Initial calibration and fixed-circuit resource rates

A Initial calibration and fixed-circuit resource rates

A.1 Initial independent equilibrium and uniformity

If the initial carrier positions are iid with law $w(0)$, the distinguished carrier has exactly that law, and

The preceding tracking proof applies conditionally on the initial carrier configuration and declared field, with the independent Poisson comparison clocks left random, and then averages. Its low-mass estimate already includes $\mathbb E\eta_N$; Jensen's inequality gives the corresponding square-root estimate. The martingale population bound gives

No independence between the later tag and census is needed: the coupling preserves the entire microscopic marginal and uses its adapted full-state intensities.

For the autonomous circuit, write $s_{nx}=|(V_n\psi)_x|^2$. Then

On the first pass $[0,T]$ each time envelope is unimodal, since with $u=\sin^2(\Omega t)$ it is proportional to $u^n(1-u)^{\ell-n}$. Each weight has at most two entries or boundary contacts with a level, even if a maximum equals that level. Count sublevel entries directly; no common regular value across unknown inputs is needed. Hence the node bound can be taken as

uniformly over normalized inputs on this fixed graph. Currents and their variations are uniformly bounded as well: each fine current is a fixed bounded bilinear coefficient of $\psi$ times an explicit smooth clock envelope. Alternatively bounded $\|H_F\|$ and $\|\dot\Psi\|\le\|H_F\|/\hbar$ give uniform bounds on $J,\dot J$. Initial calibration, all birth budgets, $R_{\delta,N}$ and the resulting path error are consequently uniform over these inputs. For a fixed finite reference included as a configuration coordinate this statement uses that full graph. For a declared spectator fibre, the norm estimates do not depend on the fibre dimension; this does not assert equality to a different finer-sector Bell law.

On $[0,2T]$ the spin clock reverses and returns, so at most four level entries per weight give $b(\varepsilon)\le5D\varepsilon+2C_0T\sqrt\varepsilon$. The path theorem still applies across the node at $T$ and all reversed currents. The unitary echo also unwrites the circuit's records; it is a reversal test of the event law, not an archive-protection theorem beyond the first pass. Exact zero-current intervals in the general programme are included in the same estimates. Finite pilot queues can produce delayed events there, but their total path discrepancy is already in (31); they are not declared absent at finite resources.

A.2 An explicit fixed-circuit resource rate

The qualitative limit in Theorem 8.1 uses the preceding kinetic proof. For the fixed clock circuit, its constants can also be controlled uniformly over unknown inputs. The following refinement, retained from the earlier manuscript and monograph [4, 5], records the cutoff dependence explicitly and supplies its complete proof. It is not a new stochastic law or a rate for growing circuits. The clock-weight shape was proved in Proposition 9.2.

Proposition A.1 (Uniform fixed-circuit kinetic rate)

Fix the complete finite graph, the clock Hamiltonian $H_F$, its first-pass horizon $T$, and $0<\kappa_-\le\kappa_e\le\kappa_+$. Suppose the initial census has expected $\ell^1$ error $O(N^{-1/2})$ and the tag starts from $\nu\le Cw(0)$ with fixed $C$. The iid equilibrium and the known basis-ready preparations satisfy this condition. With

$$\mu_N=N^{1/2},\qquad \delta=N^{-1/7},\qquad \varepsilon=N^{-1/35},$$

the signed-queue path error obeys

$$\epsilon_{\rm kin}(N,T)=O(N^{-1/70}).$$(56)

For equilibrium use, the constants are uniform over all normalized inputs on this fixed material space. A reference included in the configuration basis is part of the fixed graph; a declared spectator fibre does not enlarge the operator-norm constants. With the gas and recombination scales of Theorem 8.1, the full pilot path error is also $O(N^{-1/70})$.

Proof

Write $D=|V|$, $m=|E|$, $H_* =\|H_F\|$, and let $\Var$ denote total variation in physical time. Each edge current is a bounded quadratic form in the normalized input. Safe input-independent bounds are

$$J_*\le 2H_*/\hbar,\qquad L\le 2mH_*T/\hbar,\qquad \sum_e\Var(J_e)\le 4mH_*^2T/\hbar^2.$$

The last inequality follows by differentiating each current expectation and using $\|\dot\Psi\|\le H_*/\hbar$. All source variation, birth and census-jump budgets in Theorem 7.1 are therefore uniform.

