Version 1 · earlier version; the current one is v4 · External review, round 1: major errors

# Speed of the made field: front, formation of the static mark and long range

- **Subproject:** 07-ilang-weight
- **Package:** 08-field-speed
- **Version:** v1
- **Mode:** new
- **Date:** 2026-10-21

## Setup and assumptions

- **Class** (question; M8). $N_f\ge1$ identical field objects ($c=+1$) on the torus $X_f=\mathbb Z_L^D$, $D\in\{1,2,3\}$, $L$ even, $|X_f|=L^D$; no pair terms among them; own term $C_1=-t_fA$, $A$ the nearest-neighbour adjacency, $t_f>0$; $\Phi_x$ the number of field objects at $x$, $n=N_f/|X_f|$. $\Lambda=2D\mathbb 1-A$ has the eigenvalues $\omega(\mathbf k)=\sum_\mu4\sin^2(k_\mu/2)$, $\mathbf k\in(2\pi/L)\mathbb Z_L^D$.
- **Source** (assumption). The term $g_s(\lambda)\Phi_{x_s}$, $g_s(\lambda)=0$ for $\lambda<0$: (i) sudden, $g_s(\lambda)=g_s$ for $\lambda\ge0$; (ii) gradual, $g_s(\lambda)=g_sf(\lambda/T)$, $f\in C^\infty(\mathbb R)$ nondecreasing, $f=0$ on $(-\infty,0]$, $f=1$ on $[1,\infty)$. (i) is the evolution (1.9) for $\lambda\ge0$. (ii) has a $\lambda$-dependent coefficient, which A5 excludes for contracts; it is used as the question states it, as an effective description of a source switched on from outside, with $i\,\mathrm d\Psi/\mathrm d\lambda=C(\lambda)\Psi$.
- **Start** (assumption; M9, A11). At $\lambda\le0$ the field is in its lowest state without source.
- **Made field.** $\delta\rho(x,\lambda)=\langle\Phi_x\rangle_\lambda-n$; the static mark $\delta\rho_{\rm st}$ is the same quantity in the lowest state with constant $g_s$. Both are fixed by the state and by the operators $\Phi_x$, which are symmetric in the identical objects (M6).
- **Distances.** $r=x-x_s$ with $|r_\mu|\le L/2$; $\rho=\lVert r\rVert_\infty$, $|r|$ the Euclidean length, $d=\sum_\mu|r_\mu|$ the graph distance. Along a lattice axis $d=\rho=|r|$.
- **Regime** (A13). First order in $g_s$ at fixed $L$ and $\lambda$, remainder $O(g_s^2)$; every large-distance statement is a leading order, stated with its limit.
- **Inputs.** (4.4) of 06-ilang-source/04-profile:
$$M^{(2)}(x_1,x_2)=-\frac{2n\,g_1g_2}{t_f}\,\Lambda^+(x_1,x_2),\qquad \Lambda^+(x,y)=\frac1{\lvert X_f\rvert}\sum_{\mathbf k\neq0}\frac{e^{i\mathbf k\cdot(x-y)}}{\omega(\mathbf k)},\qquad \sum_{x_2\in X_f}M^{(2)}(x_1,x_2)=0;$$
(4.4) of 07-ilang-weight/04-weight-of-bond, with $\ell(\mathbf r)=\Lambda^+(x,x+\mathbf r)$: $\ell(0)-\ell(\mathbf r)=\frac{\ln|\mathbf r|}{2\pi}+\frac{2\gamma_E+3\ln2}{4\pi}+O(|\mathbf r|^{-2})-\frac{|\mathbf r|^2}{4L^2}$ for $D=2$, and $=G_0-\frac1{4\pi|\mathbf r|}+O(|\mathbf r|^{-3})-\frac{|\mathbf r|^2}{6L^3}$, $G_0=0.2527310$, for $D=3$.

## Derivation

### Step 1. Exact reduction and first-order response

Without pair terms and with $c=+1$, $C(\lambda)=\sum_{i=1}^{N_f}h_i(\lambda)$ on the symmetric subspace, $h(\lambda)=-t_fA+g_s(\lambda)|x_s\rangle\langle x_s|$. The lowest state of $-t_fA$ is $\psi_0=|X_f|^{-1/2}\sum_x|x\rangle$, nondegenerate (Perron–Frobenius, connected graph), with eigenvalue $-2Dt_f$; the lowest admissible state is $\psi_0^{\otimes N_f}$. The evolution of a sum of one-object terms keeps a product of identical states a product, $\Psi(\lambda)=\psi(\lambda)^{\otimes N_f}$ with $i\dot\psi=h(\lambda)\psi$. Hence $\langle\Phi_x\rangle_\lambda=N_f|\psi(x,\lambda)|^2$ exactly: the made field of $N_f$ objects is $N_f$ times that of one, at every order in $g_s$. First-order Dyson expansion around $e^{2iDt_f\lambda}\psi_0$, with $\langle x_s|\psi_0\rangle=|X_f|^{-1/2}$:

$$\delta\rho(x,\lambda)=2n\int_0^\lambda g_s(\lambda')\operatorname{Im}\mathcal U(x-x_s,\lambda-\lambda')\,\mathrm d\lambda'+O(g_s^2),\qquad \mathcal U(r,s)=\langle x_s+r|e^{-ist_f\Lambda}|x_s\rangle=\frac1{|X_f|}\sum_{\mathbf k}e^{i\mathbf k\cdot r-it_f\omega(\mathbf k)s}.\tag{8.1}$$

