Version 3 · current

# What a body presents to a field object: the source at first order in the coupling

- **Subproject:** 06-ilang-source
- **Package:** 02-what-a-body-presents
- **Version:** v3
- **Mode:** external regeneration
- **Date:** 2026-10-19

## Changes from previous version

- Step 8 ("Coefficients") and (2.14): part (iii) is restricted to additive changes between two starts ($\Delta s^{(i)}_x$, $\Delta S_i$) and to the ratio $\Delta S_1/\Delta S_2$; new part (iv): the change of a ratio, $\Delta(S_1/S_2)$, is in general not the same in every split, with a control (I1).
- Result item 13: (iii) restated and (iv) added accordingly (I1).

## Response to verification

- I1: Accepted and fixed in Step 8 ("Coefficients", (2.14)(iii)–(iv), control of (iv)) and Result item 13. A start-independent shift $W_i$ cancels in additive changes but not in the change of a ratio. (2.14)(iii) now claims split-independence only for $\Delta s^{(i)}_x$, $\Delta S_i$ and $\Delta S_1/\Delta S_2$ (where $\Delta S_2\neq0$). (iv) states that $\Delta(S_1/S_2)$ is in general not split-independent; the control is the counterexample of the verification, realized with two branch starts at $p=0$. Ratios evaluated in the component split remain functions of $C$ and the starts under the conditions of (2.14)(ii).

## Setup and assumptions

**Objects and contract** (question, "Setting"; each an assumption of the package, M8). A Model D body $b$ on a chain of $n$ position labels (A1), a medium $c$, a field object $f$ of any finite type ($d_f=\dim\mathcal H_f$) with self-adjoint $C_f$ and $\Phi_x$. Each object is the only instance of its type, so (1.11) imposes nothing and $\mathcal H_{\rm adm}=\mathcal H$. The total contract is $C=C_{\rm D}+C_{bc}+C_f+C_{bf}$ with $C_{bc}=\sum_x\lvert x\rangle\langle x\rvert\otimes\kappa xX$ and $C_{bf}=C^{(a)}_{bf}$ or $C^{(c)}_{bf}$ as in the question, $\gamma\in\mathbb R$.

**Notation.** $\Pi_x:=\lvert x\rangle\langle x\rvert\otimes\mathbb 1_2\otimes\mathbb 1_N$; $O:=\sigma_3\otimes(m\mathbb 1_N+D)$ on the internal labels; $A_x:=\frac\tau2\bigl(\lvert x+1\rangle\langle x\rvert\otimes(-i\sigma_1)+\lvert x\rangle\langle x+1\rvert\otimes(i\sigma_1)\bigr)\otimes\mathbb 1_N$, the bond $(x,x+1)$ of the motion term ($A_x:=0$ for an absent bond), so $A=\sum_xA_x$ and $B=\sum_x\Pi_xO$. $d_k=\frac{2\pi k}{N\lambda_0}$, $\mu_k=m+d_k$. The *rest* $R$ is the set of objects other than $f$, and $H_R$ the sum of the terms not involving $f$ (here $R=\{b,c\}$, $H_R=C_{\rm D}+C_{bc}$). $\mathring Y:=Y-\frac{\operatorname{Tr}Y}{d}\mathbb 1$ is the traceless part of an operator on a $d$-dimensional space; $\lVert\cdot\rVert$ is the operator norm, $\lVert\cdot\rVert_1$ the trace norm.

**Inputs** (04-ilang-time/05-moving-clock@v3, quoted in the question), with $\lvert p\rangle=\sum_xe^{ipx}\lvert x\rangle$: by (5.5) of 04-ilang-time/05-moving-clock,
$$C_{\rm D}\bigl(\lvert p\rangle\otimes\beta\otimes\lvert f_k\rangle\bigr)=\lvert p\rangle\otimes C_k(p)\beta\otimes\lvert f_k\rangle,\qquad C_k(p)=-\tau\sin p\,\sigma_1+\mu_k\sigma_3 ;$$
by (5.6) of 04-ilang-time/05-moving-clock, $C_k(p)$ has the eigenvalues $\pm\omega_k(p)$, $\omega_k=\sqrt{\mu_k^2+\tau^2\sin^2p}$, with the real eigenvectors $\beta_{k,+}=(\cos\frac{\theta_k}2,\sin\frac{\theta_k}2)^{T}$, $\beta_{k,-}=(-\sin\frac{\theta_k}2,\cos\frac{\theta_k}2)^{T}$, $\tan\theta_k(p)=-\tau\sin p/\mu_k$, $\lvert\theta_k\rvert<\frac\pi2$.

**Half-gap.** $M_k:=\mu_k=m+d_k$. At $\sin p=0$ the two branches of level $k$ have the costs $\pm M_k$, so $M_k$ is the half-gap of the on-site block $\mu_k\sigma_3$. (The question names $M_k$ without quoting its definition; this is the one used here.)

**Assumptions.**
- S1. Product start $\Psi(0)=\Psi_b\otimes\chi\otimes\varphi$ (for the pair, $\Psi_{12}\otimes\chi\otimes\varphi$).
- S2. First order in the coupling at fixed $\lambda$, valid for $\lVert C_{bf}\rVert\,\lvert\lambda\rvert\ll1$ with the remainder (2.4).
- S3. Normalized plane waves: on a ring of $n$ sites (the periodic chain, $p\in\frac{2\pi}{n}\mathbb Z$, a chosen description, M8), where (5.5)–(5.6) hold verbatim for $\lvert p\rangle_n:=n^{-1/2}\lvert p\rangle$. Items 2 and 3 are derived on the ring; open chains are not treated there.
- S4. Items 2 and 3 at zeroth order in the records: $\kappa=0$ or $X\chi=0$; for $\kappa>0$ a bound is given in Step 5.
- S5. Pair: $b_1,b_2$ of Model D type with their own parameters $(\tau_i,m_i,N_i,\lambda_{0,i})$, $M^{(i)}_{k}$, both on the position labels $x$ of the $\Phi_x$; $U=\sum_{x_1,x_2}\Pi^{(1)}_{x_1}\Pi^{(2)}_{x_2}\otimes u(x_1,x_2)$, $u(x_1,x_2)$ self-adjoint on the internal labels of both bodies, with $[U,D_i]=0$ so that fixed clock levels form invariant subspaces.

