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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 $1/2$.

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.