 Solution

ID: past-exam-of-the-mathematics-course-of-the-university-of-cambridge/2024/ib/paper-4/15a/ii/solution

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

Put

B = ξ tan (ϕ + α t)

. Matching the constant and

tan^{2}

ξ α = - \frac{i}{2ℏ}, ξ α = - 2 i ℏ ξ^{2} .

One convenient choice is therefore

ξ = \frac{1}{2ℏ}, α = - i,

B (t) = \frac{1}{2ℏ} tan (ϕ - i t) .

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