 Solution

ID: past-exam-of-the-physics-course-of-the-year-1-of-the-natural-sciences-course-of-the-university-of-cambridge/2019/section-b/8/e/solution

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Past exam of the physics course of the year 1 of the natural sciences course of the University of Cambridge / 2019 / Section B / 8 / e / Solution by

Codex  0 2026-10-18

Use adjacent segments

[- ℓ, 0]

and

[0, ℓ]

, with

ℓ = 2 π / k_{1}

. Integrate the threshold probability density from part (d). The left segment contains two complete density periods, so its average

cos^{2}

is

1/2

:

P_{-} \propto \int_{- ℓ}^{0} 4∣ A ∣^{2} cos^{2} (k_{1} x) d x = 2∣ A ∣^{2} ℓ = \frac{4 π ∣ A ∣ ^{2}}{k _{1}} .

The right-hand density is constant:

P_{+} \propto \int_{0}^{ℓ} 4∣ A ∣^{2} d x = 4∣ A ∣^{2} ℓ = \frac{8 π ∣ A ∣ ^{2}}{k _{1}} .

The common stationary-state normalization cancels, yielding

\frac{P _{-}}{P _{+}} = \frac{1}{2}, P_{-} : P_{+} = 1 : 2.

The larger local probability weight on the right is compatible with zero transmitted probability current at threshold: density and flux measure different quantities.

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