In a recent paper, the authors quote an older work of Bennett, Shor and others and make the following statement

While entanglement assistance can increase achievable rates for classical point-to-point channels in the zero-error and one-shot setting, it comes as a surprise that entanglement does not provide any advantage in the asymptotic setting with vanishing error.

However, in the abstract of the Bennett paper which is cited, it says

Prior entanglement between sender and receiver, which exactly doubles the classical capacity of anoiseless quantum channel, can increase the classical capacity of some noisy quantum channels by an arbitrarily large constant factor depending on the channel

My questions

1) I thought that the noisy channel coding theorem states that in the asymptotic limit, one has either zero error (if the rate was smaller than the capacity) or arbitrarily large error (if the rate was larger than the capacity). What is the asymptotic limit with vanishingly small error?

2) Why does the increased classical capacity by a constant factor due to entanglement not change anything in this limit?

EDIT: While the answer given by Serwyn is an excellent and correct one, I had a misunderstanding in the question. Essentially, Bennett et. al are talking about quantum channels in the second quote whereas the first quote is about classical channels only.


In my understanding, the key part for entanglement to increase capacity is to have a suboptimal channel. Suppose the input of you channel can take value in the set $X$, and note $G(X)$ the graph where nodes are possible inputs and there is an edge between to inputs if the range of their corresponding output is overlapping, i.e. When going through the channel they may become the same output. In a one-shot, error free communication scheme, you can't use two linked edge. So the maximum capacity is defined by the size of the maximum independent set of $G(X)$. However you may have a really good channel with large $|X|$ and low collision probabilities, but maybe only two independent inputs.

Using Quantum entenglement you can somehow relax this condition and use some possible imputs that you couldn't use before, with 0 error probability! The exact scheme and conditions are described in https://arxiv.org/pdf/0911.5300.pdf (Cubitt, Leung, Matthews and Winter)

But it is the same tool that one can use with classical long block coding. In that case using for exemple Polar Coding (long blocks), you can get arbitrarily low error and capacity arbitrarily close to theoretical maximum capacity.

Quantum entanglement increases 0-error, one shoot capacity only in case of an imperfect channel. It permits to uses ressources that couldn't be used otherwise. But that's also what assympotic long block coding does, so once you use let say Polar Coding, you have no "unperfect ressource" to improve with quantum entanglement anymore.


I will try to succinctly answer your first question given that I possess little knowledge regarding entanglement assistance.

Shannon's capacity theorem (the noisy channel coding theorem) states that for any code $\mathbf{C}$ of code rate $R \leq C$, where $C$ represents the channel capacity, an encoding and decoding scheme of rate $R$ with a probability of error $P_e < \epsilon$, where $\epsilon>0$, will exist. This probability of error will be vanishingly small only if the block length of the code is sufficiently large, i.e ${n\rightarrow\infty}$ leads to $\epsilon\rightarrow0$. Analogously, if $R>C$ then the error probability goes to $0.5$ as the block length of the code increases, i.e $$\lim_{n\to\infty}P_e = \frac{1}{2}$$.

Therefore, the asymptotic limit that defines when communications over a channel occur with an error probability $\epsilon < 0$ is the channel capacity. The condition of having a vanishingly small error probability will be determined by the block length of the code (for a null error probability you would need an infinite block length).

  • $\begingroup$ Thank you but I'm still a little unclear on the statement made in the paper. The authors talk about the asympototic limit (I assume this means $n\rightarrow\infty$) but with vanishing error (from context, I believe they mean nonzero). But as you point out, it is only for finite $n$ that $\epsilon$ is nonzero. So I'm still unsure what the authors mean. $\endgroup$ – user1936752 Sep 11 '19 at 14:08
  • $\begingroup$ I realize my previous comment sounds pedantic about infinities but apparently the two cases (zero error and vanishing error) are indeed different when it comes to the entanglement assisted capacity $\endgroup$ – user1936752 Sep 11 '19 at 14:26
  • $\begingroup$ From reading both the paper and its reference (Entanglement-Assisted Classical Capacity of Noisy Quantum Channels), my interpretation is that the authors make reference to the fact that entanglement assistance does not increase the capacity of a classical channel. I believe that the sentence ''it comes as a surprise ...'' is referring to the value of the capacity itself, and how quantum entanglement will not increase it (shift the Capacity curve); albeit, the wording (especially regarding the vanishing prob.) could have been made a little clearer. $\endgroup$ – Patrick Fuentes Sep 12 '19 at 11:00

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