Version 1 · current · External review, round 1: correct

# Body recorded by another cell: the body-dependent witness data through links

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

## Setup and assumptions

**Setting (question.md).** The objects are $p,b,c_1,c_2$, with pairwise different types and one instance each, so (1.11) imposes nothing. The part is $A=\{p\}$. The contract system is

$$
C=\sum_h\lvert h\rangle\langle h\rvert_p\otimes K_h+\sum_\beta\lvert\beta\rangle\langle\beta\rvert_b\otimes B_\beta+J,\qquad J=\sum_kP_k\otimes Q_k ,
$$

with $K_h$ on $c_1$, $B_\beta$ on $c_2$ and $J$ on $c_1c_2$, all self-adjoint and extended by the identity. The initial state is $\lvert\phi\rangle_p\otimes\lvert\beta_0\rangle_b\otimes\lvert\chi\rangle$ with $\lvert\chi\rangle=\lvert\chi_1\rangle\otimes\lvert\chi_2\rangle$ and $\phi_h\neq0$. We write $D=K_h-K_{h'}$ (on $c_1$) and $\tilde D=D-\langle D\rangle$.

**Notation.**
- $\langle X\rangle:=\langle\chi\vert X\vert\chi\rangle$, and $\langle\cdot\rangle_1,\langle\cdot\rangle_2$ are the expectations in $\lvert\chi_1\rangle,\lvert\chi_2\rangle$.
- For self-adjoint $X,Y$: $\tilde X=X-\langle X\rangle$, $\operatorname{Var}_\chi(X)=\langle\tilde X^2\rangle$, $\operatorname{Cov}_\chi(X,Y):=\tfrac12\langle\{\tilde X,\tilde Y\}\rangle=\tfrac12\langle\{\tilde X,Y\}\rangle$, and $\kappa_n(X)$ is the $n$-th cumulant of $X$ in $\lvert\chi\rangle$.
- $\mathrm i$ is the imaginary unit. Plain $i,j$ are place indices (Item 5 only).
- **Body-dependent part.** For $F\in\{W^{(\beta_0)},c_3^{(\beta_0)},c_4^{(\beta_0)},\alpha^{(\beta_0)}\}$, $\Delta F:=F-F\big|_{B_{\beta_0}=0}$, with $K_h$, $J$ and $\lvert\chi\rangle$ held fixed.

**Inputs.** (2.1), (2.2), (2.11), (3.5), (5.10) and (6.6)–(6.12), as quoted in question.md.

**Assumptions.** A1 (finite dimension), A5 ($\lambda$-independent contracts), $\lambda\ge0$. Items 1, 2 are exact. Items 3–5 are expansions for $\lambda\to0^+$. All operators are bounded (A1), so the series used below converge for small $\lambda$ and their remainders are uniform.

**Applicability of 06.** The derivation of (6.6)–(6.12) uses only three facts: the two branch operators and the operator common to both branches are self-adjoint, they act on a finite-dimensional medium, and the medium starts in a unit vector. All three hold here:
- the medium is $c_1c_2$, with finite-dimensional space $\mathcal H_{c_1}\otimes\mathcal H_{c_2}$ (A1);
- the branch operators $K_h\otimes\mathbb 1$, $K_{h'}\otimes\mathbb 1$ and the common operator $\mathbb 1\otimes B_{\beta_0}+J$ are self-adjoint;
- $\lvert\chi_1\rangle\otimes\lvert\chi_2\rangle$ is a unit vector.

Hence (6.6)–(6.12) hold with $\mathcal H_c\to\mathcal H_{c_1}\otimes\mathcal H_{c_2}$, $\lvert\chi\rangle\to\lvert\chi_1\rangle\otimes\lvert\chi_2\rangle$ and $B_{\beta_0}\to B_{\beta_0}+J$. The hypothesis of (6.7) becomes $[B_{\beta_0}+J,K_h]=[B_{\beta_0}+J,K_{h'}]=0$. Step 1 checks the block structure that (6.6) rests on.

