 Solution

ID: past-exam-of-the-quantum-physics-course-of-the-university-of-oxford/2016/4/solution

ϕ_{n} (x) = \frac{2}{a} sin \frac{nπ x}{a}, E_{n} = n^{2} E_{1}, E_{1} = \frac{π ^{2} ℏ ^{2}}{2 m a ^{2}} .

The identity

sin u cos u = \frac{1}{2} sin 2 u

resolves the initial wavefunction into

Ψ (0) = (2 ϕ_{1} + ϕ_{2}) / 5

Ψ (x, t_{0}) = \frac{2 ϕ _{1} ( x ) e ^{- i E_{1} t_{0} /ℏ} + ϕ _{2} ( x ) e ^{- 4 i E_{1} t_{0} /ℏ}}{5}

inside the well, and the wavefunction vanishes outside. The energy probabilities are

4/5

and

1/5

at every time. Consequently

⟨ H ⟩_{0} = ⟨ H ⟩_{t_{0}} = \frac{8}{5} E_{1} = \frac{4 π ^{2} ℏ ^{2}}{5 m a ^{2}} .

For the left half, each diagonal integral

\int_{0}^{a /2} ∣ ϕ_{n} ∣^{2} d x

1/2

, while

\int_{0}^{a /2} ϕ_{1} ϕ_{2} d x = \frac{4}{3 π} .

The interference term therefore yields

P_{left} (t_{0}) = \frac{1}{2} + \frac{16}{15 π} cos \frac{3 E _{1} t _{0}}{ℏ} .

This probability lies between approximately

0.1605

and

0.8395

, with period

2 π ℏ/ (3 E_{1})

. A conserved energy distribution is compatible with time-dependent position probabilities because the relative quantum phase keeps evolving.

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