 Solution

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

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

Write

T = tanh (a x)

and

S = sech^{2} (a x)

. Using

T^{'} = a S, S^{'} = - 2 a ST,

two differentiations of

ψ (x) = N e^{ik x} (a T - ik)

give

- \frac{ℏ ^{2}}{2 m} ψ^{''} - \frac{ℏ ^{2} a ^{2}}{m} S ψ = \frac{ℏ ^{2} k ^{2}}{2 m} ψ .

Thus it solves the time-independent Schrodinger equation in the stated Pöschl-Teller potential, with

E = \frac{ℏ ^{2} k ^{2}}{2 m} .

x \to - \infty

and

x \to + \infty

, respectively,

ψ (x) \sim N (- a - ik) e^{ik x}, ψ (x) \sim N (a - ik) e^{ik x} .

There is no left-moving term, so this is a reflectionless potential:

P_{ref} = 0 .

The incident and transmitted amplitudes have equal modulus because

∣ - a - ik ∣^{2} = ∣ a - ik ∣^{2} = a^{2} + k^{2} .

Hence

P_{tr} = 1 .

k = iλ

with

λ > 0

. Then

ψ (x) = N e^{- λ x} (a tanh (a x) + λ), E = - \frac{ℏ ^{2} λ ^{2}}{2 m} .

+ \infty

this decays for every positive

λ

. At

- \infty

, however, the factor tends to

λ - a

while

e^{- λ x}

grows. Normalizability therefore requires

λ = a

. In that case

ψ (x) = N a e^{- a x} (1 + tanh a x) = N a sech (a x),

which decays at both ends. The bound-state energy is

E = - \frac{ℏ ^{2} a ^{2}}{2 m} .

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