Q:

Does a quantum channel being time-translation invariant imply that its Kraus operators commute with the Hamiltonian?

Let $\mathcal E\in\mathrm{T}(\mathcal X,\mathcal Y)$ be a quantum channel (i.e. a completely positive, trace-preserving linear map sending states in $\mathrm{Lin}(\mathcal X,\mathcal X)$ into states in $\mathrm{Lin}(\mathcal Y,\mathcal Y)$).
It is well known that any such map can be written in the Kraus decomposition as:
$$\mathcal E(\rho)=\sum_a A_a\rho A_a^\dagger,$$
for a set of operators $A_a$ such that $\sum_a A_a^\dagger A_a=I$ (one can also choose these operators to be orthogonal with respect to the $L_2$ inner product structure: $\operatorname{Tr}(A_a^\dagger A_b)=\delta_{ab}p_a$).
Suppose now that $\mathcal E$ is time-translation invariant. This means that, given an underlying Hamiltonian $H$ generating a time-evolution operator $U(t)$, we have
$$\mathcal E(U(t)\rho U(t)^\dagger)=U(t)\mathcal E(\rho)U(t)^\dagger,\quad\forall t,\rho.
\tag{1}
$$
If $\mathcal E$ represented a simple unitary evolution $V$ (that is, $\mathcal E(\rho)=V\rho V^\dagger$), it would follow that $[V,H]=0$.
Does this still apply for the Kraus operators of a general $\mathcal E$? In other words, does time-translation invariance imply that $[A_a,H]=0$ for all $a$?

This question is related to this other question about how time-translation invariance implies preservation of coherence, as if the statement in this question is correct, then it should be easy to prove the statement in the other post.

A:

No. Take, for instance, the fully depolarizing channel, where $A_a=\{I,X,Y,Z\}$. Since 
$\mathcal E(\rho)=\tfrac12 I$, your condition $(1)$ holds for all $H$.  On the other hand, there is no operator which commutes with all $A_a$.

(Let me take the opportunity to advertise my list of canonical counterexamples for quantum channels ;) ).

A:

Writing the requirement explicitly
$$\mathcal{E}(U\rho U^\dagger)=U\mathcal{E}(\rho)U^\dagger $$
in terms of Kraus operators
$$\sum_a A_aU\rho U^\dagger A_a^\dagger=\sum_a U A_a\rho A_a^\dagger U^\dagger $$
Hence we want the channel with Kraus operators $ A'_a=A_aU $ and $A_a''=UA_a$ to be equal. We know that two channels are equal if and only if their Kraus representations are unitarily related, i.e. we have to find $B_{ij}$ such that
$$ A''_j=\sum_k B_{jk}A_k'$$
and
$$\sum_k B_{jk}B_{ik}^* =\delta_{ij}$$
the first condition is just 
$$ UA_j=\sum_k B_{jk}A_kU$$
or equivalently 
$$ UA_jU^\dagger=\sum_k B_{jk}A_k$$
This is a weaker requirement than commutation with $U$, as that is the particular case $B_{ij}=\delta_{ij}$.
As Norbert Schuch said in his comment, this boils down to the Kraus representation being non unique (and thus in a sense, non physical). In a sense, the channel commutes with the evolution if the evolution scrambles the Kraus operators to a set that would produce the same physical effect.

