2
$\begingroup$

We know that every CPTP map $\Phi:\mathcal X\to\mathcal Y$ can be represented via an isometry $U:\mathcal X\otimes\mathcal Z\to\mathcal Y\otimes\mathcal Z$, as $$\Phi(X) = \operatorname{Tr}_{\mathcal Z}[U(X\otimes E_{0,0})U^\dagger],\quad\text{where}\quad E_{a,b}\equiv \lvert a\rangle\!\langle b\rvert.\tag1$$ Showing this is quite easy e.g. from the Kraus representation. If $A_a:\mathcal X\to\mathcal Y$ are Kraus operators for $\Phi$, then $$U_{\alpha a,i0} \equiv \langle \alpha,a\rvert U\lvert i,0\rangle = \langle \alpha\rvert A_a\lvert i \rangle \equiv (A_a)_{\alpha,0}.\tag2$$ We can, of course, replace $E_{0,0}$ with any pure state in (1) without affecting the result.

This shows that, given any channel $\Phi$ and any pure state $\lvert\psi\rangle\in\mathcal Z$, we can represent $\Phi$ as in (1) (with $E_{0,0}\to\lvert\psi\rangle$). However, what about the more general case of $E_{0,0}\to\sigma$ with $\sigma$ not pure?

To analyse this case, consider a channel written as $$\Phi(X)=\operatorname{Tr}_{\mathcal Z}[U(X\otimes \sigma)U^\dagger]\tag3$$ for some state $\sigma=\sum_k p_k E_{k,k}\in\mathrm{Lin}(\mathcal Z)$ (appropriately choosing the computational basis for $\mathcal Z$). The relation with the Kraus operators reads in this case $$\Phi(X)_{\alpha,\beta} = \sum_{\ell k ij} p_\ell U_{\alpha \ell,i k} X_{ij} U^*_{\beta\ell,jk} = \sum_{\ell,k} (A_{\ell,k}XA_{\ell,k}^\dagger)_{\alpha\beta}\tag4$$ with $$(A_{\ell,k})_{\alpha,i} \equiv \sqrt{p_\ell} U_{\alpha\ell,ik}, \qquad A_{\ell,k} = \sqrt{p_\ell} (I\otimes \langle \ell\rvert)U(I\otimes \lvert k\rangle).\tag5$$

There is now a difference: the number of Kraus operators must be larger than the rank of $\sigma$ (which determines the number elements spanned by the index $k$ in $A_{\ell,k}$). Indeed, a different way to state this same fact is to notice that the $\Phi$ in (3) is a convex combination of several channels: $$\Phi(X) = \sum_k p_k \Phi_k(X), \qquad \Phi_k(X)\equiv \operatorname{Tr}_{\mathcal Z}[U(X\otimes E_{k,k})U^\dagger]. \tag 6$$


This leads me to the question: can any $\Phi$ be written as in (3) for any $\sigma$? More precisely, given $\Phi$ and $\sigma$, can I always find an isometry $U$ such that (3) holds?

The question arises from the fact that, because when $\sigma$ is not pure (3) leads to $\Phi$ be a convex combination of other maps as shown in (6), I would think that there should be maps that are "extremal", in the sense that they cannot be written as convex combinations of other maps, and that such maps shouldn't be writable as (3) for $\sigma$ not pure.

$\endgroup$
4
$\begingroup$

No, this is not always possible.

A counterexample is given by $\sigma=I/d'$ and $\Phi(X)=\mathrm{tr}(X)|0\rangle\langle0|$.

To see this, note that for $X=I/d$, \begin{align} 2(1-1/d) & = \|\,|0\rangle\langle0|-I/d\|_1 \\ &= \|\Phi(X)-I/d\|_1 \\ &\le \left\|U\left(X\otimes \frac{I}{d'}\right)U^\dagger-U\left(\frac{I}{d}\otimes\frac{I}{d'}\right)U^\dagger\right\|_1 \\ &\le \left\|X\otimes \frac{I}{d'}-\frac{I}{d}\otimes\frac{I}{d'}\right\|_1 \\ & =\left\|X-\frac{I}{d}\right\|_1 \\ &=0 \end{align} where in the 2nd step, I have used that the partial trace is contractive with respect to the trace norm (as it is a CP map), and in the fourth that $\|A\otimes I/d'\|_1 = \|A\|_1$.

This is clearly a contradiction and thus shows that such a representation for the chosen channel $\Phi$ cannot exist.


As always, let me take the opportunity to advertise my list of canonical (counter)examples for quantum channels.

| improve this answer | |
$\endgroup$
  • $\begingroup$ +1 for the (sanity-check) list! $\endgroup$ – keisuke.akira Jul 19 at 1:53
  • $\begingroup$ nice. I guess an equivalent way to state this is to observe this $\operatorname{Tr}_{\mathcal Z}[U(\frac{I}{d}\otimes \frac{I}{d'})U^\dagger]=\frac{I}{d}$, which means that any map representable as $\Phi(X)=\operatorname{Tr}_{\mathcal Z}[U(X\otimes I/d')U^\dagger]$ must preserve the identity: $\Phi(I/d)=I/d$. This is clearly not the case for all maps (e.g. for the replacement map you use). $\endgroup$ – glS Jul 19 at 14:03
  • $\begingroup$ do you know if this can be made more precise? Like if there is a relation between the rank of $\sigma$ and some property of $\Phi$ (intuitively, I would say something that quantifies the number of extremal maps required to express $\Phi$). It looks like there might be some nice relation of this sort (I might ask this as a separate question if the answer is nontrivial) $\endgroup$ – glS Jul 19 at 14:05
  • 1
    $\begingroup$ @glS My argument (or generalizations thereof) is to some extent quantitative. I suspect it can still be phrased for a general state $\sigma$ (basically you want to know what kind of state $\mathrm{tr}_B(I\otimes \sigma)$ can be). This was the intuition which led me to the argument: I thought that the mixedness of sigma should be reflected in the mixedness of the output. Here, the "mixedness" is measured by the distance to the maximally mixed states, but I'm sure other arguments are possible. E.g., the partial trace should not decrease the rank. $\endgroup$ – Norbert Schuch Jul 19 at 15:25

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.