Well-defined augmented syndrome and first-order rotation correction

The two codewords have disjoint magnetic supports, so their cross matrix elements of $J_z$ and $J_z^2$ vanish. For the Æ code the two first moments are $(5(-2)+2(5))\mathrm{/}7=0$ and $(2(-5)+5(2))\mathrm{/}7=0$, and both second moments are $10$. For the signed spin-four code the first moments are $(-4+2(2))\mathrm{/}3=0$ and $(2(-2)+4)\mathrm{/}3=0$, and both second moments are $8$. The sign in the latter codeword does not change its diagonal moments. Hence $P{J}_{z}P=0$ and $P{J}_z^{2}P=sP$ with the stated positive $s$.

The ladder formulas give $[J_z,J_{\pm }]=\pm J_{\pm }$ and $[J_{+},J_{-}]=2J_z$ on each magnetic basis vector. Consequently $T_0=J_z$ and $T_{\pm 1}=\mp J_{\pm }\mathrm{/}\sqrt{2}$ satisfy the rank-one tensor commutator equations. With the input block among the selected blocks, their extensions and the ambient identity belong to the order-one span. Applying [F4] for the Æ code and [F5] for the spin-four code, and restricting the resulting scalar pair products to the input block, gives the condition for $I,J_z,J_{+},J_{-}$.

For $u,v\in \mathcal{C}$, $⟨u,J_{z}v⟩=0$, so $\mathcal{C}\perp \mathcal{D}$. Also $⟨J_{z}u,J_{z}v⟩=s⟨u,v⟩$. Thus the two vectors $J_z\mathrm{\mid }\bar{\mu }⟩\mathrm{/}\sqrt{s}$ form an orthonormal basis of $\mathcal{D}$, and the prescribed map $W$ sends this basis to the orthonormal code basis. It is consequently a well-defined onto isometry and satisfies $W{J}_{z}v=\sqrt{s}v$ for every $v\in \mathcal{C}$.

Write $J_0=5$ or $4$ for the input spin. In its orthonormal magnetic basis, $\mathrm{\parallel }{J}_{z}v{\mathrm{\parallel }}^2={\sum }_m{m}^2\mathrm{\mid }{v}_m{\mathrm{\mid }}^2\le J_0^2\mathrm{\parallel }v{\mathrm{\parallel }}^2$, so $\mathrm{\parallel }{J}_z\mathrm{\parallel }\le J_0$. By [F9], $U_{\theta }=I-i\theta J_z+R_{\theta }$, and the finite constant $C_0:={\sum }_{k\ge 2}{J}_0^k\mathrm{/}k!$ bounds the tail by $\mathrm{\parallel }{R}_{\theta }\mathrm{\parallel }\le C_0{\theta }^2$ for $\mathrm{\mid }\theta \mathrm{\mid }\le 1$: bound each $k$th term by $\mathrm{\mid }\theta {\mathrm{\mid }}^k{J}_0^k\mathrm{/}k!\le {\theta }^2{J}_0^k\mathrm{/}k!$ and sum the convergent series. For every unit $\psi \in \mathcal{C}$, step 2.1 gives $a:=P{U}_{\theta }\psi =\psi +r$ and $b:=W{P}_{\mathcal{D}}{U}_{\theta }\psi =-i\theta \sqrt{s}\psi +t$, where $r=P{R}_{\theta }\psi$ and $t=W{P}_{\mathcal{D}}{R}_{\theta }\psi$. An orthogonal projection is contractive by the orthogonal decomposition $v=P_{a}v+(I-P_a)v$ and the identity $\mathrm{\parallel }v{\mathrm{\parallel }}^2=\mathrm{\parallel }{P}_{a}v{\mathrm{\parallel }}^2+\mathrm{\parallel }(I-P_a)v{\mathrm{\parallel }}^2$; $W$ preserves the norm on $\mathcal{D}$. Hence $\mathrm{\parallel }r\mathrm{\parallel },\mathrm{\parallel }t\mathrm{\parallel }\le C_0{\theta }^2$, uniformly in unit code vectors.

For the pure input $\mathrm{\mid }\psi ⟩⟨\psi \mathrm{\mid }$, [F7] identifies the sum of corrected outputs as $\mathrm{\mid }a⟩⟨a\mathrm{\mid }+\mathrm{\mid }b⟩⟨b\mathrm{\mid }$. Expanding $\mathrm{\mid }a⟩⟨a\mathrm{\mid }-\mathrm{\mid }\psi ⟩⟨\psi \mathrm{\mid }$ gives $\mathrm{\mid }\psi ⟩⟨r\mathrm{\mid }+\mathrm{\mid }r⟩⟨\psi \mathrm{\mid }+\mathrm{\mid }r⟩⟨r\mathrm{\mid }$. Thus [F11] bounds its trace norm by $2\mathrm{\parallel }r\mathrm{\parallel }+\mathrm{\parallel }r{\mathrm{\parallel }}^2$, and bounds $\mathrm{\parallel }\mathrm{\mid }b⟩⟨b\mathrm{\mid }{\mathrm{\parallel }}_1$ by $(\sqrt{s}\mathrm{\mid }\theta \mathrm{\mid }+C_0{\theta }^2{)}^2$. For $\mathrm{\mid }\theta \mathrm{\mid }\le 1$, the sum of corrected outputs differs from the pure input in trace norm by at most $K{\theta }^2$, where $K=2C_0+C_0^2+(\sqrt{s}+C_0{)}^2$ is independent of $\psi$.

For a code density operator, [F8] and [F10] give $\rho ={\sum }_j{\lambda }_j\mathrm{\mid }{e}_j⟩⟨e_j\mathrm{\mid }$. Every retained eigenvector lies in $\mathcal{C}$, since ${\lambda }_j{e}_j=\rho e_j=P\rho P{e}_j$ and ${\lambda }_j>0$. Denote the sum of the two corrected branch maps by ${\mathrm{\Phi }}_{\theta }(X)=P{U}_{\theta }X{U}_{\theta }^{\ast }P+W{P}_{\mathcal{D}}{U}_{\theta }X{U}_{\theta }^{\ast }{P}_{\mathcal{D}}{W}^{\ast }$. This map is linear, so [F11] and step 4.1 yield $\mathrm{\parallel }{\mathrm{\Phi }}_{\theta }(\rho )-\rho {\mathrm{\parallel }}_1\le {\sum }_j{\lambda }_{j}K{\theta }^2=K{\theta }^2$. The check needs no value of $\theta$; the bound is for one check as real $\theta \to 0$ and says nothing about accumulated repeated-check error. Assumption [A1] is retained from the algorithm's original complementary-manifold procedure; the present finite calculation uses no additional choice.