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

# 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:** v1
- **Mode:** new
- **Date:** 2026-10-19

## 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 $\lvert p\rangle_n:=n^{-1/2}\lvert p\rangle$; on the open chain, normalized packets on one branch and level.
- 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
$$\sum_xQ^{(a)}_x=B,\qquad\sum_xQ^{(c)}_x=A+B=C_{\rm D},\qquad\operatorname{Tr}Q_x=0 . \tag{2.2}$$
Since the $\Pi_x$ are orthogonal and $\lVert O\rVert=\mu_{N-1}$, $\lVert C^{(a)}_{bf}\rVert=\lvert\gamma\rvert\mu_{N-1}\max_x\lVert\Phi_x\rVert$. The $A_x$ with even (odd) $x$ act on disjoint pairs of sites and $\lVert A_x\rVert=\tau/2$, so $\lVert C^{(c)}_{bf}\rVert\le\lvert\gamma\rvert(\mu_{N-1}+\tau)\max_x\lVert\Phi_x\rVert$. 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; 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
$$G_f(\lambda)=C_f+\sum_\nu e^{i\nu\lambda}\,\Gamma_\nu,\qquad\Gamma_\nu:=\gamma\sum_{E-E'=\nu}\ \sum_x\langle\xi_0\vert P_EQ_xP_{E'}\vert\xi_0\rangle\,\Phi_x . \tag{2.5}$$
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 independent of $\lambda$ iff $\Gamma_\nu=0$ for every $\nu\neq0$, and then $G_f=C_f+\Gamma_0$. Two sufficient cases:
- 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 $\psi$;
- $\Phi_x=\Phi$ for all $x$ in coupling (c) with $\kappa=0$ or $X\chi=0$, and any $\Psi_b$: then $G_f=C_f+\gamma\langle C_{\rm D}\rangle\Phi$ 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 $\langle\beta_{k,+}\vert\sigma_3\vert\beta_{k,-}\rangle=-\sin\theta_k\neq0$.

*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 $\lambda$, and $G_f$ depends on $\lambda$ whenever $\sum_xe^{ipx}\Phi_x\neq0$ (for example, when the $\Phi_x$ 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). A normalized packet $\int g(p)\lvert p\rangle\otimes\beta_{k,\sigma}(p)\otimes\lvert f_k\rangle\frac{\mathrm dp}{2\pi}$ on one branch and level stays on that branch, and both operators are block-diagonal in $p$. Its total source is therefore $\int\lvert g\rvert^2S_{k,\sigma}(p)\frac{\mathrm dp}{2\pi}$, 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 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 fixes

*Generator.* For $C=H_R\otimes\mathbb 1+\mathbb 1\otimes C_f+C_{bf}$, $\operatorname{Tr}_R\bigl((\lvert\xi\rangle\langle\xi\rvert\otimes\mathbb 1)C\bigr)=\langle H_R\rangle\mathbb 1+G_f$, so
$$G_f(\lambda)\equiv\operatorname{Tr}_R\bigl((\lvert\xi(\lambda)\rangle\langle\xi(\lambda)\rvert\otimes\mathbb 1_f)\,C\bigr)\pmod{\mathbb 1_f}. \tag{2.12}$$
Only the multiple of $\mathbb 1_f$ drops out of $[G_f,V_f]$. The right side uses $C$ and $\xi$. Moves (2.11) inside $H_R$ (between $C_{\rm D}$ and $C_{bc}$) leave $\xi$ unchanged. A move of an $O(\gamma)$ operator between $C_{bf}$ and $H_R$ changes $\xi$ by $O(\gamma\lvert\lambda\rvert)$ and $G_f$ by $O(\gamma^2\lvert\lambda\rvert)$, which lies inside (2.4). So $G_f$ modulo $\mathbb 1_f$ is well defined to first order. For two starts of the same description, the difference is $G_f-G'_f\equiv\sum_x(s_x-s'_x)\mathring\Phi_x$.

*Split.* A move $Z^{bf}_f=-Z^f_f$ leaves $G_f$ unchanged. It shifts the part $\sum_xs_x\Phi_x$ that the listing attributes to the body by the state-independent operator $Z^{bf}_f$. For $Z^{bf}_f=\sum_x\zeta_x\mathring\Phi_x$ this is $s_x\to s_x+\zeta_x$, with the body operator $Q_x\to Q_x+\frac{\zeta_x}{\gamma}\mathbb 1_b$. Without a rule that fixes the form of the body operator, the source field is therefore fixed only up to state-independent shifts.