## Derivation

### Step 1. The coupling as a sum over positions

Shifting the summation index in $\sum_xA_x\otimes\frac12\Phi_{x+1}$ gives $\sum_xA_x\otimes\frac12(\Phi_x+\Phi_{x+1})=\sum_x\frac12(A_x+A_{x-1})\otimes\Phi_x$. Both couplings therefore have the form
$$C_{bf}=\gamma\sum_xQ_x\otimes\Phi_x,\qquad Q^{(a)}_x=\Pi_xO,\qquad Q^{(c)}_x=\Pi_xO+\tfrac12\bigl(A_x+A_{x-1}\bigr). \tag{2.1}$$
Every $A_x$ enters $Q_x$ and $Q_{x+1}$ with weight $\frac12$; $\operatorname{Tr}\sigma_3=0$, and $A_x$ is off-diagonal in the position. Hence
$$\sum_xQ^{(a)}_x=B,\qquad\sum_xQ^{(c)}_x=A+B=C_{\rm D},\qquad\operatorname{Tr}Q_x=0 . \tag{2.2}$$
Let $\phi:=\max_x\lVert\Phi_x\rVert$. Since the $\Pi_x$ are orthogonal and $\lVert O\rVert=\mu_{N-1}$, $\lVert C^{(a)}_{bf}\rVert=\lvert\gamma\rvert\mu_{N-1}\phi$. The motion part of $C^{(c)}_{bf}$ is $\frac{\gamma\tau}2(T+T^\dagger)$ with $T:=\sum_x\lvert x+1\rangle\langle x\rvert\otimes(-i\sigma_1)\otimes\mathbb 1_N\otimes\frac12(\Phi_x+\Phi_{x+1})$ (identity on $c$), summed over the bonds present. Distinct bonds have distinct target sites $x+1$, on every ring and on the open chain. So for $v=\sum_x\lvert x\rangle\otimes v_x$, $\lVert Tv\rVert^2=\sum_x\bigl\lVert\bigl((-i\sigma_1)\otimes\mathbb 1_N\otimes\frac12(\Phi_x+\Phi_{x+1})\bigr)v_x\bigr\rVert^2\le\phi^2\lVert v\rVert^2$, hence $\lVert T\rVert\le\phi$ and $\lVert C^{(c)}_{bf}\rVert\le\lvert\gamma\rvert(\mu_{N-1}+\tau)\phi$. Both bounds are independent of $n$.

### Step 2. Exact equation of the view of $f$

Let $\rho=\lvert\Psi\rangle\langle\Psi\rvert$ and $V_f=\operatorname{Tr}_R\rho$ (1.5). By (1.9), $i\,\mathrm d\rho/\mathrm d\lambda=[C,\rho]$. The partial trace over $R$ of a commutator with an operator on $R$ vanishes, and $\operatorname{Tr}_R[\mathbb 1\otimes C_f,\rho]=[C_f,V_f]$, so
$$i\frac{\mathrm d}{\mathrm d\lambda}V_f=[C_f,V_f]+\operatorname{Tr}_R[C_{bf},\rho]\qquad\text{(exact)}.$$

### Step 3. First-order generator and remainder

Let $\xi(\lambda):=e^{-iH_R\lambda}(\Psi_b\otimes\chi)$ be the state of the rest at $\gamma=0$, $V^0_b(\lambda)$ its view on $b$, and $\Delta:=\rho-\lvert\xi\rangle\langle\xi\rvert\otimes V_f$. Since $\operatorname{Tr}_R\bigl((Q_x\otimes\Phi_x)(\lvert\xi\rangle\langle\xi\rvert\otimes V_f)\bigr)=\langle\xi\vert Q_x\vert\xi\rangle\,\Phi_xV_f$, Step 2 becomes
$$i\frac{\mathrm d}{\mathrm d\lambda}V_f=[G_f(\lambda),V_f]+R_f(\lambda),\qquad G_f(\lambda)=C_f+\sum_xs_x(\lambda)\Phi_x,\qquad s_x(\lambda):=\gamma\langle\xi(\lambda)\vert Q_x\vert\xi(\lambda)\rangle=\gamma\operatorname{Tr}\bigl(Q_xV^0_b(\lambda)\bigr), \tag{2.3}$$
with $R_f=\operatorname{Tr}_R[C_{bf},\Delta]$. The family $s_x$ is the source field of this listing (what of it $C$ fixes: Step 8); its total is $S:=\sum_xs_x$.

*Remainder.* $\Psi^0(\lambda)=\xi(\lambda)\otimes e^{-iC_f\lambda}\varphi$ is the evolution at $\gamma=0$. By Duhamel's formula, $\Psi-\Psi^0=-i\int_0^\lambda e^{-iC(\lambda-\lambda')}C_{bf}\Psi^0(\lambda')\,\mathrm d\lambda'$, so $\lVert\Psi-\Psi^0\rVert\le\lVert C_{bf}\rVert\lvert\lambda\rvert$. For unit vectors $\lVert\psi\psi^\dagger-\phi\phi^\dagger\rVert_1\le2\lVert\psi-\phi\rVert$, partial traces do not increase the trace norm, and $\Delta=(\rho-\rho^0)-\lvert\xi\rangle\langle\xi\rvert\otimes(V_f-V^0_f)$ with $\rho^0=\lvert\xi\rangle\langle\xi\rvert\otimes V_f^0$. Hence $\lVert\Delta\rVert_1\le4\lVert C_{bf}\rVert\lvert\lambda\rvert$ and
$$\lVert R_f(\lambda)\rVert_1\le2\lVert C_{bf}\rVert\,\lVert\Delta\rVert_1\le8\lVert C_{bf}\rVert^2\lvert\lambda\rvert . \tag{2.4}$$
The difference between $V_f$ and the evolution of $V_f(0)$ by the $\lambda$-dependent contract $G_f$ obeys the same equation with source $R_f$ and starts at zero; unitary conjugation preserves $\lVert\cdot\rVert_1$, so the difference is at most $4\lVert C_{bf}\rVert^2\lambda^2$. For the pair, $C_{bf}\to C_{b_1f}+C_{b_2f}$, $s_x=\gamma_1\langle Q^{(1)}_x\rangle+\gamma_2\langle Q^{(2)}_x\rangle$, with $\xi$ the state of $R=\{b_1,b_2,c\}$ at zero coupling; nothing else changes.