The lowest state with constant $g_s$ is, at first order, $\psi_0-\sum_{\mathbf k\ne0}|\mathbf k\rangle\langle\mathbf k|g_s|x_s\rangle\langle x_s|\psi_0\rangle/(t_f\omega(\mathbf k))$, so

$$\delta\rho_{\rm st}(r)=-\frac{2ng_s}{t_f}\Lambda^+(r).\tag{8.2}$$

With $\operatorname{Im}\mathcal U=-|X_f|^{-1}\sum_{\mathbf k}\cos(\mathbf k\cdot r)\sin(t_f\omega s)$ and $\int_0^\lambda\sin(t_f\omega s)\,\mathrm ds=(1-\cos t_f\omega\lambda)/(t_f\omega)$, the sudden switching gives, for $\lambda\ge0$,

$$\delta\rho(r,\lambda)=\delta\rho_{\rm st}(r)+\frac{2ng_s}{t_f}R(r,\lambda),\qquad R(r,\lambda)=\frac1{|X_f|}\sum_{\mathbf k\ne0}\frac{\cos(\mathbf k\cdot r)\cos(t_f\omega(\mathbf k)\lambda)}{\omega(\mathbf k)}.\tag{8.3}$$

$R(r,0)=\Lambda^+(r)$, so $\delta\rho(r,0)=0$; $R$ is a finite sum of cosines and does not decay.

### Step 2. Free propagator and Kapteyn bound

By the Jacobi–Anger expansion $e^{iz\cos k}=\sum_mi^mJ_m(z)e^{imk}$ and Poisson summation over the torus,

$$\mathcal U(r,s)=\sum_{\mathbf m\in\mathbb Z^D}\mathcal U_\infty(r+L\mathbf m,s),\qquad \mathcal U_\infty(r,s)=e^{-2iDt_fs}\prod_{\mu=1}^Di^{r_\mu}J_{r_\mu}(2t_fs).\tag{8.4}$$

Kapteyn's inequality gives, for integer $m\ge0$ and $0\le u\le m$, $|J_m(u)|\le e^{-m\eta(u/m)}$ with $\eta(z)=\operatorname{artanh}\sqrt{1-z^2}-\sqrt{1-z^2}$, which decreases from $\infty$ to $0$ on $(0,1]$ and equals $\tfrac13(2(1-z))^{3/2}(1+O(1-z))$ near $z=1$. With $|J_m|\le1$ for the other factors, $|\mathcal U_\infty(r,s)|\le e^{-\rho\,\eta(2t_fs/\rho)}$ for $2t_fs\le\rho$; $R\eta(c/R)$ increases with $R$ (its derivative is $\eta+\sqrt{1-z^2}>0$).

### Step 3. The front (item 1)

**Ahead of the front.** Let $\rho\le L/2$ and $2t_f\lambda<\rho$. Every image ($\mathbf m\ne0$) has $\rho_{\mathbf m}=\lVert r+L\mathbf m\rVert_\infty\ge L-\rho\ge\rho$, so (8.1) and Step 2 give, for both switchings,

$$|\delta\rho(r,\lambda)|\le2n\lambda\max_{\lambda'}|g_s(\lambda')|\sum_{\mathbf m}e^{-\rho_{\mathbf m}\eta(2t_f\lambda/\rho_{\mathbf m})}\qquad(2t_f\lambda<\rho\le L/2),\tag{8.5}$$

which is exponentially small in $\rho$ at fixed $2t_f\lambda/\rho<1$; the $\mathbf m=0$ term dominates. For every start of the $N_f$ objects, at first order, $\Delta\langle\Phi_x\rangle=-i\int_0^\lambda g_s(\lambda')\langle[\Phi^H_x(\lambda),\Phi^H_{x_s}(\lambda')]\rangle\mathrm d\lambda'$ is a sum of one-object commutators of norm $\le2|\mathcal U(r,\lambda-\lambda')|$, so (8.5) holds with $2n$ replaced by $2N_f$.

**At the front, $D=1$.** Here $\operatorname{Im}\mathcal U_\infty=J_r(2t_fs)\sin(r\pi/2-2t_fs)$, and (8.1) is $\frac{ng_s}{t_f}\int_0^{2t_f\lambda}J_r(u)\sin(\frac{r\pi}2-u)\mathrm du$. One integration by parts leaves $\int J_r'(u)\cos(\frac{r\pi}2-u)\mathrm du$, which is $O(r^{-2/3})$ near $u=r$: $J_r'=O(r^{-2/3})$ there, and $J_r(u)e^{\pm iu}$ has no stationary point (the local frequency of $J_r$ is $\sqrt{1-r^2/u^2}<1$). With the Airy form $J_r(r+\xi(r/2)^{1/3})=(2/r)^{1/3}[\mathrm{Ai}(-\xi)+O(r^{-2/3})]$:

$$\delta\rho(r,\lambda)=\frac{ng_s}{t_f}\Bigl[\Bigl(\frac2r\Bigr)^{1/3}\mathrm{Ai}\Bigl(\frac{r-2t_f\lambda}{(r/2)^{1/3}}\Bigr)\cos\Bigl(\frac{r\pi}2-2t_f\lambda\Bigr)+O(r^{-2/3})\Bigr]\qquad(|2t_f\lambda-r|=O(r^{1/3})).\tag{8.6}$$

The envelope reaches $0.6749\,(n|g_s|/t_f)\,r^{-1/3}$ at $2t_f\lambda=r+1.019(r/2)^{1/3}$. **$D>1$.** The time-dependent part of (8.3) is $\frac{2ng_s}{t_f}\operatorname{Re}\int\frac{\mathrm d^D\mathbf k}{(2\pi)^D}\frac{e^{i\mathbf k\cdot r-it_f\omega\lambda}}{\omega}$; near $\rho=2t_f\lambda$ its stationary points are those of $\mathcal U_\infty$ ($\sin k_\mu=r_\mu/(2t_f\lambda)$), where $\omega\ge2$. So the change at the front is of the order $(n|g_s|/t_f)|\mathcal U_\infty|$: each factor with $|r_\mu|=\rho$ contributes $\rho^{-1/3}$, each other factor $\rho^{-1/2}$. The largest decay exponent is that of the axes, $p_D=\frac13+\frac{D-1}2$.

