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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?

Thanks in advance!

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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.

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  • $\begingroup$ 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? $\endgroup$ – glS Jun 19 at 6:17
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    $\begingroup$ 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). $\endgroup$ – keisuke.akira Jun 20 at 0:10
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    $\begingroup$ So basically it is neither necessary nor sufficient to be entanglement breaking? $\endgroup$ – Norbert Schuch Aug 12 at 8:47
  • $\begingroup$ @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. $\endgroup$ – keisuke.akira Aug 12 at 9:08
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    $\begingroup$ @glS Yes. And that's what I say in the Note #2 of my answer as well. $\endgroup$ – keisuke.akira Aug 13 at 19:10

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