Version 2 · earlier version; the current one is v3 · External review, round 2: major errors

# Moving clock: a ring clock carried by a body in the record background

- **Subproject:** 04-ilang-time
- **Package:** 05-moving-clock
- **Version:** v2
- **Mode:** external regeneration
- **Date:** 2026-10-10

## Changes from previous version

- **Items 3 and 4 are now first order in $\epsilon$ on the tick scale** (fixed $\lambda/\lambda_0$). v1 dropped the curvature of $\omega_k$ in $d_k$, whose accumulated phase is first order there. Step 6 now expands about the centre of the clock band and keeps it. The rate of the carried clock is $\Gamma$ of (5.20) instead of $\gamma=m/\omega_0$: $m$ is replaced by $\bar\mu=m+\langle0\vert D\vert0\rangle$.
- New equations (5.20)–(5.29). The v1 tags (5.13)–(5.19) are retired and not reused. Their formulas are (5.21), (5.22) and (5.26)–(5.29) with $\bar\mu\to m$; they hold at leading (zeroth) order in $\epsilon$ only.
- Step 7 distinguishes the tick of the first-order distribution from the behaviour of the reading near it, (5.23).
- Step 8 is new: the velocity of the item-3 start to first order, with its oscillating part. Step 9 (item 4, Step 8 of v1) uses the mean velocity (5.25).
- Step 2: the convergence of $\bar x$ and $v$ in the chain limit now rests on the bounded commutator $i[C,x]$.
- Setup [S] and Step 6(iii): $p_0$ and $\delta p$ are the mean and the standard deviation of the packet, and the spread errors are uniform second-order bounds. The error term of (5.8) is accordingly $O(\delta p^2)$.
- Steps 1, 3, 4, 5 and the equations (5.1)–(5.12) are otherwise unchanged.

## Response to verification

- **I1: Accepted and fixed in Steps 6–9.**
  - The quadratic term of $\omega_k$ is kept: Step 6(ii), (5.20). The first-order distribution is (5.21), with rate $\Gamma=\gamma\bigl[1+\frac{(N-1)\epsilon}2(1-\gamma^2)\bigr]+O((N\epsilon)^2)$. For $N=2$ this is the rate of the verifier's example (check 3).
  - Tick and ratio: (5.22). The nominal tick is separated from the finite-$\epsilon$ reading in (5.23) and the items after it.
  - The velocity is carried to the same order in Step 8, and the relations of item 4 are (5.26)–(5.29).
  - Every neglected term is listed with its order after (5.21) and (5.25). The v1 formulas are labelled leading order.
- **I2: Accepted and fixed in Step 2, items (a)–(d).** Superexponential tails are claimed for finitely supported starts only. For the packets, norm convergence gives the reading probabilities; $v$ and $\bar x$ converge by the bounded commutator $i[C,x]$.
- **I3: Accepted and fixed in Setup [S] and Step 6(iii).** The spread bound is second order in $\delta p$, uniform in $p_0$, and does not vanish at $p_0=0,\pm\pi/2$.

## Setup and assumptions

- **System.** The body $b$ and the medium $c$ are each the only instance of their type, so no swap condition (1.11) applies. The joint space is $\mathcal H=\mathcal H_b\otimes\mathcal H_c$, with $\mathcal H_c$ finite-dimensional (A1).
  - Item 1: $\mathcal H_b=\mathcal H_X\otimes\mathbb C^N$, with places $(x,a)$.
  - Items 2–4: $\mathcal H_b=\ell^2(\mathbb Z)\otimes\mathbb C^2\otimes\mathbb C^N$ (position, branch, clock label), with places $(x,s,a)$, obtained as the limit of Step 2.
  - Contract: $C=C_b\otimes\mathbb 1_c+\sum_h\lvert h\rangle\langle h\rvert\otimes K_h$ with $K_h=K_x$.
  - Start: $\lvert\phi\rangle\otimes\lvert\chi\rangle$. We take $\lambda\ge0$ (A9 restricted).
- **Clock reading** (A8). It is applied at a stated $\lambda$. Its projectors are $\Pi_a=\mathbb 1\otimes\lvert a\rangle\langle a\rvert\otimes\mathbb 1_c$, with $\lvert a\rangle\langle a\rvert$ on the clock-label factor. By (1.3), $q(a;\lambda)=\lVert\Pi_a\Psi(\lambda)\rVert^2$ with $\Psi(\lambda)=e^{-iC\lambda}(\phi\otimes\chi)$.
- **Branch label.** $\sigma_3\lvert1\rangle=\lvert1\rangle$, $\sigma_3\lvert2\rangle=-\lvert2\rangle$, $\sigma_1\lvert1\rangle=\lvert2\rangle$. Spinors are written as columns $(\beta_1,\beta_2)^T$.
- **Clock inputs (02-clock@v2).** Take (2.9) with $\alpha_m=0$, $\varepsilon=0$, $\ell_k=k$. Then $D=\sum_{k=0}^{N-1}d_k\lvert f_k\rangle\langle f_k\rvert$, $d_k=\frac{2\pi k}{N\lambda_0}$, satisfies $e^{-iD\lambda_0}=S$. By (2.12) with $c=0$, it has the minimal spread for the start $\lvert0\rangle$.
  - (2.13) is its reading distribution. It is written below with the place called $a$, since $m$ denotes the parameter of $B$.
  - Since $\langle a\vert f_k\rangle=N^{-1/2}e^{2\pi ika/N}$, (2.13) is a geometric sum:
$$
P_N(a;\vartheta):=\Bigl\lvert\frac1N\sum_{k=0}^{N-1}e^{2\pi ik(a-\vartheta)/N}\Bigr\rvert^2=\frac{\sin^2(\pi\vartheta)}{N^2\sin^2\bigl(\pi(\vartheta-a)/N\bigr)},\qquad p(a;\lambda)=P_N(a;\lambda/\lambda_0)\ \text{in (2.13)}.
\tag{5.1}
$$
- **Space inputs (03-ilang-space).**
  - By (5.10) and (5.11) of 03-ilang-space/05-recording-contract, the background point of a place $h$ is $u_h=(K_h-\langle K_h\rangle)\chi$, and the distance is $d_0(h,h')=\lVert u_h-u_{h'}\rVert$. Both depend only on $K_h$ and $\chi$. (5.10) of 03-ilang-space/05-recording-contract is stated for $\phi_h\neq0$; only $u_h$ and $d_0$ are used below.
  - From 03-ilang-space/17-motion: the definitions $\bar x=\sum_hp_hu_h$ and $v=\mathrm d\bar x/\mathrm d\lambda$; the bound (17.7), $\lVert v\rVert\le v_{\max}:=\varrho(a)\le\max_h\sum_{h'}\lvert t_{hh'}\rvert d_0(h,h')$ with $a_{hh'}=\lvert t_{hh'}\rvert d_0(h,h')$; the chain relations (17.15); and the chain value (17.20), $v_{\max}=2t\sigma_X$.
- **Weak-record regime [W]** (items 2–4, as fixed in item 2(b)).
  - Everything is computed at $\kappa=0$, except that positions are the background points $u_h=O(\kappa)$. Then $\Psi(\lambda)=(e^{-iC_b\lambda}\phi)\otimes\chi$, and the view of $b$ is that of $C_b$ alone.
  - By Duhamel's formula and Step 2(d), the exact state differs from this one by at most $\kappa\lVert X\rVert\int_0^\lambda\lVert x\,e^{-iC_b\lambda'}\phi\rVert\,\mathrm d\lambda'\le\kappa\lVert X\rVert\lambda\bigl(\lVert x\phi\rVert+\tfrac12\tau\lambda\bigr)$. [W] is valid while this is $\ll1$.
- **Starts near $p_0$ [S].** $g$ is smooth and $2\pi$-periodic, with $\int_{-\pi}^{\pi}\frac{\mathrm dp}{2\pi}\lvert g\rvert^2=1$. On the arc $(p_0-\pi,p_0+\pi]$ the weight $\lvert g\rvert^2\frac{\mathrm dp}{2\pi}$ has mean $p_0$ and variance $\delta p^2\ll1$. For every $2\pi$-periodic $C^2$ function $F$, Taylor's theorem with remainder then gives, uniformly in $p_0$,
$$\Bigl\lvert\int\frac{\mathrm dp}{2\pi}\lvert g\rvert^2F(p)-F(p_0)\Bigr\rvert\le\tfrac12\,\delta p^2\,\sup_p\lvert F''(p)\rvert .$$
- **Clock parameter.** $\epsilon:=\frac{2\pi}{N\lambda_0m}$, so $d_k=km\epsilon$ and $m\epsilon\lambda=\frac{2\pi}N\vartheta_0$ with $\vartheta_0:=\lambda/\lambda_0$. "First order in $\epsilon$" means: at fixed $N$, $\tau/m$, $p_0$ and $\vartheta_0$ (the tick scale), with a remainder $O(\epsilon^2)$. Bounds are given with their $N$-dependence; they need $N\epsilon(1+\vartheta_0)\le1$.

