Version 3 · current

# Inertia and internal energy: response at λ = 0 and averaged response

- **Subproject:** 04-ilang-time
- **Package:** 06-inertia-internal-energy
- **Version:** v3
- **Mode:** external regeneration
- **Date:** 2026-10-10

## Changes from previous version

- **Step 5e rewritten (I1).**
  - v2 gave the error of the average as $r=-\rho+R/\mathcal F$ with $R=O(T^2\dots)$, (6.19). That estimate is not valid for every interval allowed by (6.14): on intervals that start late, $\lambda_1\gg\Lambda$, the accumulated interbranch phase adds a term of order $T^3/\Lambda$. (6.19) is withdrawn; its tag is retired and not reused.
  - (6.14) keeps its quantity, its leading term and both conditions. Its error is now additive, $\mathcal E$.
  - New: (6.24), the exact first-order coefficient at a fixed interval; (6.25), an exact bound on the deviation of the velocity from the branch velocity; (6.26), an exact bound on $\mathcal E$ for every interval with $\mathcal F_{\max}T\le\pi/2$; (6.27), an order estimate of the part beyond first order, with the accumulated phase.
  - Every statement is marked as exact or as an order estimate, with its conditions (5e, item iv).
- **Step 6c (I2).** The sentence on the loss of factorization is replaced by two cases: a common eigenvector (product state, level-dependent scalar position dynamics) and a general internal start.
- **Step 6b.** The variation of the response with $\lambda$ is bounded exactly, including the record term. This is a precision; (6.17) is unchanged.
- Steps 1–4, 5a–5d, 5f, 6a, body E and Step 7 are unchanged in content, and so are the tags (6.1)–(6.17) and (6.20)–(6.23). (6.18) stays retired. Check 3 of v2 ((6.21) against (6.12)) remains valid; because of the limit of three checks it is replaced by a check of the new bounds.

## Response to verification

- **I1: Accepted and fixed in Step 5e ((6.14), (6.24)–(6.27)) and Step 5f.** The verifier's term is confirmed: the interbranch phase is $2M_k\lambda+\tau^2f^2\lambda^3/(3M_k)+\dots$, and expanding it gives a secular term of size up to $\tau^2\lvert f\rvert^3T^3/(3M_k^2\Lambda)$. The fix follows the first suggestion and does not restrict the intervals.
  - The linear coefficient $(1-\rho)$ is kept and stated as exact, (6.24).
  - The finite-force remainder is bounded by (6.26), an exact inequality for every $\lambda_1\ge0$. It bounds the endpoint oscillation by its amplitude, (6.25), and never expands the phase.
  - (6.27) shows the accumulated phase explicitly, as an order estimate. The v2 estimate is recovered only for $\lambda_1\lesssim\Lambda$.
  - The error is additive, including the record-induced part that does not depend on $F$.
- **I2: Accepted and fixed in Step 6c, and in the last bullet of Step 7.** Position–clock entanglement and level-dependent scalar position dynamics are now distinguished; body E at a fixed level is named as an example of the second without the first.

## Setup and assumptions

**Inputs** (as quoted in question.md).
- (2.9), (2.12) of 02-clock: a uniform ring clock with tick $\lambda_0$ and $\theta_m=0$ has $e^{-iC_K\lambda_0}=S$ and $C_K=\sum_k\frac{2\pi\ell_k}{N\lambda_0}\lvert f_k\rangle\langle f_k\rvert$ with $\ell_k\equiv k$. Its spread is minimal iff the $\ell_k$ are $N$ consecutive integers.
- (5.10) of 03-ilang-space/05-recording-contract: $\lvert u_h\rangle=(K_h-\langle K_h\rangle)\lvert\chi\rangle$, and $g(v,w)=\operatorname{Re}\langle v\vert w\rangle$.
- (17.15) of 03-ilang-space/17-motion: for $K_{h_m}=mX$, $u_{h_m}=m\,e$ with $e=(X-\langle X\rangle)\chi$, $\lVert e\rVert=\sigma_X>0$, $\bar x=\bar m\,e$ and $d_0(h_m,h_n)=\lvert m-n\rvert\sigma_X$.
- (20.10) of 03-ilang-space/20-cost-and-mass: for a uniform gradient, $a(0)=a_0+\mu F$ with $\mu=-\sum_{\{h,h'\}\in E}L_{hh'}(0)\,\Delta_{hh'}\,g(\Delta_{hh'},\cdot)$, where $L_{hh'}(0)=2\operatorname{Re}[t_{hh'}\phi_h^*\phi_{h'}]$ and $\Delta_{hh'}=u_h-u_{h'}$.

**Setting.** As in question.md:
- a body $b$, the only instance of its type, and a medium $c$;
- the contract $C=C_b\otimes\mathbb 1_c+\sum_h\lvert h\rangle\langle h\rvert\otimes K_h$, with $K_h$ depending only on the position label $x_h$;
- a product start $\phi\otimes\chi$, $\lambda\ge0$, and contracts that do not depend on $\lambda$.

Notation: $\hat x:=\sum_hx_h\lvert h\rangle\langle h\rvert$ on $\mathcal H_b$ and $\bar m:=\sum_hx_h\,p_h$. Clock places are labelled $a\in\mathbb Z_N$; these are the $\lvert m\rangle$ of 02-clock.

**Chains and limits (A1).**
- *Chain.* Model D, and the chain bodies of Step 6, are taken on $x\in\{-L,\dots,L\}$. The hopping terms act only between neighbours of the chain; every term diagonal in $x$ (on-site blocks, gradient, records) is unchanged.
- *Starts.* Compactly supported: $\phi_h=0$ for $\lvert x_h\rvert>W$, with $W<L$.
- *Propagation bound (standard).* The hopping part is nearest-neighbour with norm $\le\tau$ (Model D; $2\lvert t\rvert$ in Step 6), and every other term of $C$ is diagonal in $x$. Expand the evolution in powers of the hopping, in the interaction picture of the diagonal part. The terms of order $n<L-W$ are the same on the chain and on $\mathbb Z$, and the part of $\Psi(\lambda)$ at distance $>r$ from the support of the start has norm $\le\sum_{n>r}(\tau\lambda)^n/n!\le(e\tau\lambda/r)^r$.
- *Limit 1: $L\to\infty$*, at fixed start, $F$, $\kappa$ and $\lambda\le\lambda_{\max}$. The chain state converges in norm to the state on $\mathbb Z$, and the weighted tails $\sum_{\lvert x\rvert>W+r}\lvert x\rvert^jp_x$, $j=1,2$, are bounded by the same superexponentially small terms. Hence $\bar m$, $\dot{\bar m}$ and $\ddot{\bar m}$ converge uniformly on $[0,\lambda_{\max}]$. A vanishing tail weight alone would not give the convergence of $\bar m$; the compact support is what is used.
- *Limit 2: concentration*, $W\to\infty$ (Step 5c), taken after limit 1 at fixed $\lambda_1$, $\Lambda$, $F$, $\kappa$.
- *Then* first order in $F$ and leading order in $\kappa$, and last the long-interval limit $M_k\Lambda\to\infty$ (Steps 5e, 5f). The exact bounds (6.25) and (6.26) hold after limits 1 and 2, at finite $F$ and $\kappa$.

On $\mathbb Z$, the plane waves are $\lvert p\rangle=\sum_xe^{ipx}\lvert x\rangle$ with $p\in(-\pi,\pi]$, and $\Phi(x)=\int\frac{dp}{2\pi}e^{ipx}\tilde\Phi(p)$. In this representation $\hat S:=\sum_x\lvert x+1\rangle\langle x\rvert$ acts as $e^{-ip}$ and $\hat x$ acts as $i\partial_p$.