### Step 4. When the first-order generator is a contract of $f$ alone

With the spectral decomposition $H_R=\sum_EE\,P_E$ (finite, A1), $\xi(\lambda)=\sum_Ee^{-iE\lambda}P_E\xi_0$, $\xi_0=\Psi_b\otimes\chi$, and (2.3) becomes
$$G_f(\lambda)=C_f+\sum_\nu e^{i\nu\lambda}\,\Gamma_\nu,\qquad\Gamma_\nu:=\gamma\sum_{E-E'=\nu}\ \sum_x\langle\xi_0\vert P_EQ_xP_{E'}\vert\xi_0\rangle\,\Phi_x . \tag{2.5}$$
A multiple of $\mathbb 1_f$, $\lambda$-dependent or not, drops out of $[G_f,V_f]$, so what acts on the view is $\mathring G_f=\mathring C_f+\sum_\nu e^{i\nu\lambda}\mathring\Gamma_\nu$, with $\mathring\Gamma_\nu$ given by (2.5) with $\mathring\Phi_x$ in place of $\Phi_x$. Exponentials with distinct real frequencies are linearly independent functions of $\lambda\in\mathbb R$ (A9; apply it to each matrix element). Hence $G_f$ is, modulo $\mathbb 1_f$, a $\lambda$-independent contract of $f$ alone iff $\mathring\Gamma_\nu=0$ for every $\nu\neq0$, and then $G_f\equiv C_f+\Gamma_0$. Literal $\lambda$-independence of $G_f$ requires $\Gamma_\nu=0$ for every $\nu\neq0$, which is stronger (example below). Two sufficient cases for the criterion:
- a stationary start, $\xi_0$ in one eigenspace of $H_R$; for $\kappa=0$ or $X\chi=0$ this is $\Psi_b$ an eigenvector of $C_{\rm D}$, since then $H_R(\psi\otimes\chi)=(C_{\rm D}\psi)\otimes\chi$ for every $\psi$;
- $\mathring\Phi_x=\mathring\Phi$ for all $x$ in coupling (c) with $\kappa=0$ or $X\chi=0$, and any $\Psi_b$: then $G_f\equiv C_f+\gamma\langle C_{\rm D}\rangle\mathring\Phi$ by (2.2), and $\langle C_{\rm D}\rangle$ is conserved. In (a), $\langle B\rangle$ is not conserved in general: for a superposition of the two branches at one $p$ with $\sin p\ne0$ it oscillates with frequency $2\omega_k$, because $\langle\beta_{k,+}\vert\sigma_3\vert\beta_{k,-}\rangle=-\sin\theta_k\neq0$. With all $\mathring\Phi_x=\mathring\Phi$, $\mathring G_f=\mathring C_f+\gamma\langle B\rangle\mathring\Phi$ then depends on $\lambda$ if $\gamma\mathring\Phi\neq0$; for $\Phi_x=\mathbb 1_f$, $G_f=C_f+\gamma\langle B\rangle\mathbb 1_f$ depends on $\lambda$ while $\mathring G_f=\mathring C_f$ does not.

*Control.* Take the ring (S3), $\kappa=0$, and the start $\Psi_b=\frac1{\sqrt2}\bigl(\lvert p\rangle_n\otimes\beta_{k,+}(p)+\lvert0\rangle_n\otimes\beta_{k,+}(0)\bigr)\otimes\lvert f_k\rangle$ with $p\in\frac{2\pi}n\mathbb Z$ and $\sin p\neq0$. The two components have the costs $\omega_k(p)$ and $M_k$. The required matrix elements are $\langle0\vert_n\Pi_x\lvert p\rangle_n=e^{ipx}/n$, $\langle0\vert_n\frac12(A_x+A_{x-1})\lvert p\rangle_n=-\frac{\tau}{2n}\sin p\,e^{ipx}\sigma_1$ (direct computation) and $\beta_{k,+}(0)=(1,0)^{T}$. They give
$$s_x(\lambda)=\frac{\gamma}{2n}\Bigl(\langle\beta_{k,+}(p)\vert\hat Q\vert\beta_{k,+}(p)\rangle+M_k\Bigr)+\frac{\gamma g}{n}\cos\bigl(px-(\omega_k(p)-M_k)\lambda\bigr), \tag{2.6}$$
with $\hat Q=\mu_k\sigma_3$, $g^{(a)}=M_k\cos\frac{\theta_k}2$ for (a) and $\hat Q=C_k(p)$, $g^{(c)}=M_k\cos\frac{\theta_k}2-\frac\tau2\sin p\,\sin\frac{\theta_k}2$ for (c). Both $g$ are positive, since $\cos\frac{\theta_k}2>0$ and $\sin\frac{\theta_k}2$ has the sign of $-\sin p$. Since $\omega_k(p)>M_k$, $s_x$ depends on $\lambda$ for $\gamma\neq0$. With $\nu:=\omega_k(p)-M_k$ and $Y_p:=\sum_xe^{ipx}\mathring\Phi_x$, the $\lambda$-dependent part of $\sum_xs_x\mathring\Phi_x$ is $\frac{\gamma g}{2n}\bigl(e^{-i\nu\lambda}Y_p+e^{i\nu\lambda}Y_p^\dagger\bigr)$. So $G_f$ depends on $\lambda$ modulo $\mathbb 1_f$ whenever $Y_p\neq0$ (for example, when the $\mathring\Phi_x$ are linearly independent).

### Step 5. Total source of a branch start (item 2)

By (2.2), $S=\gamma\langle B\rangle$ for (a) and $S=\gamma\langle C_{\rm D}\rangle$ for (c). Both operators commute with the translations of the position label. By (5.5) they act on $\lvert p\rangle\otimes\beta\otimes\lvert f_k\rangle$ as $\mu_k\sigma_3$ and $C_k(p)$ on $\beta$ ($B$ is diagonal in $x$, and $O\lvert f_k\rangle=\mu_k\sigma_3\lvert f_k\rangle$). For the branch $\sigma\in\{+1,-1\}$, (5.6) gives $\langle\beta_{k,\sigma}\vert\sigma_3\vert\beta_{k,\sigma}\rangle=\sigma(\cos^2\frac{\theta_k}2-\sin^2\frac{\theta_k}2)=\sigma\cos\theta_k=\sigma M_k/\omega_k$ (because $\lvert\theta_k\rvert<\frac\pi2$) and $\langle\beta_{k,\sigma}\vert C_k\vert\beta_{k,\sigma}\rangle=\sigma\omega_k$. For the branch start $\lvert p\rangle_n\otimes\beta_{k,\sigma}(p)\otimes\lvert f_k\rangle$ (S3) this gives
$$S^{(a)}_{k,\sigma}(p)=\gamma\sigma\frac{M_k^2}{\omega_k(p)}=\frac{\gamma\sigma M_k^2}{\sqrt{M_k^2+\tau^2\sin^2p}},\qquad S^{(c)}_{k,\sigma}(p)=\gamma\sigma\,\omega_k(p)=\gamma\sigma\sqrt{M_k^2+\tau^2\sin^2p}, \tag{2.7}$$
uniformly distributed, $s_x=S/n$. With $L_b:=\langle C_{\rm D}\rangle=\sigma\omega_k(p)$, the cost of the body's own term in the start,
$$S^{(c)}_{k,\sigma}=\gamma L_b,\qquad S^{(a)}_{k,\sigma}=\gamma\frac{M_k^2}{L_b}=\gamma L_b\Bigl(\frac{M_k}{\omega_k(p)}\Bigr)^2. \tag{2.8}$$
With $\kappa=0$ or $X\chi=0$, $\xi(\lambda)=e^{-i\sigma\omega_k\lambda}\xi_0$, so (2.7) holds at every $\lambda$, and $G_f$ is $\lambda$-independent (Step 4). On the ring, a normalized packet $\sum_pg(p)\lvert p\rangle_n\otimes\beta_{k,\sigma}(p)\otimes\lvert f_k\rangle$ ($p\in\frac{2\pi}n\mathbb Z$, $\sum_p\lvert g(p)\rvert^2=1$) on one branch and level stays on that branch, and both operators are block-diagonal in $p$, since they commute with the translations of the ring. Its total source is therefore $\sum_p\lvert g(p)\rvert^2S_{k,\sigma}(p)$, constant in $\lambda$. For $\kappa>0$ and $X\chi\neq0$, Duhamel's formula between $H_R$ and $C_{\rm D}$ gives $\lvert S(\lambda)-S(0)\rvert\le2\lvert\gamma\rvert\,\lVert\sum_xQ_x\rVert\int_0^{\lvert\lambda\rvert}\lVert C_{bc}\,\xi^0(\lambda')\rVert\,\mathrm d\lambda'$, where $\xi^0$ is the evolution at $\kappa=0$. This correction is of first order in $\kappa$ at fixed $\lambda$.