## Derivation

### Step 1. Decoupled body (item 1)

Since $K_{(x,a)}=K_x$, the record term is $\sum_x\lvert x\rangle\langle x\rvert\otimes\mathbb 1_N\otimes K_x$. Order the factors as $(\mathcal H_X\otimes\mathcal H_c)\otimes\mathbb C^N$; this relabelling of tensor factors changes no reading. Then $C=G\otimes\mathbb 1_N+\mathbb 1\otimes D'$, with $G:=H\otimes\mathbb 1_c+\sum_x\lvert x\rangle\langle x\rvert\otimes K_x$.

The two terms act on different factors and commute. Hence $e^{-iC\lambda}=e^{-iG\lambda}\otimes e^{-iD'\lambda}$ and $\Psi(\lambda)=e^{-iG\lambda}(\phi_X\otimes\chi)\otimes e^{-iD'\lambda}\eta$. $\Pi_a$ acts on the last factor only, and $e^{-iG\lambda}$ is unitary, so
$$
q(a;\lambda)=\bigl\lvert\langle a\vert e^{-iD'\lambda}\vert\eta\rangle\bigr\rvert^2\qquad\text{for all }\lambda .
\tag{5.2}
$$
This holds exactly, for every $\chi$, every $K_x$ and every $\kappa$. The answer to each of the three questions is no: $q$ depends neither on $H$, nor on $\phi_X$, nor on the $K_x$ (nor on $\chi$). For $D'=D$ and $\eta=\lvert0\rangle$, (5.2) is (2.13).

The model of items 2–4 is not of this form: $B$ contains $\sigma_3\otimes D$, which does not commute with $A\propto\sigma_1$.

### Step 2. The infinite chain as a limit (item 2)

**Finite chains.** Take $\Lambda_L=\{-L,\dots,L\}$, keep in $A$ only the bonds inside $\Lambda_L$, and restrict $B$ and the records to $\Lambda_L$. This gives $C^{(L)}$; the same expressions on $\ell^2(\mathbb Z)$ give $C^{(\infty)}$.

**Locality bound.** Write $C^{(\cdot)}=A^{(\cdot)}\otimes\mathbb 1_c+Q$, where $Q$ (the $B$ term and the records) is diagonal in $x$. $A$ is $\frac\tau2$ times a sum of two operators of norm 1, so $\lVert A\rVert\le\tau$. Expand $e^{-iC\lambda}$ in powers of $A$ around $e^{-iQ\lambda}$ (Dyson series). The term of order $n$ moves $x$ by at most $n$ and has norm at most $(\tau\lambda)^n/n!$. For a start supported in $\lvert x\rvert\le R$, therefore, for both $C^{(L)}$ and $C^{(\infty)}$:
$$
\bigl\lVert\Pi_{\lvert x\rvert\ge L}\,e^{-iC^{(\cdot)}\lambda}(\phi\otimes\chi)\bigr\rVert\le\sum_{n\ge L-R}\frac{(\tau\lambda)^n}{n!} .
\tag{5.3}
$$
On vectors supported in $\Lambda_L$, $C^{(\infty)}-C^{(L)}$ is the hopping out of the sites $x=\pm L$. Its norm is at most $\tau\lVert\Pi_{\lvert x\rvert\ge L}\,\cdot\,\rVert$. By Duhamel's formula, the two evolved states therefore differ by at most $\tau\lambda\sum_{n\ge L-R}(\tau\lambda)^n/n!\to0$.

