 Solution

ID: past-exam-of-the-physics-a-course-of-the-university-of-cambridge/2021/paper-1/section-a/1/solution

The Born rule gives the required probability by integrating the probability density over the central interval. Put

d = 0.1 nm

P (∣ x ∣ \leq d) = \frac{2}{L} \int_{- d}^{d} cos^{2} (\frac{π x}{L}) d x = \frac{1}{L} \int_{- d}^{d} [1 + cos (\frac{2 π x}{L})] d x = \frac{2 d}{L} + \frac{1}{π} sin (\frac{2 π d}{L}) .

Substitution of

d / L = 0.1

gives

P = 0.2 + \frac{sin ( 0.2 π )}{π} ≃ 0.3871 ≃ 38.7%.

This exceeds the interval's

20%

fraction of the box length because the ground state probability density is largest at the centre; it is not a uniform position distribution.

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