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The quantum relative entropy is additive on tensor products:This follows by expanding $\log ({\sigma }_A\otimes {\sigma }_B)=\log {\sigma }_A\otimes I+I\otimes \log {\sigma }_B$ and its counterpart for $\rho$.

For $k$ equiprobable quantum code states decoded by a POVM $\{E_m\}$, testing whether the classical message label agrees with the decoded output gives a joint binary POVM with $F_0={\sum }_m|m⟩⟨m|\otimes E_m$. Its acceptance probability is the decoding success $1-ϵ$ on the joint classical-quantum state and $1/k$ on the product of its marginals. Applying quantum relative entropy monotonicity to this test gives the one-shot classical-quantum coding converse.

For every quantum channel $T$,For a partial trace, this says that discarding a subsystem cannot make two density operators more distinguishable in quantum relative entropy.

Diagonalize two density operators with eigenvalues $r_i,s_j$, and put $q_{ij}=|⟨i|j⟩{|}^2$. This overlap matrix has row and column sums one. The scalar inequality $u\ln (u/v)\ge u-v$ yields $D(\rho \Vert \sigma )\ge {\sum }_{ij}{q}_{ij}(r_i-s_j)=0$ in natural units; division by $\ln 2$ gives bits. Equality forces $r_i=s_j$ whenever $q_{ij}>0$, so $\sigma |i⟩=r_i|i⟩$ and $\rho =\sigma$. Zero eigenvalues use limits, and failure of support inclusion gives infinite quantum relative entropy.

For density operators, Klein's inequality gives $D(\rho \Vert \sigma )\ge 0$, with equality exactly when $\rho =\sigma$. If the support of a positive operator $\rho$ is not contained in that of $\sigma$, the quantum relative entropy is $+\infty$.

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.

Use logarithms to base two, so all entropies are in bits. For a discrete probability distribution $p$ and a density operator $\rho$ with eigenvalues ${\lambda }_j$, the Shannon entropy and Von Neumann entropy are respectivelyUse $0\log 0=0$. The second definition is basis-independent by the spectral theorem; for a density operator diagonal with probabilities $p_x$, it agrees with the first.

The classical relative entropy, or Kullback-Leibler divergence, of $p$ from $q$ isIt is $+\infty$ if some $p_x>0$ has $q_x=0$. The quantum relative entropy iswhen the support of $\rho$ is contained in that of $\sigma$, and is $+\infty$ otherwise. Matrix logarithms are taken on the corresponding positive supports. In a common eigenbasis this reduces to classical relative entropy; both divergences are nonnegative and are generally asymmetric.

To prove Fano's inequality, let the finite alphabet size be $m\ge 2$ and let $E$ indicate an incorrect guess. Since $E$ is determined by $(X,Y)$, the chain rule for conditional entropy first givesExpanding in the opposite order givesConditioning reduces entropy, so $H(E\mid Y)\le H(E)=h_2(p_e)$, the binary entropy function. When $E=0$, knowledge of $Y$ determines $X=f(Y)$ and the conditional entropy is zero. When $E=1$, $X$ can lie in at most the $m-1$ alphabet values other than $f(Y)$. Entropy on $m-1$ values is at most ${\log }_2(m-1)$: its deficit from that number is the nonnegative relative entropy from the uniform distribution. Averaging the two cases therefore givesCombining the two expansions provesThis proof of Fano's inequality via an error indicator works for any alphabet-valued guessing rule, not just an optimal one. If $m=1$, there is no guessing uncertainty and the claim is simply $H(X\mid Y)=0$; that case is treated separately rather than writing $\log 0$.

The concavity inequality is true, but its printed equality condition is false. At $b=1$ equality always holds, even for distinct states; for example take ${\rho }_1=|0⟩⟨0|$ and ${\rho }_2=|1⟩⟨1|$. The correct statement on $0\le b\le 1$ isFor $0<b<1$, entropy is therefore strictly concave.

