Theorem: Budgeted stability of records

Fix a total live-qubit budget $N$ and $\epsilon \ge 0$. Suppose a strategy in the structural class satisfies exact $(\mathrm{C}{)}^{\star }$ and:

1. For every configuration with at most $N$ live qubits and diagonal measured reduction,

   $$d_{\mathrm{TV}}({{\mathsf{P}}^{\mathsf{S}}}_{\psi },\mathrm{diag}{\rho }_S)\le \epsilon .$$

2. Every record source and target whose executions remain within $N$ live qubits obey

   $$d_{\mathrm{TV}}\left({\mathrm{Obs}}_{\mathsf{S}}[\psi ,t_{S,u}],{\mathrm{Obs}}_{\mathsf{S}}[\psi ,{\mathrm{defer}}_S(t_{S,u})]\right)\le \epsilon .$$

Then:

1. If $\psi$ has $n$ live qubits and $2n\le N$, then

   $$d_{\mathrm{TV}}({{\mathsf{P}}^{\mathsf{S}}}_{\psi },{\mathsf{Born}}_{\psi }^S)\le 2\epsilon$$

   for every $S$, where ${\mathsf{Born}}_{\psi }^S(\kappa )={\Vert {M^S}_{\kappa }\psi \Vert }^2$.

2. If $|S|=k$ and $2(n+k)\le N$, the joint law of

   $$\mathsf{let}x=\mathsf{Enter}({\vec{q}}_S)\mathsf{in}⟨\mathsf{Enter}(V({\vec{q}}_{L\setminus S})),x⟩$$

   is within $3\epsilon$ of its Born joint law for every $V$.

3. Under the budget in (b), if

   $$b={\Vert {M^S}_{\kappa }\psi \Vert }^2\ge \gamma >0,\epsilon \le \gamma /8,$$

   the branch has positive strategy probability, and

   $$D\left({{\mathsf{U}}^{\mathsf{S}}}_{\psi }(\kappa ),\frac{{M^S}_{\kappa }\psi }{\sqrt{b}}\right)\le \min \left\{1,\sqrt{\frac{6\epsilon }{\gamma }}\right\}.$$

Proof. For (a), the whole-register record experiment adds $n$ fresh qubits. Approximate records cost at most $\epsilon$; approximate anchoring of the record reduction costs at most another $\epsilon$. Exact composability and contraction under marginalization give every measured subset.

For (b), comparison with the deferred experiment costs $\epsilon$. Before deferred readout, the vector is

on $n+k$ qubits. Exact composability replaces the successive $B,R$ readouts by a joint readout. Part (a), applied to this intermediate vector, costs $2\epsilon$ and requires $2(n+k)\le N$. The Born weight is ${\left|⟨\lambda |V|{\xi }_{\kappa }⟩\right|}^2$. The triangle inequality gives $3\epsilon$.

For (c), write

Part (a) yields $|p-b|\le 2\epsilon$, hence $p\ge 3\gamma /4>0$. If the complement is empty, its residual ray is unique. Otherwise choose $V\phi =|0⟩$. The Born joint probability of $\kappa$ followed by $\lambda \ne 0$ is zero, so (b) gives

Apply (a) to $Vu$; its required budget is covered by the assumed larger budget. Then

using $\gamma \le 1$. Finally $D\le 1$.