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Given a system $\rho_{AB}\otimes\rho_{C}$, and a unitary interaction $U_{BC}$, due to the monotonicity of the relative entropy under the actions of the partial trace map, $$I(A:B)=I(A:U_{BC}(B,C))\ge I(A:N(C))$$, where $N()$ is the local channel on subsystem $C$, after tracing out $B$, correct?

I am aware most proofs usually trace out the initially uncorrelated system, and the upper bound would usually involve the local channel, but unless I have been misunderstanding something for a while now, I believe this should still hold.

I ask because, if we now consider a unitary $U_{AC}$, that commutates with $U_{BC}$, so either interaction could be performed first.

In this case, I have assumed that $$I(A'':B',C'')=I(A':U_{BC}(B,C'))$$ where $I(A':U_{BC}(B,C'))$ is is obtained via the mutual information of $U_{AC}(\rho_{AB}\otimes\rho_{C})$, with the local operations $U_{BC}$ unable to change the mutual information.

I would therefor expect that $$I(A':U_{BC}(B,C'))=I(A':B,C')$$ where the only interaction is between $A$ and $C$ using $U_{AC}$. As such, $$I(A':B,C')\ge I(A':N(C'))$$ where $N$ is the reduced map from the actions of $U_{BC}$. As such, I would expect $$I(A':C') \ge I(A':N(C'))$$

However, a calculation I have done, is giving me $$I(A':N(C'))>I(A':C').$$

So I would like to know two things:

  • If my first inequality, which I can find no reason to be wrong, is correct? I have always though it is, however I am now doubting myself because of the second inequality.
  • How is $I(A':N(C'))>I(A':C')?$

I have checked the results, and the marginal states are valid density matrices, and the mutual information is being calculated the correct way, so I am at a bit of a loss as to how this is happening.

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Define the relative entropy between two states $\rho$ and $\sigma$ as $$ D(\rho\|\sigma) := \mathrm{tr}[\rho(\log \rho - \log \sigma)]\,. $$ Then we can write a the mutual information between $X$ and $Y$ for the state $\rho_{XY}$ as $$ I(X:Y) = D(\rho_{XY} \|\rho_X \otimes \rho_Y)\,. $$ Notice also that the relative entropy satisfies a data processing inequality $$ D(\rho\|\sigma) \geq D(\mathcal{N}(\rho) \|\mathcal{N}(\sigma))\,. $$

Taking the quantum channel $\mathcal{N} = \mathrm{tr}_B \circ \mathcal{U}_{BC}$ from the beginning of the question applied to $\rho_{AB} \otimes \rho_C$ we find $$ \begin{aligned} I(A:BC) &= D(\rho_{AB}\otimes \rho_C \|\rho_A\otimes \rho_B \otimes \rho_C)\\ &\geq D(\mathcal{N}(\rho_{AB} \otimes \rho_C) \|\mathcal{N}(\rho_A\otimes \rho_B \otimes \rho_C)) \\ &= D(\sigma_{AC}\|\sigma_A \otimes \sigma_C)\\ &= I(A:C)_{\sigma_{AC}} \end{aligned} $$ where $\sigma_{AC} = \mathcal{N}(\rho_{AB} \otimes \rho_C)$ and the crucial thing to note is that after defining $\sigma_{AC}$ like this we have $\sigma_A \otimes \sigma_C = \mathcal{N}(\rho_A \otimes \rho_B \otimes \rho_C)$ which holds because $\mathcal{N}$ applies trivially on system $A$. So to answer your first question, yes, that inequality is true.

For your second inequality you write

As such, $$I(A':B,C')\ge I(A':N(C'))$$ where $N$ is the reduced map from the actions of $U_{BC}$. As such, I would expect $$I(A':C') \ge I(A':N(C'))$$

So you have a map $\mathcal{N} = \mathrm{tr}_B \circ \mathcal{U}_{BC'}$ which maps $BC'\to C''$. By data processing you have $I(A':BC') \geq I(A':C'')$. If you just use partial trace then data processing also gives $I(A':BC') \geq I(A':C')$. What I think you are trying to imply with your notation is that there exists a map $N : C' \to C''$ such that $\mathcal{N} = N \circ \mathrm{tr}_B$. Because then you'd apply data processing twice to find $I(A':C') \geq I(A':C'')$. But this decomposition of your map is simply not true in general. For instance imagine that you take the unitary map to be the swap unitary and take your state to just be $\rho_B \otimes \rho_C$, so you swap and then trace out giving you the map $\rho_B \otimes \rho_C \mapsto \rho_B$. Now suppose you first trace out $B$, for your construction to work you need a map $N$ that sends $\rho_C \mapsto \rho_B$ and for this to be true for any $\rho_C$ and $\rho_B$ which is clearly nonsense.

On the existence of N

The issue seems can be boiled down to an assumption of some local dynamics. We have a channel $\mathcal{N}: BC \to C$ which consists of a global unitary followed by a partial trace, i.e., $$ \mathcal{N} = \mathrm{tr}_B \circ \mathcal{U}_{BC}\,. $$ The false assumption used to justify the wrong inequality is that there exists a channel $N: C \to C$ such that $$ \mathcal{N} = N \circ \mathrm{tr}_B\,. $$ Let us show that in general this local channel $N$ does not exist. Assume $B$ and $C$ are isomorphic and take $\mathcal{U}_{BC}$ to be the swap channel, so $$\mathcal{U}_{BC}(\rho \otimes \sigma) = \sigma \otimes \rho\,$$ for all $\rho$ and $\sigma$. Then for all $\rho$ and $\sigma$ we must have that $$ \mathcal{N}(\rho \otimes \sigma) = \rho\,. $$ On the other hand, if we assume that we can also write $\mathcal{N} = N \circ \mathrm{tr}_B$ then we need for all $\sigma$ that $$ \rho = \mathcal{N}(\rho \otimes \sigma) = N(\sigma). $$ That is we need a channel $N$ such that $N(\sigma) = \rho$ which holds for all $\sigma$ and for all $\rho$. This channel clearly cannot exist because we are requiring $N(\sigma)$ to take infinitely many values, $$ \mathcal{N}(|0\rangle \langle 0| \otimes \sigma) = \sigma \implies N(\sigma) = |0 \rangle \langle 0|$$ but we also have $$\mathcal{N}(|1\rangle \langle 1| \otimes \sigma) = \sigma \implies N(\sigma) = |1 \rangle \langle 1|$$ showing $N$ cannot exist.

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    $\begingroup$ But your map is from $BC' \to C''$. Not from $C' \to C''$ which is what you would need to justify that inequality. Please try writing down an example. It's also very easy to try the inequality numerically and you will see that it fails. $\endgroup$
    – Rammus
    Jan 23 at 18:14
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    $\begingroup$ In a sense, yes. $\endgroup$
    – Rammus
    Jan 23 at 19:29
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    $\begingroup$ Yes you are wrong about the existence of $N$. Please try writing down an explicit example. $\endgroup$
    – Rammus
    Jan 24 at 12:23
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    $\begingroup$ See edits for a more concrete example. $\endgroup$
    – Rammus
    Jan 24 at 12:38
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    $\begingroup$ Yes, in the initial definition on $\mathcal{N}$ you allow the two systems to interact first which means that the reduced system on $C'$ can depend on system $B$. In the second definition you immediately throw away $B$ and so $C'$ cannot depend on $B$. My guess is that the partial trace can be done first only when the initial map is separable (but I haven't tried to prove this). $\endgroup$
    – Rammus
    Jan 24 at 16:47

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