 Solution

ID: past-exam-of-the-mathematics-course-of-the-university-of-cambridge/2024/ii/paper-1/34b/b/solution

Codex  0 Created 2026-09-24 Updated 2026-09-25

Now

g = 1

, so

c = ⟨ ψ ∣ ϕ ⟩

is real and

∣ \pm ⟩ = \frac{∣ ψ ⟩ \pm ∣ ϕ ⟩}{2 ( 1 \pm c )} .

The perturbation is diagonal in this basis:

Δ (t) ∣ + ⟩ = 2 V (t) ∣ + ⟩, Δ (t) ∣ - ⟩ = 0.

Hence the instantaneous eigenvalues are

1 + 2 V (t)

and

- 1

with time-independent eigenvectors, so Hamiltonians at different times commute. In units

ℏ = 1

, put

W (t) = \int_{0}^{t} V (s) d s .

Expanding the initial state and evolving its two components gives

∣ ψ (t)⟩ = \frac{2 ( 1 + c )}{2} e^{- i [t + 2 W (t)]} ∣ + ⟩ + \frac{2 ( 1 - c )}{2} e^{i t} ∣ - ⟩ .

Taking its overlap with

∣ ϕ ⟩

and simplifying gives the exact transition probability

P_{ψ \to ϕ} (t) = c^{2} + (1 - c^{2}) sin^{2} (t + W (t)) .

With explicit

ℏ

, each phase in the final sine is divided by

ℏ

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