 Solution

ID: past-exam-of-the-short-options-paper-shared-by-prelims-and-parts-a-and-b-of-the-physics-course-of-the-university-of-oxford/2018/s3/2/c/solution

The wavefunction is real and vanishes continuously at both endpoints. Hence

⟨ p ⟩ = - i ℏ \int Ψ Ψ^{'} d x = - \frac{i ℏ}{2} [Ψ^{2}]_{- \infty}^{\infty} = 0.

The momentum second moment is most safely evaluated from its quadratic form:

⟨ p^{2} ⟩ = ℏ^{2} \int ∣ Ψ^{'} ∣^{2} d x = 4 ℏ^{2} A^{2} \int_{- a}^{a} x^{2} d x .

Thus

⟨ p ⟩ = 0, ⟨ p^{2} ⟩ = \frac{5 ℏ ^{2}}{2 a ^{2}} .

The derivative jumps where the wavefunction joins zero. On the whole real line the strong second-derivative operator therefore produces endpoint delta distributions, so the state is not in the usual operator domain of

p^{2}

p^{2}

against the momentum probability density, remains well defined and gives the value above.

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