Exact recovery of an encoded quantum code

Let $H_L,H_P,H_O$ be nonzero finite-dimensional complex inner-product spaces. An encoding isometry $V:H_L\to H_P$ has code projector $P=V{V}^{\ast }$; An isometric quantum encoding gives an orthogonal code projector gives $V^{\ast }V=I_{H_L}$, so ${\mathrm{Ad}}_V(X)=VX{V}^{\ast }$ is trace-preserving. Let $\mathcal{N}:\mathcal{B}(H_P)\to \mathcal{B}(H_O)$ be a channel. The code is exactly correctable for $\mathcal{N}$ if there is a decoded recovery channel $\mathcal{R}:\mathcal{B}(H_O)\to \mathcal{B}(H_L)$ for which

$\mathcal{R}\circ \mathcal{N}\circ {\mathrm{Ad}}_V={\mathrm{id}}_{\mathcal{B}(H_L)}$

as linear maps, where ${\mathrm{Ad}}_V(X)=VX{V}^{\ast }$. Equivalently, the physical recovery ${\mathrm{Ad}}_V\circ \mathcal{R}$ satisfies $({\mathrm{Ad}}_V\circ \mathcal{R}\circ \mathcal{N}\circ {\mathrm{Ad}}_V)(\rho )=V\rho V^{\ast }$ for every positive semidefinite trace-one logical operator $\rho$.

It suffices to check equality on the rank-one projectors $P_{\psi }$ from Agreement on pure states implies complete agreement for unit vectors $\psi$: by that lemma, agreement on those projectors implies agreement on every operator and after tensoring with the identity on any finite reference system.

For any chosen finite Kraus representation $\mathcal{N}(X)={\sum }_a{E}_{a}X{E}_a^{\ast }$, trace preservation gives ${\sum }_a{E}_a^{\ast }{E}_a=I_{H_P}$ by Trace preservation is Kraus normalization.