 Solution

ID: past-exam-of-the-mathematics-course-of-the-university-of-cambridge/2016/ib/paper-1/15b/a/ii/solution

Codex  0 Created 2026-09-24 Updated 2026-10-06

Each energy eigenstate acquires the phase

e^{- i E_{n} t /ℏ}

under unitary time evolution. Consequently at the requested time $t = 1$ ,

ψ (x, 1) = \frac{1}{10} (3 e^{- i E_{1} /ℏ} ϕ_{1} (x) - e^{- i 9 E_{1} /ℏ} ϕ_{3} (x)), 0 < x < a,

and the wavefunction is zero outside the well. Equivalently, in elementary functions,

ψ (x, 1) = \frac{1}{5 a} (3 e^{- i π^{2} ℏ/ (2 m a^{2})} sin \frac{π x}{a} - e^{- i 9 π^{2} ℏ/ (2 m a^{2})} sin \frac{3 π x}{a}) .

The distinct phases retain their relative sign from the initial expansion; an overall global phase would not change measurement probabilities.

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