*Rule: the form (a) or (c).* Read the source from a listing 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
$$\text{a constant }c_0\neq0\text{ added to }\sigma_3\otimes(m\mathbb 1_N+D)\text{ leaves }C\text{ unchanged (after moves (2.11))}\iff\mathring\Phi_x=\mathring\Phi_y\ \text{for all }x,y . \tag{2.13}$$
The only remaining refactorization is $(\gamma_i,\Phi_x)\to(\gamma_i/a,\,a\Phi_x)$, common to all bodies coupled to the same $\Phi_x$: one $\gamma$ multiplies the same $O$ at every $x$. It changes no term and multiplies every source by $1/a$. When the zero is fixed, $s_x=\gamma\langle Q_x\rangle$ is thus fixed by $C$ and the state, up to this common factor. When the zero is a convention, $s_x\to s_x+\gamma c_0\langle\Pi_x\rangle$ and $S\to S+\gamma c_0$ for every normalized state, with $c_0$ chosen independently for each body. Differences between two starts of one body change by $\gamma c_0(\langle\Pi_x\rangle-\langle\Pi_x\rangle')$, which sums to zero over $x$. This gives
$$\begin{aligned}&\text{if }\exists\,x,y:\ \mathring\Phi_x\neq\mathring\Phi_y:\ s_x,\ S\text{ and the ratios }s^{(1)}_x/s^{(2)}_x,\ S_1/S_2\text{ are fixed by }C\text{ and the state, up to the common factor }1/a\ (\text{ratios exactly});\\&\text{if all }\mathring\Phi_x\text{ are equal: the family }s_x,\ S\text{ and the ratios are not fixed }(S\to S+\gamma c_0,\ c_0\text{ free for each body});\ \text{differences }S-S'\text{ between starts of one body are, up to }1/a.\end{aligned} \tag{2.14}$$
The relation (2.8) of $S$ to the cost is a relation to the listed $C_{\rm D}$: a move $Z^{\rm D}_b$, a constant included, changes $L_b$ but not $S$. So $S^{(c)}=\gamma L_b$ holds in the listing where the body's own term equals $\sum_xQ^{(c)}_x$. 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).

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 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, 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 independent of $\lambda$, hence a contract of $f$ alone, iff $\Gamma_\nu=0$ for every Bohr frequency $\nu\neq0$ of $H_R$. This 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 $\Phi_x=\Phi$ for all $x$ and $\kappa=0$ or $X\chi=0$, it holds for every start.
5. **(2.6)** (example, control of item 1). 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 $\lambda$ whenever $\sum_xe^{ipx}\Phi_x\neq0$.
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 $\sin^2p$. 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, $S$ is the $\lvert g(p)\rvert^2$-average of (2.7). 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 is a relation to the listed body term: it holds in the listing in which the body's own term is $\sum_xQ_x$, and a move (2.11) of a single-body operator $Z^{\rm D}_b$ into $C_{\rm D}$ changes $L_b$ by $\langle Z^{\rm D}_b\rangle$ but does not change $S$.
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 $M$.
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; first order in $\gamma$). $G_f$ modulo $\mathbb 1_f$ is fixed by $C$ and the start. So is its change between two starts, $\sum_x(s_x-s'_x)\mathring\Phi_x$. The split of $G_f$ into $C_f$ and a body part is not fixed: without a rule for the form of the body operator, $s_x$ is fixed only up to state-independent shifts.
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; same conditions, reading the source from a listing of form (a) or (c)). If some $\mathring\Phi_x\neq\mathring\Phi_y$, then the source field, its total and its value relative to another body coupled to the same $\Phi_x$ are well defined. The first two are fixed up to one factor $1/a$ (the normalization of the $\Phi_x$) that is common to all bodies; the ratios are fixed exactly where the denominator is nonzero. If all $\mathring\Phi_x$ are equal, none of the three is well defined, since $S\to S+\gamma c_0$ with $c_0$ free for each body. What is then well defined are the differences $S-S'$ between two starts of one body (up to $1/a$) and the ratios of such differences.

## 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.
- Normalized branch starts are taken on a ring or as packets; boundary corrections for long open chains are not estimated.
- 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 common factor $1/a$ (normalization of the $\Phi_x$) and which field descriptions fix it 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

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