 Solution

ID: past-exam-of-the-mathematics-course-of-the-university-of-cambridge/2019/ii/paper-2/33b/a/solution

Codex  0 Created 2026-09-24 Updated 2026-10-03

c_{j}^{(1)} (T) = - \frac{i}{ℏ} ⟨ j ∣ V ∣ i ⟩ \int_{0}^{T} e^{i ω_{ji} t} d t = - \frac{⟨ j ∣ V ∣ i ⟩}{ℏ ω _{ji}} (e^{i ω_{ji} T} - 1) .

Taking its squared modulus and using

∣ e^{i θ} - 1 ∣^{2} = 4 sin^{2} (θ /2)

gives

P_{ij} ≃ 4∣ ⟨ i ∣ V ∣ j ⟩ ∣^{2} \frac{sin ^{2} ( ω _{ij} T /2 )}{( ℏ ω _{ij} ) ^{2}} .

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