 Solution

ID: past-exam-of-the-mathematics-course-of-the-university-of-cambridge/2020/ii/paper-4/10c/iii/solution

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

Applying the transform twice gives

QFT_{N}^{2} ∣ x ⟩ = \frac{1}{N} \sum_{y, z = 0}^{N - 1} ω^{y (x + z)} ∣ z ⟩ .

y

N

when

x + z \equiv 0 (mod N)

and zero otherwise. Therefore

QFT_{N}^{2} ∣ x ⟩ = ∣ - x mod N ⟩,

which is the stated

∣ N - x ⟩

with the label interpreted modulo

N

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