Version 1 · earlier version; the current one is v3 · External review, round 1: major errors

# Mediated effect of a source on a moving body

- **Subproject:** 03-ilang-space
- **Package:** 18-mediated-effect
- **Version:** v1
- **Mode:** new
- **Date:** 2026-10-08

## Setup and assumptions

Setting, notation and definitions as in question.md: source $s$ (places $\sigma$, no hopping), moving body $b$ (places $h$, hopping $t_{hh'}$, records $K_h$), medium $c$; start $\lvert\sigma_0\rangle\lvert\beta\rangle\lvert\chi\rangle$; $\langle X\rangle=\langle\chi\vert X\vert\chi\rangle$, $\tilde X=X-\langle X\rangle$, $u_h=\tilde K_h\chi$ (5.10); $p_h(\lambda;\sigma_0)$ the place weights of $V_{\{b\}}$ (2.2); $\bar x=\sum_hp_hu_h$; influence $\delta p_h=p_h(\cdot;\sigma)-p_h(\cdot;\sigma')$. Further notation used below:
$$
\Delta:=B_\sigma-B_{\sigma'},\qquad D_x:=K_x-K_\beta\ (x\in H_T,\ D_\beta=0),\qquad \tau_x:=t^*_{h\beta}t_{hx}t_{x\beta},\qquad h\sim\beta:\ t_{h\beta}\neq0,\ h\neq\beta .
$$
Inputs used: (2.1), (2.2), (5.10), (6.4), (6.11), (6.14), (6.15), (8.15), (17.24) as quoted in question.md. Contracts are $\lambda$-independent, $\lambda\ge0$; items 2–5 are Taylor statements as $\lambda\to0^+$ (all expansions converge, A1). The diagonal entries $t_{hh}$ are allowed; they are covered by all formulas.

## Derivation

### Step 1. Reduction (item 1a)

Every term of $C$ is diagonal in the source places ($C_b$, $K_h$, $C_c$ act trivially on $s$), so
$$
C=\sum_\sigma\lvert\sigma\rangle\langle\sigma\rvert\otimes C^{(\sigma)},\qquad
C^{(\sigma)}=C_b\otimes\mathbb 1_c+\sum_h\lvert h\rangle\langle h\rvert\otimes K_h+\mathbb 1_b\otimes(C_c+B_\sigma),
$$
$$
\lvert\Psi(\lambda)\rangle=\lvert\sigma_0\rangle\otimes e^{-\mathrm iC^{(\sigma_0)}\lambda}\bigl(\lvert\beta\rangle\otimes\chi\bigr),\qquad V_{\{s\}}(\lambda)=\lvert\sigma_0\rangle\langle\sigma_0\rvert .
\tag{18.1}
$$
The source stays at $\sigma_0$ for every $\lambda$ (as in (6.4)); $b$ and $c$ are driven by the **reduced contract** $C^{(\sigma_0)}$, in which the source acts only as the medium term $B_{\sigma_0}$.

Since $\sum_x\lvert x\rangle\langle x\rvert=\mathbb 1_b$, (18.1) can be rewritten with the medium operator $R_\sigma:=K_\beta+C_c+B_\sigma$:
$$
C^{(\sigma)}=C_b\otimes\mathbb 1+\mathbb 1\otimes R_\sigma+\sum_x\lvert x\rangle\langle x\rvert\otimes D_x .
\tag{18.2}
$$
Interaction picture with respect to $\mathbb 1\otimes R_\sigma$: $e^{-\mathrm iC^{(\sigma)}\lambda}=(\mathbb 1\otimes e^{-\mathrm iR_\sigma\lambda})\,\mathcal T\exp\bigl(-\mathrm i\int_0^\lambda C_I(s)\,\mathrm ds\bigr)$ with
$$
C_I(s)=C_b\otimes\mathbb 1+\sum_x\lvert x\rangle\langle x\rvert\otimes D_x(s),\qquad D_x(s):=e^{\mathrm iR_\sigma s}D_xe^{-\mathrm iR_\sigma s} .
\tag{18.3}
$$
The prefactor is a unitary on the companion of $b$, hence it leaves $V_{\{b\}}$ unchanged ((2.2): $\psi_h\to e^{-\mathrm iR\lambda}\psi_h$). **$p_h(\lambda;\sigma)$ depends on the source only through the operator functions $D_x(s)$.**

