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For every state on ${\mathcal{H}}_A\otimes {\mathcal{H}}_B$,The upper bound follows from Subadditivity of Von Neumann entropy and the lower bound from the Araki–Lieb inequality.

Use logarithms to base two and set $0{\log }_{2}0=0$. For a density matrix with eigenvalues $({\lambda }_j)$, the Von Neumann entropy isThus it is the Shannon entropy of the spectrum. In dimension $d$, each term is nonnegative, and entropy vanishes exactly when one eigenvalue is one, namely for a pure state. The Jensen inequality applied to the strictly concave function $-x{\log }_{2}x$ gives $S(\rho )\le {\log }_{2}d$, with equality precisely at the maximally mixed state $I/d$. Unitary conjugation preserves eigenvalues, so $S(U\rho U^{†})=S(\rho )$. Product eigenvalues giveThe finite-dimensional entropy is continuous because the eigenvalues depend continuously on the matrix and $-x\log x$ is continuous also at zero.

Define the partial trace over ${\mathcal{K}}_2$ by choosing an orthonormal basis $(e_b)$ and settingEquivalently it is uniquely characterized by $\mathrm{Tr}({\rho }_{1}A)=\mathrm{Tr}(\rho (A\otimes I))$ for every operator $A$ on ${\mathcal{K}}_1$. This identity proves basis independence. Define ${\rho }_2$ similarly, with $\mathrm{Tr}({\rho }_{2}B)=\mathrm{Tr}(\rho (I\otimes B))$. Hermiticity is preserved by the displayed sum. For each $u\in {\mathcal{K}}_1$,Taking the trace in a product orthonormal basis gives $\mathrm{Tr}{\rho }_1=\mathrm{Tr}\rho =1$, and the same proof works for ${\rho }_2$. Thus both reductions are density matrices. This is the partial trace positivity from product vectors argument.

For a bipartite pure state $\rho =|\psi ⟩⟨\psi |$, expand $\psi$ in a product orthonormal basis with coefficient matrix $C$. A singular value decomposition gives the Schmidt decompositionwith orthonormal families on both sides. Taking the two partial traces makes all cross terms vanish:The spectra may have different numbers of zeros but have identical nonzero eigenvalues. Consequently the reduced entropies coincide:This common value is the entanglement entropy. In infinite-dimensional trace-class systems the same partial-trace and Schmidt arguments apply, and the equality remains valid with the value $+\infty$ allowed; the bound ${\log }_{2}d$ is specifically finite-dimensional.

For completeness, the basic mixing and correlation properties also admit short proofs. Define quantum relative entropy by $D(\tau \Vert \sigma )=\mathrm{Tr}\tau ({\log }_2\tau -{\log }_2\sigma )$ when supports are included, and by $+\infty$ otherwise. Here is its nonnegativity proof without merely citing an operator inequality. Diagonalize $\tau ,\sigma$ with eigenvalues $r_i,s_j$, and let $q_{ij}$ be the squared overlaps of their eigenbases. Both row and column sums of $q$ are one. The scalar inequality $r\ln (r/s)\ge r-s$, with zero cases taken by limits, givesEquality forces $r_i=s_j$ on every nonzero overlap, hence $\tau =\sigma$. For $\overline{\rho }={\sum }_i{p}_i{\rho }_i$, expanding the trace definition yieldswhich proves Concavity of Von Neumann entropy.

The complementary entropy bounds for a quantum mixture areTo prove the upper bound, write the spectral pure-state decompositions of the ${\rho }_i$ as one ensemble with weights $w_{ij}=p_i{\lambda }_{ij}$. Form the column matrix with columns $\sqrt{w_{ij}}|{\psi }_{ij}⟩$. Its two Gram operators have the same nonzero spectrum: one is $\overline{\rho }$, and the other, $G$, has diagonal $w$. Relative entropy against $\mathrm{diag}w$ gives $0\le D(G\Vert \mathrm{diag}w)=H(w)-S(G)$. Since $H(w)=H(p)+{\sum }_i{p}_{i}S({\rho }_i)$ and $S(G)=S(\overline{\rho })$, the bound follows. Zero-weight columns are omitted.

