Variation Principle: Problem-Solving Workshop

Introduction

Variational questions can look different on the page. One may ask for the best width of a single trial function; another may provide two basis functions and ask for a determinant. They share one idea: minimise the energy expectation over a restricted family of admissible wavefunctions. This workshop connects the calculus route, the linear-algebra route and the interpretation of the resulting energy estimate.

Core explanation

For any nonzero admissible ψ, define E[ψ] = ⟨ψ Ĥ ψ⟩/⟨ψ ψ⟩. For the ground state of the Hamiltonian under the same boundary conditions, E[ψ] ≥ E₀. The quotient makes overall amplitude irrelevant. If ψ depends on a nonlinear shape parameter α, first evaluate the integrals to obtain E(α), then minimise over the physically allowed α. If E(α) = Aα² − Bα with positive A and B, the minimum occurs at α = B/(2A), giving −B²/(4A). A derivative test and domain check still matter: a stationary point outside the allowed range is not an acceptable answer.

For a linear combination ψ = Σᵢcᵢφᵢ, the coefficients cᵢ are adjustable. Define Hᵢⱼ = ⟨φᵢ Ĥ φⱼ⟩ and Sᵢⱼ = ⟨φᵢ φⱼ⟩. Minimising the quotient with respect to coefficients produces the generalised eigenvalue equation Hc = ESc. For a nonzero coefficient vector, det(H − ES) = 0. The lowest resulting root gives the best ground-state energy available in that basis, assuming the overlap matrix represents linearly independent admissible basis functions. If the basis is orthonormal, S is the identity and the equation simplifies to Hc = Ec.

Matrix off-diagonal elements matter. Ignoring H₁₂ can miss mixing and an energy lowering. Ignoring S₁₂ when the basis functions overlap changes the equation itself. If two basis functions are nearly redundant, S can become poorly conditioned; a numerical calculation may then be unstable even though the variational principle remains true. A larger trial subspace can match or improve the minimum because all old combinations are still accessible.

An excited-state estimate needs extra conditions. The lowest root of the full basis subspace approximates the ground state, not an arbitrary excited state. Higher roots can represent excited states within a suitable symmetry sector, but a simple upper-bound statement for a particular excited level requires appropriate orthogonality or a more careful min–max argument. In an exam, label the root you are using and avoid assigning every root without checking its state character.

Step-by-step reasoning

Decide whether the trial is one shape with adjustable parameter or a linear combination. For a shape parameter, normalise or retain the norm denominator, calculate E(α), differentiate and check the minimum. For a basis expansion, build both H and S, write Hc = ESc, solve det(H − ES) = 0, and identify the lowest root for a ground-state bound. In either path, inspect dimensions, boundary conditions and the sign of the final energy relative to known limits.

Visual explanation

Draw a small trial-function family as a curved line through a vast space of possible wavefunctions. Its lowest point lies above or touches the exact ground-energy plane. Next draw a two-dimensional plane spanned by φ₁ and φ₂; minimising over that plane gives a better or equal low point. A matrix diagram shows H and S as two labelled squares feeding the equation det(H − ES) = 0.

Real-world analogy

Choosing a single adjustable trial is like finding the best recipe by changing one ingredient amount. A linear basis allows combinations of several prepared ingredients, so it offers more choices. The larger menu cannot make the best achievable option worse, but it can still miss a recipe outside its ingredients. The analogy illustrates constrained optimisation, not the physics of quantum amplitudes.

Real-world example

Molecular orbital calculations combine atom-centred functions to approximate an electron's spatial state. Their coefficients are found from matrix equations involving Hamiltonian and overlap integrals. Although practical chemistry uses many basis functions and often self-consistent electron interactions, the two-function secular determinant is the same structural step in miniature.

Why?

The stationarity condition comes from requiring that a small allowed change in the trial state does not lower its expectation energy at the optimum. For a one-parameter family, this is dE/dα = 0. For independent linear coefficients, it becomes a system of homogeneous equations. The determinant must vanish because a homogeneous system otherwise has only the trivial c = 0 solution, which cannot represent a state.

Common misconception

Setting det(H − EI) = 0 is correct only when the basis is orthonormal. If two basis functions overlap, S belongs in the equation. Another error is to interpret any root as a guaranteed exact physical energy. The roots are energies optimised within the chosen subspace; the lowest is an upper bound for the exact ground energy, and the higher roots need more careful interpretation.

Worked example

Take an orthonormal two-function basis with H = [[−2, −0.2], [−0.2, −1]] in some consistent energy unit. The secular equation is (−2 − E)(−1 − E) − 0.04 = 0, or E² + 3E + 1.96 = 0. Its roots are [−3 ± √(9 − 7.84)]/2 = [−3 ± √1.16]/2, about −2.039 and −0.961. The lower root lies below the diagonal-only estimate −2 because the basis functions mix. It remains at or above the exact ground energy of the full Hamiltonian.

Quick check

1. What changes in the secular equation if two basis functions are individually normalised but not orthogonal? Answer: Their diagonal overlaps are 1, but S₁₂ and S₂₁ are nonzero. Use det(H − ES) = 0 with those off-diagonal overlaps; replacing S by I would give the wrong coefficient equations.

Exam focus

Write the Rayleigh quotient first. State why a trial is admissible and whether it is normalised. In matrix problems, calculate S as well as H unless orthonormality is given. Select the lowest root for a ground-state estimate and state the inequality E var ≥ E₀. Show enough algebra to distinguish a genuine minimum from an untested stationary value.

Advanced insight

The Rayleigh–Ritz method underpins basis-set electronic-structure calculations. For a fixed Hermitian Hamiltonian, increasing a nested basis cannot increase its lowest variational eigenvalue. In multi-electron applications, the Hamiltonian may itself depend on an approximate mean field or chosen nuclear geometry, so careful comparisons hold the physical problem and assumptions fixed.

Summary

The variation principle minimises an energy quotient over allowed trial states. A nonlinear parameter leads to ordinary calculus; linear coefficients lead to Hc = ESc and det(H − ES) = 0. The lowest result is an upper bound for the ground energy of the specified Hamiltonian. Overlap, boundary conditions and the meaning of each root must be handled explicitly.

Practice questions

1. Minimise E(α) = 2α² − 8α + 1 for α > 0. Answer: dE/dα = 4α − 8 gives α = 2. The second derivative is 4 > 0, and E(2) = 8 − 16 + 1 = −7 in the stated units. It is the best energy within that family. 2. If H₁₂ = 0 and an orthonormal two-function basis has diagonal energies −1.5 and −0.7, what are its secular roots and lowest variational estimate? Answer: The determinant factors as (−1.5 − E)(−0.7 − E) = 0, giving −1.5 and −0.7. The lowest basis estimate is −1.5. 3. Why cannot a trial function with divergent norm establish a variational bound? Answer: Its Rayleigh quotient is not a well-defined finite expectation for a normalisable bound state. The theorem requires an admissible state in the Hamiltonian's domain; a divergent norm fails that requirement.