**Dependence.** $S_{k,\sigma}(p)$ is odd in $\sigma$. It depends on $p$ only through $\sin^2p$, and on $k$ only through $M_k=m+\frac{2\pi k}{N\lambda_0}$. As $\sin^2p$ goes from $0$ to $1$, $\lvert S^{(a)}\rvert$ decreases from $\lvert\gamma\rvert M_k$ to $\lvert\gamma\rvert M_k^2/\sqrt{M_k^2+\tau^2}$, while $\lvert S^{(c)}\rvert$ increases from $\lvert\gamma\rvert M_k$ to $\lvert\gamma\rvert\sqrt{M_k^2+\tau^2}$. Both moduli increase with $M_k$.

### Step 6. Bound pair (item 3)

Take $\Psi_{12}$ an eigenvector of the pair contract $C_{\rm pair}:=C_{{\rm D},1}+C_{{\rm D},2}+U$ with eigenvalue $E_{12}$ (the cost of the pair), on the clock levels $(k_1,k_2)$ (S5), with $\kappa=0$. The start is stationary, so $G_f$ is $\lambda$-independent (Step 4). By Step 3 and (2.2), $S=\gamma_1\langle\sum_xQ^{(1)}_x\rangle+\gamma_2\langle\sum_xQ^{(2)}_x\rangle$. On level $k_i$, $B_i$ acts as $M^{(i)}_{k_i}\sigma^{(i)}_3$, and $\langle C_{{\rm D},1}\rangle+\langle C_{{\rm D},2}\rangle=E_{12}-\langle U\rangle$. Therefore
$$S^{(a)}_{12}=\gamma_1M^{(1)}_{k_1}\langle\sigma^{(1)}_3\rangle+\gamma_2M^{(2)}_{k_2}\langle\sigma^{(2)}_3\rangle,\qquad S^{(c)}_{12}=\gamma_1\langle C_{{\rm D},1}\rangle+\gamma_2\langle C_{{\rm D},2}\rangle\ \overset{\gamma_1=\gamma_2=\gamma}{=}\ \gamma\bigl(E_{12}-\langle U\rangle\bigr), \tag{2.9}$$
where the expectation values are taken in $\Psi_{12}$. $U$ contributes no term of its own: the binding enters only through the state. The total is exactly the sum of what the two parts present in the joint state, because $\sum_xQ_x$ is a sum of one-body operators.

*Control.* Take $U=u\,\sigma^{(1)}_1\sigma^{(2)}_1$ ($u\in\mathbb R$, $u\neq0$, identity in positions and clocks), the ring with $p_1=p_2=0$, and levels $(k_1,k_2)$. By (5.5), $C_{k}(0)=\mu_k\sigma_3$ and $\langle A_i\rangle=0$, so (a) and (c) coincide. Put $M:=M^{(1)}_{k_1}+M^{(2)}_{k_2}$. On ${\rm span}\{\lvert{\uparrow\uparrow}\rangle,\lvert{\downarrow\downarrow}\rangle\}$, $C_{\rm pair}$ is $\begin{pmatrix}M&u\\u&-M\end{pmatrix}$. Its upper eigenvector $(\cos\frac\phi2,\sin\frac\phi2)$, $\tan\phi=u/M$, $\lvert\phi\rvert<\frac\pi2$, has $\langle\sigma^{(i)}_3\rangle=\cos\phi$ and $\langle\sigma^{(1)}_1\sigma^{(2)}_1\rangle=\sin\phi$, hence
$$E_{12}=\sqrt{M^2+u^2},\qquad\langle U\rangle=\frac{u^2}{E_{12}},\qquad S_{12}=\bigl(\gamma_1M^{(1)}_{k_1}+\gamma_2M^{(2)}_{k_2}\bigr)\frac{M}{\sqrt{M^2+u^2}} . \tag{2.10}$$
The free parts in the same branch, level and $p=0$ present $\gamma_iM^{(i)}_{k_i}$ (2.7). So $S_{12}$ differs from the sum of the free sources whenever $u\neq0$ and $\gamma_1M^{(1)}_{k_1}+\gamma_2M^{(2)}_{k_2}\neq0$, while $E_{12}$ exceeds the sum $M$ of the free costs. The product of the free branch states is a start of the same description, with $s_x(0)=\frac1n\sum_i\gamma_iM^{(i)}_{k_i}$; the eigenvector has $s_x=S_{12}/n$. So on the ring the difference is a change between two starts at $\lambda=0$, $\Delta G_f=\frac1n\bigl(S_{12}-\sum_i\gamma_iM^{(i)}_{k_i}\bigr)\sum_x\mathring\Phi_x$, which by (2.12) is fixed by $C$ and the two starts. The same block, with $u=u(x_1,x_2)$, arises for held bodies at fixed positions (A15).