**How the limit is taken.** First $L\to\infty$, at fixed $\lambda$, $\kappa$ and start; then $\kappa\to0$ [W], then the small spread [S], then the expansion in $\epsilon$. On $\Lambda_L$ the start is the normalized restriction $\phi^{(L)}$ of $\phi$.
- (a) *Finitely supported starts.* By (5.3) the tails are superexponentially small in $L-R$, and the evolved states converge in norm.
- (b) *Starts with $\lVert x\phi\rVert<\infty$.* These include the packets used below: their $\phi_x$ decay faster than any power of $x$, because $g\beta$ is smooth and periodic. $\phi^{(L)}\to\phi$ in norm, and unitarity carries an approximation of the start to every $\lambda$. With (a), the evolved states converge in norm; hence every reading probability ($q$, $p_h$) converges.
- (c) *Velocity.* Norm convergence does not control the unbounded label $x$. But $x$ commutes with $Q$, and $[T,x]=-T$ for $T=\sum_x\lvert x+1\rangle\langle x\rvert$, so $i[C^{(\cdot)},x]=-\frac\tau2(T+T^\dagger)\otimes\sigma_1\otimes\mathbb 1_N\otimes\mathbb 1_c$ (bonds inside $\Lambda_L$ for $C^{(L)}$), of norm $\le\tau$ for every $L$. $\mathrm d\bar x_{\rm lab}/\mathrm d\lambda$ (Step 4) is its expectation. These operators converge strongly as $L\to\infty$, so by (b) the velocity converges at every $\lambda$, and it is bounded by $\tau$.
- (d) *Mean position.* $\bar x_{\rm lab}(\lambda)=\bar x_{\rm lab}(0)+\int_0^\lambda(\mathrm d\bar x_{\rm lab}/\mathrm d\lambda')\,\mathrm d\lambda'$. The first term converges since $\sum_x\lvert x\rvert\lvert\phi_x\rvert^2<\infty$, the second by (c) and bounded convergence. The same commutator gives $\lVert x\Psi(\lambda)\rVert\le\lVert x\phi\rVert+\tau\lambda$, uniformly in $L$.

**Plane waves.** On $\ell^2(\mathbb Z)$ we use the standard Fourier facts.
- The plane waves are $\lvert p\rangle=\sum_xe^{ipx}\lvert x\rangle$, $p\in(-\pi,\pi]$, with $\langle p\vert p'\rangle=2\pi\delta(p-p')$.
- $T\lvert p\rangle=e^{-ip}\lvert p\rangle$.
- A vector $\psi=\int\frac{\mathrm dp}{2\pi}\lvert p\rangle\otimes G(p)$ has $\lVert\psi\rVert^2=\int\frac{\mathrm dp}{2\pi}\lVert G\rVert^2$.
- The position label acts as $G\mapsto i\partial_pG$, for smooth periodic $G$ (integration by parts).

### Step 3. Anticommutation and spectrum (item 2(a))

In terms of $T$, $A=\frac\tau2\bigl(-i(T-T^\dagger)\bigr)\otimes\sigma_1\otimes\mathbb 1_N$ and $B=\mathbb 1\otimes\sigma_3\otimes M$, with $M:=m\mathbb 1_N+D$. On the position and clock factors the factors of $A$ and $B$ commute, and $\sigma_1\sigma_3+\sigma_3\sigma_1=0$. Hence
$$
AB+BA=0,\qquad C_b^2=A^2+B^2 .
\tag{5.4}
$$
Further, $-i(T-T^\dagger)\lvert p\rangle=-2\sin p\,\lvert p\rangle$ and $M\lvert f_k\rangle=\mu_k\lvert f_k\rangle$. So
$$
C_b\bigl(\lvert p\rangle\otimes\beta\otimes\lvert f_k\rangle\bigr)=\lvert p\rangle\otimes C_k(p)\beta\otimes\lvert f_k\rangle,\qquad C_k(p):=-\tau\sin p\,\sigma_1+\mu_k\sigma_3,\qquad\mu_k:=m+d_k .
\tag{5.5}
$$
By (5.4), $C_k(p)^2=(\mu_k^2+\tau^2\sin^2p)\mathbb 1$; moreover $\operatorname{Tr}C_k=0$. Write $C_k=\omega_k(\sin\theta_k\sigma_1+\cos\theta_k\sigma_3)$, with $\cos\theta_k=\mu_k/\omega_k>0$. The half-angle formulas then give the generalized eigenvectors $\lvert p\rangle\otimes\beta_{k,\pm}(p)\otimes\lvert f_k\rangle$ of $C_b$, with eigenvalues $\pm\omega_k(p)$:
$$
\omega_k(p)=\sqrt{\mu_k^2+\tau^2\sin^2p},\quad
\beta_{k,+}=\begin{pmatrix}\cos\frac{\theta_k}2\\ \sin\frac{\theta_k}2\end{pmatrix},\quad
\beta_{k,-}=\begin{pmatrix}-\sin\frac{\theta_k}2\\ \cos\frac{\theta_k}2\end{pmatrix},\quad
\tan\theta_k(p)=-\frac{\tau\sin p}{\mu_k},\ \ \lvert\theta_k\rvert<\frac\pi2 .
\tag{5.6}
$$
The two branches are separated by the gap $2\mu_k\ge2m>0$. The spinors $\beta_{k,\pm}$ are real, of unit length, and smooth and $2\pi$-periodic in $p$.

### Step 4. Background positions and velocity (item 2(b))

Every place has $u_{(x,s,a)}=\kappa x(X-\langle X\rangle)\chi=\kappa x\,e$, with $\lVert e\rVert=\sigma_X$. Hence $d_0(h,h')=\kappa\sigma_X\lvert x-x'\rvert$: this is (17.15) of 03-ilang-space/17-motion with $X\to\kappa X$. By (5.11) of 03-ilang-space/05-recording-contract, all places with the same $x$ are one background point. With $P(x;\lambda):=\sum_{s,a}p_{(x,s,a)}(\lambda)$,
$$
\bar x(\lambda)=\kappa\,\bar x_{\rm lab}(\lambda)\,e,\qquad\bar x_{\rm lab}:=\sum_xx\,P(x;\lambda),\qquad v=\kappa\,\frac{\mathrm d\bar x_{\rm lab}}{\mathrm d\lambda}\,e .
\tag{5.7}
$$
**Start.** $\phi=\psi_{k,+}\otimes\lvert f_k\rangle$, with $\psi_{k,+}:=\int\frac{\mathrm dp}{2\pi}g(p)\lvert p\rangle\otimes\beta_{k,+}(p)$. Under [W], (5.6) gives the evolved amplitude $G(p,\lambda)=g(p)e^{-i\omega_k(p)\lambda}\beta_{k,+}(p)$.

