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Put $u=\epsilon /2$. The state is well defined for $0\le \epsilon \le 2$, with eigenvalues $1-u,u$. The first state is pure and has zero Von Neumann entropy, so the exact difference isThis is the entropy of a binary diagonal qubit mixture. Its trace distance from the pure state is $u$, not $\epsilon$. An explicit elementary upper bound follows from $-(1-u)\ln (1-u)\le u$:At zero the limit is zero. For small $\epsilon$, the sharp exact answer behaves as $(\epsilon /2){\log }_2(2/\epsilon )+\epsilon /(2\ln 2)+O({\epsilon }^2)$, displaying the logarithmic continuity of entropy near a pure state.

Let $q_y=⟨y|\rho |y⟩$. Positivity and normalization of the density operator imply $q_y\ge 0$ and ${\sum }_y{q}_y=1$. The completely dephasing channel produceswhich is already a spectral decomposition. Its Von Neumann entropy is thereforeThe quantum output entropy is exactly the Shannon entropy of the basis-measurement probabilities, including possible zero eigenvalues via $0{\log }_{2}0=0$.

Dilate the second quantum channel by an isometry $W:B_1⟶B_2{E}_2$, and set ${\tau }_{R{B}_2{E}_2}=(I_R\otimes W){\sigma }_{R{B}_1}(I_R\otimes W^{†})$. Its $R{B}_2$ marginal is the final output. Isometry invariance of Von Neumann entropy gives $S(B_1{)}_{\sigma }=S(B_2{E}_2{)}_{\tau }$ and $S(R{B}_1{)}_{\sigma }=S(R{B}_2{E}_2{)}_{\tau }$. The difference between the two values of coherent information is thereforeNonnegativity follows from Strong subadditivity of Von Neumann entropy. This proves the data-processing inequality for coherent information:The dilated state here need not be pure. The proof uses only isometry invariance and strong subadditivity, so it applies even when the first channel has already entangled the input with its own discarded environment.

By spectral decomposition, the Von Neumann entropy is the Shannon entropy of the eigenvalues:Throughout, logarithms have base two, so the Von Neumann entropy and Shannon entropy are measured in bits. The continuous convention $0{\log }_{2}0=0$ includes zero eigenvalues.

Use the Schmidt decomposition of the bipartite pure state:The two reduced density matrices, obtained by partial trace, areTheir nonzero eigenvalues are identical, even if the two ambient dimensions differ. Zero eigenvalues contribute no Von Neumann entropy, soThis common Von Neumann entropy is the entanglement entropy of the bipartite pure state.

Take a purification of a density operator $|\Psi {⟩}_{ABCD}$ of ${\rho }_{ABC}$. Apply Strong subadditivity of Von Neumann entropy to the reduced state on $ACD$, with $A$ the conditioning system:Complementary subsystems of a pure state have the same Von Neumann entropy, by the Schmidt decomposition. Therefore $S(AD)=S(BC)$ and $S(ACD)=S(B)$. Substitution givesThis is weak monotonicity of quantum entropy. The purification of a density operator is only a proof device; no purity assumption is imposed on ${\rho }_{ABC}$.

Choose an orthonormal basis of the joint $d^2$-dimensional Hilbert space whose first vector is the given purification of a density operator $|\rho {⟩}_{QR}$. Apply rank-one dephasing to ${\sigma }_{QR}$ in this basis, and denote the resulting diagonal probabilities by $p_1,\dots ,p_{d^2}$. Their first entry isThe preceding entropy increase under nonselective projective measurement and entropy bound with one prescribed probability yieldTherefore the quantum Fano inequality isThe quantity $f^2$ is the entanglement fidelity of $\mathcal{N}$ on ${\rho }_Q$; it is already a squared overlap, so it is $f^2$, rather than $f$, that enters the binary entropy. The argument is an instance of the entropy bound from overlap with a pure state in dimension $d^2$. At $f^2=1$ the output is the original pure state and its Von Neumann entropy is zero. For $d=1$ the system is trivial and the same zero-entropy conclusion holds without evaluating ${\log }_2(d^2-1)$.

Write $X$ and $Y$ for the classical quantum registers $\tilde{X}$ and $\tilde{Y}$. By the quantum mutual information formula $I(U:V)=S(U)+S(V)-S(UV)$, expanding the right-hand side of the quantum mutual information balance identity givesAll Von Neumann entropies here are evaluated in $\omega$.

