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A decomposable positive map between complex matrix algebras has the form $\Phi ={\Phi }_1+{\Phi }_2\circ T$, where ${\Phi }_1,{\Phi }_2$ are completely positive maps and $T$ is the matrix transpose. Such a map cannot detect a positive partial transpose entangled state: applying it to one subsystem gives a sum of two positive operators.

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.

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.