Proposed augmented Jz syndrome check for an uncalibrated lift

PDF p. 16: Remark: uncalibrated lifts

Choose either of the displayed encodings: the spin-five Æ code or the spin-four Aydin–Barg code. Let $J_0$ be respectively $5$ or $4$, let $H_0$ be its spin block, and let $\mathrm{\mid }\bar{0}⟩,\mathrm{\mid }\bar{1}⟩$ be its displayed orthonormal codewords. Put $\mathcal{C}=\mathrm{span}\{\mathrm{\mid }\bar{0}⟩,\mathrm{\mid }\bar{1}⟩\}$, let $P_{\mathcal{C}}$ project orthogonally onto it, and let $J_z\mathrm{\mid }{J}_0,m⟩=m\mathrm{\mid }{J}_0,m⟩$. Set $s=10$ for the Æ code and $s=8$ for the Aydin–Barg code. Define $\mathcal{D}=J_z\mathcal{C}$ and $W:\mathcal{D}\to \mathcal{C}$ by $W(J_z\mathrm{\mid }\bar{\mu }⟩\mathrm{/}\sqrt{s})=\mathrm{\mid }\bar{\mu }⟩$ for $\mu =0,1$.

Let $H$ be the finite physical space of the original manifold check, containing $H_0$. In operator products, regard $W$ as an operator on $H$ with range in $\mathcal{C}$, extended by zero on ${\mathcal{D}}^{\perp }$. In the ideal manifold check, replace the original outcome for $H_0$ by the three orthogonal outcomes with projectors $P_{\mathcal{C}}$, $P_{\mathcal{D}}$, and $P_R=P_{H_0}-P_{\mathcal{C}}-P_{\mathcal{D}}$, where $P_{\mathcal{D}}$ projects onto $\mathcal{D}$. Retain the original complementary-manifold outcome $Q=I-P_{H_0}$. For input density operator $\rho$ after the modeled evolution, the unnormalized output for a projector $P_a$ is $P_a\rho P_a$; its probability is its trace, and its conditional state is the output divided by that probability when positive. On the $\mathcal{D}$ outcome apply $W$, giving the unnormalized output $W{P}_{\mathcal{D}}\rho P_{\mathcal{D}}{W}^{\ast }$. Leave the $\mathcal{C}$ and $R$ outcomes as projected; on $Q$, retain the original check's conditional procedure. A zero-probability outcome has no conditional state.

J equals five absorption-emission code

J equals four Aydin–Barg encoding

Finite spin blocks, spherical tensors and Clebsch–Gordan phases

Projective code-manifold syndrome check

Finite instruments and conditional states

Given: The stated definitions, hypotheses and cited interfaces.

Goal: Establish the well-definedness of the projectors, branch maps and conditional-state prescription.

1.1 By [F1]–[F3], the codewords have disjoint magnetic supports, so their off-diagonal matrix elements of $J_z$ and $J_z^2$ vanish. For the Æ code the diagonal first moments are $(5(-2)+2(5))\mathrm{/}7=0$ and $(2(-5)+5(2))\mathrm{/}7=0$; both second moments equal $10$. For the Aydin–Barg code the diagonal first moments are $(-4+2(2))\mathrm{/}3=0$ and $(2(-2)+4)\mathrm{/}3=0$; both second moments equal $8$. Therefore $P_{\mathcal{C}}{J}_z{P}_{\mathcal{C}}=0$ and $P_{\mathcal{C}}{J}_z^2{P}_{\mathcal{C}}=s{P}_{\mathcal{C}}$ with $s>0$. The first identity makes $\mathcal{C}$ orthogonal to $\mathcal{D}$; the second makes $J_z\mathrm{\mid }\bar{0}⟩\mathrm{/}\sqrt{s}$ and $J_z\mathrm{\mid }\bar{1}⟩\mathrm{/}\sqrt{s}$ orthonormal. Hence they form a basis of $\mathcal{D}$ and the stated $W$ is a well-defined isometry onto $\mathcal{C}$. [given, F1, F2, F3, F4]

2.1 Consequently $P_R$ is the orthogonal projector onto the remainder of $H_0$, and the four listed projectors are mutually orthogonal and sum to $I_H$. Each projection outcome map $X\mapsto P_{a}X{P}_a$ has the single Kraus operator $P_a$, so [F5] establishes its complete positivity. In the finite magnetic basis, direct summation gives $\mathrm{tr}(AB)={\sum }_{i,j}{A}_{ij}{B}_{ji}=\mathrm{tr}(BA)$ for operators on $H$. Applying this to each projection and using $P_a^2=P_a$ gives $\mathrm{tr}(P_{a}X{P}_a)=\mathrm{tr}(X{P}_a)$ for every $X$. Since ${\sum }_a{P}_a=I_H$, the sum of the four traces is $\mathrm{tr}X$. Thus these maps form a finite instrument in the sense of [F4]. [F3, F4, F5, step 1.1, algebra]

3.1 The corrected $\mathcal{D}$ map has the single Kraus operator $K_{\mathcal{D}}=W{P}_{\mathcal{D}}$ on $H$, so it too is completely positive by [F5]. Step 1.1 gives $K_{\mathcal{D}}^{\ast }{K}_{\mathcal{D}}=P_{\mathcal{D}}$. The finite trace identity of step 2.1 therefore gives $\mathrm{tr}(K_{\mathcal{D}}X{K}_{\mathcal{D}}^{\ast })=\mathrm{tr}(X{P}_{\mathcal{D}})=\mathrm{tr}(P_{\mathcal{D}}X{P}_{\mathcal{D}})$ for every $X$. Replacing the uncorrected $\mathcal{D}$ branch by this map preserves total trace and gives another instrument. The probabilities and positive-weight conditional states are consequently those specified in [F4]. Assumption [A1] is inherited only when invoking the original check's conditional procedure on $Q$; this finite matrix verification makes no additional use of choice. That $Q$ procedure remains separate, and no decoder for it is constructed here. [F1, F2, F3, F4, F5, A1, step 1.1, step 2.1, algebra] ∎

The separate physics-theorem Well-defined augmented syndrome and first-order rotation correction analyzes the single-check isolated-rotation property of this procedure. The article leaves the accumulated effect of higher-order terms under the complete repeated-check channel open.