Past exam of the mathematics course of the University of Cambridge / 2025 / iii / Paper 324 / 1 / b / ii

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- Solution ii

Solution

The multiplicative order

r

makes the states

∣ a^{k} mod N ⟩

,

0 \leq k < r

, distinct and cyclic under

U_{a}

. Therefore

U_{a} ∣ ψ_{s} ⟩ = \frac{1}{r} k = 0 \sum r - 1 e^{- 2 πi s k / r} ∣ a^{k + 1} mod N ⟩ = e^{2 πi s / r} \frac{1}{r} j = 0 \sum r - 1 e^{- 2 πi s j / r} ∣ a^{j} mod N ⟩ .

Thus each

∣ ψ_{s} ⟩

is an eigenvector with

U_{a} ∣ ψ_{s} ⟩ = e^{2 πi s / r} ∣ ψ_{s} ⟩ .

These are the Fourier eigenvectors of the cyclic modular-multiplication orbit.

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