**Mean position.** By Step 2,
$$\bar x_{\rm lab}=\int\frac{\mathrm dp}{2\pi}G^\dagger i\partial_pG=\int\frac{\mathrm dp}{2\pi}\bigl(ig^*\partial_pg+\lambda\lvert g\rvert^2\partial_p\omega_k+i\lvert g\rvert^2\beta_{k,+}^T\partial_p\beta_{k,+}\bigr).$$
The last term vanishes, because $\beta_{k,+}$ is real and of unit length. So $\bar x_{\rm lab}$ is exactly linear in $\lambda$, with the group velocity as slope; the last equality is [S]:
$$
\frac{\mathrm d\bar x_{\rm lab}}{\mathrm d\lambda}=\int_{-\pi}^{\pi}\frac{\mathrm dp}{2\pi}\lvert g(p)\rvert^2\,\partial_p\omega_k(p)=\partial_p\omega_k(p_0)+O(\delta p^2).
\tag{5.8}
$$
With $\hat e:=e/\sigma_X$:
$$
v=\kappa\sigma_X\,w_k(p_0)\,\hat e,\qquad w_k(p):=\partial_p\omega_k(p)=\frac{\tau^2\sin p\cos p}{\sqrt{\mu_k^2+\tau^2\sin^2p}} .
\tag{5.9}
$$

### Step 5. Largest velocity and the bound (17.7) of 03-ilang-space/17-motion (item 2(b))

**Largest velocity.** With $u=\sin^2p_0$, $w_k^2=\tau^4u(1-u)/(\mu_k^2+\tau^2u)$. It vanishes at $u=0$ and $u=1$, and its $u$-derivative vanishes iff $\tau^2u^2+2\mu_k^2u-\mu_k^2=0$. The root in $[0,1]$ is $u_*=\mu_k/(\mu_k+W_k)$, with $W_k:=\sqrt{\mu_k^2+\tau^2}$. There $\mu_k^2+\tau^2u_*=\mu_kW_k$ and $1-u_*=W_k/(\mu_k+W_k)$, so $w_k^2=\tau^4/(\mu_k+W_k)^2$:
$$
\max_{p_0}\lVert v\rVert=\kappa\sigma_X\,(W_k-\mu_k)=\kappa\sigma_X\frac{\tau^2}{\sqrt{\mu_k^2+\tau^2}+\mu_k},\qquad\text{attained at }\sin^2p_0=\frac{\mu_k}{\mu_k+W_k}.
\tag{5.10}
$$
**The bound.**
- $A$ has $\lvert t_{hh'}\rvert=\tau/2$ between $(x,s,a)$ and $(x\pm1,\bar s,a)$, $\bar s\ne s$.
- The off-diagonal entries of $B$ come from $D$, which is not diagonal on the clock places. They join places with equal $x$, where $d_0=0$.
- Hence $a_{hh'}=\frac\tau2\kappa\sigma_X$ on the edges $(x,s,a)$–$(x\pm1,\bar s,a)$, and $a_{hh'}=0$ otherwise.
- This graph is the disjoint union of $2N$ chains $\{(x,s,a):x+s\equiv c\ (\mathrm{mod}\ 2)\}$, along which $d_0=\kappa\sigma_X\lvert x-x'\rvert$.
- $a$ depends only on $\lvert t_{hh'}\rvert$ and $d_0$. Each chain is therefore that of (17.20) of 03-ilang-space/17-motion, with $t=\tau/2$ and $X\to\kappa X$.

$\varrho(a)$ is the maximum over the components, so
$$
v_{\max}=\tau\,\kappa\sigma_X=\max_h\sum_{h'}\lvert t_{hh'}\rvert\,d_0(h,h') .
\tag{5.11}
$$
On $\Lambda_L$, $\varrho(a)=\tau\kappa\sigma_X\cos\frac{\pi}{2L+2}\to\tau\kappa\sigma_X$.

In units of $\kappa\sigma_X$: $v$ is $w_k(p_0)$, its largest value is $W_k-\mu_k$, and the bound is $\tau$. The bound is not reached, since $(W_k-\mu_k)/\tau=\sqrt{(W_k-\mu_k)/(W_k+\mu_k)}<1$.

### Step 6. Clock reading of the moving clock, to first order in $\epsilon$ (item 3)

**Start.** $\phi=\psi_{0,+}\otimes\lvert0\rangle$: the clock label is exactly in the place $\lvert0\rangle$, and the branch spinor is that of level $0$. Write $\beta_0:=\beta_{0,+}$, and $\beta_{0,-}$ for the other spinor of level $0$. Since $\lvert0\rangle=N^{-1/2}\sum_k\lvert f_k\rangle$, (5.5), [W] and Step 2 give
$$
q(a;\lambda)=\int\frac{\mathrm dp}{2\pi}\lvert g(p)\rvert^2\,\lVert\Phi_a(p,\lambda)\rVert^2,\qquad
\Phi_a:=\frac1N\sum_{k=0}^{N-1}e^{2\pi ika/N}\,e^{-iC_k(p)\lambda}\beta_0(p).
\tag{5.12}
$$
**Order counting.** A phase $d_k\lambda=2\pi k\vartheta_0/N$ is $O(1)$ and is kept exactly. A term $\propto d_k^n\lambda/m^{n-1}$ of a phase is of order $\epsilon^{n-1}$. The quadratic term of $\omega_k$ in $d_k$ is therefore first order and is kept; the cubic term is second order.

*(i) Branch mixing: second order.*
- $\beta_{k,\pm}$ is $\beta_{0,\pm}$ rotated by $-\delta_k$, with $\delta_k:=(\theta_0-\theta_k)/2$. Since $\lvert\partial_\mu\theta\rvert=\tau\lvert\sin p\rvert/(\mu^2+\tau^2\sin^2p)\le1/(2\mu)$, we have $\lvert\delta_k\rvert\le d_k/(4m)\le\delta:=(N-1)\epsilon/4$.
- Expanding $\beta_0=\cos\delta_k\,\beta_{k,+}+\sin\delta_k\,\beta_{k,-}$ and rotating back gives $e^{-iC_k\lambda}\beta_0=\bigl(e^{-i\omega_k\lambda}+2i\sin^2\delta_k\sin\omega_k\lambda\bigr)\beta_0+i\sin2\delta_k\sin\omega_k\lambda\,\beta_{0,-}$.
- Since $\beta_0\perp\beta_{0,-}$, the term of first order in $\delta_k$ enters $\lVert\Phi_a\rVert^2$ only squared. With $\mathcal A_a(p,\lambda):=\frac1N\sum_ke^{2\pi ika/N-i\omega_k(p)\lambda}$: $\bigl\lvert\lVert\Phi_a\rVert^2-\lvert\mathcal A_a\rvert^2\bigr\rvert\le8\delta^2+4\delta^4\le(N\epsilon)^2$, uniformly in $\lambda$ and $p$.
- The same holds for any other reference level used to define "the positive branch".

