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Give Alice the first qubit $A$ and Bob the second $B$. Compare the Bell state $|{\Phi }^{+}⟩$ with a preparation that tosses a fair classical bit and gives both parties its value. Their density operators, written in the computational basis ordered $00,01,10,11$, areThey have identical diagonal quantum state populations. The off-diagonal entries of the Bell state retain quantum coherence between $00$ and $11$; the classical mixture has none in this basis.

If both parties perform a Pauli measurement of $Z$, the Born rule gives $P(00)=P(11)=1/2$ and $P(01)=P(10)=0$ for both preparations. Perfect agreement here therefore cannot establish entanglement: an ordinary shared random bit already explains it.

Now both measure $X$, whose normalized eigenstates are $|+⟩=(|0⟩+|1⟩)/\sqrt{2}$ and $|-⟩=(|0⟩-|1⟩)/\sqrt{2}$. Changing basis givesThe Bell state again has matching outcomes, each with probability $1/2$. For either product state $|00⟩$ or $|11⟩$, each of the four $X$ outcome pairs has probability $1/4$, so their classical mixture does too. Equivalently, the Pauli correlator $⟨X\otimes X⟩$ is one for the Bell state and zero for this mixture, while $⟨Z\otimes Z⟩$ is one for both.

Taking a partial trace gives ${\rho }_A={\rho }_B=I_2/2$ for both preparations. Thus every measurement confined to Alice's system has the same statistics in the two cases, and so does every measurement confined to Bob's. They reveal the difference by comparing records through classical communication. Conditional on Alice's result, Bob can predict a matching $X$ result for the Bell state; without that result, his outcomes remain evenly split. An uncommunicated local quantum channel, including a measurement with its outcome discarded, leaves his reduced density matrix unchanged. This is quantum no-signalling, even when conditional states change. ^[ref]

The Von Neumann entropy makes the joint-versus-local distinction numerical. With base-two logarithms, the Bell state has eigenvalues $1,0,0,0$, giving $S(AB)=0$, whereas the mixture has $1/2,1/2,0,0$, giving $S(AB)=1$. Both have $S(A)=S(B)=1$. Their quantum mutual information $I(A:B)=S(A)+S(B)-S(AB)$ is consequently two and one bits, respectively. The value two is not two readable classical bits in either local qubit: the matching $Z$ records have ordinary mutual information one bit in both cases. Random records can come from a globally pure state. ^[ref]

These two matching-axis measurements distinguish the specified preparations; they alone are not a test of a Bell inequality. A local hidden-variable theory could assign shared random answers separately to these two axes. Bell theorem concerns incompatible correlations across a suitable larger collection of measurement settings. Bell's 1964 paper established the obstruction for local hidden-variable descriptions of the predicted correlations. ^[ref]

In a purified Stinespring dilation, coherent information satisfies $I_c=S(RE)-S(E)=S(\rho )-I(R:E)$. Nonnegativity of quantum mutual information implies $I_c\le S(\rho )$. Equality means that the reference and environment are uncorrelated, so the environment learns nothing about the purified input reference.

A local quantum channel on either subsystem cannot increase quantum mutual information. Indeed, apply the data-processing inequality for quantum relative entropy to $D({\rho }_{AB}\Vert {\rho }_A\otimes {\rho }_B)$; the channel maps the product of marginals to the product of the new marginals.

For input dimension $D$, an input of entropy $S(\rho )$ gives output entropy $h_2(p)+(1-p)S(\rho )$ and reference-output entropy $h_2(p)+pS(\rho )$. These follow from entropy of an orthogonal quantum mixture and purity of the input purification. Thus the quantum mutual information is $2(1-p)S(\rho )$, maximized at $\rho =I_D/D$. The entanglement-assisted classical capacity is consequently $2(1-p){\log }_{2}D$ bits per use; for a binary input it is $2(1-p)$.

This is the supremum of reliable classical bits per use of a memoryless quantum channel when sender and receiver share unlimited prior quantum entanglement. If $|{\psi }_{\rho }{⟩}_{RA}$ purifies the input, let ${\omega }_{RB}=({\mathrm{id}}_R\otimes \mathcal{N})(|{\psi }_{\rho }⟩⟨{\psi }_{\rho }|)$. The single-letter capacity formula is ${\max }_{\rho }[S(\rho )+S(\mathcal{N}(\rho ))-S({\omega }_{RB})]$, the maximum quantum mutual information between reference and output. The preshared resource is independent of the message.

