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For the quantum depolarizing channel applying the identity with probability $1-p$ and each nonidentity Pauli operator with probability $p/3$, the entanglement-assisted classical capacity is the displayed number of bits per use. Here $H_4$ is Shannon entropy of the four listed probabilities. Its unassisted product-state capacity is $1-h_2(2p/3)$. Both vanish at $p=3/4$. If $q$ instead denotes replacement by $I/2$, then $p=3q/4$ and complete depolarization is at $q=1$. The original entanglement-assisted capacity paper gives the depolarizing-channel formula.

With Bloch vector retention $\eta \to 0$, the unassisted product-state capacity of a quantum depolarizing channel is ${\eta }^2/(2\ln 2)+O({\eta }^4)$. Its entanglement-assisted capacity is $3{\eta }^2/(2\ln 2)+O({\eta }^3)$, obtained by expanding Shannon entropy around the uniform four-outcome distribution. Consequently their limiting ratio is three, independent of the linear noise-parameter convention.

Conjugation by one Pauli matrix fixes that same matrix and reverses the other two. For example, $XXX=X$, $YXY=-X$, and $ZXZ=-X$. Applying this to each component of the Bloch vector givesHence the quantum depolarizing channel isThe Kraus operators $\sqrt{1-p}I$ and $\sqrt{p/3}{\sigma }_j$ establish complete positivity for $0\le p\le 1$. At $p=3/4$ every state becomes maximally mixed. For $p>3/4$ the factor is negative: it reverses the vector as well as reducing its length. Its most negative value is $-1/3$, consistent with the complete-positivity range of the isotropic qubit map.

Fix the noise convention explicitly: $p$ is the total probability of a nonidentity Pauli error. The memoryless qubit quantum depolarizing channel iswith independent identical use on successive qubits, namely ${\mathcal{D}}_p^{\otimes n}$ on an $n$-qubit block. Its normalized Kraus operators are $\sqrt{1-p}I,\sqrt{p/3}X,\sqrt{p/3}Y,\sqrt{p/3}Z$.

Write the Bloch sphere representation of a density matrix as $\rho =(I+\mathbf{r}\cdot \mathbf{\sigma })/2$, with $|\mathbf{r}|\le 1$. Its eigenvalues are $(1\pm |\mathbf{r}|)/2$, proving exactly this positivity condition. Conjugation by one Pauli operator preserves that coordinate of the Bloch vector and reverses the other two. Summing the three conjugations givessoThe Bloch ball contracts isotropically, collapses to the maximally mixed state at $p=3/4$, and contracts with reversal for $p>3/4$. That last interval still represents a physical Pauli mixture. If instead $q$ is the probability of replacing the state by $I/2$, then $\mathcal{D}(\rho )=(1-q)\rho +qI/2$, with $p=3q/4$ and $\eta =1-q$; the usual replacement-probability range $0\le q\le 1$ stops at complete depolarization.

Let $h_2(x)=-x{\log }_{2}x-(1-x){\log }_2(1-x)$, with endpoint limits. In bits per use, the unassisted product-state classical capacity and the entanglement-assisted capacity of a qubit depolarizing channel arewhere $H_4$ is Shannon entropy of the listed four probabilities. Equivalently $C_E=2+(1-p){\log }_2(1-p)+p{\log }_2(p/3)$. The first quantity is a classical capacity with product quantum inputs and potentially collective output measurements, not the quantum capacity. These are the requested capacity formulae; a proof of their coding theorems is not needed here. The maximally entangled input gives a Bell-diagonal joint state with those four probabilities, explaining the entropy appearing in $C_E$. The original entanglement-assisted capacity calculation gives the depolarizing-channel expression.

At $p=0$ the capacities are one and two. Both vanish at $p=3/4$. On the full Pauli-error interval they increase again beyond that point, reaching $1-h_2(2/3)$ and $2-{\log }_{2}3$ at $p=1$; on the replacement-probability plot both decrease to zero at $q=1$. The two panels make this convention distinction explicit.

For the high-noise capacity ratio for qubit depolarization, put $\eta =1-4p/3\to 0$. Taylor expansion of binary entropy around one half givesThe four probabilities for $C_E$ are $1/4+3\eta /4$ and three copies of $1/4-\eta /4$. For a perturbation ${\delta }_i$ of the uniform four-outcome distribution, with ${\sum }_i{\delta }_i=0$, entropy expansion givesHere ${\sum }_i{\delta }_i^2=3{\eta }^2/4$, hence $C_E=3{\eta }^2/(2\ln 2)+O({\eta }^3)$. ThereforeThe reciprocal ratio is $1/3$, and the same enhancement factor occurs as $q\to 1$ in the replacement convention.

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.

Conjugation by a Pauli operator preserves its matching Bloch vector component and reverses the other two. Summing all three conjugations therefore sends $\mathbf{s}$ to $-\mathbf{s}$. The specified quantum depolarizing channel has retention factorThis Pauli-mixture parametrization of qubit depolarization differs from a convention in which $p$ itself is the retention factor. For $0\le p\le 1$, $-1/3\le \eta \le 1$. A pure input gives output eigenvalues $(1\pm |\eta |)/2$, and hence entropy $h_2((1+\eta )/2)$ by symmetry of binary entropy. Mixed inputs have a smaller output radius and at least this much entropy. Every average output has entropy at most one. ThusTwo equiprobable orthogonal pure inputs have antipodal output Bloch vectors, maximally mixed average output, and this same minimal individual output entropy. They attain the bound. Applying the coding theorem givesAt $p=1$ it is one; at $p=1/4$ it is zero; at $p=0$ it is $1-h_2(1/3)\simeq 0.0817$. Negative retention reverses the labels of the two signals but does not eliminate their distinguishability.

Write a qubit density operator in Bloch vector form, $\rho =(I+\mathbf{r}\cdot \mathbf{\sigma })/2$, with $|\mathbf{r}|\le 1$. The quantum depolarizing channel maps $\mathbf{r}$ to $p\mathbf{r}$, so its output eigenvalues are $(1\pm p|\mathbf{r}|)/2$. The output Von Neumann entropy is minimized on pure inputs, where it equals $h_2((1+p)/2)$.

For any input ensemble, its output Holevo quantity is at most the maximum qubit entropy minus this minimum output entropy:Equality is attained by two equiprobable orthogonal pure inputs: their average output is $I/2$, while each output has the minimum entropy. ThusThis uses the question's retention parameter $p$; in the usual convex-mixture range, $0\le p\le 1$. The expression also holds throughout the full completely positive qubit range $-1/3\le p\le 1$, since $h_2((1+p)/2)=h_2((1+|p|)/2)$.

The supplied additivity of Holevo capacity gives ${\chi }^{\ast }({\Lambda }_{\mathrm{dep}}^{\otimes n})=n{\chi }^{\ast }({\Lambda }_{\mathrm{dep}})$. The Holevo-Schumacher-Westmoreland theorem expresses the unassisted classical capacity of a quantum channel as the regularized Holevo capacity, soTherefore entangled inputs across channel uses cannot increase this classical capacity.