### Step 2. Sufficient condition and invariance (item 1b)

Suppose
$$
[\Delta,\;K_h+C_c+B_{\sigma'}]=0\qquad\text{for every }h\in H_T .
\tag{18.4}
$$
Taking differences over $h$ gives $[\Delta,D_x]=0$, and $h=\beta$ gives $[\Delta,R_{\sigma'}]=0$. Then $e^{\mathrm iR_\sigma s}=e^{\mathrm iR_{\sigma'}s}e^{\mathrm i\Delta s}$ and $D_x^{(\sigma)}(s)=D_x^{(\sigma')}(s)$ for all $s,x$, so by Step 1 the whole view $V_{\{b\}}$ coincides for $\sigma$ and $\sigma'$: $\delta p_h(\lambda)=0$ and $\delta\bar x(\lambda)=0$ for every $\lambda$. Special cases: $\Delta\propto\mathbb 1$; or $\Delta$ commuting separately with all $K_h$, $C_c$ and $B_{\sigma'}$.

*Invariance.* The total $C$ is convention-independent (A5), hence so are its blocks $C_{\sigma h,\sigma h}:=(\langle\sigma h\rvert\otimes\mathbb 1_c)C(\lvert\sigma h\rangle\otimes\mathbb 1_c)=t_{hh}+K_h+C_c+B_\sigma$. Thus $\Delta=C_{\sigma h,\sigma h}-C_{\sigma' h,\sigma' h}$ (any $h$) is invariant, and (18.4) reads $[\Delta,C_{\sigma'h,\sigma'h}]=0$ ($t_{hh}$ is a number): the condition is invariant. The separate commutators $[\Delta,K_h]$, $[\Delta,C_c]$ are *not* (moving a medium term $E$ from $C_c$ into all $K_h$ changes them by $\pm[\Delta,E]$). The influence is invariant, since $p_h$ is computed from the state (18.1), which depends only on the total $C$. The same holds for $R_\sigma$, $D_x$ and $D_x(s)$, which are differences or sums of the blocks.

### Step 3. Leading source dependence of $D_x(s)$

Treat items 2 and 3 together: one medium object with $C_c=0$ is the case $r=0$ below (no $J_k$). $R_\sigma=R_{\sigma'}+\Delta$ and $D_x(s)=\sum_k\frac{(\mathrm is)^k}{k!}\mathrm{ad}_{R_\sigma}^k(D_x)$. In the chain, $D_x$ acts on $c_1$ and $\Delta$ on $c_{r+1}$. Each $\mathrm{ad}_{J_k}$ enlarges the support by at most one cell; $K_\beta$ and $B_{\sigma'}$ do not enlarge it; $\Delta$ commutes with every operator on $c_1\cdots c_r$. So the shortest $\Delta$-dependent word is $\mathrm{ad}_\Delta\mathrm{ad}_{J_r}\cdots\mathrm{ad}_{J_1}$, which is (8.15) with $B_{\beta_0}\to\Delta$, $D\to D_x$:
$$
\delta D_x(s):=D_x^{(\sigma)}(s)-D_x^{(\sigma')}(s)=\frac{s^{r+1}}{(r+1)!}\,Y_r^{(x)}+O(s^{r+2}),\qquad
Y_r^{(x)}:=\mathrm i^{\,r+1}\bigl[\Delta,[J_r,[\cdots[J_1,D_x]\cdots]]\bigr],
\tag{18.5}
$$
with $Y_0^{(x)}=\mathrm i[\Delta,D_x]$ and $Y_r^{(\beta)}=0$. Since $D_x(s)$ is self-adjoint for every $s$, every Taylor coefficient of $\delta D_x$ is self-adjoint; in particular $\langle Y_r^{(x)}\rangle\in\mathbb R$. By Step 2, $\delta D_x$ and hence $Y_r^{(x)}$ do not depend on the A5 convention.

