The diamond distance between two channels $\Phi_0$ and $\Phi_1$ is defined in this answer.

$$ \| \Phi_0 - \Phi_1 \|_{\diamond} = \sup_{\rho} \: \| (\Phi_0 \otimes \operatorname{Id}_k)(\rho) - (\Phi_1 \otimes \operatorname{Id}_k)(\rho) \|_1 $$ where $\operatorname{Id}_k$ denotes the identity channel from $M_k(\mathbb{C})$ (the set of $k\times k$ complex matrices) 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 $M_{nk}(\mathbb{C}) = M_n(\mathbb{C}) \otimes M_{k}(\mathbb{C})$.

Let $N_1$ and $N_2$ be two completely positive trace nonincreasing maps that satisfy

$$\|N_1 - N_2\|_\diamond\leq \varepsilon.$$

For any channel $N_{A\rightarrow B}$, we define its Stinespring dilation to be an isometry $V_{A\rightarrow BE}$ such that $\text{Tr}_E(V\rho V^\dagger) = N(\rho)$.

Can one show that there exist Stinespring dilations $V_1$ and $V_2$ of $N_1$ and $N_2$ respectively such that we also have a bound on

$$\|V_1 - V_2\|_\diamond$$

in terms of $\varepsilon$?

  • $\begingroup$ Can you be a bit more precise on how you are defining $V_i$? $\endgroup$
    – Rammus
    Commented Nov 16, 2020 at 16:12
  • $\begingroup$ @Rammus, I have added the definition of the Stinespring dilation. Please let me know if something is still unclear. Thanks! $\endgroup$ Commented Nov 16, 2020 at 16:22
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    $\begingroup$ Sorry, thanks. I was just a little confused at the notation $\|V_1 - V_2\|_{\diamond}$ because the $V_i$ are isometries and not channels but I guess this is shorthand for the diamond norm between the channels $\rho \mapsto V_i \rho V_i^{\dagger}$? $\endgroup$
    – Rammus
    Commented Nov 17, 2020 at 11:48
  • $\begingroup$ @Rammus yes, that's what I meant $\endgroup$ Commented Nov 17, 2020 at 11:49
  • $\begingroup$ The issue is that the Stinespring dilation isn't unique i.e.~ there are many isometric channels $V_i\rho V_i^\dagger$ that come from the same completely positive map. $\endgroup$
    – Condo
    Commented Nov 17, 2020 at 15:13

1 Answer 1


Yes, in fact there exists Stinespring dilations such that

$$\frac{\|N_1-N_2\|_{cb}}{\sqrt{\|N_1\|_{cb}}+\sqrt{\|N_2\|_{cb}}}\leq \|V_1-V_2\|\leq \sqrt{\|N_1-N_2\|_{cb}}$$ where the distance between the isometries is the in terms of the operator norm and $N_1,N_2$ are unital completely positive maps and $V_1,V_2$ are their Stinespring isometries.

A unital completely positive map $N_i(X)=V_i^*\pi(X)V_i$ is dual to a CPTP map, so that is something to be aware of here...

See https://arxiv.org/pdf/0710.2495.pdf

Note: above the $cb$-norm is the completely bounded "operator" norm, where as the completely bounded trace norm (or diamond norm) is more common in quantum information. I believe that this does bound the diamond norm since, at least for finite dimensional maps (completely postive maps between matrix algebras) we have $$\|V_1-V_2\|_\diamond =\sup_{dim(H)}\|(V_1-V_2)\otimes id_{B(H)}\|_1\\\leq d\cdot \sup_{dim(H)}\|(V_1-V_2)\otimes id_{B(H)}\|_\infty=d\|V_1-V_2\|_{cb}\leq d\sqrt{\|N_1-N_2\|_{cb}},$$ where $d$ is the dimension of the space on which $(V_1-V_2)\otimes id_{B(H)}$ acts, and $B(H)$ is the set of bounded operators on $H$.

  • $\begingroup$ Thank you for the answer. It seems as though the paper claims this result for the cb norm, not the diamond norm? And the connection between those two norms is that $\|\Phi\|_\diamond = \|\Phi^*\|_{cb}$ where $\Phi^*$ is the adjoint channel of $\Phi$. Could you clarify how the adjoint channels were replaced with the original channels in your answer? arxiv.org/pdf/0901.4709.pdf is a reference that shows the connection between the cb norm and diamond norm (bottom of page 4) $\endgroup$ Commented Nov 16, 2020 at 16:29
  • $\begingroup$ the diamond norm is the same thing as the completely bounded trace norm. The $cb$-norm is the "completely bounded operator norm". In general $\|X\|_\infty\leq \|X\|_1$ so I think the same would hold for the $cb$-versions. $\endgroup$
    – Condo
    Commented Nov 16, 2020 at 16:39
  • $\begingroup$ I don't want to say for sure but I would expect that the same inequality would be true if you replaced each operator norm with the trace norm (maybe some constants that depend on dimensions would come into play aswell since the sum of the singular values can't be greater than the dimension times the largest singular value.) $\endgroup$
    – Condo
    Commented Nov 16, 2020 at 16:48
  • $\begingroup$ So to clarify, we have $\|T_1 - T_2\|_\diamond \geq \|T_1 - T_2\|_{cb} \geq \|V_1 - V_2\|_\infty^2$. Does this bound $\|V_1 - V_2\|_\diamond$ though since we only have a bound on the operator norm for $(V_1 - V_2)$? Thanks for the answer though, it certainly is a useful one! $\endgroup$ Commented Nov 16, 2020 at 16:57
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    $\begingroup$ Yes it does, thank you very much! I will leave it open for a bit longer to see if anyone else has comments or tighter bounds $\endgroup$ Commented Nov 17, 2020 at 10:48

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