### Step 7. Replacements that leave $C$ unchanged (item 4)

Every operator $Y$ on $\mathcal H_b\otimes\mathcal H_c\otimes\mathcal H_f$ has a unique decomposition $Y=\sum_{S\subseteq\{b,c,f\}}Y_S$, in which $Y_S$ acts as the identity outside $S$ and is traceless in every object of $S$. (Expand in product bases of $\{\mathbb 1\}\cup\{\text{traceless}\}$; this is linear algebra, M5.) A term on the pair $\{o,o'\}$ has components only for $S\subseteq\{o,o'\}$. Let two listings with one term per pair give the same $C$. The component on a two-object set $\{o,o'\}$ comes only from the difference of the two $\{o,o'\}$ terms, so that difference has no two-object component. The single-object components of the differences add up to zero for each object, up to constants whose total is zero. Hence the replacements that leave $C$ unchanged are exactly
$$\begin{aligned}&C_{\rm D}\to C_{\rm D}+Z^{\rm D}_b,\quad C_{bc}\to C_{bc}+Z^{bc}_b+Z^{bc}_c,\quad C_{bf}\to C_{bf}+Z^{bf}_b+Z^{bf}_f,\quad C_f\to C_f+Z^{f}_f,\quad C_{cf}:=Z^{cf}_c+Z^{cf}_f,\\&Z^{\rm D}_b+Z^{bc}_b+Z^{bf}_b=0,\qquad Z^{bc}_c+Z^{cf}_c=0,\qquad Z^{f}_f+Z^{bf}_f+Z^{cf}_f=0\quad(\text{up to constants of total zero}),\end{aligned} \tag{2.11}$$
where $Z^{\cdot}_o$ is any self-adjoint operator on $\mathcal H_o$: a constant, a place-dependent operator such as $\sum_xz_x\Pi_x$, or an internal operator. $C_{cf}$ is a new $c$–$f$ term, absent from the listing. In words, the only freedom is to move single-object operators between the terms that contain the object. The two-object components $C_{\{b,c\}}$ and $C_{\{b,f\}}$ are fixed by $C$. By (2.2), $\operatorname{Tr}Q_x=0$, so $C_{\{b,f\}}=\gamma\sum_xQ_x\otimes\mathring\Phi_x$, and the trace parts of the $\Phi_x$ give the single-$b$ operator $\gamma\sum_x\frac{\operatorname{Tr}\Phi_x}{d_f}Q_x$.

### Step 8. What the total contract and the start fix

*Component listing.* By Step 7, $C$ is the sum of its components $C_S$, each fixed by $C$. Collect them as $C=c_\varnothing\mathbb 1+H^\circ_R+C_{\{f\}}+K$, where $H^\circ_R$ is the sum of the components on nonempty subsets of $R$ and $K$ the sum of those on sets that contain $f$ and meet $R$. In the listing of the question $C_{\rm D}$, $C_{bc}$ and $C_f$ contribute nothing to $K$, and $\operatorname{Tr}Q_x=0$ (2.2). Hence $K=C_{\{b,f\}}=\gamma\sum_xQ_x\otimes\mathring\Phi_x$, $C_{\{f\}}=\mathring C_f$, $C_{bf}=K+Z_b\otimes\mathbb 1_f$ with $Z_b:=\gamma\sum_x\frac{\operatorname{Tr}\Phi_x}{d_f}Q_x$, and $H^\circ_R=H_R+Z_b$ up to a constant. The map $Y\mapsto d_f^{-1}\operatorname{Tr}_fY\otimes\mathbb 1_f$ does not increase the operator norm, so $\lVert Z_b\rVert\le\lVert C_{bf}\rVert$ and $\lVert K\rVert\le2\lVert C_{bf}\rVert$. For the pair, $K=\sum_iC_{\{b_i,f\}}$, one such term per body.

*Generator.* $(H^\circ_R,C_{\{f\}},K)$ is a listing of $C$ determined by $C$ alone. Steps 2–3, applied to it, give the first-order generator
$$G^\circ_f(\lambda):=C_{\{f\}}+\operatorname{Tr}_R\bigl((\lvert\xi^\circ\rangle\langle\xi^\circ\rvert\otimes\mathbb 1_f)K\bigr)=\mathring C_f+\sum_xs^\circ_x(\lambda)\mathring\Phi_x,\qquad\xi^\circ(\lambda):=e^{-iH^\circ_R\lambda}\xi_0,\quad s^\circ_x:=\gamma\langle\xi^\circ\vert Q_x\vert\xi^\circ\rangle, \tag{2.12}$$
and the remainder (2.4) with $\lVert C_{bf}\rVert$ replaced by $\lVert K\rVert$ (for the pair, $s^\circ_x=\sum_i\gamma_i\langle\xi^\circ\vert Q^{(i)}_x\vert\xi^\circ\rangle$). The right side uses only $C$ and $\xi_0$ ($\xi^\circ$ up to a phase, which drops out). So $G^\circ_f$ is invariant under every replacement (2.11), place-dependent and large ones included. In any listing $L$ of $C$, the terms containing $f$ sum to $C_{\{f\}}+K-Y\otimes\mathbb 1_f$ up to a constant, where $Y$ collects the single-object operators of $R$ moved into them, and $H^L_R=H^\circ_R+Y$. The generator of $L$ (Step 3) is the partial trace of these terms against $\lvert\xi^L\rangle\langle\xi^L\rvert$, hence $C_{\{f\}}+\operatorname{Tr}_R((\lvert\xi^L\rangle\langle\xi^L\rvert\otimes\mathbb 1_f)K)$ modulo $\mathbb 1_f$. Single-$f$ operators moved between terms cancel exactly; $Y$ changes only $\xi^L$, by $\lVert\xi^L-\xi^\circ\rVert\le\lVert Y\rVert\lvert\lambda\rvert$ (Duhamel). For the listing of the question $Y=-Z_b$, so $\lvert s_x-s^\circ_x\rvert\le2\lvert\gamma\rvert\lVert Q_x\rVert\lVert C_{bf}\rVert\lvert\lambda\rvert$, and $G_f\equiv G^\circ_f$ modulo $\mathbb 1_f$ up to $2\lVert K\rVert\lVert Z_b\rVert\lvert\lambda\rvert\le4\lVert C_{bf}\rVert^2\lvert\lambda\rvert$, the order of (2.4). For two starts $\xi_0,\xi'_0$ of the rest, $G^\circ_f-G^{\circ\prime}_f=\sum_x(s^\circ_x-s^{\circ\prime}_x)\mathring\Phi_x$ is fixed by $C$ and the two starts, with no further convention.

