 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/s18/1/4/solution

Put

β = m α / ℏ^{2}

. The delta interaction reduces the integral equation to

ψ_{s} = e^{ik x} + i β e^{ik ∣ x ∣} ψ_{s} (0) / k

. At zero,

ψ_{s} (0) = (1 - i β / k)^{- 1}

. Hence

t = \frac{k}{k - i β}, r = \frac{i β}{k - i β}, T = \frac{k ^{2}}{k ^{2} + β ^{2}}, R = \frac{β ^{2}}{k ^{2} + β ^{2}} .

Their sum is one, as required for a real potential. The explicit wave is the indicated incident-plus-reflected form for

x < 0

and

t e^{ik x}

for

x > 0

; it is continuous and has derivative jump

- 2 β ψ (0)

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