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The unassisted classical capacity is the supremum of classical bit rates achievable with asymptotically vanishing message error. It allows entanglement among different channel inputs, but no preshared sender-receiver entanglement. The Holevo-Schumacher-Westmoreland theorem gives its regularized entropy expression.

The Holevo capacity of a quantum channel is the maximum one-use output Holevo quantity,It is the unassisted classical communication limit for product state encodings with collective output measurements. Arbitrary entangled inputs require the regularized expression in the Holevo-Schumacher-Westmoreland theorem.

The Holevo-Schumacher-Westmoreland theorem gives the classical coding formulaIts ensemble-coding statement makes every rate below a one-use output Holevo quantity achievable with product input encodings and collective output decoding. Under this product-input restriction, the capacity isFor any such input ensemble, rates below its output Holevo quantity are achievable with asymptotically vanishing decoding error. Maximization gives the product-state classical capacity. The converse follows from the Holevo bound, entropy bounds on product outputs, and Fano's inequality.

Define the binary entropy $h_2(x)=-x{\log }_{2}x-(1-x){\log }_2(1-x)$. For an input Bloch radius $r\le 1$, the output eigenvalues are $(1\pm (1-q)r)/2$. Its entropy isOn $[1/2,1]$, $h_2$ is decreasing, so this is minimized by pure inputs $r=1$. Throughout the completely positive interval $0\le q\le 4/3$, the minimum equals $h_2(q/2)$, using symmetry $h_2(x)=h_2(1-x)$. The average output entropy is at most $1$, because it is a qubit. Hence every ensemble satisfiesTwo equally likely orthogonal pure inputs attain this bound: their average output is $I/2$, and each individual output has entropy $h_2(q/2)$. ThereforeThis gives the Holevo capacity of a qubit depolarizing channel in the noise-weight convention used here.

For completeness, correlated classical code labels do not evade this upper bound. Each product input to $n$ uses has product output entropy at least $n{h}_2(q/2)$, whereas the average output entropy is at most $n$. Its Holevo quantity is consequently at most $n[1-h_2(q/2)]$. The Holevo bound and Fano's inequality rule out a reliable rate larger than this value. Combined with HSW achievability, this proves the capacity under the stated product-input restriction. It also remains valid in the extended physical parameter range arising from the original $p$ assumption.

The printed normalization has its factors reversed. For the displayed map $\Phi (X)={\sum }_k{A}_{k}X{A}_k^{†}$, the condition ${\sum }_k{A}_k{A}_k^{†}=I$ says $\Phi (I)=I$, which is unitality. Trace preservation instead requiresThese are distinct trace-preserving and unital Kraus conditions. To disprove the printed equivalence, take $0<\gamma <1$ andThey obey ${\sum }_j{B}_j{B}_j^{†}=I$, butThus their completely positive map is not trace preserving. They are the adjoints of the operators of an amplitude damping channel. That channel is trace preserving but sends $I$ to $\mathrm{diag}(1+\gamma ,1-\gamma )$, so it cannot satisfy the printed normalization in any Kraus representation either. The corrected characterization is proved next.

Work in finite-dimensional input and output Hilbert spaces, with a linear map $\Phi$ on operators. Suppose first that $\Phi (X)={\sum }_k{A}_{k}X{A}_k^{†}$ and ${\sum }_k{A}_k^{†}{A}_k=I$. For every auxiliary dimension $m$ and every positive operator $W$ on the extended space,Each summand is positive because conjugation preserves positivity. This is exactly the definition of a completely positive map. Cyclicity of the trace givesso the map is also trace preserving.

