Consider two channels, $\Phi,\Psi\in\mathrm C(\mathcal X)$ acting on some space $\mathcal X$, and suppose they commute, that is, $$\Phi(\Psi(\rho))=\Psi(\Phi(\rho))$$ for all states $\rho$. Can anything be said about the structure, e.g. in terms of their Kraus operators, that this implies? For example, does there have to be a specific relation between their Kraus operators?

Suppose $\Phi(X)=\sum_a A_a X A_a^\dagger$ and $\Psi(X)=\sum_b B_b X B_b^\dagger$. Then the question relates to the channels with Kraus operators $\{A_a B_b\}_{ab}$ and $\{B_b A_a\}_{ab}$. There are a few cases in which the solution is obvious, e.g. if the Kraus operators commute with each others, $[A_a,B_b]=0$, but I'm not sure how to proceed in the more general case. Is there anything general that can be said about the problem?


1 Answer 1



The condition $[A_a, B_b]=0$ is sufficient, but not necessary for $\Phi$ and $\Psi$ to commute. Indeed, the standard Kraus representations of many commuting pairs of channels have non-commuting Kraus operators, e.g. bit- and phase-flip channel, depolarizing and unitary channel, amplitude damping and phase-flip channel etc. However, Kraus operators of commuting channels necessarily satisfy $(3)$ below.

Relation between Kraus operators of commuting channels

Let $K(\Phi)\in L(\mathcal{X}\otimes\mathcal{X})$ denote the unique linear operator such that

$$ K(\Phi)\mathrm{vec}(X) = \mathrm{vec}(\Phi(X)),\tag1 $$

for all $X\in L(\mathcal{X})$, c.f. equation $(2.61)$ on page $77$ in The Theory of Quantum Information by John Watrous. Then $\Phi$ and $\Psi$ commute if and only if $K(\Phi)$ and $K(\Psi)$ commute. By Proposition $2.20$ on page $80$, we have

$$ K(\Phi)=\sum_a A_a\otimes \overline{A_a}, \quad K(\Psi)=\sum_b B_b\otimes \overline{B_b}\tag2 $$

which means that $\Phi$ and $\Psi$ commute if and only if

$$ \sum_{a,b}A_aB_b \otimes \overline{A_aB_b} = \sum_{a,b}B_bA_a \otimes \overline{B_bA_a}.\tag3 $$

This allows us to state sufficient - but not necessary - conditions on Kraus operators that are somewhat more general than the requirement $[A_a,B_b]=0$. Specifically, if for every pair $a, b$ we have

$$ A_a B_b = e^{i\theta_{a,b}}B_b A_a\tag4 $$

for some $\theta_{a,b}\in[0,2\pi)$ then $\Phi$ and $\Psi$ commute.

  • $\begingroup$ how did you get from (2) to (3)? We have $[K(\Phi),K(\Psi)]=\sum_{ab}([A_a,B_b]\otimes[\bar A_a, \bar B_b])=0$. But how do you "decouple" the two pairs of products from this? $\endgroup$
    – glS
    Nov 14, 2021 at 10:13
  • $\begingroup$ Good question! I have added a lemma to explain this step. $\endgroup$ Nov 14, 2021 at 17:03
  • $\begingroup$ thanks. There might be a typo in (3) though, unless I'm misunderstanding something. There should only be a single sum I think: $K(\Phi)K(\Psi)=K(\Phi\circ\Psi)=\sum_{ab} (A_aB_b\otimes \bar A_a \bar B_b)$, which is not in general a simple tensor product $\endgroup$
    – glS
    Nov 14, 2021 at 17:05
  • $\begingroup$ Yes, you're right. Good catch! That's a fairly embarrassing mistake. Fixed. Note that it affects the conclusion: now we have a free phase angle for each term in $(4)$. $\endgroup$ Nov 14, 2021 at 17:16
  • $\begingroup$ I'm still not entirely convinced though. I agree that what you show gives a sufficient condition for commutativity, but why do you conclude that it is necessary? You are giving an iff condition to have each term in the sums in (3) equal. However, it could be the case that the individual terms are not equal, but their sum still is. How do you exclude such cases? $\endgroup$
    – glS
    Nov 14, 2021 at 17:29

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