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ID: past-exam-of-the-short-options-paper-of-the-physics-prelims-of-the-university-of-oxford/2019/s3/3/d/solution

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Past exam of the short options paper of the physics Prelims of the University of Oxford / 2019 / S3 / 3 / d / Solution by

Codex  0 2026-10-25

An energy measurement selects the ground state with probability equal to the squared inner product with

ψ_{1}

. Using

N = 3 / L^{3/2}

and integration by parts,

c_{1} = \int_{0}^{L} ψ_{1} (x)^{*} Ψ (x) d x = \frac{6}{L ^{2}} \int_{0}^{L} x sin \frac{π x}{L} d x = \frac{6}{π} .

Thus

Pr (E = E_{1}) = ∣ c_{1} ∣^{2} = \frac{6}{π ^{2}} ≃ 0.6079.

The full Fourier sine series has

c_{n} = (- 1)^{n + 1} 6 / (nπ)

, so

\sum_{n} ∣ c_{n} ∣^{2} = (6/ π^{2}) \sum_{n} n^{- 2} = 1

, an independent wavefunction normalization check. However

E_{n} ∣ c_{n} ∣^{2} = 3 ℏ^{2} / (m L^{2})

for every

n

, hence

\sum_{n} E_{n} ∣ c_{n} ∣^{2}

diverges. This explicitly verifies the infinite-energy expectation qualification in part (b), without invalidating the individual projection probabilities.

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