**Criterion (front).** Fix $\varepsilon>0$ and $p>p_D$. The change arrives at $r$ at $\lambda_a(r)=\min\{\lambda\ge0:|\delta\rho(r,\lambda)|\ge\varepsilon\,(n|g_s|/t_f)\,\rho^{-p}\}$, sudden switching. By (8.5), with $2n|g_s|\lambda<n|g_s|\rho/t_f$, the threshold is not reached while $\rho\,\eta(2t_f\lambda/\rho)>(p+1)\ln\rho+\ln(1/\varepsilon)+o(1)$, that is while $1-2t_f\lambda/\rho>\frac12[3((p+1)\ln\rho+\ln(1/\varepsilon))/\rho]^{2/3}(1+o(1))$. Within $O(\rho^{1/3})$ behind $\rho=2t_f\lambda$ the change reaches the order $\rho^{-p_D}$, above $\varepsilon\rho^{-p}$ for large $\rho$. Hence

$$\lambda_a(r)=\frac{\rho}{2t_f}+O\Bigl(\frac{\rho^{1/3}(\ln\rho)^{2/3}}{t_f}\Bigr)\quad(\rho\to\infty,\ \rho\le L/2):\qquad v_f=2t_f\ \text{place labels per unit }\lambda\ \text{in }\lVert\cdot\rVert_\infty .\tag{8.7}$$

In graph distance $d/D\le\rho\le d$, so $\lambda_a\in[\frac d{2Dt_f},\frac d{2t_f}](1+o(1))$: speed $2t_f$ along the axes, $2Dt_f$ along the diagonals. Code (ring $L=16384$ with $\varepsilon=0.1$, $p=\frac12$; cube $L=160$ along an axis with $p=\frac32$):
- Fitted speeds: $2.036\,t_f$ (ring, $\rho=64$–$384$) and $2.06\,t_f$ (cube, $\rho=12$–$64$).
- On the ring, $2t_f\lambda_a/\rho$ rises from $0.87$ at $\rho=16$ to $0.971$ at $\rho=384$, as the correction in (8.7) allows.
- Ahead of the front, $|\delta\rho|$ stays below $0.02$ of the bound (8.5).
- The front peak decays as $\rho^{-0.340}$ on the ring (prediction $-\frac13$; peak $\times\rho^{1/3}=0.34$–$0.35$ against $0.337$ in units $2ng_s/t_f$) and as $\rho^{-1.322}$ on the cube (prediction $-\frac43$).
- (8.6) deviates from the exact change by $0.44,0.33,0.27,0.24$ of the peak at $r=64,128,256,384$, decreasing like $r^{-1/3}$.

### Step 4. Behind the front: scaling form

Take $L\to\infty$ first, so that $\mathcal U\to\mathcal U_\infty$ and the $\mathbf k$-sums become integrals; then (8.3) reads $\delta\rho=-\frac{2ng_s}{t_f}\int\frac{\mathrm d^D\mathbf k}{(2\pi)^D}\cos(\mathbf k\cdot r)\frac{1-\cos t_f\omega\lambda}{\omega}$, convergent at $\mathbf k=0$ for every $D$. Let $|r|\to\infty$ at fixed $\zeta=|r|^2/(4t_f\lambda)$ (behind the front: $\zeta<|r|/2$, true for large $|r|$). At leading order only $|\mathbf k|=O(1/|r|)$ contributes, where $\omega=\mathbf k^2(1+O(\mathbf k^2))$: elsewhere the integrand oscillates on the scale $1/|r|$, and the other critical points of $\omega$ ($k_\mu\in\{0,\pi\}$) contribute $O((t_f\lambda)^{-D/2})=O(|r|^{-D})$, smaller by $|r|^{-2}$. With $(1-e^{-ia\mathbf k^2})/\mathbf k^2=i\int_0^ae^{-ia'\mathbf k^2}\mathrm da'$ and the Fresnel integral $\int\frac{\mathrm d^D\mathbf k}{(2\pi)^D}e^{i\mathbf k\cdot r-ia'\mathbf k^2}=(4\pi ia')^{-D/2}e^{i|r|^2/4a'}$:

$$\delta\rho(r,\lambda)\simeq-\frac{2ng_s}{t_f}\operatorname{Re}W_D(|r|,it_f\lambda),\qquad W_D(r,\Theta)=\int_0^\Theta\frac{e^{-r^2/4s}}{(4\pi s)^{D/2}}\mathrm ds,\tag{8.8}$$

continued along $s=ia'$, $0\le a'\le t_f\lambda$. Substituting $u=r^2/4s$, $\partial_rW_D(r,\Theta)=-\Gamma(\frac D2,\frac{r^2}{4\Theta})/(2\pi^{D/2}r^{D-1})$, and $r^2/(4it_f\lambda)=-i\zeta$. As $\Theta\to\infty$ the gradient of (8.8) tends to that of the long-range part $\delta\rho^\infty_{\rm st}$ of the static mark, $\partial_{|r|}\delta\rho^\infty_{\rm st}=\frac{2ng_s}{t_f}\frac{\Gamma(D/2)}{2\pi^{D/2}|r|^{D-1}}$. Hence

$$\frac{\partial_{|r|}\delta\rho(r,\lambda)}{\partial_{|r|}\delta\rho^\infty_{\rm st}(r)}\simeq g_D(\zeta)=\frac{\operatorname{Re}\Gamma(D/2,-i\zeta)}{\Gamma(D/2)}=\begin{cases}\operatorname{Re}\operatorname{erfc}\sqrt{-i\zeta},&D=1,\\ \cos\zeta,&D=2,\\ \operatorname{Re}\bigl[\operatorname{erfc}\sqrt{-i\zeta}+2\sqrt{-i\zeta/\pi}\;e^{i\zeta}\bigr],&D=3,\end{cases}\tag{8.9}$$

