Version 1 · earlier version; the current one is v2 · External review, round 1: minor issues

# Motion of a hopping body in the background of a recording medium

- **Subproject:** 03-ilang-space
- **Package:** 17-motion
- **Version:** v1
- **Mode:** new
- **Date:** 2026-10-08

## Setup and assumptions

- Objects: body $b$ (type $T$, places $h\in H_T$, $d_T\ge2$) and medium $c$, each the only instance of its type. $\mathcal H=\mathcal H_T\otimes\mathcal H_c$ is finite-dimensional (A1), except in the chain limit of Step 9.
- Contract: $C=C_b\otimes\mathbb 1_c+\sum_h\lvert h\rangle\langle h\rvert\otimes K_h$, $C_b=\sum_{h,h'}t_{hh'}\lvert h\rangle\langle h'\rvert$, $t_{hh'}=t_{h'h}^*$, $K_h=K_h^\dagger$, all independent of $\lambda$ (A5). $t$ is the matrix $(t_{hh'})$ and $(t^n)_{hh'}$ are the entries of its powers (diagonal included). $E$ is the set of edges $\{h,h'\}$ ($h\ne h'$, $t_{hh'}\neq0$) of the hopping graph. Start $\lvert\phi\rangle\otimes\lvert\chi\rangle$, $\lambda\ge0$.
- Notation: $\langle X\rangle=\langle\chi\vert X\vert\chi\rangle$, $\epsilon_h:=\langle K_h\rangle$, $\lvert u_h\rangle=(K_h-\epsilon_h)\lvert\chi\rangle$, so $\langle\chi\vert u_h\rangle=0$, and $d_0(h,h')=\lVert u_h-u_{h'}\rVert$. $\mathcal H_c$ carries $g(v,w)=\operatorname{Re}\langle v\vert w\rangle$ with norm $\lVert\cdot\rVert$.
- Inputs: (2.1) and (2.2) give $\lvert\Psi\rangle=\sum_h\lvert h\rangle\otimes\lvert\psi_h\rangle$, $(V_b)_{hh'}=\langle\psi_{h'}\vert\psi_h\rangle$ and $p_h=\lVert\psi_h\rVert^2$. (2.11) gives $W(h,h')=\lvert(V_b)_{hh'}\rvert^2/(p_hp_{h'})$. (11.6), (11.8) and (11.10) give $d_0$ in the setting of 11. (5.10) and (5.11) give the background $(H_T/{\sim_0},d_0)\cong(\{u_h\},\lVert\cdot\rVert)\subset(\mathcal H_c,g)$.
- Convention: $W(h,h';\lambda)\,p_hp_{h'}$ stands for $\lvert(V_b)_{hh'}\rvert^2$. This expression is also defined when $p_hp_{h'}=0$, and then it vanishes by Cauchy–Schwarz.
- Items 1(a) and 3 are statements for $\lambda\to0^+$; items 2, 4 and 5 are exact.

## Derivation

### Step 1. Branch equations

Inserting (2.1) into (1.9) and projecting on $\langle h\rvert\otimes\mathbb 1_c$:

$$
i\,\frac{\mathrm d}{\mathrm d\lambda}\lvert\psi_h\rangle=\sum_{h'}t_{hh'}\lvert\psi_{h'}\rangle+K_h\lvert\psi_h\rangle,\qquad \lvert\psi_h(0)\rangle=\phi_h\lvert\chi\rangle .
\tag{17.1}
$$

### Step 2. Item 1(a): the background does not depend on the hopping

The setting is that of 11 with $A=\{b\}$, $\bar A=\{c\}$, $C_A=C_b$, $C_{\bar A}=0$ and $C_\partial=\sum_h\lvert h\rangle\langle h\rvert\otimes K_h$. Each $A_n=\lvert h\rangle\langle h\rvert$ is diagonal and $B_n=K_h$. The two types differ, so no type has objects on both sides of the cut. By (11.8) and (11.10), for a start with all $\phi_h\neq0$:

$$
d_0(h,h')=\lim_{\lambda\to0^+}\frac{\alpha(h,h';\lambda)}{\lambda}=\bigl\lVert u_h-u_{h'}\bigr\rVert,\qquad u_h=(K_h-\epsilon_h)\chi,\quad\text{independent of the }t_{hh'} .
\tag{17.2}
$$