**Approximations, used in item 3 and Step 6 only.**
- (W) Weak records: Step 5a, and the record terms of (6.26), (6.27) in Step 5e.
- (F) First order in the force: Steps 5d–5f and 6.
- (Cst) Start concentrated at $p_0=0$: Step 5c.

Their ranges of validity are stated in (6.14), (6.26) and (6.27).

## Derivation

### Step 1. The coefficient (20.10) for any diagonal and any start

Let $\Pi_h:=\lvert h\rangle\langle h\rvert\otimes\mathbb 1_c$. Then $p_h=\langle\Psi\vert\Pi_h\vert\Psi\rangle$, and (1.9) gives $\ddot p_h(0)=-\langle[C,[C,\Pi_h]]\rangle_{\Psi(0)}$.

Split $C=C_{\rm d}+C_{\rm o}$ with $C_{\rm d}:=\sum_h\Pi_hC\Pi_h$. Since $[C_{\rm d},\Pi_h]=0$,
$$a(0)=-\sum_hu_h\langle[C_{\rm o},[C_{\rm o},\Pi_h]]+[C_{\rm d},[C_{\rm o},\Pi_h]]\rangle .$$
This is affine in $C_{\rm d}$, hence in the entries $t_{hh}$. Its linear part depends only on $C_{\rm o}$ and on the start, not on the base values of the $t_{hh}$.

Change $t_{hh}\to t_{hh}+\delta_h$ with $\delta_h-\delta_{h'}=-g(F,\Delta_{hh'})$. From a base with all $\Omega_h$ equal this is the uniform gradient of 20, and by (20.10) $a(0)$ changes by $\mu F$. By affinity the change is the same from any base:
$$
a(0)\big|_{t_{hh}+\delta_h}=a(0)\big|_{t_{hh}}+\mu F,\qquad \mu\ \text{as in (20.10) of 03-ilang-space/20-cost-and-mass}.
\tag{6.1}
$$
Both sides of (6.1) are polynomials in the start amplitudes. The identity therefore extends by continuity from starts with all $\phi_h\neq0$ to every start.

Model D and the bodies of Step 6 need both extensions: their diagonals are not uniform, and the starts used have zero amplitudes.

### Step 2. Item 1: a body with internal labels

For $h=(x,a)$ and $h'=(x',a')$ we have $t_{hh'}=\langle a\vert T_{xx'}\vert a'\rangle$ and $\phi_h=(\phi_x)_a$. Since $K_h$ depends only on $x$, (5.10) gives $u_h=u_x$. Hence $\Delta_{hh'}=\Delta_{xx'}:=u_x-u_{x'}$, which vanishes for $x=x'$.

Inserting this into (20.10):
$$
L_{hh'}(0)=2\operatorname{Re}\bigl[(\phi_x)_a^*\langle a\vert T_{xx'}\vert a'\rangle(\phi_{x'})_{a'}\bigr],\qquad
L_{xx'}(0):=\sum_{a,a'}L_{(x,a)(x',a')}(0)=2\operatorname{Re}\langle\phi_x\vert T_{xx'}\vert\phi_{x'}\rangle .
\tag{6.2}
$$
Pairs with $t_{hh'}=0$ add zero, so the sum may run over all $a,a'$. Then
$$
\mu=-\sum_{\{x,x'\},\,x\neq x'}L_{xx'}(0)\,\Delta_{xx'}\,g(\Delta_{xx'},\cdot\,).
\tag{6.3}
$$
Edges between places with the same $x$ carry edge cost, but they have $\Delta_{hh'}=0$ and do not contribute.

**Decoupled body** ($T_{xx'}=t_{xx'}\mathbb 1_N$ for $x\neq x'$, arbitrary $T_{xx}$, also position-dependent). For the start $\phi_X\otimes\eta$ with $\lVert\eta\rVert=1$, we have $\phi_x=\phi_X(x)\eta$ and $L_{xx'}(0)=2\operatorname{Re}[t_{xx'}\phi_X(x)^*\phi_X(x')]$. Hence
$$
\mu=-\sum_{\{x,x'\},\,x\neq x'}2\operatorname{Re}\bigl[t_{xx'}\phi_X(x)^*\phi_X(x')\bigr]\,\Delta_{xx'}\,g(\Delta_{xx'},\cdot\,).
\tag{6.4}
$$
This is the value of (20.10) for the body without internal labels, with hopping $t_{xx'}$ and start $\phi_X$. By (6.1) it holds for every choice of the $T_{xx}$, and it depends neither on $\eta$ nor on the $T_{xx}$.

### Step 3. Model D: clock, branches, rest cost, positions, gradient

**Clock.** $S\lvert f_k\rangle=e^{-2\pi ik/N}\lvert f_k\rangle=e^{-id_k\lambda_0}\lvert f_k\rangle$, so $e^{-iD\lambda_0}=S$. By (2.9), $D$ is the uniform ring clock with $\theta_m=0$ and $\ell_k=k$; by (2.12) with $c=0$, its spread is minimal. On branch $s$ it enters as $(\sigma_3)_{ss}D$.

In the place basis, use $\sum_{j=0}^{N-1}jz^j=N/(z-1)$ for $z^N=1\neq z$ (the derivative of the geometric sum):
$$
D_{aa}=\frac{\pi(N-1)}{N\lambda_0},\qquad D_{aa'}=\frac{2\pi}{N\lambda_0}\,\frac{1}{e^{2\pi i(a-a')/N}-1}\neq0\quad(a\neq a').
\tag{6.5}
$$

**Branches.** Every term of $C_b$, and also $\hat x$, commutes with $\mathbb 1\otimes\mathbb 1\otimes\lvert f_k\rangle\langle f_k\rvert$, so the clock level $k$ is conserved. On level $k$, $A$ acts on $\lvert p\rangle\otimes v$ as $\frac\tau2(-ie^{-ip}+ie^{ip})\sigma_1=-\tau\sin p\,\sigma_1$:
$$
C_b\bigl(\lvert p\rangle\otimes v\otimes\lvert f_k\rangle\bigr)=\lvert p\rangle\otimes C_k(p)v\otimes\lvert f_k\rangle,\qquad
C_k(p)=-\tau\sin p\,\sigma_1+M_k\sigma_3,
\tag{6.6}
$$
with eigenvalues $\pm\omega_k(p)$, $\omega_k(p)=\sqrt{M_k^2+\tau^2\sin^2p}$.

**Rest cost.** $C_k(0)=M_k\sigma_3$, so
$$
M_k:=m+d_k=m+\frac{2\pi k}{N\lambda_0}\ \ (\text{rest cost of level }k,\ \text{eigenvector }\lvert1\rangle),\qquad\text{gap }2M_k,\qquad \omega_k''(0)=\frac{\tau^2}{M_k}.
\tag{6.7}
$$

**Positions and gradient.**
- $K_x=\kappa xX$ is (17.15) with $X$ replaced by $\kappa X$. Hence $u_{(x,s,a)}=\kappa x\,e$, $\bar x=\kappa\bar m\,e$ and $\Delta_{hh'}=\kappa(x-x')e$.
- The record means $\epsilon_h=\kappa x\langle X\rangle$ depend on position. Let $F$ be the force of the uniform gradient of $\delta_h+\epsilon_h$, where $\delta_h$ is the change of $t_{hh}$.
- $F=0$ thus means that the $\delta_h$ cancel the record means, the analogue of "all $\Omega_h$ equal" in 20. The $B$-diagonal is the same at every $x$ and does not enter, by Step 1.

Then
$$
\delta_h+\epsilon_h=-\mathcal F\,x+\text{const},\qquad \mathcal F:=\kappa\,g(e,F),\qquad \bar x=\kappa\,\bar m\,e .
\tag{6.8}
$$
Only $g(e,F)$ enters, as in (20.10). If $F$ is meant as the force of the $\delta_h$ alone, replace $\mathcal F$ by $\mathcal F-\kappa\langle X\rangle$; the coefficients of $F$ below are unchanged.