*(ii) Dispersion about the band centre.* Let $\bar d:=\frac1N\sum_kd_k=\frac{N-1}2m\epsilon=\langle0\vert D\vert0\rangle$, $\bar\mu:=m+\bar d$ and $j:=k-\frac{N-1}2$. Then $d_k-\bar d=m\epsilon j$, and the values of $j$ are symmetric about $0$. Taylor's theorem in $d$ about $\bar d$, with $\partial_d\omega=\mu/\omega$, $\partial_d^2\omega=\tau^2\sin^2p/\omega^3$ and $\lvert\partial_d^3\omega\rvert=3\mu\tau^2\sin^2p/\omega^5\le1/m^2$, gives
$$
\omega_k=\bar\omega+\Gamma\,m\epsilon j+\tfrac12\bar c\,(m\epsilon j)^2+\bar\rho_k,\qquad
\bar\omega(p):=\sqrt{\bar\mu^2+\tau^2\sin^2p},\quad\Gamma(p):=\frac{\bar\mu}{\bar\omega(p)},\quad\bar c:=\frac{\tau^2\sin^2p}{\bar\omega^3}\le\frac1m,\quad\lvert\bar\rho_k\rvert\le\frac{(m\epsilon\lvert j\rvert)^3}{6m^2}.
\tag{5.20}
$$
- With $m\epsilon\lambda=\frac{2\pi}N\vartheta_0$: $\omega_k\lambda=\bar\omega\lambda+\frac{2\pi}Nj\,\Gamma\vartheta_0+\zeta j^2+\bar\rho_k\lambda$, where $\zeta:=\frac{\pi\epsilon}N\,m\bar c\,\vartheta_0\le\frac{\pi\epsilon}N\vartheta_0$ and $\lvert\bar\rho_k\rvert\lambda\le\frac\pi{24}(N\epsilon)^2\vartheta_0$ (using $\lvert j\rvert<N/2$).
- The factors $e^{-i\bar\omega\lambda}$ and $e^{i\pi(N-1)a/N}$ are common to all $k$ and drop out of the modulus. With $y:=a-\Gamma\vartheta_0$, $\lvert\mathcal A_a\rvert=\bigl\lvert\frac1N\sum_je^{2\pi ijy/N}e^{-i\zeta j^2}e^{-i\bar\rho_k\lambda}\bigr\rvert$.
- Expand $e^{-i\zeta j^2}=1-i\zeta j^2+O(\zeta^2j^4)$. By the symmetry $j\to-j$, both $\mathcal D(y):=\frac1N\sum_je^{2\pi ijy/N}=\frac{\sin\pi y}{N\sin(\pi y/N)}$ and $\mathcal E(y):=\frac1N\sum_jj^2e^{2\pi ijy/N}$ are real. Hence $\lvert\mathcal D-i\zeta\mathcal E\rvert^2=\mathcal D^2+\zeta^2\mathcal E^2$: about the band centre the curvature has no first-order effect. Its whole first-order effect is the change of the rate from $\gamma=m/\omega_0$ to $\Gamma$.
- By (5.1), $\mathcal D(y)^2=P_N(a;\Gamma\vartheta_0)$. The terms of order $\zeta^2$ change $\lvert\mathcal A_a\rvert^2$ by at most $\frac14(N\epsilon\vartheta_0)^2$, and the phases $\bar\rho_k\lambda$ by at most $\frac\pi{12}(N\epsilon)^2\vartheta_0$ to leading order.

*(iii) Spread.* By (i) and (ii), $\lVert\Phi_a(p,\lambda)\rVert^2=P_N(a;\Gamma(p)\vartheta_0)$ up to the stated terms, uniformly in $p$. Apply [S] to $F(p)=P_N(a;\Gamma(p)\vartheta_0)$. From $P_N=\mathcal D^2$, $\lvert\mathcal D'\rvert\le\frac\pi2$ and $\lvert\mathcal D''\rvert\le\frac{\pi^2}3$: $\lvert\partial_\vartheta P_N\rvert\le\pi$ and $\lvert\partial_\vartheta^2P_N\rvert\le\frac{7\pi^2}6$. With $r:=\tau/\bar\mu$ and $\Gamma=(1+r^2\sin^2p)^{-1/2}$: $\lvert\Gamma'\rvert\le\frac{r^2}2$ and $\lvert\Gamma''\rvert\le r^2+\frac34r^4$. The replacement $\Gamma(p)\to\Gamma(p_0)$ therefore costs at most
$$\mathcal R_{\rm S}:=\frac{\delta p^2}2\Bigl[\frac{7\pi^2}{24}\,r^4\vartheta_0^2+\pi\bigl(r^2+\tfrac34r^4\bigr)\vartheta_0\Bigr],$$
uniformly in $p_0$; the bound does not vanish at the stationary points $p_0=0,\pm\frac\pi2$ of $\Gamma$.

**Result.**
$$
q(a;\lambda)=P_N\Bigl(a;\Gamma(p_0)\frac{\lambda}{\lambda_0}\Bigr)+\mathcal R,\qquad
\Gamma(p_0)=\frac{\bar\mu}{\sqrt{\bar\mu^2+\tau^2\sin^2p_0}}=\gamma\Bigl[1+\frac{(N-1)\epsilon}2\bigl(1-\gamma^2\bigr)\Bigr]+O\bigl((N\epsilon)^2\bigr),\quad\gamma:=\frac m{\omega_0(p_0)} .
\tag{5.21}
$$
This is (2.13) with $\lambda$ replaced by $\Gamma(p_0)\lambda$, and $\bar\mu=m+\frac{\pi(N-1)}{N\lambda_0}$. The neglected terms $\mathcal R$ are:
- in $\epsilon$, second order only: $\lvert\mathcal R_\epsilon\rvert\le(N\epsilon)^2(1+\vartheta_0)^2$, from branch mixing ($\le(N\epsilon)^2$), the quadratic dispersion at second order ($\le\frac14(N\epsilon\vartheta_0)^2$) and the cubic dispersion ($\lesssim\frac\pi{12}(N\epsilon)^2\vartheta_0$);
- in the spread: $\mathcal R_{\rm S}=O(\delta p^2)$;
- in the records: $O(\kappa)$, from [W].

No term of first order in $\epsilon$ is neglected. The v1 form $P_N(a;\gamma\lambda/\lambda_0)$ differs from (5.21) by up to $\pi(\Gamma-\gamma)\vartheta_0=O(N\epsilon\,\vartheta_0)$: it is the leading order only.

### Step 7. Tick and ratio (item 3)

By (5.1), $P_N(\cdot\,;\vartheta)$ is concentrated on a single place iff $\vartheta\in\mathbb Z$, and then on $a\equiv\vartheta$. To first order, (5.21) is the distribution (2.13) of a uniform ring clock with tick $\lambda_0/\Gamma(p_0)$: starting from $a=0$, it gives $a=1$ with certainty first at $\Gamma(p_0)\lambda=\lambda_0$. Since $\Gamma(0)=1$, with $\omega:=\omega_0(p_0)$:
$$
\lambda_0(p_0)=\frac{\lambda_0}{\Gamma(p_0)},\qquad
R(p_0):=\frac{\lambda_0(p_0)}{\lambda_0(0)}=\frac{\bar\omega(p_0)}{\bar\mu}=\sqrt{1+\frac{\tau^2}{\bar\mu^2}\sin^2p_0}=\frac\omega m\Bigl[1-\frac{(N-1)\epsilon}2\,\frac{\tau^2\sin^2p_0}{\omega^2}\Bigr]+O\bigl((N\epsilon)^2\bigr)\ \ge1 .
\tag{5.22}
$$
Relative to its rest rate, the carried clock advances at the rate $\Gamma(p_0)=1/R\le1$. The rate equals $1$ only for $\sin p_0=0$. It is smallest at $p_0=\pm\pi/2$, where it equals $\bar\mu/\bar W$, with $\bar W:=\sqrt{\bar\mu^2+\tau^2}$. At leading order, $R=\omega/m$ (v1).

