3
$\begingroup$

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.

$\endgroup$
5
$\begingroup$

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 ;) ).

$\endgroup$
  • $\begingroup$ It is an interesting question what the right condition on the $A_a$ is -- I'm pretty certain there should be one. I'll try to think about it if I find time. A key point is that the Kraus decomposition is not unique, this is exactly what is happening in this example. A proof would have to take this into account. It might be easier to assess this using the Choi state. There, the corresponding condition should be sth. like that the Choi state has to be invariant under conjugation by $U\otimes U^\dagger$, or something the like. $\endgroup$ – Norbert Schuch Oct 14 '19 at 13:34
  • $\begingroup$ ah! yes I should have checked the simple examples first, good point. About the non-uniqueness of the Kraus decomp, I guess one can remove some of of the arbitrariness by choosing operators that are the eigenvectors (up to some (un)vectorisation shenanigans) of the Choi, which amounts to imposing the Kraus ops to be orthogonal. If I were to make a wild guess, it might be that the unitary connecting the operators (as per the other answer) then only mixes ops in the same degenerate eigenspace of the Choi. I'll think about this when I get the time $\endgroup$ – glS Oct 15 '19 at 12:05
  • $\begingroup$ @glS I guess the best way to proceed constructively is not to choose a basis, but to take the differential version of the condition derived in the other answer. Then this boils down to having the commutator $[H,A_a]=\sum_b G_{ab} A_b$, with $G$ anti-hermitian. Basically this is a condition that the $A_a$ span a certain algebra under commutation. I wanted to think a bit about it when I find some time. But I suspect this structure is mathematically well known. $\endgroup$ – Norbert Schuch Oct 15 '19 at 12:26
3
$\begingroup$

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.

$\endgroup$
  • $\begingroup$ The interesting question here is of course what happens when the $U$ are a family generated by a Hamiltonian. $\endgroup$ – Norbert Schuch Oct 14 '19 at 14:33
  • $\begingroup$ @NorbertSchuch This is probably ignorance on my part, but what difference would that make? In that case $U(t)=e^{-iHt}$. If $H$ is Hermitian, commuting with $U$ is the same as commuting with $H$. There would be a time dependence on the $B_{jk}$ I guess, but I don't see how much can be said about that. Unless I'm totally missing the point. $\endgroup$ – user2723984 Oct 14 '19 at 14:36
  • $\begingroup$ It is a set of conditions. Maybe there is a solution for individual $U$'s, but not for certain families -- or the class of solutions for the family is significantly more restrictive. (Just imagine having a multi-parameter family: Commuting with one unitary is clearly different from commuting with all unitaries.) $\endgroup$ – Norbert Schuch Oct 14 '19 at 14:39

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.