### Step 4. Item 2: response at $\lambda=0$

The start has the amplitudes $\phi_{(x,s,a)}=\phi_X(x)\,\delta_{s1}\,N^{-1/2}e^{2\pi ika/N}$. The edges of $C_b$ are as follows; the records, $m\mathbb 1$ and the gradient are diagonal.
- **(A)** $\{(x,s,a),(x+1,s',a)\}$ with $s\neq s'$, and $t_{(x+1,s',a),(x,s,a)}=-i\tau/2$.
- **(B)** $\{(x,s,a),(x,s,a')\}$ with $a\neq a'$, and $t=(\sigma_3)_{ss}D_{aa'}$, which is nonzero by (6.5).

Every (A)-edge, and every (B)-edge with $s=2$, has an end with $s=2$, where $\phi=0$; there $L_{hh'}(0)=0$.

On the (B)-edges with $s=1$, let $n:=a-a'$ and $\vartheta:=2\pi n/N$. Since $\operatorname{Re}[e^{-ik\vartheta}/(e^{i\vartheta}-1)]=-\sin((k+\tfrac12)\vartheta)/(2\sin\tfrac\vartheta2)$,
$$
L_{(x,1,a)(x,1,a')}(0)=-\frac{2\pi}{N^2\lambda_0}\,\lvert\phi_X(x)\rvert^2\,\frac{\sin\bigl((2k+1)\pi n/N\bigr)}{\sin(\pi n/N)},\qquad L_{hh'}(0)=0\ \text{on all other edges.}
\tag{6.9}
$$
The (B)-edges lie inside one point and do not enter (6.3). On the (A)-blocks, $L_{x,x+1}(0)=2\operatorname{Re}[\phi_X(x)^*\phi_X(x+1)\,\tfrac{i\tau}{2}\langle1\vert\sigma_1\vert1\rangle]=0$. Hence
$$
\mu=0\qquad\text{for every }\phi_X,\ k,\ N,\ \tau,\ m,\ \kappa .
\tag{6.10}
$$

### Step 5. Item 3: averaged response

**5a. Records (W).** $C$ contains $c$ only through $\hat x\otimes X$, so $X$ is conserved. Decompose $\chi=\sum_\xi\Pi_\xi\chi$ over the eigenspaces of $X$, with weights $w_\xi:=\lVert\Pi_\xi\chi\rVert^2$. The view of $b$ is then exactly the $w_\xi$-mixture of the pure evolutions by $C_b-\mathcal F_\xi\hat x$ (+const), with
$$
\mathcal F_\xi:=\mathcal F-\kappa(\xi-\langle X\rangle),\qquad \sum_\xi w_\xi\mathcal F_\xi=\mathcal F,\qquad \sum_\xi w_\xi(\mathcal F_\xi-\mathcal F)^2=\kappa^2\sigma_X^2 .
\tag{6.11}
$$
The part of $\ddot{\bar m}$ that is linear in the sector force sums exactly to its value at $\mathcal F$, as for the single evolution by $C_b-\mathcal F\hat x$. The weak-record approximation drops the terms of higher order in the $\mathcal F_\xi$; their size is given in (6.26) and (6.27). It is used nowhere else.

**5b. Equations of motion.** By (1.9), $\frac{d}{d\lambda}\langle O\rangle=\langle i[C,O]\rangle$. In the $p$-representation on level $k$, $[f(p),\hat x]=-if'(p)$. Hence $\hat v:=i[C_k-\mathcal F\hat x,\hat x]=\partial_pC_k$ and $[\hat x,\hat v]=i\partial_p\hat v$, so that
$$
\dot{\bar m}=\langle\hat v\rangle,\quad \hat v=-\tau\cos\hat p\,\sigma_1;\qquad
\ddot{\bar m}=\langle\hat a_k\rangle,\quad \hat a_k=i[C_k,\hat v]-i\mathcal F[\hat x,\hat v]=2\tau M_k\cos\hat p\,\sigma_2+\mathcal F\tau\sin\hat p\,\sigma_1 .
\tag{6.12}
$$

**5c. Drift and concentrated start (Cst).**
- With $\hat x=i\partial_p$, (1.9) reads $(\partial_\lambda+\mathcal F\partial_p)\tilde\Phi=-iC_k(p)\tilde\Phi$.
- Along $p=q+\mathcal F\lambda$ this gives $\tilde\Phi(q+\mathcal F\lambda,\lambda)=w_q(\lambda)$ with $i\dot w_q=C_k(q+\mathcal F\lambda)w_q$. This is the standard drift $\dot p=\mathcal F$ of the plane-wave label.
- Hence $\ddot{\bar m}(\lambda)=\int\frac{dq}{2\pi}\,w_q^\dagger\,\hat a_k(q+\mathcal F\lambda)\,w_q$.
- *Start.* $\phi^{(W)}=\phi_X^{(W)}\otimes\lvert1\rangle\otimes\lvert f_k\rangle$ with $\phi_X^{(W)}(x)=c_W\varphi(x/W)$, where $\varphi$ is smooth with support in $[-1,1]$ and $c_W$ normalizes. Then $w_q(0)=\tilde\phi_W(q)\lvert1\rangle$ and $\lvert\tilde\phi_W(q)\rvert^2\frac{dq}{2\pi}\to\delta(q)$ for $W\to\infty$: the start is concentrated at $p_0=0$.
- *Branch.* The real eigenvector of the positive branch is $\lvert+_k(q)\rangle=\lvert1\rangle+O(\tau q/M_k)$, by (6.7). The start therefore lies in the positive branch up to a part of norm $O(\tau/(M_kW))$. The start $\tilde\phi_W(q)\lvert+_k(q)\rangle$, exactly in the positive branch, differs from it by the same amount and has the same limit, because $\hat a_k$ is bounded.
- *Limit 2.* The integrand is $\lvert\tilde\phi_W(q)\rvert^2$ times a function of $q$ that is continuous and bounded, uniformly on bounded $\lambda$-intervals. For $W\to\infty$ the integral tends to the value of the two-level problem $i\dot w=C_k(\mathcal F\lambda)w$ with $w(0)=\lvert1\rangle$.

**5d. First order (F).** $C_k(\mathcal F\lambda)=M_k\sigma_3-\tau\mathcal F\lambda\,\sigma_1+O(\mathcal F^3)$. To first order, $w_1=e^{-iM_k\lambda}$ and $w_2=e^{iM_k\lambda}\gamma$, with $\dot\gamma=i\tau\mathcal F\lambda\,e^{-2iM_k\lambda}$ and $\gamma(0)=0$. Then
$$\langle\sigma_2\rangle=2\operatorname{Im}(w_1^*w_2)=2\tau\mathcal F\int_0^\lambda\lambda'\cos\bigl(2M_k(\lambda-\lambda')\bigr)d\lambda'=\tau\mathcal F\sin^2(M_k\lambda)/M_k^2 .$$
The term $\mathcal F\tau\sin(\mathcal F\lambda)\langle\sigma_1\rangle$ is $O(\mathcal F^2)$.

Also $a_0(\lambda)=0$ at leading order in $\kappa$: at $F=0$ the state stays in the positive branch, and $\langle+\vert\sigma_2\vert+\rangle=0$ for real vectors. Using (6.12) and $\bar x=\kappa\bar m e$:
$$
a(\lambda)=\frac{2\tau^2}{M_k}\,\sin^2(M_k\lambda)\;\kappa^2\,g(e,F)\,e+O(F^2)\qquad(\text{fixed }\lambda).
\tag{6.13}
$$
The solution of the two-level problem is analytic in $\mathcal F$ at fixed $\lambda$, so the coefficient of $F$ in (6.13) is exact for every fixed $\lambda\ge0$.

