 Solution

ID: past-exam-of-the-mathematics-course-of-the-university-of-cambridge/2022/ib/paper-1/14b/d/solution

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

∣ ψ ∣^{2} = ∣ N ∣^{2} exp [- 2 Re F x^{2} + 2 Re G x] .

Re F > 0

∣ ψ ∣^{2} = C (t) exp [- 2 Re F (x - h)^{2}],

where

h (t) = \frac{Re G ( t )}{2 Re F ( t )} .

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