The reason is visible in (17.1): $\psi_h/\phi_h=\chi-i\lambda(\tau_h+K_h)\chi+O(\lambda^2)$ with the scalar $\tau_h=\sum_{h'}t_{hh'}\phi_{h'}/\phi_h$. At first order the hopping adds only a multiple of $\chi$, and $Q_\chi$ removes it in (11.6). **Answer: no.** The background is that of the pure recording contract (5.10)–(5.11).

### Step 3. Item 1(b): well-definedness and two identities

(i) $b$ and $c$ are the only instances of their types, so Law 5 creates no ambiguity. The places are part of the type (A2).
(ii) The $p_h(\lambda)$ are diagonal elements of the view, so they are invariant under (1.2).
(iii) The partial matrix elements of $C$ are $(\langle h\rvert\otimes\mathbb 1)C(\lvert h'\rangle\otimes\mathbb 1)=t_{hh'}\mathbb 1$ for $h\neq h'$ and $t_{hh}\mathbb 1+K_h$ for $h=h'$. So $C$ fixes $K_h$ only up to $K_h\to K_h+c_h\mathbb 1$ (with $t_{hh}\to t_{hh}-c_h$), and $u_h$ is invariant under this change.
(iv) The phase split of the start, $\phi\otimes\chi=(e^{-i\varphi}\phi)\otimes(e^{i\varphi}\chi)$, sends every $u_h$ to $e^{i\varphi}u_h$. This is one common real-linear isometry of $(\mathcal H_c,g)$, the same freedom that the embedding (5.11) already carries.

Hence $s$ is well defined. $\bar x$ and $v$ are well defined up to the common isometry $e^{i\varphi}$, which acts on them and on the background points $u_h$ together. Every $g$-invariant built from $\bar x,v,u_h$ is well defined, e.g. $\lVert\bar x-u_h\rVert$, $\lVert v\rVert$ and $g(v,u_h-u_{h'})$. The weights $p_h$ are analytic in $\lambda$ (finite dimension), so $v$ exists.

The identities follow from $\sum_hp_h=1$ and $\sum_hp_h(u_h-\bar x)=0$. First expand $\sum_hp_h\lVert(u_h-\bar x)+(\bar x-u_k)\rVert^2$. Then use $\sum_{h,h'}p_hp_{h'}\lVert u_h-u_{h'}\rVert^2=2\sum_hp_h\lVert u_h\rVert^2-2\lVert\bar x\rVert^2=2s^2$:

$$
\bigl\lVert\bar x(\lambda)-u_k\bigr\rVert^2=\sum_hp_h(\lambda)\,d_0(h,k)^2-s(\lambda)^2,\qquad
s(\lambda)^2=\frac12\sum_{h,h'}p_h(\lambda)\,p_{h'}(\lambda)\,d_0(h,h')^2 .
\tag{17.3}
$$

### Step 4. Item 2(a): currents

By (17.1), $\dot p_h=2\operatorname{Re}\langle\psi_h\vert\dot\psi_h\rangle=2\operatorname{Im}\bigl[\sum_{h'}t_{hh'}\langle\psi_h\vert\psi_{h'}\rangle+\langle\psi_h\vert K_h\vert\psi_h\rangle\bigr]$. The last term is real, so its imaginary part vanishes. With $\langle\psi_h\vert\psi_{h'}\rangle=(V_b)_{h'h}$ (2.2):

$$
\frac{\mathrm dp_h}{\mathrm d\lambda}=\sum_{h'}J_{h'\to h},\qquad J_{h'\to h}(\lambda):=2\operatorname{Im}\bigl[t_{hh'}\,(V_b(\lambda))_{h'h}\bigr]=-J_{h\to h'} .
\tag{17.4}
$$

Antisymmetry: $t=t^\dagger$ and $V_b=V_b^\dagger$ give $t_{h'h}(V_b)_{hh'}=\bigl(t_{hh'}(V_b)_{h'h}\bigr)^*$. Hence $J_{h\to h}=0$, and $J_{h'\to h}=0$ off the edges. The $K_h$ enter only through the coherences $(V_b)_{h'h}$.