**5e. Average.** Let $0\le\lambda_1<T:=\lambda_1+\Lambda$ and $\overline{(\cdot)}:=\frac1\Lambda\int_{\lambda_1}^{T}(\cdot)\,d\lambda$. Everything in 5e holds after limits 1 and 2. By 5a and 5c, $\bar m=\sum_\xi w_\xi\,\bar m_{\mathcal F_\xi}$, where $\bar m_f$ is the mean label of the two-level problem $i\dot w=C_k(f\lambda)w$, $w(0)=\lvert1\rangle$, so that $\dot{\bar m}_f=w^\dagger\hat v(f\lambda)\,w$. Two exact properties:
- $\dot{\bar m}_f(\lambda)$ is entire in $f$, because the equation is linear with coefficients entire in $f$;
- it is odd in $f$, because $\sigma_3C_k(q)\sigma_3=C_k(-q)$, $\sigma_3\hat v(q)\sigma_3=-\hat v(-q)$ and $\sigma_3\lvert1\rangle=\lvert1\rangle$.

**(i) Coefficient of $F$ at a fixed interval (exact).** By 5d, $\ddot{\bar m}_f=\frac{2\tau^2}{M_k}\sin^2(M_k\lambda)\,f+O(f^3)$. With (6.11), the part of $\overline{\ddot{\bar m}}$ that is linear in the sector forces is $\frac{\tau^2}{M_k}(1-\rho)\,\mathcal F$, where
$$
\rho:=\frac{\sin(2M_kT)-\sin(2M_k\lambda_1)}{2M_k\Lambda},\qquad \lvert\rho\rvert\le\frac{1}{M_k\Lambda},\qquad \overline{2\sin^2(M_k\lambda)}=1-\rho .
\tag{6.24}
$$
At first order in $F$ and leading order in $\kappa$, the coefficient of $F$ in $\overline a$ is therefore exactly $\frac{\tau^2}{M_k}(1-\rho)\,\kappa^2e\,g(e,\cdot)$, for every interval and every $\lambda_1\ge0$.

**(ii) Finite $F$ and $\kappa$ (exact bound).** Exactly, $\overline{\ddot{\bar m}_f}=[\dot{\bar m}_f(T)-\dot{\bar m}_f(\lambda_1)]/\Lambda$.

*Following of the branch.* Write $C_k(q)=\omega_k\,\vec y\cdot\vec\sigma$ with the unit vector $\vec y(q)=(-\sin\Theta,0,\cos\Theta)$, $\tan\Theta=\tau\sin q/M_k$, and let $\vec z:=w^\dagger\vec\sigma\,w$, a unit vector with $\vec z(0)=\vec y(0)$. By (1.9), $\dot{\vec z}=2\omega_k\,\vec y\times\vec z$. With $\hat v=\partial_qC_k$ and $\vec y_\perp:=\partial_\Theta\vec y$, at $q=f\lambda$:
$$\dot{\bar m}_f=\omega_k'\,(\vec y\cdot\vec z)+\omega_k\Theta'\,(\vec y_\perp\cdot\vec z),\qquad \omega_k\Theta'=\frac{\tau M_k\cos q}{\omega_k},\qquad \omega_k'=\frac{\tau^2\sin q\cos q}{\omega_k}.$$
- In the frame that turns with $\vec y$ about the 2-axis, at the rate $\dot\Theta=f\Theta'$, $\vec z$ precesses about an axis that is tilted from $\vec y$ towards the 2-axis by the angle $\zeta$, with $\tan\zeta=\lvert\dot\Theta\rvert/(2\omega_k)=\lvert f\rvert\tau M_k\lvert\cos q\rvert/(2\omega_k^3)$.
- Precession does not change the angle $\alpha$ between $\vec z$ and this axis; only the motion of the axis does. Hence $\lvert\dot\alpha\rvert\le\lvert\dot\zeta\rvert$.
- For $\lvert q\rvert\le\pi/2$, $\zeta$ decreases from $\zeta(0)\le\lvert f\rvert\tau/(2M_k^2)$, and $\alpha(0)=\zeta(0)$. Hence $\alpha\le2\zeta(0)-\zeta$, and the angle $\beta$ between $\vec z$ and $\vec y$ obeys $\beta\le\alpha+\zeta\le\lvert f\rvert\tau/M_k^2$.

With $\lvert\omega_k'\rvert\le\tau$, $\lvert\omega_k\Theta'\rvert\le\tau$, $1-\vec y\cdot\vec z\le\beta^2/2$ and $\lvert\vec y_\perp\cdot\vec z\rvert\le\beta$:
$$
\bigl\lvert\dot{\bar m}_f(\lambda)-\omega_k'(f\lambda)\bigr\rvert\le\frac{\lvert f\rvert\tau^2}{M_k^2}\Bigl(1+\frac{\lvert f\rvert\tau}{2M_k^2}\Bigr)\qquad\text{for }\lvert f\rvert\lambda\le\frac\pi2 .
\tag{6.25}
$$
The oscillating part of the velocity is bounded by its amplitude. The bound does not use the phase of the oscillation, so it contains no term that grows with $\lambda$.

*Branch velocity.* By the mean value theorem, $[\omega_k'(fT)-\omega_k'(f\lambda_1)]/\Lambda=f\,\omega_k''(q^*)$ with $\lvert q^*\rvert\le\lvert f\rvert T$. From $\omega_k''=(\tau^2\cos2q-\omega_k'^2)/\omega_k$, with $1-\cos2q\le2q^2$, $0\le\frac1{M_k}-\frac1{\omega_k}\le\frac{\tau^2q^2}{2M_k^3}$ and $\omega_k'^2/\omega_k\le\tau^4q^2/M_k^3$:
$$\Bigl\lvert\omega_k''(q)-\frac{\tau^2}{M_k}\Bigr\rvert\le\frac{\tau^2}{M_k}\,\Gamma_k\,q^2,\qquad \Gamma_k:=2+\frac{3\tau^2}{2M_k^2}.$$

*Sum over the sectors.* Define $\mathcal E$ by the first equation of (6.14); the linear parts sum to $\mathcal F$ by (6.11):
$$
\overline{a}=\frac{\tau^2}{M_k}\,\kappa\,\bigl[\mathcal F+\mathcal E\bigr]\,e=\frac{\tau^2}{M_k}\,\kappa^2\,g(e,F)\,e+\frac{\tau^2}{M_k}\,\kappa\,\mathcal E\,e,\qquad
\frac{1}{2M_k}\ll\Lambda,\qquad \mathcal F_{\max}\,T\ll\min\Bigl(1,\frac{M_k}{\tau}\Bigr),
\tag{6.14}
$$
with $\mathcal F_{\max}:=\max_\xi\lvert\mathcal F_\xi\rvert$ and $\Sigma_j:=\sum_\xi w_\xi\lvert\mathcal F_\xi\rvert^j$. The three estimates above give
$$
\lvert\mathcal E\rvert\le\Gamma_k\,T^2\,\Sigma_3+\frac{2}{M_k\Lambda}\Bigl(\Sigma_1+\frac{\tau}{2M_k^2}\,\Sigma_2\Bigr)\qquad\text{for every }0\le\lambda_1<T\text{ with }\mathcal F_{\max}T\le\frac\pi2 ,
\tag{6.26}
$$
where $\Sigma_2=\mathcal F^2+\kappa^2\sigma_X^2$ by (6.11), $\Sigma_1\le\Sigma_2^{1/2}$ and $\Sigma_3\le\mathcal F_{\max}\Sigma_2$.

