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Given some partial states $\rho_{AB}$ and $\sigma_{AC}$, is there a general procedure to construct a state $\delta_{ABC}$ such that the following sum of trace distances

$$||\text{Tr}_C(\delta_{ABC}) - \rho_{AB}||_1 + ||\text{Tr}_B(\delta_{ABC}) - \sigma_{AC}||_1$$

is minimal? That is we want a joint state that under partial traces is as close as possible to some given targets. In particular, when there is entanglement in both $AB$ and $AC$, I'm not sure how to proceed.

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I would probably set this up as a semi-definite program and throw it at the computer. Basically, your problem is a linear problem in the coefficients of the matrix $\delta_{ABC}$ except for the constraint that $\delta_{ABC}$ is positive semi-definite. This is exactly what semi-definite programming is designed to do.

Note: the calculation below uses a different 1-norm (my bad). See the comment by Norbert, below.

It's a bit of a pain to set it all up in the right way (which is why I'm not directly stating the full formula!). You need to do things like let $$ M1=\text{Tr}_C(\delta_{ABC})-\rho_{AB} $$ and the define variables $x_{ij}$ such that $$ M1_{ij}\leq x_{ij},\qquad -M1_{ij}\leq x_{ij} $$ with the intent of getting (when we're minimising $x_{ij}$) $x_{ij}=|M1_{ij}|$. Then, you set $$ \sum_{j}x_{ij}\leq x\qquad\forall i. $$ In effect, $x=\|M_1\|_1$. If you do something similar with a matrix $M2$, variables $y_{ij}$ and $y$, your final problem is to minimise $x+y$.

Don't forget to include the constraint $\text{Tr}(\rho_{ABC})=1$. It might be implicit in previous constraints, I'm not sure.

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    $\begingroup$ How does this give the trace norm? It seems you are summing all elements in absolute value. (This is a 1-norm, but a different one.) Though you should be able to express the trace norm with semidefinite constraints (using e.g. the dual characterization through the operator norm, i.e. maximizing $\mathrm{tr}(MX)$ subject to $I \le X \le I$). $\endgroup$ Mar 9 '20 at 13:27
  • $\begingroup$ @NorbertSchuch Fair point. I was going for sum of absolute values. I saw the equation and filtered out the words around it. $\endgroup$
    – DaftWullie
    Mar 9 '20 at 13:40
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    $\begingroup$ Well, density matrices beg for trace norms, don't they? $\endgroup$ Mar 9 '20 at 13:47
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    $\begingroup$ Sure. I've just been using the other matrix norm for a lot of calculations recently and was in the zone! $\endgroup$
    – DaftWullie
    Mar 10 '20 at 11:48
  • $\begingroup$ To be honest, I am not sure this can be done with an SDP. Computing the trace norm alone is already an SDP, but it is a maximization. However, if on top of that you have to minimize the trace norm, this is a $\min\max$, which is no longer an SDP. $\endgroup$ Mar 10 '20 at 12:00

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