### Step 5. Item 2(b): bound on the current

$\lvert\operatorname{Im}z\rvert\le\lvert z\rvert$ and (2.11) give

$$
\lvert J_{h'\to h}(\lambda)\rvert\le2\lvert t_{hh'}\rvert\sqrt{W(h,h';\lambda)\,p_h(\lambda)\,p_{h'}(\lambda)}\;\le\;2\lvert t_{hh'}\rvert\sqrt{p_hp_{h'}} .
\tag{17.5}
$$

The first bound is attained when $t_{hh'}(V_b)_{h'h}$ is purely imaginary. If an edge is completely witnessed ($W=0$), no current flows across it.

### Step 6. Item 2(c): velocity and maximal speed

From $v=\sum_h\dot p_hu_h=\sum_{h,h'}J_{h'\to h}u_h$, symmetrize with (17.4): $v=\tfrac12\sum_{h,h'}J_{h'\to h}(u_h-u_{h'})$. Each edge contributes twice with the same term:

$$
v(\lambda)=\sum_{\{h,h'\}\in E}2\operatorname{Im}\bigl[t_{hh'}(V_b(\lambda))_{h'h}\bigr]\,(u_h-u_{h'}) .
\tag{17.6}
$$

Each term is independent of the orientation of its edge. Only edges with $d_0(h,h')>0$ contribute.

For the bound, let $a_{hh'}:=\lvert t_{hh'}\rvert d_0(h,h')$. This matrix is real, symmetric and nonnegative, and it vanishes off $E$. Let $q_h:=\sqrt{p_h}$, so that $\sum_hq_h^2=1$. Use the triangle inequality, (17.5), the Rayleigh quotient, and the standard bound "largest eigenvalue $\le$ maximal row sum" for nonnegative matrices:

$$
\lVert v(\lambda)\rVert\le\sum_{\{h,h'\}\in E}2\lvert t_{hh'}\rvert d_0(h,h')\sqrt{W(h,h';\lambda)p_hp_{h'}}\le\sum_{h,h'}a_{hh'}q_hq_{h'}\le v_{\max}:=\varrho(a)\le\max_h\sum_{h'}\lvert t_{hh'}\rvert\,d_0(h,h') .
\tag{17.7}
$$

Here $\varrho(a)$ is the largest eigenvalue of $a$. The bound holds for every start and every $\lambda$, so $\lVert\bar x(\lambda_2)-\bar x(\lambda_1)\rVert\le v_{\max}\lvert\lambda_2-\lambda_1\rvert$.

### Step 7. Item 3: localized start $\phi=\lvert\beta\rangle$

The series $\Psi(\lambda)=\sum_n\frac{(-i\lambda)^n}{n!}C^n(\lvert\beta\rangle\otimes\chi)$ converges, since $\mathcal H$ is finite-dimensional. Define its components $c_n(h):=(\langle h\rvert\otimes\mathbb 1)C^n(\lvert\beta\rangle\otimes\chi)$. They obey $c_{n+1}(h)=\sum_{h'}t_{hh'}c_n(h')+K_hc_n(h)$ with $c_0(h)=\delta_{h\beta}\chi$. For $h\neq\beta$:

$$
\begin{aligned}
c_1(h)&=t_{h\beta}\chi,\qquad c_2(h)=(t^2)_{h\beta}\chi+t_{h\beta}(K_h+K_\beta)\chi,\\
\langle\chi\vert c_3(h)\rangle&=(t^3)_{h\beta}+\sum_{h'}t_{hh'}t_{h'\beta}(\epsilon_h+\epsilon_{h'}+\epsilon_\beta)+t_{h\beta}\langle K_h^2+K_hK_\beta+K_\beta^2\rangle .
\end{aligned}
\tag{17.8}
$$

