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ID: past-exam-of-the-options-ii-course-of-the-university-of-oxford/2012/j/j2/b/solution

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Past exam of the options II course of the University of Oxford / 2012 / J / J2 / b / Solution by

Codex  0 2026-10-31

The Born probability is the integral of

∣ ψ_{n} ∣^{2}

over the indicated interval. Changing variables to

ξ

cancels the dimensional normalization factors and gives

P_{0} (ℓ) = \frac{1}{π} \int_{- ℓ}^{ℓ} e^{- ξ^{2}} d ξ, P_{1} (ℓ) = \frac{2}{π} \int_{- ℓ}^{ℓ} ξ^{2} e^{- ξ^{2}} d ξ .

Since

\frac{d}{d ξ} (ξ e^{- ξ^{2}}) = e^{- ξ^{2}} - 2 ξ^{2} e^{- ξ^{2}}

, integration gives

P_{1} (ℓ) = P_{0} (ℓ) - \frac{2}{π} ℓ e^{- ℓ^{2}} .

ℓ \to \infty

, the correction term tends to zero and

P_{0} (ℓ) \to 1

by normalization. Hence

P_{1} (ℓ) \to 1

. This is expected because the expanding interval exhausts the whole position space and the state is normalized: the particle is somewhere on the line with total probability one.

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