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?

  • $\begingroup$ Convexity? The span of the output only gets larger under mixing? $\endgroup$ Commented Mar 5, 2021 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
    Commented Mar 5, 2021 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. $\endgroup$ Commented Mar 5, 2021 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. $\endgroup$ Commented Mar 5, 2021 at 19:29
  • $\begingroup$ @NorbertSchuch Great! I'm glad that we're on the same page. $\endgroup$
    – mathwizard
    Commented Mar 6, 2021 at 2:25

1 Answer 1


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:

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


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