Insert $\psi_h=-i\lambda c_1-\tfrac{\lambda^2}{2}c_2+\tfrac{i\lambda^3}{6}c_3+O(\lambda^4)$ into $p_h=\lVert\psi_h\rVert^2$. The orders are: $\lambda^2$, $\lVert c_1\rVert^2$; $\lambda^3$, $\operatorname{Im}\langle c_1\vert c_2\rangle$; $\lambda^4$, $\tfrac14\lVert c_2\rVert^2-\tfrac13\operatorname{Re}\langle c_1\vert c_3\rangle$. In the $\lambda^3$ term, $\operatorname{Im}[\lvert t_{h\beta}\rvert^2(\epsilon_h+\epsilon_\beta)]=0$. Hence, for $h\neq\beta$:

$$
p_h(\lambda)=\lambda^2\lvert t_{h\beta}\rvert^2+\lambda^3\operatorname{Im}\bigl[t_{h\beta}^*(t^2)_{h\beta}\bigr]+\lambda^4p_h^{(4)}+O(\lambda^5),\qquad p_\beta=1-\sum_{h\neq\beta}p_h .
\tag{17.9}
$$

The $\lambda^3$ terms cancel in $p_\beta$: their sum is $\operatorname{Im}\bigl[(t^3)_{\beta\beta}-t_{\beta\beta}(t^2)_{\beta\beta}\bigr]=0$, because $t=t^\dagger$. For $h$ at graph distance $n\ge2$ from $\beta$, every contribution to $c_k(h)$ with $k\le n$ consists of hops only, because a $K$ insertion uses up a power without moving. Hence $p_h=\lambda^{2n}\lvert(t^n)_{h\beta}\rvert^2/(n!)^2+O(\lambda^{2n+1})$, and the $K_h$ enter $p_h$ not before $\lambda^{2n+1}$.

**Order at which the records enter.** At orders $\lambda^2$ and $\lambda^3$, $p_h$ is independent of the $K_h$. They enter first at $\lambda^4$, both in $p_\beta$ and in the neighbours of $\beta$. Evaluate the $\lambda^4$ coefficient with (17.8), using $\operatorname{Re}\langle K_hK_\beta\rangle=\tfrac12\langle K_hK_\beta+K_\beta K_h\rangle$ and the orthogonal split $(K_h-K_\beta)\chi=(\epsilon_h-\epsilon_\beta)\chi+(u_h-u_\beta)$:

$$
p_h^{(4)}=\underbrace{\tfrac14\lvert(t^2)_{h\beta}\rvert^2-\tfrac13\operatorname{Re}\bigl[t^*_{h\beta}(t^3)_{h\beta}\bigr]}_{K\text{-independent}}
\;\underbrace{-\;\tfrac{1}{12}\lvert t_{h\beta}\rvert^2d_0(h,\beta)^2}_{\text{record spread}}
\;\underbrace{-\;\tfrac{1}{12}\lvert t_{h\beta}\rvert^2(\epsilon_h-\epsilon_\beta)^2+\tfrac16\sum_{h'}\operatorname{Re}\bigl(t^*_{h\beta}t_{hh'}t_{h'\beta}\bigr)(\epsilon_h+\epsilon_\beta-2\epsilon_{h'})}_{\text{record means}} .
\tag{17.10}
$$

