Consider a channel $C$ with Kraus operators $\{K_k\}$ and a unitary U. How can I check that $C$ implements $U$ ?

One can write that their Choi matrices are equal i.e:

\begin{equation} \sum_{i,j}|i\rangle\langle j|\otimes \sum_{k}K_k|i\rangle\langle j|K_k^\dagger = \sum_{i,j}|i\rangle\langle j|\otimes U|i\rangle\langle j|U^\dagger \end{equation} but I'm not sure if this can be simplified to obtain a condition on the $K_k$'s.


1 Answer 1


Provided the Kraus operators are linearly independent, this is only possible if there is only one nonzero Kraus operator, which then is exactly the unitary.

If we relax the constraint of linearly independentness, the Kraus operators should be scalar multiples of each other.

The most straightforward way to see this is by realising that the Choi operator of a unitary channel is rank $1$, and that therefore the rank of the Choi operator of the channel $C$ should also be rank $1$. The eigenvectors of the Choi operator are in direct one-to-one correspondence with the Kraus operators (by the vectorisation mapping). Therefore, the only way for the two Choi operators to coincide is when the Kraus operators are either all zero except for one, or if they are scalar multiples of each other.

  • $\begingroup$ If you'd like a more fleshed-out answer please let me know @Nichola $\endgroup$
    – JSdJ
    Commented Aug 16, 2022 at 18:05

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