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

Number the wires from top to bottom as $1,\dots ,5$. The first two controlled-NOT gates encode $|\psi ⟩=u|0⟩+v|1⟩$ into the bit-flip repetition codeon data wires $1,2,3$. The lower two zero ancillas store an error syndrome. Reading the PDF's four detection CNOTs givesThe syndrome and the data qubit to be flipped are thereforeThis is coherent syndrome extraction for the three-qubit repetition code. The recovery Toffoli gates implement these conditions; the intervening Pauli X gates temporarily turn zero-valued controls into one-valued controls and then restore the syndrome wires. Finally, inverse encoding returns the logical state to the top wire and resets data wires 2 and 3 to zero.

For the specified error $X_1$, the encoded data becomes $u|100⟩+v|011⟩$. Both terms have syndrome $01$, so detection factors out $|01{⟩}_{45}$ without revealing the logical amplitudes. Recovery flips data wire 1, restoring $u|000⟩+v|111⟩$, and decoding yieldsThe lower four wires need not all return to zero: their syndrome record is independent of the logical state, which is what the correction requirement needs.

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.

The point needing proof is that a corrected error's residual state cannot retain dependence on the logical input. Let $W_j$ denote the complete linear map from the logical input to all five output wires for error $E_j$. Protection gives $W_j|0⟩=|0⟩|{\eta }_{j,0}⟩$ and $W_j|1⟩=|1⟩|{\eta }_{j,1}⟩$. Applying protection to $|+⟩$ and comparing its two top-qubit components yieldsso $|{\eta }_{j,0}⟩=|{\eta }_{j,1}⟩=|{\eta }_{j,+}⟩$. Denote the common vector by $|{\eta }_j⟩$. Linearity then gives $W_j|\psi ⟩=|\psi ⟩|{\eta }_j⟩$ for every input.

The encoding and all operations surrounding the error are fixed. Replacing the error by $E=\alpha E_1+\beta E_2$ consequently replaces the overall map by $\alpha W_1+\beta W_2$, givingBecause $E$ and the surrounding circuit are unitary, the output is normalized for every normalized input; the residual vector in parentheses thus has norm one. This proves linearity of coherent quantum error correction and hence protection against $E$. Orthogonality of $|{\eta }_1⟩$ and $|{\eta }_2⟩$ is not required for this proof.

Use phase-flip protection by Hadamard conjugation. Insert a Hadamard gate on each of the three data wires immediately after the two encoding CNOTs and before the physical error. Insert another Hadamard on each data wire immediately after the error and before syndrome detection. Leave the two syndrome ancillas and every detection, recovery and decoding gate unchanged.

The first layer changes the encoded state to $\alpha |+++⟩+\beta |---⟩$, the three-qubit phase-flip repetition code. The two layers surrounding the error turn the effective error seen by the old circuit intoSince $H^2=I$ and $HZH=X$, the four physical cases become respectively $I,X_1,X_2,X_3$. The original coherent syndrome and recovery circuit corrects all four, as its syndrome table shows. Thus two Hadamard layers on the three data wires, straddling the error, give the required phase-flip protection. The final top wire is the original $|\psi ⟩$, not $H|\psi ⟩$, because the second layer returns the data to the original coding basis before recovery and decoding.