It may or may not come at a surprise that the natural representation of channels can only be unitary if the channel itself is unitary. More precisely, we will prove the following statement:
Theorem. Let $\Phi\in L(\mathbb C^{n\times n})$ be given such that $K(\Phi)$ is unitary. The following statements are equivalent.
- $\Phi$ is a channel
- $\Phi$ is completely positive
- $\Phi=U(\cdot)U^*$ for some unitary $U$
Proof. As 3. $\Rightarrow$ 1. $\Rightarrow$ 2. is trivial we only have to prove "2. $\Rightarrow$ 3.". The key here are the well known identities $K(\Psi\Xi)=K(\Psi)K(\Xi)$ and $K(\Psi^\dagger)=K(\Psi)^\dagger$ which hold for all linear maps $\Psi,\Xi$, where $\Psi^\dagger$ is the dual map of $\Psi$ with respect to the Hilbert-Schmidt inner product $\langle A,B\rangle_{\rm HS}:={\rm tr}(A^\dagger B)$ (i.e. $\Psi^\dagger$ is the unique linear map which satisfies $\langle A,\Psi(B)\rangle_{\rm HS}=\langle \Psi^\dagger(A),B\rangle_{\rm HS}$ for all $A,B$). Thus $K(\Phi)$ being unitary implies
$$
K({\rm id})={\bf1}=K(\Phi)^\dagger K(\Phi)=K(\Phi^\dagger\circ\Phi)\,;
$$
in particular this shows that $\Phi$ is invertible and its inverse $\Phi^\dagger$ is completely positive (because $\Phi$ is). But this is only possible if the Kraus rank of $\Phi$ is $1$, which is a special case of Theorem 3.1 in Chapter 2 of Davies' book "Quantum Theory of Open Systems" (1976). This, in turn, means that $K(\Phi)=K(X(\cdot)X^*)=(X^*)^T\otimes X=\overline{X}\otimes X$ (in the second step we used a standard vectorization identity) for some $X\in\mathbb C^{n\times n}$. Obviously, $\overline{X}\otimes X$ is unitary if and only if $X$ is unitary which concludes the proof. $\square$