### Step 4. Dyson expansion and order counting

Expand (18.3) in the hops. The $n$-hop part of $\psi_h$ (notation (2.1)) is
$$
\psi_h^{[n]}=(-\mathrm i)^n\sum_{w}t_w\int_{0<s_1<\cdots<s_n<\lambda}V_{w_n}(\lambda,s_n)\cdots V_{w_0}(s_1,0)\,\chi ,\qquad V_x(b,a)=\mathcal T e^{-\mathrm i\int_a^bD_x},
$$
summed over walks $w=(\beta=w_0,\dots,w_n=h)$ with $t_w=\prod_jt_{w_{j+1}w_j}$. Every hop and every insertion of $D$ adds one power of $\lambda$; by (18.5) an insertion of $\delta D$ adds $r+2$ powers. For $h$ at graph distance $n$ from $\beta$, $\psi_h^{[k]}=0$ for $k<n$, so $\delta p_h=O(\lambda^{2n+r+2})$. At this order only $\delta\psi_h^{[n]}$ with one $\delta D$ and no $D$ contributes. The time spent on $w_j$ is the interval $(s_j,s_{j+1})$ ($s_0=0$, $s_{n+1}=\lambda$), and the Beta integral gives
$$
\delta\psi_h^{[n]}=-\mathrm i(-\mathrm i)^n\sum_wt_w\sum_{j=0}^n a_j\,Y_r^{(w_j)}\chi,\qquad a_j=\binom{r+1+j}{j}\frac{\lambda^{n+r+2}}{(n+r+2)!}.
$$
With $\psi_h^{[n]}=(-\mathrm i)^n(t^n)_{h\beta}\frac{\lambda^n}{n!}\chi+\dots$, we get $\delta\lVert\psi_h^{[n]}\rVert^2=2\operatorname{Re}\langle\psi_h^{[n]}\vert\delta\psi_h^{[n]}\rangle$, and therefore
$$
\delta p_h=\frac{2\lambda^{2n+r+2}}{n!\,(n+r+2)!}\operatorname{Im}\Bigl[(t^n)^*_{h\beta}\sum_wt_w\sum_{j=1}^{n-1}\binom{r+1+j}{j}\langle Y_r^{(w_j)}\rangle\Bigr]+O(\lambda^{2n+r+3}) .
\tag{18.6}
$$
The terms $j=0$ ($Y^{(\beta)}=0$) and $j=n$ (common to all $w$, giving $\operatorname{Im}(\lvert(t^n)_{h\beta}\rvert^2\cdot\text{real})=0$) drop. Equivalently, the bracket is $\sum_{w,w'}\operatorname{Im}(t^*_{w'}t_w)\,\mu_w$ with real $\mu_w$. So (18.6) vanishes for $n=1$, and whenever all products $t^*_{w'}t_w$ of shortest walks are real (real hopping, or a unique shortest path); then $\delta p_h=O(\lambda^{2n+r+3})$.

