What is the first-order energy correction E_n^(1) for a state |n⟩ in time-independent perturbation theory?
AThe eigenvalue of H' (the perturbation Hamiltonian) in isolation
BThe expectation value of H' evaluated in the unperturbed state: ⟨n^(0)|H'|n^(0)⟩
CThe overlap integral ⟨n^(0)|n^(1)⟩ between the unperturbed and first-order corrected states
DThe sum of all off-diagonal matrix elements of H' connecting state n to all other states
The first-order energy correction is E_n^(1) = ⟨n^(0)|H'|n^(0)⟩ — simply the expectation value of the perturbation Hamiltonian in the unperturbed state. The elegance is that you need no new wavefunction at this order: just integrate the known unperturbed state against the perturbation. This is the most important practical result in perturbation theory, used whenever you need a first estimate of how a small perturbation (a magnetic field, a weak potential, a relativistic correction) shifts an energy level.
Question 2 Multiple Choice
In the first-order state correction, which factor determines how strongly an unperturbed state |m^(0)⟩ mixes into the perturbed state |ψ_n^(1)⟩?
AThe magnitude of the energy difference E_n^(0) − E_m^(0); states far in energy mix more strongly
BWhether m and n have the same parity; only states of opposite parity can mix
CThe matrix element ⟨m^(0)|H'|n^(0)⟩ and the inverse of the energy gap; large matrix element and small gap means strong mixing
DThe population of state m in thermal equilibrium at the system temperature
The mixing coefficient for state m into the corrected state n is ⟨m^(0)|H'|n^(0)⟩ / (E_n^(0) − E_m^(0)). Two factors matter: (1) The matrix element must be nonzero — if symmetry (a selection rule) forbids ⟨m|H'|n⟩, that state contributes nothing. (2) The energy denominator E_n − E_m must be large — states nearly degenerate with state n (small denominator) mix in most strongly, and the formula actually diverges when the gap goes to zero, signaling the breakdown of non-degenerate perturbation theory.
Question 3 True / False
If the matrix element ⟨m^(0)|H'|n^(0)⟩ = 0 for all m ≠ n due to a symmetry selection rule, then the first-order correction to the state |ψ_n⟩ vanishes.
TTrue
FFalse
Answer: True
The first-order state correction is a sum over all other states m, weighted by ⟨m|H'|n⟩/(E_n − E_m). If every off-diagonal matrix element ⟨m|H'|n⟩ = 0 (e.g., because H' has odd parity and the states are all even, or by some other symmetry), every term in the sum vanishes and |ψ_n^(1)⟩ = 0. The energy still receives a first-order correction (the diagonal element ⟨n|H'|n⟩ may be nonzero), but the wavefunction is unmodified at first order.
Question 4 True / False
Time-independent perturbation theory remains valid even when two unperturbed energy levels are nearly degenerate, because the first-order energy correction formula handles this case correctly.
TTrue
FFalse
Answer: False
Near-degeneracy is precisely where non-degenerate perturbation theory breaks down. The first-order state correction contains the denominator E_n^(0) − E_m^(0): when two levels are close in energy, this denominator approaches zero and the mixing coefficient diverges, making the perturbative expansion uncontrolled. Exact degeneracy makes the formula undefined. Degenerate perturbation theory handles this by first diagonalizing H' within the degenerate subspace, choosing a good basis before applying the expansion.
Question 5 Short Answer
Why is the first-order energy correction E_n^(1) = ⟨n^(0)|H'|n^(0)⟩ particularly elegant, and what does it tell you physically about the relationship between the perturbation and the unperturbed wavefunction?
Think about your answer, then reveal below.
Model answer: It is elegant because you obtain the first-order energy shift using only the unperturbed wavefunction — no correction to the state is needed at this order. Physically, it says that to first approximation, the energy shift equals the average value of the perturbation as experienced by the unperturbed probability distribution |ψ_n^(0)|². The wavefunction has not yet 'responded' to the perturbation; it simply samples the perturbation according to the unperturbed probability density.
This interpretation connects naturally to the Hellmann-Feynman theorem and to classical intuition: if you turn on a weak external field slowly, the first-order energy change is what the original state would feel on average. The fact that no new wavefunction is needed makes first-order energy corrections computationally inexpensive and explains why they appear everywhere in atomic and molecular physics: Zeeman splitting, Stark effect, fine structure corrections all start with this expectation value calculation.