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Purify the input on $RA$ and dilate a channel by $A\to BE$. The joint state $RBE$ is pure, so complementary entropies obey $S(B)=S(RE)$ and $S(RB)=S(E)$. Therefore coherent information is $I_c=S(B)-S(RB)=S(R\mid E)=-S(R\mid B)$. The positive environment conditional entropy and negative receiving-system conditional entropy are two equivalent forms.

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 quantum channel on the receiving system cannot increase coherent information: $I(A⟩B{)}_{\rho }\ge I(A⟩B^{\prime }{)}_{\sigma }$. Dilate the channel to a linear isometry of Hilbert spaces $B\to B^{\prime }E$. The difference is $I(A:E|B^{\prime })$ of the dilated state, nonnegative by Strong subadditivity of Von Neumann entropy.

The channel acts only on the input system $Q$; the reference $R$ is untouched. WriteThe coherent information isThe reference marginal remains ${\rho }_R$, whose nonzero eigenvalues agree with those of $\rho$. Unlike a classical information quantity, coherent information can be negative.

The lower diagram uses a Stinespring dilation: append an environment in a fixed pure state, apply a joint unitary operator, and discard the environment by partial trace. Before discarding it, the global $RBE$ state is pure.

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.

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.

Choose a Stinespring dilation $V:A⟶B_{1}E$ of the quantum channel. Applying it to the input purification gives the pure vectorIts $R{B}_1$ marginal is the output state in the definition of coherent information. Complementary subsystems of a pure state have the same nonzero eigenvalues, by the Schmidt decomposition. Thus $S(B_1)=S(RE)$ and $S(R{B}_1)=S(E)$. It follows thatHere $S(R\mid E)$ denotes quantum conditional entropy, evaluated on the complementary output $RE$. This is coherent information as an environment conditional entropy. Equivalently $I_c=-S(R\mid B_1)$; the environment expression has the positive sign, while the receiving-system expression has the negative sign.

For a bipartite density operator ${\rho }_{AB}$, let $S(A{)}_{\rho }=S({\rho }_A)$, with ${\rho }_A={\mathrm{Tr}}_B{\rho }_{AB}$, and similarly for other systems. The quantum conditional entropy and coherent information areUnlike classical conditional entropy, quantum conditional entropy can be negative. For a Bell state, $S(AB)=0$ and $S(B)=1$, giving $H(A|B)=-1$ and coherent information equal to one bit.

Use two results: the isometric Stinespring dilation of a quantum channel, and Strong subadditivity of Von Neumann entropy. Let $V:B\to B^{\prime }E$ dilate the given operation, and defineAn linear isometry of Hilbert spaces preserves the nonzero eigenvalues, so $S(B^{\prime }E{)}_{\omega }=S(B{)}_{\rho }$ and $S(A{B}^{\prime }E{)}_{\omega }=S(AB{)}_{\rho }$. Subtracting the two coherent information expressions givesThe 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 the data-processing inequality for coherent information isThe lost coherent information is precisely the quantum conditional mutual information between the reference $A$ and discarded environment $E$, conditional on the retained output $B^{\prime }$.

For the pure state ${\sigma }_{RQE}$ induced by a purification and Stinespring dilation, the coherent information is

For a tripartite density operator, the quantum conditional mutual information is $I(A:C|B)=S(AB)+S(BC)-S(B)-S(ABC)$. It is nonnegative by Strong subadditivity of Von Neumann entropy. It measures the coherent information lost when a channel environment is discarded after an isometric dilation.