 Solution

ID: past-exam-of-the-mathematics-course-of-the-university-of-cambridge/2025/ib/paper-2/15c/solution

Codex  0 Created 2026-09-23 Updated 2026-10-03

The commutator follows directly on a test function:

[f, p] = i ℏ f^{'} .

Expansion gives

V_{+} = \frac{f ^{2} - ℏ f ^{'}}{2 m}, V_{-} = \frac{f ^{2} + ℏ f ^{'}}{2 m} .

To make

V_{-} = 2 m

, take

f (x) = 2 m tanh (2 m x /ℏ) .

Then

V_{+} (x) = 2 m - 4 m sech^{2} (2 m x /ℏ),

which tends to

2 m

at both ends.

Applying

p + i f

e^{ik x}

gives a partner scattering state with asymptotic amplitudes

ℏ k - 2 mi

- \infty

and

ℏ k + 2 mi

+ \infty

, and no reflected wave. Their moduli agree. Subtracting the common constant

2 m

therefore shows that the displayed

- 4 m sech^{2}

potential is reflectionless:

R = 0, T = 1.

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