 Solution

ID: past-exam-of-the-mathematics-course-of-the-university-of-cambridge/2020/ib/paper-2/14a/a/solution

Codex  0 Created 2026-09-24 Updated 2026-09-29

j = \frac{ℏ}{2 mi} (ψ^{*} \frac{d ψ}{d x} - ψ \frac{d ψ ^{*}}{d x}) .

D e^{ik x}

carries current

(ℏ k / m) ∣ D ∣^{2}

to the right, while

D e^{- ik x}

carries the negative of this current. Thus the asymptotic terms are interpreted as an incident wave of amplitude

A

B

, and a transmitted wave of amplitude

C

. Their currents are

j_{in} = \frac{ℏ k}{m} ∣ A ∣^{2}, j_{ref} = - \frac{ℏ k}{m} ∣ B ∣^{2}, j_{tr} = \frac{ℏ k}{m} ∣ C ∣^{2} .

Current conservation for a real potential gives

j_{in} + j_{ref} = j_{tr}

, and therefore

P_{ref} = \frac{∣ B ∣ ^{2}}{∣ A ∣ ^{2}}, P_{tr} = \frac{∣ C ∣ ^{2}}{∣ A ∣ ^{2}},

with

P_{ref} + P_{tr} = 1

A normalizable bound state has

E < 0

κ = \frac{- 2 m E}{ℏ} > 0,

its decaying asymptotic behaviour is

ψ (x) \sim D_{-} e^{κ x} (x \to - \infty), ψ (x) \sim D_{+} e^{- κ x} (x \to + \infty) .

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