The classical-quantum state is block diagonal. If ${\lambda }_{xj}$ are the eigenvalues of ${\rho }_x$, its eigenvalues are $p_x{\lambda }_{xj}$. Inserting these into the definition of Von Neumann entropy gives the displayed sum of classical Shannon entropy and mean conditional quantum entropy. Its quantum mutual information is therefore the Holevo quantity $S({\sum }_x{p}_x{\rho }_x)-{\sum }_x{p}_{x}S({\rho }_x)$.

Attach an orthogonal classical label register to an ensemble. Its quantum mutual information with the quantum system equals the ensemble Holevo quantity. A local quantum channel changes the ensemble states but preserves their labels and probabilities. Data processing for quantum mutual information then proves that the output Holevo quantity cannot exceed the input value.

For a memoryless quantum channel $\mathcal{N}$, successive uses are described by ${\mathcal{N}}^{\otimes n}$. Its entanglement-assisted classical capacity is the supremum of classical bits per use achievable with error tending to zero when sender and receiver may use unlimited preshared quantum entanglement, independent of the message. Only the channel is used to transmit the message.

For input state ${\rho }_A$, choose a purification $|{\psi }_{\rho }{⟩}_{RA}$ and set ${\omega }_{RB}=({\mathrm{id}}_R\otimes \mathcal{N})(|{\psi }_{\rho }⟩⟨{\psi }_{\rho }|)$. The capacity formula isPurifications differ only by an isometry on the reference, so this expression depends only on the input density matrix. The quantity optimized is quantum mutual information.

The binary quantum erasure channel sends a qubit to a qubit subspace plus an orthogonal erasure flag:On the nonerased branch it transmits the entire state, including its coherences. Put $s=S(\rho )$. The entropy of an orthogonal quantum mixture givesThe reference-output state has two orthogonal branches:The first branch is pure; the second has entropy $S({\rho }_R)=s$. Thus $S({\omega }_{RB})=h_2(p)+ps$ andA qubit has entropy at most one bit, with equality at $\rho =I_2/2$. We concludeThe endpoint capacities are two and zero. As an operational check, superdense coding encodes four classical messages in one member of a shared Bell pair. A successful channel use lets the receiver identify the message; an erased qubit yields the flag. This induces a four-symbol classical erasure channel of capacity $(1-p){\log }_{2}4=2(1-p)$, agreeing with the entropy calculation.

Expand the difference between the two quantum mutual informations:This is precisely the nonnegative expression in Strong subadditivity of Von Neumann entropy. ThusThe reduced state on $AB$ is obtained by the partial trace over $C$, so this proves that discarding a system cannot increase its quantum mutual information with the retained reference.

Here a quantum operation is interpreted as a deterministic quantum channel, so it is completely positive and matrix trace preserving. Write its Kraus representation with ${\sum }_k{A}_k^{†}{A}_k=I$ and define the isometry in its Stinespring dilationThen $V^{†}V=I$ and $\Phi (\rho )={\mathrm{Tr}}_E(V\rho V^{†})$. Acting on $B$ gives ${\sigma }_{A{B}^{\prime }E}=(I_A\otimes V){\rho }_{AB}(I_A\otimes V^{†})$. An isometry preserves the nonzero eigenvalues of a density operator. Hence $S(A{B}^{\prime }E)=S(AB)$, $S(B^{\prime }E)=S(B)$, and $S(A)$ is unchanged. It follows that $I(A:B^{\prime }E{)}_{\sigma }=I(A:B{)}_{\rho }$.

Now discard $E$ and apply part 1. This proves data processing for quantum mutual information:The matrix trace-preserving convention matters. Postselection can increase conditional quantum mutual information: take a uniform classical bit $A$, copy it to $B$ with probability $\epsilon$, and otherwise put $B$ in an erasure state independent of $A$. Initially $I(A:B)=\epsilon$ bits. Projecting onto the nonerased sector and conditioning on success gives a perfectly correlated bit pair with quantum mutual information one. That success branch is matrix trace decreasing, and its renormalized action is not a deterministic channel. Retaining the full outcome flag and averaging all branches restores the ordinary channel inequality.

