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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.

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.

Every random unitary channel satisfiesIt is therefore a unital quantum channel.