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?


1 Answer 1


You could try solving this numerically using semidefinite programming.

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 $$ \begin{aligned} &\min_{Y,Z,\sigma}\quad \frac12\mathrm{Tr}[Y+Z] \\ &\mathrm{s.t.} \quad \begin{pmatrix} Y & \sigma_{AB} - \rho_{AB}' \\ \sigma_{AB} - \rho_{AB}' & Z \end{pmatrix} \geq 0 \\ &\qquad \mathrm{Tr}_A[\sigma_{AB}] = \rho_B \\ &\qquad \mathrm{Tr}_B[\sigma_{AB}] = \rho_A \\ &\qquad \sigma_{AB} \geq 0. \end{aligned} $$

SDPs are fast to solve as long as the dimension of the system is not too large. Plus they should converge to the optima which you'll be able to read out from the solver.


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