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?

  • $\begingroup$ I think this question is similar to physics.stackexchange.com/questions/434702/… but I did not really understand the answer $\endgroup$
    – StarBucK
    Apr 4, 2020 at 12:46
  • $\begingroup$ @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! $\endgroup$ Apr 4, 2020 at 14:04
  • 1
    $\begingroup$ @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 $A$ $\endgroup$ Apr 4, 2020 at 14:05

2 Answers 2


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.

  • $\begingroup$ 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? $\endgroup$
    – Noiralef
    Apr 5, 2020 at 10:10
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    $\begingroup$ 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 $\endgroup$
    – glS
    Apr 5, 2020 at 10:16
  • $\begingroup$ Thank you, in the end I accepted the other answer as it was a more concise argument, but this was helful as well! $\endgroup$ Apr 6, 2020 at 11:23

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