 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/2018/s18/1/d/ii/solution

Replace the slowly varying interior wave by

C

and

e^{ik ∣ x - x^{'} ∣}

by one when both arguments lie within the range. The integral equation gives

C = 1 - \frac{im}{k ℏ ^{2}} γ C, γ = \int_{- \infty}^{\infty} U (x) d x .

Hence

C = t ≃ \frac{1}{1 + imγ / ( k ℏ ^{2} )}, r ≃ C - 1.

The integrated strength has dimensions of energy times length. For

U = - α δ (x)

γ = - α

and this is exactly the result in part c. For a regular weak potential the approximation sums the important ratio

mγ / (k ℏ^{2})

without requiring it to be small, which is why it remains useful beyond the naive first Born expansion at small momentum.

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