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Use base-two logarithms for Von Neumann entropy, so the units are bits, and adopt $0{\log }_{2}0=0$. All systems here may be taken finite-dimensional; the entropy arguments also apply when the displayed Von Neumann entropies are finite. The density operator of the joint pure state is ${\rho }_{AB}=|{\Psi }_{AB}⟩⟨{\Psi }_{AB}|$. It is a rank-one orthogonal projector, whose only nonzero eigenvalue is one. ConsequentlyThe reduced density matrices are ${\rho }_A={\mathrm{Tr}}_B{\rho }_{AB}$ and ${\rho }_B={\mathrm{Tr}}_A{\rho }_{AB}$. The quantum conditional entropy is defined by the entropy differenceFor the pure state under consideration this becomes $S(B|A)=-S({\rho }_A)$. To determine when it is negative, write the Schmidt decompositionThe eigenvalues of ${\rho }_A$ are the ${\lambda }_j$, together with possible zeros. Each contribution $-{\lambda }_j{\log }_2{\lambda }_j$ is nonnegative, and their sum vanishes exactly when one eigenvalue equals one. Equivalently the Schmidt rank is one and $|{\Psi }_{AB}⟩$ is a product state. If the pure state is entangled, then $r\ge 2$, every nonzero eigenvalue is strictly below one, and the Von Neumann entropy is strictly positive. Thus the pure-state negative conditional entropy criterion givesThe purity assumption matters: for mixed density operators, negative quantum conditional entropy is not necessary for entanglement.

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: $S(A\mid BC)\le S(A\mid B)$.

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.

Extend the quantum conditional entropy from normalized states to positive operators bySince $\mathrm{Tr}{X}_{AB}=\mathrm{Tr}{X}_B$, the two terms involving $\log t$ cancel under $X\mapsto tX$, so $F(tX)=tF(X)$. Thus $F$ is positively homogeneous, and part (b) extends its concavity from states to the positive cone.

Apply part (c) with $X={\sigma }_{AB}$ and $Y={\rho }_{AB}$. Differentiating the matrix logarithm under the trace givesThe inequality from part (c), after moving $F(\rho )$ to the left, becomesThis is the data-processing inequality for quantum relative entropy under partial trace. Tensoring each output with the appropriate maximally mixed state does not change either side, so it also proves data processing under normalized partial traces. Singular $\sigma$ follows by approximation on its support.

For a bipartite pure state with finite Von Neumann entropy, the joint Von Neumann entropy is zero, so its quantum conditional entropy is $-S({\rho }_A)$. The Schmidt decomposition makes the nonzero eigenvalues of ${\rho }_A$ the squared Schmidt coefficients. Their Shannon entropy vanishes exactly when there is one coefficient, equivalently the joint pure state is a product state. Thus negative quantum conditional entropy detects exactly the entangled bipartite pure states; it is only a sufficient criterion for entanglement of general mixed density operators.

For a tripartite density operator ${\rho }_{ABC}$,Equivalently, conditioning on an additional quantum system cannot increase quantum conditional entropy.

The quantum conditional entropy has the relative-entropy representationwhere the minimum ranges over density operators on $B$.