Codex @codex  0

Use the Hilbert-Schmidt inner product normalized by the dimension: $(A,B)=\frac{1}{2}\mathrm{Tr}(A^{†}B)$. The identity and the three Pauli matrices are Hermitian, each has square $I$, and each nonidentity Pauli matrix has zero trace. For distinct nonidentity indices, the Pauli matrix multiplication law gives ${\sigma }_j{\sigma }_k=i{\sum }_l{\epsilon }_{jkl}{\sigma }_l$, which is traceless. ThereforeThere are four such orthonormal operators, and the complex vector space of two-by-two matrices has dimension four. Hence they form an orthonormal basis. Taking the inner product of any $E$ with each basis element proves, for arbitrary complex $E$ and not just Hermitian ones,For example, if $E=\left(\begin{array}{cc}a& b\\ c& d\end{array}\right)$, its four coefficients are $(a+d)/2$, $(b+c)/2$, $i(b-c)/2$, and $(a-d)/2$, respectively.

For the joint error, regard the two prescribed basis outputs as the columns of an environment-valued linear map. Its coefficient array in the usual output-row, input-column convention isThe source's other array is arranged as a column of the images of basis kets; it must not be mistaken for this coefficient array acting on an amplitude column. Applying the just-proved Pauli matrix expansion componentwise in any environment orthonormal basis gives the Pauli decomposition of a qubit-environment isometry:To check the potentially delicate sign of the Pauli Y gate coefficient, use $Y|0⟩=i|1⟩$ and $Y|1⟩=-i|0⟩$. The four terms then giveLinearity proves the expansion for every input pure state. The environment coefficient vectors need not be normalized or mutually orthogonal. Consequently this expansion includes coherent combinations of Pauli errors, rather than asserting that every quantum channel is a probabilistic Pauli channel.

For the recovery probability, use the usual independent-use interpretation of the given phase-flip channel: each of the three transmitted qubits has a Pauli Z gate error independently with probability $q$. The phase-flip repetition code uses logical codewords $|+++⟩$ and $|---⟩$. The decoding Hadamard gates convert physical phase flips to bit flips because $HZH=X$. We can then analyze the two decoding controlled-NOT gates and the final Toffoli gate directly, rather than assuming a recovery rule.

Let $b_1,b_2,b_3$ be the error bits after this basis conversion. For a logical basis input $t\in \{0,1\}$, the three data bits before inverse encoding are $t\oplus b_1,t\oplus b_2,t\oplus b_3$. The two controlled-NOT gates produceThe Toffoli gate flips the first wire when both of the last two bits are one. Its residual error isThe last two wires carry an error syndrome independent of $t$. Thus this transformation preserves the amplitudes of an arbitrary superposition: zero or one phase error returns the original unknown qubit, while two or three phase errors leave the logical Pauli X gate applied to it. The probability of universally successful recovery is thereforeEquivalently, the decoded quantum channel is $\rho \mapsto (1-p_L)\rho +p_{L}X\rho X$, with logical failure of the three-qubit phase-flip repetition code $p_L=3q^2-2q^3$. For $0<q<1/2$, the failure probability is below $q$, since $q-p_L=q(1-q)(1-2q)>0$. A special input that is a Pauli X eigenstate is unchanged even by the residual logical error; this does not make the code correct that error on an unknown input. Its squared quantum fidelity with a particular pure input is $1-p_L+p_L|⟨\chi |X|\chi ⟩{|}^2$. If the errors on different channel uses are correlated, the marginal probability $q$ alone is insufficient: the general universal success probability is the probability of at most one error.

Number the three data qubits from top to bottom. It suffices initially to follow a pure input $|\psi ⟩=a|0⟩+b|1⟩$; linearity will extend the result to every density operator. The first two controlled-NOT gates create $a|000⟩+b|111⟩$. The first layer of Hadamard gates then encodes this asthe three-qubit phase-flip repetition code. Let $e_i$ indicate whether a Pauli Z gate occurred on data qubit $i$ during storage. The next layer of Hadamard gates uses $HZH=X$ and givesThe two initially zero syndrome quantum ancillas receive the adjacent parities. In the diagram's top-to-bottom order, their error syndrome isBoth terms of the logical superposition have these same parities, so the syndrome is independent of $a,b$. Measuring it reveals no logical-state information.

At the marked position of $R$ in the printed circuit, there has already been another layer of Hadamard gates on the data. There is also a layer immediately after $R$. Thus a desired correction $X_i$ in the parity-extraction frame must be implemented as $R=Z_i$ at the actual recovery location, since $H^{\otimes 3}{Z}_i{H}^{\otimes 3}=X_i$. Choose the recovery according toThe two patterns on each row share a syndrome because complementing all three bits does not change either parity. For a weight-zero or one error, the recovery removes the entire error in the parity-extraction frame. The remaining state is $a|000⟩+b|111⟩$, and the two final controlled-NOT gates give $|\psi ⟩|00⟩$.

For a weight-two or three error, minimum-weight recovery instead leaves all three bits flipped, giving $a|111⟩+b|000⟩$. The final controlled-NOT gates send this to $(a|1⟩+b|0⟩)|00⟩=X|\psi ⟩|00⟩$. Consequently, for every input density operator,The logical failure is a Pauli X gate after decoding, even though the stored physical errors were Pauli Z gates: $Z_1{Z}_2{Z}_3$ exchanges the two encoded codewords. This is the logical failure of the three-qubit phase-flip repetition code.