 Solution

ID: past-exam-of-the-advanced-quantum-physics-course-of-the-university-of-cambridge/2021/section-b/4/a/solution

i ℏ \partial_{t} ∣ ψ ⟩ = (H_{0} + V) ∣ ψ ⟩

and differentiate the definition of the interaction picture state:

i ℏ \frac{d}{d t} (e^{i H_{0} t /ℏ} ∣ ψ ⟩) = - H_{0} e^{i H_{0} t /ℏ} ∣ ψ ⟩ + e^{i H_{0} t /ℏ} (H_{0} + V) ∣ ψ ⟩ = e^{i H_{0} t /ℏ} V e^{- i H_{0} t /ℏ} ∣ ψ_{I} ⟩ .

The free Hamiltonian operator terms cancel because

H_{0}

commutes with its own matrix exponential. Therefore

i ℏ \frac{d}{d t} ∣ ψ_{I} (t)⟩ = V_{I} (t) ∣ ψ_{I} (t)⟩, V_{I} (t) = e^{i H_{0} t /ℏ} V (t) e^{- i H_{0} t /ℏ} .

This interaction picture transformation is exact.

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