## Derivation

### Step 1. Exact evolution (Item 1)

The projectors $\lvert h\rangle\langle h\rvert_p\otimes\lvert\beta\rangle\langle\beta\rvert_b$ commute with every term of $C$: the first two terms are diagonal in the place bases of $p$ and $b$, and $J$ acts on neither object. For every $\lvert\xi\rangle\in\mathcal H_{c_1}\otimes\mathcal H_{c_2}$,

$$
C\bigl(\lvert h\rangle_p\lvert\beta\rangle_b\otimes\lvert\xi\rangle\bigr)=\lvert h\rangle_p\lvert\beta\rangle_b\otimes H_{h\beta}\lvert\xi\rangle,\qquad H_{h\beta}:=K_h+B_\beta+J .
\tag{8.1}
$$

By (1.9), applied term by term, and writing $H_h:=H_{h\beta_0}$ (tensor factors labelled by object),

$$
\lvert\Psi(\lambda)\rangle=\lvert\beta_0\rangle_b\otimes\sum_h\phi_h\lvert h\rangle_p\otimes\lvert\chi_h(\lambda)\rangle,\qquad \lvert\chi_h(\lambda)\rangle:=e^{-\mathrm iH_h\lambda}\lvert\chi\rangle .
\tag{8.2}
$$

The factor of $b$ is $\lvert\beta_0\rangle$ for every $\lambda$. So the body stays at $\beta_0$: its view is $\lvert\beta_0\rangle\langle\beta_0\rvert$, and every reading of $b$ gives $\beta_0$ with probability one.

By (2.1), $\lvert\psi_h\rangle=\phi_h\lvert\beta_0\rangle\otimes\lvert\chi_h(\lambda)\rangle$. Then (2.2) gives $(V_A)_{hh'}=\phi_h\phi_{h'}^*\langle\chi_{h'}(\lambda)\vert\chi_h(\lambda)\rangle$ and $p_h=\lvert\phi_h\rvert^2>0$, so $H_V$ contains every place. By (2.11),

$$
W^{(\beta_0)}(h,h';\lambda)=\bigl\lvert G^{(\beta_0)}_{hh'}(\lambda)\bigr\rvert^2,\qquad
G^{(\beta_0)}_{hh'}(\lambda)=\langle\chi\vert e^{\mathrm i(K_{h'}+B_{\beta_0}+J)\lambda}e^{-\mathrm i(K_h+B_{\beta_0}+J)\lambda}\vert\chi\rangle .
\tag{8.3}
$$

This is (6.6) under the substitution above.

### Step 2. No link, no effect (Item 2)

Let $J=0$. The common operator $\mathbb 1\otimes B_{\beta_0}$ acts on $c_2$, so it commutes with $K_h\otimes\mathbb 1$ and $K_{h'}\otimes\mathbb 1$, and the hypothesis of (6.7) holds. By (6.7), $G^{(\beta_0)}_{hh'}=\langle\chi_1\vert e^{\mathrm iK_{h'}\lambda}e^{-\mathrm iK_h\lambda}\vert\chi_1\rangle$, which does not contain $B_{\beta_0}$ and is therefore also its value at $B_{\beta_0}=0$:

$$
J=0:\qquad \Delta W^{(\beta_0)}(h,h';\lambda)=0\qquad\text{for every }\lambda\ge0 .
\tag{8.4}
$$

### Step 3. Expansion of $W^{(\beta_0)}$ to fourth order (Item 3)

Let $M:=K_{h'}+B_{\beta_0}+J$, so that $K_h+B_{\beta_0}+J=M+D$. Use BCH to third order, $\ln(e^Se^T)=S+T+\tfrac12[S,T]+\tfrac1{12}\bigl([S,[S,T]]+[T,[T,S]]\bigr)+\dots$, with $S=\mathrm i\lambda M$ and $T=-\mathrm i\lambda(M+D)$. The three correction terms are $\tfrac{\lambda^2}{2}[M,D]$, $\tfrac{\mathrm i\lambda^3}{12}[M,[M,D]]$ and $\tfrac{\mathrm i\lambda^3}{12}\bigl([M,[M,D]]+[D,[M,D]]\bigr)$. Hence

