Bounding diamond norm distance using probability of error in transmission of classical information

Question

Let us consider an encode, noisy channel and a decoder such that classical messages $m\in\mathcal{M}$ can be transmitted with some small error. That is, for a message $m$ that is sent by Alice, Bob guesses $\hat{m}$ and the average error satisfies $\frac{1}{|\mathcal{M}|}\sum_{m}Pr[m\neq \hat{m}] \leq \varepsilon$. One could also consider the worst case error that is, $\max_m Pr[m\neq\hat{m}] \leq \varepsilon'$.

Are either of these quantities related to a channel distance measure and if yes, how? The idea is that the composite channel where we encode, use a noisy channel and decode $\mathcal{D}\circ\mathcal{N}\circ\mathcal{E}$ must be effectively an identity channel. The most commonly used distance measure is the diamond distance so can we bound

$$\|\mathcal{D}\circ\mathcal{N}\circ\mathcal{E} - \mathcal{I}\|_\diamond$$

using either $\varepsilon$ or $\varepsilon'$?

Note that while $\mathcal{N}$ may be quantum, the composite channel $\mathcal{D}\circ\mathcal{N}\circ\mathcal{E}$ is in fact classical. Also note that if we know an upper bound for $\|\mathcal{D}\circ\mathcal{N}\circ\mathcal{E} - \mathcal{I}\|_\diamond$, then we can easily bound the worst case error $\max_m Pr[m\neq\hat{m}]$ and hence also the average error $\frac{1}{|\mathcal{M}|}\sum_{m}Pr[m\neq \hat{m}]$. The other way around seems far less obvious.

Answer 1

Intuition

The expression $\|\mathcal{A} - \mathcal{I}\|_\diamond$ quantifies how close the channel $\mathcal{A}$ is to the identity channel $\mathcal{I}$ which is the channel that preserves quantum information perfectly. In order for a channel to transfer quantum information well, it must preserve both diagonal and off-diagonal elements of the input density matrix. On the other hand, in order for a channel to transfer classical information well, it is sufficient for it to preserve the diagonal elements of the density matrix.

Consequently, we would expect that a channel's ability to transfer classical information is a poor guide to its ability to transfer quantum information and therefore a poor guide to how close it is to the identity channel in diamond norm distance.

We can quantify the above reasoning by analyzing the case of strong dephasing noise. This will rule out the possibility of a bound

$$ \|\mathcal{D}\circ\mathcal{N}\circ\mathcal{E} - \mathcal{I}\|_\diamond \le B(\varepsilon) $$

with $\lim_{\varepsilon\to 0} B(\varepsilon)=0$.

Representing classical information

For the expression $\|\mathcal{D}\circ\mathcal{N}\circ\mathcal{E} - \mathcal{I}\|_\diamond$ to make sense the domain of $\mathcal{E}$ and the codomain of $\mathcal{D}$ need to be the space of linear maps on some Hilbert space $\mathcal{H}$. This raises the question about the way in which the classical messages in $\mathcal{M}$ are represented in $\mathcal{H}$. Since the elements of $\mathcal{M}$ are reliably distinguishable, the appropriate choice is to use an orthonormal basis. Therefore, we assume that a message $m\in\mathcal{M}$ is represented using a computational basis vector $|m\rangle\in\mathcal{H}$. This includes the asumption that $\dim\mathcal{H} \ge |\mathcal{M}|$.

Dephasing noise

Consider the completely dephasing channel defined by

$$ \left(\mathcal{N}_{DF}(\rho)\right)_{ij} = \delta_{ij}\rho_{ij} $$

where $\delta_{ij}=1$ if $i=j$ and $\delta_{ij}=0$ otherwise. Thus, $\mathcal{N}_{DF}$ preserves the diagonal elements and forgets the off-diagonal elements of the input density matrix.

