Code restriction and decoded channel types

Let $H_L,H_P,H_O$ be nonzero finite-dimensional complex inner-product spaces, and write $\mathcal{L}(H)$ for the complex-linear endomorphisms of $H$. Let $V:H_L\to H_P$ be a linear isometry, $P=V{V}^{\ast }$ its code projector, $\mathcal{N}:\mathcal{L}(H_P)\to \mathcal{L}(H_O)$ a quantum channel, and $\mathcal{R}:\mathcal{L}(H_O)\to \mathcal{L}(H_L)$ a quantum channel, called a decoded recovery or decoder. A quantum channel is a complex-linear, completely positive, trace-preserving map; completely positive means that its amplification by the identity on every finite-dimensional ancilla preserves positive semidefinite operators.

Throughout this definition, a code means the isometric encoding $V$ on the arbitrary physical space $H_P$, with code subspace $\mathrm{im}V$; no tensor decomposition is required. This generalizes the ordered-register code convention, with which it agrees whenever $H_P$ is supplied as a fixed ordered register tensor product with at least one factor and specified ordered orthonormal factor bases.

Define the encoding channel by ${\mathrm{Ad}}_V(X)=VX{V}^{\ast }$ and the restricted noise channel by $\mathcal{N}\circ {\mathrm{Ad}}_V$. The decoded logical channel is $\mathcal{E}=\mathcal{R}\circ \mathcal{N}\circ {\mathrm{Ad}}_V$. The code is exactly correctable for $\mathcal{N}$ when a decoder can be chosen with $\mathcal{E}={\mathrm{id}}_{\mathcal{L}(H_L)}$.

A physical recovery supported on the code is a channel $\mathcal{T}:\mathcal{L}(H_O)\to \mathcal{L}(H_P)$ such that $P\mathcal{T}(Y)P=\mathcal{T}(Y)$ for every $Y\in \mathcal{L}(H_O)$. It corresponds to the decoder $Y\mapsto V^{\ast }\mathcal{T}(Y)V$; conversely a decoder $\mathcal{R}$ corresponds to the physical recovery ${\mathrm{Ad}}_V\circ \mathcal{R}$. If $H_P$ is supplied as the tensor-product space of an ordered physical register, the one-factor case has a single physical factor; this terminology imposes no additional channel condition.

By [F5], $V^{\ast }V=I$. For every finite ancilla, the amplification of $X\mapsto VX{V}^{\ast }$ is conjugation by $V\otimes I$, which preserves self-adjointness and sends positive semidefinite operators to positive semidefinite operators by [F7]: its quadratic form at $z$ is the original quadratic form at $(V^{\ast }\otimes I)z$. Its trace equals the input trace by [F6] and $V^{\ast }V=I$. Thus it is a channel by [F2] and [F4]. Compositions of channels are linear and trace-preserving; their finite amplifications are compositions of positive maps, so they are completely positive. This justifies both the restricted noise channel and the logical channel.

For a code-supported physical recovery $\mathcal{T}$, the map $Y\mapsto V^{\ast }\mathcal{T}(Y)V$ is completely positive by the same conjugation argument and composition. Its trace is $\mathrm{tr}(\mathcal{T}(Y)P)$ by [F6], equal to $\mathrm{tr}(\mathcal{T}(Y))=\mathrm{tr}(Y)$ because $P\mathcal{T}(Y)P=\mathcal{T}(Y)$ and $P^2=P$. It is therefore a decoder. Conversely, ${\mathrm{Ad}}_V\circ \mathcal{R}$ is a channel, and $PV=V$ and $V^{\ast }P=V^{\ast }$ show that its outputs are supported on the code. The two assignments are inverse: $V^{\ast }V=I$ recovers $\mathcal{R}$, while $V{V}^{\ast }\mathcal{T}(Y)V{V}^{\ast }=P\mathcal{T}(Y)P=\mathcal{T}(Y)$ recovers $\mathcal{T}$.