 Solution

ID: past-exam-of-the-quantum-theory-course-of-the-university-of-oxford/2021/1/b/solution

The identity

cos^{2} x = \frac{1}{2} + \frac{1}{4} e^{2 i x} + \frac{1}{4} e^{- 2 i x}

gives the momentum eigenstate expansion

Ψ (x, 0) = A 2 π (\frac{1}{2} ψ_{0} + \frac{1}{4} ψ_{2} + \frac{1}{4} ψ_{- 2}) .

1 = ∣ A ∣^{2} (2 π) (\frac{1}{4} + \frac{1}{16} + \frac{1}{16}) = ∣ A ∣^{2} \frac{3 π}{4} .

Hence

∣ A ∣ = 2/ 3 π

. The wavefunction normalization leaves a global phase free; the positive real choice used in the requested expansion gives

A = \frac{2}{3 π}, Ψ (x, 0) = \frac{2}{3} (ψ_{0} (x) + \frac{1}{2} ψ_{2} (x) + \frac{1}{2} ψ_{- 2} (x)) .

E_{⋆} = 2 ℏ^{2} / m

Ω = E_{⋆} /ℏ = 2ℏ/ m

Ψ (x, t) = \frac{2}{3} (ψ_{0} (x) + \frac{e ^{- i Ω t}}{2} ψ_{2} (x) + \frac{e ^{- i Ω t}}{2} ψ_{- 2} (x)) = \frac{1 + e ^{- i Ω t} cos ( 2 x )}{3 π} .

t = 0

1 + cos (2 x) = 2 cos^{2} x

recovers the initial wavefunction. Its probability density, useful below, is

∣Ψ (x, t) ∣^{2} = \frac{1 + c o s ^{2} ( 2 x ) + 2 c o s ( 2 x ) c o s ( Ω t )}{3 π} .

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