 Solution

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

At the midpoint the second sine vanishes, whereas the derivative of the first sine vanishes. Thus the wavefunction and its spatial derivative simplify to

Ψ (a /2, t) = \frac{e ^{- i E_{1} t /ℏ}}{a}, \partial_{x} Ψ (a /2, t) = - \frac{2 π}{a ^{3/2}} e^{- i E_{2} t /ℏ} .

Substitution in the probability current gives

j (a /2, t) = \frac{2 π ℏ}{m a ^{2}} sin Ω t .

The continuity equation, integrated over the left half, gives

P_{L}^{'} (t) = j (0, t) - j (a /2, t)

j (0, t) = 0

, and the preceding probability formula gives

P_{L}^{'} (t) = - \frac{4Ω}{3 π} sin Ω t = - \frac{2 π ℏ}{m a ^{2}} sin Ω t = - j (a /2, t) .

Positive midpoint current removes probability from the left half. This is precisely the midpoint current of a two-mode box state.

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