The quadratic terms combine as $\tfrac14\langle(K_h+K_\beta)^2\rangle-\tfrac13\operatorname{Re}\langle K_h^2+K_hK_\beta+K_\beta^2\rangle=-\tfrac1{12}\langle(K_h-K_\beta)^2\rangle$. The linear terms come from $\tfrac12\operatorname{Re}[t^*_{h\beta}(t^2)_{h\beta}](\epsilon_h+\epsilon_\beta)$ and $-\tfrac13\sum_{h'}\operatorname{Re}(t^*_{h\beta}t_{hh'}t_{h'\beta})(\epsilon_h+\epsilon_{h'}+\epsilon_\beta)$. The "means" part is exactly the change of the $K$-independent part under the shift $t_{hh}\to t_{hh}+\epsilon_h$. The genuine record part, $-\tfrac1{12}\lvert t_{h\beta}\rvert^2d_0(h,\beta)^2\le0$, slows the transfer to neighbours that lie at a different background point. For $p_\beta$, the $K$-dependent part is minus the sum of these parts over $h\neq\beta$.

**Position, spread, velocity.** Insert (17.9) into $\bar x-u_\beta=\sum_{h\neq\beta}p_h(u_h-u_\beta)$. For $s^2$, use (17.3) with $k=\beta$, where $\lVert\bar x-u_\beta\rVert^2=O(\lambda^4)$:

$$
\bar x(\lambda)-u_\beta=\lambda^2w_2+\lambda^3w_3+O(\lambda^4),\qquad w_2=\sum_h\lvert t_{h\beta}\rvert^2(u_h-u_\beta),\quad w_3=\sum_h\operatorname{Im}\bigl[t^*_{h\beta}(t^2)_{h\beta}\bigr](u_h-u_\beta),
\tag{17.11}
$$

$$
v(\lambda)=2\lambda\,w_2+3\lambda^2w_3+O(\lambda^3),\qquad v(0)=0,
\tag{17.12}
$$

$$
s(\lambda)^2=\lambda^2\sum_h\lvert t_{h\beta}\rvert^2\,d_0(h,\beta)^2+O(\lambda^3).
\tag{17.13}
$$

Generically, the leading orders are $\lambda^2$ for $\bar x-u_\beta$, $\lambda$ for $v$ and $\lambda^2$ for $s^2$. If $w_2=0$, the next order is set by $w_3$. Up to these orders the records act only through the positions $u_h$. Their effect on the weights first moves $\bar x$ at $\lambda^4$ (and $v$ at $\lambda^3$), with record-spread part $-\tfrac{\lambda^4}{12}\sum_h\lvert t_{h\beta}\rvert^2d_0(h,\beta)^2(u_h-u_\beta)$.

### Step 8. Item 4: commuting records

Let $D_j:=\sum_h\kappa_h(j)\lvert h\rangle\langle h\rvert$. Then $C=\sum_j(C_b+D_j)\otimes P_j$. The $P_j$ are orthogonal projectors with $\sum_jP_j=\mathbb 1$, so $e^{-iC\lambda}=\sum_jU_j(\lambda)\otimes P_j$ and $\Psi(\lambda)=\sum_jU_j\phi\otimes P_j\chi$. The partial trace over $c$ kills the cross terms, since $P_{j'}P_j=\delta_{jj'}P_j$:

$$
V_b(\lambda)=\sum_jw_j\,U_j(\lambda)\lvert\phi\rangle\langle\phi\rvert U_j(\lambda)^\dagger,\qquad U_j(\lambda)=e^{-i(C_b+D_j)\lambda},\qquad w_j=\langle\chi\vert P_j\vert\chi\rangle .
\tag{17.14}
$$

The view is the $w_j$-weighted mixture of record-free hopping evolutions with on-site terms $\kappa_h(j)$.