**Nominal tick and actual reading.** From $\mathcal D(y)^2=1-\frac{\pi^2(N^2-1)}{3N^2}y^2+O(y^4)$ and (5.21), with $\xi:=\lambda/\lambda_0(p_0)-1$:
$$
1-q(1;\lambda)=\frac{\pi^2(N^2-1)}{3N^2}\,\xi^2+O(\xi^4)-\mathcal R .
\tag{5.23}
$$
- At $\lambda=\lambda_0(p_0)$ the reading gives $a=1$ with probability $1-O((N\epsilon)^2)$: with certainty to first order, not exactly.
  - The dispersion part of the deficit is $\zeta^2\bigl(\overline{j^4}-\overline{j^2}^{\,2}\bigr)=\frac{\pi^2\epsilon^2(N^2-1)(N^2-4)}{180N^2}\frac{m^2}{\bar\mu^2}(1-\Gamma^2)^2$. For $N\ge3$ and $\sin p_0\ne0$ it is nonzero: the levels $\omega_k$ are not equally spaced, so no $\lambda$ near $\lambda_0(p_0)$ gives exact certainty (cf. (2.9)).
  - Branch mixing adds a deficit of the same order, for every $N$; it is not computed.
- By (5.23), $q(1;\lambda)=1-O((N\epsilon)^2)$ in the whole window $\lvert\xi\rvert\lesssim N\epsilon$. The condition "certain up to the neglected order" thus locates the tick only to relative $O(N\epsilon)$, the size of the first-order term of (5.22).
- That term is fixed by the rate $\Gamma$ in (5.21): wherever $\partial_\vartheta P_N\ne0$, $q$ depends on $\Gamma$ at first order, with a second-order remainder. $\lambda_0(p_0)$ is the tick of the first-order distribution (5.21) in this sense.

### Step 8. Velocity of the item-3 start, to first order in $\epsilon$ (item 4)

The item-3 start is a superposition of all levels, and at level $k\ne0$ it is not purely in the positive branch (Step 6(i)). By Step 2(c), $\mathrm d\bar x_{\rm lab}/\mathrm d\lambda$ is the expectation of $i[C_b,x]$, which acts on $\lvert p\rangle\otimes\cdot\otimes\lvert f_k\rangle$ as $\partial_pC_k(p)=-\tau\cos p\,\sigma_1$. By (5.6), $\beta_{k,\pm}^T(\partial_pC_k)\beta_{k,\pm}=\pm w_k$ and $\beta_{k,+}^T(\partial_pC_k)\beta_{k,-}=-\tau\cos p\cos\theta_k$. The amplitude of level $k$ is $N^{-1/2}g\,(\cos\delta_k\,e^{-i\omega_k\lambda}\beta_{k,+}+\sin\delta_k\,e^{i\omega_k\lambda}\beta_{k,-})$. Hence, exactly under [W],
$$
\frac{\mathrm d\bar x_{\rm lab}}{\mathrm d\lambda}=\int\frac{\mathrm dp}{2\pi}\lvert g\rvert^2\,\frac1N\sum_{k=0}^{N-1}\Bigl[\cos2\delta_k\,w_k-\sin2\delta_k\,\tau\cos p\,\cos\theta_k\,\cos(2\omega_k\lambda)\Bigr].
\tag{5.24}
$$
- *Constant part.* $\cos2\delta_k=1-O((N\epsilon)^2)$. Write $w_k=w(p;\mu_k)$, with $w(p;\mu):=\tau^2\sin p\cos p/\sqrt{\mu^2+\tau^2\sin^2p}$ and $\lvert\partial_\mu^2w\rvert\le2\tau/\mu^2$. Since $\frac1N\sum_k(d_k-\bar d)=0$, the level average is $w(p;\bar\mu)$ up to $\tau\epsilon^2\frac{N^2-1}{12}$. [S] replaces $p$ by $p_0$.
- *Oscillating part.* It is of first order: to that order it equals $w_0\gamma^2\frac1N\sum_k\frac{d_k}m\cos(2\omega_k\lambda)$, and it cancels the first-order change of the constant part at $\lambda=0$. Its frequencies are $2\omega_k\ge2m$, so its mean vanishes. Its contribution to $\bar x_{\rm lab}(\lambda)-\bar x_{\rm lab}(0)$ is bounded for all $\lambda$ by $\frac1N\sum_k\frac{\tau d_k}{2m}\frac1{2\omega_k}\le\frac{(N-1)\epsilon\,\tau}{8m}$.

**The velocity used in item 4** is the mean of $v$ over $[0,\lambda]$, with $\lambda$ of the order of the tick. With $\frac1{m\lambda}=\frac{N\epsilon}{2\pi\vartheta_0}$:
$$
\bar v:=\frac{\bar x(\lambda)-\bar x(0)}{\lambda}=\kappa\sigma_X\,\bar w(p_0)\,\hat e+\mathcal R_v,\qquad\bar w(p):=w(p;\bar\mu)=\frac{\tau^2\sin p\cos p}{\sqrt{\bar\mu^2+\tau^2\sin^2p}} .
\tag{5.25}
$$
- Neglected, in units of $\kappa\sigma_X$: $\tau\,O((N\epsilon)^2)$ from the constant part; at most $\tau\frac{N(N-1)\epsilon^2}{16\pi\vartheta_0}$ from the oscillating part; $\frac12\delta p^2\sup_p\lvert\partial_p^2\bar w\rvert$ from [S]. No term of first order in $\epsilon$ is neglected.
- To first order, $\bar w$ is the velocity (5.9) of item 2(b) at $\mu_k\to\bar\mu$, the average of the $w_k$ over the levels. A single level differs from it at first order: $w_k=\bar w\,[1-\Gamma^2(d_k-\bar d)/\bar\mu]+O(\epsilon^2)$.
- The instantaneous $v(\lambda)$ of 03-ilang-space/17-motion differs from $\bar v$ at first order, by the oscillating part of (5.24).

