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Every separable quantum state has positive partial transpose. Here the partial transpose iswith eigenvalues $1/2,1/2,\alpha /2,-\alpha /2$. Any nonzero $\alpha$ therefore gives a negative eigenvalue, excluding a separable quantum state. At $\alpha =0$, the explicit decompositionis a mixture of product states. ConsequentlyOnly necessity of the positive partial transpose criterion is needed here; sufficiency at zero follows directly from the decomposition.

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 $1/2$.

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.

The positive partial transpose criterion states that every separable quantum state satisfiesConsequently, a negative eigenvalue of the partial transpose proves entanglement. Positivity of the partial transpose is also sufficient for separability in dimensions $2\otimes 2$ and $2\otimes 3$, equivalently $3\otimes 2$, but it is not sufficient in general higher dimensions.

At the proposed boundary $p_0=1/(d+1)$, the identity from part (iv) yieldsBoth $\sigma$ and $D/d$ are convex combinations of product states, so ${\rho }_{p_0}$ is a separable quantum state. For $0\le p\le p_0$, the state ${\rho }_p$ is a convex combination of ${\rho }_{p_0}$ and the maximally mixed product state $I/d^2$, and is therefore separable.

For the converse, the partial transpose of the maximally entangled projector is $F/d$, where $F$ is the swap operator. ThereforeOn the antisymmetric subspace, $F$ has eigenvalue $-1$, so the corresponding eigenvalue of ${\rho }_p^{T_B}$ iswhich is negative exactly when $p>1/(d+1)$. The positive partial transpose criterion then proves that ${\rho }_p$ is entangled. Thus

The positive partial transpose criterion says separability implies ${\rho }_{AB}^{T_B}\ge 0$. A nonpositive partial transpose therefore proves entanglement. Positive partial transpose is also sufficient for separability in dimensions $2\otimes 2$ and $2\otimes 3$, but not in general higher dimensions.

Partial transposition interchanges the positions occupied by $x$ and $y$. Positivity of ${\rho }^{T_B}$ therefore requiresBecause the system is $2\otimes 2$, the positive partial transpose criterion is necessary and sufficient. Combining these inequalities with validity of the original state gives

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.

Every positive linear map between $M_2(\mathbb{C})$ and $M_2(\mathbb{C})$, or between $M_2(\mathbb{C})$ and $M_3(\mathbb{C})$ in either direction, is a decomposable positive map. The low-dimensional result is the ingredient that makes the positive partial transpose criterion sufficient in two-qubit and qubit-qutrit systems. Woronowicz's low-dimensional positive-map paper establishes the $M_2$ to $M_3$ case; taking adjoints gives the reverse direction.

A tensor-product operator factors as $A\otimes B$ on a tensor product of two spaces. If both factors are positive semidefinite operators, their tensor-product operator is positive: a tensor-product basis of eigenvectors has eigenvalues $a_i{b}_j\ge 0$. This positivity underlies the necessary direction of the positive partial transpose criterion.