Correlated preparations with identical marginals diverge

Let $\mathrm{\mid }{\psi }_{\pm }⟩=(\mathrm{\mid }00⟩\pm \mathrm{\mid }11⟩)\mathrm{/}\sqrt{2}$ and ${\omega }_{\pm }=\mathrm{\mid }{\psi }_{\pm }⟩⟨{\psi }_{\pm }\mathrm{\mid }$; each ${\omega }_{\pm }$ is positive semidefinite with trace one. Using [F1], the cross terms $\mathrm{\mid }00⟩⟨11\mathrm{\mid }$ and $\mathrm{\mid }11⟩⟨00\mathrm{\mid }$ vanish under ${\mathrm{Tr}}_E$ because their environment indices differ, while the diagonal terms give ${\mathrm{Tr}}_E{\omega }_{+}={\mathrm{Tr}}_E{\omega }_{-}=I_2\mathrm{/}2$.

Define $U\mathrm{\mid }a,b⟩=\mathrm{\mid }a,a\oplus b⟩$ for $a,b\in \{0,1\}$, where $\oplus$ is addition modulo two. This CNOT map permutes the orthonormal product basis and is therefore unitary. It sends $\mathrm{\mid }{\psi }_{\pm }⟩$ to $\mathrm{\mid }\pm ⟩\otimes \mathrm{\mid }0⟩$, where $\mathrm{\mid }\pm ⟩=(\mathrm{\mid }0⟩\pm \mathrm{\mid }1⟩)\mathrm{/}\sqrt{2}$. Thus ${\mathrm{Tr}}_E(U{\omega }_{\pm }{U}^{\ast })=\mathrm{\mid }\pm ⟩⟨\pm \mathrm{\mid }$, and the outputs are distinct because their $(0,1)$ matrix entries are $1\mathrm{/}2$ and $-1\mathrm{/}2$. Equal system marginals therefore do not determine the final system state for these correlated preparations, refuting the claim. The product-preparation theorem [F2] is unaffected.