 Solution

ID: past-exam-of-the-mathematics-course-of-the-university-of-cambridge/2018/ii/paper-3/15d/b/iv/solution

Codex  0 Created 2026-09-24 Updated 2026-10-03

Here

N = 12

r = 4

A = N / r = 3

, and

x_{0} = 1

ω = e^{2 πi /12}

QFT_{12} ∣ α ⟩ = \frac{1}{2} l = 0 \sum 3 ω^{3 l} ∣3 l ⟩ = \frac{1}{2} (∣0 ⟩ + i ∣3 ⟩ - ∣6 ⟩ - i ∣9 ⟩) .

Consequently,

Pr (c) = {\frac{1}{4}, 0, c \in {0, 3, 6, 9}, otherwise .

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