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

Use the Hadamard gate identity $H=(X+Z)/\sqrt{2}$ 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 $\alpha |000⟩-\beta |111⟩$, with syndrome $00$. No bit-flip recovery is triggered, and decoding leaves $Z|\psi ⟩$ on the top wire.

By linearity of the full quantum circuit, its output for $H_1$ is thereforewhere 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 isThe 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.