Quantum marginal problem - constructing a global state from reduced states

If perturbations are sufficiently small and $rho_{AB}$ has sufficiently broad support then a desired global state $rho_{AB}'$ exists. Define

$$ rho_{AB}' = rho_{AB} + (rho_A' - rho_A) otimes rho_B + rho_A otimes (rho_B' - rho_B) tag1. $$

Note that $rho_{AB}'$ is Hermitian and trace one, but may not be positive. However, $rho_{AB}'$ is positive if $rho_{AB}$ has broad support and perturbations are not too large. For example, if

$$ |rho_A' - rho_A|_2 + |rho_B' - rho_B|_2 leq lambda_{min}(rho_{AB}) $$

where $lambda_{min}(X)$ denotes the smallest eigenvalue of operator $X$, then for any $|psirangle$

$$ begin{align} langlepsi|rho_{AB}'|psirangle & = langlepsi|rho_{AB}|psirangle + langlepsi|(rho_A' - rho_A) otimes rho_B|psirangle + langlepsi|rho_A otimes (rho_B' - rho_B)|psirangle & geq lambda_{min}(rho_{AB}) - |rho_A' - rho_A|_2 - |rho_B' - rho_B|_2 geq 0. end{align} $$

If $lambda_{min}(rho_{AB}') = 0$, then it may be possible to restrict the support of the perturbations so that an analogous inequality holds.

The reduced states of $rho_{AB}'$ are

$$ begin{align} {rm tr}_A(rho_{AB}') & = {rm tr}_A(rho_{AB}) + rho_B , {rm tr}(rho_A' - rho_A) + (rho_B' - rho_B) , {rm tr}(rho_A) & = rho_B + rho_B' - rho_B & = rho_B' end{align} $$

and similarly ${rm tr}_B(rho_{AB}') = rho_A'$.

Finally, the distance

$$ begin{align} |rho_{AB}' - rho_{AB}|_1 & = |(rho_A' - rho_A) otimes rho_B + rho_A otimes (rho_B' - rho_B)|_1 & leq |(rho_A' - rho_A) otimes rho_B|_1 + | rho_A otimes (rho_B' - rho_B)|_1 & = |rho_A' - rho_A|_1 + |rho_B' - rho_B|_1 & leq 2delta end{align} $$

and

$$ lim_{deltarightarrow 0}varepsilon(delta) = lim_{deltarightarrow 0} 2delta = 0 $$

as required.

Note that the above construction fails for very pure highly entangled states since in this case the reduced states are nearly completely mixed and so the two product terms in $(1)$ will contain negative elements that are not compensated for by $rho_{AB}$.