with the principal root. From $\gamma(s,w)=w^s/s-w^{s+1}/(s+1)+\dots$: $1-g_1=\sqrt{2\zeta/\pi}\,(1+O(\zeta))$, $1-g_2=\frac{\zeta^2}2(1+O(\zeta^2))$, $1-g_3=-\frac{4\zeta^{3/2}}{3\sqrt{2\pi}}(1+O(\zeta))$. For $\zeta\gg1$, $g_1\to0$, $g_2$ oscillates between $\pm1$, and $g_3=2\sqrt{\zeta/\pi}\cos(\zeta-\pi/4)+o(1)$: there the made field is a ripple of local wavenumber $|r|/(2t_f\lambda)$, not the static mark. In $D=3$ the value itself obeys $\delta\rho/\delta\rho^\infty_{\rm st}\simeq\operatorname{Re}\operatorname{erfc}\sqrt{-i\zeta}=1-\sqrt{2\zeta/\pi}(1+O(\zeta))$. Code (window before the wrap time): for $\zeta\le0.5$ the exact gradient ratio follows (8.9) within $0.006$ on the ring ($|r|=6$–$40$, $L=16384$), and within $0.04$ to $0.008$ on the cube along an axis ($|r|=3$–$9$, $L=160$), decreasing with $|r|$.

### Step 5. Formation of the static mark, sudden switching (item 2 (i))

**Criterion (formation).** The long-range part at distance $|r|$ is formed to accuracy $\varepsilon$ from $\lambda_f$ on if $|1-\partial_{|r|}\delta\rho/\partial_{|r|}\delta\rho_{\rm st}|\le\varepsilon$ for every $\lambda\ge\lambda_f$ in the regime of Step 4 ($\delta\rho_{\rm st}$ replaced by its long-range part $\delta\rho^\infty_{\rm st}$). By (8.9) this holds iff $\zeta\le\zeta_D(\varepsilon)$, the largest $\zeta$ below which $|1-g_D|\le\varepsilon$:

$$\lambda_{\rm form}(|r|;\varepsilon)=\frac{|r|^2}{4t_f\,\zeta_D(\varepsilon)},\qquad \zeta_1=\frac{\pi\varepsilon^2}2\bigl(1+O(\varepsilon^2)\bigr),\quad \zeta_2=\sqrt{2\varepsilon}\bigl(1+O(\varepsilon)\bigr),\quad \zeta_3=\Bigl(\frac{3\sqrt{2\pi}\,\varepsilon}4\Bigr)^{2/3}\bigl(1+O(\varepsilon^{2/3})\bigr),\tag{8.10}$$

independent of $L$. In $D=3$ the value at $|r|$ comes within $\varepsilon$ of $\delta\rho^\infty_{\rm st}$ at $\lambda=|r|^2/(2\pi\varepsilon^2t_f)$. In $D=1,2$ the level does not form on the infinite lattice: at fixed $|r|$, $\operatorname{Re}W_1=\sqrt{t_f\lambda/2\pi}-|r|/2+o(1)$ and $\operatorname{Re}W_2=\frac1{4\pi}\bigl(\ln\frac{4t_f\lambda}{|r|^2}-\gamma_E\bigr)+o(1)$ grow without bound. Code: on the ring ($L=16384$, $\varepsilon=0.2$, $\zeta_1=0.0604$ from (8.9), leading order $0.0628$), $\lambda_{\rm form}$ at $|r|=6$–$40$ lies within $3.4\%$ of (8.10), with fitted exponent $1.98$. On the cube along an axis ($L=160$, $\varepsilon=0.1$, $\zeta_3=0.407$, leading order $0.328$), it lies within $5\%$, with exponent $2.03$ and prefactor $0.586$ against $1/(4\zeta_3)=0.614$.

**The torus.** By (8.4), $\delta\rho_L(r,\lambda)=\sum_{\mathbf m}\delta\rho_\infty(r+L\mathbf m,\lambda)$ exactly at first order; by (8.5) the image terms are exponentially small for $\lambda<(L-\rho)/(2t_f)$, the wrap time. So (8.8)–(8.10) hold on the torus for $\lambda$ below the wrap time, and the formation (8.10) happens before it when $|r|^2<2\zeta_D(\varepsilon)(L-\rho)$. After it the wrapped changes add to (8.3) a part that does not decay. Only the $\lambda$-average converges, exactly: $\omega(\mathbf k)>0$ for $\mathbf k\ne0$ and $\frac1\Lambda\int_0^\Lambda\cos(t_f\omega\lambda)\mathrm d\lambda=\frac{\sin t_f\omega\Lambda}{t_f\omega\Lambda}$, so

$$\Bigl|\frac1\Lambda\int_0^\Lambda\delta\rho(r,\lambda)\,\mathrm d\lambda-\delta\rho_{\rm st}(r)\Bigr|\le\frac{2n|g_s|}{t_f}\,\frac{S_2}{t_f\Lambda}\xrightarrow{\Lambda\to\infty}0,\qquad S_2=\frac1{|X_f|}\sum_{\mathbf k\ne0}\frac1{\omega(\mathbf k)^2}=O(L^{4-D}),\tag{8.11}$$

since $S_2$ is dominated by $|\mathbf k|=O(1/L)$, where $\omega^{-2}=O(L^4)$; $S_2=\frac{L^3}{720}(1+O(L^{-2}))$ for $D=1$. The averaged field is within $\varepsilon|\delta\rho_{\rm st}(r)|$ for $\Lambda\ge S_2/(\varepsilon t_f|\Lambda^+(r)|)$: of order $L^2/(60\varepsilon t_f)$ ($D=1$, $|r|\ll L$, $\Lambda^+\simeq L/12$), $L^2/(\varepsilon t_f\ln(L/|r|))$ ($D=2$) and $L|r|/(\varepsilon t_f)$ ($D=3$). The slowest modes saturate the bound up to a factor of order one. Code (ring $L=1024$, units $2ng_s/t_f$): at $\Lambda=1,4,16\,L^2/t_f$ the running average of the exact evolution deviates from its $\Lambda\to\infty$ limit by $1.34,0.25,0.025$, against the bound $1.42,0.36,0.089$.

