# Perturbation theory and approximation checks > Write H=H0+λV and expand observables or states about a solvable reference problem. Small λ is useful only relative to relevant spectral scales. Course: [Advanced physics methods and spacetime](https://studybrix.com/en/courses/advanced-physics-methods/index.md) Level: advanced Prerequisites: Eigenstates, inner products and nondegenerate quantum mechanics. ## Notes ### Setup Write H=H0+λV and expand observables or states about a solvable reference problem. Small λ is useful only relative to relevant spectral scales. ### First order For a normalized nondegenerate state, the first-order energy coefficient is ⟨n∣V∣n⟩. Include λ when calculating the first-order energy shift. ### Degeneracy Degenerate levels require diagonalizing the perturbation in the relevant subspace rather than applying a single arbitrary diagonal element. ### Validation Check dimensions, symmetry, limiting behaviour and the size of corrections. Near-degeneracy can spoil a naive small-coupling expansion. ## Practice questions ### 1. For λ=0.1 and ⟨n∣V∣n⟩=3 eV, find the first-order energy shift. Answer: 0.3 eV Explanation: Multiply the matrix element by λ. ### 2. Which quantity gives the nondegenerate first-order energy coefficient? A. Every off-diagonal entry added without denominators B. ⟨n∣V∣n⟩ C. The state norm cubed D. Zero in all systems Answer: ⟨n∣V∣n⟩ Explanation: The diagonal expectation in the unperturbed state is the first-order coefficient. ### 3. An unperturbed energy is 5 eV and first-order shift is −0.2 eV. Find the first-order corrected energy. Answer: 4.8 eV Explanation: Add the correction to the unperturbed value. ### 4. What should be done first within a degenerate subspace? A. Divide by zero energy gaps B. Diagonalize the perturbation restricted to that subspace C. Discard normalization D. Choose any one basis vector and ignore mixing Answer: Diagonalize the perturbation restricted to that subspace Explanation: The appropriate combinations resolve first-order splitting. ### 5. Why can a tiny coupling produce a significant change near two nearly degenerate levels? Sample answer / marking guidance: State-mixing corrections involve energy-gap denominators. A small gap can make the ratio large despite a small numerator. Explanation: Credit relative scales rather than coupling size alone. ### 6. For a normalized state with components (1/√2,1/√2) and V=diag(2,4) eV, find ⟨V⟩. Answer: 3 eV Explanation: The weighted diagonal mean is 2/2+4/2. ### 7. How can symmetry make a first-order correction vanish without making all higher corrections vanish? Sample answer / marking guidance: A symmetry can force the diagonal expectation of V to zero, while allowed off-diagonal couplings can still contribute at higher order. Explanation: Credit the distinction between diagonal and off-diagonal terms. ### 8. State two checks on a claimed perturbative energy correction. Sample answer / marking guidance: Its units must be energy, it should vanish as the perturbation strength tends to zero, and it should be small compared with relevant separation scales in the asserted regime. Explanation: Award credit for any two appropriate checks. ## Further reading - https://ocw.mit.edu/courses/8-09-classical-mechanics-iii-fall-2014/ - https://openstax.org/details/books/university-physics-volume-3 Original AI-assisted educational draft; not independently teacher-reviewed. References support further study, not copied exercises, affiliation or curriculum accreditation.