 Solution

ID: past-exam-of-the-theoretical-physics-ii-course-of-the-university-of-cambridge/2014/paper-1/2/a/solution

Ψ (x, t) = \int K (x, t ∣ x^{'}, t^{'}) Ψ (x^{'}, t^{'}) d x^{'}

U = e^{- i H (t - t^{'}) /ℏ}

. Inserting a complete orthonormal energy eigenbasis gives

K (x, t ∣ x^{'}, t^{'}) = α \sum φ_{α} (x) φ_{α}^{*} (x^{'}) e^{- i E_{α} (t - t^{'}) /ℏ} .

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