### Step 6. Formation, gradual switching (item 2 (ii))

Integrating (8.1) by parts in $\lambda'$ ($g_s(0)=0$), and with $F(\nu)=\int_0^1f'(u)e^{i\nu u}\mathrm du$:

$$\delta\rho(r,\lambda)-\frac{g_s(\lambda)}{g_s}\delta\rho_{\rm st}(r)=\frac{2n}{t_f|X_f|}\sum_{\mathbf k\ne0}\frac{\cos(\mathbf k\cdot r)}{\omega}\int_0^\lambda g_s'(\lambda')\cos\bigl(t_f\omega(\lambda-\lambda')\bigr)\mathrm d\lambda',\qquad =\frac{2ng_s}{t_f|X_f|}\sum_{\mathbf k\ne0}\frac{\cos(\mathbf k\cdot r)}{\omega}\operatorname{Re}\bigl[e^{-it_f\omega\lambda}F(t_f\omega T)\bigr]\ (\lambda\ge T).\tag{8.12}$$

The sudden switching is $F\equiv1$, which gives back (8.3). Relative to its static weight, mode $\mathbf k$ deviates by at most $\min\{1,c_f/(t_f\omega T)^2\}$ during the ramp, $c_f=\lVert f''\rVert_\infty+\lVert f'''\rVert_1$ (two more integrations by parts; the boundary terms vanish because $\sin0=0$ and $f'(0)=f''(0)=0$; the bound $1$ holds because $f$ is nondecreasing). After the ramp it deviates by $|F(t_f\omega T)|\le C_m(t_f\omega T)^{-m}$ for every $m$, because $f'$ is smooth with all derivatives zero at $0$ and $1$. So mode $\mathbf k$ follows iff $t_f\omega(\mathbf k)T\gg1$. **Criterion (following).** In the regime of Step 4, $E_D$ is the largest relative deviation, over $0\le\lambda\le2T$, of the gradient at $|r|$ from $(g_s(\lambda)/g_s)\,\partial_{|r|}\delta\rho_{\rm st}$. With $\mathbf k=\mathbf q/|r|$ it is a function $E_D(\nu;f)$ of $\nu=|r|^2/(t_fT)$ alone. The unfollowed modes $|\mathbf k|\lesssim k_T=(t_fT)^{-1/2}$ carry the relative weight $O(\nu^{D/2})$, and the lag of the modes $k_T\lesssim|\mathbf k|\lesssim1/|r|$, $|r|\int_{k_T}\mathrm dk\,k^{D-1}(t_fk^2T)^{-2}$ relative to $|r|^{1-D}$, is $O(\nu^{D/2})$ as well. As $\nu\to\infty$ the switching is sudden on the scale $|r|$, and $E_D$ tends to the deviation of (8.9) at $\zeta\ge\nu/8$, which is of order one. Hence, for $T\ll L^2/t_f$,

$$E_D(\nu;f)=O(\nu^{D/2})\ (\nu\to0),\qquad T_{\rm form}(|r|;\varepsilon)=\frac{|r|^2}{t_f\,\nu_D(\varepsilon;f)},\qquad \nu_D\propto\varepsilon^{2/D}\quad(\varepsilon\to0),\tag{8.13}$$

where $\nu_D(\varepsilon;f)$ is the largest $\nu$ below which $E_D\le\varepsilon$ (its constants depend on $f$); the gradient follows within $\varepsilon$ for every $T\ge T_{\rm form}$. **The whole field.** The modes $|\mathbf k|<k_T=(t_fT)^{-1/2}$ are not followed. In $D=1$ the modes $|\mathbf k|=2\pi/L$ alone carry the fraction $6/\pi^2$ of $\Lambda^+(0)$, so the level follows only for $T\gtrsim L^2/(4\pi^2t_f)$. In $D=2$ the unfollowed share of $\Lambda^+(r)$ is $\ln(k_TL/2\pi)/\ln(L/|r|)$ at leading logarithmic order, so the value follows to $\varepsilon$ for $T\ge(L/2\pi)^2(|r|/L)^{2\varepsilon}/t_f$. In $D=3$ the unfollowed share of $1/(4\pi|r|)$ is $2|r|k_T/\pi$, so $T\ge4|r|^2/(\pi^2\varepsilon^2t_f)$. For $T\gg L^2/(4\pi^2t_f)$ every mode follows. Code (ring $L=1024$ with $\varepsilon=0.2$; cube $L=128$ along an axis with $\varepsilon=0.1$): for $\nu<0.5$, $E\propto\nu^{0.52}$ on the ring and $\nu^{1.37}$ on the cube (predictions $\frac12$ and $\frac32$). The fitted $T_{\rm form}$ scales as $|r|^{1.90}$ ($|r|=4$–$16$) and $|r|^{2.18}$ ($|r|=3$–$7$).

