 Solution

ID: past-exam-of-the-mathematics-course-of-the-university-of-cambridge/2022/ii/paper-1/35d/c/solution

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

Put

κ = \frac{2 m ( V _{0} - E )}{ℏ} .

An even solution is proportional to

cosh (κ x)

inside the potential barrier. Normalize the right exterior region as

ψ (x) = e^{- ik x} + (r + t) e^{ik x} (x > a /2) .

Continuity of

ψ

and

ψ^{'}

x = a /2

gives, with

γ = κ tanh (κa /2)

\frac{ik ( ( r + t ) e ^{ika /2} - e ^{- ika /2} )}{( r + t ) e ^{ika /2} + e ^{- ika /2}} = γ .

Solving,

r + t = e^{- ika} \frac{ik + γ}{ik - γ}, γ = κ tanh (κa /2) .

Its modulus is one, as expected for the even-parity eigenvalue of the unitary S-matrix.

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