Finite pilot resources: exact inventories and retained records

A carrier permutation takes channel $(e,b,a)$ to $(e,b,\sigma(a))$, which has the same frequency and hence the same mark probability. Recombination marks are unchanged. The joint ordered-time and channel-mark law is invariant under this relabelling. In the spatial realization this can equally be implemented by a measure-preserving permutation of equal-area channel cells; no symmetry of their geometric shapes is needed. The deterministic reaction map commutes with this relabelling, because eligibility and all service rules depend on the current carrier vertex, not its name. Carrier-independent exporter ties retain their order; contact ties and coincidences with deterministic export times have probability zero. The common ready configuration is invariant. Consequently all $N$ summands in $\mathbb E[n_A\mathbf1_D]=\sum_a\mathbb P(X_a\in A,D)$ coincide, proving the formula. This proves the required marginal exchangeability; a stronger invariance assertion about every coordinate of the complete microstate is unnecessary.

The quantity $n+BZ+Bu-Nw$ is constant between events and at each export, service and recombination. Its initial value is zero. On a monotone edge the scaled action crosses successive nonnegative integers, triggering exactly one positive export at each crossing. This proves both claims.

For a queued packet the predictable service intensity is $\kappa_e\mu_N n_{\rm origin}/N$. It is bounded below by the exponent's rate until that packet is served. Its survival probability is therefore bounded by the corresponding exponential, by the compensator or thinning construction. A union bound proves (43). A single addition of (39), applied to the complete experiment, transfers any union of such failure events to the physical gas; it need not be added separately at each stage.

During the copy, $w_{v,1}(s)=W\sin^2(\omega s)$, so the copy current is nonnegative and its total action is $W$. Its final residue therefore equals the production residue $u_{\rm prod}=\{NW\}$. At the final field configuration $w_{v,0}=0$ and the census at $(v,0)$ reads

Thus every queued copy packet always has an eligible origin carrier during the final hold, even though the coherent origin weight has become zero. Lemma 11.3 with $c_N=1/N$ proves (46). No source packets remain on the successful production event, and no coherent current emits further packets after the copy. Once $Z_f=0$, no event can alter the copied bit under idle continuation. The writer is an ordinary coherent Hamiltonian on the enlarged graph, exactly the interaction type allowed by P3.

The production census is $n_0=N(1-w_p)+Z_e+u_e\ge N/4$, since no negative packet is emitted. Lemma 11.3 bounds production failure by the first term of (49). Conditional on its complement, Lemma 11.4 supplies the second term. The full spatial-gas comparison adds the third term once. On successful production and copy drainage, the bit-one census is deterministically $\lfloor Nw_p(t)\rfloor$. The good event is carrier-label invariant, so Lemma 11.1 proves (50); averaging proves (51).

For the open interval, let $y=N\sin^2(2gt)$ and $x=N\sin^2(gt)$. The increasing $y$ reaches $1$ strictly before $\pi/(6g)$ because its terminal value is $3N/4>1$. At that crossing $x=1/(4\cos^2(gt))\le1/3$. Just beyond it, $1<y<2$, $x<1$ and $x+y<2$. Hence $\lfloor x\rfloor=0$, $\lfloor y\rfloor=1$, $\lfloor(x+y)/2\rfloor=0$, giving the ideal difference $1/(2N)$. The two errors yield (52).

The proposed law is the same for both preparations. The triangle inequality makes one of its two distances at least half the distance between the actual bit laws, which is at least the probability gap in (52).

Alice's field evolves to $-i\sum_a|a\rangle|x_a\rangle_A|x_a\rangle_B/\sqrt2$, with monotone integrated actions $1/2$ on her two productive edges. At the end of her rotation the ready-vertex census is

Whenever an Alice packet remains queued, at least one carrier is eligible. Lemma 11.3 bounds Alice failure by $2B_*e^{-\kappa_-\mu_Nh_A/N}$. For odd $N$, a ready carrier remains even after drainage; it is retained, not discarded or renormalized.

In Alice slice $a$, Bob's wave has squared norm $1/2$ and internal input $x_a$. Its productive action is $W_a=\tfrac12 w_{p_a}(t_B)$, yielding the stated values. Conditional on Alice drainage, the census at $(a,0,0)$ during Bob's hold is

Here $w_{a,0,0}\ge1/8$, $0\le u_{A,a}<1$, and $u_{B,a}\ge0$. If $Z_{B,a}\ge1$, the right side is positive, so the integer count is at least one. Bob's production failure is bounded by the second exponential term in (55).

On successful production drainage, $n_{a,1,0}=\lfloor NW_a\rfloor$ and $w_{a,1,0}=W_a$. Hamiltonian (54) is the direct sum of identical ordinary copy rotations in the Alice slices. Each has copy action $W_a$, and at its final origin the production and copy residues cancel exactly as in Lemma 11.4. Each queued copy packet has at least one eligible carrier, giving the third exponential term. On complete success the number of carriers with $m=1$ is the deterministic sum of floors in (56). Label exchangeability and one full-history gas comparison prove the claimed law. The branch $a=r$ has zero field after Alice's rotation and exports no Bob packets; any odd-$N$ stranded carrier remains with $m=0$ and is included in the law.

