Past exam of the mathematics course of the University of Cambridge / 2020 / ii / Paper 4 / 10C / iii

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- Solution iii

Solution

Applying the transform twice gives

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

By the finite geometric series, the inner sum over

y

is

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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