For $0<\delta\le1$ put $\alpha=\mu_N/N\le1$ and $a_0=\kappa_-\delta$, $a_1=\kappa_+$. In the proof of Theorem 7.1, the instantaneous companion root has $\Var(f)=O(\delta^{-2})$ and $|f(0)|=O(\delta^{-1})$. Its deterministic integrated tracking cost is consequently $O((\mu_N\delta^3)^{-1})$. The square-jump estimate there gives

$$V'\le-2\mu_Na_0V+ \mu_N\alpha a_1(J_*/a_0+2c_0\alpha+\sqrt V).$$

Using $\alpha a_1\sqrt V\le a_0V+\alpha^2a_1^2/(4a_0)$ and $V(0)=0$ yields

$$\sup_{t\le T}V(t)\le \frac{\alpha a_1J_*}{a_0^2} +\frac{2c_0a_1\alpha^2}{a_0} +\frac{\alpha^2a_1^2}{4a_0^2} \le C\frac{\alpha+\alpha^2}{\delta^2}.$$

The deterministic export discrepancy is $O(\alpha)$, so (23) and (25) have the explicit forms

$$\begin{align} R_{\delta,N}&\le C\left[ \frac{1}{\mu_N\delta^3} +\frac{\sqrt{\mu_N/N}+\mu_N/N}{\delta}\right],\tag{57}\\ D_{\delta,N}&\le C\sqrt{\delta+(\mu_N\delta)^{-1} +\E\eta_N}. \notag\end{align}$$

Here and below constants depend on the fixed graph, programme and response bounds, not on the input or the two cutoffs.

For iid equilibrium, direct multinomial variance gives $\E\|x^N(0)-w(0)\|_1\le\sqrt{D/N}$. For a separately initialized dominated tag and $N-1$ independent equilibrium carriers, add at most $2/N$. The deterministic basis-ready census has zero initial error. The tracking proof applies conditionally on the initial census and then averages; Jensen's inequality handles the square root in the low-mass estimate. Its population martingale obeys

$$\epsilon_{x,N}\le\E\|x^N(0)-w(0)\|_1 +C_B\epsilon_{F,N}+C\sqrt{(L+1)/N}.$$

With $\mu_N=N^{1/2}$ and $\delta=N^{-1/7}$, the leading deterministic term of (57) is $N^{-1/14}$, its square-root noise term is $N^{-3/28}$, and $D_{\delta,N}=O(N^{-1/14})$. Equation (26) therefore gives

$$\epsilon_{F,N}+\epsilon_{x,N}=O(N^{-1/14}).$$

For the clock family (37), each fine weight is a nonnegative input-dependent coefficient times a unimodal binomial envelope. Count entries into a sublevel set directly: each weight has at most two boundary entries on $[0,T]$, including tangencies. A coefficient may put its maximum exactly at $\varepsilon$, so a common regular value is not assumed. Initial small-weight mass, jump influx and these entries give the uniform node budget

$$b_C(\varepsilon)\le C\bigl(3D\varepsilon+C_0T\sqrt\varepsilon\bigr).$$

Use this budget in (29). At $\varepsilon=N^{-1/35}$ the node term and the $\epsilon_{x,N}/\varepsilon^2$ term are $O(N^{-1/70})$; the $(\epsilon_{x,N}+\epsilon_{F,N})/\varepsilon$ term is $O(N^{-3/70})$. This proves (56). The gas and recombination contributions are respectively $O(N^{-2})$ and $O(N^{-1/2})$, so they do not worsen this conservative rate.

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On $[0,2T]$ at most four level entries per weight replace the node bound by $C(5D\varepsilon+2C_0T\sqrt\varepsilon)$; the same rate applies to the reversed full path. That second pass unwrites the records and is not an extension of the first-pass archive theorem. Merely assuming $\mathbb E\|x^N(0)-w(0)\|_1\to0$ without a rate does not imply (56). Nor does this estimate apply uniformly to a growing graph, increasing fine reference, changing Hamiltonian or unbounded storage time.