We give the matrix inequality needed for the proof. Define the quantum relative entropy $D(\tau \Vert \sigma )=\mathrm{tr}\tau ({\log }_2\tau -{\log }_2\sigma )$ if the support of $\tau$ is contained in that of $\sigma$, and $+\infty$ otherwise. Diagonalize the two states with eigenvalues $r_i,s_j$, and put $q_{ij}=|⟨i|j⟩{|}^2$. Each row and column of $q$ sums to one. The elementary scalar inequalityfollows from $x\ln x-x+1\ge 0$, with equality exactly at $x=1$; use limits at zero. ConsequentlyEquality forces $r_i=s_j$ whenever $q_{ij}>0$. Hence every eigenvector $|i⟩$ of $\tau$ is also a $\sigma$-eigenvector with eigenvalue $r_i$, so $\tau =\sigma$. Conversely identical states give zero relative entropy. This proves nonnegativity of quantum relative entropy and its equality condition, including states with zero eigenvalues under the support convention.

Now set $\rho =b{\rho }_1+(1-b){\rho }_2$. For $0<b<1$ its support contains both component supports. Direct expansion givesSince both coefficients are positive, equality holds exactly when both component states equal $\rho$, equivalently ${\rho }_1={\rho }_2$. At either endpoint the mixture has only one component and equality is automatic. This proves the corrected strict concavity of Von Neumann entropy.

For a density matrix $\rho$ on a $d$-dimensional Hilbert space, its Von Neumann entropy, in bits, iswhere the ${\lambda }_a$ are its eigenvalues and $0{\log }_{2}0=0$. Each summand is nonnegative, proving $S\ge 0$, with equality exactly when one eigenvalue is one and the rest zero, namely for a pure state.

We first prove the nonnegativity of quantum relative entropy needed below. Diagonalize $\rho$ with eigenvalues ${\lambda }_a$ and eigenvectors $u_a$, and $\sigma$ with eigenvalues ${\mu }_b$ and eigenvectors $v_b$. Put $B_{ab}=|⟨u_a,v_b⟩{|}^2$ and $q_a=⟨u_a|\sigma |u_a⟩={\sum }_b{B}_{ab}{\mu }_b$. Each row of $B$ sums to one. Concavity of the ordinary logarithm gives ${\sum }_b{B}_{ab}{\log }_2{\mu }_b\le {\log }_2{q}_a$. ConsequentlyThe final Gibbs inequality follows directly from ${\sum }_{{\lambda }_a>0}{\lambda }_a\log (q_a/{\lambda }_a)\le \log {\sum }_{{\lambda }_a>0}{q}_a\le 0$. Support failure gives $D=+\infty$; zero eigenvalues can otherwise be handled by restriction or regularization. Equality requires $q_a={\lambda }_a$ and equality in the logarithm inequality on every positive-weight row. Thus $u_a$ is a $\sigma$ eigenvector of eigenvalue ${\lambda }_a$ wherever ${\lambda }_a>0$. The remaining diagonal entries of $\sigma$ vanish and positivity makes the remaining rows and columns zero. Hence equality holds exactly when $\rho =\sigma$.

Taking $\sigma =I/d$ gives $D(\rho \Vert I/d)={\log }_{2}d-S(\rho )$, provingThe lower equality is attained exactly by pure states and the upper exactly by the maximally mixed state $I/d$.

For $\bar{\rho }={\sum }_i{p}_i{\rho }_i$, each positive-weight state has support contained in that of $\bar{\rho }$. Expanding the quantum relative entropy givesThis proves Concavity of Von Neumann entropy, with equality precisely when every positive-weight ${\rho }_i$ is the same state.

