 Solution

ID: past-exam-of-the-mathematics-course-of-the-university-of-cambridge/2022/ii/paper-4/33a/c/solution

Codex  0 Created 2026-09-24 Updated 2026-09-25

⟨ k, - ∣ x ∣0 ⟩ = 2 \frac{K}{π} \int_{0}^{\infty} x e^{- K x} sin (k x) d x = \frac{4 K k K / π}{( K ^{2} + k ^{2} ) ^{2}},

where differentiating the elementary Laplace integral for

sin (k x)

evaluates the last integral. Hence

∣ ⟨ k, - ∣ x ∣0 ⟩ ∣^{2} = \frac{16 K ^{3} k ^{2}}{π ( K ^{2} + k ^{2} ) ^{4}} .

Put

E_{f} = E_{0} + ℏ ω > 0

and

k_{f} = 2 m E_{f} /ℏ

\frac{s i n ^{2} ( Ω _{k} t /2 )}{( Ω _{k} /2 ) ^{2}} ⟶ 2 π t δ (Ω_{k}) .

Since

δ (Ω_{k}) = \frac{m}{ℏ k _{f}} δ (k - k_{f}),

the total escape probability becomes

P_{free} (t) = \int_{0}^{\infty} ∣ c_{k} (t) ∣^{2} d k \sim \frac{2 π F ^{2} t}{ℏ ^{2}} \frac{m}{ℏ k _{f}} \frac{16 K ^{3} k _{f}^{2}}{π ( K ^{2} + k _{f}^{2} ) ^{4}} = \frac{32 F ^{2} t m K ^{3} k _{f}}{ℏ ^{3} ( K ^{2} + k _{f}^{2} ) ^{4}} .

For the attractive delta potential,

∣ E_{0} ∣ = \frac{ℏ ^{2} K ^{2}}{2 m}, \frac{k _{f}^{2}}{K ^{2}} = \frac{E _{f}}{∣ E _{0} ∣} .

Substitution gives the requested leading-order result:

P_{free} (t) = \frac{8ℏ F ^{2} t}{m E _{0}^{2}} \frac{E _{f} /∣ E _{0} ∣}{( 1 + E _{f} /∣ E _{0} ∣ ) ^{4}} .

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