Section 3 4 October 2026

Canonical edge ownership and conservative export

Jeremy Rodgers · Independent Researcher
doi:10.5281/zenodo.23131075

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3 Canonical edge ownership and conservative export

Use the common action

\begin{align}S&=\int\left[\frac{i\hbar}{2}(\Psi^\dagger\dot\Psi-\dot\Psi^\dagger\Psi) +\hbar\sum_e\Pi_e\dot\chi_e-h(\Psi,\chi,t)\right]\dd t,\tag{3}\\ h&=\sum_r\Psi_r^*H_{rr}\Psi_r+ \sum_{e=(r,q)}\left(e^{i\chi_e}\Psi_q^*H_{qr}\Psi_r+\mathrm{c.c.}\right). \tag{4}\end{align}

Block sectors can replace scalars throughout with their inner products. Passive connection ownership means that $h$ has no $\Pi$ dependence. At $\chi(0)=0$ , Hamilton's equations give

i\hbar\dot\Psi=H\Psi,\qquad \dot\chi=0,\qquad \dot\Pi_e=-\hbar^{-1}\partial_{\chi_e}h=J_e,\qquad \dot w=BJ.

(5)

Proposition 3.1 (Primitive bond ownership)

Within real quadratic energies additive over single vertices and primitive binary bonds, with the displayed canonical action unit and endpoint covariance

\Psi_r\mapsto e^{i\alpha_r}\Psi_r,\qquad \chi_e\mapsto\chi_e+\alpha_q-\alpha_r,

fixing $H_{qr}$ at $\chi=0$ fixes its bond torque to $J_e$ .

Proof

A binary cross term is $\Psi_q^*T_e(\chi_e)\Psi_r+\mathrm{c.c.}$ . Covariance gives $T_e(\chi+\delta)=e^{i\delta}T_e(\chi)$ , so $T_e(\chi)=e^{i\chi}H_{qr}$ . Differentiate the action. Vertex-diagonal terms have zero bond torque. Every finite Hermitian $H$ supplies a realization.

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The assumptions matter. A triangle term $-\hbar k\|\Psi\|^2\sin(\chi_{12}+\chi_{23}+\chi_{31})$ is gauge invariant and gives an additional divergence-free action current $k$ at $\chi=0$ while changing no coherent Hamiltonian there. Primitive bond additivity excludes this concrete rival. Gauge invariance alone does not. The conserved source moment map has sign $B\Pi-w$ .

Take $k_e(0)=u_e(0)=0$ and $u_e=N(\Pi_e-\Pi_e(0))-k_e$ . At a first hit $u_e=s\in\{-1,1\}$ , put a packet of species $s$ in the next unused slot, mark its dedicated exporter blank/fuel cell spent with the retained sign and slot identifier, advance $k_e$ by $s$ , and set $u_e$ to zero. Both species may remain simultaneously present. No cancellation is part of export. Tie events use a fixed ordering; gas ties with deterministic export times have probability zero. If $L_e\ge\int_0^T|J_e|\dd t$ , then

\left\|\frac{k_e}{N}-\int_0^\cdot J_e\dd t\right\|_\infty\le\frac1N, \qquad \#\mathrm{exports}_e\le NL_e.

(6)

Indeed the first error is $-u_e/N$ and each full excursion consumes at least $1/N$ of action variation. For an admissible nonzero initial residue $|u_e(0)|<1$ , use $u_e(t)=u_e(0)+N(\Pi_e(t)-\Pi_e(0))-k_e(t)$ . The export discrepancy is then at most $2/N$ , with at most one extra birth. Choose $B_e=\lceil NL_e\rceil+1$ slots before the experiment. A bound from $H,T$ alone can be used, so the apparatus need not know an unknown input vector. The exporter uses finite increments of a canonical coordinate; it does not evaluate the Bell escape rate or supply a stochastic production clock.