Past exam of the mathematics course of the University of Cambridge / 2019 / iii / Paper 323 / 1

A bipartite density operator ${\rho }_{AB}$ has positive partial transpose when ${\rho }_{AB}^{T_B}\ge 0$ in a product basis. Positivity is independent of the choice of local bases, although the matrix representing the partial transpose changes.

If the state is a separable quantum state, write ${\rho }_{AB}={\sum }_j{p}_j{\rho }_j^A\otimes {\rho }_j^B$. Taking its partial transpose givesIndeed, the matrix transpose preserves the nonnegative eigenvalues of each local density operator; a tensor product of positive semidefinite operators is positive, and so is their nonnegative sum. This proves the necessary direction of the positive partial transpose criterion.

Use two precise finite-dimensional results. The positive-map separability criterion says that a state on ${\mathbb{C}}^2\otimes {\mathbb{C}}^3$ is separable exactly when $({\mathrm{id}}_2\otimes \Phi )(\sigma )\ge 0$ for every positive linear map $\Phi :M_3(\mathbb{C})\to M_2(\mathbb{C})$. The Størmer-Woronowicz decomposability theorem says that every such map is a decomposable positive map, sowith ${\Phi }_1,{\Phi }_2$ completely positive maps and $T$ the matrix transpose on $M_3$.

If $\sigma$ has positive partial transpose, both $\sigma$ and ${\sigma }^{T_B}$ are positive. HenceThe positive-map separability criterion now proves that $\sigma$ is a separable quantum state. The dimension-specific decomposability theorem is essential: the conclusion does not extend to arbitrary bipartite dimensions.

The partial transpose of the spin singlet state projector is the Hermitian operatorFor every separable quantum state $\tau$, the positive partial transpose criterion and the trace-adjoint identity for partial transpose giveFor the specified Bell state, however,Thus $W$ is an entanglement witness detecting $|{\Phi }^{+}⟩$. In fact, it detects any state whose overlap with that Bell state exceeds $1/2$.

Use the normalized Choi state $C_{\Lambda }=(\Lambda \otimes \mathrm{id})(|{\Phi }_d⟩⟨{\Phi }_d|)$, with $|{\Phi }_d⟩=d^{-1/2}{\sum }_i|ii⟩$. If $\Lambda$ is an entanglement-breaking channel, its action on this particular bipartite input makes $C_{\Lambda }$ a separable quantum state.

Conversely, suppose $C_{\Lambda }={\sum }_a{p}_a{\sigma }_a\otimes {\tau }_a$, with local density operators ${\sigma }_a,{\tau }_a$. The Choi reconstruction formula givesBecause $\Lambda$ is trace preserving, ${\mathrm{Tr}}_A{C}_{\Lambda }=I/d$, so ${\sum }_a{E}_a=I$. Thus the $E_a$ form a POVM and the channel is a measure-and-prepare channel.

For any bipartite input ${\rho }_{AR}$, define the positive, possibly unnormalized reference operatorsIts output is ${\sum }_a{\sigma }_a\otimes R_a$. Since ${\sum }_a\mathrm{Tr}{R}_a=1$, normalizing each nonzero $R_a$ expresses this as a convex combination of product states. Hence every output is separable, proving the separable Choi-state criterion for entanglement breaking.

Apply the measure-and-prepare channel locally to an arbitrary bipartite density operator ${\rho }_{AR}$. The resulting state iswhereTo see that this is the correct output, the partial trace is cyclic for operators acting only on the traced subsystem, so the displayed expression equals ${\mathrm{Tr}}_A[(E_a\otimes I){\rho }_{AR}]$. Put $q_a=\mathrm{Tr}{R}_a$. The POVM completeness relation implies ${\sum }_a{q}_a=1$, and terms with $q_a=0$ vanish. Thereforeis a separable quantum state. Thus the channel is an entanglement-breaking channel, for every reference-system dimension.

Expanding the normalized maximally entangled state in the definition of the Choi state givesHere the bar denotes componentwise complex conjugation in the basis defining the Choi state. Every term is a positive tensor-product operator, and the whole operator has trace one because $\tilde{\Lambda }$ is assumed to be a quantum channel. It is therefore a separable quantum state, proving entanglement breaking by the separable Choi-state criterion for entanglement breaking.

If the vectors are unit vectors, the displayed $p_j$ are already the product-state weights. If they are not normalized, absorb their squared norms into the weights and normalize the nonzero vectors. Trace preservation requiresthe probability-distribution condition alone would not guarantee this. Equivalently, the Kraus operators are $K_j=\sqrt{d{p}_j}|a_j⟩⟨b_j|$, exhibiting the rank-one Kraus representation of an entanglement-breaking channel.