This paper shows the impossibility of perfect error correction for strictly contractive quantum channels, i.e., for channels such that $||\mathcal{E}(\rho)-\mathcal{E}(\sigma) ||\leq k ||\rho-\sigma||$, for $0\leq k <1$.

The requirement for perfect error correction of a subspace $K$ is that there exists a channel $S$ such that $S$ is the inverse of the restriction of $\mathcal{E}$ to the subspace $K$.

The proof of impossibility uses the fact that this would require $||S\mathcal{E}(|u\rangle\langle u|)-S\mathcal{E}(|v\rangle\langle v|)|| = |||u\rangle\langle u|-|v\rangle\langle v|||$, for some basis vectors $u,v$, which would contradict strict contractivity.

My confusion is concerning how this contradiction argument doesn't seem take into consideration the fact that we should restrict to the subspace $K$. In other words, if $P$ is the projector onto the subspace $K$, is it generally true that if $\mathcal{E}$ is strictly contractive, then $||P(\mathcal{E}(\rho))-P(\mathcal{E}(\sigma)) ||<||P(\rho)-P(\sigma)||$?

Thank you in advance.

  • 1
    $\begingroup$ I'm not sure I understand the confusion. The projector is also contractive (and non-invertible) so the proof holds. In general, as long as the input-output spaces have the same dimension then the only invertible CP-maps are unitaries (but if you relax this condition then one can have invertible maps). $\endgroup$ Jul 28, 2020 at 22:27
  • $\begingroup$ But the projector is applied on both sides. So I mean the question is, could it be that the channel is contractive when we look at the entire space but not contractive if we only restrict to a subspace? $\endgroup$ Jul 29, 2020 at 7:29
  • $\begingroup$ No, that is negated by noting that the projector itself is a CP-map: consider $P(\rho) = \sum\limits_{j} \Pi_{j} \rho \Pi_{j}$ and notice that it is already in a Kraus form (and hence automatically CP); although it is not TP (unless it is simply a dephasing operator). Therefore, $\left\Vert P \circ \mathcal{E} (\rho - \sigma) \right\Vert_{1} \leq \left\Vert \mathcal{E} (\rho - \sigma) \right\Vert_{1} \leq \left\Vert \rho - \sigma \right\Vert_{1} $. So, no, the projector will only ``contract'' the $1$-norm distance even more. $\endgroup$ Jul 29, 2020 at 7:46
  • $\begingroup$ I understand that the projector will cause the trace distance to contract compared to the original space, but what happens when we compare $||P\circ \mathcal{E}(\rho-\sigma)||$ to $||P(\rho -\sigma)||$ not $||\rho -\sigma||$? $\endgroup$ Jul 29, 2020 at 7:51

1 Answer 1


I am no longer confused about this, since now I see in this equation we are already restricting to a subspace $||S\mathcal{E}(|u\rangle\langle u|)-S\mathcal{E}(|v\rangle\langle v|)|| = |||u\rangle\langle u|-|v\rangle\langle v|||$, and the contracting map has to contract every subspace.


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