Let $\Phi_1,\Phi_2$ be CPT maps with Kraus decomposition \begin{equation} \Phi_1=\sum_{k=1}^{d_1}M_k\rho M_k^\dagger, \quad \Phi_2=\sum_{k=1}^{d_2} N_k\rho N_k^\dagger, \quad \text{s.t.} \quad \sum_{k=1}^{d_1}M_k^\dagger M_k=\sum_{k=1}^{d_2} N_k^\dagger N_k=\mathbb I. \end{equation} and let $p\in[0,1]$ . Consider the following (convex) combinations of the two maps: \begin{equation} \Phi=p\Phi_1+(1-p)\Phi_2. \end{equation} I'd like to find a Kraus decomposition for this map. Inserting the above gives $$ \Phi(\rho)=p\left(\sum_{k=1}^{d_1}M_k\rho M_k^\dagger\right)+(1-p)\left(\sum_{k=1}^{d_2} N_k\rho N_k^\dagger\right), $$ The two representations, in general, have a different number of elements; I guess I can always fix this by taking $D:=\max(d_1,d_2)$ and extend the smallest one with null operators. After doing this, one can rewrite the sum as $$ \Phi=p\left(\sum_{k=1}^DM_k\rho M_k^\dagger-N_k\rho N_k^\dagger\right)+\sum_{k=1}^DN_k\rho N_k^\dagger $$ which strikes me something we need to do to exploit the convexity property. To make progress, I've thought about inserting a few identities by the normalization conditions above. However, I can't quite make it work.

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    $\begingroup$ the Kraus operators are, modulo vectorisation, the eigenvectors of the Choi representation (or more generally, decompositions of the Choi in terms of positive semidefinite operators). You are therefore essentially asking about how to decompose the sum of two positive operators in terms of the decompositions of the individual operators. If you don't require a specific number of operators in the final decomposition, the trivial answer is obtained by simply concatenating the individual decomposition, up to some rescaling. $\endgroup$
    – glS
    Apr 12, 2021 at 20:20

1 Answer 1


Why not just concatenate the Kraus operators from each channel, i.e., take $\{Z_{k}\}_{1 \leq k \leq d_1 + d_1}$ where $$ Z_k = \begin{cases} \sqrt{p} M_k \qquad \quad &\text{if } 1 \leq k \leq d_1 \\ \sqrt{1-p}N_{k - d_1} &\text{if } d_1 < k \leq d_1 + d_2 \end{cases} $$

You can check that these operators satisfy completeness property for Kraus operators and that they characterise the action of the convex combination channel $p \Phi_1 + (1-p) \Phi_2$.

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    $\begingroup$ This is a very straightforward but elegant solution! $\endgroup$
    – JSdJ
    Apr 12, 2021 at 20:53
  • $\begingroup$ The above solution is indeed simple and effective but what if $d_1 + d_2 > N^2$? $\endgroup$
    – trurl
    Sep 3, 2022 at 12:21
  • $\begingroup$ @trurl There's no rule that the number of Kraus operators must be smaller than N^2. But if it bothers you, you can always use the isometric degree of freedom in the Kraus representation. (Or go via the Choi-Jamiolkowski isomorphism: The Swiss knife of QI.) $\endgroup$ Sep 3, 2022 at 20:58

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