What information is required to calculate the first-order energy correction for a non-degenerate energy level?
AThe exact perturbed wave function of the state, obtained by solving the full Schrödinger equation
BThe unperturbed wave function of the state and the perturbation operator H'
CAll other unperturbed wave functions in addition to the state of interest
DThe second-order correction must be computed before the first-order correction is accessible
The first-order energy correction is E⁽¹⁾ = ⟨ψ⁽⁰⁾|H'|ψ⁽⁰⁾⟩ — an expectation value of the perturbation operator computed using only the unperturbed wave function for that state. You do not need to solve a new differential equation or know any other states. This is the remarkable economy of the method: you reuse the solution you already have to estimate how much the energy shifts. Options C and D describe what is needed for the first-order wave function correction (C) or higher-order energy corrections — not the first-order energy correction itself.
Question 2 Multiple Choice
A student applies standard perturbation theory to a system where two unperturbed energy levels are separated by a gap much smaller than the perturbation strength. What problem arises?
AThe perturbation series converges faster when levels are close, making first-order corrections exact
BThe energy denominators in the wave function correction terms become very large, causing the expansion to blow up and become unreliable
CThe zeroth-order wave functions for close-lying levels become non-orthogonal, violating the method's assumptions
DThe expectation value ⟨ψ⁽⁰⁾|H'|ψ⁽⁰⁾⟩ becomes imaginary when energy gaps are small
The first-order wave function correction involves a sum over all other states with terms proportional to ⟨ψₖ⁽⁰⁾|H'|ψₙ⁽⁰⁾⟩ / (Eₙ⁽⁰⁾ − Eₖ⁽⁰⁾). When two levels are nearly degenerate (Eₙ ≈ Eₖ), the denominator approaches zero and the correction term diverges — the series breaks down. This is why degenerate perturbation theory must be used instead, diagonalizing H' within the degenerate subspace before applying the standard expansion.
Question 3 True / False
To apply first-order perturbation theory, you is expected to solve the full perturbed Schrödinger equation to obtain corrected wave functions before computing energy corrections.
TTrue
FFalse
Answer: False
This is exactly what perturbation theory avoids. The first-order energy correction E⁽¹⁾ = ⟨ψ⁽⁰⁾|H'|ψ⁽⁰⁾⟩ uses only the unperturbed wave function — no new differential equation is solved. You compute an integral using the solution you already know. This is the central practical value of perturbation theory: you can estimate energy shifts without solving an intractable new problem, as long as the perturbation is genuinely small.
Question 4 True / False
Perturbation theory gives more reliable results when the perturbation H' is large relative to the spacing between unperturbed energy levels.
TTrue
FFalse
Answer: False
The opposite is true. Perturbation theory is a power series in the perturbation strength; it converges only when the perturbation is small. The critical condition is that the perturbation must be small compared to the energy gaps between levels — specifically, the matrix elements ⟨ψₖ⁽⁰⁾|H'|ψₙ⁽⁰⁾⟩ must be small compared to |Eₙ⁽⁰⁾ − Eₖ⁽⁰⁾|. When levels are nearly degenerate or the perturbation is large, the expansion fails and other methods (degenerate perturbation theory or variational methods) must be used.
Question 5 Short Answer
Explain in physical terms what the first-order energy correction ⟨ψ⁽⁰⁾|H'|ψ⁽⁰⁾⟩ is calculating, and why using the unperturbed wave function is justified.
Think about your answer, then reveal below.
Model answer: The first-order energy correction is the average value of the perturbation experienced by the particle as if it were still in its unperturbed state. Because the perturbation is small, the wave function changes only slightly — to first order, the particle still occupies the unperturbed orbital, so computing the expectation value of H' with that orbital is a good approximation to the true energy shift. The correction captures how much energy the perturbation adds 'on average' given the existing probability distribution.
This physical interpretation matters because it shows why the method works: if the perturbation is small, the state barely changes, so the unperturbed wave function is a good proxy for the true state when computing the average perturbation energy. The analogy is a guitar string tuned slightly off — you can estimate the frequency shift from the change in tension without re-deriving the physics of wave propagation. Higher-order corrections account for the fact that the wave function itself deforms in response to the perturbation, which matters progressively more as the perturbation grows.