 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/c/solution

Write

L_{0} ψ = - U ψ

and use the outgoing inverse of

L_{0}

. Adding the homogeneous incoming wave yields

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

For the attractive delta potential, let

η = m α / (k ℏ^{2})

. The equation reduces to

ψ (x) = e^{ik x} + i η e^{ik ∣ x ∣} ψ (0)

and

ψ (0) = 1/ (1 - i η)

. Separating the two half-lines gives

t = \frac{1}{1 - i η}, r = \frac{i η}{1 - i η}, ψ = {e^{ik x} + r e^{- ik x}, t e^{ik x}, x < 0, x > 0.

Both sides have the same group velocity, so

T = \frac{1}{1 + η ^{2}}, R = \frac{η ^{2}}{1 + η ^{2}}, T + R = 1.

The same amplitudes satisfy continuity and the exact delta-potential derivative jump

ψ^{'} (0^{+}) - ψ^{'} (0^{-}) = - 2 m α ψ (0) / ℏ^{2}

, providing an independent local check on the sign.

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