\section{Proofs of the bounds}
\label{app:bounds}

Lengths are in units of $\sigma$ unless stated.

\paragraph{Proof of Theorem~\ref{thm:parallel}.} $h(x)=f(\sqrt x)=\pi^{-1/2}\int_0^1e^{-xu^2/4}du$ is a mixture of
decaying exponentials: decreasing, convex and log-convex. Put $x=|r|^2$, $y=|w|^2$, $P=|r-w|^2$,
$Q=|r+w|^2$, so $P+Q=2x+2y$, and $N=2h(x)-h(P)-h(Q)$, $s=|N|/(2D(|w|))$. If $N>0$: convexity gives
$h(P)+h(Q)\geq2h(x+y)$, log-convexity $h(x+y)h(0)\geq h(x)h(y)$, so
$N\leq2[h(x)-h(x+y)]\leq2(f(|r|)/f_0)\,D(|w|)$, and $|r|\geq d$. If $N<0$: (i) $-N$ is the second difference of
$t\mapsto f(|r+tw|)$ at $t=-1,0,1$, hence $\leq|w|^2\Lambda_+$, with $\Lambda_+$ the largest Hessian
eigenvalue of $f$ over space, which is $\max f''$ because $f'<0$; (ii) with $P\leq Q$ the parallelogram law
gives $Q\geq x$, so $-N\leq h(P)-h(x)\leq f(d)$. For $|w|\geq L_{1/2}$, (ii) gives $-N/(2D)\leq f(d)/f_0$.
For $|w|\leq L_{1/2}$, (i) gives $-N/(2D)\leq\Lambda_+|w|^2/(2D(|w|))\leq\Lambda_+L_{1/2}^2/f_0$, because
$L^2/D(L)$ is increasing ($D(L)/L^2$ is a difference quotient of the convex $h$). The transversal family with
$L\to\infty$ approaches $f(d)/f_0$ from below. \qed

\paragraph{Proof of Proposition~\ref{prop:ceilings}.} Write $N=\psi(b_1)-\psi(b_2)$ with
$\psi(b)=f(|b-a_1|)-f(|b-a_2|)$. (i) $|\psi|\leq g_dL_A$ with $g_d=\sup_{r\geq d}|f'|$. (ii)
$|\psi|\leq m_d(L_A)$, $m_d(L)=\sup_{r\geq d}[f(r)-f(r+L)]$. The same with $A$ and $B$ exchanged. (iii) $N$
is a double difference, so $|N|\leq L_AL_B\sup\|\nabla^2f\|$ over the parallelogram of difference vectors, on
which $|x|\geq d$ at slab separation and $|x|\geq d-(L_A+L_B)/2$ at closest approach; $\|\nabla^2f\|=
\max(|f''|,|f'|/r)\leq\kappa_0$. Then
\begin{equation}
  s\leq\frac{\min\{2g_dL_A,\,2g_dL_B,\,2m_d(L_A),\,2m_d(L_B),\,HL_AL_B\}}{2\sqrt{D(L_A)D(L_B)}},
\end{equation}
with $H$ the supremum in (iii), maximised over
$(L_A,L_B)$ with cell bounds (numerators at the upper corner, denominators at the lower corner of each cell
of a geometric grid of ratio $1.01$, a last cell for $L>400\,\sigma$), using that $L/\sqrt{D(L)}$, $m_d$ and
$D$ are monotone. The suprema over $r$ are taken on grids, and the whole evaluation is in floating point
without interval arithmetic. The bound is non-increasing in $d$. It is $0.70456$ at $2.75\,\sigma$ (slab)
and $0.70457$ at $3.78\,\sigma$ (closest approach), $0.36\%$ below $1/\sqrt2$, and exceeds $1/\sqrt2$ at
$2.735\,\sigma$ ($0.70875$) and $3.765\,\sigma$ ($0.71019$). \qed

\paragraph{The numerical search over two-branch geometries.} Differential evolution followed by
Nelder--Mead, over several parametrisations of the two branch pairs with the separation constraint built in,
found nothing above $f(d)/f_0$ at slab separation, or above $\max(f(d)/f_0,\text{straddle limit})$ at
closest approach. This supports, but does not prove, that the closed forms of Table~\ref{tab:twobranch} are
the suprema over all two-branch geometries.