*Splits.* A source field is a split of $G^\circ_f$ into a start-independent contract of $f$ and a start-dependent body part. The *component split* takes $C_{\{f\}}$ and $F^\circ:=\sum_xs^\circ_x\mathring\Phi_x$, with one part $F^\circ_i$ for each body, from its component $C_{\{b_i,f\}}$. It is a function of $C$ and the start. Its zero is a convention: $\operatorname{Tr}_RK=0$, so $F^\circ$ would vanish for a view of $R$ proportional to $\mathbb 1$. Any other split with a start-independent field part differs from it by a start-independent operator $W$ on $\mathcal H_f$. Every $W$ is realized by a listing: the move (2.11) $C_{bf}\to C_{bf}+\mathbb 1_b\otimes W$, $C_f\to C_f-W$ leaves $C$ and $G^\circ_f$ unchanged and adds $W$ to the body part $\operatorname{Tr}_R((\lvert\xi\rangle\langle\xi\rvert\otimes\mathbb 1_f)C_{bf})$ of the listing. For $W=\sum_x\zeta_x\mathring\Phi_x$ this is $s_x\to s_x+\zeta_x$ and $S\to S+\sum_x\zeta_x$. With several bodies coupled to $f$, each has its own $W_i$, and $C_f\to C_f-\sum_iW_i$. Since $\operatorname{Tr}Q_x=0$, the listing of the question realizes the component split at first order: its body part is $\sum_xs_x\mathring\Phi_x$ modulo $\mathbb 1_f$, with $s_x=s^\circ_x$ up to the bound above.

*Zero of the body operator.* Consider listings whose $b$–$f$ term has the form (a), resp. (c), with $O$ replaced by $a'O+c_0$ and with $\mathring\Phi_x$ unchanged up to one common factor. Assume $\gamma\neq0$ and $\mathring\Phi_y\neq0$ for some $y$. Two such listings of the same $C$ differ in $C_{\{b,f\}}$ by the two-object component of $\gamma\sum_x\Pi_x\bigl((a'-1)O+c_0\bigr)\otimes\mathring\Phi_x$, which (2.11) requires to vanish. Taking $\operatorname{Tr}_f$ removes the single-$b$ part, since $\mathring\Phi_x$ is traceless. Multiplying by $\Pi_y\otimes\mathbb 1$ then gives $((a'-1)O+c_0)\otimes\mathring\Phi_y=\mathbb 1\otimes\mathring Z$ for every $y$, for one traceless $\mathring Z$ on $\mathcal H_f$. Since $O$ is not a multiple of $\mathbb 1$, $a'=1$, and then $c_0\mathring\Phi_y=\mathring Z$ for all $y$. Conversely, if all $\mathring\Phi_x$ are equal, $c_0\sum_x\Pi_x\otimes\mathring\Phi=c_0\mathbb 1_b\otimes\mathring\Phi$ is a single-$f$ operator, which can be moved into $C_f$. Hence
$$\text{a constant }c_0\neq0\text{ added to }\sigma_3\otimes(m\mathbb 1_N+D)\text{ leaves }C\text{ unchanged (after moves (2.11))}\iff\mathring\Phi_x=\mathring\Phi_y\ \text{for all }x,y . \tag{2.13}$$
(2.13) is a statement about $C$; it does not fix a split.

*Coefficients.* The numbers $s_x$, $S$ are coefficients with respect to the $\mathring\Phi_x$ of the stated coupling. The rescaling $(\gamma_i,\Phi_x)\to(\gamma_i/a,\,a\Phi_x)$, common to all bodies coupled to the same $\Phi_x$, changes no term and multiplies every coefficient by $1/a$. The coefficients of an operator in the span of the $\mathring\Phi_x$ are unique if the $\mathring\Phi_x$ are linearly independent; if all $\mathring\Phi_x$ equal one $\mathring\Phi\neq0$, only their sum is. A shift $W_i$ is the same for every start, so it cancels in the additive change $\Delta F_i$ of the part of body $i$ between two starts, hence in $\Delta s^{(i)}_x$, $\Delta S_i$ and $\Delta S_1/\Delta S_2$. It does not cancel in the change of a ratio, since $S_1/S_2\to(S_1+w_1)/(S_2+w_2)$ with start-independent $w_i$. Hence, for ratios wherever the denominator is nonzero,
$$\begin{aligned}&\text{(i) }s_x,\ S,\ s^{(1)}_x/s^{(2)}_x\text{ and }S_1/S_2\text{ read from a listing are not fixed by }C\text{ and the start};\\&\text{(ii) in the component split they are, up to }1/a\text{ (ratios exactly), if the }\mathring\Phi_x\text{ are linearly independent; }S^\circ\text{ and }S^\circ_1/S^\circ_2\text{ are if all }\mathring\Phi_x=\mathring\Phi\neq0;\\&\text{(iii) the additive changes between two starts are the same in every split: }\Delta s^{(i)}_x\text{ and }\Delta S_i\text{ if the }\mathring\Phi_x\text{ are linearly independent, }\Delta S_i\text{ if all }\mathring\Phi_x=\mathring\Phi\neq0;\\&\qquad\text{so is }\Delta S_1/\Delta S_2\text{ where }\Delta S_2\neq0;\text{ hence they are fixed by }C\text{ and the two starts, the changes up to }1/a,\text{ the ratio exactly};\\&\text{(iv) the change of a ratio between two starts, }\Delta(S_1/S_2),\text{ is in general not the same in every split.}\end{aligned} \tag{2.14}$$
*Control of (iv).* Take two bodies coupled to $f$ (S5 with $U=0$), $\kappa=0$, the ring, $\gamma_1,\gamma_2\neq0$, all $\mathring\Phi_x=\mathring\Phi\neq0$. In both starts the bodies are at $p_1=p_2=0$ on levels $k_1,k_2$ and body 2 is on branch $+1$; body 1 is on branch $+1$ in the first start and $-1$ in the second. At $\lambda=0$, $\xi^\circ=\xi$, and (2.7), applied to each body of the product start, gives $(S^\circ_1,S^\circ_2)=(q_1,q_2)$ and $(-q_1,q_2)$, with $q_i:=\gamma_iM^{(i)}_{k_i}\neq0$. So, with $\Delta$ the value in the first start minus that in the second, $\Delta(S_1/S_2)=2q_1/q_2$. The move (2.11) of $\mathbb 1_{b_2}\otimes w\mathring\Phi$ from $C_f$ into $C_{b_2f}$, with $w\notin\{0,-q_2\}$, leaves $C$ unchanged. It changes $\Delta(S_1/S_2)$ to $2q_1/(q_2+w)\neq2q_1/q_2$, while $\Delta S_1=2q_1$ and $\Delta S_2=0$ stay the same.
The relation (2.8) of $S$ to the cost holds in the listing of the question and is not invariant: a move $Z^{\rm D}_b$ between $C_{\rm D}$ and $C_{bc}$, a constant included, changes $L_b$ but not $S$, and a move $\mathbb 1_b\otimes W$ changes $S$ but not $L_b$. This is consistent with M6: the cost of a single object is not well defined.

## Result

Class throughout: the setting of the question (Model D body, medium with $K_x=\kappa xX$, field object of any finite type with any $C_f$ and $\Phi_x$, coupling (a) or (c)); further conditions are stated per item. All of these are chosen descriptions (M8). The values of the source field, its total and their ratios in items 2 and 5–9 are those of the listing of the question; items 11 and 13 state what of them $C$ and the starts fix.

