# Quantum marginal problem - constructing a global state from reduced states

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

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 $$\rho_{AB}$$.