For the upper entropy bound for a quantum mixture, diagonalize each component, ${\rho }_i={\sum }_a{\lambda }_{ia}|v_{ia}⟩⟨v_{ia}|$, and set $q_{ia}=p_i{\lambda }_{ia}$. Form the rectangular matrix $B$ whose columns are $\sqrt{q_{ia}}{v}_{ia}$. Then $B{B}^{†}=\bar{\rho }$, while the Gram matrix $\Gamma =B^{†}B$ has diagonal $q_{ia}$. The two matrices have the same nonzero eigenvalues: $B$ and $B^{†}$ map corresponding positive-eigenvalue eigenspaces isomorphically. Thus $S(\Gamma )=S(\bar{\rho })$.

If there are $L$ columns, let $W=\mathrm{diag}(1,\omega ,\dots ,{\omega }^{L-1})$, with $\omega =e^{2\pi i/L}$. ThenAll the summands have the same Von Neumann entropy. The just-proved concavity therefore implies $S(\bar{\rho })=S(\Gamma )\le S(\mathrm{diag}\Gamma )=H(q)$. Finally,We have proved both requested bounds:For the upper equality, strictness in the concavity argument forces $\Gamma$ to be diagonal. Its entries between two positive-weight component eigenvectors then force those vectors to be orthogonal. Thus upper equality holds exactly when the supports of the positive-weight component states are mutually orthogonal. In that case their classical label is perfectly distinguishable, explaining the extra $H(p)$ bits.

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 ${\rho }_{AB}={\rho }_A\otimes {\rho }_B$.

Omit indices of probability zero. The support of a positive operator of every remaining ${\rho }_i$ lies within that of $\overline{\rho }$, because $\overline{\rho }$ is a positive sum containing $p_i{\rho }_i$. Thus $D({\rho }_i\Vert \overline{\rho })$ is finite.

First suppose $\mathrm{supp}\overline{\rho }\subseteq \mathrm{supp}\sigma$, so all quantities are finite. Expand the trace definitions of quantum relative entropy:Therefore Donald's identity isIf the support condition on $\sigma$ fails, at least one positive-weight ${\rho }_i$ also fails it. The left side and the final term on the right are then $+\infty$, while the first right-hand sum remains finite. The equality is valid in this extended sense, without taking any undefined difference of infinities.

The unread measurement is the pinching map $\mathcal{P}(\rho )={\rho }^{\prime }={\sum }_i{P}_i\rho P_i$, a nonselective projective measurement. Its output is block diagonal and commutes with each $P_i$, so $\log {\rho }^{\prime }$ also commutes with the projectors on its support.

First the quantum relative entropy $D(\rho \Vert {\rho }^{\prime })$ is finite. If $v$ lies in the kernel of ${\rho }^{\prime }$, positivity givesEvery summand is zero; summing ${\rho }^{1/2}{P}_{i}v=0$ gives ${\rho }^{1/2}v=0$. Thus $\ker {\rho }^{\prime }\subseteq \ker \rho$, equivalently $\mathrm{supp}\rho \subseteq \mathrm{supp}{\rho }^{\prime }$.

The block-diagonal logarithm obeys ${\sum }_i{P}_i(\log {\rho }^{\prime })P_i=\log {\rho }^{\prime }$ on this support. Cyclicity of the trace therefore givesConsequently the relative entropy of a pinched state is exactly the entropy increase:By the proved nonnegativity of quantum relative entropy and its equality condition,Equivalently, equality holds exactly when all off-diagonal blocks $P_i\rho P_j$ for $i\ne j$ already vanish, or when $\rho$ commutes with every measurement projector. This argument proves the equality condition even when the projectors are degenerate and the states singular.

Let $\mathcal{P}$ be the pinching map determined by the projectors, and put $\sigma =\mathcal{P}(\rho )$. This is a nonselective projective measurement; its projectors are its Kraus operators. We will prove the exact relative entropy of a pinched state identity.

First the support of $\rho$ is contained in that of $\sigma$. Indeed, if $v$ lies in the kernel of $\sigma$, positivity givesEvery summand is zero. Since ${\sum }_i{P}_i=I$, also ${\rho }^{1/2}v=0$, and thus $v$ is in the kernel of $\rho$. Therefore the logarithm of $\sigma$ may be used on its support without an infinite relative-entropy term.