$$
e^{\mathrm iM\lambda}e^{-\mathrm i(M+D)\lambda}=e^{-\mathrm i\lambda\Omega},\qquad
\Omega=D+\lambda\Omega_1+\lambda^2\Omega_2+O(\lambda^3),\qquad
\Omega_1=\tfrac{\mathrm i}{2}[M,D],\quad \Omega_2=-\tfrac16[M,[M,D]]-\tfrac1{12}[D,[M,D]] ,
\tag{8.5}
$$

where $\Omega_1$ and $\Omega_2$ are self-adjoint. The cumulant expansion $\ln\langle e^{-\mathrm i\lambda\Omega}\rangle=\sum_n(-\mathrm i\lambda)^n\kappa_n(\Omega)/n!$ has real $\kappa_n$, so the odd orders are imaginary. Since $W=\lvert G\rvert^2$, $\ln W=2\operatorname{Re}\ln G$, and

$$
\ln W^{(\beta_0)}=-\lambda^2\operatorname{Var}_\chi(\Omega)+\tfrac{\lambda^4}{12}\kappa_4(\Omega)+O(\lambda^6) .
\tag{8.6}
$$

Insert (8.5):
- $\operatorname{Var}_\chi(\Omega)=\operatorname{Var}_\chi(D)+2\lambda\operatorname{Cov}_\chi(D,\Omega_1)+\lambda^2\bigl[\operatorname{Var}_\chi(\Omega_1)+2\operatorname{Cov}_\chi(D,\Omega_2)\bigr]+O(\lambda^3)$;
- $\kappa_4(\Omega)=\kappa_4(D)+O(\lambda)$.

Exponentiating,

$$
W^{(\beta_0)}=1-\lambda^2\operatorname{Var}_\chi(D)-2\lambda^3\operatorname{Cov}_\chi(D,\Omega_1)
+\lambda^4\Bigl[\tfrac12\operatorname{Var}_\chi(D)^2+\tfrac1{12}\kappa_4(D)-\operatorname{Var}_\chi(\Omega_1)-2\operatorname{Cov}_\chi(D,\Omega_2)\Bigr]+O(\lambda^5).
\tag{8.7}
$$

**Check against (6.10).** Use $[M,D]=[M,\tilde D]$ and $\{\tilde D,[M,\tilde D]\}=[M,\tilde D^2]$. The $\lambda^3$ coefficient of (8.7) is then $-\tfrac{\mathrm i}{2}\langle[M,\tilde D^2]\rangle=\operatorname{Im}\langle M\tilde D^2\rangle$, which is (6.10) with $B_{\beta_0}\to B_{\beta_0}+J$.

### Step 4. Body-dependent part of $c_3^{(\beta_0)}$ (Item 3a)

By (6.11) with $B_{\beta_0}\to B_{\beta_0}+J$,
$$c_3^{(\beta_0)}=\operatorname{Im}\langle K_{h'}\tilde D^2\rangle+\operatorname{Im}\langle J\tilde D^2\rangle+\operatorname{Im}\langle B_{\beta_0}\tilde D^2\rangle .$$
Only the last term contains $B_{\beta_0}$. Since $B_{\beta_0}$ acts on $c_2$ and $\tilde D^2$ on $c_1$,

$$
\Delta c_3^{(\beta_0)}=\operatorname{Im}\langle\chi\vert B_{\beta_0}\tilde D^2\vert\chi\rangle=\frac{\langle\chi\vert[B_{\beta_0},\tilde D^2]\vert\chi\rangle}{2\mathrm i}=0 .
\tag{8.8}
$$

In the product state this is explicit: $\langle B_{\beta_0}\tilde D^2\rangle=\langle B_{\beta_0}\rangle_2\langle\tilde D^2\rangle_1$ is real.