### Step 9. Item 5: the chain, its background and the limit

Let $e:=(X-\langle X\rangle)\chi$, so $\lVert e\rVert=\sigma_X>0$. Then $u_{h_m}=m\,e$, and

$$
d_0(h_m,h_n)=\lvert m-n\rvert\,\sigma_X,\qquad \bar x=\bar m\,e,\quad \bar m:=\sum_mm\,p_{h_m},\qquad s^2=\sigma_X^2\sum_m(m-\bar m)^2p_{h_m} .
\tag{17.15}
$$

The background is an equally spaced line of distinct points. Item 4 applies with $P_j\to\Pi_\xi$, $w_\xi=\langle\chi\vert\Pi_\xi\vert\chi\rangle$ (finitely many $\xi$, since $\mathcal H_c$ is finite) and $D_\xi=\xi M$, where $M:=\sum_mm\lvert h_m\rangle\langle h_m\rvert$.

**Limit.** Take the finite chains $\{h_{-L},\dots,h_L\}$, with $C_b$ and $K_{h_m}$ restricted to them. Let $L\to\infty$ at fixed $\lambda$. In the interaction picture with respect to the diagonal $\xi M$, each $U_\xi$ is a Dyson series in the hopping. Its $n$-th term has norm $\le(2t\lambda)^n/n!$ and moves a place by at most $n$ steps. For starts supported on $\{h_0,h_1\}$, terms of order $<L$ do not feel the ends. So the finite and infinite amplitudes differ by at most $2\sum_{n\ge L}(2t\lambda)^n/n!\to0$. The amplitude at $h_m$ is bounded by $2\sum_{n\ge\lvert m\rvert-1}(2t\lambda)^n/n!$ uniformly in $L$. By dominated convergence, $p_{h_m}$, $\bar m$, $s^2$ and $v$ converge to their values on $\ell^2(\mathbb Z)$. There $S:=\sum_m\lvert h_m\rangle\langle h_{m+1}\rvert$ is unitary and $C_b=t(S+S^\dagger)$.

**Standard results (Wannier–Stark chain).** $[M,S]=-S$ gives $e^{i\xi M\lambda}Se^{-i\xi M\lambda}=e^{-i\xi\lambda}S$. The interaction-picture generators commute, so $U_\xi=e^{-i\xi M\lambda}\exp\bigl[-it(\zeta S+\zeta^*S^\dagger)\bigr]$ with $\zeta=\int_0^\lambda e^{-i\xi\lambda'}\mathrm d\lambda'$. The Bessel generating function $e^{(z/2)(w-w^{-1})}=\sum_nJ_n(z)w^n$ then gives

$$
\langle h_m\vert U_\xi(\lambda)\vert h_k\rangle=e^{-i\xi\lambda(m+k)/2}\,(-i)^{m-k}J_{m-k}\bigl(z_\xi(\lambda)\bigr),\qquad z_\xi(\lambda)=\frac{4t}{\xi}\sin\frac{\xi\lambda}{2}\quad(z_0=2t\lambda).
\tag{17.16}
$$

Heisenberg equations: $\tfrac{\mathrm d}{\mathrm d\lambda}S_H=-i\xi S_H$, so $S_H=e^{-i\xi\lambda}S$, and $\tfrac{\mathrm d}{\mathrm d\lambda}M_H=it(S_H-S_H^\dagger)$. Integrating,

$$
U_\xi^\dagger M\,U_\xi=M+\frac{t}{\xi}\Bigl[(1-e^{-i\xi\lambda})S+(1-e^{i\xi\lambda})S^\dagger\Bigr]\qquad(\xi=0:\ M+it\lambda(S-S^\dagger)).
\tag{17.17}
$$

### Step 10. Item 5(a): start $\lvert h_0\rangle\otimes\lvert\chi\rangle$

By (17.14) and (17.16):

$$
p_{h_m}(\lambda)=\sum_\xi w_\xi\,J_m\bigl(z_\xi(\lambda)\bigr)^2 .
\tag{17.18}
$$

