to the ground-state energy, so improvements can be judged directly by how low the energy goes. 4. When does perturbation theory work poorly? Answer: When the perturbation is large compared with the spacing of the unperturbed levels, or when states are nearly degenerate, so the series converges slowly or diverges.