Finally the support of $\rho$ is contained in $\mathrm{supp}{\rho }_1\otimes \mathrm{supp}{\rho }_2$. Indeed, a vector in a marginal kernel gives a sum of nonnegative product-vector expectations equal to zero; positivity forces the corresponding product vectors into the kernel of $\rho$. The operator identity $\log ({\rho }_1\otimes {\rho }_2)=\log {\rho }_1\otimes I+I\otimes \log {\rho }_2$ on this support and the partial-trace identities giveThis proves Subadditivity of Von Neumann entropy, with equality exactly for a product state. Purify ${\rho }_{12}$ by adjoining a third system, using its spectral decomposition. Equality of pure-state marginal entropies gives $S({\rho }_{13})=S({\rho }_2)$ and $S({\rho }_3)=S({\rho }_{12})$. Subadditivity on systems $1,3$ then gives $S({\rho }_2)\le S({\rho }_1)+S({\rho }_{12})$; swapping $1,2$ gives the Araki–Lieb inequality $\boxed{|S({\rho }_1)-S({\rho }_2)|\le S({\rho }_{12})}$. These finite-dimensional proofs establish the usual basic entropy bounds, rather than attributing all mixed-state behavior to pure-state Schmidt decompositions.

For a classical error-correcting code of length $n$, alphabet size $q$, minimum Hamming distance $d$ and $M$ codewords, the Singleton bound isFor a linear $[n,k,d{]}_q$ code this is $n-k\ge d-1$. Deleting any $d-1$ coordinates is injective on the codewords, since two words agreeing on the remaining coordinates would differ in fewer than $d$ positions; this also explains the classical bound.

For an n-qubit quantum code carrying $k>0$ logical qubits and having quantum code distance $d$, the quantum Singleton bound isMore generally replace $k$ by ${\log }_{2}K$ for code dimension $K>1$. The proof applies to degenerate codes too.

Let $r=d-1$. Purify the maximally mixed encoded input by a reference $R$ of dimension $K$:Distance $d$ makes erasure of any subset of at most $r$ qubits correctable. The allowed quantum erasure correction condition says that a correctable erased set $A$ is decoupled from $R$:This is the matrix-element correction condition in another form: all operators on $A$ have constant diagonal matrix elements and zero off-diagonal matrix elements in the logical basis, so tracing the other physical qubits in $|\Omega ⟩⟨\Omega |$ yields the displayed product.

First justify the sizes needed for the partition. If $2r>n$, partition all physical qubits into $A,B$, both of size at most $r$. Both sets are correctable. Purity of $RAB$ and the Schmidt decomposition giveAdding forces $k=0$, a contradiction. Thus $2r\le n$, without assuming the bound we aim to prove.

Choose disjoint sets $A,B$ of size $r$, and let $C$ contain the remaining $n-2r$ qubits. The joint state $RABC$ is pure, so decoupling and Subadditivity of Von Neumann entropy giveAdding and cancelling gives $k\le S(C)$. The Von Neumann entropy of $D$-dimensional states is at most ${\log }_{2}D$: for positive eigenvalues $p_i$, concavity of the logarithm gives ${\sum }_i{p}_i{\log }_2(1/p_i)\le {\log }_2{\sum }_i{p}_i/p_i\le {\log }_{2}D$. Here $D=2^{n-2r}$, so $k\le n-2r$, proving the quantum bound. The code is assumed to encode information ($K>1$); a one-dimensional code requires a separate distance convention, since simply re-preparing a known state makes its erasure-correction condition vacuous.