### Step 9. Rate and velocity (item 4)

Below, $\bar w:=\lvert\bar w(p_0)\rvert=\lVert\bar v\rVert/(\kappa\sigma_X)$, and by (5.11) $\lVert\bar v\rVert/v_{\max}=\bar w/\tau$. The relations are identities between the first-order expressions (5.22) and (5.25). They are exact in $p_0$ and $\tau/\bar\mu$, and hold up to the second-order terms of (5.21) and (5.25).

**Exact relation on the chain.** By (5.22), $\tau^2\sin^2p_0=\bar\mu^2(R^2-1)$ and $\tau^2\cos^2p_0=\bar W^2-\bar\mu^2R^2$. Inserting these in $\bar w^2=\tau^4\sin^2p_0\cos^2p_0/\bar\omega^2$ gives
$$
\bar\mu^2R^4-(\bar W^2+\bar\mu^2-\bar w^2)R^2+\bar W^2=0 .
\tag{5.26}
$$
The discriminant is $\bigl((\bar W-\bar\mu)^2-\bar w^2\bigr)\bigl((\bar W+\bar\mu)^2-\bar w^2\bigr)$. It is nonnegative iff $\bar w\le\bar W-\bar\mu$, consistent with (5.10) at $\mu_k\to\bar\mu$. The roots are
$$
R=\frac{\sqrt{(\bar W+\bar\mu)^2-\bar w^2}\mp\sqrt{(\bar W-\bar\mu)^2-\bar w^2}}{2\bar\mu},\qquad\text{sign }-\text{ for }\sin^2p_0\le\frac{\bar\mu}{\bar\mu+\bar W},\quad\text{sign }+\text{ for }\sin^2p_0\ge\frac{\bar\mu}{\bar\mu+\bar W}.
\tag{5.27}
$$
- Squaring confirms the roots; their product is $\bar W/\bar\mu$.
- $R$ increases with $\sin^2p_0$, while $\bar w$ increases up to $\sin^2p_0=\bar\mu/(\bar\mu+\bar W)$ and then decreases (Step 5); this fixes the sign.
- At $\bar w=0$: $R=1$ ($p_0=0,\pi$) and $R=\bar W/\bar\mu$ ($p_0=\pm\pi/2$). At $\bar w=\bar W-\bar\mu$ both roots equal $\sqrt{\bar W/\bar\mu}$.
- So $R$ is a two-valued function of $\lVert\bar v\rVert$. The branch through $p_0=0$ has the sign $-$.
- Near the turning point $\bar w=\bar W-\bar\mu$, $\partial R/\partial\bar w$ diverges: there (5.27) is accurate to $O(N\epsilon)$ only, while (5.26) keeps its accuracy.

**Small $p_0$.** $\bar\omega=\bar\mu+\frac{\tau^2p_0^2}{2\bar\mu}+O(p_0^4)$ and $\bar w=\frac{\tau^2\lvert p_0\rvert}{\bar\mu}+O(\lvert p_0\rvert^3)$. Let $v_{\rm top}:=\kappa\sigma_X(\bar W-\bar\mu)$ be the largest velocity (5.10) at $\mu_k\to\bar\mu$. Then
$$
R=1+\frac{\lVert\bar v\rVert^2}{2v_{\max}^2}+O(p_0^4)=1+\frac{\bar W-\bar\mu}{2(\bar W+\bar\mu)}\,\frac{\lVert\bar v\rVert^2}{v_{\rm top}^2}+O(p_0^4).
\tag{5.28}
$$
**Comparison.** Exactly, $1-\bar w^2/\tau^2=(\bar\mu^2+\tau^2\sin^4p_0)/\bar\omega^2$. Hence
$$
R^2\Bigl(1-\frac{\lVert\bar v\rVert^2}{v_{\max}^2}\Bigr)=1+\frac{\tau^2}{\bar\mu^2}\sin^4p_0 .
\tag{5.29}
$$
- In units of the bound $v_{\max}$ of (17.7) of 03-ilang-space/17-motion, the coefficient in (5.28) is $\frac12$ for all $\tau$, $m$, $N$ and $\epsilon$ (to the order kept). In units of the largest velocity $v_{\rm top}$ it depends on $\tau/\bar\mu$.
- By (5.29), $R=(1-\lVert\bar v\rVert^2/v_{\max}^2)^{-1/2}\sqrt{1+\tau^2\sin^4p_0/\bar\mu^2}$. The correction factor is $1+O(p_0^4)$.
  - At fixed $\lVert\bar v\rVert/v_{\max}<1$ on the branch through $p_0=0$, the factor tends to $1$ as $\bar\mu/\tau\to0$, since there $\sin p_0=O(\bar\mu/\tau)$.
- Always $\lVert\bar v\rVert/v_{\max}\le(\bar W-\bar\mu)/\tau<1$: the bound is never reached, and $R$ stays finite. On the branch through $p_0=0$, $R\le\sqrt{\bar W/\bar\mu}$, with equality at $v_{\rm top}$.
- With $\bar\mu\to m$, (5.26)–(5.29) are the v1 relations; those hold at leading order in $\epsilon$ only, for $R$ and for $\lVert\bar v\rVert$ alike.