\paragraph{Proof of Theorem~\ref{thm:nodistance}.} Smearing lowers the self-energy, so $\ip{q}{q}\leq16/(3\pi a)$
per unit-charge uniform disc; $f(r)\geq\operatorname{erf}(d/2\sigma)/r$ for $r\geq d$; and for unsmeared
coaxial discs $E(0)-E(d)=\int\frac{d^2k}{(2\pi)^2}\frac{2\pi}{k}|\hat q|^2(1-e^{-kd})\leq2\pi d\int q^2
=2d/a^2$. The far copy makes the pattern neutral without changing the ratio; this proves~\eqref{eq:discbound}. For
the first statement at arbitrary $d$ and $C$, let $A$ carry a uniform planar lattice patch (with in-plane
distances $\Delta_{jj'}$) and its far copy, and $B$ the same patch translated by $d$ along the normal. The
same Fourier estimate, with the smeared kernel, gives
$1-\rmax\leq(d/2\sigma^2)\sum_{jj'}e^{-\Delta^2_{jj'}/4\sigma^2}/\sum_{jj'}f(\Delta_{jj'})$; the numerator grows
like the number of points and the denominator like its square times $f$ of the patch size, so the right side
tends to zero for growing patches. As wavefunctions, discs of small thickness inherit a strict inequality by
continuity of the bounded smeared kernel. \qed

\paragraph{Derivation of the extent law.} For charges on coaxial unit discs at gap $\delta$ ($\sigma\to0$),
the mutual energy is, up to a common constant, $E(\delta)=\int_0^\infty e^{-k\delta}\hat q(k)^2dk$, with
$\hat q$ the Hankel transform of the charge. Writing an axisymmetric charge through its Abel transform,
$\hat q(k)=\int_0^1g(t)\cos(kt)\,dt$ with $g$ extended evenly to $(-1,1)$, gives
$E(0)=(\pi/2)\int_0^1g^2$ and $E(\delta)=(\pi/2)\int_0^1g\,T_\delta g$, with $T_\delta$ of
Eq.~\eqref{eq:love}, the operator of Love's integral equation~\cite{Love1949}. The canonical correlation is
the top eigenvalue of $T_\delta$ on even functions. The uniform disc is $g(t)=\sqrt{1-t^2}$. Its Rayleigh
quotient is $\Phi(a/d)$, so the uniform pattern is a witness if and only if
\begin{equation}
  \Phi(a/d)>\sth,\qquad \Phi(x)=\frac{3\pi}{16}\int_0^\infty\Big(\frac{2J_1(u)}{u}\Big)^2e^{-u/x}\,du .
  \label{eq:Phi}
\end{equation}

