Past exam of the mathematics course of the University of Cambridge / 2020 / ii / Paper 2 / 34A / a

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Solution

Let

H_{0} ∣ r ⟩ = E_{r} ∣ r ⟩

. In the interaction picture,

δ H_{I} (t) = e^{i H_{0} t /ℏ} δH (t) e^{- i H_{0} t /ℏ},

and the state obeys

i ℏ \frac{d}{d t} ∣ ψ_{I} (t)⟩ = δ H_{I} (t) ∣ ψ_{I} (t)⟩ .

The first term of the Dyson series, or equivalently first-order time-dependent perturbation theory, gives

∣ ψ_{I} (t)⟩ = [1 - \frac{i}{ℏ} \int_{0}^{t} δ H_{I} (t^{'}) d t^{'}] ∣ ψ (0)⟩ + O (δ H^{2}) .

Thus, if

∣ ψ (0)⟩ = \sum_{r} a_{r} ∣ r ⟩

,

∣ ψ_{I} (t)⟩ = s \sum [a_{s} - \frac{i}{ℏ} r \sum a_{r} \int_{0}^{t} e^{i (E_{s} - E_{r}) t^{'} /ℏ} ⟨ s ∣ δH (t^{'}) ∣ r ⟩ d t^{'}] ∣ s ⟩ + O (δ H^{2}) .

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