 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/s18/2/e/solution

Multiply the expression for

ψ

F^{*} (x)

and integrate. Cancelling the nonzero

C

gives

1 = \frac{α m}{κ ℏ ^{2}} \int \int F^{*} (x) F (y) e^{- κ ∣ x - y ∣} d x d y .

The kernel is real and symmetric, so relabelling the integration variables puts the expression in the printed order:

κ = \frac{α m}{ℏ ^{2}} \int \int F (x) F^{*} (y) e^{- κ ∣ x - y ∣} d x d y, E = - \frac{ℏ ^{2} κ ^{2}}{2 m}, κ > 0.

This is the scalar consistency condition for the rank-one potential, rather than an independent normalization equation. The double integral is real and nonnegative because the Fourier transform of

e^{- κ ∣ x ∣}

2 κ / (q^{2} + κ^{2}) > 0

. Thus attractive

α > 0

is needed for a nontrivial solution. Under the usual integrability assumptions the corresponding free resolvent expectation decreases with

κ

, giving at most one bound state for this rank-one attraction.

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