In quantum information theory, the distance between two quantum channels is often measured using the diamond norm. There are also a number of ways to measure distance between two quantum states, such as the trace distance, fidelity, etc. The Jamiołkowski isomorphism provides a duality between quantum channels and quantum states.

This is interesting, to me at least, because the diamond norm is notoriously hard to calculate, and the Jamiołkowski isomorphism would seem to imply some correlation between distance measures of quantum channels and quantum states. So, my question is this: Is there any known relation between the distance in the diamond norm and the distance between the associated states (in some measure)?

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    $\begingroup$ I am not sure what you mean by “the diamond norm is notoriously hard to calculate.” If you are given a quantum channel as an explicit matrix (of its Choi-Jamiołkowski representation, say), the square of its diamond norm can be formulated as a semidefinite program; see Section 20.4 of lecture note by John Watrous. In that sense, the diamond norm has an efficient means to calculate. $\endgroup$ – Tsuyoshi Ito Feb 12 '11 at 22:12
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    $\begingroup$ @Tsuyoshi: I was just refering to the implicit optimization. I didn't mean computationally hard, but rather awkward to work with. $\endgroup$ – Joe Fitzsimons Feb 12 '11 at 22:24
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    $\begingroup$ These are very nice lecture notes, as an aside. $\endgroup$ – Suresh Venkat Feb 13 '11 at 5:05
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    $\begingroup$ @Suresh @Tsuyoshi: Yes, they are great notes, but I don't think they answer this particular question. $\endgroup$ – Joe Fitzsimons Feb 13 '11 at 15:02
  • $\begingroup$ @TsuyoshiIto : for some reason, the last section in QIP slides is 20.3, do you have a more complete lectures set? $\endgroup$ – Artem Oboturov May 3 '16 at 14:03

For a quantum channel $\Phi$, let us write $J(\Phi)$ to denote the associated state: $$ J(\Phi) = \frac{1}{n} \sum_{1\leq i,j \leq n} \Phi(|i \rangle \langle j|) \otimes |i \rangle \langle j|. $$ Here we are assuming that the channel maps $M_n(\mathbb{C})$ (i.e., $n\times n$ complex matrices) to $M_m(\mathbb{C})$ for whatever choice of positive integers $n$ and $m$ you like. The matrix $J(\Phi)$ is sometimes called the Choi matrix or Choi-Jamiolkowski representation of $\Phi$, but it is more frequent that those terms are used when the $\frac{1}{n}$ normalization is omitted.

Now, suppose that $\Phi_0$ and $\Phi_1$ are quantum channels. We may define the "diamond norm distance" between them as $$ \| \Phi_0 - \Phi_1 \|_{\Diamond} = \sup_{\rho} \: \| (\Phi_0 \otimes \operatorname{Id}_k)(\rho) - (\Phi_1 \otimes \operatorname{Id}_k)(\rho) \|_1 $$ where $\operatorname{Id}_k$ denotes the identity channel from $M_k(\mathbb{C})$ to itself, $\| \cdot \|_1$ denotes the trace norm, and the supremum is taken over all $k \geq 1$ and all density matrices $\rho$ chosen from $M_{nk}(\mathbb{C}) = M_n(\mathbb{C}) \otimes M_{k}(\mathbb{C})$. The supremum always happens to be achieved for some choice of $k\leq n$ and some rank 1 density matrix $\rho$.

(Note that the above definition does not work for arbitrary mappings, only those of the form $\Phi = \Phi_0 - \Phi_1$ for completely positive maps $\Phi_0$ and $\Phi_1$. For general mappings, the supremum is taken over all matrices with trace norm 1, as opposed to just density matrices.)

If you don't have any additional assumptions on the channels, you cannot say too much about how these norms relate aside from these coarse bounds: $$ \frac{1}{n} \| \Phi_0 - \Phi_1 \|_{\Diamond} \leq \| J(\Phi_0) - J(\Phi_1) \|_1 \leq \| \Phi_0 - \Phi_1 \|_{\Diamond}. $$ For the second inequality, one is essentially settling for the specific choice $$ \rho = \frac{1}{n} \sum_{1\leq i,j \leq n} |i \rangle \langle j| \otimes |i \rangle \langle j| $$ rather than taking the supremum over all $\rho$. The first inequality is a bid tougher, but it would be a reasonable assignment question for a graduate course on quantum information. (At this point I should thank you for your question, because I fully intend to use this question in the Fall offering of my quantum information theory course.)