True: quantum entropy is subadditive. Write $\sigma ={\rho }_1\otimes {\rho }_2$. Positivity ensures the support of $\rho$ lies in that of $\sigma$. Indeed, if $|v⟩\in \ker {\rho }_1$, then ${\sum }_j⟨v,j|\rho |v,j⟩=0$; all summands are nonnegative, so ${\rho }^{1/2}|v,j⟩=0$ for every $j$. Thus $\rho$ annihilates $\ker {\rho }_1\otimes {\mathcal{K}}_2$, and similarly the kernel from the other factor.

On the resulting support, ${\log }_2({\rho }_1\otimes {\rho }_2)={\log }_2{\rho }_1\otimes I+I\otimes {\log }_2{\rho }_2$. Using the defining property of the partial trace and nonnegativity of quantum relative entropy proved in part (i),HenceEquality holds exactly for the product state $\rho ={\rho }_1\otimes {\rho }_2$. This proves Subadditivity of Von Neumann entropy directly.

Let a density operator on a Hilbert space ${\mathcal{H}}_S$ have the spectral decompositionIntroduce an auxiliary Hilbert space ${\mathcal{H}}_R$ with an orthonormal basis containing $r$ vectors $|j{⟩}_R$, and setThis vector has norm one. Its partial trace isThis explicitly constructs a purification of a density operator. The same construction works for an arbitrary trace-class density operator: its nonzero eigenvalues form a finite or countable summable family, and the displayed vector converges in the Hilbert space norm. The entropy inequality below is understood with finite Von Neumann entropies, avoiding undefined differences of infinities.

Apply this construction with $S=AB$ to obtain a joint pure state on $ABC$. The Schmidt decomposition across $AB:C$, $A:BC$ and $B:AC$ respectively givesNow apply Subadditivity of Von Neumann entropy to the reduced density matrices on $BC$ and $AC$:These inequalities say $S(A)-S(B)\le S(C)$ and $S(B)-S(A)\le S(C)$. Replacing $S(C)$ by $S(AB)$ proves the Araki–Lieb inequality:The subadditivity used here concerns the actual reduced density operators, rather than statistical independence. In particular it follows from nonnegativity of quantum relative entropy, sinceand similarly for $AC$. Thus no independence assumption on the purification of a density operator is needed.

Under the intended assumption that one use of the quantum channel transmits one qubit without noise, Alice can communicate two classical bits with certainty if she and Bob already share a Bell pair. She uses superdense coding. The prior distribution of the Bell pair must be a resource available before the permitted channel use; distributing it through that same channel would consume another use.

Take the shared Bell state to be $|{\Phi }^{+}⟩=(|00⟩+|11⟩)/\sqrt{2}$, with Alice holding the first qubit. For a two-bit message $(a,b)\in \{0,1{\}}^2$, Alice applies the local unitary operation $U_{ab}=X^a{Z}^b$ to her qubit. The four encoded pure states are, up to irrelevant global phases,They are orthogonal: for any matrix $V$ on the first qubit, $⟨{\Phi }^{+}|(V\otimes I)|{\Phi }^{+}⟩=\mathrm{Tr}V/2$, and the four Pauli operators are orthogonal in the trace inner product. Explicitly,Alice sends her qubit through the quantum channel once. Bob now holds both qubits and performs a Bell-basis measurement. Its four orthogonal projectors identify $(a,b)$ with certainty, givingBefore the transmission, Bob's reduced density matrix is $I/2$ for every message, so shared entanglement alone does not convey the message.

Without shared entanglement, a noiseless single-qubit transmission cannot perfectly distinguish four messages: perfect discrimination requires orthogonal state supports, and its two-dimensional Hilbert space has room for at most two nonzero mutually orthogonal supports. The Holevo bound also limits its accessible classical information to one bit. For the usual protocol starting with a shared two-qubit pure state and using local unitary encodings, exact two-bit transmission requires a maximally mixed Bob marginal. Indeed all encoded joint states are pure and have the same ${\rho }_B$, while Subadditivity of Von Neumann entropy gives their average entropy at most $1+S({\rho }_B)$. Four equally likely perfectly distinguishable messages have average Von Neumann entropy two, forcing $S({\rho }_B)=1$ and hence a maximally entangled state.