1. **(2.1), (2.2)** (definition, exact). Both couplings are $C_{bf}=\gamma\sum_xQ_x\otimes\Phi_x$ with $Q^{(a)}_x=\Pi_xO$ and $Q^{(c)}_x=\Pi_xO+\frac12(A_x+A_{x-1})$. They satisfy $\sum_xQ^{(a)}_x=B$, $\sum_xQ^{(c)}_x=C_{\rm D}$ and $\operatorname{Tr}Q_x=0$.
2. **(2.3)** (formula; every product start $\Psi_b\otimes\chi\otimes\varphi$; first order in $\gamma$ at fixed $\lambda$). The view of $f$ obeys $i\,\mathrm dV_f/\mathrm d\lambda=[G_f(\lambda),V_f]+R_f$, with $G_f=C_f+\sum_xs_x(\lambda)\Phi_x$. The source field of the listing of the question is $s_x(\lambda)=\gamma\langle\xi(\lambda)\vert Q_x\vert\xi(\lambda)\rangle$, where $\xi(\lambda)=e^{-iH_R\lambda}(\Psi_b\otimes\chi)$ is the state of the rest at zero coupling. For the pair, $s_x=\gamma_1\langle Q^{(1)}_x\rangle+\gamma_2\langle Q^{(2)}_x\rangle$.
3. **(2.4)** (theorem; same class; every $\lambda$). $\lVert R_f(\lambda)\rVert_1\le8\lVert C_{bf}\rVert^2\lvert\lambda\rvert$, with $\lVert C_{bf}\rVert\le\lvert\gamma\rvert(\mu_{N-1}+\tau)\max_x\lVert\Phi_x\rVert$ for both couplings on every ring and open chain, independently of $n$. The view differs from its evolution under $G_f$ by at most $4\lVert C_{bf}\rVert^2\lambda^2$ in trace norm. The first-order generator is accurate for $\lVert C_{bf}\rVert\lvert\lambda\rvert\ll1$.
4. **(2.5)** (criterion; same class). $G_f$ is, modulo $\mathbb 1_f$, a $\lambda$-independent contract of $f$ alone iff $\mathring\Gamma_\nu=0$ for every Bohr frequency $\nu\neq0$ of $H_R$, where $\mathring\Gamma_\nu$ is $\Gamma_\nu$ with $\mathring\Phi_x$ in place of $\Phi_x$. Literal $\lambda$-independence of $G_f$, $\Gamma_\nu=0$ for every $\nu\neq0$, is a stronger condition. The criterion holds for every start in one eigenspace of $H_R$ (for $\kappa=0$ or $X\chi=0$: $\Psi_b$ an eigenvector of $C_{\rm D}$). In coupling (c) with $\mathring\Phi_x=\mathring\Phi$ for all $x$ and $\kappa=0$ or $X\chi=0$, it holds for every $\Psi_b$.
5. **(2.6)** (example, control of item 1; $\gamma\neq0$). On a ring with $\kappa=0$, take the start $\frac1{\sqrt2}(\lvert p\rangle_n\beta_{k,+}(p)+\lvert0\rangle_n\beta_{k,+}(0))\lvert f_k\rangle$ with $\sin p\neq0$. Its source field oscillates with frequency $\omega_k(p)-M_k$ in both couplings, so $G_f$ depends on $\lambda$ modulo $\mathbb 1_f$ whenever $\sum_xe^{ipx}\mathring\Phi_x\neq0$.
6. **(2.7)** (formula; branch start of branch $\sigma=\pm1$, level $k$, label $p$ on a ring; zeroth order in the records, $\kappa=0$ or $X\chi=0$; first order in $\gamma$). The total sources are $S^{(a)}_{k,\sigma}(p)=\gamma\sigma M_k^2/\omega_k(p)$ and $S^{(c)}_{k,\sigma}(p)=\gamma\sigma\omega_k(p)$, constant in $\lambda$ and uniform in $x$. Both are odd in $\sigma$ and depend on $p$ only through $\sin^2p$. For $\gamma\neq0$, as $\sin^2p$ goes from $0$ to $1$, $\lvert S^{(a)}\rvert$ decreases from $\lvert\gamma\rvert M_k$, while $\lvert S^{(c)}\rvert$ increases from $\lvert\gamma\rvert M_k$. Both depend on $k$ only through $M_k$, and their moduli increase with it. For a packet on one branch and level on the ring, $S$ is the $\lvert g(p)\rvert^2$-weighted sum of (2.7) over $p\in\frac{2\pi}n\mathbb Z$. For $\kappa>0$, $X\chi\neq0$, the correction is of first order in $\kappa$ at fixed $\lambda$, with the bound of Step 5.
7. **(2.8)** (formula; same conditions as (2.7)). $S^{(c)}=\gamma L_b$ and $S^{(a)}=\gamma M_k^2/L_b$, with $L_b=\langle C_{\rm D}\rangle=\sigma\omega_k(p)$. This relation holds in the listing of the question and is not invariant: moving a single-body operator $Z^{\rm D}_b$ between $C_{\rm D}$ and $C_{bc}$ changes $L_b$ by $\langle Z^{\rm D}_b\rangle$ but not $S$, and moving $\mathbb 1_b\otimes\sum_x\zeta_x\mathring\Phi_x$ from $C_f$ into $C_{bf}$ changes $S$ by $\sum_x\zeta_x$ but not $L_b$.
8. **(2.9)** (formula; every eigenvector of $C_{\rm pair}$ on fixed levels, every binding of the form S5; $\kappa=0$; first order in $\gamma_i$). The total source of the pair is $S^{(a)}_{12}=\sum_i\gamma_iM^{(i)}_{k_i}\langle\sigma^{(i)}_3\rangle$ and $S^{(c)}_{12}=\sum_i\gamma_i\langle C_{{\rm D},i}\rangle$; for $\gamma_1=\gamma_2=\gamma$, $S^{(c)}_{12}=\gamma(E_{12}-\langle U\rangle)$. The binding term contributes no term of its own, and the total is exactly the sum of what the parts present in the joint state.
9. **(2.10)** (example, control of item 3; $U=u\sigma^{(1)}_1\sigma^{(2)}_1$, $p_1=p_2=0$ on a ring, or held bodies). The pair presents $(\gamma_1M^{(1)}_{k_1}+\gamma_2M^{(2)}_{k_2})M/\sqrt{M^2+u^2}$. This differs from the sum of the free sources whenever $u\neq0$ and $\gamma_1M^{(1)}_{k_1}+\gamma_2M^{(2)}_{k_2}\neq0$, while its cost $\sqrt{M^2+u^2}$ exceeds $M$. On the ring the difference is a change between two starts of the same description at $\lambda=0$; it reaches $f$ as $\frac1n\bigl(S_{12}-\sum_i\gamma_iM^{(i)}_{k_i}\bigr)\sum_x\mathring\Phi_x$, fixed by $C$ and the two starts.
10. **(2.11)** (theorem; whole setting). The replacements that leave $C$ unchanged are exactly the moves of single-object operators (constants, place-dependent and internal operators) between the terms that contain the object, including a new $c$–$f$ term. The two-object components, in particular $C_{\{b,f\}}=\gamma\sum_xQ_x\otimes\mathring\Phi_x$, are fixed by $C$.
11. **(2.12)** (theorem; whole setting; first order in $\gamma$). The component listing $(H^\circ_R,C_{\{f\}},K)$ is fixed by $C$. Its first-order generator $G^\circ_f(\lambda)=\mathring C_f+\sum_xs^\circ_x(\lambda)\mathring\Phi_x$, with $s^\circ_x=\gamma\langle\xi^\circ\vert Q_x\vert\xi^\circ\rangle$ and $\xi^\circ$ evolved by $H^\circ_R$, is a function of $C$ and the start, hence invariant under every replacement (2.11). The view of $f$ follows it with the remainder (2.4), $\lVert C_{bf}\rVert$ replaced by $\lVert K\rVert\le2\lVert C_{bf}\rVert$. The generator (2.3) of the listing of the question equals it modulo $\mathbb 1_f$ up to $4\lVert C_{bf}\rVert^2\lvert\lambda\rvert$ in operator norm. The change between two starts, $\sum_x(s^\circ_x-s^{\circ\prime}_x)\mathring\Phi_x$, is fixed by $C$ and the two starts.
12. **(2.13)** (criterion; couplings (a), (c) with $\gamma\neq0$ and some $\mathring\Phi_y\neq0$). A constant added to $\sigma_3\otimes(m\mathbb 1_N+D)$ is a convention iff all $\mathring\Phi_x$ are equal (the field does not resolve positions). Otherwise the zero of the body operator is fixed by $C$.
13. **(2.14)** (criterion; whole setting; first order in $\gamma$; ratios where the denominator is nonzero). (i) The source field, its total and its ratio to another body coupled to the same $\Phi_x$, read from a listing, are not fixed by $C$ and the start: the move (2.11) of $\mathbb 1_b\otimes W$ from $C_f$ into $C_{bf}$, with any self-adjoint $W$ on $\mathcal H_f$ chosen for each body, adds $W$ to the body part, and $s_x\to s_x+\zeta_x$ for $W=\sum_x\zeta_x\mathring\Phi_x$. (ii) The component split, a convention computed from $C$ in which the body part $F^\circ_i$ of each body comes from its component $C_{\{b_i,f\}}$ (zero fixed by $\operatorname{Tr}_RK=0$), is a function of $C$ and the start, and the listing of the question realizes it at first order. In it, the coefficients $s^\circ_x$ with respect to the stated $\mathring\Phi_x$, their total $S^\circ$ and the ratios between bodies are fixed by $C$ and the start, up to the common factor $1/a$ (ratios exactly), if the $\mathring\Phi_x$ are linearly independent; if all $\mathring\Phi_x$ equal one $\mathring\Phi\neq0$, $S^\circ$ and $S^\circ_1/S^\circ_2$ are. (iii) The additive changes between two starts are the same in every split: $\Delta s^{(i)}_x$ and $\Delta S_i$ if the $\mathring\Phi_x$ are linearly independent, $\Delta S_i$ if all $\mathring\Phi_x=\mathring\Phi\neq0$; under the same conditions so is $\Delta S_1/\Delta S_2$ where $\Delta S_2\neq0$. Hence they are fixed by $C$ and the two starts, the changes up to $1/a$, the ratio exactly. (iv) The change of a ratio between two starts, $\Delta(S_1/S_2)$, is in general not the same in every split: in the component split it is a function of $C$ and the two starts under the conditions of (ii), but in the control of Step 8 (two branch starts at $p=0$, all $\mathring\Phi_x=\mathring\Phi\neq0$) a move of $\mathbb 1_{b_2}\otimes w\mathring\Phi$ changes it from $2q_1/q_2$ to $2q_1/(q_2+w)$.

