Quantum marginal problem - constructing a global state from reduced states

Question

Consider a bipartite state $\rho_{AB}$ with reduced states $\rho_A = \text{Tr}_B(\rho_{AB})$ and $\rho_B = \text{Tr}_A(\rho_{AB})$.

Suppose one obtains states $\rho'_{A}$ and $\rho'_{B}$ such that $\|\rho'_A - \rho_A\|_1 \leq \delta$ and $\|\rho'_B - \rho_B\|_1 \leq \delta$. That is, we have perturbed the reduced states slightly. In the specific problem I am looking at $\rho'_A$ commutes with $\rho_A$ and $\rho'_B$ commutes with $\rho_B$ but perhaps this is not relevant.

Does there exist a global state $\rho'_{AB}$ with marginals $\rho'_A$ and $\rho'_B$ such that $\|\rho'_{AB} - \rho_{AB}\|_1\leq \varepsilon(\delta)$, where $\lim_{\delta\rightarrow 0}\varepsilon(\delta) = 0$?

Answer 1

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 $|\psi\rangle$

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

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

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

and

$$ \lim_{\delta\rightarrow 0}\varepsilon(\delta) = \lim_{\delta\rightarrow 0} 2\delta = 0 $$

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 $\rho_{AB}$.