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ID: past-exam-of-the-mathematics-course-of-the-university-of-cambridge/2010/iii/paper-49/3/b/solution

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Past exam of the mathematics course of the University of Cambridge / 2010 / iii / Paper 49 / 3 / b / Solution by

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

Use the Hadamard gate identity

H = (X + Z) / 2

on the first data wire. The

X_{1}

branch is corrected as above and has syndrome

01

. The

Z_{1}

branch changes the encoded state to

α ∣000 ⟩ - β ∣111 ⟩

, with syndrome

00

. No bit-flip recovery is triggered, and decoding leaves

Z ∣ ψ ⟩

on the top wire.

By linearity of the full quantum circuit, its output for

H_{1}

is therefore

\frac{1}{2} (∣ ψ ⟩ \otimes ∣0001 ⟩ + Z ∣ ψ ⟩ \otimes ∣0000 ⟩),

where the four-bit strings are ordered as wires

2, 3, 4, 5

. The two syndrome states are orthogonal. Taking their partial trace removes the cross terms, so the reduced density matrix is

ρ = \frac{1}{2} (∣ ψ ⟩ ⟨ ψ ∣ + Z ∣ ψ ⟩ ⟨ ψ ∣ Z) = (∣ α ∣^{2} 0 0 ∣ β ∣^{2}) .

The top output undergoes a completely dephasing channel in the computational basis. The bit-flip component is corrected, but the undetected phase component destroys logical coherence after the syndrome is ignored.

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