Encoded error instruments and full-space CPTP error maps

If $K_j\in {\mathcal{E}}_s$ are Kraus operators of a full-space channel, their restrictions $A_j=K_{j}V$ satisfy ${\sum }_j{A}_j^{\ast }{A}_j=V^{\ast }({\sum }_j{K}_j^{\ast }{K}_j)V=I_{H_L}$. Partitioning a finite Kraus family into recorded outcomes gives an encoded instrument; each branch is completely positive and the sum is trace preserving on the logical input. No assertion that each branch has trace one is needed.

In the code [F2], no nonidentity scalar-phase stabilizer has weight one or two. Indeed, an inner-block stabilizer has binary $X,Z$ labels in the row code $D$; every nonzero word of $D$ has weight four, so a nonidentity inner stabilizer has support at least four. If a full stabilizer has a nontrivial outer stabilizer label modulo the inner groups, that outer label has at least four occupied inner blocks, and each nontrivial inner logical restriction has weight at least three, giving weight at least twelve. Thus distinct phase-free Paulis $E_a,E_b$ of weight at most one have $V^{\ast }{E}_a^{\ast }{E}_{b}V=0$: their product has weight at most two, is not a scalar-phase stabilizer, and is detected by some check. Also $V^{\ast }{E}_a^{\ast }{E}_{a}V=I$. Hence the maps $E_{a}V$ have pairwise orthogonal ranges and the restriction $E\mapsto EV$ is injective on ${\mathcal{E}}_1$.

List the phase-free weight-at-most-$s$ Pauli operators as $F_1,\dots ,F_M$. Their normalized mixed channel has Kraus operators $F_a\mathrm{/}\sqrt{M}$. Let $R_l$ be Kraus operators of a fixed decoded recovery correcting it. Then $R_l{F}_{a}V\mathrm{/}\sqrt{M}$ form a Kraus representation of the identity logical channel. Every Kraus operator of the identity channel is a scalar multiple of $I_{H_L}$: its output on every pure input must lie in that input's one-dimensional span, and linearity on basis vectors and their sums makes the scalar independent of the input. Thus $R_l{F}_{a}V=c_{la}{I}_{H_L}$ for scalars $c_{la}$.

Put $B=(I+X_1)\mathrm{/}\sqrt{2}$. Step 1.2 gives $V^{\ast }{X}_{1}V=0$, so $(BV{)}^{\ast }(BV)=V^{\ast }(I+X_1)V=I$. Therefore $A=BV$ is an isometry and its encoded channel is trace preserving. Suppose a full-space channel with Kraus operators $K_j\in {\mathcal{E}}_1$ agreed with this channel on the code. For any logical vector $\psi$, positivity of the rank-one output sum forces every $K_{j}V\psi$ to be proportional to $BV\psi$. Since $BV$ is an isometry, linearity on a basis and its pairwise sums gives scalars $c_j$, independent of $\psi$, with $K_{j}V=c_{j}BV$. The injectivity from step 1.2 then gives $K_j=c_{j}B$.

For any encoded instrument write $A_{bj}={\sum }_a{\alpha }_{bja}{F}_{a}V$. Step 1.3 gives $R_l{A}_{bj}={\gamma }_{lbj}{I}_{H_L}$, where ${\gamma }_{lbj}={\sum }_a{\alpha }_{bja}{c}_{la}$. Complete trace preservation on the encoded input gives ${\sum }_{l,b,j}\mathrm{\mid }{\gamma }_{lbj}{\mathrm{\mid }}^2=1$. Hence recovery of the full instrument is the identity logical channel; each recorded branch is a nonnegative scalar multiple of that channel, without requiring a normalized physical error map. Tensoring every displayed equality with an external reference identity proves its reference-preserving version. For several blocks use tensor Pauli labels and the product recovery; tensoring the scalar identities from step 1.3 yields the same proof for their joint span.

Trace preservation of the supposed extension would require $I={\sum }_j{K}_j^{\ast }{K}_j=({\sum }_j\mathrm{\mid }{c}_j{\mathrm{\mid }}^2)(I+X_1)$. This is impossible: $I+X_1$ annihilates the nonzero $-1$ eigenspace of $X_1$. Thus no such extension exists. The encoded map can also occur as a normalized measured branch: prepare a control in $\mathrm{\mid }+⟩$, apply controlled $X_1$, and measure that control in the $X$ basis. The plus branch has physical Kraus operator $(I+X_1)\mathrm{/}2$; on code inputs its probability is $1\mathrm{/}2$ by step 1.2, so its normalized output is exactly ${\mathrm{Ad}}_{BV}$. The full two-branch instrument is trace preserving.

If coefficients in these expansions are operators on fault memory, the corrected Kraus operators instead have the form $I_{H_L}\otimes M_{lbj}$. Trace preservation makes their full sum a channel acting on that memory alone. It leaves the logical/reference marginal unchanged even for correlations with memory, since a local trace-preserving operation does not change a remote marginal. Its individual outcomes may have subnormalized probabilities and may condition correlations; its memory output need not equal its input. This proves the stated interpretation with memory and completes all assertions.