### Step 5. Neighbours and $\beta$: order $r+5$

Let $h\sim\beta$. By the counting of Step 4, the $\lambda^{r+5}$ terms of $\delta p_h$ come from (i) $\lVert\psi_h^{[1]}\rVert^2$ with one $\delta D$ and one $D$, and (ii) $2\operatorname{Re}\langle\psi_h^{[1]}\vert\psi_h^{[2]}\rangle$ with one $\delta D$ and no $D$. Write $\psi_h^{[1]}=-\mathrm it_{h\beta}\Phi$, $\Phi=\lambda\chi-\mathrm i\frac{\lambda^2}{2}D_h\chi+\dots$, $Y:=Y_r^{(h)}$, and use the simplex integrals $\int_0^\lambda s\,f=\frac{(r+2)\lambda^{r+3}}{(r+3)!}$, $\int_{0<a<b<\lambda}a\,f(a)\,\mathrm d a\,\mathrm d b=\frac{(r+2)\lambda^{r+4}}{(r+4)!}$, $\int_{0<a<b<\lambda}a\,f(b)\,\mathrm da\,\mathrm db=\frac{(r+2)(r+3)}{2}\frac{\lambda^{r+4}}{(r+4)!}$ for $f(s)=s^{r+1}/(r+1)!$:
$$
\begin{aligned}
\delta\Phi&=-\mathrm i\tfrac{(r+2)\lambda^{r+3}}{(r+3)!}Y\chi-\tfrac{(r+2)\lambda^{r+4}}{(r+4)!}\bigl[D_hY+\tfrac{r+3}{2}YD_h\bigr]\chi+\dots,\\
\text{(i)}:\ \lvert t_{h\beta}\rvert^2\,2\operatorname{Re}\langle\Phi\vert\delta\Phi\rangle&=\lvert t_{h\beta}\rvert^2\tfrac{(r+2)\lambda^{r+5}}{(r+4)!}\bigl[(r+4)-(r+5)\bigr]\operatorname{Re}\langle D_hY\rangle .
\end{aligned}
$$
The $\lambda^{r+4}$ term $2\operatorname{Re}\langle\lambda\chi\vert{-\mathrm i}c\,Y\chi\rangle$ vanishes because $\langle Y\rangle$ is real; the same holds for the next Taylor term of $\delta D$. For (ii), $\psi_h^{[2]}=-\sum_xt_{hx}t_{x\beta}\frac{\lambda^2}{2}\chi+\mathrm i\sum_xt_{hx}t_{x\beta}\frac{(r+2)\lambda^{r+4}}{(r+4)!}\bigl[\tfrac{r+3}{2}Y_r^{(h)}+Y_r^{(x)}\bigr]\chi+\dots$, which gives $2\operatorname{Re}\langle\psi^{[1]}\vert\psi^{[2]}\rangle\supset\sum_x\operatorname{Re}\tau_x\frac{(r+2)\lambda^{r+5}}{(r+4)!}\bigl[(r+4)-(r+3)\bigr]\langle Y_r^{(h)}\rangle-2\sum_x\operatorname{Re}\tau_x\frac{(r+2)\lambda^{r+5}}{(r+4)!}\langle Y_r^{(x)}\rangle$. Hence
$$
\delta p_h=\frac{(r+2)\lambda^{r+5}}{(r+4)!}\Bigl[\sum_x\operatorname{Re}\tau_x\,\bigl\langle Y_r^{(h)}-2Y_r^{(x)}\bigr\rangle-\lvert t_{h\beta}\rvert^2\operatorname{Re}\langle D_hY_r^{(h)}\rangle\Bigr]+O(\lambda^{r+6})\qquad(h\sim\beta),
\tag{18.7}
$$
where $x$ runs over all places with $t_{hx}t_{x\beta}\ne0$ (common neighbours, and $x=h,\beta$ if $t_{hh},t_{\beta\beta}\ne0$). Since $\sum_hp_h=1$ (2.1) and places at distance $\ge2$ contribute at order $\ge r+6$ (18.6),
$$
\delta p_\beta=-\lambda^{r+5}\sum_{h\sim\beta}\delta p_h^{(r+5)}+O(\lambda^{r+6}),
\tag{18.8}
$$
where $\delta p_h^{(r+5)}$ is the bracket coefficient of (18.7).

### Step 6. Item 2: one medium object, $C_c=0$ ($r=0$)

With $\langle\mathrm i[\Delta,A]\rangle=-2\operatorname{Im}\langle\Delta A\rangle$ and $\operatorname{Re}\langle D\,\mathrm i[\Delta,D]\rangle=-\operatorname{Im}\langle\Delta D^2\rangle$ (self-adjoint $A,D,\Delta$), (18.7) becomes, for $h\sim\beta$,
$$
\delta p_h=\lambda^5\Bigl[\frac{\lvert t_{h\beta}\rvert^2}{12}\operatorname{Im}\langle\Delta D_h^2\rangle+\frac16\sum_x\operatorname{Re}\tau_x\operatorname{Im}\bigl\langle\Delta(2D_x-D_h)\bigr\rangle\Bigr]+O(\lambda^6),
\tag{18.9}
$$
$\delta p_\beta=O(\lambda^5)$ by (18.8), and for distance $n\ge2$ the lowest possible order is $2n+2$, with coefficient (18.6) at $r=0$: $\frac{4}{n!(n+2)!}\operatorname{Im}\bigl[(t^n)^*_{h\beta}\sum_wt_w\sum_{j=1}^{n-1}(j+1)\operatorname{Im}\langle D_{w_j}\Delta\rangle\bigr]$ (otherwise $\ge2n+3$). The orders 2 to 4 are source-independent. At $r=0$, (18.5) holds with any $C_c$, so (18.9) holds also for $C_c\neq0$.

