Questions: The Variational Method: Application

Question 1

A physicist uses a Gaussian trial wavefunction for the hydrogen atom and obtains E_trial = −11.5 eV. The exact ground-state energy is −13.6 eV. What is the correct interpretation?

• A. The calculation must contain an error — variational energies should converge to the exact answer
• B. The Gaussian family does not contain the true ground state; −11.5 eV is the best approximation within that family, and the variational bound guarantees the true energy is at or below −11.5 eV
• C. The true ground-state energy lies between −11.5 eV and −13.6 eV
• D. The variational method has failed because the estimate is wrong by 15%

Answer: (B)

The variational principle guarantees ⟨H⟩ ≥ E₀ for any trial state. Getting −11.5 eV > −13.6 eV is not an error — it is the expected behavior when your trial family doesn't include the true ground state. The 15% error reflects the price of the wrong functional form (Gaussian vs. the true exponential). The bound tells us the exact energy is somewhere at or below −11.5 eV — and indeed, −13.6 eV < −11.5 eV, consistent with the bound. Option C confuses 'between' with 'below': the true energy is below the estimate, not between it and some other value.

Question 2

What happens to the variational energy estimate when you add more free parameters to your trial wavefunction?

• A. It can increase or decrease unpredictably, depending on how the parameters interact with the Hamiltonian
• B. It increases — more parameters introduce more uncertainty in the approximation
• C. It stays the same — variational energy depends only on the Hamiltonian, not the parameterization
• D. It decreases (or stays the same) — a more flexible trial family includes more candidate states, so the minimum over parameters can only be at most as high as the previous best

Answer: (D)

Adding parameters expands the trial family. The new minimum is taken over a strictly larger set of states than before, so it can only be equal to or lower than the previous minimum — never higher. This is why 'more flexible trial function → tighter upper bound on E₀' is a reliable rule. A one-parameter family sits inside every two-parameter generalization of it (just fix one parameter at its previous optimal value). Therefore expanding the family never raises the variational energy.

Question 3

If the true ground-state wavefunction is a member of the parameterized trial family, the variational method will recover the exact ground-state energy.

• T. True
• F. False

Answer: True

The variational energy E(α, β, ...) = ⟨ψ(α,β,...)|Ĥ|ψ(α,β,...)⟩ achieves its minimum at E₀ when the minimization reaches the true ground state. If ψ_true is in the family (i.e., there exist parameter values that reproduce it exactly), then the minimum over parameters will find those values and return E₀. This is demonstrated by the hydrogen atom example: using exp(−αr) as the trial function includes the true ground state at α = 1/a₀, and the variational minimum recovers −13.6 eV exactly.

Question 4

The variational method can yield an energy estimate lower than the true ground-state energy if the trial wavefunction is a poor approximation.

• T. True
• F. False

Answer: False

This is categorically forbidden by the variational principle. For any normalized trial state |ψ_trial⟩, ⟨ψ_trial|Ĥ|ψ_trial⟩ ≥ E₀. The proof: expand ψ_trial in the exact energy eigenbasis — the resulting expectation value is a weighted average of energies, all ≥ E₀, so the sum is ≥ E₀. No matter how bad the trial wavefunction is, ⟨H⟩ can never go below the true ground-state energy. This guarantee is the entire foundation of the variational method's reliability.

Question 5

Why does the variational principle guarantee that ⟨H⟩ ≥ E₀ for any trial wavefunction? What does this guarantee imply about how to improve your estimate?

Model answer: Expand the trial wavefunction in the exact energy eigenbasis: |ψ_trial⟩ = Σcₙ|n⟩. Then ⟨H⟩ = Σ|cₙ|²Eₙ. Since E₀ ≤ Eₙ for all n, and the |cₙ|² form a probability distribution summing to 1, ⟨H⟩ = Σ|cₙ|²Eₙ ≥ E₀Σ|cₙ|² = E₀. To improve: expand the trial family (add parameters or use more flexible functional forms) — the minimum over a larger family can only be lower, tightening the bound on E₀.

The guarantee means: (1) you can never accidentally undershoot the true ground-state energy, making the method safe from false negatives, and (2) 'lower variational energy is always better' is a rigorous statement, not just a heuristic. This turns the quantum mechanical problem (find the lowest eigenvalue of a complicated Hamiltonian) into an optimization problem (minimize a functional over a parameterized family), which computers are very good at.