Theorem: Exact recovery and a constant complement

Let $\mathcal{N}:\mathcal{B}(H)\to \mathcal{B}(K)$ be a finite-dimensional channel. The following are equivalent:

1. a channel $\mathcal{R}$ satisfies $\mathcal{RN}={\mathrm{id}}_{\mathcal{B}(H)}$;

2. a complementary channel is constant: ${\mathcal{N}}^c(X)=\mathrm{Tr}(X)\sigma$;

3. its Kraus operators satisfy

   $${A_{\alpha }}^{†}{A}_{\beta }={\lambda }_{\alpha \beta }{I}_H$$

   for a positive matrix $\lambda$;

4. after unitary mixing of Kraus operators,

   $$\mathcal{N}(X)=\sum _j{p}_j{V}_{j}X{V}_j^{†},{V_i}^{†}{V}_j={\delta }_{ij}{I}_H,p_j>0,\sum _j{p}_j=1.$$

All recovery identities hold completely under references.

Proof. A Kraus dilation gives

so (ii) and (iii) are equivalent.

Suppose $\mathcal{R}$ has Kraus operators $B_{\nu }$. The composed Kraus operators $B_{\nu }{A}_{\alpha }$ represent the identity channel. Its Choi matrix has rank one, so every $B_{\nu }{A}_{\alpha }=c_{\nu \alpha }I$. Consequently

This proves (i)$\Rightarrow$(iii).

Diagonalize the positive matrix $\lambda$ by a unitary mixing of the Kraus operators. Nonzero transformed operators have products ${L_i}^{†}{L}_j=p_i{\delta }_{ij}I$. Set $V_i=L_i/\sqrt{p_i}$, obtaining (iv). Their range projections are orthogonal. With $P={\sum }_i{V}_i{V}_i^{†}$, the map

is completely positive and trace preserving for any fixed state ${\tau }_H$. It recovers every matrix unit and hence every operator. Tensoring the identity proves complete recovery.