# Does $N(U\rho U^\dagger)=U' N(\rho)U'^\dagger$ for unitaries $U,U'$ and a channel $N$ imply $UK_i=K_i U'$?

Let $$H_A, H_B$$ be Hilbert spaces and let a channel $$N_{A\rightarrow B}$$ be a CPTP map between them. If there exist that unitaries $$U\in H_A$$ and $$U'\in H_B$$ such that for all $$\rho\in H_A$$

$$N(U\rho U^\dagger) = U' N(\rho) U'^\dagger$$

then does it imply that $$UK_i = K_iU'$$ for every Kraus operator $$K_i$$ of the channel? Note that the reverse implication is obviously true.

No. Take a depolarizing channel, which can be written as $$\mathcal{D}_p(X) = p\, \mathrm{tr}(X) \frac{I}{d} + (1-p) X\,$$ Note that it does commute with any unitary $$U$$, $$\mathcal{D}_p(U X U^\dagger) = U\mathcal{D}_p( X ) U^\dagger$$. However, $$\mathcal{D}_p$$ has a Kraus decomposition that involves all Pauli operators, and clearly the only operator that commutes with all Paulis is the identity.

Note: Keywords in this context are "covariant" or "equivariant channels", see also the answer by Norbert Schuch.

As Markus Heinrich points out, the answer is "no".

On the other hand, for an abelian symmetry group, you can find a set of Kraus operators such that each Kraus operator individually satisfies $$U_g K_i = e^{i \phi^i(g)} K_i U'_g\ ,$$ where $$\phi^i(g)$$ is a 1D representation of the symmetry group, which however can depend on $$g$$.

This is shown in Sec. 3.1 of our paper Symmetry Protected Topological Order in Open Quantum Systems.

Indeed, having symmetric Kraus operators as you ask for (equivalently, $$\phi^i(g)$$ does not depend on $$i$$) has much stronger consequences on the channel (e.g., $$N(U_g)=U_g$$), which are discussed subsequently in the paper.