# Constructing a state with constraints on reduced states

Suppose $$\rho'_{AB} \approx_\varepsilon \rho_{AB}$$ in trace distance. Is there an explicit construction of some state $$\tilde{\rho}_{AB}$$ using $$\rho'_{AB}, \rho'_A, \rho'_B, \rho_A$$ and $$\rho_B$$ (but not $$\rho_{AB}$$) such that

$$\tilde{\rho}_A = \rho_A, \tilde{\rho}_B = \rho_B$$

while ensuring that $$\tilde{\rho}_{AB}$$ is as close in trace distance to $$\rho'_{AB}$$ as possible?

To get some obvious answers out of the way, we have the trivial solution of $$\tilde{\rho} = \rho_A\otimes\rho_B$$. This satisfies the marginal condition but probably is not optimal in terms of the trace distance between $$\tilde{\rho}$$ and $$\rho'$$.

A much cleverer attempt is shown in this answer to my previous question. Take

$$\tilde{\rho}_{AB} = \rho'_{AB} + (\rho_A - \rho'_A) \otimes \rho'_B + \rho'_A \otimes (\rho_B - \rho'_B)$$

but this construction is not guaranteed to be positive semidefinite for entangled pure states. Apart from that, it satisfies the requirement on the marginals and keeps $$\tilde{\rho}$$ close to $$\rho'$$.

Is there anything else one can try?

We know the trace norm of an operator $$X$$ can be formulated as \begin{aligned} \|X\|_1 &= \min_{Y,Z}\quad \frac12\mathrm{Tr}[Y+Z] \\ &\quad \mathrm{s.t.} \quad \begin{pmatrix} Y & X \\ X^* & Z \end{pmatrix} \geq 0. \end{aligned} Furthermore, we can write your problem as \begin{aligned} &\min_{\sigma}\quad \|\sigma_{AB} - \rho_{AB}'\|_1 \\ &\mathrm{s.t.}\quad \mathrm{Tr}_A[\sigma_{AB}] = \rho_B \\ &\qquad \mathrm{Tr}_B[\sigma_{AB}] = \rho_A \\ &\qquad \sigma_{AB} \geq 0. \end{aligned} Combining the two problems we can solve for a given $$\rho_{AB}$$ and $$\rho_{AB}'$$ the following SDP \begin{aligned} &\min_{Y,Z,\sigma}\quad \frac12\mathrm{Tr}[Y+Z] \\ &\mathrm{s.t.} \quad \begin{pmatrix} Y & \sigma_{AB} - \rho_{AB}' \\ \sigma_{AB} - \rho_{AB}' & Z \end{pmatrix} \geq 0 \\ &\qquad \mathrm{Tr}_A[\sigma_{AB}] = \rho_B \\ &\qquad \mathrm{Tr}_B[\sigma_{AB}] = \rho_A \\ &\qquad \sigma_{AB} \geq 0. \end{aligned}