Separable quantum state

A bipartite quantum state is separable when its density operator is a convex combination ${\sum }_j{p}_j{\rho }_j^A\otimes {\rho }_j^B$ of product states; otherwise it has entanglement. Operationally, Alice and Bob can prepare a separable quantum state by sharing a classical random label $j$ and each preparing the corresponding local state.

Separable does not mean uncorrelated. The mixture in Bell correlations versus a classical mixture is separable with two terms of weight $1/2$, yet the parties' $Z$ records agree perfectly. A single product state would give independent local outcomes. Allowing a mixture of products captures shared classical randomness; excluding this possibility would incorrectly classify ordinary correlations as entanglement.

A separable positive operator is a finite sum ${\eta }_{AB}={\sum }_y{\alpha }_y\otimes {\beta }_y$ with both factors positive operators. If its trace is one, normalizing the nonzero factors gives a separable quantum state. Conversely, a nonnegative scalar multiple of a separable quantum state is a separable positive operator.

A finite-dimensional bipartite density operator ${\rho }_{AB}$ is a separable quantum state exactly when $({\mathrm{id}}_A\otimes \Phi )({\rho }_{AB})\ge 0$ for every positive linear map from operators on $B$ to operators on $A$. In combination with the Størmer-Woronowicz decomposability theorem, it makes the positive partial transpose criterion sufficient for qubit-qutrit states. The criterion is established in the Horodeckis' separability paper.

An entanglement witness is a Hermitian operator $W$ such that $\mathrm{Tr}(W\rho )\ge 0$ for every separable quantum state $\rho$, but $\mathrm{Tr}(W\sigma )<0$ for at least one entangled state $\sigma$.

In a fixed product basis, the partial transpose on subsystem $B$ sendsIt preserves Hermiticity and trace but need not preserve positivity on entangled inputs.

For $|{\Phi }_d⟩=d^{-1/2}{\sum }_j|jj⟩$, the partial transpose of its pure state projector is $S/d$, where $S$ is the swap operator. Expand in matrix units: $|i⟩⟨j|\otimes |i⟩⟨j|$ becomes $|i⟩⟨j|\otimes |j⟩⟨i|$.

A bipartite density operator has positive partial transpose when ${\rho }^{T_B}\ge 0$. Every separable quantum state has this property, since transposing one factor of a positive product state preserves positivity and the separable sum preserves it as well. The property does not depend on which local basis defines the transpose.

Every density operator $\sigma$ with positive partial transpose has squared quantum fidelity at most $1/d$ with a fixed maximally entangled state on $d\otimes d$. The partial transpose of a maximally entangled projector and the trace-adjoint identity for partial transpose express that squared overlap as $\mathrm{Tr}(S{\sigma }^{T_B})/d$. Since ${\sigma }^{T_B}$ is positive of trace one and the swap operator has operator norm one, trace duality gives the bound. A matching product state attains equality.

For bipartite operators $X,Y$ in a fixed product basis,Expand both operators in products of matrix units. On the transposed factor, the identity is $\mathrm{Tr}(AT)=\mathrm{Tr}(A^T{T}^T)$; bilinearity gives the bipartite result. It allows a negative partial transpose eigenvector to define an entanglement witness.

Every separable state has positive partial transpose. The condition is also sufficient for separability on $2\otimes 2$ and $2\otimes 3$ systems.

A channel is entanglement breaking when applying it to one side of every bipartite state always produces a separable state. Equivalently, its Choi matrix is separable, and equivalently it has a measure-and-prepare representation.

A quantum channel is entanglement breaking exactly when its Choi state is a separable quantum state. If $C_{\Lambda }={\sum }_a{p}_a{\sigma }_a\otimes {\tau }_a$, the Choi reconstruction formula gives a measure-and-prepare channel with POVM $E_a=d{p}_a{\tau }_a^T$. Conversely, applying an entanglement-breaking channel to a maximally entangled state produces a separable Choi state. Horodecki, Shor and Ruskai's entanglement-breaking channel paper develops the equivalences.

An entanglement-breaking channel admits a Kraus representation with rank-one Kraus operators $K_j=|a_j⟩⟨b_j|$. Its Choi state is a sum of positive tensor-product operators, hence separable. Conversely, spectral decompositions of each prepared state and each POVM effect in a measure-and-prepare channel produce rank-one Kraus operators. Trace preservation imposes ${\sum }_j\Vert a_j{\Vert }^2|b_j⟩⟨b_j|=I$.