The specification “quantum channel” alone does not determine its dimension or noise, so the one-qubit, noiseless interpretation is needed for this particular conclusion. A noiseless four-dimensional channel could send four orthogonal states without any shared entanglement; a completely depolarizing qubit channel cannot transmit the message even with a shared Bell pair, since its output is independent of Alice's encoding. Shared entanglement is therefore the extra resource for the intended single-qubit channel, not a guarantee for every unspecified quantum channel.

Use base-two logarithms, so Von Neumann entropy is measured in bits, and set $0\log 0=0$. Write $D(\rho \Vert \sigma )=\mathrm{Tr}\rho ({\log }_2\rho -{\log }_2\sigma )$ for the quantum relative entropy, the quantity denoted $S(\rho \Vert \sigma )$ in the paper. It is $+\infty$ unless the support of a positive operator $\rho$ lies within that of $\sigma$.

We first establish nonnegativity of quantum relative entropy, including its equality condition. Let $\rho ={\sum }_{i:r_i>0}{r}_i|u_i⟩⟨u_i|$ and $\sigma ={\sum }_j{s}_j|v_j⟩⟨v_j|$, and suppose the support condition holds. Put $q_i=⟨u_i|\sigma |u_i⟩>0$. Concavity of the scalar logarithm givesTerms with $s_j=0$ have zero weight here. Working first with natural logarithms, the inequality $-\ln x\ge 1-x$ yieldssince the $q_i$ sum to at most $\mathrm{Tr}\sigma =1$. Equality forces $q_i=r_i$ for every positive-weight eigenvector, no remaining weight of $\sigma$ outside their span, and equality in the logarithm's strict concavity. The latter makes each $u_i$ an eigenvector of $\sigma$ with eigenvalue $q_i$. Thus $D(\rho \Vert \sigma )=0$ exactly when $\rho =\sigma$.

For the bipartite density operator,Indeed, a projector onto the kernel of ${\rho }_A$, tensored with $I_B$, has zero expectation in ${\rho }_{AB}$. Positivity implies that ${\rho }_{AB}$ annihilates its range: $\mathrm{Tr}({\rho }_{AB}P)=\Vert {\rho }_{AB}^{1/2}P{\Vert }_{\mathrm{HS}}^2=0$. The same argument applies to $B$. Hence the quantum relative entropy below is finite. On this support,Using the partial trace to evaluate the two local terms givesThis proves Subadditivity of Von Neumann entropy:The proof also identifies equality exactly with the product state ${\rho }_{AB}={\rho }_A\otimes {\rho }_B$.

Introduce an orthonormal classical-label basis $|i⟩$ and the classical-quantum stateIts reduced density operators are ${\Gamma }_X={\sum }_i{p}_i|i⟩⟨i|$ and ${\Gamma }_A=\overline{\rho }={\sum }_i{p}_i{\rho }_i$. If ${\lambda }_{ij}$ are the eigenvalues of ${\rho }_i$, the eigenvalues of ${\Gamma }_{XA}$ are $p_i{\lambda }_{ij}$. Thus the entropy of an orthogonal quantum mixture iswhere $H(p)=-{\sum }_i{p}_i{\log }_2{p}_i$ is the Shannon entropy. Apply the proved Subadditivity of Von Neumann entropy to $X$ and $A$ and cancel $H(p)$. This gives Concavity of Von Neumann entropy:Zero-probability terms may simply be omitted. The equality condition from the preceding solution implies that equality holds precisely when all the positive-probability states are the same: the block equations $p_i{\rho }_i=p_i\overline{\rho }$ follow from ${\Gamma }_{XA}={\Gamma }_X\otimes {\Gamma }_A$.

Use base-two logarithms, so Von Neumann entropy is measured in bits. Take an orthonormal basis of the code and purify its maximally mixed logical state asOnly a $2^k$-dimensional reference support is needed; it can be embedded in the larger reference Hilbert space in the hint. Its reduced density matrix is maximally mixed on that support, so $S(R)=k$.