The operator $\sigma$ is block diagonal, so each $P_i$ commutes with $\log \sigma$ on that support. Cyclicity of the trace givesExpand the quantum relative entropy:Its nonnegativity now proves the entropy increase under nonselective projective measurement:The equality case in nonnegativity of quantum relative entropy says equality holds exactly when $\rho =\sigma$. Equivalently,Thus equality means the state already has no coherence between distinct measurement blocks. It need not be diagonal inside a block of rank greater than one.

The quantum relative entropy is $S(\rho \Vert \sigma )=\mathrm{Tr}\rho ({\log }_2\rho -{\log }_2\sigma )$ when the support of a positive operator $\rho$ is contained in that of $\sigma$, and $+\infty$ otherwise. Write spectral decompositions $\rho ={\sum }_i{r}_i|i⟩⟨i|$ and $\sigma ={\sum }_j{t}_j|u_j⟩⟨u_j|$, and set $q_i=⟨i|\sigma |i⟩$.

The weights $w_{ij}=|⟨i|u_j⟩{|}^2$ sum to one. Concavity of the scalar logarithm gives the diagonal logarithm concavity boundFor $r_i>0$, support inclusion removes overlaps with zero eigenvalues of $\sigma$; alternatively take a positive regularization and pass to the limit. The two diagonal vectors $r$ and $q$ are probability distributions. ThereforeThis proves nonnegativity of quantum relative entropy without assuming the states commute. Equality in the strict logarithmic concavity condition makes each occupied $|i⟩$ an eigenvector of $\sigma$; classical equality gives $q_i=r_i$ and no weight outside the support, hence equality occurs exactly for $\rho =\sigma$.

For any tripartite density operator, Strong subadditivity of Von Neumann entropy isUse quantum relative entropy $D(\rho \Vert \sigma )=\mathrm{Tr}\rho ({\log }_2\rho -{\log }_2\sigma )$, with the usual support condition. The equivalent relative-entropy comparison isIndeed the two sides expand respectively as $S(A)+S(BC)-S(ABC)$ and $S(A)+S(B)-S(AB)$. Subtracting cancels $S(A)$ and leaves exactly the strong-subadditivity gap. Marginal-product supports contain the support of the joint state, so these expressions are finite; singular marginals can also be handled by full-rank regularization and a limit.

Finally, tracing out $C$ sends the numerator and denominator of the first relative entropy to those of the second. The data-processing inequality for quantum relative entropy therefore proves the comparison. This is strong subadditivity from relative-entropy monotonicity, rather than an assumption that classical entropy proofs automatically apply to quantum states.

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 $+\infty$.

For the distribution and POVM from part 2, apply the data-processing inequality for quantum relative entropy to the measurement channel. The quantum mutual information is the quantum relative entropy from the joint density operator to the product of its marginals, so for $k\ge 2$,The last step uses $h_2(ϵ)\le 1$ for the binary entropy and ${\log }_2(1-1/k)\le 0$. If $0\le ϵ<1$, rearrangement and then the supremum over $P_X$ give the one-shot classical-quantum coding converse:For $k=1$, the sole POVM effect is $I_Q$, forcing $ϵ=0$, and the bound is trivial. For $ϵ=1$, division by $1-ϵ$ is undefined: the undivided inequality remains valid and gives no size restriction. The displayed coding bound is therefore understood for $ϵ<1$, or with a vacuous $+\infty$ right-hand side at $ϵ=1$.

An alternative uses the original decoder directly. Let $M$ be the uniform message and $\hat{M}$ its measured estimate. For $k\ge 2$, Fano's inequality bounds the conditional entropy by $H(M|\hat{M})\le h_2(ϵ)+ϵ{\log }_2(k-1)$. The Holevo bound givesThus the same coding bound follows by bounding the classical mutual information obtainable from the POVM.