# Mutual information of Choi state=0, what would that imply about the quantum channel?

Classically, if the mutual information between the input and output of some channel or circuit $$= 0$$, it means the output is independent of the input, and the circuit is in a way 'useless'.

For the quantum case, defining the mutual information between an input $$\rho_A$$ and the output $$\rho_B$$, where $$\rho_B = \mathcal{E}_{B|A}(\rho_A)$$ is not so straightforward. Let the Choi state be $$\rho_{A'B}=\mathbb{I}_{A'}\otimes \mathcal{E}_{B|A}(\Omega_{A'A})$$, where $$\Omega_{A'A}$$ is a maximally entangled state. If $$I(A';B)=0$$, can one make a similar conclusion about 'independence' of input and output or 'usefulness' of the circuit in the classical case? Or what can one conclude here?

Here's a guess: they might be related to entanglement-breaking channels (also known as measure-and-prepare channels, quantum-classical channels, etc.). Any channel of the form, $$\Phi(\rho) = \sum\limits_{k} \operatorname{Tr}\left( M_{k} \rho \right) \sigma_{k} , \text{ where } M_{k}\geq0,\sum\limits_{k}^{} M_{k} = \mathbb{I},$$ are POVM elements and $$\{ \sigma_{k} \}$$ are quantum states is called EB. One can show that the Choi states of such channels are always seperable (in fact, $$\mathcal{I}^{A} \otimes \Phi^{B} (\Gamma^{AB})$$ is seperable for any entangled input density matrix -- hence the name, entanglement-breaking.)

As an example, consider a (simplified) EB channel of the form, $$\Phi(\rho) = \operatorname{Tr}\left( \rho \right) \sigma$$. Then, note that its Choi state is, $$\mathcal{I} \otimes \Phi \left( | \Omega \rangle \langle \Omega| \right) = \frac{1}{d} \sum\limits_{j,k}^{} | j \rangle \langle k | \otimes \Phi (| j \rangle \langle k | ) = \frac{1}{d} \sum\limits_{j,k}^{} | j \rangle \langle k | \otimes \delta_{jk} \sigma = \frac{\mathbb{I}}{d} \otimes \sigma.$$ Since the quantum mutual information of the input state $$I(A:B) = S_{\mathrm{rel}}(\rho^{AB} || \rho^{A} \otimes \rho^{B})$$, where $$S_{\mathrm{rel}}(\cdot || \cdot)$$ is the quantum relative entropy (see for ex. Nielsen and Chuang); we have, that for a "simple" measure-and-prepare channel, the quantum mutual information of the input-output states in the Choi representation is zero.

Note #1: The Choi states of other EB channels are seperable too, but it may not be product, in which case the QMI is not zero -- I'm not sure how to classify such channels in general. Of course, the convex combinations of two (or more) channels of the form $$\Phi_{j}(\rho) = \operatorname{Tr}\left( \rho \right) \sigma_{j}$$ will also satisfy this property, but I'm not sure how far this can generalize.

Note #2: To remark on the broader question, yes, convex combinations of channels of the form $$\Phi_{j}(\rho) = \operatorname{Tr}\left( \rho \right) \sigma_{j}$$ are, in fact, useless since their input-output states are independent (note that this is a convex subset of EB channels, and this is not true for all EB channels). Unfortunately, this set is simply a sufficient condition for the QMI of the Choi state to be zero (and not a necessary condition). And so the question remains unanswered.

• but for more general EB channels $\Phi(\rho)=\sum_k \operatorname{Tr}(M_k \rho)\sigma_k$ the Choi is separable but not product, so don't you have nonzero mutual information due to the classical correlations?
– glS
Jun 19 '20 at 6:17
• Yes, I mention that in the Note #1 to my answer. The set of channels formed by convex combinations of the "simple" EB channels have zero QMI, but this is not true for the more general set of EB channels (as I mention in my answer). Jun 20 '20 at 0:10
• So basically it is neither necessary nor sufficient to be entanglement breaking? Aug 12 '20 at 8:47
• @NorbertSchuch Unfortunately, yes. This subset of EB channels was the only example I could think of which had a similar "uselessness" as the classical scenario. Aug 12 '20 at 9:08
• @glS Yes. And that's what I say in the Note #2 of my answer as well. Aug 13 '20 at 19:10