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.


2 Answers 2



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 $\varepsilon=0$.

  • $\begingroup$ Thank you, that's very illuminating. Regarding your main point, I suppose my question is a bit silly for expecting the channel to be close to an identity channel since that should only be true in the quantum case as you point out. I think the correct expectation would be to bound $\|\mathcal{D}\circ\mathcal{N}\circ\mathcal{E} - \mathcal{I}\circ\mathcal{P}\|_\diamond$ using $\varepsilon'$, where $\mathcal{P}$ is the dephasing map that throws away all the off diagonal parts of the input. Does this make more sense? $\endgroup$ Commented Jul 31, 2021 at 12:26
  • $\begingroup$ Unfortunately, the expression $\|\mathcal{D}\circ\mathcal{N}\circ\mathcal{E} - \mathcal{I}\circ\mathcal{P}\|_\diamond$ won't work either. In this case when you set $\mathcal{N} := \mathcal{I}$ (rather $\mathcal{N} := \mathcal{P}$ as I did in the answer; note that both substitutions satisfy the requirement of perfect preservation of classical information, i.e. $\varepsilon=0$) than you get very similar calculation and the same lower bound. $\endgroup$ Commented Jul 31, 2021 at 15:10
  • $\begingroup$ The issue lies not in what channel we're comparing to, but in what metric we use. The diamond norm takes the supremum over all quantum states and therefore any channels that differ in their treatment of quantum information are separated in the diamond norm distance. $\endgroup$ Commented Jul 31, 2021 at 15:11
  • $\begingroup$ Sorry maybe I'm missing something but when we consider $\mathcal{I}\circ\mathcal{P}$, all quantum states first get dephased by the $\mathcal{P}$ map. Similarly, for the encoder we use, we can also choose, without loss of generality, have a $\mathcal{P}$ map acting first on its inputs. In other words, quantum information becomes classical before being processed by either $\mathcal{I}$ or by $\mathcal{D}\circ\mathcal{N}\circ\mathcal{E}$. In particular, the third line of your lower bound will now become $(\mathcal{I}\circ\mathcal{P})\vert +\rangle\langle +\vert = \frac{I}{d}$, isn't it? $\endgroup$ Commented Aug 1, 2021 at 10:03
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    $\begingroup$ Yes. Although ultimately the appropriate level of generality depends on what you're trying to study. If you are interested in how well the channel transfers quantum information, then you should use the diamond norm and should not force dephasing at encoding. If you are interested in how well your channel transfers classical information then you are free to force dephasing (since it does nothing then), but then the diamond is not appropriate. Instead, you should use classical metrics like total variation distance. $\endgroup$ Commented Aug 1, 2021 at 17:54

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 $\varepsilon'$.

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    $\begingroup$ Note that $(\mathcal{P}\otimes\mathcal{I})(\rho)$ is separable for any $\rho$. Therefore, applying $\mathcal{P}$ to both terms in $\|\mathcal{D}\circ\mathcal{N}\circ\mathcal{E}\circ\mathcal{P} - \mathcal{I}\circ\mathcal{P}\|_\diamond$ is equivalent to restricting the supremum in definition of the norm to diagonal density matrices on the first subsystem. This in turn is equivalent to restricting to classical probability distributions. Thus, diamond norm is unnecessary and conceals classical character of the expression. Classical distance can be bounded by $\varepsilon$. Quantum distance can't. $\endgroup$ Commented Aug 1, 2021 at 19:59
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    $\begingroup$ Thank you for the answer and the comments! $\endgroup$ Commented Aug 2, 2021 at 3:56

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