\paragraph{Proof of Theorem~\ref{thm:ratebounds}.} (i) $C\ip{q_1}{q_1}+C\ip{q_2}{q_2}\geq2C|\ip{q_1}{q_2}|/\rmax$, so the
numerator of~\eqref{eq:rate} is $\geq-2(1-\theta)\sqrt{1+C^2}|\ip{q_1}{q_2}|$, and
$|\ip{q_1}{q_2}|\leq f(d)\|q_1\|_1\|q_2\|_1\leq\tfrac12f(d)(\sum|q_1|^2/p+\sum|q_2|^2/r)$ by Cauchy--Schwarz.
(ii) For charges at depths $z\leq0$ and $z'\geq d$, $\ip{q_1}{q_2}=\int\frac{d^2k}{(2\pi)^2}\frac{2\pi}{k}
e^{-kd}\bar A_1A_2$ with $A_1(k)=\int dz\,e^{kz}\hat q_1(k,z)$, $|A_1|\leq\|q_1\|_1$, and
$\ip{q_1}{q_1}\geq\int\frac{d^2k}{(2\pi)^2}\frac{2\pi}{k}|A_1|^2$, because $e^{-k|z-z'|}-e^{k(z+z')}$ is $2k$ times
the Dirichlet Green function of $-\partial_z^2+k^2$ on the half-line, a positive kernel. Then mode by mode
the numerator is at least $-\int\frac{d^2k}{(2\pi)^2}\frac{2\pi}{k}[\sqrt{1+C^2}e^{-kd}-C]_+(|A_1|^2+|A_2|^2)$,
and $\int_0^\infty[\sqrt{1+C^2}e^{-kd}-C]_+dk=B(C)/d$. With smearing, for planar supports, $e^{-kz}$ is replaced
by $w(k,z)=\tfrac12[e^{-kz}\operatorname{erfc}(\sigma k-z/2\sigma)+e^{kz}\operatorname{erfc}(\sigma k+z/2\sigma)]$,
which gives the finite-$\sigma$ value of Table~\ref{tab:constants}. \qed

\paragraph{Gaussian sheets.} For the explicit states of Section~\ref{sec:rate}, a Gaussian sheet of rms
width $b$ facing the sheet of the other mass at gap $d$ ($\sigma\to0$, zero thickness), the same mode
formula gives the rate of the witness proportional to the sheet profile. Maximised over the width, it is
\begin{equation}
  \kappa_G=\max_{y}\ \frac{\sqrt\pi}{2}\,y\big[\sqrt{1+C^2}\,e^{y^2}\operatorname{erfc}(y)-C\big],\qquad y=\frac{d}{2b}.
  \label{eq:kappaG}
\end{equation}

\paragraph{Proof of Theorem~\ref{thm:negativity}.} Let $\Lambda_\theta$ be $\Lambda$ with $F$ replaced by $\theta F$ in
the Hamiltonian and in the correlated noise, local dephasing unchanged, and
$\rho_t^\theta=\rho_0\circ e^{t'\Lambda_\theta}$. Its canonical correlations are $\theta s_i\leq\sth$, so by
Remark~\ref{rem:general} $\rho_t^\theta$ is separable and $(\rho_t^\theta)^{T_1}\geq0$. Now
$\rho_t=\rho_t^\theta\circ M$ with $M=\exp[t'(1-\theta)(i\epsilon-C\delta)]$, so by Weyl's inequality
$\lambda_{\min}(\rho_t^{T_1})\geq-\|\rho_t^\theta\circ(M-1)\|_{\rm HS}$, the Hilbert--Schmidt norm being
invariant under partial transposition. Entrywise $|\rho_t^\theta|=(p_ap_{a'}r_br_{b'})^{1/2}
e^{-t'C(D_A+D_B+\theta\delta)}$ and $|M-1|\leq t'(1-\theta)(\epsilon^2+C^2\delta^2)^{1/2}
\max(1,e^{-t'(1-\theta)C\delta})$; every exponent is non-positive since
$D_A+D_B+\delta=\tfrac12\|g_a-g_{a'}+g_b-g_{b'}\|^2\geq0$. For the corollary: terms with $a\neq a'$, $b\neq b'$
have $C^2\delta^2+\epsilon^2\leq4C^2f(d)^2$ (for $C\geq1/\sqrt3$) and exponent $\geq2D_\ell-2f(d)$; terms with
one index pair equal have $\delta=0$, $|\epsilon|\leq f(d)$, exponent $\geq D_\ell$ and weight $\sum p^2$ or
$\sum r^2$; and $\sup_tte^{-\alpha t}=1/(e\alpha)$. For $\ell,d\gg\sigma$, $D_\ell\to f_0=1/\sqrt\pi$ and
$\sum p^2,\sum r^2\to0$. \qed