**(iii) The part beyond first order (order estimate).** Let $\mathcal N_f$ be the part of $\frac{M_k}{\tau^2}\overline{\ddot{\bar m}_f}$ beyond first order in $f$. Exactly, $\mathcal E=-\rho\,\mathcal F+\mathcal N$ with $\mathcal N:=\sum_\xi w_\xi\,\mathcal N_{\mathcal F_\xi}$. At a fixed interval, $\mathcal N_f=\mathcal N^{(3)}f^3+O(f^5)$ by the two properties above, so
$$\mathcal N=\mathcal N^{(3)}\Bigl[\mathcal F^3+3\mathcal F\kappa^2\sigma_X^2-\kappa^3\bigl\langle(X-\langle X\rangle)^3\bigr\rangle\Bigr]+\dots$$
These are a term of higher order in $F$, a record term linear in $F$, and a record-induced term that does not depend on $F$. All three are additive and of order $\kappa^3$ at fixed $F$ and fixed interval.

The size of $\mathcal N$ on long or late intervals follows from the precession of (ii). To leading order in $\zeta$, the oscillating part of $\dot{\bar m}_f$ is $-\frac{f\tau^2}{2M_k^2}\sin\Upsilon_f(\lambda)$, with the precession phase
$$\Upsilon_f(\lambda)=\int_0^\lambda\sqrt{4\omega_k^2+\dot\Theta^2}\;d\lambda'=2M_k\lambda+\frac{\tau^2f^2\lambda^3}{3M_k}+\dots$$
Its first order, $\Upsilon_f\to2M_k\lambda$, gives the term $-\rho f$. The accumulated phase $\Upsilon_f-2M_k\lambda$ is not small on late intervals. Under both conditions of (6.14):
$$
\mathcal E=-\rho\,\mathcal F+\mathcal N,\qquad
\mathcal N=O\bigl(\Gamma_kT^2\Sigma_3\bigr)+O\Bigl(\frac{1}{M_k\Lambda}\sum_\xi w_\xi\lvert\mathcal F_\xi\rvert\,\min\Bigl\{1,\frac{\tau^2\mathcal F_\xi^2T^3}{M_k}\Bigr\}\Bigr).
\tag{6.27}
$$
- The first term is the variation of the branch curvature along the drift.
- The second term is the accumulated interbranch phase at the two ends of the interval. For $\tau^2f^2T^3\ll M_k$ its leading form in one sector is $-\frac{\tau^2f^3}{6M_k^2\Lambda}\bigl[T^3\cos(2M_kT)-\lambda_1^3\cos(2M_k\lambda_1)\bigr]$. For larger phases it saturates at the endpoint term of (6.26).
- The ratio of the second term to the first is at most of order $\tau^2T/(\Gamma_kM_k^2\Lambda)$. For $\lambda_1\lesssim\Lambda$ the second term is at most of the order of the first. On late intervals, $\tau^2T\gg\Gamma_kM_k^2\Lambda$, it dominates. An estimate $O(T^2\Sigma_3)$ alone holds only for $\lambda_1\lesssim\Lambda$.

**(iv) Conditions and status.**
- *Exact, for every interval:* the average as velocity change; the first-order coefficient (6.24); the decomposition $\mathcal E=-\rho\mathcal F+\mathcal N$, with $\mathcal N$ of third order in the sector forces at a fixed interval.
- *Exact inequalities:* (6.25), and (6.26) for every interval with $\mathcal F_{\max}T\le\pi/2$. They need no condition on $\lambda_1$, on $\Lambda$ or on $\mathcal F_{\max}\tau/M_k^2$.
- *Order estimate, under both conditions of (6.14):* the estimate of $\mathcal N$ in (6.27).
- *First condition*, $\Lambda\gg1/(2M_k)$: the interval is long compared with the inverse gap. It makes the endpoint term of (6.26) small, whatever the phases at the two ends.
- *Second condition*, $\mathcal F_{\max}T\ll\min(1,M_k/\tau)$: it is the condition for the first-order treatment, and it makes $\Gamma_k(\mathcal F_{\max}T)^2\ll1$. The drift $\mathcal F_\xi\lambda$ of every sector stays where $\omega_k''\approx\omega_k''(0)$. $T$ is counted from the start at $\lambda=0$, because the drift accumulates from the start: the condition limits the end of the interval, not only its length.
- *Every interval that satisfies both conditions* has $\lvert\mathcal E\rvert\le\varepsilon_{\rm av}\,\Sigma_1\le\varepsilon_{\rm av}(\mathcal F^2+\kappa^2\sigma_X^2)^{1/2}$ with $\varepsilon_{\rm av}:=\Gamma_k(\mathcal F_{\max}T)^2+\frac{2}{M_k\Lambda}\bigl(1+\frac{\mathcal F_{\max}\tau}{2M_k^2}\bigr)\ll1$. The two conditions together imply $\mathcal F_{\max}\tau\ll M_k^2$ (no branch transitions).
- *Records.* The bound is relative to $\Sigma_1$, not to $\lvert\mathcal F\rvert$. The leading term of (6.14) dominates the value of $\overline a$ if in addition $\varepsilon_{\rm av}\,\kappa\sigma_X\ll\lvert\mathcal F\rvert$. Otherwise the record-induced part of $\mathcal E$ that does not depend on $F$ can be comparable. That part vanishes identically if the distribution of $X$ in $\chi$ is symmetric about $\langle X\rangle$, since $\dot{\bar m}_f$ is odd. At a fixed interval $\mathcal N=O(\kappa^3)$, which is the weak-record regime of the question.
- *Values, not derivatives.* (6.26) bounds the value of $\overline a$. It does not bound derivatives with respect to $F$ at finite $F$ and $\kappa$: once the phases $\Upsilon_{\mathcal F_\xi}$ differ by more than 1, the derivative of the endpoint term need not be small. The coefficient of $F$ is therefore taken at a fixed interval, as in (i).
- Finite-packet corrections vanish in limit 2, which is taken before. The chain limit requires $L-W\gg\tau T$.

**5f. Averaged inverse inertia.** $\mu_{\rm av}(k)$ is the coefficient of $F$ in $\overline a$ at first order in $F$ and leading order in $\kappa$, in the long-interval limit $M_k\Lambda\to\infty$. By (6.24), at a fixed interval this coefficient is exactly $(1-\rho)$ times the following, and $\lvert\rho\rvert\le1/(M_k\Lambda)$ for every $\lambda_1$; the limit therefore exists and does not depend on $\lambda_1$:
$$
\mu_{\rm av}(k)=\frac{\tau^2}{M_k}\,\kappa^2\,e\,g(e,\cdot\,)=\frac{\tau^2}{m+d_k}\,\kappa^2\,e\,g(e,\cdot\,).
\tag{6.15}
$$
At finite $F$ and $\kappa$, (6.14) reads $\overline a=\mu_{\rm av}(k)F+\frac{\tau^2}{M_k}\kappa\,\mathcal E\,e$, with $\mathcal E$ bounded by (6.26).
$$
M_k\,\mu_{\rm av}(k)=\tau^2\kappa^2\,e\,g(e,\cdot\,)\ \ \text{for every }k,\qquad \frac{M_k}{\tau^2}=\frac{m}{\tau^2}+\frac{2\pi k}{N\lambda_0\,\tau^2}.
\tag{6.16}
$$
So $\mu_{\rm av}(k)$ decreases strictly with $d_k$, and it is inversely proportional to the rest cost (6.7).

