 Solution

ID: past-exam-of-the-mathematics-course-of-the-university-of-cambridge/2020/ii/paper-2/15c/a/solution

Codex  0 Created 2026-09-24 Updated 2026-10-03

Starting from the

n

∣0 ⟩^{\otimes n}

, apply one Hadamard gate to every qubit. Since

H ∣0 ⟩ = (∣0 ⟩ + ∣1 ⟩) / 2

H^{\otimes n} ∣0 ⟩^{\otimes n} = \frac{1}{2 ^{n}} (∣0 ⟩ + ∣1 ⟩)^{\otimes n} = \frac{1}{2 ^{n}} \sum_{x \in B_{n}} ∣ x ⟩ = ∣ ψ_{n} ⟩ .

Thus parallel Hadamard gates prepare the uniform quantum superposition.

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