**Item 2(b).** The source part of (6.11) for the pair $(h,\beta)$ with $B_{\beta_0}\to\Delta$ is $s_{h\beta}:=\operatorname{Im}\langle\Delta\tilde D_h^2\rangle$. From $D_h^2=\tilde D_h^2+2\langle D_h\rangle D_h-\langle D_h\rangle^2$:
$$
\operatorname{Im}\langle\Delta D_h^2\rangle=s_{h\beta}+2(\epsilon_h-\epsilon_\beta)\operatorname{Im}\langle\Delta D_h\rangle .
\tag{18.10}
$$
So the coefficient (18.9) is $\frac{\lvert t_{h\beta}\rvert^2}{12}s_{h\beta}$ plus the record-mean term $\frac{\lvert t_{h\beta}\rvert^2}{6}(\epsilon_h-\epsilon_\beta)\operatorname{Im}\langle\Delta D_h\rangle$ plus the $\tau_x$ term. These are independent data, so the coefficient is **not** a function of $s_{h\beta}$: in Step 9, $s_{h\beta}=2\sigma\ne0$ while the coefficient vanishes. It reduces to $\frac{\lvert t_{h\beta}\rvert^2}{12}s_{h\beta}$ exactly when $(\epsilon_h-\epsilon_\beta)\operatorname{Im}\langle\Delta D_h\rangle=0$ and the $\tau_x$ term vanishes (e.g. no triangle through $h,\beta$ and $t_{hh}=t_{\beta\beta}=0$).

**Item 2(c).** $\sum_h\delta p_h=0$, and only $\beta$ and its neighbours carry $\lambda^{r+5}$ terms. Hence
$$
\delta\bar x=\lambda^{r+5}\sum_{h\sim\beta}\delta p_h^{(r+5)}\,(u_h-u_\beta)+O(\lambda^{r+6}),
\tag{18.11}
$$
i.e. the lowest order is $\lambda^5$ for item 2, with $\delta p_h^{(5)}$ from (18.9). A common shift of all $u_h$ (A5 convention $K_h\to K_h+E$) drops out.

### Step 7. Item 3: locality

By (18.6)–(18.8), the lowest order at which $\delta p_h$ can be nonzero is $\lambda^{r+5}$, at $\beta$ and at its neighbours. It is governed by the nested commutators $Z_r^{(x)}=[\Delta,[J_r,[\cdots[J_1,D_x]\cdots]]]$ of (8.15), with $x=h$ and the intermediate places $x$ of (18.7). Places at distance $n$ follow at order $2n+r+2$ (only with non-real loop products, (18.6)) or later. Each cell between the record cell $c_1$ and the source cell $c_{r+1}$ delays the influence by exactly one order. If $Z_r^{(x)}=0$ for all relevant $x$, the order is higher.

