Transpose (Petz) recovery on a code

PDF p. 9: Text before Eq. (S1.9); PDF p. 2: Checking-time discussion

Let $H_P,H_O$ be nonzero finite-dimensional complex inner-product spaces. Write $\mathcal{L}(H)$ for the linear operators on $H$, and let $P$ be the orthogonal projector onto a nonzero code subspace of $H_P$. Let $\mathcal{N}:\mathcal{L}(H_P)\to \mathcal{L}(H_O)$ be a quantum channel, meaning a linear, completely positive, trace-preserving map, with finite Kraus representation $\mathcal{N}(X)={\sum }_k{K}_{k}X{K}_k^{†}$. Here $†$ denotes the adjoint.

Put $A=\mathcal{N}(P)$ and let $S$ project onto its support $(\ker A{)}^{\perp }$. The symbol $A^{-1\mathrm{/}2}$ denotes the inverse of the positive square root on this support and zero on the kernel. The support-normalized transpose map is ${\mathcal{T}}_0(Y)={\sum }_k{R}_{k}Y{R}_k^{†}$, where $R_k=P{K}_k^{†}{A}^{-1\mathrm{/}2}$. Its output lies in the code and it preserves trace on operators $Y=SYS$.

To make a recovery channel on all of $\mathcal{L}(H_O)$, choose a code state $\sigma$, meaning a positive semidefinite trace-one operator with $P\sigma P=\sigma$, and define $\mathcal{T}(Y)={\mathcal{T}}_0(Y)+\mathrm{tr}((I-S)Y)\sigma$. We call any such completion a transpose recovery. The completion choice does not affect $\mathcal{T}(\mathcal{N}(X))$ for $X=PXP$.

The channel and Kraus conventions in [F2] and [F3] give $A={\sum }_k{K}_{k}P{K}_k^{†}$, a positive semidefinite operator. Thus the support and inverse square root are well-defined by [F4]. A code state exists because the code is nonzero: the projector onto any unit code vector is one. The maps in the definition are therefore well-defined. The support normalization, complete positivity, trace-preserving completion and its independence on noise outputs of code-supported inputs follow from [F5] with $E_a=K_k$. In particular, the formula applies to every finite-dimensional code and channel satisfying these hypotheses; the article describes its use more generally and cites Refs. [30,34–36].