You can achieve either inequality for an appropriate choice of channels $\Phi_0$ and $\Phi_1$, even under the additional assumption that the channels are perfectly distinguishable (meaning $\| \Phi_0 - \Phi_1 \|_{\Diamond} = 2$).

  • $\begingroup$ Thanks John, that answers my question perfectly, and has saved me a lot of time. $\endgroup$ – Joe Fitzsimons Feb 15 '11 at 0:04

You might also want to look into Distance measures to compare real and ideal quantum processes arXiv:quant-ph/0408063 which gives an overview of distance measures for quantum channels and their relationships.

They use the term S distance for the diamond distance and J distance for the trace distance of the Jamiołkowski operators associated to the channels.


I like to think of the first inequality that Watrous wrote in terms of probabilistic channel teleportation. If you interpret the diamond norm as a measure of the smallest error probability in discriminating channels $\Phi_0$ and $\Phi_1$, and the trace norm as the equivalent for their Jamiolkowski states, you can always implement the optimal strategy for the channels from their corresponding states with $\frac{1}n$ success probability. Putting this rigorously may be a way of proving the inequality.

Also, this way of thinking shows that if the channels can be teleported deterministically (such as Pauli channels), then their diamond norm equals the Jamiolkowski trace distance.


Following up on the line of thinking presented by Alex Monras, there is actually a quite generic argument for this kind of bound that goes beyond diamond norm and applies to many other channel distance measures. The proof applies to diamond distance, negative root fidelity, relative entropy, Petz- and sandwiched Renyi relative quasi-entropies, etc. I understand it in terms of quantum steering, which is related to but different from teleportation.

Here it is. Let $D$ be a generalized divergence defined on quantum states $\rho$ and $\sigma$, which obeys the data-processing inequality:

$$ D(\rho \Vert \sigma) \geq D(\mathcal{N}(\rho) \Vert \mathcal{N}(\sigma)), $$

where $\mathcal{N}$ is a quantum channel. Suppose that $D$ obeys the following direct-sum property:

$$ D(\rho_{XB} \Vert \sigma_{XB}) = \sum_x p(x) D(\rho_x \Vert \sigma_x) $$

on classical--quantum states:

$$ \rho_{XB} := \sum_x p(x) |x \rangle \langle x| \otimes \rho_x, \qquad \sigma_{XB} := \sum_x p(x) |x \rangle \langle x| \otimes \sigma_x . $$

Suppose also that it is weakly faithful, in the sense that $D(\rho\Vert \rho) = 0$ for all states $\rho$. (Data processing and weakly faithful imply that the minimum value of $D$ is zero.)

Now let $\mathcal{N}$ and $\mathcal{M}$ be quantum channels and define the generalized channel divergence (see https://arxiv.org/abs/1709.01111) as

$$ D(\mathcal{N} \Vert \mathcal{M}) := \sup_{\psi_{RA}} D(\mathcal{N}_{A\to B}(\psi_{RA}) \Vert \mathcal{M}_{A \to B}(\psi_{RA})), $$

where the optimization is with respect to all pure bipartite states $\psi_{RA}$ with system $R$ isomorphic to system $A$ (due to data processing, purification, and Schmidt decomposition, this optimization suffices -- no need to consider mixed states $\rho_{RA}$ with arbitrarily large reference $R$).


Then the following inequality holds

$$ \frac{1}{d} D(\mathcal{N} \Vert \mathcal{M}) \leq D(\mathcal{N}_{A\to B}(\Phi_{RA}) \Vert \mathcal{M}_{A \to B}(\Phi_{RA})), $$

where $d$ is the dimension of the channel inputs and $\Phi_{RA}$ denotes the maximally entangled state:

\begin{align} \Phi_{RA} & := |\Phi\rangle \langle\Phi|_{RA}\\ |\Phi\rangle_{RA} & := \frac{1}{\sqrt{d}} \sum_{i} |i\rangle_R |i\rangle_A. \end{align}


Let $\psi_{RA}$ be an arbitrary pure bipartite state. Then there exists an operator $Z_R$ such that

$$ \psi_{RA} = d Z_R \Phi_{RA} Z_R^\dagger, $$

with $\mathrm{Tr}[Z_R^\dagger Z_R] = 1$ and where $d$ is the dimension of system $A$. This is a key equation that indicates how one can steer the state $\psi_{RA}$ from the maximally entangled state $\Phi_{RA}$.