### Step 8. Item 4: cost

Write $B_\sigma=\langle B_\sigma\rangle\mathbb 1+\tilde B_\sigma$. By (18.1), $C^{(\sigma)}=C^{(\sigma)}\big|_{B_\sigma\to\tilde B_\sigma}+\langle B_\sigma\rangle\mathbb 1$, and the last term gives only the common phase $e^{-\mathrm i\langle B_\sigma\rangle\lambda}$ (1.2). So $p_h(\lambda;\sigma)$ depends on $B_\sigma$ only through $\tilde B_\sigma$, and $\delta p_h$ is unchanged under $\Delta\to\Delta+c\mathbb 1$. The cost difference of the two starts is $L(\sigma)-L(\sigma')=\langle\Delta\rangle$ ((1.8); $L$ is conserved under (1.9)), and it is A5-invariant (Step 2). Therefore
$$
\delta p_h\ \text{and}\ \delta\bar x\ \text{are independent of}\ L(\sigma)-L(\sigma')=\langle\Delta\rangle ;
\tag{18.12}
$$
the coefficients (18.7), (18.9) contain $\Delta$ only through commutators. Answer: **no**. Counterexamples: $\Delta=c\mathbb 1$ has cost difference $c\ne0$ and zero influence (exactly); Step 9 has $\langle\sigma\sigma_y\rangle=0$, so zero cost difference, and nonzero influence.

### Step 9. Item 5: the qubit example

Here $\Delta=\sigma\sigma_y$ ($\sigma'=0$), $r=0$, $t$ is real, the graph is a chain (bipartite, $t_{hh}=0$, no triangles) and $\beta=h_m$. All records including the source are Pauli vectors in one plane $\Pi$: $Q_{h_i}:=K_{h_i}+B_\sigma=\vec q_i\cdot\vec\sigma$ with $\vec q_i=(i,\sigma,i)$.

*Parity.* (a) Complex conjugation in the place and $\sigma_z$ bases maps $C^{(\sigma)}\to C^{(-\sigma)}$ and fixes $\lvert h_m\rangle\chi$, so $p_h(\lambda;\sigma)=p_h(-\lambda;-\sigma)$. (b) In (17.24) with $K_h\to Q_h$, each $c_k(h)$ is a sum of real hopping products times words in the $Q$'s applied to $\chi$. On a bipartite chain all walks $\beta\to h$ have the parity of the distance, so the number of $Q$'s in $\langle c_k\vert c_l\rangle$ has the parity of $k+l$. Since $(\vec a\cdot\vec\sigma)(\vec b\cdot\vec\sigma)=\vec a\cdot\vec b+\mathrm i(\vec a\times\vec b)\cdot\vec\sigma$ with $\vec a\times\vec b\perp\Pi$, an odd product of $Q$'s equals $\vec v\cdot\vec\sigma$ with real $\vec v$ and has a real expectation. For odd $N=k+l$ the factor $\mathrm i^k(-\mathrm i)^l$ is imaginary, so the real number $p_h^{(N)}$ vanishes. Hence $p_h$ is even in $\lambda$, and by (a) even in $\sigma$: **the influence is exactly even in $\sigma$.**

*Order 5.* For $h=h_{m\pm1}$, $D_h=\pm X$ and $X^2=2\cdot\mathbb 1$, so (18.9) gives $\operatorname{Im}\langle\Delta D_h^2\rangle=2\sigma\operatorname{Im}\langle\sigma_y\rangle=0$. In (18.10), $s_{h\beta}=2\sigma$ (as in (6.14)) is cancelled by the record-mean term $2(\pm1)(\mp\sigma)=-2\sigma$.

