1
$\begingroup$

Consider quantum channels $\Phi : M_n \rightarrow M_{d_1}$ and $\Psi : M_n \rightarrow M_{d_2}$ with $d_1\leq d_2$. We say that $\Phi$ is isometrically extended by $\Psi$ (denoted $\Phi \leq_{\text{iso}} \Psi$) if $\exists$ an isometry $V: \mathbb{C}^{d_1} \rightarrow \mathbb{C}^{d_2}$ such that

$$\forall \, \rho\in M_n: \qquad \Psi (\rho) = V\Phi (\rho) V^\dagger . $$

Now, define the minimal output dimension of a channel $\Psi : M_n \rightarrow M_{m}$ as follows:

$$\text{min out }(\Psi) = \text{min} \{d \in \mathbb{N} : \exists \, \text{channel } \Phi : M_n \rightarrow M_d \text{ such that } \Phi \leq_{\text{iso}}\Psi \}.$$

I claim that for any given channel $\Psi : M_n \rightarrow M_m$, $$\text{min out }(\Psi) = \text{rank}\,\Psi (\mathbb{I_n}),$$ where $\mathbb{I_n}\in M_n$ is the identity matrix. Can someone help me prove this?

$\endgroup$
6
  • $\begingroup$ Convexity? The span of the output only gets larger under mixing? $\endgroup$ – Norbert Schuch Mar 5 at 16:12
  • 1
    $\begingroup$ @NorbertSchuch I'm not really sure what you mean by this. Would you mind elaborating a bit? In any case, I've provided an answer below which seems to work. $\endgroup$ – mathwizard Mar 5 at 17:21
  • $\begingroup$ Well, pretty much what you say in your answer. I'd say the proof uses convex combinations and that they can't decrease the support. $\endgroup$ – Norbert Schuch Mar 5 at 18:51
  • $\begingroup$ I suggest that you split the equality you wish to prove into two inequalities and handle them separately. You're well on your way to one of the inequalities, and to finish it off you could use the fact that the projection onto the image of $\Psi(\mathbb{I}_n)$ can be expressed as $V V^{\dagger}$ for $V$ being an isometry from $\mathbb{C}^r$ to $\mathbb{C}^m$, where $r$ is the rank of $\Psi(\mathbb{I}_n)$. The reverse inequality should be comparatively simpler. $\endgroup$ – John Watrous Mar 5 at 19:29
  • $\begingroup$ @NorbertSchuch Great! I'm glad that we're on the same page. $\endgroup$ – mathwizard Mar 6 at 2:25
2
$\begingroup$

The proof of the claim hinges on the fact that for a given channel $\Psi: M_n\rightarrow M_m$, the following inclusion holds for all positive semi-definite $\rho\in M_n$ (see the answer provided here):

$$ \text{range} \, \Psi(\rho) \subseteq \text{range}\, \Psi (\mathbb{I_n}).$$

Once we know the above fact, it becomes clear that all the outputs of $\Psi$ can be thought to operate on the smaller subspace $\text{range}\, \Psi(\mathbb{I}_n) \subseteq \mathbb{C}^m$, which can be isometrically embedded into $\mathbb{C}^m$. It is then straightforward to construct a channel $\Phi : M_n \rightarrow M_r \simeq \mathcal{B}(\text{range}\, \Psi(\mathbb{I_n}))$ such that $\Phi\leq_{\text{iso}} \Psi$, where $r=\text{rank} \, \Psi(\mathbb{I_n})$. Hence, we have established the following inequality: $$ \text{min out } (\Psi) \leq \text{rank} \, \Psi(\mathbb{I_n}).$$

For the reverse inequality, observe that since ranks of matrices stay invariant under isometric conjugations, any channel $\Phi:M_n\rightarrow M_d$ with $\Phi\leq_{\text{iso}}\Psi$ must be defined with an output dimension $d\geq \text{rank}\, \Psi(\mathbb{I_n})$. In other words, we obtain:

$$ \text{min out }(\Psi) \geq \text{rank }\Psi(\mathbb{I_n}).$$

$\endgroup$

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.