Detection-code errors retain partial logical information

A downward jump from $J=2$ lands in $J=1$, orthogonal to every code vector, so $P{L}_{\delta }P=0$ and a $J$ measurement flags that transition. The normalized $J=2$-to-$J=1$ matrices are the $B_{\delta }$ of [F9], multiplied by $\sqrt{\gamma }$ under [F2]–[F3]. Their actions give error spaces $\mathrm{span}\{\mathrm{\mid }1,-1⟩,\mathrm{\mid }1,1⟩\}$ from $a$ and the full $\mathrm{span}\{\mathrm{\mid }1,-1⟩,\mathrm{\mid }1,0⟩,\mathrm{\mid }1,1⟩\}$ from $b$. In particular the logical restriction of $B_0^{†}{B}_0$ is $\mathrm{diag}(0,2\mathrm{/}3)$, which is not scalar. Thus the single-error detection condition holds while the pair-product correction condition fails. The two diagonal one-decay output populations differ, so they retain logical-basis information rather than replacing both inputs by the same state.

The source-stated fixed-jump model uses $H_S+H_{\mathrm{\Delta }}$ while retaining the same $L_{\delta }$; this is the physical modeling input explicitly stated after Eq. (11), not a consequence of the lift postulate alone. The scalar loss identity in [F3] restricts to $\gamma I$ on both $V_2$ and $V_1$ and is unchanged by this Hamiltonian-only lift. On $J\le 2$, these matrices and the detection isometry are exactly those in [F10]. No higher $J$ can be reached from this single-$J$ input by the diagonal Hamiltonian or downward jumps. The orthogonal-sector calculation [F10]–[F12] therefore gives $F_D(t)\ge (1+3e^{-\gamma t})\mathrm{/}4+\gamma t{e}^{-\gamma t}\mathrm{/}8$, for every real $\mathrm{\Delta }$ within the stipulated effective model. This argument derives neither the microscopic accuracy of retaining jumps nor the article's accompanying relative-amplitude estimate.

Fix $t>0$ and write $A=C_t(\mathrm{\mid }0⟩⟨0\mathrm{\mid })$ and $B=C_t(\mathrm{\mid }1⟩⟨1\mathrm{\mid })$. Since the two input projectors are $(I_2+Z)\mathrm{/}2$ and $(I_2-Z)\mathrm{/}2$, [F10] gives $A=\mathrm{diag}(1\mathrm{/}2,0,1\mathrm{/}2)$ and $B=\mathrm{diag}(1\mathrm{/}6,2\mathrm{/}3,1\mathrm{/}6)$. Thus the conditional one-decay probability of magnetic label zero is zero for $a$ and $2\mathrm{/}3$ for $b$, and $\mathrm{Tr}(AB)=1\mathrm{/}6$. By [F20], $C_t$ has a finite Kraus representation; since it is a channel, [F16] permits choosing a finite family $K_r:{\mathbb{C}}^2\to V_1$ representing it with ${\sum }_r{K}_r^{†}{K}_r=I_2$. Put $u_r=K_r\mathrm{\mid }0⟩$ and $v_r=K_r\mathrm{\mid }1⟩$. Then $A={\sum }_r{u}_r{u}_r^{†}$ and $B={\sum }_s{v}_s{v}_s^{†}$. Expanding these finite sums gives $\mathrm{Tr}(AB)={\sum }_{r,s}\mathrm{Tr}(u_r{u}_r^{†}{v}_s{v}_s^{†})={\sum }_{r,s}\mathrm{\mid }{u}_r^{†}{v}_s{\mathrm{\mid }}^2=1\mathrm{/}6$, since the trace of $u(u^{†}v)v^{†}$ is $(u^{†}v)(v^{†}u)$. Hence some $⟨0\mathrm{\mid }{K}_r^{†}{K}_s\mathrm{\mid }1⟩=u_r^{†}{v}_s$ is nonzero, so that pair product is not scalar. This uses the stated channel outputs, without subtracting any asserted component from a time integral.

Include each $K_r$ in the $V_1$ output block by its isometric inclusion ${\iota }_1$. The finite family $\sqrt{p_0}{W}_t$, $\sqrt{p_1}{\iota }_1{K}_r$, $\sqrt{p_2}\mathrm{\mid }g⟩⟨0\mathrm{\mid }$, $\sqrt{p_2}\mathrm{\mid }g⟩⟨1\mathrm{\mid }$ represents the full logical-input restricted channel in [F10]. The sector weights are nonnegative and sum to one; the family's adjoint-product sum is consequently $(p_0+p_1+p_2)I_2=I_2$, using $W_t^{†}{W}_t=I_2$ and ${\sum }_r{K}_r^{†}{K}_r=I_2$. Its one-decay pair products are $p_1{K}_r^{†}{K}_s$, because ${\iota }_1^{†}{\iota }_1=I_{V_1}$. Since $p_1=\gamma t{e}^{-\gamma t}>0$, step 2.1 supplies a pair with a nonzero off-diagonal logical entry. Applying [F15] to this finite restricted channel with the identity isometry on ${\mathbb{C}}^2$ rules out exact recovery. Exact correction of the original physical code would give a decoder for this same restricted channel, so it too is impossible at every finite $t>0$.

The first term of the bound in step 1.2 is precisely the minimal-code fidelity in [F5]. For every finite positive time the extra term is positive. The continuous-fidelity strict-crossing lemma [F13] therefore applies to the same threshold convention as [F6], and proves strictly longer checking time when $1\mathrm{/}4<F_0<1$, including $0.9999$. The same lemma gives equality of the two zero infima at target one and the two infinite values for $0<F_0\le 1\mathrm{/}4$. This comparison is an analytic consequence of the fixed-jump benchmark, not a claim to have reproduced the plotted numerical curves.