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The displayed qubit quantum channel is completely positive for $0\le q\le 4/3$. Indeed its Pauli channel probabilities are $1-3q/4$ for the identity and $q/4$ for each nonidentity Pauli operator. In the range $q>1$ the coefficient $1-q$ is negative, so the displayed expression is not a convex mixture of the unchanged and maximally mixed states, even though the Pauli representation is a valid random unitary channel. Its Bloch vector is contracted and reversed by the factor $1-q$. The Holevo capacity of a qubit depolarizing channel remains $1-h_2(q/2)$ throughout this interval.

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.

There is a normalization error in the original PDF: the first term contains $I$ rather than $\rho$. For a normalized qubit density operator, that printed expression has trace $2(1-p)+p=2-p$, so it is not a quantum channel except at $p=1$. The intended depolarizing channel is the Pauli channel that leaves the input unchanged with probability $1-p$ and applies each nonidentity Pauli matrix with probability $p/3$, for $0\le p\le 1$. Its Kraus operators areand ${\sum }_{k=0}^3{K}_k^{†}{K}_k=I$, proving that it is trace-preserving and a completely positive map.

Write the input in its Bloch vector representation,Conjugation by a Pauli matrix keeps its own Bloch vector component and reverses the other two. More explicitly, ${\sigma }_k{\sigma }_j{\sigma }_k={\sigma }_j$ for $j=k$ and $-{\sigma }_j$ for $j\ne k$. HenceThe corrected quantum channel therefore givesThus its action isThe Bloch sphere is sent to a sphere of radius $|1-4p/3|$ about the origin of the Bloch ball. At $p=0$ the action is the identity. For $0<p<3/4$ the Bloch vector shrinks without changing direction. At $p=3/4$ every input becomes the maximally mixed density operator $I/2$. For $3/4<p\le 1$, the Bloch vector reverses direction and shrinks, reaching factor $-1/3$ at $p=1$. The reversal is compatible with complete positivity because this remains a probabilistic mixture of Pauli matrices. The parameter here is the total nonidentity-error probability, not the retention parameter used in some definitions of the quantum depolarizing channel.

For completeness, retain the PDF literally and call its output $F_p(\rho )$. The same calculation givesThis has no normalized-output Bloch ball interpretation as printed. If it is manually normalized, thenThat different transformation sends $p=0$ to $I/2$, rather than to the input, and agrees with the intended answer only at $p=1$. The literal and corrected conventions therefore cannot be silently identified.

The PDF's first Pauli term contains an undefined ${\rho }_x$. For the stated depolarizing channel, use ${\sigma }_x\rho {\sigma }_x$, consistently with the other two terms. This is the minimal typographical correction needed to define the intended Pauli channel.

Write $\rho =\frac{1}{2}(I+\mathbf{s}\cdot \mathbf{\sigma })$ using its Bloch vector. The Pauli matrices satisfy ${\sigma }_j{\sigma }_k{\sigma }_j={\sigma }_k$ for $j=k$ and $-{\sigma }_k$ for $j\ne k$. Thus conjugating by each nonidentity Pauli matrix leaves one Bloch component unchanged and reverses the other two. Summing the three conjugates givesSubstitution yieldsso the Pauli-mixture parametrization of qubit depolarization, with $p$ here the total error probability, givesThere is also a genuine range error in the printed claim. For its full $0<p<1$ assumption, $0<q<4/3$, and $0<q<1$ holds exactly when $0<p<3/4$. For example $p=9/10$ gives $q=6/5$; on $|0⟩⟨0|$ the output is $\mathrm{diag}(2/5,3/5)$, with negative $z$ Bloch component. No $q\in (0,1)$ could produce this output from that input, since its component would be $1-q>0$.

The corrected channel is still completely positive in the larger range: its original Pauli probabilities are $1-p,p/3,p/3,p/3$. This is depolarizing noise weight above one; the formula with $q>1$ is not a convex mixture with weights $1-q,q$, but it has a valid random unitary channel representation. The subsequent parts can therefore be solved both in the printed $q\in (0,1)$ regime and in the actual parameter range.

The four index pairs giveThe last global phase cancels under conjugation. Hence the Pauli channel probabilities are

Every linear map from a qubit into the tensor product of that qubit and an environment can be expanded in the Pauli matrix orthonormal basis of operators. If $V|0⟩=|0⟩|e_{00}⟩+|1⟩|e_{01}⟩$ and $V|1⟩=|0⟩|e_{10}⟩+|1⟩|e_{11}⟩$, then the four coefficient vectors areThese environment vectors need not be orthogonal, so this is an operator expansion rather than a claim that the quantum channel is a classical Pauli channel. Physical isometries additionally preserve the inner products of the two input basis states.

The tensor products of single-qubit Pauli matrices form an orthogonal basis of the operator space on $n$ qubits, so every linear error operator has a Pauli expansion. Using $X^a{Z}^b$, with scalar phases absorbed into the coefficients, its computational basis action is $E|x⟩={\sum }_{a,b}{c}_{a,b}(-1{)}^{b\cdot x}|x+a⟩$, with binary arithmetic. This describes coherent errors as well as individual Pauli errors; it does not assume that the noise is a probabilistic Pauli channel. A physical noise map has Kraus operators of this form when the usual quantum channel description applies, for instance when the initial system and environment are uncorrelated.