*Order 6.* The Δ-dependent $\lambda^6$ terms of $p_{h_{m\pm1}}$ come only from the one-hop part: $\psi^{[2]}=0$, and in $2\operatorname{Re}\langle\psi^{[1]}\vert\psi^{[3]}\rangle$ one $\delta D$ without $D$ gives (real)$\times\mathrm i\langle Y\rangle$, whose real part is zero. The one-hop part is $t^2\lVert\Phi\rVert^2$ with $\Phi=\int_0^\lambda U(s)\chi\,\mathrm ds$, $U(s)=e^{\mathrm iQ_hs}e^{-\mathrm iQ_\beta s}$, and
$$
\lVert\Phi\rVert^2=\sum_{n,n'}\frac{\lambda^{n+n'+2}\langle U_{n'}\chi\vert U_n\chi\rangle}{(n+1)!(n'+1)!},\qquad U_n=\sum_k\binom nk(\mathrm iQ_h)^k(-\mathrm iQ_\beta)^{n-k}.
$$
With $d=\pm1$, $A_h=\lvert\vec q_h\rvert^2$, $A_\beta=\lvert\vec q_\beta\rvert^2$, $W=\sigma_z-\sigma_x$, $Q_hQ_\beta=\vec q_h\cdot\vec q_\beta+\mathrm i\sigma dW$, and $\langle X\rangle=\langle W\rangle=1$, $\langle\sigma_y\rangle=0$, $W^2=2$:
$U_1=\mathrm idX$, $U_2=-2+2\mathrm i\sigma dW$, $U_3=-\mathrm i[(A_h+3A_\beta)Q_h-(A_\beta+3A_h)Q_\beta]$, $U_4=A_h^2+6A_hA_\beta+A_\beta^2-4(A_h+A_\beta)Q_hQ_\beta$. Their $\sigma$-dependent parts are $\operatorname{Re}\langle U_4\rangle\ni8\sigma^2$, $\operatorname{Re}\langle U_1\chi\vert U_3\chi\rangle\ni-8\sigma^2$ and $\lVert U_2\chi\rVert^2=4+8\sigma^2$. The $\lambda^6$ coefficient $\frac{2\operatorname{Re}\langle U_4\rangle}{5!}+\frac{2\operatorname{Re}\langle U_1\chi\vert U_3\chi\rangle}{2!\,4!}+\frac{\lVert U_2\chi\rVert^2}{3!^2}$ has the $\sigma$-part $\sigma^2\bigl(\frac{16}{120}-\frac{16}{48}+\frac{8}{36}\bigr)=\frac{\sigma^2}{45}$, the same for $d=\pm1$. Distance-2 places: order 6 vanishes by (18.6) (real $t$), order 7 by parity. Hence
$$
\delta p_{h_{m\pm1}}=\frac{t^2\sigma^2}{45}\lambda^6+O(\lambda^8),\qquad
\delta p_{h_m}=-\frac{2t^2\sigma^2}{45}\lambda^6+O(\lambda^8),\qquad \delta p_{h_{m\pm k}}=O(\lambda^8)\ (k\ge2).
\tag{18.13}
$$
Only one-hop amplitudes enter, so (18.13) holds for every $2\le m\le n-1$.

*Position.* $u_{h_i}=\mathrm i(X-1)\lvert0\rangle=\mathrm i\lvert1\rangle$, so $\delta\bar x=\lvert1\rangle\sum_ii\,\delta p_{h_i}$. At order 6 this is $\frac{t^2\sigma^2}{45}[(m+1)+(m-1)-2m]=0$, and odd orders vanish. Hence
$$
\delta\bar x=O(\lambda^8),\qquad\text{even in }\sigma .
\tag{18.14}
$$

*Comparison with (6.14), (6.15).* On the background of $b$, the source acts at first order in $\sigma$ and reverses with its sign: $W$ changes by $2\sigma(i-j)^2\lambda^3$, and the derived distance $\alpha(h_i,h_j)=\lambda\lvert i-j\rvert(1-\sigma\lambda)$ changes at relative order $\lambda$. On the motion of $b$ there is no effect at first order in $\sigma$, at any order in $\lambda$. The leading effect is $\propto\sigma^2\lambda^6$, at relative order $\lambda^4$ with respect to the hopping weight $t^2\lambda^2$. It is the same for both signs of $\sigma$, symmetric between $h_{m\pm1}$, and moves no weight from $h_m$ to one side; the position shift starts at $\lambda^8$ at the earliest. The motion is therefore not determined by the background deformation: the (6.11)-based guess $\frac{t^2}{12}s_{h\beta}\lambda^5=\frac{\sigma t^2}{6}\lambda^5$ is cancelled exactly by (18.10).