## Consistency checks

1. **Dimensions.** $\tau$, $m$, $d_k$, $\omega_k$ and $C$ have the unit of $1/\lambda$. $s_x\Phi_x$ has the unit of $\gamma\Phi_x\cdot O$, the same as $C_{bf}$, so $G_f$ has the unit of $C$. In (2.4), $\lVert C_{bf}\rVert^2\lvert\lambda\rvert$ has the unit of $C$, like $[G_f,V_f]$. In (2.7)–(2.10), $S$ has the unit of $\gamma M_k$. Passed.
2. **Body without motion, $\tau\to0$.** $\omega_k\to M_k$, $\theta_k\to0$ and $Q^{(c)}_x\to Q^{(a)}_x$. Both lines of (2.7) give $\gamma\sigma M_k$, (2.8) gives $\gamma M_k^2/L_b=\gamma L_b$, and the two forms of (2.9) coincide. Passed.
3. **Sum over both branches.** $S_{k,+}(p)+S_{k,-}(p)=0$ in both couplings. This equals $\gamma$ times the trace of the $2\times2$ block ($\mu_k\sigma_3$ resp. $C_k(p)$) and agrees with $\operatorname{Tr}Q_x=0$ in (2.2). Passed.

## Open issues

- $M_k$ is taken as $\mu_k=m+d_k$, the half-gap of the on-site block; the question does not quote its definition.
- Items 2 and 3 are derived on rings of $n$ sites; open chains and their boundary corrections are not treated.
- With records ($\kappa>0$, $X\chi\neq0$) only a bound on the change of the total source is given; the $O(\kappa)$ correction itself is not computed.
- The source field, its total and their ratios are values of a split; the listing of the question realizes the component split, a convention computed from $C$. Whether a later package singles out a split, and which field descriptions fix the common factor $1/a$ (normalization of the $\Phi_x$), are left to the packages on the field's response and profile.
- For the pair, only eigenvectors of the pair contract and one control are treated; there is no general relation between the pair's source and those of its free parts.

## Methods used

- von Neumann equation for the view (partial trace)
- first-order (mean-field) perturbation theory at fixed $\lambda$; Duhamel formula; trace-norm bounds
- spectral decomposition; linear independence of exponentials
- plane-wave block diagonalization; two-level eigenproblem
- decomposition of operators on tensor products into traceless components; component listing
- weighted-shift operator-norm estimate

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