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

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.