Suppose I have a pair of bipartite states $\rho_{AR}$ and $\sigma_{BR}$. $R$ is a reference system that we do not have access to.

It is clear that we cannot always have a channel $N_{A\rightarrow B}$ such that $(N\otimes I_R)(\rho_{AR}) = \sigma_{AR}$. If this were the case, we could, for example, generate entanglement through a local operation on $A$ and $B$ which is not possible.

Given a $\rho_{AR}$ and $\sigma_{BR}$, is there a way to know if there exists a channel $N_{A\rightarrow B}$ with the desired property?

  • $\begingroup$ just to be clear, you are asking under what conditions, given $\rho_{AR}$ and $\sigma_{BR}$, there is a channel $N$ such that $(N\otimes I_R)(\rho_{AR})=\sigma_{AR}$? $\endgroup$
    – glS
    Jan 26, 2022 at 12:49
  • $\begingroup$ Yes, that's right $\endgroup$ Jan 26, 2022 at 13:08

1 Answer 1


One necessary (but I doubt sufficient) condition is that the partial trace that removes everything but the reference system must be the same for the initial and final state. First, assume that $N$ is trace-preserving for system $A$. Then, using some decomposition $$\rho_{AR}=\sum_{klmn}p_{klmn}|k\rangle_A\langle l|\otimes |m\rangle_R\langle n|,$$ we have \begin{aligned} \text{Tr}_B(\sigma_{BR})&=\text{Tr}_B[(N\otimes \mathbb{I})\rho_{AR}]\\ &=\sum_{klmn}p_{klmn}\text{Tr}_B[N(|k\rangle_A\langle l|)\otimes |m\rangle_R\langle n|]\\ &=\sum_{klmn}p_{klmn}\text{Tr}[N(|k\rangle_A\langle l|) ]|m\rangle_R\langle n|\\ &=\sum_{klmn}p_{klmn}\text{Tr}[|k\rangle_A\langle l| ]|m\rangle_R\langle n|\\ &=\text{Tr}_A(\rho_{AR}). \end{aligned} If $N$ is not normalized, then the trace of $\rho_{AR}$ and $\sigma_{BR}$ will be different, so they won't both be "states." If you're ok with that then we get stuck at the step $$\text{Tr}_B(\sigma_{BR})=\sum_{klmn}p_{klmn}\text{Tr}[N(|k\rangle_A\langle l|) ]|m\rangle_R\langle n|$$ and the relationships between pairs of related states will be even more complicated.

In general, we can use the same basis labels to express either system $A$ or $B$, because we can always extend one basis if the other has a larger dimension. In that sense, we can always write $$\sigma_{BR}=\sum_{klmn}s_{klmn}|k\rangle\langle l|\otimes |m\rangle_R\langle n|$$ and drop the basis label for system $B$. The channel then necessitates $$\sum_{klmn}s_{klmn}|k\rangle\langle l|\otimes |m\rangle_R\langle n|=\sum_{klmn}p_{klmn}N(|k\rangle\langle l|)\otimes |m\rangle_R\langle n|,$$ so the two states must obey the mutual relationship $$\sum_{kl}s_{klmn}|k\rangle\langle l|=\sum_{kl}p_{klmn}N(|k\rangle\langle l|)$$ for all $m$ and $n$ (this trivially includes cases when $p_{klmn}$ or $s_{klmn}$ vanish for a particular pair of $m$ and $n$ values). That is the necessary and sufficient condition that you are looking for.

Can we make more sense of this requirement? Say we know everything about our channel, such that we know how it acts on all basis states $$N(|k\rangle \langle l|)=\sum_{k^\prime l^\prime}n^{(kl)}_{k^\prime l^\prime}|k^\prime\rangle \langle l^\prime|.$$ Then the necessary and sufficient condition is that $$s_{klmn}=\sum_{k^\prime l^\prime}p_{k^\prime l^\prime mn}n^{(k^\prime l^\prime)}_{kl}$$ for all $k$, $l$, $m$, and $n$. For every state $\rho_{AB}$, there is a corresponding paired state $\sigma_{BR}$. The converse is not true, unless you can invert the relationship $s_{klmn}=\sum_{k^\prime l^\prime}p_{k^\prime l^\prime mn}n^{(k^\prime l^\prime)}_{kl}$, which will only happen for certain channels $N$. Which channels? That is a good question, beyond the scope here! Possibly related to this question.


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