*Two normalizations of the inertia.* Let $n:=e/\sigma_X$, a unit vector of $(\mathcal H_c,g)$, and $d_1:=\kappa\sigma_X$, the background distance of neighbouring points by (17.15) with $X\to\kappa X$. Then $\bar x=d_1\bar m\,n$ and $\mathcal F=d_1\,g(n,F)$, and
$$
\mu_{\rm av}(k)=\frac{1}{I_k}\,n\,g(n,\cdot\,),\qquad
I_k:=\frac{M_k}{\tau^2\kappa^2\sigma_X^2}=\frac{I_k^{\rm lab}}{d_1^{\,2}},\qquad
I_k^{\rm lab}:=\frac{M_k}{\tau^2}\quad\bigl(\overline{\ddot{\bar m}}=\mathcal F/I_k^{\rm lab}\bigr).
\tag{6.20}
$$
- $I_k$ is the **background inertia**: $\mu_{\rm av}(k)$ has the single nonzero eigenvalue $1/I_k$, on $n$, and vanishes on the $g$-orthogonal complement of $e$.
- $I_k^{\rm lab}$ is the **label inertia**: the response of the dimensionless position label $\bar m$ to the label force $\mathcal F$.
- Both are proportional to the rest cost $M_k=m+d_k$. The clock cost $d_k$ adds to each with the same coefficient as the rest term $m$.

### Step 6. Decoupled body on the chain

**6a. The whole class.** Take $x\in\mathbb Z$ (limits as in the Setup), $T_{x+1,x}=t\mathbb 1_N$ with $t\neq0$, and arbitrary self-adjoint on-site blocks $T_{xx}$; records and gradient as in (6.8). Let
$$\omega(p):=te^{-ip}+t^*e^{ip},\qquad V:=\sum_x\lvert x\rangle\langle x\rvert\otimes T_{xx},\qquad G_x:=T_{x+1,x+1}-T_{xx},$$
so that $C_b=\omega(\hat p)\otimes\mathbb 1_N+V$ with $\omega(\hat p)=t\hat S+t^*\hat S^\dagger$. Since $V$ commutes with $\hat x$, the steps of 5b give $\hat v=i[C_b-\mathcal F\hat x,\hat x]=\omega'(\hat p)=-it\hat S+it^*\hat S^\dagger$. With $[V,\hat S]=\sum_x\lvert x+1\rangle\langle x\rvert\otimes G_x$:
$$
\hat v=\omega'(\hat p)\otimes\mathbb 1_N,\qquad
\ddot{\hat x}=\mathcal F\,\omega''(\hat p)+\hat a_{\rm free},\qquad
\hat a_{\rm free}:=i[V,\hat v]=\sum_x\bigl(t\,\lvert x+1\rangle\langle x\rvert+t^*\lvert x\rangle\langle x+1\rvert\bigr)\otimes G_x .
\tag{6.21}
$$
Step 5a applies to the whole class, since it uses only that $c$ enters through $\hat x\otimes X$. For the start $\phi_X\otimes\eta$, (6.21) gives the following.
- **At $\lambda=0$.** The state does not depend on $\mathcal F$, and $\hat a_{\rm free}$ does not contain $\mathcal F$. The coefficient of $F$ in $a(0)$ is therefore $\kappa^2\langle\phi_X\vert\omega''(\hat p)\vert\phi_X\rangle\,e\,g(e,\cdot)$. By (6.2), $\langle\omega''(\hat p)\rangle_0=-2\sum_x\operatorname{Re}[t^*\phi_X(x)^*\phi_X(x+1)]=-\sum_xL_{x,x+1}(0)$, and $\Delta_{x,x+1}=-\kappa e$: this is $\mu$ of (6.4), for every choice of the $T_{xx}$. Only $a_0=\kappa\langle\hat a_{\rm free}\rangle_0\,e$ depends on $\eta$ and on the $T_{xx}$.
- **At $\lambda>0$.** The coefficient of $\mathcal F$ in $\ddot{\bar m}(\lambda)$ is $\langle\omega''(\hat p)\rangle_\lambda+\partial_{\mathcal F}\langle\hat a_{\rm free}\rangle_\lambda$, taken at $\mathcal F=0$.
- **Criterion.** Since $t\neq0$, $\hat a_{\rm free}=0$ iff $G_x=0$ for all $x$, i.e. iff the on-site blocks are the same at every position. Exactly then the velocity commutes with $C_b$.

**6b. Uniform on-site blocks.** Let $T_{xx}=D'$ for all $x$, with any self-adjoint clock term $D'$. Then $\hat a_{\rm free}=0$ and
$$C_b-\mathcal F\hat x=(\omega(\hat p)-\mathcal F\hat x)\otimes\mathbb 1_N+\mathbb 1\otimes D' .$$
The two parts commute, and $\bar m$ involves only the first. By (6.21), $\ddot{\hat x}=\mathcal F\omega''(\hat p)$ exactly, and $\frac{d}{d\lambda}\omega''(\hat p)=\mathcal F\omega'''(\hat p)$ with $\lVert\omega'''(\hat p)\rVert\le2\lvert t\rvert$. Applied in each sector of 5a and summed with (6.11), this gives exactly
$$\bigl\lvert\ddot{\bar m}(\lambda)-\mathcal F\,\langle\omega''(\hat p)\rangle_0\bigr\rvert\le2\lvert t\rvert\,\lambda\,\bigl(\mathcal F^2+\kappa^2\sigma_X^2\bigr).$$
The response does not oscillate. At first order in $F$ and leading order in $\kappa$ its coefficient is the same at every $\lambda$, hence equal to its average over every interval; no condition of the first kind in (6.14) is needed. So
$$
\mu_{\rm av}=\mu=\kappa^2\,\langle\phi_X\vert\omega''(\hat p)\vert\phi_X\rangle\,e\,g(e,\cdot\,)\ \xrightarrow{\ \text{concentrated at }p_0\ }\ -2\operatorname{Re}\bigl(t^*e^{ip_0}\bigr)\,\kappa^2\,e\,g(e,\cdot\,).
\tag{6.17}
$$
This is independent of the clock level, of $\eta$ and of $D'$.

**6c. Position-dependent on-site blocks.** Now $G_x\neq0$ for some $x$.
- *What holds.* (6.4): $\mu$ is the same as in 6b and depends neither on $\eta$ nor on the $T_{xx}$. Also (6.21) and Step 5a hold.
- *What fails.* $[C_b,\hat v]=0$, the constancy of the response in $\lambda$, and with them the argument for $\mu_{\rm av}=\mu$.
- *Scalar differences*, $T_{xx}=D'+\nu_x\mathbb 1_N$. The factorization of 6b persists with $\omega(\hat p)$ replaced by $\omega(\hat p)+\nu(\hat x)$. The response at every $\lambda$ is that of the body without internal labels with site costs $\nu_x$. It is independent of the clock level, of $\eta$ and of $D'$, but it need not be constant in $\lambda$ or equal to $\mu$ (for $\nu_x=(-1)^xM$ it is body E below with $m\mathbb 1_N+D$ replaced by $M\mathbb 1_N$).
- *Common eigenvector.* Let $\eta$ be a common eigenvector of all on-site blocks, $T_{xx}\eta=\nu_x^{(\eta)}\eta$. Then $C_b$ and $\hat x$ map the states $\phi_X\otimes\eta$ to states of the same form. In every sector of 5a the state stays a product $\phi_X(\lambda)\otimes\eta$, and the position evolves as for the body without internal labels with site costs $\nu_x^{(\eta)}$. The response depends on $\eta$ through these scalars, unless their differences $\nu_{x+1}^{(\eta)}-\nu_x^{(\eta)}$ are the same for all $\eta$ (scalar differences). The clock level can thus enter the position dynamics without any entanglement between position and clock. Body E below at a fixed level is of this kind, with $\eta=\lvert f_k\rangle$ and $\nu_x^{(\eta)}=(-1)^xM_k$.
- *General internal start.* If $\eta$ is not a common eigenvector, the state need not stay a product: position and internal labels can become entangled, and generically do. This is a possibility for arbitrary internal starts, not a consequence of non-scalar differences alone.
- In both cases the term $\partial_{\mathcal F}\langle\hat a_{\rm free}\rangle_\lambda$ can depend on the $T_{xx}$ and on $\eta$, and no formula for the response in terms of $t$ alone holds for the class. The following body shows this.

