Projective code-manifold syndrome check

PDF p. 15: Paragraph before Eq. (S4.5) and Eqs. (S4.5), (S4.7)–(S4.9); PDF p. 16: First paragraph

Adopt the finite lifted model and all hypotheses of its first-order fast-check channel, including the Axiom of Choice, Countable Choice and Dependent Choice. Its physical space is the orthogonal sum of the blocks spanned by the basis vectors $\mathrm{\mid }J,m⟩$ with $J=0,\dots ,N$ and $m=-J,\dots ,J$. A density operator is a positive semidefinite operator of trace one. Let $J_0\in \{1,\dots ,N\}$, let $P={\mathrm{\Pi }}_{J_0}$ be the orthogonal projector onto the $J_0$ block, and let the input density operator satisfy $P{\rho }_{0}P={\rho }_0$. Choose a check interval $\tau \ge 0$. The exact model channel for that interval is ${\mathcal{N}}_{\tau }$.

Measure the two projectors $P$ and $Q=I-P$ on the output of ${\mathcal{N}}_{\tau }$. The ideal measurement instrument has unnormalized outputs $B_{\mathrm{n}\mathrm{u}\mathrm{l}\mathrm{l}}=P{\mathcal{N}}_{\tau }({\rho }_0)P$ and $B_{\mathrm{d}\mathrm{e}\mathrm{c}\mathrm{a}\mathrm{y}}=Q{\mathcal{N}}_{\tau }({\rho }_0)Q$. For either outcome $a$, its probability is $p_a=\mathrm{Tr}{B}_a$; if $p_a>0$, retain the conditional state $B_a\mathrm{/}{p}_a$. A zero-probability outcome has no conditional state.

For the first-order branch description, put $K_{\mathrm{n}\mathrm{u}\mathrm{l}\mathrm{l}}=I+\tau G$ and $K_{\delta }=\sqrt{\tau }{L}_{\delta }$ for $\delta \in \{-1,0,1\}$, using the model's diagonal $G$ and downward jumps $L_{\delta }$. Replace ${\mathcal{N}}_{\tau }$ by the displayed first-order Kraus sum. The resulting unnormalized null branch is $K_{\mathrm{n}\mathrm{u}\mathrm{l}\mathrm{l}}{\rho }_0{K}_{\mathrm{n}\mathrm{u}\mathrm{l}\mathrm{l}}^{†}$ and the resulting unnormalized decay branch is ${\sum }_{\delta }{K}_{\delta }{\rho }_0{K}_{\delta }^{†}$; normalize a branch by its trace only when that trace is positive. Here $\gamma >0$ is the inherited equal-rate scale, and for the normalized excited-block input the first-order decay-branch weight is $\mathrm{Tr}({\sum }_{\delta }{K}_{\delta }{\rho }_0{K}_{\delta }^{†})=\gamma \tau$. Thus for $\tau >0$ its conditional state in this approximation is ${\sum }_{\delta }{K}_{\delta }{\rho }_0{K}_{\delta }^{†}\mathrm{/}(\gamma \tau )$; at $\tau =0$ this branch has zero weight and no conditional state. These are branch expressions for the approximation, which need not itself be trace preserving at finite $\tau$.

If an encoded input and a suitable decoder are supplied, apply that decoder after the decay outcome. Here a decoder is a quantum channel from the post-measurement physical space to the logical space; suitable means that it recovers the encoded information for the specified transition-error channel under the stated code assumptions. The check itself does not construct such a decoder or establish that an arbitrary code has one. For an initial state supported in the $J_0=0$ block, the model has no downward jump, and its density operator remains unchanged under the exact evolution.

First-order fast-check channel

Normalized jumps give a scalar total decay operator

Code restriction and decoded channel types

Calibrated recovery cycles and accumulated-error contract

Finite instruments and conditional states

Orthogonal projection is linear, and an orthonormal basis (eᵢ) of W gives P_Wv=∑ᵢ⟨ v,eᵢ⟩ eᵢ

Finite projective measurements

Given: The stated definitions, hypotheses and cited interfaces.

Goal: Justify the stated construction and its scope.

1.1 In the orthonormal block basis, [F5] makes $P$ diagonal with entry one on the $J_0$ block and zero elsewhere. Thus $P^{\ast }=P$, $P^2=P$, and $Q=I-P$ is the complementary orthogonal projector, with $Q^{\ast }=Q$, $Q^2=Q$, $PQ=QP=0$ and $P+Q=I$. The pair is a finite projective measurement by [F6]. Each map $X\mapsto PXP$ or $X\mapsto QXQ$ is completely positive because conjugation by the corresponding projector remains positive after every finite identity amplification. By [F7] and idempotence, $\mathrm{Tr}(PXP)+\mathrm{Tr}(QXQ)=\mathrm{Tr}(XP)+\mathrm{Tr}(XQ)=\mathrm{Tr}X$ for every endomorphism $X$. Thus these maps form an instrument in the sense of [F2], and the stated probabilities and conditional states follow. The exact channel precedes this instrument, so its two probabilities sum to one. [given, F1, F2, F3, F4, F5, F6, F7]

2.1 By [F1], the diagonal $K_{\mathrm{n}\mathrm{u}\mathrm{l}\mathrm{l}}$ preserves the $J_0$ block and the jumps take it into the orthogonal $J_0-1$ block. Consequently $P{K}_{\mathrm{n}\mathrm{u}\mathrm{l}\mathrm{l}}{\rho }_0{K}_{\mathrm{n}\mathrm{u}\mathrm{l}\mathrm{l}}^{†}P=K_{\mathrm{n}\mathrm{u}\mathrm{l}\mathrm{l}}{\rho }_0{K}_{\mathrm{n}\mathrm{u}\mathrm{l}\mathrm{l}}^{†}$, while projecting any jump term with $P$ gives zero. Conversely, projection with $Q$ removes the null term and leaves the jump terms. The approximation bound in [F1] concerns unnormalized channel outputs; it does not by itself give a uniform bound on normalized conditional states when an outcome probability is small. [F1, F2, F3, F4, step 1.1]

3.1 Since $J_0\ge 1$, the supported input satisfies ${\mathrm{\Pi }}_{\mathrm{e}\mathrm{x}\mathrm{c}}{\rho }_0={\rho }_0$ and $\mathrm{Tr}{\rho }_0=1$. By the finite trace cyclicity in [F7], $\mathrm{Tr}({\sum }_{\delta }{K}_{\delta }{\rho }_0{K}_{\delta }^{†})=\tau {\sum }_{\delta }\mathrm{Tr}({\rho }_0{L}_{\delta }^{†}{L}_{\delta })=\gamma \tau \mathrm{Tr}({\rho }_0{\mathrm{\Pi }}_{\mathrm{e}\mathrm{x}\mathrm{c}})=\gamma \tau$ using [F8]. This is the weight of the approximate decay branch selected in step 2.1. Since $\gamma >0$, it is positive precisely when $\tau >0$, giving the stated division by $\gamma \tau$; at $\tau =0$ every $K_{\delta }$ is zero. The exact branch probability remains defined by the exact instrument, and is not identified with this first-order weight. [given, F1, F2, F7, F8, step 2.1, algebra]

4.1 The decoder is an additional input of the type in [F3], as in the outcome-dependent cycle of [F4]. Its suitability is a separate hypothesis, so the procedure makes no unconditional correction claim. On the one-dimensional ground block the jumps vanish by [F1], and the diagonal Hamiltonian acts as a scalar; its unitary phase cancels in the density operator. Hence the exact ground-block state is unchanged. [F1, F2, F3, F4, step 1.1] ∎