 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/s18/1/2/solution

Write the full equation as

L ψ_{s} = - U ψ_{s}

. Apply the outgoing inverse of

L

and add the incident homogeneous solution with unit left-incident amplitude. This gives

ψ_{s} (x) = e^{ik x} - \frac{im}{k ℏ ^{2}} \int e^{ik ∣ x - x^{'} ∣} U (x^{'}) ψ_{s} (x^{'}) d x^{'} .

For sufficiently short-range potentials this is the outgoing Lippmann-Schwinger equation. The minus sign follows from the definition

L = H_{0} - E

, rather than

E - H_{0}

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