Fixed-time convergence of calibrated recovery cycles

Fix a finite reference space $A$ of positive dimension and write ${\widehat{C}}_h=C_h\otimes {\mathrm{id}}_A$ and $I$ for the identity on the corresponding logical-reference matrix space. Each extension is a channel: a further identity extension is an amplification of $C_h$, hence positive by complete positivity; its trace is the sum of the traces of diagonal reference blocks, each preserved by $C_h$. It therefore contracts trace norm by [F7]. For equal intervals $\tau =T\mathrm{/}N$, [F2] with comparison channel $I$ and [F1] give $\mathrm{\parallel }{\widehat{C}}_{\tau }^N-I{\mathrm{\parallel }}_{1\to 1}\le N\mathrm{\parallel }{\widehat{C}}_{\tau }-I{\mathrm{\parallel }}_{1\to 1}\le NK{\tau }^2=KT\tau$. Re-encoding as in [F3] makes these powers precisely the repeated recovered cycles, with the same fixed data and constants. A zero-dimensional reference has no density inputs and only trivial operator estimates.

For any logical-reference density operator $\rho$, positivity and trace one give $\mathrm{\parallel }\rho {\mathrm{\parallel }}_1=1$. Put $\sigma ={\widehat{C}}_{\tau }^N(\rho )$, which is again a density operator because the map is a channel. Step 1.1 gives $\mathrm{\parallel }\sigma -\rho {\mathrm{\parallel }}_1\le KT\tau$. Define infidelity here as $1-F(\rho ,\sigma {)}^2$, using the squared-fidelity convention in [F8]. Then $0\le 1-F(\rho ,\sigma {)}^2=(1+F(\rho ,\sigma ))(1-F(\rho ,\sigma ))\le \mathrm{\parallel }\rho -\sigma {\mathrm{\parallel }}_1\le KT\tau$. This proves vanishing infidelity as well as trace-norm convergence, in particular for pure logical inputs and pure states on a logical-reference space. No fidelity value on a single input is used to infer a norm bound.

More generally let $h_1,\dots ,h_m$ be positive interval lengths with ${\sum }_{k=1}^m{h}_k=T$ and ${\max }_k{h}_k\le \tau \le {\tau }_0$, and put $P={\widehat{C}}_{h_m}\cdots {\widehat{C}}_{h_1}$, with the rightmost map applied first. Cancellation of adjacent terms gives the finite identity $P-I={\sum }_{k=1}^m{\widehat{C}}_{h_m}\cdots {\widehat{C}}_{h_{k+1}}({\widehat{C}}_{h_k}-I)$, with the factor preceding the parentheses equal to $I$ when $k=m$. Each preceding product contracts trace norm by step 1.1. Apply the trace-norm triangle inequality to this identity on any matrix $X$ and use [F1] to get $\mathrm{\parallel }(P-I)(X){\mathrm{\parallel }}_1\le K{\sum }_k{h}_k^2\mathrm{\parallel }X{\mathrm{\parallel }}_1$. Taking the induced norm and using $h_k^2\le \tau h_k$ gives $\mathrm{\parallel }P-I{\mathrm{\parallel }}_{1\to 1}\le K{\sum }_k{h}_k^2\le KT\tau$. Applying step 2.1 with $P(\rho )$ in place of $\sigma$ gives the same infidelity bound. Thus a shorter final interval preserves both convergence statements.

For the partition in step 3.1, define two logical maps for interval $k$. Let $D_{k,0}=R{\mathcal{U}}_{h_k}({\mathcal{N}}_{h_k}^{(0)}+{\mathcal{N}}_{h_k}^{(1)})\mathcal{V}$ and $D_{k,1}=R{\mathcal{U}}_{h_k}({\sum }_{n\ge 2}{\mathcal{N}}_{h_k}^{(n)})\mathcal{V}$, using the same compensation ${\mathcal{U}}_h$, decoder $R$ and encoding $\mathcal{V}$ as in the complete-cycle lemma. Both maps are completely positive by [F1] and [F4], and $D_{k,0}+D_{k,1}=C_{h_k}$. They therefore form a two-outcome instrument by [F5]; outcome $1$ means at least two jumps in this interval, and outcome $0$ means fewer than two. Their finite reference extensions ${\widehat{D}}_{k,b}=D_{k,b}\otimes {\mathrm{id}}_A$ are completely positive, and their sum ${\widehat{C}}_{h_k}$ is trace preserving by step 1.1. This instrument only groups mathematical trajectory contributions; the implemented decoder is the same in both outcomes and requires no observation of the jump record.

On the finite outcome set $\mathrm{\Omega }=\{0,1{\}}^m$ define $J_{\omega }={\widehat{D}}_{m,{\omega }_m}\cdots {\widehat{D}}_{1,{\omega }_1}$. Compositions are completely positive, since every amplification is a composition of positive maps. Distributing the finite sum gives ${\sum }_{\omega \in \mathrm{\Omega }}{J}_{\omega }=P$, a channel by step 3.1. Thus $(J_{\omega }{)}_{\omega \in \mathrm{\Omega }}$ is a finite instrument. For the initial density operator $\rho$, [F5] supplies weights $p_{\omega }=\mathrm{Tr}(J_{\omega }(\rho ))\ge 0$ with ${\sum }_{\omega }{p}_{\omega }=1$. Define a probability measure on all subsets of $\mathrm{\Omega }$ by $\mathbb{P}(B)={\sum }_{\omega \in B}{p}_{\omega }$; finite sums give additivity, and countable additivity holds because any disjoint family of nonempty subsets of this finite set has only finitely many members. This constructs the required joint law without assuming independent intervals.

Let $A_k=\{\omega :{\omega }_k=1\}$. Summing over the later outcomes replaces later branch maps by their trace-preserving channel sums, which disappear under the trace; summing over earlier outcomes gives ${\rho }_{k-1}={\widehat{C}}_{h_{k-1}}\cdots {\widehat{C}}_{h_1}(\rho )$, with ${\rho }_0=\rho$. Consequently $\mathbb{P}(A_k)=\mathrm{Tr}({\widehat{D}}_{k,1}({\rho }_{k-1}))$. Each ${\rho }_{k-1}$ is a density operator, so the input-uniform bound [F1] gives $\mathbb{P}(A_k)\le K_2{h}_k^2$. Applying [F6] in the finite probability space of step 5.1 yields $\mathbb{P}({\bigcup }_k{A}_k)\le K_2{\sum }_k{h}_k^2\le K_{2}T\tau$. In particular equal intervals give $N{K}_2{\tau }^2=K_{2}T\tau$, and any shorter final interval obeys the same bound. Together with steps 1.1–2.1 this proves every stated fixed-time limit with the original constants and arbitrary finite references.