### Step 5. Body-dependent part of $c_4^{(\beta_0)}$ (Item 3b)

The first two terms of the $\lambda^4$ bracket in (8.7) do not contain $M$, so they are body-free. Operators on $c_2$ commute with operators on $c_1$, so $[B_{\beta_0},D]=0$ and $[B_{\beta_0},[K_{h'},D]]=0$. Hence

$$
[M,D]=[K_{h'}+J,D],\qquad [M,[M,D]]=[K_{h'}+J,[K_{h'}+J,D]]+[B_{\beta_0},[J,D]] .
\tag{8.9}
$$

So $\Omega_1$ and $[D,[M,D]]$ are body-free, and the only body-dependent operator in (8.7) is $\Delta\Omega_2=-\tfrac16[B_{\beta_0},[J,D]]$. It enters linearly. With $\operatorname{Cov}_\chi(D,Y)=\tfrac12\langle\{\tilde D,Y\}\rangle$,

$$
\Delta c_4^{(\beta_0)}(h,h')=-2\operatorname{Cov}_\chi(D,\Delta\Omega_2)=\frac16\,\langle\chi\vert\{\tilde D,[B_{\beta_0},[J,D]]\}\vert\chi\rangle .
\tag{8.10}
$$

**Product form.** Three identities give it:
- since $D$ acts on $c_1$, $[B_{\beta_0},[J,D]]=\sum_k[P_k,D]\otimes[B_{\beta_0},Q_k]$;
- $\{\tilde D,[P_k,\tilde D]\}=[P_k,\tilde D^2]$;
- for self-adjoint $X,Y$, $\langle[X,Y]\rangle=2\mathrm i\operatorname{Im}\langle XY\rangle$.

Then $\langle[P_k,\tilde D^2]\rangle_1\langle[B_{\beta_0},Q_k]\rangle_2=(2\mathrm i)^2\operatorname{Im}\langle P_k\tilde D^2\rangle_1\operatorname{Im}\langle B_{\beta_0}Q_k\rangle_2$, and

$$
\Delta c_4^{(\beta_0)}(h,h')=-\frac23\sum_k\operatorname{Im}\langle\chi_1\vert P_k\tilde D^2\vert\chi_1\rangle\,\operatorname{Im}\langle\chi_2\vert B_{\beta_0}Q_k\vert\chi_2\rangle .
\tag{8.11}
$$

By (8.10), the result does not depend on the choice of decomposition of $J$.

### Step 6. Witness angle at order $\lambda^3$ (Item 3c)

By (5.10), $u_h-u_{h'}=\tilde D\lvert\chi\rangle$, so $d_0^2=\operatorname{Var}_\chi(D)$. Let $d_0>0$. By (3.5) with $\alpha\in[0,\pi/2]$, $\sin\alpha=\sqrt{1-W}=\lambda d_0\bigl(1-\lambda c_3/d_0^2-\lambda^2c_4/d_0^2+O(\lambda^3)\bigr)^{1/2}$. Expanding the square root and using $\arcsin s=s+s^3/6+O(s^5)$,

$$
\alpha^{(\beta_0)}=\lambda d_0-\lambda^2\frac{c_3^{(\beta_0)}}{2d_0}+\lambda^3\Bigl[\frac{d_0^3}{6}-\frac{(c_3^{(\beta_0)})^2}{8d_0^3}-\frac{c_4^{(\beta_0)}}{2d_0}\Bigr]+O(\lambda^4),
\tag{8.12}
$$

whose $\lambda^2$ term is (6.12). Here $d_0$ is body-free, and $c_3^{(\beta_0)}$ is body-free by (8.8). Hence

$$
\Delta\alpha^{(\beta_0)}(h,h';\lambda)=-\lambda^3\,\frac{\Delta c_4^{(\beta_0)}(h,h')}{2d_0(h,h')}+O(\lambda^4).
\tag{8.13}
$$