Now define the following (steering) quantum channel:

$$ \mathcal{P}_{R \to XR}(\omega_R) := |0\rangle \langle 0|_X \otimes Z_R \omega_R Z_R^\dagger + |1\rangle \langle 1|_X \otimes \sqrt{I_R - Z_R^\dagger Z_R} \omega_R \sqrt{I_R - Z_R^\dagger Z_R} . $$

Consider that

\begin{align} \mathcal{P}_{R \to XR}(\Phi_{RA}) & = \frac{1}{d} |0\rangle\langle 0|_X \otimes \psi_{RA} + \left(1-\frac{1}{d}\right) |1\rangle\langle 1|_X \otimes \textrm{``other state''} . \end{align}

Applying the steering channel to the state $\mathcal{N}_{A\to B}(\Phi_{RA})$, we find that

\begin{align} & \mathcal{P}_{R \to XR}(\mathcal{N}_{A\to B}(\Phi_{RA}))\\ & = \mathcal{N}_{A\to B}(\mathcal{P}_{R \to XR}(\Phi_{RA}))\\ & = \frac{1}{d} |0\rangle\langle 0|_X \otimes \mathcal{N}_{A\to B}(\psi_{RA}) + \left(1-\frac{1}{d}\right) |1\rangle\langle 1|_X \otimes \textrm{``other state 1''} . \end{align}


\begin{align} & \mathcal{P}_{R \to XR}(\mathcal{M}_{A\to B}(\Phi_{RA}))\\ & = \mathcal{M}_{A\to B}(\mathcal{P}_{R \to XR}(\Phi_{RA})) \\ & = \frac{1}{d} |0\rangle\langle 0|_X \otimes \mathcal{M}_{A\to B}(\psi_{RA}) + \left(1-\frac{1}{d}\right)|1\rangle\langle 1|_X \otimes \textrm{``other state 2''} . \end{align}

Now applying the data-processing inequality, the direct-sum property, and the weakly faithful property, we find that

\begin{align} & D(\mathcal{N}_{A\to B}(\Phi_{RA}) \Vert \mathcal{M}_{A \to B}(\Phi_{RA})) \\ & \geq D(\mathcal{P}_{R \to XR}(\mathcal{N}_{A\to B}(\Phi_{RA})) \Vert \mathcal{P}_{R \to XR}(\mathcal{M}_{A \to B}(\Phi_{RA}))) \\ & = \frac{1}{d} D(\mathcal{N}_{A\to B}(\psi_{RA}) \Vert \mathcal{M}_{A \to B}(\psi_{RA})) + \left(1 - \frac{1}{d}\right) D(\mathrm{``other state 1''}\Vert \mathrm{``other state 2''})\\ & \geq \frac{1}{d} D(\mathcal{N}_{A\to B}(\psi_{RA}) \Vert \mathcal{M}_{A \to B}(\psi_{RA})). \end{align}

Since the state $\psi_{RA}$ is arbitrary, we conclude the claimed bound.


If it should prove useful, this is a derivation of the inequality $$ \frac{1}{n} \| \Phi_0 - \Phi_1 \|_{\Diamond} \leq \| J(\Phi_0) - J(\Phi_1) \|_1 $$ in John Watrous' answer. We will make use of the (basis-dependent) operator-vector correspondence denoted as $\operatorname{Vec}$: for any operator $A:M_n(\mathbb{C}) \to M_{n'}(\mathbb{C})$, we define a corresponding vector in $M_{n'}(\mathbb{C}) \otimes M_n(\mathbb{C})$ via $$ \operatorname{Vec}(A) := (A \otimes I) \sum_{i=1}^{n} |i\rangle \otimes |i\rangle. $$ The $\operatorname{Vec}$ operation is invertible, i.e. for any $|\psi\rangle \in M_{n'}(\mathbb{C}) \otimes M_n(\mathbb{C})$ there exists a unique $A:M_n(\mathbb{C}) \to M_{n'}(\mathbb{C})$ such that $\operatorname{Vec}(A)=|\psi\rangle$. Additionally, it is not hard to verify that $\langle \operatorname{Vec}(A),\operatorname{Vec}(B) \rangle = \operatorname{Tr}[A^\dagger B]$ and $\operatorname{Vec}(A) = (I \otimes A^T) \sum_{j=1}^{n'} |j\rangle \otimes |j\rangle$. (In the above expressions, $|i\rangle$ and $|j\rangle$ denote fixed choices of orthonormal bases for $M_n(\mathbb{C})$ and $M_{n'}(\mathbb{C})$ respectively, and the transpose $A^T$ is taken w.r.t. these bases.)