We use three facts: quantum mutual information is nonnegative by nonnegativity of quantum relative entropy; a local quantum channel cannot increase it by data processing for quantum mutual information; and the quantum mutual information of a classical-quantum state is its ensemble's Holevo quantity. Write $\chi (\mathcal{T})$ for the Holevo capacity, the supremum of the output Holevo quantity over finite input ensembles.

The $XB$ marginal is an output ensemble for $\mathcal{E}$, with inputs $\rho (x{)}_A={\mathrm{Tr}}_C\rho (x{)}_{AC}$. Consequently $I(X:B{)}_{\omega }\le \chi (\mathcal{E})$. Also ${\omega }_{XBD}$ is obtained from ${\sigma }_{XYD}$ by a quantum channel on $XY$ which retains $X$ and prepares $B$ from $Y$. HenceTo bound the latter even for entangled states $\rho (x{)}_{AC}$, exhibit the conditional input ensemble after a local measurement. Setwhen $q(y|x)>0$; zero-weight outcomes can be omitted. The numerator is a positive operator, its trace is $q(y|x)$, and the $q(y|x)$ sum to one. ThusThis is a classical-quantum state with an output ensemble for $\mathcal{N}$, so $I(XY:D{)}_{\sigma }\le \chi (\mathcal{N})$. Combining these bounds with $I(B:D{)}_{\omega }\ge 0$ yieldsThe last step takes the supremum over all input ensembles on $AC$. The left-hand channel direction is $B\leftarrow A$, as established in part (i); the printed $A\leftarrow B$ in the last inequality is a typographical reversal. Independent product ensembles also give the reverse inequality, so this proves Holevo-capacity additivity for entanglement-breaking channels.

Use logarithms to base two, so Von Neumann entropy and quantum relative entropy are measured in bits. For density operators $\rho ,\sigma$,The quantum relative entropy formula applies when the support of a positive operator $\rho$ is contained in that of $\sigma$; otherwise $D(\rho \Vert \sigma )=+\infty$. Use $0{\log }_{2}0=0$ in the Von Neumann entropy formula.

The binary density operator $\omega [p]$ is diagonal with eigenvalues $1-p,p$. Its Von Neumann entropy is the binary entropy:The two diagonal density operators commute, so their quantum relative entropy isA term with zero numerator has value zero; a positive numerator and zero denominator give $+\infty$. In particular, $D(\omega [0]\Vert \omega [0])=D(\omega [1]\Vert \omega [1])=0$, whereas $q=0,p>0$ or $q=1,p<1$ gives $+\infty$.

Memorylessness means that the ensemble's average density operator is ${\rho }^{(n)}={\pi }^{\otimes n}$. Its spectral decomposition isThe product eigenvalues giveZero eigenvalues contribute zero to the Von Neumann entropy and are excluded from logarithmic typicality tests.

The quantum typical subspace is the span of eigenvectors whose eigenvalues satisfyIts orthogonal projection $P_{\epsilon }^{(n)}$ selects precisely the classical typical set for the eigenvalue distribution $q$. The typical-set cardinality bounds and the weak law of large numbers therefore givefor any fixed $\delta >0$ and all sufficiently large $n$.

Assume no previously shared entanglement. Alice's encoded ensemble $\{p_x,{\rho }_x\}$ lives in a Hilbert space of dimension $2^n$. Its Holevo quantity satisfiesusing nonnegativity of Von Neumann entropy and the maximum entropy of a quantum state. The data-processing inequality for quantum relative entropy implies ${\chi }_{\mathrm{out}}\le {\chi }_{\mathrm{in}}$ for the transmission quantum channel. Finally, the Holevo bound applied to Bob's quantum measurement givesThis argument also works when the output system has larger dimension than the input. Prior shared entanglement changes the communication resource: superdense coding can transmit two classical bits per sent qubit, so the absence of that resource is part of this bound's setting.

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.

The two-sided eternal black hole is dual to the thermofield double stateTracing out the left CFT givesThus the geometric period $\beta$ is the boundary inverse temperature, and $S_{\mathrm{BH}}$ is the Von Neumann entropy $S_R=-\mathrm{Tr}({\rho }_R\log {\rho }_R)$, equivalently the entanglement entropy between the two CFTs.

The modular Hamiltonian of this thermal state isFor any first-order state variation with $\mathrm{Tr}\delta \rho =0$,This is the first law of entanglement entropy.