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For real $-1\le \alpha \le 1$, this two-qubit density operator has equal $00$ and $11$ populations and off-diagonal coherences $\alpha /2$. It is a pure state exactly when $\alpha =\pm 1$, and a separable quantum state exactly when $\alpha =0$. Its partial transpose has eigenvalues $1/2,1/2,\alpha /2,-\alpha /2$, directly witnessing entanglement for every nonzero admissible coherence.

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 linear map $\Phi$ is a completely positive map if ${\mathrm{id}}_R\otimes \Phi$ sends every positive operator to a positive operator for every finite-dimensional auxiliary system $R$. Positivity of $\Phi$ alone tests only inputs without an auxiliary system and is weaker.

For the Kraus representation, let $\Omega \ge 0$ be any operator on the auxiliary system and the input system. ThenFor any vector $v$, its expectation is ${\sum }_k⟨(I_R\otimes A_k^{†})v,\Omega (I_R\otimes A_k^{†})v⟩\ge 0$. Every amplification is therefore positive, proving the Kraus-form map is completely positive. No normalization condition on the $A_k$ is needed for this positivity proof. To make it a deterministic quantum channel, one additionally requires ${\sum }_k{A}_k^{†}{A}_k=I$; a physical outcome branch instead obeys ${\sum }_k{A}_k^{†}{A}_k\le I$.

Transposition is positive on one system: for any positive $\rho$ and any $v$, $v^{†}{\rho }^{T}v$ is the complex conjugate of ${\bar{v}}^{†}\rho \bar{v}$, hence is real and nonnegative. It fails complete positivity already for a qubit. Apply partial transposition to the second subsystem of the Bell state $|{\Phi }^{+}⟩=(|00⟩+|11⟩)/\sqrt{2}$. The resulting operator isOn the antisymmetric vector $(|01⟩-|10⟩)/\sqrt{2}$, its eigenvalue is $-1/2$. Thus the partial transpose is not positive on this entangled input, andThe same witness embeds into any larger input dimension; in the exceptional one-dimensional case transposition is simply the identity.

First explain the stated local conversion. The four equally weighted correlated local Pauli operations $I\otimes I$, $X\otimes X$, $Z\otimes Z$, and $Y\otimes Y$ remove coherences in the Bell states basis, since those states have distinct joint parity eigenvalues. This leaves four Bell probabilities. One-party Pauli operations permute the Bell projectors, so the largest probability can be placed on the singlet ${\Phi }_{-}$. Call it $p$; necessarily $p\ge 1/4$.

The two common unitaries found above generate permutations of the three triplet projectors: $S\otimes S$ exchanges ${\Psi }_{+}$ and ${\Psi }_{-}$, and $HSH\otimes HSH$ exchanges ${\Psi }_{+}$ and ${\Phi }_{+}$. Averaging the six triplet permutations replaces their weights by their common mean, while collective-unitary covariance of the two-qubit singlet fixes the singlet projector. This is Bell twirling followed by triplet symmetrization, producing the specified two-qubit Werner state. Shared randomness coordinates the local choices, so the procedure is allowed by local operations and classical communication. Common unitaries alone would preserve the original singlet weight, which need not initially be at least $1/4$; the preceding one-party permutation is what ensures that range.

For the eigenvalue calculation, put $P_s=|{\Phi }_{-}⟩⟨{\Phi }_{-}|$, $a=(1-p)/3$, and $b=p-a=(4p-1)/3$. Then $\rho =aI+b{P}_s$. In the ordered computational basis $00,01,10,11$, transposing the second subsystem givesThe $00,11$ block has eigenvalues $a-b/2$ and $a+b/2$, while the other two basis vectors also have eigenvalue $a+b/2$. Thus the partial transpose spectrum isThe last three are positive throughout the given range, so the partial transpose is not positive exactly when $p>1/2$. A separable quantum state is a convex sum of product density operators, each still positive after transposing one factor. The negative eigenvalue therefore proves entanglement.

For completeness, separability on the other side can be proved explicitly rather than inferred solely from a theorem name. Let $P_{j,\pm }=(I\pm {\sigma }_j)/2$ be the one-qubit Pauli eigenprojectors. At $p=1/2$, the mixture of six antiparallel product states isusing $P_s=(I-{\sum }_j{\sigma }_j\otimes {\sigma }_j)/4$. It is manifestly separable. For $1/4\le p\le 1/2$,with nonnegative coefficients summing to one; $I/4$ is also separable. ConsequentlyAt the boundary $p=1/2$ the partial transpose has a zero eigenvalue and the explicit decomposition shows separability.

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.

Let $\Gamma =T_B$ denote the partial transpose. The partial transpose of a maximally entangled projector iswhere $S$ is the swap operator, satisfying $S^{†}S=I$ and $\Vert S{\Vert }_{\mathrm{op}}=1$. If $\sigma$ has positive partial transpose, then ${\sigma }^{\Gamma }$ is a positive operator of trace one, hence $\Vert {\sigma }^{\Gamma }{\Vert }_1=1$.

For a pure state ${\varphi }^{+}=|\Phi ⟩⟨\Phi |$, the unsquared quantum fidelity obeys $F({\varphi }^{+},\sigma {)}^2=⟨\Phi |\sigma |\Phi ⟩$. Using the trace-adjoint identity for partial transpose and part (i),The overlap is nonnegative, so taking its square root proves the PPT maximally entangled overlap bound:The product state $|00⟩⟨00|$ attains the bound, so it is sharp.

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 swap operator exchanges two tensor factors: $F(|\psi ⟩\otimes |\varphi ⟩)=|\varphi ⟩\otimes |\psi ⟩$. It has eigenvalue $+1$ on the symmetric subspace and $-1$ on the antisymmetric subspace, and the partial transpose of $|{\Phi }^{+}⟩⟨{\Phi }^{+}|$ is $F/d$.

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.

This collective-unitary-invariant two-qubit state has singlet weight $p$ and equal weights on the three triplet Bell states. With $P_s$ the spin singlet state projector, its partial transpose has eigenvalues $(1-2p)/2$ once and $(1+2p)/6$ three times. It is entangled exactly when $p>1/2$. At $p=1/2$, it is the equally weighted mixture of six antiparallel product states along the three Pauli axes. In the interval $1/4\le p\le 1/2$, convex mixing of this boundary state with $I/4$ explicitly proves separability. The weight $p$ here is the singlet probability, not the alternative coefficient multiplying a pure singlet in a white-noise mixture.