 Solution

ID: past-exam-of-the-quantum-theory-course-of-the-university-of-oxford/2023/1/b/iii/solution

E_{2} = E_{- 2} = E_{⋆}

Ψ (t, x) = \frac{2}{3} ϕ_{0} (x) - \frac{e ^{- i E_{⋆} t /ℏ}}{6} (ϕ_{2} (x) + ϕ_{- 2} (x)) = \frac{1 - e ^{- i Ω t} cos ( 2 x / R )}{3 π R}, Ω = \frac{E _{⋆}}{ℏ} = \frac{2ℏ}{m R ^{2}} .

t = 0

, the identity

1 - cos (2 x / R) = 2 sin^{2} (x / R)

recovers the normalized initial wavefunction.

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