Stabilizer distance detection and correction

Let $E$ be a Pauli of weight less than $d(S)$. The code space is the common fixed space by [F3], and [F4] gives $P=Pg=gP$ for each $g\in S$. If $E$ anticommutes with some $g\in S$, then $PEP=PgEP=-PEgP=-PEP$ and hence $PEP=0$. If instead $E$ commutes with every $g$, then $E\in N(S)$. For $k>0$, its weight is below the minimum in [F1], forcing $E\in Z_{n}S$; for $k=0$, the same conclusion follows from $N(S)=Z_{n}S$. Writing $E=zs$ gives $PEP=zP$. Thus every such Pauli has scalar code-space compression, and every nonzero one not in the scalar-phase stabilizer is detected.

Enumerate the $m\ge 1$ tensor Paulis of weight at most $t$ with factors in $\{I,X,Y,Z\}$ as $E_1,\dots ,E_m$; scalar phases do not affect correction. For any pair, the support of $E_a^{\ast }{E}_b$ lies in the union of their supports, so its weight is at most $2t<d(S)$. Step 1.1 gives $P{E}_a^{\ast }{E}_{b}P=C_{ab}P$ for a scalar $C_{ab}$. Let $V$ be the inclusion of the nonzero code space, so $P=V{V}^{\ast }$ by [F4]. Each $E_a$ is unitary by the matrices in [F6] and their tensor action [F7], so $F_a=E_a\mathrm{/}\sqrt{m}$ are Kraus operators of the channel $\mathcal{N}(A)=m^{-1}{\sum }_a{E}_{a}A{E}_a^{\ast }$: each summand is a unitary conjugation and ${\sum }_a{F}_a^{\ast }{F}_a=I$. Applying [F5] to $P{F}_a^{\ast }{F}_{b}P=(C_{ab}\mathrm{/}m)P$ gives a recovery $\mathcal{R}$. Its off-diagonal formula gives $\mathcal{R}(E_{a}VX{V}^{\ast }{E}_a^{\ast })=C_{aa}X=X$, since $E_a^{\ast }{E}_a=I$. Thus this single recovery corrects every Pauli in the family.

On one qubit every matrix $A=\left(\begin{array}{cc}a& b\\ c& d\end{array}\right)$ has the expansion $A=\frac{a+d}{2}I+\frac{b+c}{2}X+\frac{c-b}{2i}Y+\frac{a-d}{2}Z$, as direct substitution of [F6] verifies. In the ordered one-qubit basis, the matrix units satisfy $\mathrm{\mid }0⟩⟨0\mathrm{\mid }=(I+Z)\mathrm{/}2$, $\mathrm{\mid }1⟩⟨1\mathrm{\mid }=(I-Z)\mathrm{/}2$, $\mathrm{\mid }0⟩⟨1\mathrm{\mid }=(X+iY)\mathrm{/}2$, and $\mathrm{\mid }1⟩⟨0\mathrm{\mid }=(X-iY)\mathrm{/}2$. For bit strings $a,b\in \{0,1{\}}^n$, [F7] identifies the global matrix unit $\mathrm{\mid }a⟩⟨b\mathrm{\mid }$ with the tensor product of the corresponding one-qubit matrix units, so each global matrix unit is a linear combination of tensor Pauli operators. Every endomorphism has its matrix-unit expansion $A={\sum }_{a,b}⟨A{e}_b,e_a⟩\mathrm{\mid }a⟩⟨b\mathrm{\mid }$ in this ordered computational basis; hence tensor Pauli operators span the full operator space. An operator supported on a fixed set of at most $t$ qubits is a linear combination of the tensor Paulis that are identity off that set, so each term has weight at most $t$. For Kraus operators $K_{\mu }={\sum }_a{u}_{\mu a}{E}_a$ in the stated span, $P{K}_{\mu }^{\ast }{K}_{\nu }P=\left({\sum }_{a,b}\overline{u_{\mu a}}{u}_{\nu b}{C}_{ab}\right)P$ by step 2.1. The code is nonzero by [F4], so [F5] applies to this actual finite channel and gives exact recovery.