Theorem: Exact recovery and constant complement

For a finite-dimensional isometry $W:A\to Z\otimes F$, the following are equivalent:

1. there is a CPTP map $\mathcal{D}:Z\to A$ with $\mathcal{DN}={\mathrm{id}}_A$;

2. there is a density operator ${\tau }_F$ such that

   $$\mathcal{C}(X)=\mathrm{Tr}(X){\tau }_F\text{for every }X\in \mathcal{B}({\mathcal{H}}_A);$$

3. for a Kraus family $\{K_{\alpha }\}$ of $\mathcal{N}$,

   $${K_{\alpha }}^{†}{K}_{\beta }=c_{\alpha \beta }{I}_A$$

   with scalar coefficients $c_{\alpha \beta }$.

If these conditions hold and $\mathrm{rank}{\tau }_F=r$, then

Proof. Dilate an exact decoder by $T:Z\to A\otimes G$. The composite $(T\otimes I_F)W$ dilates the identity channel. The Choi matrix of the identity has rank one, so every Kraus operator of that composite is proportional to $I_A$. Consequently

for one fixed normalized vector $\chi$. Tracing $AG$ proves (eq:constantcomplement).

Conversely, diagonalize ${\tau }_F={{\sum }_{j=1}}^r{\lambda }_j|j⟩⟨j|$, with ${\lambda }_j>0$, and write

The constant-complement identities on the matrix units $|i⟩⟨i^{\prime }|$ give

Thus there are $dr$ orthonormal vectors in $Z$, proving (eq:rankcapacity). On their span $L$, define $U|{\varphi }_{ij}⟩=|i{⟩}_A|j{⟩}_J$. For a fixed state ${\sigma }_A$, the map

is CPTP and satisfies $\mathcal{DN}(X)=X$.

Finally, in the environment basis associated with the Kraus family,

Constancy on every operator is equivalent to (eq:KL).