### Step 7. Chain of cells (Item 4)

The probe and body projectors still commute with $C$, because the links act on cells only. So Step 1 holds with $J\to J_{\rm ch}:=\sum_{k=1}^{L-1}J_k$ and $\lvert\chi\rangle=\lvert\chi_1\rangle\otimes\cdots\otimes\lvert\chi_L\rangle$. Set $M:=K_{h'}+B_{\beta_0}+J_{\rm ch}$ and $U(\lambda):=e^{\mathrm iM\lambda}e^{-\mathrm i(M+D)\lambda}$. Then $\mathrm dU/\mathrm d\lambda=-\mathrm iD(\lambda)U$ with $U(0)=\mathbb 1$ and $D(s):=e^{\mathrm iMs}De^{-\mathrm iMs}$, so the Dyson series gives

$$
G^{(\beta_0)}_{hh'}=1-\mathrm i\!\int_0^\lambda\!\langle D(s)\rangle\,ds-\int_0^\lambda\!ds_1\!\int_0^{s_1}\!ds_2\,\langle D(s_1)D(s_2)\rangle+\dots,\qquad
D(s)=\sum_{n\ge0}\frac{(\mathrm is)^n}{n!}\operatorname{ad}_M^n(D).
\tag{8.14}
$$

**Support counting.** $\operatorname{ad}_M^n(D)$ is a sum of words $\operatorname{ad}_{Y_n}\cdots\operatorname{ad}_{Y_1}(D)$ with letters $Y_\ell\in\{K_{h'},B_{\beta_0},J_1,\dots,J_{L-1}\}$. A commutator of operators with disjoint supports vanishes, and $\operatorname{supp}[Y,Z]\subseteq\operatorname{supp}Y\cup\operatorname{supp}Z$. By induction from $\operatorname{supp}D=\{c_1\}$:
- a nonzero word has support in $\{c_1,\dots,c_{1+m}\}$, where $m$ is the number of link letters in it;
- the letter $B_{\beta_0}$ (support $\{c_{r+1}\}$) gives a nonzero commutator only after at least $r$ link letters.

So the words that contain $B_{\beta_0}$ have $n\ge r+1$. For $n=r+1$ the only such word is $J_1,\dots,J_r,B_{\beta_0}$. Hence

$$
\delta D(s):=D(s)-D(s)\big|_{B_{\beta_0}=0}=\frac{(\mathrm is)^{r+1}}{(r+1)!}Z_r+O(s^{r+2}),\qquad
Z_r:=[B_{\beta_0},[J_r,[\cdots[J_1,D]\cdots]]],
\tag{8.15}
$$

with $Z_0=[B_{\beta_0},D]$. Each $\operatorname{ad}$ of a self-adjoint letter exchanges self-adjoint and anti-self-adjoint operators, so $\mathrm i^{r+1}Z_r$ is self-adjoint.

**Order of $\Delta W$.** In (8.14), the first-order term gives $\delta G=-\mathrm i\,a\,\lambda^{r+2}+O(\lambda^{r+3})$ with real $a=\langle\mathrm i^{r+1}Z_r\rangle/(r+2)!$. Every higher Dyson term that contains $\delta D$ is $O(\lambda^{r+3})$. With $G_0:=G\big|_{B_{\beta_0}=0}=1+O(\lambda)$,

$$
\Delta W^{(\beta_0)}=2\operatorname{Re}\bigl(G_0^*\,\delta G\bigr)+\lvert\delta G\rvert^2=2\operatorname{Re}\bigl(-\mathrm i\,a\,\lambda^{r+2}\bigr)+O(\lambda^{r+3})=O(\lambda^{3+r}).
\tag{8.16}
$$

- For $L=2$, $r=1$ this agrees with Steps 4–5.
- For $r=0$, $Z_0=[B_{\beta_0},D]\neq0$ in general, and the $\lambda^3$ coefficient is the term $\operatorname{Im}\langle B_{\beta_0}\tilde D^2\rangle$ of (6.11), taken with medium $c_1\cdots c_L$ and $B_{\beta_0}\to B_{\beta_0}+J_{\rm ch}$.
- Steps 4, 5 (operator form (8.10)) and 7 use only supports, not the product form of $\lvert\chi\rangle$.

