# Do channels which do not affect a subsystem always have the form $\mathcal E=\mathcal T_{\mathcal A}\otimes\text{id}_B$?

consider a bipartite state on two Hilbert spaces $$\rho_{AB}$$. Usually, I think of a channel $$\mathcal E$$ as acting only on $$A$$ if

$$\mathrm{Tr}_A(\mathcal E(\rho_{AB}))=\mathrm{Tr}_A(\rho_{AB}) \tag A$$

i.e., the action of the channel on a system does not influence the reduced state on the other system. My question is: is this equivalent to

$$\mathcal E=\mathcal{T_A}\otimes\mathrm{id}_B$$

where $$\mathcal T_A:A\rightarrow A$$?

That the second line imples the first is obvious, the other direction doesn't look obvious to me, there could be some action on $$B$$ that somehow gets canceled by taking the trace somehow. Is it true?

• I think this question is similar to physics.stackexchange.com/questions/434702/… but I did not really understand the answer Commented Apr 4, 2020 at 12:46
• @StarBucK You're right! It's very similar. The answer is pretty long, I will read it when I have time and report back, thanks! Commented Apr 4, 2020 at 14:04
• @Noiralef No, I mean $\mathrm {Tr}_A$, the equation means that the "remaining" state on $B$ when we trace $A$ away is not affected by a channel acting only on $A$ Commented Apr 4, 2020 at 14:05

An simple argument goes via the Choi-Jamiolkowski isomorphism:

The Choi state is obtained by applying the channel to the left half of a maximally entangled state, $$\vert\Omega\rangle = \sum_{ij} \vert i,j\rangle_{AB}\vert i',j'\rangle_{A'B'}\ .$$ The Choi state is then $$\sigma_{\mathcal E}= (\mathcal E_{AB}\otimes I_{A'B'})\,(\vert\Omega\rangle\langle\Omega\vert)\ .$$ Now consider the channel you describe above, namely $$\mathrm{tr}_A\circ \mathcal E$$. Its Choi state is $$\chi=\mathrm{tr}_A\,\sigma$$. On the other hand, your equation (A) tells us that $$\mathrm{tr}_A\,\sigma = \chi = \rho_{A'}\otimes \vert\omega\rangle\langle\omega\vert_{BB'}\ ,$$ where $$\vert\omega\rangle = \sum_i \vert i\rangle_B\vert i\rangle_{B'}$$ (since (A) says that $$\mathrm{tr}_A\circ \mathcal E$$ acts as the identity channel on $$B$$. This implies that $$\sigma$$ must be of the form $$\sigma = \rho_{AA'}\otimes \vert\omega\rangle\langle\omega\vert_{BB'}\ ,$$ which is the Choi state of a channel of the form $$\mathcal T_A\otimes I_B\ .$$

The idea is to prove that (A) implies that the Kraus operators $$A_a$$ of $$\mathcal E$$ are local, that is, they have the form $$A_a=\tilde A_a\otimes I$$. For the purpose, I will leverage here the representation of the channel in terms of an enlarged unitary, and use it to prove the result on the Kraus operators.

## Enlarged unitary representation

Every map CPTP $$\mathcal E$$ can be written as $$\newcommand{\ketbra}[1]{|#1\rangle\!\langle #1|}\newcommand{\calU}{{\mathcal{U}}}\newcommand{\calE}{{\mathcal{E}}}\newcommand{\calH}{{\mathcal{H}}}\newcommand{\tr}{\operatorname{tr}}\calE(\rho)=\tr_E[\calU(\rho\otimes\ketbra{0_E}) \calU^\dagger]$$ for some (unique) unitary $$\calU$$. Explicitly, denoting with $$K(\calE)$$ the natural representation of $$\calE$$, the relation between $$\calE$$ and $$\calU$$ can be written as $$K(\calE)_{ij}^{k\ell} = \sum_a \calU_{ia}^{k0} (\calU^*)_{ja}^{\ell0}.$$ Feel free to completely ignore the placement of up/down indices, I only do it to make the formula a bit easier to read. If this looks similar to the Kraus decomposition of the channel, that's not a coincidence. The relation between the unitary $$\calU$$ and the Kraus operators $$A_a$$ is $$(A_a)_i^k = \calU_{ia}^{k0}$$.