Use base-two logarithms, so the Von Neumann entropy is $S(\rho )=-\mathrm{Tr}(\rho {\log }_2\rho )$, with $0\log 0=0$. The quantum mutual information is $I(A:B)=S(A)+S(B)-S(AB)$; this is the quantity denoted by $S({\rho }_A:{\rho }_B)$ in the question, using the given joint state.

Strong subadditivity of Von Neumann entropy states that for every tripartite density operator,Equivalently the conditional quantum mutual information $I(A:C\mid B)$ is nonnegative. We apply this property to the original state or to a dilation of the relevant quantum channel.

For probability distributions $p$ and $q$ on a finite alphabet, their classical relative entropy in bits isA zero-$p$ term contributes zero, including when $q(x)=0$; a term with $p(x)>0$ and $q(x)=0$ makes the result $+\infty$. In that infinite case nonnegativity is immediate. Otherwise $q(x)>0$ on the support $J_p$ of $p$. For every $u>0$, $\ln u\le u-1$: the function $u-1-\ln u$ has derivative $1-1/u$ and its minimum zero at $u=1$. Apply this with $u=q(x)/p(x)$ to obtainThe last inequality uses normalization of $q$. Thus $\boxed{D(p\Vert q)\ge 0}$. Equality requires $q(x)=p(x)$ wherever $p(x)>0$ and no $q$-mass outside that support, hence occurs precisely when $p=q$.

For density matrices, usewhen the support of a positive operator $\rho$ is contained in that of $\sigma$, and set it to $+\infty$ otherwise. The data-processing inequality for quantum relative entropy states that for any quantum channel $\Phi$,Thus the relative entropy of the two output states cannot exceed that of the inputs.

For the states with a classical flag, omit indices of weight zero. On a positive-weight block, the matrix logarithms obeyon the relevant supports. The logarithms of $p_i$ cancel. Taking the trace block by block therefore gives the relative entropy of classically flagged states:If support inclusion fails in a positive-weight block, both sides are infinite. A zero-weight block makes no contribution, even if its unweighted relative entropy is infinite.

Tracing out the flag is a quantum channel with operators $A_i=⟨i|\otimes I$, since ${\sum }_i{A}_i^{†}{A}_i=I$. It sends the two flagged states to ${\sum }_i{p}_i{\rho }_i$ and ${\sum }_i{p}_i{\sigma }_i$. Applying the data-processing inequality for quantum relative entropy to this partial trace proves joint convexity of quantum relative entropy:

For a bipartite state, set ${\rho }_A={\mathrm{Tr}}_B{\rho }_{AB}$ and ${\rho }_B={\mathrm{Tr}}_A{\rho }_{AB}$. Its quantum mutual information iswhere $S(\tau )=-\mathrm{Tr}\tau {\log }_2\tau$ is the Von Neumann entropy. The support of ${\rho }_{AB}$ lies in $\mathrm{supp}{\rho }_A\otimes \mathrm{supp}{\rho }_B$. To justify this when marginals are singular, let $Q_A$ project onto $\ker {\rho }_A$. Then $\mathrm{Tr}[(Q_A\otimes I){\rho }_{AB}]=\mathrm{Tr}(Q_A{\rho }_A)=0$; positivity implies $(Q_A\otimes I){\rho }_{AB}=0$. Applying the same argument on $B$ proves the claimed support inclusion.

On that support, product eigenvectors showThe definition of the partial trace therefore givesConsequently $\boxed{S(A:B)=S({\rho }_{AB}\Vert {\rho }_A\otimes {\rho }_B)}$.

For completeness, nonnegativity of quantum relative entropy also follows from the classical proof above. Diagonalize $\rho$ in a basis $|j⟩$, with eigenvalues $r_j$, and put $q_j=⟨j|\sigma |j⟩$. These are two classical probability distributions. Scalar concavity of the logarithm applied to the spectral weights of $\sigma$ gives $⟨j|{\log }_2\sigma |j⟩\le {\log }_2{q}_j$ whenever $r_j>0$ and support inclusion holds. HenceThe support-failure case is again infinite. Applying this to the product of the marginals proves $\boxed{S(A:B)\ge 0}$ without requiring an additional entropy inequality.