Put $x=N\sin^2(gt_B)/2$, $y=N\sin^2(2gt_B)/2$. Since $y$ increases to $3N/8>1$ for $N\ge3$, it crosses $1$ before the interval ends; at the crossing $x\le1/3$. Just afterwards $\lfloor x\rfloor=0$, $\lfloor y\rfloor=1$ and $\lfloor(x+y)/2\rfloor=0$. The difference of the ideal numerators is therefore $1$, proving (57).

Apply the same triangle-inequality argument as in Corollary 11.6 to Bob's two retained marginal laws. Marginalization contracts total variation, so the lower bound also holds for a proposed complete joint law.

The inventory must include the changed initial constant. For a vector of initial residues $U$, it is

In the single-edge target coordinate this reads $n_1=x-Z-u+U$. Writing $k$ for signed cumulative exports, the unchanged exporter obeys $u=U+x-k$, so $n_1=k-Z$. At a drained endpoint $n_1=k$. Exchangeability is unchanged by a carrier-independent initial residue.

During $0\le x\le a<1$, no negative threshold can be reached and at most one positive export occurs. Including a hit at the endpoint,

Equality to $a/N$ for every $a\in(0,1)$ forces $\mathbb P(U\ge b)=1-b$ for every $b\in(0,1)$. Letting $b\downarrow0$ gives $\mathbb P(U>0)=1$, and these tail probabilities uniquely specify the uniform law on $[0,1)$.

Now reverse the action after reaching $a$. On the exported branch $U\ge1-a>0$, the residue decreases from $U+a-1$ to $U-1>-1$, so it never reaches the negative threshold. On the other branch it decreases from $U+a<1$ to $U\ge0$, again without an export. Hence the final count remains $k=\mathbf1_{\{U\ge1-a\}}$. At most one positive packet needs to drain; whenever it remains alive, all $N$ carriers are still at the origin. Its non-drain probability after an idle hold $h$ is at most $e^{-\kappa\mu_Nh}$ in the Poisson comparison. The finite-gas comparison adds $\delta_{\rm g}$. The returned finite-hold endpoint therefore approaches $a/N$ with controlled error, whereas $w_1=0$. A two-level forward Rabi pulse followed by its reverse realizes the stated weight path; no change to the carrier or contact law is used.

Work first in the Poisson comparison and write $n=n_1$, $Z=k-n$, $r=w-k/N\in[0,1/N)$ and $Y=w-n/N=(Z/N)+r\ge0$. Between deterministic exports the carrier count has the exact birth generator

The exporter does not itself change $n$. Every queued packet has service hazard $\kappa\mu_N n_0/N\ge\kappa\mu_N c$, because $n_0/N=1-w+Y\ge c$.

We give the required second-moment estimate explicitly. For each packet, construct an independent baseline clock of rate $a=\kappa\mu_Nc$, with a supplementary state-dependent clock of rate $\kappa\mu_N(n_0/N-c)$. On either service, choose uniformly among the current origin carriers. Every eligible packet–carrier pair then has total rate $\kappa\mu_N/N$, so this construction has precisely the original carrier/packet generator. A packet alive at $t$ must have survived its first baseline ring after its deterministic export time $\tau_i$. Therefore $Z(t)$ is dominated by

Writing $K=\|J\|_\infty$, Stieltjes integration and $|\lfloor Nw\rfloor-Nw|\le1$ give

Since the indicators are independent, $\mathbb E\widetilde Z^2\le m^2+m$. Consequently, uniformly in $t$,

No independence of actual packet lifetimes has been assumed; only their comparison clocks are independent.

Let $d(t)=\mathbb E Y(t)$. The exact generator and $1-n/N=1-w+Y$ imply, almost everywhere,

Since $0\le d\le1$, (61) yields

Here $m_*:=NK/(\kappa\mu_Nc)+1$. The bounded jumps in $k$ and $r$ cancel in $Y$; $d$ is absolutely continuous and the almost-everywhere equation suffices. Variation of constants from $d(0)=0$ gives

Set $F=J/[\kappa(1-w)]$. Integration by parts of the $J$ term, using the derivative of the exponential with respect to $s$, proves the explicit bound

All terms vanish. Exchangeability identifies $\mathbb E n/N=\mathbb P(Q=1)$. The gas comparison adds at most $\mu_N\delta_{\mathrm g}$ to (63), proving (58).

At $t=\pi/(6g)$, for $p=1,0,1/2$ respectively, the limits are

The Born terms cancel between the equal-density-matrix ensembles. Thus $\mu_N(P_X-P_Z)\to(F_1+F_0)/2-F_{1/2}=5\sqrt3g/(6\kappa)$.