### Step 7. Long range and finite speed (item 3)

The long-range part at distance $|r|$ is carried by $|\mathbf k|\sim1/|r|$, whose frequencies are $t_f\omega\sim t_f/|r|^2$; at $\lambda=\kappa|r|/t_f$ their phases are $O(\kappa/|r|)\to0$. Quantitatively: if the criterion of Step 5 held from $\lambda_f\le\kappa|r|/t_f$, it would hold at $\lambda=K|r|/t_f$ for a fixed $K\ge\max(\kappa,1)$. There the continuum form applies (the relevant $|\mathbf k|=1/(2K)$ is small) with $\zeta=|r|/(4K)\to\infty$. But on $[\zeta_D(\varepsilon),\zeta]$, $|1-g_D|$ exceeds $\tfrac12$ for large $\zeta$: $g_1\to0$, $g_2=-1$ at $\zeta=\pi$, and $g_3$ is unbounded. For the gradual switching, $T=\kappa|r|/t_f$ gives $\nu=|r|/\kappa\to\infty$. Hence:

$$\text{in the class, for every fixed }\kappa>0\text{ and }\varepsilon<\tfrac12:\quad \lambda_{\rm form}(|r|;\varepsilon)>\kappa|r|/t_f\ \text{ and }\ T_{\rm form}(|r|;\varepsilon)>\kappa|r|/t_f\ \text{ for all large }|r|.\tag{8.14}$$

The long-range part forms only on $\lambda\propto|r|^2/t_f$, although the front moves with the finite speed $2t_f$.

**Control** (wider class: field objects with more places and pair terms among them). One field object per place $x$, with places $0,\dots,M-1$ carrying truncated oscillator operators $q_x,p_x$, and the contract $\sum_xp_x^2/2+\frac\kappa2\sum_{\langle xy\rangle}(q_x-q_y)^2+g_s(\lambda)q_{x_s}$. The pair terms join neighbouring field objects, and the source couples to $q_{x_s}$. For states with negligible weight on the top places ($M$ large compared with the occupations reached), (1.9) gives the linear equations $\ddot{\langle q_x\rangle}=-\kappa(\Lambda\langle q\rangle)_x-g_s\delta_{x,x_s}$ for every start. For the sudden switching from $\langle q\rangle=\langle p\rangle=0$, the $\mathbf k\ne0$ part is $-\frac{g_s}{\kappa\omega}(1-\cos\sqrt{\kappa\omega}\,\lambda)$. The static mark $-\frac{g_s}\kappa\Lambda^+$ has the same long-range part as (8.2), while the $\mathbf k=0$ part drifts uniformly. In $D=3$, with $\int\frac{\mathrm d^3\mathbf k}{(2\pi)^3}e^{i\mathbf k\cdot r}\frac{\cos ck\lambda}{k^2}=\frac{\theta(|r|-c\lambda)}{4\pi|r|}$:

$$\langle q\rangle(r,\lambda)-\langle q\rangle_{\mathbf k=0}(\lambda)\simeq-\frac{g_s}{4\pi\kappa|r|}\,\theta\bigl(\sqrt\kappa\,\lambda-|r|\bigr)\qquad(|r|\gg1).\tag{8.15}$$

The long-range part forms at the front $|r|=\sqrt\kappa\lambda$, linearly in $|r|$; the largest group velocity of $\sqrt{\kappa\omega(\mathbf k)}$ is $\sqrt\kappa$. A mode follows a gradual switching iff $\sqrt{\kappa\omega(\mathbf k)}\,T\gg1$, and $|\mathbf k|\sim1/|r|$, so here the answers of items 2 and 3 differ: $\lambda_{\rm form},T_{\rm form}\propto|r|$.

**Control for item 1** (wider class: own terms of unbounded range). On the ring, $C_1'=-t_f\sum_{x\ne y}|x-y|^{-\alpha}|x\rangle\langle y|$ ($|x-y|$ the ring distance, $\alpha>1$) has the uniform lowest state. For $t_fs\ll1$, $\operatorname{Im}\mathcal U(r,s)=t_fs|r|^{-\alpha}(1+O((t_fs)^2))$ uniformly in $r$, since powers of a kernel decaying like $|r|^{-\alpha}$, $\alpha>1$, decay like $|r|^{-\alpha}$. So at a fixed small $\lambda_1$,

$$\delta\rho(r,\lambda_1)=\frac{n\,g_s\,t_f\lambda_1^2}{|r|^{\alpha}}\bigl(1+O((t_f\lambda_1)^2)\bigr),\tag{8.16}$$

which exceeds the front threshold $\varepsilon(n|g_s|/t_f)\rho^{-p}$ for all large $\rho$ when $p>\alpha$. The arrival time stays bounded, and the change ahead of any line $\lambda=\rho/v$ is only power-law small. There is no finite front speed.

### Step 8. Speeds (item 4)

The motion term of a Model D body is, in $k$-space, $A(k)=-\tau\sin k\,\sigma_1\otimes\mathbb 1_N$, with group velocities $\mp\tau\cos k$, largest $\tau$. Its propagator has the entries $J_r(\tau\lambda)$ (Jacobi–Anger with $\sin k$), so the body's own front is at $|x|=\tau\lambda$ with the bound of Step 2; with the rest and clock term the bands are $\pm\sqrt{\tau^2\sin^2k+m_a^2}$, $m_a>0$ the levels of $m\mathbb 1_N+D$, and their group velocities are smaller than $\tau$. The body's position label is the field place of the couplings ($\sum_x|x\rangle\langle x|\otimes G\otimes\Phi_x$), along one axis of $X_f$. With (8.7):

$$\frac{v_f}{\tau}=\frac{2t_f}{\tau},\qquad\text{a body moves more slowly than the front of the field iff }\ \tau<2t_f .\tag{8.17}$$

## Result

Class: the field class of the setup ($N_f\ge1$, $D=1,2,3$, torus of even side $L$), source at $x_s$, start in the lowest state; first order in $g_s$ unless stated.