## Result

- **(5.2)** Decoupled body: $q(a;\lambda)=\lvert\langle a\vert e^{-iD'\lambda}\vert\eta\rangle\rvert^2$, exactly, for all $\lambda$. It does not depend on $H$, $\phi_X$, the $K_x$ or $\chi$.
- **(5.3)** The infinite chain is the limit $L\to\infty$, taken first at fixed $\lambda$. It is controlled by the locality bound (5.3) and, for $\bar x$ and $v$, by $\lVert i[C,x]\rVert\le\tau$.
- **(5.4)** $AB+BA=0$ and $C_b^2=A^2+B^2$.
- **(5.6)** For each level $d_k$, the eigenvalues are $\pm\omega_k(p)=\pm\sqrt{(m+d_k)^2+\tau^2\sin^2p}$, with eigenvectors $\lvert p\rangle\otimes\beta_{k,\pm}(p)\otimes\lvert f_k\rangle$.
- **(5.9)** $v=\kappa\sigma_X\,w_k(p_0)\hat e$, with $w_k=\tau^2\sin p_0\cos p_0/\omega_k(p_0)$.
- **(5.10)** Largest velocity: $\kappa\sigma_X(\sqrt{\mu_k^2+\tau^2}-\mu_k)$.
- **(5.11)** Bound (17.7) of 03-ilang-space/17-motion: $v_{\max}=\tau\kappa\sigma_X$, strictly larger.
- **(5.20)** $\bar\mu=m+\langle0\vert D\vert0\rangle=m+\frac{\pi(N-1)}{N\lambda_0}$, $\bar\omega=\sqrt{\bar\mu^2+\tau^2\sin^2p}$, rate $\Gamma=\bar\mu/\bar\omega$.
- **(5.21)** To first order in $\epsilon$: $q(a;\lambda)=P_N(a;\Gamma(p_0)\lambda/\lambda_0)$, i.e. (2.13) with $\lambda\to\Gamma(p_0)\lambda$; remainder $O((N\epsilon)^2(1+\lambda/\lambda_0)^2)+O(\delta p^2)+O(\kappa)$.
- **(5.22)** $\lambda_0(p_0)=\lambda_0/\Gamma(p_0)$ and $R=\lambda_0(p_0)/\lambda_0(0)=\sqrt{1+(\tau/\bar\mu)^2\sin^2p_0}=\frac\omega m\bigl[1-\frac{(N-1)\epsilon}2\frac{\tau^2\sin^2p_0}{\omega^2}\bigr]+O((N\epsilon)^2)$.
- **(5.23)** Near the tick, $1-q(1;\lambda)=\frac{\pi^2(N^2-1)}{3N^2}\xi^2+O((N\epsilon)^2)$: the certainty is exact only to first order, and the certainty condition alone fixes the tick only to relative $O(N\epsilon)$.
- **(5.24)** Velocity of the item-3 start: a constant part and a first-order part oscillating at the frequencies $2\omega_k$.
- **(5.25)** Mean velocity over the tick: $\bar v=\kappa\sigma_X\bar w(p_0)\hat e$, $\bar w=\tau^2\sin p_0\cos p_0/\bar\omega(p_0)$, to first order in $\epsilon$.
- **(5.26)** Exact relation on the chain: $\bar\mu^2R^4-(\bar W^2+\bar\mu^2-\bar w^2)R^2+\bar W^2=0$.
- **(5.27)** $R=\bigl[\sqrt{(\bar W+\bar\mu)^2-\bar w^2}\mp\sqrt{(\bar W-\bar\mu)^2-\bar w^2}\bigr]/(2\bar\mu)$, a two-valued function of $\bar w=\lVert\bar v\rVert/(\kappa\sigma_X)$.
- **(5.28)** Small $p_0$: $R=1+\lVert\bar v\rVert^2/(2v_{\max}^2)+O(p_0^4)$; the coefficient is universal only in units of the bound $v_{\max}$.
- **(5.29)** $R^2(1-\lVert\bar v\rVert^2/v_{\max}^2)=1+(\tau/\bar\mu)^2\sin^4p_0$.

## Consistency checks

1. **Dimensions.** $[\tau]=[m]=[d_k]=[\bar\mu]=[C]=[\lambda]^{-1}$, so $\epsilon$, $\Gamma$, $R$, $\zeta$, $m\bar c$ and $\bar w/\tau$ are dimensionless. $[v]=[\kappa\sigma_X][C]$, consistent with (5.9)–(5.11) and (5.25)–(5.29). Passed.
2. **$p_0=0$ and $\tau\to0$.**
   - At $p=0$, $C_k(0)=\mu_k\sigma_3$ and $\beta_0(0)=(1,0)^T$ is an eigenvector of every $C_k(0)$. (5.12) then gives exactly $P_N(a;\lambda/\lambda_0)$, i.e. (2.13), with no $\epsilon$ expansion. In (5.21): $\Gamma(0)=1$, $\delta_k=0$, $\bar c=0$ and $\bar\rho_k=0$, so no $\epsilon$-remainder. This agrees with (5.2).
   - Also $w_k(0)=0$ and $\sin2\delta_k=0$ in (5.24), so $v=0$; and $R=1$ by (5.27) with sign $-$.
   - For $\tau\to0$ the same holds for every $p_0$, and (5.10) and (5.11) tend to $0$. Passed.
3. **$N=2$.** Without branch mixing, (5.12) gives exactly $q(0;\lambda)=\cos^2\bigl((\omega_1-\omega_0)\lambda/2\bigr)$, and (5.21) gives $P_2(0;\vartheta)=\cos^2(\pi\vartheta/2)$ with $\vartheta=\Gamma\lambda/\lambda_0$. With $d_1=\pi/\lambda_0$ they agree iff $(\omega_1-\omega_0)/d_1=\Gamma$. The secant slope differs from the midpoint derivative $\Gamma$ by at most $\frac{d_1^2}{24}\sup\lvert\partial_d^3\omega\rvert\le\frac{\epsilon^2}{24}$, and $\Gamma=\gamma[1+\frac\epsilon2(1-\gamma^2)]+O(\epsilon^2)$ is the rate of the verifier's example. The v1 rate $\gamma$ fails this check at first order. No chirp remains, in agreement with the factor $N^2-4$ in Step 7. Passed.

## Open issues

- Terms of second order in $\epsilon$ are not computed: branch mixing, and the dispersion beyond (5.20). They make the certainty at the tick inexact at $O((N\epsilon)^2)$ and limit (5.21) to $N\epsilon\,\lambda/\lambda_0\ll1$, i.e. to fewer than about $1/(N\epsilon)$ ticks.
- The first-order term of the tick (5.22) is defined through the rate of the first-order distribution (5.21). The certainty condition of item 3, applied to the actual reading, resolves the tick only to relative $O(N\epsilon)$, by (5.23). Whether a later package needs a sharper operational definition of the tick is left to it.
- "In the branch of positive eigenvalue" is read as the product start with the spinor of level $0$, so that the clock label is exactly in $\lvert0\rangle$. The start projected on the positive branch of every level gives the same (5.21) and (5.25) to this order, without the oscillating part of (5.24); its clock label is in $\lvert0\rangle$ only up to $O((N\epsilon)^2)$ in the reading. This alternative is not worked out here.
- Effects of the records beyond leading order in $\kappa$ are out of scope: the drift from $\langle X\rangle$, and the decoherence of the view.
- (5.10) of 03-ilang-space/05-recording-contract assumes $\phi_h\ne0$ for all $h$, but the item-3 start has $\phi_h=0$ for $a\ne0$. Only $u_h$ and $d_0$ are used here, and they do not depend on $\phi$.
- The comparison of two carried clocks between two meetings is left to a later package.

## Methods used

- tensor-product factorization of commuting generators
- Dyson series and Duhamel formula (locality bound)
- bounded commutator $i[C,x]$ (velocity operator), strong convergence
- Fourier analysis on $\mathbb Z$, plane waves, position as $i\partial_p$
- anticommuting (Clifford) terms, $2\times2$ diagonalization, half-angle formulas
- group velocity, Hellmann–Feynman
- Taylor expansion with remainder bounds about the band centre, symmetry of the level sum
- geometric sums (Dirichlet kernel)
- spectral radius of path adjacency matrices

Author: Ivan Honis · ivan.honis@ndot.io · License: CC BY 4.0 · Human–AI note