Since $J_{-m}^2=J_m^2$, we get $\bar m=0$. By (17.17), $\sum_mm^2p_{h_m}=\sum_\xi w_\xi\lVert U_\xi^\dagger MU_\xi h_0\rVert^2=\sum_\xi w_\xi\tfrac{2t^2}{\xi^2}\lvert1-e^{i\xi\lambda}\rvert^2$. Hence

$$
\bar x(\lambda)=u_{h_0}=0,\qquad s(\lambda)^2=4t^2\sigma_X^2\sum_\xi w_\xi\,\frac{1-\cos\xi\lambda}{\xi^2}\qquad\Bigl(\xi=0\text{ term: }\tfrac{\lambda^2}{2}\Bigr).
\tag{17.19}
$$

The bound (17.7) for this chain: $a=t\sigma_X(S+S^\dagger)$, whose largest spectral value and row sum are both $2t\sigma_X$. On the finite chain the bound is $2t\sigma_X\cos\frac{\pi}{2L+2}$:

$$
v_{\max}=2t\,\sigma_X .
\tag{17.20}
$$

For this start, $v\equiv0$.

### Step 11. Item 5(b): start $2^{-1/2}(\lvert h_0\rangle+e^{i\theta}\lvert h_1\rangle)\otimes\lvert\chi\rangle$

Here $\langle\phi\vert M\vert\phi\rangle=\tfrac12$ and $\langle\phi\vert S\vert\phi\rangle=\tfrac12e^{i\theta}$. By (17.14), $\bar m=\sum_\xi w_\xi\langle\phi\vert U_\xi^\dagger MU_\xi\vert\phi\rangle$, and with (17.17):

$$
\bar x(\lambda)=\Bigl[\frac12+t\sum_\xi w_\xi\,\frac{\cos\theta-\cos(\theta-\xi\lambda)}{\xi}\Bigr]e\qquad\bigl(\xi=0\text{ term: }-\lambda\sin\theta\bigr),
\tag{17.21}
$$

$$
v(\lambda)=t\sum_\xi w_\xi\sin(\xi\lambda-\theta)\;e=t\,\operatorname{Im}\Bigl[e^{-i\theta}\langle\chi\vert e^{iX\lambda}\vert\chi\rangle\Bigr]\,(X-\langle X\rangle)\lvert\chi\rangle .
\tag{17.22}
$$

Hence $\lVert v\rVert\le t\sigma_X=v_{\max}/2$. At $\lambda=0$, (17.22) agrees with (17.6): $J_{h_0\to h_1}=2\operatorname{Im}[t\cdot\tfrac12e^{-i\theta}]=-t\sin\theta$.

### Step 12. Item 5(c): bounded spread

Every term of (17.19) is nonnegative. A term with $\xi\neq0$ is at most $2w_\xi/\xi^2$, while the $\xi=0$ term $w_0\lambda^2/2$ is unbounded iff $w_0>0$. Therefore

$$
\sup_\lambda s(\lambda)<\infty\iff\Pi_0\lvert\chi\rangle=0\ \ (\text{i.e. }\chi\perp\ker X),\qquad\text{then}\quad s^2\le8t^2\sigma_X^2\sum_{\xi\neq0}\frac{w_\xi}{\xi^2},\quad \Bigl\lVert\bar x_{(b)}-\tfrac12e\Bigr\rVert\le2t\sigma_X\sum_{\xi\ne0}\frac{w_\xi}{\lvert\xi\rvert}.
\tag{17.23}
$$

The last bound follows from (17.21), since each term with $\xi\neq0$ is at most $2w_\xi/\lvert\xi\rvert$. If $0\notin\operatorname{spec}X$, the condition holds automatically. **Answer:** under this condition, $\bar x(\lambda)$ of (b) stays bounded (it is quasi-periodic).

