 Solution

ID: past-exam-of-the-chemistry-course-of-the-year-3-of-the-natural-sciences-course-of-the-university-of-cambridge/2022/paper-2b/section-b/33/b/solution

Expand the trial wavefunction in a complete orthonormal set of exact eigenfunctions,

ψ = \sum_{n} c_{n} ϕ_{n}

, with

\hat{H} ϕ_{n} = E_{n} ϕ_{n}

and

E_{n} \geq E_{0}

. The coefficients may depend on any chosen variational parameters. Orthogonality removes the cross terms in both integrals, giving

⟨ ψ ∣ ψ ⟩ = \sum_{n} ∣ c_{n} ∣^{2}, ⟨ ψ ∣ \hat{H} ∣ ψ ⟩ = \sum_{n} E_{n} ∣ c_{n} ∣^{2} .

Consequently

E [ψ] - E_{0} = \frac{\sum _{n} ( E _{n} - E _{0} ) ∣ c _{n} ∣ ^{2}}{\sum _{n} ∣ c _{n} ∣ ^{2}} \geq 0.

This proves the quantum variational principle. The denominator is positive because the trial function is nonzero. Equality requires every occupied exact state to have

E_{n} = E_{0}

, allowing for ground-state degeneracy. With a continuous part of the spectrum, the corresponding nonnegative spectral integral gives the same conclusion.

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