Physics theoremStatement: AI-adaptedProof: AI-generated

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Null-syndrome state and first-order expansion

PDF p. 15: Eq. (S4.5) and two following display lines

Statement

Adopt the finite lifted model and all assumptions of its first-order fast-check channel, including the Axiom of Choice, Countable Choice and Dependent Choice. The physical space is the orthogonal sum of blocks labelled $J = 0, \dots, N$ . Let $J_{0} \in {1, \dots, N}$ , let $P = Π_{J_{0}}$ be the orthogonal projection onto that block, and let $ρ_{0}$ be a density operator supported there: $P ρ_{0} P = ρ_{0}$ . A density operator is a positive semidefinite operator of trace one. Let $τ \geq 0$ , $K_{n u l l} = I + τ (- i H + V)$ , $V = - \frac{1}{2} \sum_{δ} L_{δ}^{†} L_{δ}$ , and $K_{δ} = \sqrt{τ} L_{δ}$ for $δ = - 1, 0, 1$ , with the model's Hermitian diagonal Hamiltonian $H$ and downward jumps $L_{δ}$ .

For the first-order Kraus approximation, projection onto the null outcome removes every jump term, and its unnormalized null output is $B = P K_{n u l l} ρ_{0} K_{n u l l}^{†} P = K_{n u l l} ρ_{0} K_{n u l l}^{†}$ . Whenever $Tr B > 0$ , its conditional state is $B / Tr B$ ; at zero trace no conditional state is assigned. With $[ρ_{0}, H] = ρ_{0} H - H ρ_{0}$ , the numerator has the first-order expansion $B = ρ_{0} + τ (- i H + V) ρ_{0} + τ ρ_{0} (i H + V) + O (τ^{2}) = ρ_{0} + i [ρ_{0}, H] τ + V ρ_{0} τ + ρ_{0} V τ + O (τ^{2})$ as $τ ↓ 0$ at fixed model parameters, in matrix spectral norm.

For the exact channel $N_{τ}$ , the unnormalized null output is $P N_{τ} (ρ_{0}) P = B + O (τ^{2})$ in the same norm at fixed model parameters. This unnormalized estimate alone is not a uniform normalized-state estimate when the outcome weight is small.

Facts & Assumptions

[F1]

The syndrome-check procedure measures $P$ and $I - P$ after the channel. Its unnormalized null output is $P N_{τ} (ρ_{0}) P$ , and an outcome is normalized by its trace only when that trace is positive. Projective code-manifold syndrome check

[F2]

In the finite lifted model, $H$ and $V$ are diagonal in the block basis, each $L_{δ}$ takes the $J_{0}$ block to the orthogonal $J_{0} - 1$ block, and the exact channel differs from the first-order Kraus sum by $O (τ^{2})$ in induced spectral-norm map norm at fixed model parameters. The supplier carries the physical, spectral and choice assumptions stated above. First-order fast-check channel

Proof

Given: The finite model, initial density operator and check interval in the statement, with the hypotheses of [F1] and [F2].

Goal: Derive the approximate null branch and its first-order expansion, and compare its unnormalized output with the exact-channel null output.

1.1givenF1F2algebra

By block diagonality in [F2], $K_{n u l l} P = P K_{n u l l} P$ . Since $P ρ_{0} P = ρ_{0}$ , it follows that $P K_{n u l l} ρ_{0} K_{n u l l}^{†} P = K_{n u l l} ρ_{0} K_{n u l l}^{†}$ . Each downward jump takes the input block into an orthogonal block, so $P K_{δ} P = 0$ and $P K_{δ} ρ_{0} K_{δ}^{†} P = 0$ . Thus the projected approximate Kraus sum is $B$ . The normalization and zero-weight convention follow from [F1].

1.2givenF1F2algebra

Let $I$ be the finite orthonormal-basis index set of the model and $I_{0}$ its $J_{0}$ subset. Write $∥ y ∥_{2}^{2} = \sum_{k \in I} ∣ y_{k} ∣^{2}$ and $∥ X ∥ = \sup_{y \neq 0} ∥ X y ∥_{2} / ∥ y ∥_{2}$ for the matrix spectral norm used in [F2]. Since $P$ selects the coordinates in $I_{0}$ , $∥ P y ∥_{2}^{2} = \sum_{k \in I_{0}} ∣ y_{k} ∣^{2} \leq ∥ y ∥_{2}^{2}$ for every vector $y$ . The definition of $∥ X ∥$ therefore gives $∥ P X P u ∥_{2} \leq ∥ X P u ∥_{2} \leq ∥ X ∥ ∥ P u ∥_{2} \leq ∥ X ∥ ∥ u ∥_{2}$ for every vector $u$ (including $P u = 0$ ). Taking the supremum over nonzero $u$ proves $∥ P X P ∥ \leq ∥ X ∥$ for every matrix $X$ .

2.1givenF2step 1.1algebra

Set $G = - i H + V$ . Because $H$ and $V$ are Hermitian, $G^{†} = i H + V$ . Multiplying the finite matrices gives $B = ρ_{0} + τ G ρ_{0} + τ ρ_{0} G^{†} + τ^{2} G ρ_{0} G^{†}$ . At fixed model parameters the last term is $O (τ^{2})$ in spectral norm. Expanding the linear terms gives the displayed commutator expression.

3.1F1F2step 1.1step 1.2algebra∎

Let ${\tilde{N}}_{τ}$ denote the first-order Kraus sum. The induced-norm estimate in [F2], applied to the fixed finite matrix $ρ_{0}$ , bounds $∥ N_{τ} (ρ_{0}) - {\tilde{N}}_{τ} (ρ_{0}) ∥$ by $O (τ^{2})$ . By step 1.2, projecting this difference on both sides by $P$ preserves the $O (τ^{2})$ bound. Combining this with step 1.1 yields the exact-channel unnormalized estimate. Dividing by a trace requires a positive denominator and further lower-bound control for a uniform normalized estimate.

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Used by · 2

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Sources

Monolithic Quantum Error Correction in the Presence of Distinguishability · PDF p. 15, Eq. (S4.5) and the two following display lines