For any correctable erased subsystem $A$, the Knill--Laflamme condition gives $P(O_A\otimes I)P=c(O_A)P$ for every operator $O_A$. Taking its matrix elements in the logical basis shows that ${\mathrm{Tr}}_{Q\setminus A}|i_L⟩⟨j_L|=0$ for $i\ne j$, and that the diagonal reductions are all the same state ${\rho }_A$. ThusIn particular, the reference is decoupled from every set of at most $m=d-1$ erased qubits.

For an information-carrying code $k>0$, first justify that two disjoint sets of size $m$ fit. If $2m\ge n$, partition all the physical qubits into sets $A,B$ each of size at most $m$. Both are correctable. Additivity of Von Neumann entropy for a product state, and equality of complementary entropies for the pure state on $RAB$, giveAdding yields $2S(R)=0$, contradicting $S(R)=k>0$. Hence $2m<n$.

Now choose disjoint $A,B$ of size $m$, and let $C$ contain the remaining $n-2m$ qubits. The state on $RABC$ is pure. Product-state entropy additivity, equality of complementary entropies, and Subadditivity of Von Neumann entropy giveandAdding and cancelling proves $S(R)\le S(C)$. Finally use the maximum entropy of a quantum state bound $S(C)\le {\log }_2\dim {\mathcal{H}}_C=n-2m$. Therefore the quantum Singleton bound isThis proof uses quantum erasure correction, not nondegeneracy, so it applies to degenerate codes as well. A code encoding no logical information requires a separate distance convention; the argument above concerns the usual $k>0$ coding problem.

Using base-two logarithms, define the quantum conditional entropy byThe Subadditivity of Von Neumann entropy, $S(AB)\le S(A)+S(B)$, immediately gives $S(A\mid B)\le S(A)$.

For the lower bound, purify ${\rho }_{AB}$ by an auxiliary system $C$. Complementary subsystems of a pure state have the same nonzero reduced-state eigenvalues, by the Schmidt decomposition. Hence $S(BC)=S(A)$ and $S(C)=S(AB)$. Applying Subadditivity of Von Neumann entropy to $BC$ givesRearranging yields $S(A\mid B)\ge -S(A)$. This is the relevant side of the Araki–Lieb inequality, here derived using purification.

Finally the maximum entropy of a quantum state gives $S(A)\le {\log }_2\dim {\mathcal{H}}_A$. For example, this follows directly fromby nonnegativity of quantum relative entropy. Combining the two bounds provesBoth signs can be attained: a maximally mixed $A$ independent of $B$ has the positive extreme, while a maximally entangled state with a sufficiently large $B$ has the negative extreme.

For a bipartite density operator ${\rho }_{AB}$, the Subadditivity of Von Neumann entropy isHere $S(A)=S({\mathrm{Tr}}_B{\rho }_{AB})$ and likewise for $B$, using the partial trace. For a tripartite density operator, the Strong subadditivity of Von Neumann entropy isEquivalently, quantum conditional entropy cannot increase upon adding a conditioning system: $S(A\mid BC)\le S(A\mid B)$.

Introduce an orthonormal classical label register $X$ and the classical-quantum stateThe blocks have orthogonal supports because their label vectors are orthogonal, regardless of whether the original ${\rho }_i$ have orthogonal supports. Part (a) gives $S(XQ)=H(p)+{\sum }_i{p}_{i}S({\rho }_i)$. The marginals are ${\omega }_X={\sum }_i{p}_i|i⟩⟨i|$ and ${\omega }_Q={\sum }_i{p}_i{\rho }_i$, so $S(X)=H(p)$. Apply the Subadditivity of Von Neumann entropy and cancel $H(p)$:Thus $S({\sum }_i{p}_i{\rho }_i)\ge {\sum }_i{p}_{i}S({\rho }_i)$. This is the Concavity of Von Neumann entropy.