While the above notation is for maps acting on a single system, the formulae translate seamlessly to a bipartite scenario by "duplicating" each index: $$i\to(i_1,i_2)$$, $$j\to (j_1,j_2)$$ etc.

## What (A) implies for $$\mathcal U$$

Note that if (A) holds for all states $$\rho_{AB}$$, it holds in particular for all pure product states $$|\psi\rangle|\phi\rangle$$. On these states, $$\calU$$ must act in such a way that tracing out the subsystem $$A$$ still gives a pure state on $$B$$ (because $$\tr_A(\ketbra\psi\otimes\ketbra\phi)=\ketbra\phi$$ is pure). We must thus have

$$\calU\, |\psi\rangle_A|\phi\rangle_B|0\rangle_E = |\Psi_{AE}\rangle\otimes|\phi\rangle_B,$$ where $$|\Psi_{AE}\rangle$$ is some state in $$\calH_A\otimes\calH_E$$ (the joint space of $$A$$ and the auxiliary space $$E$$). More explicitly, this expression reads $$\calU_{ima}^{kn0} \psi_{k}\phi_n = \Psi_{ia} \phi_m.$$ If we choose $$|\psi\rangle,|\phi\rangle$$ to be elements of the computational basis, we conclude that $$\calU_{ima}^{kn0} = \delta_{nm} \Psi^k_{ia}$$, where $$\sum_{ia}\Psi^k_{ia}|i,a\rangle$$ is some pure state in $$\calH_A\otimes\calH_E$$ for each $$k$$. Now all that remains is to see why this implies the original statement about the form of $$\calE$$.

## Conclude that $$\mathcal E$$ is local

We proved that the Kraus operators of $$\calE$$ have the form $$(A_a)_{im}^{kn} = \delta_{mn} \Psi^{k}_{ia} \Longleftrightarrow A_a = \tilde A_a\otimes I$$ for some set of operators $$\tilde A_a$$ acting on $$\calH_A$$. This is enough to prove the statement: a map is local if and only if its Kraus operators are. If we want to be more explicit, we can write the expressions componentwise as: $$\calE(\rho) = \sum_a A_a \rho A_a^\dagger \Longleftrightarrow (\calE(\rho))_{im,jn} = (A_a)_{im}^{k\ell} (A_a^*)_{jn}^{pq} \rho_{k\ell,pq} = \Psi^k_{ia}(\Psi^*)^p_{ja}\rho_{km,pn}.$$ We can then define the operator $$\mathcal T$$ with components $$(\mathcal T)_{ij}^{kp} = \sum_a \Psi^k_{ia}(\Psi^*)^p_{ja},$$ and observe that $$\calE = \mathcal T\otimes I$$.

TL;DR: Consider the action of the map on pure product states, and observe that (A) implies that its Kraus operators are local.

• Just a technical question: you are saying that if the partial trace of $\rho_{AB}$ is a pure state then $\rho_{AB}$ must have been a pure product state. Is it easy to see why that is true? Commented Apr 5, 2020 at 10:10
• that is not true, a counterexample is $\rho_{AB}=\rho\otimes |1\rangle\!\langle 1|$. What I'm saying (if I understand what you are referring to) is that if the partial trace of a pure state $|\Psi_{AB}\rangle$ is pure, then the state must have been a (pure) product state $|\Psi_{AB}\rangle=|\psi_A\rangle\otimes|\psi_B\rangle$. To see it, compute the Schmidt decomposition of $\Psi_{AB}$. If it's not a product state, there are multiple Schmidt coefficients, and thus the partial state is not pure
– glS
Commented Apr 5, 2020 at 10:16
• Thank you, in the end I accepted the other answer as it was a more concise argument, but this was helful as well! Commented Apr 6, 2020 at 11:23