Exact correction of a protected subsystem

Let $H_G,H_L,H_P,H_O$ be nonzero finite-dimensional complex inner-product spaces with the fixed factor order $H_G\otimes H_L$. Let $J:H_G\otimes H_L\to H_P$ be an isometric encoding and set $P=J{J}^{\ast }$. The factors $H_G$ and $H_L$ are the gauge and protected logical subsystems. Let $\mathcal{N}:\mathcal{B}(H_P)\to \mathcal{B}(H_O)$ be a channel.

For any finite-dimensional residual space $H_Q$, use the first-factor partial trace ${\mathrm{Tr}}_G$ of Partial trace over the first tensor factor on $H_G\otimes H_Q$. In particular, for a reference $H_R$, take $H_Q=H_L\otimes H_R$ and leave that factor order unchanged.

A decoded recovery is a channel $\mathcal{R}:\mathcal{B}(H_O)\to \mathcal{B}(H_G\otimes H_L)$. The subsystem $H_L$ is exactly correctable for $\mathcal{N}$ under $J$ when some decoded recovery satisfies, for every finite-dimensional reference $H_R$ and every positive semidefinite trace-one operator $\rho$ on $(H_G\otimes H_L)\otimes H_R$,

$(({\mathrm{Tr}}_G\otimes {\mathrm{id}}_{\mathcal{B}(H_R}))\circ ((\mathcal{R}\circ \mathcal{N}\circ {\mathrm{Ad}}_J)\otimes {\mathrm{id}}_{\mathcal{B}(H_R)}))(\rho )=({\mathrm{Tr}}_G\otimes {\mathrm{id}}_{\mathcal{B}(H_R)})(\rho ),$

where ${\mathrm{Ad}}_J(X)=JX{J}^{\ast }$. The recovery may change the gauge state. The one-operator Kraus map ${\mathrm{Ad}}_J$ is completely positive by A finite Kraus sum is completely positive and trace-preserving by $J^{\ast }J=I$ and Trace preservation is Kraus normalization. Thus $\mathrm{\Phi }=\mathcal{R}\circ \mathcal{N}\circ {\mathrm{Ad}}_J$ is a channel by Finite-dimensional channels compose and tensor, and it has a finite Kraus representation by Kraus and Choi characterizations of complete positivity. Explicitly, if $(R_b)$ and $(E_a)$ represent $\mathcal{R}$ and $\mathcal{N}$, then $(R_b{E}_{a}J{)}_{b,a}$ represents $\mathrm{\Phi }$. Taking a one-dimensional reference in the displayed recovery condition gives precisely the marginal hypothesis of Preserving the logical marginal forces a gauge-only channel. By that lemma and its converse with arbitrary finite references, this is equivalently a corrected decoded channel of the form ${\mathcal{T}}_G\otimes {\mathrm{id}}_{\mathcal{B}(H_L)}$ for a channel ${\mathcal{T}}_G$ on the gauge factor. For a unit vector $g_0\in H_G$, resetting the decoded gauge and re-encoding uses $X\mapsto J(P_{g_0}\otimes {\mathrm{Tr}}_G(\mathcal{R}(X)))J^{\ast }$, where $P_{g_0}(x)=⟨x,g_0⟩g_0$; the same supplier proves this is a channel. If $\dim H_G=1$, the marginal condition is ordinary exact recovery of the code subspace in the sense of Exact recovery of an encoded quantum code.