Implement the quantum channel by a unitary dilation with a pure initial environment. The final state on $RBE$ is pure. The Schmidt decomposition across either bipartition gives equal complementary Von Neumann entropies, so $S(RB)=S(E)$ and $S(B)=S(RE)$. By the Subadditivity of Von Neumann entropy,The reference is untouched and purifies the input, hence $S(R)=S(\rho )$. Therefore $\boxed{I_c(\Phi ,\rho )\le S(\rho )}$.

Subadditivity of Von Neumann entropy applied to $RE$ gives $S(RE)\le S(R)+S(E)$. Together with the environment expression for coherent information,The channel does not act on $R$. Its reduced state has the same nonzero eigenvalues as the original input $\rho$, because $RA$ initially purifies that input. HenceMore precisely, $I_c=S(\rho )-I(R:E)$: the coherent information upper bound by input entropy is a consequence of nonnegative quantum mutual information with the environment. Equality holds when $R$ and $E$ are uncorrelated.

Assume a finite-dimensional memoryless quantum channel, no shared entanglement, and a code of $M_n$ messages with $n^{-1}{\log }_2{M}_n\to R$. Use a uniform message $M$ to relate average and maximum error. Each permitted input is a product state ${\rho }_m^{(n)}={⨂}_{i=1}^n{\rho }_{m,i}$, giving output ${\sigma }_m^{(n)}={⨂}_i\Lambda ({\rho }_{m,i})$. The factors may depend on both the message and the channel use.

Let ${\chi }^{\ast }(\Lambda )$ denote the one-use Holevo capacity in the question. The Subadditivity of Von Neumann entropy for the average output, and entropy additivity for each product output, implyThis is the key step of the product-input classical-capacity converse; it does not assume that the average output itself is a product.

Bob may measure all outputs jointly. The Holevo bound still gives $I(M:\hat{M})\le {\chi }_n$. Let $P_e$ be his average message-error probability. Fano's inequality givesSince the maximum error $P_{\max }$ is at least $P_e$,when $R>{\chi }^{\ast }(\Lambda )$. Thus the maximum error cannot vanish asymptotically.

The Umegaki relative entropy iswhen the support of $\rho$ is contained in that of $\sigma$, and $+\infty$ otherwise. Its additivity of quantum relative entropy isits superadditivity of quantum relative entropy isand its data-processing inequality for quantum relative entropy is $D(T(\rho )\Vert T(\sigma ))\le D(\rho \Vert \sigma )$ for every quantum channel $T$.

Additivity follows from the logarithm of a tensor product,and the analogous identity for ${\sigma }_A\otimes {\sigma }_B$. For superadditivity, subtract the two marginal relative entropies from the joint one. The reference-state terms cancel, leavingThis is the nonnegativity of quantum mutual information, equivalently Subadditivity of Von Neumann entropy.

Combining parts (iii) and (iv), then multiplying by $1+\epsilon$, givesThe dimension bound for quantum conditional entropy isIndeed, Subadditivity of Von Neumann entropy gives $H(A|B{)}_{\tau }\le S({\tau }_A)\le \log d_A$, while the Araki–Lieb inequality gives $H(A|B{)}_{\tau }\ge -S({\tau }_A)\ge -\log d_A$. HenceInterchanging $\rho$ and $\sigma$ proves the continuity bound for quantum conditional entropy:

For a code with product state inputs to $n$ uses of a memoryless quantum channel, Subadditivity of Von Neumann entropy bounds its output Holevo quantity by $n{\chi }^{\ast }(\Lambda )$. The Holevo bound and Fano's inequality then give, for asymptotic rate $R$, average and hence maximum error at least $1-{\chi }^{\ast }(\Lambda )/R$ in the limit inferior. In particular, no code above that rate can have vanishing maximum error, even with a collective decoder measurement.

The quantum mutual information of a bipartite state isIts nonnegativity is equivalent to Subadditivity of Von Neumann entropy.