Resilience of classical information

Choosing the encoding and decoding operations to be the identity map on $L(\mathcal{H})$, we have for every $m\in\mathcal{M}$

$$ (\mathcal{D}\circ\mathcal{N}_{DF}\circ\mathcal{E})(|m\rangle\langle m|) = |m\rangle\langle m|. $$

Thus, despite the strong dephasing noise in $\mathcal{N}_{DF}$, the classical messages pass through it undisturbed and so

$$\max_m Pr[m\neq\hat{m}] \le \varepsilon = 0.$$

Lower bound on diamond norm

However, the diamond distance is bounded from below by

$$ \begin{align} \|\mathcal{D}\circ\mathcal{N}_{DF}\circ\mathcal{E} - \mathcal{I}_{L(\mathcal{H})}\|_\diamond &=\|\mathcal{N}_{DF} - \mathcal{I}_{L(\mathcal{H})}\|_\diamond \\ &= \sup_{k,\rho}\|(\mathcal{N}_{DF}\otimes \mathcal{I}_k)(\rho) - (\mathcal{I}_{L(\mathcal{H})}\otimes \mathcal{I}_k)(\rho)\|_1 \\ &\ge \|\mathcal{N}_{DF}(|{+_d}\rangle\langle+_d|) - |{+_d}\rangle\langle+_d|\|_1 \\ &= \left\|\frac{I}{d} - |{+_d}\rangle\langle+_d|\right\|_1 \\ &\ge \left\|\frac{I}{d} - |{+_d}\rangle\langle+_d|\right\|_2 \\ &=\sqrt{\frac{d-1}{d}} \end{align} $$

where $|{+_d}\rangle=\frac{1}{\sqrt{d}}\sum_{i=1}^d|i\rangle$, $d=\dim\mathcal{H}$ and $\|\,.\|_2$ denotes the Frobenius norm. Thus, the diamond norm distance $\|\mathcal{D}\circ\mathcal{N}\circ\mathcal{E} - \mathcal{I}\|_\diamond$ is bounded from below by an expression independent of the associated bound $\varepsilon$ on the error probability in the transmission of classical information. In particular, the diamond norm distance $\|\mathcal{D}\circ\mathcal{N}\circ\mathcal{E} - \mathcal{I}\|_\diamond$ may be fairly large even when $\varepsilon=0$.

Answer 2

Let us pick a basis for the Hilbert space $\mathcal{H}$ of dimension $d$ to be $\{\vert i\rangle\}$ and let $\mathcal{P}: \mathcal{H}\rightarrow\mathcal{H}$ be the dephasing map given by

$$\mathcal{P}: \rho \rightarrow \sum_i \vert i\rangle\langle i\vert \rho \vert i\rangle\langle i\vert $$

The states $\vert i\rangle\langle i\vert$ will be the chosen states to encode the classical bit $i$ from some alphabet of size $d$. I now claim that

$$\|\mathcal{D}\circ\mathcal{N}\circ\mathcal{E}\circ\mathcal{P} - \mathcal{I}\circ\mathcal{P}\|_\diamond \leq \varepsilon'\tag{1}$$

Note that since $\varepsilon'$ was defined only for classical messages. The channels $\mathcal{D}\circ\mathcal{N}\circ\mathcal{E}\circ\mathcal{P}$ performs exactly the same as $\mathcal{D}\circ\mathcal{N}\circ\mathcal{E}$ for classical messages so this quantity is a reasonable one to look at.

Now we have $$\begin{align} &\|\mathcal{D}\circ\mathcal{N}\circ\mathcal{E}\circ\mathcal{P} - \mathcal{I}\circ\mathcal{P}\|_\diamond = \sup_{|\mathcal{K}|,\rho}\|\left((\mathcal{D}\circ\mathcal{N}\circ\mathcal{E}\circ\mathcal{P})\otimes\mathcal{I_K}\right)\rho - \left((\mathcal{I}\circ\mathcal{P})\otimes\mathcal{I_K}\right)(\rho)\|_1\\ &= \|\left((\mathcal{D}\circ\mathcal{N}\circ\mathcal{E})\otimes\mathcal{I_K}\right)\sum_j p_j\vert j\rangle\langle j\vert\otimes\sigma^j - (\mathcal{I}\otimes\mathcal{I_K})\sum_j p_j\vert j\rangle\langle j\vert\otimes\sigma^j\|_1\\ &\leq \sum_j p_j \|\left((\mathcal{D}\circ\mathcal{N}\circ\mathcal{E})\otimes\mathcal{I_K}\right)\vert j\rangle\langle j\vert\otimes\sigma^j - \vert j\rangle\langle j\vert\otimes\sigma^j\|_1\\ &\leq \sum_j p_j \varepsilon' = \varepsilon' \end{align}$$

The second line holds because the dephasing map on the space $\mathcal{H}$ and the identity on $\mathcal{K}$ results in some classical-quantum state. The first inequality is due to convexity of the 1-norm and the last line follows since we know that the alphabet $j$ is transmitted with error at most $\varepsilon'$.