Write $\overline{\rho }={\sum }_i{p}_i{\rho }_i$. The Holevo quantity iswhere $S$ is Von Neumann entropy, with a consistent logarithm base. Attach an orthogonal classical label register $X$ to form the classical-quantum stateThe entropy of an orthogonal quantum mixture givesHere $H(p)$ is the Shannon entropy of the classical label. Hence $I(X:B)=\chi (\mathcal{E})$, where $I$ denotes quantum mutual information.

Let $\Phi :B\to C$ be the proposed quantum channel. Using its Kraus representation, define an linear isometry $V={\sum }_{\ell }{A}_{\ell }\otimes |\ell {⟩}_E$, since $V^{†}V=I$. This is a Stinespring representation. Apply it to $B$, obtaining a state ${\eta }_{XCE}$; tracing out $E$ applies $\Phi$ to every ensemble member. An linear isometry preserves nonzero eigenvalues and therefore Von Neumann entropy, soTheir difference isThe Strong subadditivity of Von Neumann entropy states that this expression is nonnegative for every tripartite state. ConsequentlyThus the Holevo quantity under a quantum channel decreases because discarding the dilation environment cannot improve correlations with the classical label. The probabilities remain unchanged throughout.

Let the quantum channel send $B$ to $B^{\prime }$. From its Kraus representation, choose a Stinespring representation $V:B\to B^{\prime }E$ and formThe channel output is ${\sigma }_{A{B}^{\prime }}={\mathrm{Tr}}_E{\sigma }_{A{B}^{\prime }E}$. Isometric invariance of Von Neumann entropy givesTherefore the quantum mutual information satisfies $I(A:B{)}_{\rho }=I(A:B^{\prime }E{)}_{\sigma }$. Expanding its loss after discarding the environment givesThe final expression is the quantum conditional mutual information $I(A:E\mid B^{\prime })$. The Strong subadditivity of Von Neumann entropy states exactly that it is nonnegative. HenceThis proves data processing for quantum mutual information for any mixed input and any local quantum channel, not just the atomic channel in earlier parts. The correlations lost are exactly the conditional correlations with the discarded environment.

A deterministic quantum operation here is a CPTP map. By Stinespring dilation, there is an isometry $V:B\to B^{\prime }E$ with $\Lambda (\tau )={\mathrm{Tr}}_E(V\tau V^{†})$. Let ${\omega }_{A{B}^{\prime }E}=(I\otimes V){\rho }_{AB}(I\otimes V^{†})$. An isometry preserves all nonzero eigenvalues, soThe quantum mutual information is $I(A:B)=S(A)+S(B)-S(AB)$. The difference after discarding the environment iswhere the last inequality is Strong subadditivity of Von Neumann entropy, in the form $S(A{B}^{\prime })+S(B^{\prime }E)\ge S(B^{\prime })+S(A{B}^{\prime }E)$. Thus data processing for quantum mutual information givesSubsystem $A$ is unchanged; its prime simply labels the output. Keeping the environment would preserve the mutual information, while tracing it out cannot increase it.

Take a Stinespring dilation $V:B⟶B^{\prime }E$ of the local quantum channel and define ${\tau }_{A{B}^{\prime }E}=(I_A\otimes V){\rho }_{AB}(I_A\otimes V^{†})$. Tracing out $E$ gives the prescribed output, with $A^{\prime }=A$. An isometry preserves the nonzero eigenvalues of a density operator; therefore $S(B^{\prime }E{)}_{\tau }=S(B{)}_{\rho }$, $S(A{B}^{\prime }E{)}_{\tau }=S(AB{)}_{\rho }$, and $S(A{)}_{\tau }=S(A{)}_{\rho }$.

Consequently $I(A:B^{\prime }E{)}_{\tau }=I(A:B{)}_{\rho }$. The loss of quantum mutual information isThe last inequality is Strong subadditivity of Von Neumann entropy, or nonnegativity of quantum conditional mutual information. HenceThis mutual-information loss as conditional mutual information shows exactly which correlations are discarded into the environment. No purity assumption on the original state is needed.

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.

Both output states have the same reference marginal, and $S(R)=S(\rho )$. Expand quantum mutual information to obtain the mutual information and coherent information identityThe input-entropy term is identical in the two expressions. Subtract them and use the preceding data-processing inequality for coherent information:Thus the final quantum channel cannot increase the reference-output quantum mutual information, as required by data processing for quantum mutual information.