“Know how to solve every problem that has been solved.” “What I cannot create, I do not understand.” — Richard Feynman

The Variational Principle by Hand — Why the bound is for the ground state only

Exercises

Quantum Chemistry I › Unit 2 · The variational game › The Variational Principle by Hand › all problems

Check Problem 8 of 8

Why the bound is for the ground state only

Problem

The variational principle bounds the ground state. Explain why naively minimizing

⟨ H ⟩

does not bound the first excited state, and what extra condition repairs it.

Solution

Expand the trial in exact eigenstates: $⟨ H ⟩ = \sum_{i} ∣ c_{i} ∣^{2} E_{i}$ with $\sum_{i} ∣ c_{i} ∣^{2} = 1$ . Since $E_{0}$ is the smallest eigenvalue, this weighted average can never drop below it — equality only when the trial is the ground state.

For the first excited state there is no such floor at $E_{1}$ : a trial aimed at $E_{1}$ can always lower its energy by mixing in ground-state character ( $c_{0} \neq = 0$ ), sliding $⟨ H ⟩$ below $E_{1}$ toward $E_{0}$ .

The repair is orthogonality: constrain $⟨ ψ_{trial} ∣ ψ_{0} ⟩ = 0$ , forcing $c_{0} = 0$ and restoring the bound at $E_{1}$ . Done systematically for a basis, this is the Hylleraas–Undheim / MacDonald theorem: the $k$ -th secular root bounds the $k$ -th exact state.