 Solution

ID: past-exam-of-the-quantum-theory-course-of-the-university-of-oxford/2014/1/c/solution

(ψ_{1} + ψ_{2}) / 2

. The orthonormality of the two energy eigenstates makes its squared norm equal to one. Their separate stationary state phases give

Ψ (x, t) = \frac{1}{a} [sin \frac{π x}{a} e^{- i E_{1} t /ℏ} + sin \frac{2 π x}{a} e^{- i E_{2} t /ℏ}] .

Write

Ω = (E_{2} - E_{1}) /ℏ = 3 π^{2} ℏ/ (2 m a^{2})

ρ (x, t) = \frac{1}{a} [sin^{2} \frac{π x}{a} + sin^{2} \frac{2 π x}{a} + 2 sin \frac{π x}{a} sin \frac{2 π x}{a} cos Ω t] .

Each squared sine integrates to

a /4

on the left half. For the quantum interference term, the product-to-sum identity gives

\int_{0}^{a /2} sin \frac{π x}{a} sin \frac{2 π x}{a} d x = \frac{1}{2} \int_{0}^{a /2} (cos \frac{π x}{a} - cos \frac{3 π x}{a}) d x = \frac{2 a}{3 π} .

Hence the Born rule gives

P_{L} (t) = \int_{0}^{a /2} ρ (x, t) d x = \frac{1}{2} + \frac{4}{3 π} cos Ω t .

The quantum interference moves probability between the halves even though each component separately is a stationary state.

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