- **(8.1)** Exactly, $\langle\Phi_x\rangle_\lambda=N_f|\psi(x,\lambda)|^2$: the made field of $N_f$ objects is $N_f$ times that of one, at every order. At first order $\delta\rho$ is the time integral (8.1) of the free propagator, for either switching.
- **(8.2)** The static mark is $-(2ng_s/t_f)\Lambda^+(r)$.
- **(8.3)** Sudden switching, $\lambda\ge0$: $\delta\rho=\delta\rho_{\rm st}+(2ng_s/t_f)R$. On the torus $R$ is a finite sum of cosines and does not decay.
- **(8.4)** The propagator is the image sum of products of Bessel functions, with $|\mathcal U_\infty|\le e^{-\rho\eta(2t_fs/\rho)}$ for $2t_fs\le\rho$.
- **(8.5)** Item 1, change ahead of the front: for $2t_f\lambda<\rho\le L/2$, $|\delta\rho|\le2n\lambda\max|g_s|\sum_{\mathbf m}e^{-\rho_{\mathbf m}\eta(2t_f\lambda/\rho_{\mathbf m})}$. This is exponentially small in $\rho$ at fixed $2t_f\lambda/\rho<1$, for both switchings, and holds for every start with $2N_f$ in place of $2n$.
- **(8.6)** Item 1, change at the front: for $D=1$, at leading order in $r$ and for $|2t_f\lambda-r|=O(r^{1/3})$, an Airy envelope of maximum $0.6749\,(n|g_s|/t_f)\,r^{-1/3}$. Along an axis in $D$ dimensions the change at the front is of order $\rho^{-1/3-(D-1)/2}$.
- **(8.7)** Item 1: the front speed is finite. With the criterion $|\delta\rho|\ge\varepsilon(n|g_s|/t_f)\rho^{-p}$, $p>\frac13+\frac{D-1}2$, the sudden switching gives $\lambda_a=\rho/(2t_f)+O(\rho^{1/3}(\ln\rho)^{2/3}/t_f)$ as $\rho\to\infty$, $\rho\le L/2$. The speed is $v_f=2t_f$ place labels per unit $\lambda$ in $\lVert\cdot\rVert_\infty$: $2t_f$ along the axes in graph distance, $2Dt_f$ along the diagonals.
- **(8.8), (8.9)** Behind the front ($L\to\infty$ first; $|r|\to\infty$ at fixed $\zeta=|r|^2/(4t_f\lambda)$; leading order) the made field is $-(2ng_s/t_f)\operatorname{Re}W_D(|r|,it_f\lambda)$. Its gradient is $g_D(\zeta)=\operatorname{Re}\Gamma(D/2,-i\zeta)/\Gamma(D/2)$ times that of the long-range static part.
- **(8.10)** Item 2 (i), sudden switching. The long-range part (the gradient) at distance $|r|$ is formed to accuracy $\varepsilon$ from $\lambda_{\rm form}=|r|^2/(4t_f\zeta_D(\varepsilon))$, with $\zeta_1=\frac{\pi\varepsilon^2}2(1+O(\varepsilon^2))$, $\zeta_2=\sqrt{2\varepsilon}(1+O(\varepsilon))$, $\zeta_3=(3\sqrt{2\pi}\varepsilon/4)^{2/3}(1+O(\varepsilon^{2/3}))$. This scale is independent of $L$ and holds on the torus up to the wrap time $(L-\rho)/(2t_f)$. In $D=3$ the value at $|r|$ forms at $|r|^2/(2\pi\varepsilon^2t_f)$ (relative to $\delta\rho^\infty_{\rm st}$). In $D=1,2$ the level grows without bound on the infinite lattice.
- **(8.11)** Item 2 (i), $\lambda$-average on the torus. The running average of $\delta\rho$ tends to $\delta\rho_{\rm st}$, with error at most $(2n|g_s|/t_f)S_2/(t_f\Lambda)$, $S_2=O(L^{4-D})$. It is within $\varepsilon|\delta\rho_{\rm st}(r)|$ for $\Lambda\ge S_2/(\varepsilon t_f|\Lambda^+(r)|)$, of order $L^2/(60\varepsilon t_f)$, $L^2/(\varepsilon t_f\ln(L/|r|))$ and $L|r|/(\varepsilon t_f)$ for $D=1,2,3$. The instantaneous field does not converge as $\lambda\to\infty$, since $R$ is a nonconstant finite sum of cosines.
- **(8.12)** Gradual switching: the first-order deviation from the adiabatic target $(g_s(\lambda)/g_s)\delta\rho_{\rm st}$. Per mode it is at most $\min\{1,c_f/(t_f\omega T)^2\}$ during the ramp and at most $C_m(t_f\omega T)^{-m}$ after it.
- **(8.13)** Item 2 (ii). For $T\ll L^2/t_f$, at leading order, the gradient at $|r|$ follows the static one within $\varepsilon$ on $[0,2T]$ for every $T\ge T_{\rm form}=|r|^2/(t_f\nu_D(\varepsilon;f))$, with $\nu_D\propto\varepsilon^{2/D}$ as $\varepsilon\to0$ and deviation $O((|r|^2/t_fT)^{D/2})$. The whole field follows for $T\gg L^2/(4\pi^2t_f)$. The level needs $T\gtrsim L^2/(4\pi^2t_f)$ in $D=1$, and the value at $|r|$ needs $T\ge(L/2\pi)^2(|r|/L)^{2\varepsilon}/t_f$ in $D=2$ (leading logarithmic order) and $T\ge4|r|^2/(\pi^2\varepsilon^2t_f)$ in $D=3$.
- **(8.14)** Item 3, negative. In the class, at first order (with $L\to\infty$ first, or before the wrap time), the long-range part does not form on any scale linear in $|r|$. For every fixed $\kappa>0$ and $\varepsilon<\frac12$, $\lambda_{\rm form}$ and $T_{\rm form}$ exceed $\kappa|r|/t_f$ for all large $|r|$; the long-range part forms on $|r|^2/t_f$.
- **(8.15)** Control for items 2 and 3, in a wider class: field objects with more places and nearest-neighbour pair terms (harmonic lattice, $D=3$, coupling to $q_{x_s}$). For every start with zero means and $M$ large, the long-range part forms at the front $|r|=\sqrt\kappa\lambda$, linearly in $|r|$.
- **(8.16)** Control for item 1, in a wider class: own term with hopping $\propto|x-y|^{-\alpha}$, $\alpha>1$, on the ring. At fixed small $\lambda_1$ the change is $ng_st_f\lambda_1^2|r|^{-\alpha}$; for $p>\alpha$ the arrival time stays bounded, so there is no finite front speed.
- **(8.17)** Item 4: $v_f/\tau=2t_f/\tau$. A body moving along an axis of the field moves more slowly than the front iff $\tau<2t_f$ (both in place labels per unit $\lambda$).