Conversely, suppose $\Phi$ is completely positive and trace preserving, and let $|i⟩$ be an orthonormal input basis. Use the unnormalized entangled vector $|\Omega ⟩={\sum }_i|i⟩\otimes |i⟩$, with the reference factor first. Complete positivity makes the corresponding unnormalized Choi matrix positive:Its spectral decomposition can be written $J_{\Phi }={\sum }_k|v_k⟩⟨v_k|$, absorbing the nonnegative eigenvalues into the vectors. Define linear operators $A_k$ uniquely byExpanding the outer products and comparing their $(i,j)$ reference blocks yieldsThe matrix units span all operators, so linearity proves the Kraus representation $\Phi (X)={\sum }_k{A}_{k}X{A}_k^{†}$ for every $X$. Finally, trace preservation givesfor all $X$. Taking matrix units as test operators shows that the matrix in parentheses is zero. This proves both directions of the corrected CPT characterization, including the normalization rather than assuming it.

For an input quantum state ensemble $\{p_x,{\rho }_x\}$, write ${\sigma }_x=\Phi ({\rho }_x)$ and $\bar{\sigma }={\sum }_x{p}_x{\sigma }_x$. Its output Holevo quantity isThe Holevo capacity is the optimized one-use quantitywhere the supremum ranges over finite input ensembles and $S$ is the Von Neumann entropy in bits.

For $n$ uses of a memoryless quantum channel, the map is ${\Phi }^{\otimes n}$. An $n$-use code consists of a finite message set ${\mathcal{M}}_n$ of size $M_n$, an encoding state ${\rho }_m^{(n)}$ on the $n$ inputs for each message, and a decoding POVM $\{D_m^{(n)}:m\in {\mathcal{M}}_n\}$ on the output space, with $D_m^{(n)}\ge 0$ and ${\sum }_m{D}_m^{(n)}=I$. The encoding states may be entangled across uses. The measurement outcome is the receiver's decoded message, with conditional probabilityFor uniformly distributed messages the average error and rate areA rate $R$ is achievable if there is a sequence of such codes with $P_e^{(n)}\to 0$ and ${\mathrm{liminf}}_n{R}_n\ge R$. The unassisted classical capacity of a quantum channel is the supremum of achievable rates; the sender and receiver do not share prior entanglement.

The operational content of the Holevo-Schumacher-Westmoreland theorem is that ${\chi }^{\ast }(\Phi )$ is the capacity for product-state input encodings with collective output measurements. Allowing arbitrary entangled block inputs gives the regularized formulaThe limit equals the supremum because superadditivity of Holevo capacity follows by taking product ensembles for two blocks and using additivity of the Von Neumann entropy on product states. The coding theorem makes every rate below the one-block Holevo quantity achievable through many independently encoded blocks, with collective decoding.

The converse can be seen explicitly from the Holevo bound and Fano's inequality. For a uniform transmitted message $M$ and decoded message $\hat{M}$,The first inequality is obtained by subtracting the Fano bound on $H(M\mid \hat{M})$ from $H(M)={\log }_2{M}_n$. Dividing by $n$ and letting the average error vanish bounds every achievable rate by the regularized expression. The block-coding achievability just described supplies the reverse bound.

Under the assumed additivity of Holevo capacity across tensor powers, ${\chi }^{\ast }({\Phi }^{\otimes n})=n{\chi }^{\ast }(\Phi )$ for every $n$. ThusThe stated conditional assertion is true: under this additivity assumption, entangled input states cannot increase the asymptotic unassisted classical rate beyond the rate already achievable with product inputs. The decoder may still use collective measurements across channel outputs.

For a finite-dimensional memoryless quantum channel $\Lambda$, the Holevo-Schumacher-Westmoreland theorem identifies its product-state classical capacity with the optimized output Holevo quantity:With product codewords and a collective measurement on the outputs, every rate below this value is achievable with vanishing error; no larger rate can be reliable under that product-input restriction. For a fixed classical-quantum output alphabet the corresponding optimized entropy difference gives its classical coding capacity. If entangled inputs across channel uses are also allowed, the general unassisted capacity is the regularized value $C(\Lambda )={\lim }_{n\to \infty }{n}^{-1}{\chi }^{\ast }({\Lambda }^{\otimes n})$. The requested calculation below concerns the single-use optimized product-input expression, so no unproved additivity assumption is needed.

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.