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Past exam of the mathematics course of the University of Cambridge / 2021 / ib / Paper 2 / 15C / b / Solution Created 2026-09-24 Updated 2026-09-25

Put

k = \frac{2 m E}{ℏ}, K = \frac{2 m ( E + V _{0} )}{ℏ} .

The Time-independent Schrodinger equation has the forms

ψ (x) = ⎩ ⎨ ⎧ A e^{ik x} + B e^{- ik x}, C e^{i K x} + D e^{- i K x}, F e^{ik x}, x < 0, 0 \leq x \leq a, x > a .

The first region contains the incident and reflected waves, while the final region contains only the transmitted wave.

Continuity of

ψ

and

ψ^{'}

at both edges of the finite square well gives the standard transmission coefficient

T = \frac{∣ F ∣ ^{2}}{∣ A ∣ ^{2}} = [1 + \frac{( K ^{2} - k ^{2} ) ^{2}}{4 k ^{2} K ^{2}} sin^{2} (K a)]^{- 1} .

When

V_{0} = 3 E

, one has

K = 2 k

, and therefore

\frac{( K ^{2} - k ^{2} ) ^{2}}{4 k ^{2} K ^{2}} = \frac{9}{16}, K a = \frac{a 8 m E}{ℏ} .

Hence the transmission probability is

T = \frac{16}{16 + 9 sin ^{2} ( a 8 m E /ℏ )} .

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