### Step 8. Qubit example (Item 5)

**Basic quantities.** $D=(i-j)X_1$. Since $\{\sigma_x,\sigma_z\}=0$, $X_1^2=2\cdot\mathbb 1$, and $\langle0\vert X_1\vert0\rangle=1$. Hence

$$
\tilde D=(i-j)(X_1-\mathbb 1),\qquad \tilde D^2=(i-j)^2(3-2X_1),\qquad \operatorname{Var}_\chi(D)=(i-j)^2,\qquad d_0(h_i,h_j)=\lvert i-j\rvert .
\tag{8.17}
$$

**$c_3$.** By (6.10) with $B_{\beta_0}\to B_{\beta_0}+J$, three terms contribute:
- $K_{h_j}\tilde D^2=j(i-j)^2X_1(3-2X_1)$ is a self-adjoint function of $X_1$, so its expectation is real;
- the $B_{\beta_0}$ term vanishes by (8.8);
- $\langle J\tilde D^2\rangle$ contains the factor $\langle0\vert\sigma_x\vert0\rangle=0$.

Hence

$$
c_3^{(\beta_0)}(h_i,h_j)=0 .
\tag{8.18}
$$

**$\Delta c_4$.** Take $P_1=g\sigma_y$ and $Q_1=\sigma_x$. Using $\sigma_y\sigma_x=-\mathrm i\sigma_z$ and $\sigma_y\sigma_z=\mathrm i\sigma_x$:
- $\sigma_y(3-2X_1)=3\sigma_y+2\mathrm i\sigma_z-2\mathrm i\sigma_x$, whose expectation in $\lvert0\rangle$ is $2\mathrm i$, so $\operatorname{Im}\langle P_1\tilde D^2\rangle_1=2g(i-j)^2$;
- $\operatorname{Im}\langle0\vert\beta_0\sigma_y\sigma_x\vert0\rangle=\operatorname{Im}(-\mathrm i\beta_0)=-\beta_0$.

By (8.11),

$$
\Delta c_4^{(\beta_0)}(h_i,h_j)=-\tfrac23\cdot2g(i-j)^2\cdot(-\beta_0)=\tfrac43\,g\,\beta_0\,(i-j)^2 .
\tag{8.19}
$$

The operator form (8.10) gives the same value: $[B_{\beta_0},[J,D]]=4g\beta_0(i-j)(\sigma_x-\sigma_z)\otimes\sigma_z$ and $\langle\{\tilde D,\cdot\}\rangle=8g\beta_0(i-j)^2$.

**Witness angle.** By (8.13) and (8.17), for $i\neq j$,

$$
\Delta\alpha^{(\beta_0)}(h_i,h_j;\lambda)=-\tfrac23\,g\,\beta_0\,\lvert i-j\rvert\,\lambda^3+O(\lambda^4).
\tag{8.20}
$$

**Contrast with $J=g\,\sigma_z\otimes\sigma_z$.** Take $P_1=g\sigma_z$ and $Q_1=\sigma_z$. Then $\lvert\chi_1\rangle=\lvert\chi_2\rangle=\lvert0\rangle$ are eigenvectors of $\sigma_z$, so both factors of (8.11) vanish: $\langle0\vert\sigma_z\tilde D^2\vert0\rangle=\langle0\vert\tilde D^2\vert0\rangle$ is real, and $\langle0\vert\sigma_y\sigma_z\vert0\rangle=\mathrm i\langle0\vert\sigma_x\vert0\rangle=0$. Hence

$$
J=g\,\sigma_z\otimes\sigma_z:\qquad \Delta c_4^{(\beta_0)}(h_i,h_j)=0,\qquad \Delta W^{(\beta_0)}=O(\lambda^5),\qquad \Delta\alpha^{(\beta_0)}=O(\lambda^4).
\tag{8.21}
$$

