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Let $\Phi:\mathcal{L}(H)\rightarrow \mathcal{L}(K)$ be a completely positive map sending positive self-adoint operators on a finite-dimensional Hilbert space $H$ to positive self-adoint operators on a finite-dimensional Hilbert space $K$. Let $\Phi^\dagger:\mathcal{L}(K)\rightarrow \mathcal{L}(H)$ be the adjoint of $\Phi$, obtained by taking the adjoints of all Kraus operators.

When $\Phi^\dagger\Phi = id_{\mathcal{L}(H)}$, i.e. when $\Phi$ is an isometry in the dagger compact category of finite-dimensional Hilbert spaces and completely positive maps, then it can be shown that $\Phi$ is necessarily in the following form \begin{equation} \Phi = \rho \mapsto \sum_i q_i V_i \rho V_i^\dagger \end{equation} where $q_i \in \mathbb{R}^+$ and $V_i: H \rightarrow K$ are isometries---in the dagger compact category of finite-dimensional Hilbert spaces and complex linear maps---with pairwise orthogonal images, i.e. $V_j^\dagger V_i = \delta_{ij} id_{H}$.

I already know how to prove the result, but I have the distinct feeling that I have seen it somewhere before, either by itself or as a consequence of some more general characterisation of completely positive maps. Hence the question: is anyone aware of some piece of literature where this result is proven, or from which this result follows as a direct consequence?

(I spend hours trying to track it down, to no avail, but in the process I have discovered a question from a few years ago that is related to it, namely When is this map completely positive?)

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I would suggest you look at "Isometries on Banach algebras, vol. 1 and 2" by Fleming and Jamison.

What you are describing is very similar to an application of theorems of Schur (Theorem 10.2.2 in the above books) and Kadison (Theorem 6.2.5).

Namely, they show that every linear isometry between matrix algebras, $\varphi : M_n(\mathbb C) \rightarrow M_m(\mathbb C)$ has the form $$ \varphi(a) = UaV \quad \textrm{or} \quad \varphi(a) = Ua^TV, \quad \forall a\in M_n(\mathbb C) $$ where $U$ and $V$ are unitaries in $M_m(\mathbb C)$.

If you require that $\varphi$ is a completely positive as well then there can be no transpose and $V = U^\dagger$.

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