(Non)Additivity, Part II

In the last post I reported that regularization is a fact of life for capacities of quantum channels, at least when trying the same trick that worked for classical information, random coding. The need for regularization comes from the nonadditivity of the random-coding-capacities \chi(\mathcal{N}), P^1(\mathcal{N}), and Q^1(\mathcal{N}) for classical, private, and quantum information respectively.

But what about the additivity of the true capacities? For instance, (ir)regardless of Q^1, the quantum capacity Q could be additive. Of course, because of the regularization it certainly is additive for two identical channels: Q(\mathcal{N}\otimes \mathcal{N})=2Q(\mathcal{N}). But what about two different channels? Same question for classical and private capacities. Are they additive for dissimilar channels?

The quantum channel is spectacularly nonadditive, as shown by Smith and Yard [arxiv]. They give the most extreme example possible, two channels which are individually worthless for sending quantum information that can nevertheless be usefully combined. 0 + 0 > 0, if you will.

The recipe for making this work is quite interesting. It calls for one part “private Horodecki channel” \mathcal{N}_H, which is a special channel which can transmit secret classical information but no quantum information, and an erasure channel \mathcal{N}_E, which just throws the input away with probability one half. Each channel takes in two qubits and outputs two qubits (actually, the erasure channel has another output, because it is nice enough to tell you when the input was erased.) Now pull out the recipe for random coding and follow the directions, combining these two channels into one. The yield is Q^1(\mathcal{N}_H\otimes \mathcal{N}_E), which Smith and Yard show is positive.

So the quantum capacity is not additive, but what about the other two? To my knowledge, the additivity of the classical capacity is open. The additivity of the private capacity is not certain, but some facts have recently come to light. Smith and Smolin [arxiv] recently showed that when combining a certain class of channels (called retro- or echo-correctable) with erasure channels, either \mathcal{P} is not additive or else \chi isn’t, at least not for these channels. They believe that \chi is additive in this case, and so the private capacity is nonadditive, but the jury’s still out.

In summary, none of the random-coding capacities are additive, meaning regularization is unavoidable and the various capacities over quantum channels don’t have simple expressions similar to the capacity of a classical channel. Nor do the true capacities of a quantum channel seem to be additive. It seems reasonable to expect this to be the case (this is more convincing knowing the answer, of course!*): An additive capacity implies that the usefulness of the channel for the given purpose is independent of what other channels you have handy, but the phenomena of entanglement suggests that quantum information is not so easily divided. And so random coding, with its product-state codewords, may not be the best way of approaching the problem.
Here’s a table of the results

Additive? Capacity Random-Coding Capacity
Classical ? no [arxiv]
Private * [arxiv] no [arxiv]
Quantum no [arxiv] no [arxiv,arxiv]

*Which is precisely the point. As Penrose said: “The great thing about physical intuition is that it can be adjusted to fit the facts.” (from the preface to Quantum Field Theory by Mark Srednicki.)

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3 Comments

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3 responses to “(Non)Additivity, Part II

  1. Thanks for these posts Joe. I admit to sometimes getting a little confused about additivity per se vs. additivity in the random coding scenario, despite having it explained to me several times in talks. Now I have a place to go to when that happens.

    • joerenes

      Thanks for reading! I have often been confused on this point myself, which is why I decided to write it down. I hope to do more writing stuff down here in the future.

  2. Pingback: QIP 2009 Day 5 Liveblogging - The Quantum Pontiff

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