On the second-order asymptotics for entanglement-assisted communication

Abstract

The entanglement-assisted classical capacity of a quantum channel is known to provide the formal quantum generalization of Shannon’s classical channel capacity theorem, in the sense that it admits a single-letter characterization in terms of the quantum mutual information and does not increase in the presence of a noiseless quantum feedback channel from receiver to sender. In this work, we investigate second-order asymptotics of the entanglement-assisted classical communication task. That is, we consider how quickly the rates of entanglement-assisted codes converge to the entanglement-assisted classical capacity of a channel as a function of the number of channel uses and the error tolerance. We define a quantum generalization of the mutual information variance of a channel in the entanglement-assisted setting. For covariant channels, we show that this quantity is equal to the channel dispersion and thus completely characterize the convergence toward the entanglement-assisted classical capacity when the number of channel uses increases. Our results also apply to entanglement-assisted quantum communication, due to the equivalence between entanglement-assisted classical and quantum communication established by the teleportation and super-dense coding protocols.

Appendices

Appendix 1: Heisenberg–Weyl operators

For any $x,z \in \{0,1,\ldots , d\}$ the Heisenberg–Weyl Operators X(x) and Z(z) are defined through their actions on the vectors of the qudit computational basis $\{|j\rangle \}_{j \in \{0,1,\ldots , d-1\}}$ as follows:

$$\begin{aligned} X(x) |j\rangle&= |j \oplus x\rangle ,\end{aligned}$$

(6.1)

$$\begin{aligned} Z(z)|j\rangle&= e^{2\pi i z j / d} |j\rangle , \end{aligned}$$

(6.2)

where $j \oplus x = (j + x) \, \mathrm{mod }\, d$. Also note that if $d = 1$, then both X(x) and Z(z) are equal to the identity operator.

Appendix 2: The method of types

In our proofs we employ the notion of types [13], and hence we briefly recall certain relevant definitions and properties here.

Let $\mathcal {X}$ denote a discrete alphabet and fix $n \in \mathbb {N}$. The type (or empirical probability distribution) $P_{x^n}$ of a sequence $x^n \in \mathcal {X}^n$ is the empirical frequency of occurrences of each letter of $\mathcal {X}$, i.e., $P_{x^n}(a) := \frac{1}{n} \sum _{i=1}^n \delta _{x_i,a}$ for all $a \in \mathcal {X}$. Let $\mathcal {P}_n$ denote the set of all types. The number of types, $|\mathcal {P}_n|$, satisfies the bound [14, Thm. 11.1.1]

$$\begin{aligned} |\mathcal {P}_n| \le (n+1)^{|\mathcal {X}|}. \end{aligned}$$

(7.1)

For any type $t \in \mathcal {P}_n$, the type class $\mathcal {T}^t$ of t is the set of sequences of type t, i.e.

$$\begin{aligned} \mathcal {T}^t := \{x^n \in \mathcal {X}^n \, : \, P_{x^n} = t \}. \end{aligned}$$

(7.2)

The number of types in a type class $\mathcal {T}^t$ satisfies the following lower bound [13, Lm. II.2]:

$$\begin{aligned} |\mathcal {T}^t| \ge \frac{2^{nH(t)}}{(n+1)^{|\mathcal {X}|}}, \end{aligned}$$

(7.3)

where $H(t) := - \sum _{a \in \mathcal {X}} t(a) \log t(a)$, is the Shannon entropy of the type.

Let q be any probability distribution on $\mathcal {X}$. For any sequence $x^n = (x_1, x_2, \ldots , x_n) \in \mathcal {X}^n$, let $q^n(x^n) = \prod _{i=1}^n q(x_i)$. Then, we have

$$\begin{aligned} q^n(x^n) = 2^{-n\left( H(t) + D(t\Vert q)\right) }, \qquad \text {where} \quad t = P_{x^n} \end{aligned}$$

(7.4)

is the type of $x^n$ and $D(t\Vert q):= \sum _{a \in \mathcal {X}} t(a) \log \frac{t(a)}{q(a)}$ is the Kullback-Leibler divergence of the probability distributions t and q. From (7.1), (7.3) and (7.4) it follows that for any sequence $x^n \in \mathcal {X}^n$ of type t,

$$\begin{aligned} (n+1)^{|\mathcal {X}|} 2^{nD(t\Vert q)} q^n(x^n) = 2^{-nH(t)}(n+1)^{|\mathcal {X}|}\ge \frac{1}{|\mathcal {T}^t|}. \end{aligned}$$

(7.5)

Finally, for any $\mu > 0$ we have [14, Eq. (11.98)]

$$\begin{aligned} \mathop {\mathop {\sum }\limits _{x^n \in \mathcal {X}^n}}\limits _{D(P_{x^n}\Vert q) > \mu } q^n(x^n) \le 2^{-n\left( \mu - |\mathcal {X}|\frac{ \log (n+1)}{n}\right) }. \end{aligned}$$

(7.6)