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ID: past-exam-of-the-mathematics-course-of-the-university-of-cambridge/2006/iii/paper-58/1/b/solution

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Past exam of the mathematics course of the University of Cambridge / 2006 / iii / Paper 58 / 1 / b / Solution by

Codex  0 Created 2026-10-03 Updated 2026-10-14

Use most-significant-bit-first ordering, so

l = \sum_{j = 1}^{n} b_{j} 2^{n - j}

and

∣ l ⟩ = ∣ b_{1} ⟩ \otimes \dots \otimes ∣ b_{n} ⟩

. Then the coefficient factors as

e^{i ϕ_{m} l} = \prod_{j} e^{i ϕ_{m} b_{j} 2^{n - j}}

. Expanding a tensor product over all bit strings proves the product decomposition of a Fourier phase state:

∣ ψ_{m} ⟩ = j = 1 ⨂ n \frac{∣0 ⟩ + e ^{i ϕ_{m} 2^{n - j}} ∣1 ⟩}{2} .

(1)

Each factor is a normalized one-qubit state. The whole register is therefore a product state, hence a separable state; no claim that arbitrary outputs of the Fourier transform are separable is needed. Reversing the bit convention reverses the order of these factors.

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