## Result

- **Item 1.** Exact state (8.2); the body stays at $\beta_0$. The witness data are (8.3), i.e. (6.6) with $B_{\beta_0}\to B_{\beta_0}+J$ on the medium $c_1c_2$.
- **Item 2.** For $J=0$, $\Delta W^{(\beta_0)}=0$ exactly, for every $\lambda$ (8.4).
- **Item 3.**
  - (a) $\Delta c_3^{(\beta_0)}=\operatorname{Im}\langle B_{\beta_0}\tilde D^2\rangle=0$ (8.8).
  - (b) $\Delta c_4^{(\beta_0)}=\tfrac16\langle\{\tilde D,[B_{\beta_0},[J,D]]\}\rangle=-\tfrac23\sum_k\operatorname{Im}\langle P_k\tilde D^2\rangle_1\operatorname{Im}\langle B_{\beta_0}Q_k\rangle_2$ (8.10), (8.11).
  - (c) $\Delta\alpha^{(\beta_0)}=-\lambda^3\Delta c_4^{(\beta_0)}/(2d_0)+O(\lambda^4)$ (8.13); the full angle is (8.12).
- **Item 4.** On a chain, with the body recorded $r$ links away from the probe's cell, $\Delta W^{(\beta_0)}=O(\lambda^{3+r})$ (8.16). The leading body-dependent operator is the nested commutator $Z_r$ (8.15).
- **Item 5.**
  - $c_3^{(\beta_0)}(h_i,h_j)=0$ (8.18).
  - $\Delta c_4^{(\beta_0)}=\tfrac43g\beta_0(i-j)^2$ (8.19).
  - $\Delta\alpha^{(\beta_0)}=-\tfrac23g\beta_0\lvert i-j\rvert\lambda^3+O(\lambda^4)$ (8.20).
  - For $J=g\,\sigma_z\otimes\sigma_z$, $\Delta c_4^{(\beta_0)}=0$ (8.21).
- All expected formulas of question.md are reproduced, with no discrepancy in sign or factor.

## Consistency checks

1. **$J=0$.** (8.10) and (8.11) vanish because $[J,D]=0$, and $\Delta c_3=0$ by (8.8). This agrees with the exact statement (8.4) obtained from (6.7). On the chain, without links no word in (8.15) reaches $c_{r+1}$ for $r\ge1$.
2. **$B_{\beta_0}=\beta_0\mathbb 1$.** In (8.3) the factors $e^{\pm\mathrm i\beta_0\lambda}$ cancel, so $\Delta W=0$ for every $\lambda$. Consistently, $[\beta_0\mathbb 1,\cdot]=0$ in (8.10), and $\operatorname{Im}\langle\beta_0Q_k\rangle_2=0$ in (8.11).
3. **Body recorded by $c_1$ ($r=0$).** Then $[B_{\beta_0},D]\neq0$ in general. The support bound (8.16) allows $O(\lambda^3)$, and the coefficient is the term $\operatorname{Im}\langle B_{\beta_0}\tilde D^2\rangle=\langle[B_{\beta_0},\tilde D^2]\rangle/(2\mathrm i)$ of (6.11). This is also what the general $\lambda^3$ coefficient $\operatorname{Im}\langle M\tilde D^2\rangle$ of Step 3 gives.

## Open issues

- For $r\ge2$ only the bound (8.16) is shown. The coefficient of $\lambda^{3+r}$, and whether it is generically nonzero, are not computed.
- For $J=g\,\sigma_z\otimes\sigma_z$, the first nonzero body-dependent order is not determined.
- The operator results (8.8), (8.10) and (8.16) do not use the product form of $\lvert\chi\rangle$; only (8.11) does. Non-product medium states are outside the scope.

## Methods used

- Block diagonalization by conserved place projectors
- Baker–Campbell–Hausdorff formula to third order
- Cumulant expansion of $\langle e^{-\mathrm i\lambda\Omega}\rangle$
- Interaction picture and Dyson series
- Support counting of nested commutators
- Pauli-matrix algebra; expansion of $\arcsin$