**Body E.** Let $t=-\tfrac{i\tau}{2}$ and $T_{xx}=(-1)^x(m\mathbb 1_N+D)$, with records and gradient as in Model D. Define $J\lvert x\rangle\otimes\lvert a\rangle:=\lvert x\rangle\otimes\lvert s(x)\rangle\otimes\lvert a\rangle$, with $s(x)=1$ for even $x$ and $s(x)=2$ for odd $x$.
- $A$ of Model D maps $\lvert x,s\rangle$ to $-\tfrac{i\tau}{2}\lvert x+1,\bar s\rangle+\tfrac{i\tau}{2}\lvert x-1,\bar s\rangle$, where $\bar s$ is the other branch label, and $\overline{s(x)}=s(x\pm1)$. $B$ multiplies $\lvert x,s(x)\rangle$ by $(-1)^x(m\mathbb 1_N+D)$. Hence $C_b^{\rm D}J=JC_b^{\rm E}$.
- $J$ preserves $x$, so it also intertwines $\hat x$, the gradient and the records. It is a unitary from the places of E onto the places of Model D with $x+s$ odd, which span an invariant subspace of $C$.
- Therefore the place weights, $\bar x$ and $a(\lambda)$ of E with start $\phi$ are those of Model D with start $J\phi$, for all $\lambda$, $F$ and $\kappa$. (The places with $x+s$ even carry the body with $T_{xx}=-(-1)^x(m\mathbb 1_N+D)$: Model D is the direct sum of two bodies of this class.)

Take the start $\phi=\phi_X\otimes\lvert f_k\rangle$ with $\phi_X(x)\propto\varphi(x/W)$ on even $x$ and $\phi_X(x)=0$ on odd $x$. In the terms of E, it sits at the plane-wave label $0$ of the two-site cell, in the branch of positive eigenvalue $M_k$: at this label that branch lives on the even sites. Its image is $J\phi=\phi_X\otimes\lvert1\rangle\otimes\lvert f_k\rangle$, a start of item 2.
- *$\mu$.* Every edge between points joins an even and an odd site, so $L_{x,x+1}(0)=0$ in (6.4). This agrees with (6.10) for $J\phi$.
- *$\mu_{\rm av}$.* $\tilde\phi_X(q+\pi)=\tilde\phi_X(q)$, so $J\phi$ is concentrated at $q\in\{0,\pi\}$. Since $\sigma_3C_k(q)\sigma_3=C_k(q+\pi)$, $\sigma_3\hat a_k(q)\sigma_3=\hat a_k(q+\pi)$ and $\sigma_3\lvert1\rangle=\lvert1\rangle$, the integrand of 5c has period $\pi$ in $q$. The limit is the two-level problem of 5d, and Steps 5d–5f hold unchanged.

Hence
$$
\text{body E, level }k:\qquad \mu=0,\qquad \mu_{\rm av}(k)=\frac{\tau^2}{m+d_k}\,\kappa^2\,e\,g(e,\cdot\,)\neq\mu .
\tag{6.22}
$$
In the class of item 1 with position-dependent $T_{xx}$, $\mu_{\rm av}$ can thus depend on the clock level and differ from $\mu$, although $\mu$ itself depends on neither $\eta$ nor the $T_{xx}$.

### Step 7. Item 4: comparison

$$
\begin{aligned}
&\text{Model D (and body E):}&&\mu=0\neq\mu_{\rm av}(k)=\frac{\tau^2}{M_k}\,\kappa^2e\,g(e,\cdot\,);\\
&\text{decoupled body, }T_{xx}=D'\text{ at every }x:&&\mu=\mu_{\rm av}\quad\text{(6.17), independent of the clock};\\
&\text{decoupled body, }T_{xx}\text{ position-dependent:}&&\text{no general relation; }\mu\neq\mu_{\rm av}\text{ for body E (6.22)}.
\end{aligned}
\tag{6.23}
$$
Of the two bodies of the question, $\mu$ and $\mu_{\rm av}$ coincide for the decoupled body, provided its clock term is the same at every position; they differ for Model D.

**Reason.** $\mu$ is the coefficient of the term $-i\mathcal F[\hat x,\hat v]$ of the acceleration, evaluated in the start; it is fixed by the edge costs between points, which is the content of (20.10). The averaged response contains in addition the response of the free acceleration $i[C_b,\hat v]$ to the gradient.
- **Uniform on-site blocks.** The velocity commutes with $C_b$, so the free acceleration vanishes identically (6.21). The term $\mathcal F\omega''(\hat p)$ is the whole response, already at $\lambda=0$. The clock term commutes with position and hopping, and drops out.
- **Model D.** The velocity $-\tau\cos\hat p\,\sigma_1$ connects the two branches and does not commute with the term $\sigma_3\otimes(m\mathbb 1+D)$ that carries the clock.
  - At $\lambda=0$ the $s=1$ start has no edge cost between points, which gives (6.10).
  - The response comes from interference between the admixture of the negative branch that the gradient induces, of amplitude $\propto\mathcal F\tau/M_k^2$, and the free acceleration $2\tau M_k\cos\hat p\,\sigma_2$. It builds up as $\sin^2(M_k\lambda)$, and its average is the branch curvature $\tau^2/M_k$.
  - The clock cost enters through the gap $2M_k$.
- **Position-dependent on-site blocks.** The same mechanism acts through $G_x\neq0$ in (6.21). Model D is itself of this kind in each of its two invariant subspaces (body E): the dividing line is whether the term that carries the clock commutes with the velocity. It is neither whether the hopping blocks are multiples of $\mathbb 1_N$, nor whether position and clock become entangled (6c).

