 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/2017/s18/1/3/solution

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Past exam of the short options paper shared by Prelims and Parts A and B of the physics course of the University of Oxford / 2017 / S18 / 1 / 3 / Solution by

Codex  0 2026-10-25

x \to - \infty

, the scattered kernel is

e^{- ik x} e^{ik x^{'}}

x \to + \infty

, it is

e^{ik x} e^{- ik x^{'}}

. Therefore

ψ_{s} \sim e^{ik x} + r e^{- ik x}

on the left and

t e^{ik x}

on the right, with

r = - \frac{im}{k ℏ ^{2}} \int e^{ik x^{'}} U (x^{'}) ψ_{s} (x^{'}) d x^{'}, t = 1 - \frac{im}{k ℏ ^{2}} \int e^{- ik x^{'}} U (x^{'}) ψ_{s} (x^{'}) d x^{'} .

Equal asymptotic wavenumbers give

R = ∣ r ∣^{2}

,

T = ∣ t ∣^{2}

. Squaring the first amplitude proves the requested expression. These are current ratios, which would require a velocity factor if the two asymptotic potentials differed.

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