As stated in John Watrous' answer, the diamond norm distance between channels can be computed as $$ \| \Phi_0 - \Phi_1 \|_{\Diamond} = \sup_{\rho} \: \| ((\Phi_0 - \Phi_1) \otimes \operatorname{Id}_k)[\rho] \|_1, $$ and the supremum is attained by some (normalized) pure state $|\psi\rangle \in M_{n}(\mathbb{C}) \otimes M_n(\mathbb{C})$. By the operator-vector correspondence, there exists $A:M_n(\mathbb{C}) \to M_n(\mathbb{C})$ such that $|\psi\rangle = \operatorname{Vec}(A) = (I \otimes A^T) \sum_{j=1}^{n} |j\rangle \otimes |j\rangle$ and $\operatorname{Tr}[A^\dagger A] = 1$. Denoting $|\Psi\rangle := \sum_{i=1}^{n} |i\rangle \otimes |i\rangle/\sqrt{n} = \sum_{j=1}^{n} |j\rangle \otimes |j\rangle /\sqrt{n}$ for brevity (note the domain and codomain of $A$ are isomorphic here), this gives \begin{align} \| \Phi_0 - \Phi_1 \|_{\Diamond} &= \| ((\Phi_0 - \Phi_1) \otimes \operatorname{Id}_k)\left[(I \otimes A^T)n|\Psi\rangle\langle\Psi|(I \otimes A^T)^\dagger\right] \|_1 \\ &= n\| (I \otimes A^T) \left(((\Phi_0 - \Phi_1) \otimes \operatorname{Id}_k)\left[|\Psi\rangle\langle\Psi|\right]\right) (I \otimes A^T)^\dagger\|_1 \\ &= n\| (I \otimes A^T) (J(\Phi_0) - J(\Phi_1)) (I \otimes A^T)^\dagger\|_1, \end{align} where we have used the trick of commuting the channels $\Phi_x$ with the $A^T$ operators since they act on different systems. We then apply Hölder's inequality twice, followed by monotonicity of Schatten norms: \begin{align} \| \Phi_0 - \Phi_1 \|_{\Diamond} &\leq n\| (I \otimes A^T) \|_\infty \| (J(\Phi_0) - J(\Phi_1)) (I \otimes A^T)^\dagger\|_1 \\ &\leq n\| (I \otimes A^T) \|_\infty \|(I \otimes A^T)^\dagger\|_\infty \| J(\Phi_0) - J(\Phi_1) \|_1 \\ &= n\| A^T \|_\infty \|\bar{A}\|_\infty \| J(\Phi_0) - J(\Phi_1) \|_1 \\ &\leq n\| A^T \|_2 \|\bar{A}\|_2 \| J(\Phi_0) - J(\Phi_1) \|_1 \\ &= n \sqrt{\operatorname{Tr}[\bar{A} A^T]} \sqrt{\operatorname{Tr}[A^T \bar{A}]} \| J(\Phi_0) - J(\Phi_1) \|_1 \\ &= n\| J(\Phi_0) - J(\Phi_1) \|_1 . \end{align}

I also attempted a more "operational" argument using gate teleportation as in https://physics.stackexchange.com/a/274739/213379 --- essentially, given the state $J(\Phi_x)$, one crude attempt to distinguish whether $x=0$ or $1$ would be to use it to probabilistically simulate the channel $\Phi_x$ in the optimal distinguishing strategy between the two channels, keeping the output of that optimal strategy if the simulation succeeds and otherwise making a random guess. (Note that whether the simulation succeeded is "heralded", in that it can be determined from the outcome of the entangled measurement.) This yields a lower bound on $\| J(\Phi_0) - J(\Phi_1) \|_1$ in terms of $\| \Phi_0 - \Phi_1 \|_{\Diamond} $, since each of these terms is linearly related to the corresponding distinguishing probabilities. Unfortunately, the probability of successful channel simulation using the Choi state is $1/n^2$ instead of $1/n$, which instead yields a weaker bound of $\| J(\Phi_0) - J(\Phi_1) \|_1 \geq \| \Phi_0 - \Phi_1 \|_{\Diamond}/n^2$. (A little care is still needed to derive this, because distinguishing probability is linearly related to the trace/diamond norms, not proportional to them.)

  • $\begingroup$ Credit to a colleague who showed me this derivation. $\endgroup$ – helloworld Jun 24 '20 at 17:51

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