## Consistency checks

1. **Static limit.** The $\lambda$-average (8.11) gives (8.2), and $\Theta\to\infty$ in the gradient of (8.8) gives the gradient of its long-range part. By Hellmann–Feynman ($\partial E/\partial g_2=\langle\Phi_{x_2}\rangle$), $M^{(2)}(x_1,x_2)=g_2\,\delta\rho_{{\rm st},1}(x_2)$, which is (4.4) of 06-ilang-source/04-profile. The gradient limit of (8.9) is the $d$-dependence of (4.4) of 07-ilang-weight/04-weight-of-bond: $-\ln|r|/2\pi$ for $D=2$, $1/(4\pi|r|)$ for $D=3$. Code: the exact $\lambda\to\infty$ average differs from $-\Lambda^+$ by $1.1\cdot10^{-3}$ (ring) and $3.7\cdot10^{-4}$ (cube) of $|\Lambda^+(0)|$, the $O(g_s)$ second order. Passed.
2. **Sum rule.** $\sum_x\delta\rho(x,\lambda)=0$ at every $\lambda$, from (8.3) ($\sum_r\cos\mathbf k\cdot r=0$ for $\mathbf k\ne0$), as $\sum_{x_2}M^{(2)}=0$ in (4.4). Code: $|\sum_x\delta\rho|\le9\cdot10^{-11}$ in units $2ng_s/t_f$. Passed.
3. **Dimensions.** $g_s$, $t_f$ and $\tau$ have the units of $C$, and $\lambda$ the inverse units. $\delta\rho$, $g_s/t_f$, $t_f\lambda$, $\zeta$ and $\nu$ are numbers; $v_f$ and $\tau$ are place labels per unit $\lambda$; the scales (8.10), (8.13) are $\propto|r|^2/t_f$. Passed.

## Open issues

- On the torus, the size of the wrapped changes after the wrap time is not derived for the sudden switching; the instantaneous formation statements hold up to the wrap time, or with $L\to\infty$ first.
- The constants $\nu_D(\varepsilon;f)$ of (8.13) are not given in closed form, and $D=2$ is not checked numerically (the code covers the ring and the cube, as the question requires).
- Second order in $g_s$, and starts other than the lowest state: only (8.5) is derived for every start.
- Since the long-range part forms only on $|r|^2/t_f$, it is open how far out the long-range mark of a moving body is formed. Moving sources are out of scope.
- The meaning of (8.17) rests on identifying the body's position label with the field place along an axis. The relation to the distances of 03-ilang-space is out of scope.
- Whether a field with linear long-wavelength frequencies, as in (8.15), is selected by anything in the language is item 5, and is not addressed (M10).

## Code

- `code/v1/check_front_formation.py` (role: check). It evolves one field object ($N_f=1$, which suffices by (8.1)) exactly from its lowest state, with the source switched on suddenly and gradually ($f$ the $C^\infty$ step $e^{-1/u}/(e^{-1/u}+e^{-1/(1-u)})$). The evolution runs in the sector invariant under the point group of the torus fixing $x_s$, which the state never leaves. Constant contracts are propagated by Chebyshev expansion, which agrees with dense diagonalization to $2\cdot10^{-15}$ (cube $L=16$). The ramp uses the exponential midpoint rule with step $0.2/t_f$; a step of $0.05/t_f$ changes the gradient by $5\cdot10^{-5}$ relative. Lattices: rings $L=1024$, $16384$; cubic tori $L=32$, $128$, $160$. Couplings: $g_s/t_f=10^{-5}$ (ring), $10^{-3}$ (cube); the exact evolution differs from (8.3) by $3.7\cdot10^{-5}$ and $7.9\cdot10^{-4}$ of $|\Lambda^+(0)|$. It checks (8.3), (8.5)–(8.7), (8.9)–(8.11) and (8.13), with the fits quoted in Steps 3–6.
- Outputs: `code/v1/output/check_front_formation.json` (all numbers) and `code/v1/output/check_front_formation.log` (run log).

## Methods used

- Reduction of non-interacting identical objects to one object; first-order Dyson expansion; Hellmann–Feynman
- Fourier analysis on tori, Poisson summation, Jacobi–Anger expansion
- Kapteyn's inequality, Airy asymptotics of Bessel functions, stationary phase
- Fresnel–Gaussian integrals, analytic continuation of the integrated heat kernel, incomplete Gamma function
- Integration by parts in time (adiabatic following), Riemann–Lebesgue decay of smooth ramps
- Ehrenfest equations of a quadratic contract (control)
- Numerical check: exact evolution in the symmetric sector, Chebyshev propagation, exponential midpoint rule

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