 Solution

ID: past-exam-of-the-chemistry-course-of-the-year-3-of-the-natural-sciences-course-of-the-university-of-cambridge/2022/paper-4/59/a/solution

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Past exam of the chemistry course of the year 3 of the natural sciences course of the University of Cambridge / 2022 / Paper 4 / 59 / a / Solution by

Codex  0 2026-10-20

Let

H^{(0)} ∣ n ⟩ = E_{n}^{(0)} ∣ n ⟩

with a nondegenerate reference eigenvalue, and choose intermediate normalization

⟨ n ∣ n^{(1)} ⟩ = 0

. The first-order equation is

(H^{(0)} - E_{n}^{(0)}) ∣ n^{(1)} ⟩ = - (H^{(1)} - E_{n}^{(1)}) ∣ n ⟩ .

Projection on

∣ j ⟩

,

j \neq = n

, gives

⟨ j ∣ n^{(1)} ⟩ = H_{jn}^{(1)} / (E_{n}^{(0)} - E_{j}^{(0)})

. The second-order projected equation gives

E_{n}^{(2)} = ⟨ n ∣ H^{(1)} ∣ n^{(1)} ⟩

, hence

E_{n}^{(2)} = - \sum_{j \neq = n} \frac{∣ H _{nj}^{(1)} ∣ ^{2}}{E _{j}^{(0)} - E _{n}^{(0)}} .

The sum runs over a complete orthonormal reference basis except the reference state; continuum states require integration if present. Degenerate states with zero denominators require degenerate perturbation theory first. These are expansion coefficients in

E_{n} = E_{n}^{(0)} + λ E_{n}^{(1)} + λ^{2} E_{n}^{(2)} + \dots

.

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