You apply the non-degenerate perturbation theory formula for first-order state corrections to two states |n⟩ and |m⟩ with the same unperturbed energy E⁰ₙ = E⁰ₘ. What goes wrong?
ANothing goes wrong — the formula still gives finite corrections when the states are orthogonal
BThe formula gives zero corrections for all degenerate states, which is too conservative
CThe denominator E⁰ₙ − E⁰ₘ = 0, making the correction terms diverge, so the formula is undefined
DThe first-order energy correction ⟨n|H'|n⟩ is not well-defined for degenerate states
The non-degenerate state correction formula mixes |m⟩ into |n⟩ with coefficient ⟨m|H'|n⟩/(E⁰ₙ − E⁰ₘ). When E⁰ₙ = E⁰ₘ, this denominator is zero and the coefficient diverges — the formula breaks down entirely. Option D is wrong because ⟨n|H'|n⟩ is perfectly well-defined; it's the state corrections (mixing with degenerate partners) that fail, not the energy corrections by themselves.
Question 2 Multiple Choice
In degenerate perturbation theory, what determines the 'good states' — the correct zeroth-order states to use?
AAny orthonormal basis within the degenerate subspace works equally well; the choice is arbitrary
BThe eigenvectors of H' restricted to the degenerate subspace — the states that diagonalize the perturbation within that subspace
CThe states with the largest matrix element ⟨ψᵢ|H'|ψᵢ⟩
DThe original spherical harmonics, which are always the correct starting point
The good states are those linear combinations of the degenerate subspace that make H' diagonal within that subspace — i.e., the eigenvectors of the n×n matrix Wᵢⱼ = ⟨ψᵢ|H'|ψⱼ⟩. These are the states H' 'prefers' and the states for which the first-order energy corrections are well-defined. Option A is the central misconception: not just any basis works. In the Stark effect, for instance, the good states are specific mixtures of 2s and 2p, not the individual spherical harmonics.
Question 3 True / False
When applying degenerate perturbation theory, any linear combination of the degenerate states is a valid 'good state' that can be used to compute perturbative corrections.
TTrue
FFalse
Answer: False
False — this is the key misconception degenerate perturbation theory corrects. The 'good states' are specifically those combinations that diagonalize H' within the degenerate subspace. An arbitrary combination will have off-diagonal matrix elements with its degenerate partners, meaning the perturbation mixes them and the non-degenerate formula still diverges. The perturbation itself selects the preferred basis — you must diagonalize W to find it. Once you have the good states, the corrections are finite and well-defined.
Question 4 True / False
If the perturbation H' completely lifts an n-fold degeneracy at first order, all n eigenvalues of the degenerate subspace matrix W are distinct.
TTrue
FFalse
Answer: True
True by definition. The eigenvalues of W are the first-order energy corrections. If all n eigenvalues are distinct, then after adding H', all n levels sit at different energies — the original degeneracy has been completely resolved. If some eigenvalues are equal, that degeneracy persists at first order and you must go to higher order to resolve it. 'Complete lifting' is exactly the condition of n distinct eigenvalues.
Question 5 Short Answer
Explain why finding 'good states' resolves the breakdown of non-degenerate perturbation theory in the degenerate case.
Think about your answer, then reveal below.
Model answer: Non-degenerate perturbation theory fails because it tries to mix degenerate states with each other using a formula that has zero in the denominator. The good states are eigenvectors of H' within the degenerate subspace — they diagonalize the perturbation there. Because they diagonalize H', the off-diagonal matrix elements between good states vanish: ⟨good state i|H'|good state j⟩ = 0 for i ≠ j. So when the non-degenerate formula tries to mix them, the numerator is zero (not just the denominator), and the 0/0 problem is resolved — there is no mixing between degenerate good states needed, and mixing with states outside the subspace uses the original non-degenerate formula with nonzero denominators.
The geometric intuition is that H' 'breaks' the symmetry of the degenerate subspace by preferring certain directions. The good states align with those preferred directions, so H' can be treated as if it simply assigns different energies to each good state — no mixing needed within the subspace.