## Result

- **(6.1)** The inverse inertia $\mu$ of (20.10) of 03-ilang-space/20-cost-and-mass is the coefficient of $F$ for a uniform gradient added to any diagonal, and for every product start.
- **Item 1.**
  - (6.2): $L_{hh'}(0)$ in terms of the blocks, and $L_{xx'}(0)=2\operatorname{Re}\langle\phi_x\vert T_{xx'}\vert\phi_{x'}\rangle$.
  - (6.3): $\mu=-\sum_{x\neq x'}L_{xx'}(0)\Delta_{xx'}g(\Delta_{xx'},\cdot)$. Edges inside one point do not contribute.
  - (6.4): for the decoupled body, $\mu$ depends neither on $\eta$ nor on the $T_{xx}$, uniform or position-dependent.
- **Item 2 (Model D).**
  - (6.7): rest cost $M_k=m+d_k$.
  - (6.9): edge costs at $\lambda=0$, nonzero only on the clock edges of branch $s=1$.
  - (6.10): $\mu=0$.
- **Item 3.**
  - (6.13): $a(\lambda)=\frac{2\tau^2}{M_k}\sin^2(M_k\lambda)\,\kappa^2g(e,F)e$ at first order in $F$.
  - (6.24), exact for every interval: at first order in $F$ and leading order in $\kappa$, the coefficient of $F$ in the average is $\frac{\tau^2}{M_k}(1-\rho)\kappa^2e\,g(e,\cdot)$, with $\lvert\rho\rvert\le1/(M_k\Lambda)$.
  - (6.14): the average is $\frac{\tau^2}{M_k}\kappa^2g(e,F)e+\frac{\tau^2}{M_k}\kappa\,\mathcal E\,e$, for intervals with $1/(2M_k)\ll\Lambda$ (long compared with the inverse gap) and $\mathcal F_{\max}T\ll\min(1,M_k/\tau)$ (first-order treatment), where $T$ is the end of the interval.
  - (6.25), exact: the velocity differs from the branch velocity by at most $\frac{\lvert f\rvert\tau^2}{M_k^2}\bigl(1+\frac{\lvert f\rvert\tau}{2M_k^2}\bigr)$ while $\lvert f\rvert\lambda\le\pi/2$.
  - (6.26), exact for every interval with $\mathcal F_{\max}T\le\pi/2$ and every $\lambda_1$: $\lvert\mathcal E\rvert\le\Gamma_kT^2\Sigma_3+\frac{2}{M_k\Lambda}\bigl(\Sigma_1+\frac{\tau}{2M_k^2}\Sigma_2\bigr)$. Under both conditions of (6.14), $\lvert\mathcal E\rvert\ll\Sigma_1$.
  - (6.27), order estimate: $\mathcal E=-\rho\mathcal F+\mathcal N$, where $\mathcal N$ contains the curvature term $O(\Gamma_kT^2\Sigma_3)$ and the accumulated-phase term, which exceeds the first on late intervals.
  - (6.15): $\mu_{\rm av}(k)=\frac{\tau^2}{m+d_k}\kappa^2e\,g(e,\cdot)$, the coefficient at first order in $F$, leading order in $\kappa$, in the long-interval limit, for every $\lambda_1$.
  - (6.16), (6.20): $\mu_{\rm av}(k)$ is inversely proportional to the rest cost. Background inertia $I_k=M_k/(\tau^2\kappa^2\sigma_X^2)$, label inertia $I_k^{\rm lab}=M_k/\tau^2$; the clock cost $d_k$ adds to each like $m$.
  - (6.21): for the decoupled chain body with arbitrary $T_{xx}$, $\ddot{\hat x}=\mathcal F\omega''(\hat p)+\hat a_{\rm free}$, and $\hat a_{\rm free}=0$ iff the on-site blocks are uniform.
  - (6.17): for uniform on-site blocks, $\mu_{\rm av}=\mu=\kappa^2\langle\omega''\rangle e\,g(e,\cdot)$, independent of the clock level.
  - (6.22): for position-dependent on-site blocks there is no such statement; for body E, $\mu=0$ and $\mu_{\rm av}(k)=\frac{\tau^2}{m+d_k}\kappa^2e\,g(e,\cdot)$.
- **Item 4, (6.23).** $\mu$ and $\mu_{\rm av}$ coincide for the decoupled body with uniform on-site blocks, whose velocity commutes with $C_b$. They differ for Model D, and for body E, where the term that carries the clock does not commute with the velocity.

## Consistency checks

1. **Dimensions and $\tau\to0$.** $C$, $\tau$, $m$, $d_k$ and $\mathcal F$ have the dimension $1/[\lambda]$. $I_k^{\rm lab}=M_k/\tau^2$ has the dimension $[\lambda]$, so $\mathcal F/I_k^{\rm lab}$ has the dimension $1/[\lambda]^2$ of $\ddot{\bar m}$. $\mu_{\rm av}F=\frac{\tau^2}{M_k}\,\kappa e\cdot\kappa g(e,F)$ has the dimension of $\kappa e/[\lambda]^2$, that of $\ddot{\bar x}$. In (6.26), $T^2\Sigma_3$, $\Sigma_1/(M_k\Lambda)$ and $\tau\Sigma_2/(M_k^3\Lambda)$ have the dimension of $\mathcal F$, and $\Gamma_k$ is a number. In (6.9), $2\pi/(N^2\lambda_0)$ is a cost. As $\tau\to0$, $\hat v=0$, (6.13), (6.15) and the bound (6.25) vanish and $I_k\to\infty$. Passed.
2. **$\lambda=0$ by an independent route.** For the item-2 start with any $\phi_X$, $\langle1\vert\sigma_2\vert1\rangle=\langle1\vert\sigma_1\vert1\rangle=0$ in (6.12) gives $a(0)=0$ for every $F$. This agrees with (6.10), obtained from (20.10) through edge costs, and with (6.13) at $\lambda=0$. For body E, (6.21) gives the same: $\langle\omega''(\hat p)\rangle_0$ and $\langle\hat a_{\rm free}\rangle_0$ are sums of bond terms $\phi_X(x)^*\phi_X(x+1)$, which vanish for $\phi_X$ on even sites. Passed.
3. **The bounds (6.25), (6.26) against the first-order solution and the secular term.** Integrating (6.13) gives $\dot{\bar m}_f=\frac{\tau^2f}{M_k}\lambda-\frac{\tau^2f}{2M_k^2}\sin(2M_k\lambda)+O(f^3)$, and $\omega_k'(f\lambda)=\frac{\tau^2f}{M_k}\lambda+O(f^3)$. The difference has modulus $\le\frac{\lvert f\rvert\tau^2}{2M_k^2}$, half the bound (6.25). Accordingly $\lvert\rho\rvert\le\frac{1}{M_k\Lambda}$ uses half of the endpoint term $\frac{2}{M_k\Lambda}$ of (6.26). The secular term of (6.27) has size at most $\frac{\tau^2\lvert f\rvert^3T^3}{3M_k^2\Lambda}$; it fits into the other half, $\frac{\lvert f\rvert}{M_k\Lambda}$, iff $\frac{\tau^2f^2T^3}{3M_k}\le1$, which is exactly where the expansion of the phase is valid. For a short interval at the start, $\lambda_1=0$ and $M_k\Lambda\ll1$: $1-\rho\approx\frac23(M_k\Lambda)^2\to0$, in agreement with $a(0)=0$, and (6.26) allows this because its endpoint term is then large. Passed.

## Open issues

- A ticking clock, i.e. a superposition of levels, is not treated. Since the level is conserved, the averaged response is the level-weighted mean of (6.15); whether this mixture is the inertia of one body is left open.
- (6.26) is exact but not sharp: its endpoint term carries twice the first-order amplitude. (6.27) is an order estimate, without constants. Neither controls derivatives with respect to $F$ at finite $\kappa$. Both hold for the concentrated start, after limit 2; finite-packet corrections and wide packets are not treated (for wide packets the average is expected to be $\langle\omega_k''\rangle$, which is not derived here).
- No statement is made for $\mathcal F_{\max}T>\pi/2$ (Bloch oscillation) or for $\mathcal F_{\max}\tau\gtrsim M_k^2$ (branch transitions), where (6.25) is not useful.
- Position-dependent on-site blocks: only scalar differences, common eigenvectors and body E are solved. Which position-dependent blocks make $\mu_{\rm av}$ depend on the clock level is not classified, and a general member of the class has no gap that fixes the averaging interval.
- Second order in $\kappa$ (decoherence by the records, the record-induced acceleration at $F=0$) is only bounded by (6.26) and estimated by (6.27), not computed.
- The meaning of $\tau$ and of $d_1\tau$ as speeds, and of the inertias (6.20), belongs to criteria 4 and 6; it is outside this package (M1, M3).

## Methods used

- Heisenberg equations of motion (double commutators)
- Affine dependence of the acceleration on the diagonal of the contract
- Fourier transform on $\mathbb Z$, plane waves, two-branch dispersion
- Method of characteristics (drift of the plane-wave label in a uniform gradient)
- First-order perturbation theory in the force; parity and analyticity in the force
- Bloch-vector precession in a co-rotating frame (exact bound on the following of a branch)
- Averaged acceleration as velocity change over the interval; mean value theorem
- Conserved record observable: exact mixture decomposition of the view
- Invariant subspaces and unitary equivalence (staggered chain and two-branch chain)
- Propagation bound for nearest-neighbour hopping; limit of finite chains
- Geometric sums

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