N = 2^{m}

, the Hadamards first give

(H^{\otimes m} \otimes I) ∣0 ⟩^{\otimes m} ∣ ψ ⟩ = \frac{1}{N} \sum_{k = 0}^{N - 1} ∣ k ⟩ ∣ ψ ⟩ .

Since

U^{k} ∣ ψ ⟩ = e^{2 πik θ} ∣ ψ ⟩

and

θ = j / N

∣Φ ⟩ = \frac{1}{N} k = 0 \sum N - 1 e^{2 πijk / N} ∣ k ⟩ ∣ ψ ⟩ .

ii

Solution

The first register is

∣ ϕ_{j} ⟩ = \frac{1}{N} k = 0 \sum N - 1 e^{2 πijk / N} ∣ k ⟩ .

For

j, j^{'}

⟨ ϕ_{j^{'}} ∣ ϕ_{j} ⟩ = \frac{1}{N} \sum_{k = 0}^{N - 1} e^{2 πi (j - j^{'}) k / N} .

The geometric sum is one when

j = j^{'}

and zero otherwise. There are

N

such orthonormal vectors in an

N

-dimensional space, so they form an orthonormal basis.

The operator

F

maps the computational orthonormal basis

{∣ j ⟩}

to the orthonormal basis

{∣ ϕ_{j} ⟩}

. It therefore preserves inner products and is surjective, so

F^{†} F = F F^{†} = I

and

F

is unitary.

First apply

F^{†}

, which sends

∣ ϕ_{j} ⟩

∣ j ⟩

. Then measure in the computational basis. Because the phase has an exact

m

p_{j} = 1 .

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