Linear Variation Functions

Introduction

So far, variational parameters have sat inside a function, like the exponent in e^(−ζr). Changing such a parameter changes the shape non-linearly, and each new trial family needs its own calculus. There is a far more systematic approach: choose a set of fixed functions and let the trial function be any mixture of them. The mixing coefficients become the parameters. This "linear variation" method turns quantum mechanics into linear algebra, and it underlies molecular orbital theory and practically all of computational chemistry.

Core explanation

The trial function. Choose N fixed basis functions φ₁, φ₂, …, φ N and write

ψ = c₁φ₁ + c₂φ₂ + … + c Nφ N = Σᵢ cᵢφᵢ

For simplicity take the functions and coefficients as real. The basis functions need not be orthogonal or normalised.

The energy. The variational energy is the Rayleigh ratio

E = ∫ψĤψ dτ / ∫ψψ dτ = Σᵢ Σⱼ cᵢcⱼHᵢⱼ / Σᵢ Σⱼ cᵢcⱼSᵢⱼ

where Hᵢⱼ = ∫φᵢĤφⱼ dτ and Sᵢⱼ = ∫φᵢφⱼ dτ. Both are just numbers once the basis is chosen; because Ĥ is Hermitian, Hᵢⱼ = Hⱼᵢ, and Sᵢᵢ = 1 for normalised functions.

Minimising. Write the energy as E × Σ cᵢcⱼSᵢⱼ = Σ cᵢcⱼHᵢⱼ and differentiate with respect to each coefficient c k, setting ∂E/∂c k = 0. The result is a set of N simultaneous linear equations:

Σⱼ (H kj − E S kj) cⱼ = 0, for k = 1, 2, …, N

These are the secular equations . In matrix form they read HC = ESC .

When do solutions exist? A set of homogeneous linear equations has a non-trivial solution (not all cⱼ zero) only if the determinant of the coefficients is zero:

det Hᵢⱼ − E Sᵢⱼ = 0

Expanding this determinant gives a polynomial of degree N in E, so there are N roots E₁ ≤ E₂ ≤ … ≤ E N. Each root, substituted back, gives a set of coefficients and so a wavefunction.

What the roots mean. The lowest root E₁ is the best estimate of the ground-state energy obtainable from this basis, and it lies at or above the true ground-state energy. The Hylleraas–Undheim–MacDonald theorem goes further: the k-th root is an upper bound to the k-th exact energy. The linear method therefore approximates excited states as well.

Improving the basis. Adding an extra basis function can never raise any root: the old trial space is contained within the new one, so the minimum can only go down or stay the same. As the basis approaches completeness the roots converge on the exact energies.

Linear versus non-linear. Non-linear parameters such as exponents require iterative searching; linear coefficients are found in one step by solving a matrix eigenvalue problem, which computers do extremely efficiently. Practical calculations therefore fix the exponents in advance and optimise only linear coefficients.

Formulae

ψ = Σᵢ cᵢφᵢ

Hᵢⱼ = ∫φᵢĤφⱼ dτ, Sᵢⱼ = ∫φᵢφⱼ dτ

Secular equations: Σⱼ (Hᵢⱼ − E Sᵢⱼ) cⱼ = 0

Secular determinant: det H − ES = 0

Step-by-step reasoning

1. Choose a set of basis functions suited to the problem. 2. Compute every Hᵢⱼ and Sᵢⱼ. 3. Form the determinant det Hᵢⱼ − E Sᵢⱼ . 4. Set it to zero and solve for the N energies. 5. For each energy, solve the secular equations for the coefficient ratios. 6. Fix the remaining constant by normalisation.

Visual explanation

Think of the basis functions as axes in an abstract space. Every trial function is a point in the space spanned by those axes, and the energy is a landscape over it. The linear method finds the lowest point of that landscape exactly. Adding a basis function adds a new axis, enlarging the space and possibly revealing an even lower point.

Real-world analogy

A painter with a fixed set of tubes of paint can mix any shade by choosing the proportions. The tubes are the basis functions and the proportions are the coefficients. With only three tubes some colours are out of reach; more tubes let the painter match the target more closely.

Real-world example

Every molecular orbital calculation, from simple Hückel theory for benzene to large computations on drug molecules, expresses orbitals as linear combinations of atom-centred basis functions and solves HC = ESC. The coefficients show how much each atomic orbital contributes to a molecular orbital.

Why?

Why do we get a determinant condition rather than a direct formula for E? The secular equations are homogeneous — every term contains a coefficient — so c = 0 always satisfies them. Physically meaningful, non-zero solutions exist only at special energies, and these are exactly the zeros of the determinant.

Common misconception

"Only the lowest root has meaning; the others are mathematical artefacts." Each higher root is an upper bound to the corresponding exact excited-state energy, and its coefficients describe an approximate excited state or unoccupied orbital.

Worked example

Question: Two orthonormal basis functions give H₁₁ = −3.0, H₂₂ = −1.0 and H₁₂ = −1.0 (arbitrary energy units). Find the two variational energies.

Reasoning: With S = identity, the determinant is (−3.0 − E)(−1.0 − E) − (−1.0)² = 0. Expanding: E² + 4.0E + 3.0 − 1.0 = 0, so E² + 4.0E + 2.0 = 0. E = [−4.0 ± √(16 − 8)]/2 = −2.0 ± 1.41.

Answer: E₁ ≈ −3.41 and E₂ ≈ −0.59. Mixing pushes the lower level down and the upper level up compared with H₁₁ and H₂₂.

Quick check

1. What happens to the lowest variational energy when an extra basis function is added to a linear trial function? Answer: It either decreases or stays the same, because the new trial space contains every function from the old space.

Exam focus

Be able to derive the secular equations from the Rayleigh ratio, state the determinant condition, and solve a 2 × 2 case by expanding the determinant as a quadratic. Examiners often ask you to explain why the energies are upper bounds and how the result changes when the basis is enlarged.

Advanced insight

In practice one often transforms to an orthonormal basis first, turning HC = ESC into an ordinary eigenvalue problem H′C′ = EC′ that standard routines diagonalise efficiently. If basis functions become nearly linearly dependent, S has tiny eigenvalues and the calculation becomes numerically unstable, a real problem with very large basis sets.

Summary

A linear variation function is a weighted sum of fixed basis functions, with the weights as variational parameters. Minimising the Rayleigh ratio gives the secular equations Σⱼ (Hᵢⱼ − ESᵢⱼ)cⱼ = 0, solvable only when det H − ES = 0. The N roots are upper bounds to the N lowest exact energies, and enlarging the basis can only lower them.

Practice questions

1. Write the secular equations for a basis of two functions. Answer: (H₁₁ − ES₁₁)c₁ + (H₁₂ − ES₁₂)c₂ = 0 and (H₂₁ − ES₂₁)c₁ + (H₂₂ − ES₂₂)c₂ = 0. 2. How many energy roots does a basis of six functions produce? Answer: Six, because the secular determinant expands to a polynomial of degree six in E. 3. Why must the secular determinant be zero for a physically acceptable solution? Answer: The equations are homogeneous, so a solution with coefficients not all zero exists only when the determinant of the coefficient matrix vanishes. 4. For two orthonormal functions with H₁₁ = H₂₂ = −2.0 and H₁₂ = −0.5, find the energies. Answer: (−2.0 − E)² = 0.25, so −2.0 − E = ±0.5, giving E = −2.5 and −1.5.