## Result

- **Reduction (18.1)–(18.3):** the source stays at $\sigma_0$; $b,c$ evolve under $C^{(\sigma_0)}$; $p_h$ depends on the source only through $D_x(s)=e^{\mathrm iR_\sigma s}(K_x-K_\beta)e^{-\mathrm iR_\sigma s}$, $R_\sigma=K_\beta+C_c+B_\sigma$.
- **No influence (18.4):** $[B_\sigma-B_{\sigma'},K_h+C_c+B_{\sigma'}]=0$ for all $h$ is sufficient. It is A5-invariant, as is the influence; the separate commutators with $K_h$ and $C_c$ are not.
- **Leading orders (18.6)–(18.9), (18.11):** at $\beta$ and its neighbours, $\delta p_h=O(\lambda^{r+5})$ with coefficient (18.7) ($r=0$: (18.9)); $\delta\bar x=O(\lambda^{r+5})$ with coefficient (18.11). Distance $n\ge2$: order $2n+r+2$ (18.6), which needs non-real loop products of shortest walks; otherwise $\ge2n+r+3$.
- **2(b) (18.10):** the coefficient equals $\frac{\lvert t_{h\beta}\rvert^2}{12}s_{h\beta}$ plus a record-mean term and a triangle term, so it is not a function of the (6.11) source part.
- **Locality (Step 7):** order $\lambda^{r+5}$, governed by $Z_r^{(x)}=[\Delta,[J_r,\dots[J_1,K_x-K_\beta]]]$, one order per intermediate cell.
- **Cost (18.12):** no; the influence is independent of the cost difference $\langle B_\sigma-B_{\sigma'}\rangle$.
- **Example (18.13), (18.14):** $\delta p_{h_{m\pm1}}=t^2\sigma^2\lambda^6/45$, $\delta p_{h_m}=-2t^2\sigma^2\lambda^6/45$, $\delta\bar x=O(\lambda^8)$; the influence is exactly even in $\sigma$, unlike the background effect (6.14), (6.15), which is odd in $\sigma$.

## Consistency checks

1. **Dimensions.** $[t]=[K]=[\Delta]=[J]=[C]=[\lambda]^{-1}$. The coefficient of (18.7) scales as $[C]^{r+5}$ (both terms), that of (18.6) as $[C]^{2n+r+2}$. In (18.13), restoring a record scale $K_{h_i}=\mathrm i\kappa X$ gives $t^2\sigma^2\kappa^2/45\sim[C]^6$.
2. **Limits.** $t\to0$: $p_h=\delta_{h\beta}$ exactly, and every coefficient is $\propto t^2$. $\Delta\propto\mathbb 1$, or (18.4) holds: all $Y_r^{(x)}=0$ and the influence vanishes. $\sigma\to0$ in (18.13): zero.
3. **Special case (17.10).** For $C_c=0$ the reduced contract is the 17 contract with records $K_h+B_\sigma$. (17.10) depends on the records only through $d_0(h,\beta)$ and differences of $\epsilon$'s, which are invariant under this common shift, so $\delta p_h^{(4)}=0$, as Step 6 states. In addition, the first term of (18.9) agrees with a direct Taylor expansion of $\lVert\Phi\rVert^2$, whose $\lambda^5$ coefficient is $\frac1{12}\operatorname{Im}\langle Q_\beta D_h^2\rangle$.

## Open issues

- In the example, the leading coefficient of $\delta\bar x$, at order $\lambda^8$ or later, is not computed. It needs the $\lambda^8$ terms of the 2-, 4- and 6-hop parts, and for $m\le3$ or $m\ge n-2$ also the chain ends. Even its non-vanishing is open.
- (18.6), (18.7) give the lowest *possible* orders. Nonzero instances: (18.9) for generic records. For (18.6), a square $\beta,x,y,h$ with $n=2$ gives $\frac16\operatorname{Im}(t_x^*t_y)\operatorname{Im}\langle(K_y-K_x)\Delta\rangle$, where $t_x=t_{hx}t_{x\beta}$ and $t_y=t_{hy}t_{y\beta}$. Exact vanishing conditions beyond these are not classified.
- (18.4) is sufficient, not necessary.

## Methods used

- block decomposition in the conserved source place
- interaction picture, Dyson (time-ordered) expansion
- adjoint expansion and support counting of nested commutators
- simplex and Beta integrals
- Pauli algebra; antiunitary (complex-conjugation) symmetry and a parity argument