## Result

- **1(a)** No: $d_0$ is independent of the $t_{hh'}$ (17.2).
- **1(b)** $s$ is well defined. $\bar x$ and $v$ are well defined up to the common isometry $e^{i\varphi}$ of the background embedding, so all their $g$-invariants are well defined. Identities (17.3).
- **2** Continuity equation with antisymmetric current $J_{h'\to h}=2\operatorname{Im}[t_{hh'}(V_b)_{h'h}]$ (17.4). Current bound $\lvert J_{h'\to h}\rvert\le2\lvert t_{hh'}\rvert\sqrt{Wp_hp_{h'}}$ (17.5). Velocity (17.6). Maximal speed $\lVert v\rVert\le\varrho(\lvert t_{hh'}\rvert d_0(h,h'))\le\max_h\sum_{h'}\lvert t_{hh'}\rvert d_0(h,h')$ (17.7).
- **3** Leading orders (17.9), (17.11)–(17.13). The $K_h$ enter $p_h$ first at $\lambda^4$, with coefficient (17.10); the genuine record part is $-\tfrac1{12}\lvert t_{h\beta}\rvert^2d_0(h,\beta)^2$.
- **4** Closed form (17.14).
- **5** $p_{h_m}$ (17.18); $\bar x=0$ and $s^2$ (17.19); bound $2t\sigma_X$ (17.20); $\bar x$ and $v$ (17.21)–(17.22). Spread bounded iff $\Pi_0\chi=0$, and then $\bar x$ of (b) is bounded (17.23).

## Consistency checks

1. **Dimensions and conservation.** $[t]=[K]=[\lambda]^{-1}$, so $[u_h]=[d_0]=[\lambda]^{-1}$, as $d_0=\lim\alpha/\lambda$ requires. Further, $[v]=[\lambda]^{-2}=[t\,d_0]$ in (17.7), $\lambda^4\lvert t\rvert^2d_0^2$ in (17.10) is dimensionless, and so is $z_\xi$. By antisymmetry, $\sum_h\dot p_h=\sum_{h,h'}J_{h'\to h}=0$. The $\lambda^3$ terms of (17.9) cancel in $p_\beta$.
2. **Trivial limits.** No hopping: $J=0$, $v=0$, $v_{\max}=0$, and all coefficients in (17.9)–(17.13) vanish, consistent with $p_h=\delta_{h\beta}$. All $K_h=K$ equal: the two terms of $C=C_b\otimes\mathbb 1+\mathbb 1\otimes K$ commute, so the state stays a product and the view is record-free. Accordingly $d_0=0$, $v_{\max}=0$, $\bar x$ is constant, and the $K$-dependent part of (17.10) vanishes.
3. **Item 3 against item 5.** Expanding (17.18) for $m=1$ with $J_1^2=z^2/4-z^4/16+\dots$ and $z_\xi=2t\lambda-t\xi^2\lambda^3/12+\dots$ gives $p_{h_1}=t^2\lambda^2-\bigl(t^4+\tfrac{t^2}{12}\langle X^2\rangle\bigr)\lambda^4+O(\lambda^6)$. (17.10) with $(t^2)_{h_1h_0}=0$, $(t^3)_{h_1h_0}=3t^3$, no common neighbour and $d_0^2+(\epsilon_1-\epsilon_0)^2=\sigma_X^2+\langle X\rangle^2=\langle X^2\rangle$ gives the same result.

## Open issues

- Item 3: for places at graph distance $n\ge2$, only the leading term and the bound "$K$ not before $\lambda^{2n+1}$" are given. The leading order of $\bar x-u_\beta$ when $w_2=w_3=0$ is not determined in general (in 5(a) it vanishes identically).
- The bound (17.7) is not shown to be attained. In 5(b), $\lVert v\rVert\le v_{\max}/2$.
- Position and velocity are measured in the fixed background built from the start $\chi$ (Goal definition). How the witness geometry at finite $\lambda$ deviates from $\lambda d_0$ is not addressed.

## Methods used

- Branch (component) form of the evolution, continuity equation for place weights
- Cauchy–Schwarz, Rayleigh quotient, row-sum bound for nonnegative matrices
- Taylor/Dyson expansion in $\lambda$, partial trace
- Joint spectral decomposition of commuting operators
- Interaction and Heisenberg pictures, Bessel generating function (Wannier–Stark chain)
- Finite-chain limit via Dyson bounds and dominated convergence