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

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.

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 $$ \sigma = \rho_{A'}\otimes \vert\omega\rangle\langle\omega\vert_{BB'}\ , $$ which is the Choi state of a channel of the form $$ \mathcal T_A\otimes I_B\ . $$