\section{Proof of the separable half, and wavefunctions}
\label{app:proofs}

\paragraph{Proof of Theorem~\ref{thm:criterion}(b).} Absorb the masses into $g$. Choose orthonormal bases $e_i$ of
$V_A$ and $e_i'$ of $V_B$ with $\ip{e_i}{e_j'}=s_i\delta_{ij}$ (singular value decomposition; $s_1=\rmax$,
$s_i=0$ beyond the smaller rank). Fix reference branches $a_0$, $b_0$ and put
$x_i(a)=\ip{e_i}{g_a-g_{a_0}}$, $y_i(b)=\ip{e_i'}{g_b-g_{b_0}}$. Since the $g_a-g_{a'}$ lie in $V_A$,
\begin{equation}
  D_A=\tfrac12\sum_i\big(x_i(a)-x_i(a')\big)^2,\qquad
  \delta=\sum_is_i\big(x_i(a)-x_i(a')\big)\big(y_i(b)-y_i(b')\big),
\end{equation}
and $F(a,b)=\sum_is_ix_i(a)y_i(b)+\varphi_A(a)+\varphi_B(b)+\text{const}$. Hence
$\Lambda=\sum_i\Lambda_i+i[\varphi_A(a)-\varphi_A(a')]+i[\varphi_B(b)-\varphi_B(b')]$ with $\Lambda_i$ as in
Eq.~\eqref{eq:pairs}; the phases $\varphi$ are local unitaries. For each $i$ take real Gaussian variables
$(\eta_i,\zeta_i,\eta_i',\zeta_i')$, mean zero, independent for different $i$, with covariance
\begin{equation}
  t\,s_i\begin{pmatrix}|J|&J\\J&|J|\end{pmatrix},\qquad J=\begin{pmatrix}0&\tfrac12\\ \tfrac12&C\end{pmatrix},
\end{equation}
positive semidefinite because $|J|\pm J\geq0$. Define the random diagonal operators
$U_A=\sum_a\exp[\sum_ix_i(a)(\eta_i+i\zeta_i)]\,|a\rangle\langle a|$ and $U_B$ likewise with $y_i$,
$\eta_i'$, $\zeta_i'$. With $\mathbb E\,e^Z=e^{\mathbb E[Z^2]/2}$ the cross covariances
$\mathbb E[\zeta_i\zeta_i']=ts_iC$ and $\mathbb E[\eta_i\zeta_i']=\mathbb E[\zeta_i\eta_i']=ts_i/2$ give
\begin{equation}
  \begin{split}
  &\mathbb E\big[u_A(a)u_B(b)\bar u_A(a')\bar u_B(b')\big]\\
  &\qquad=\exp\Big\{t\sum_is_i\big[i(x_iy_i-x_i'y_i')-C(x_i-x_i')(y_i-y_i')\big]\Big\}\,L_A(a,a')L_B(b,b'),
  \end{split}
\end{equation}
using $(x+x')(y-y')+(x-x')(y+y')=2(xy-x'y')$, with local factors $L$. Dividing,
$e^{t\Lambda}=\mathbb E[u\bar u]\,T_A(a,a')T_B(b,b')$ up to the local phases, with
\begin{equation}
  T_A(a,a')=d(a)\overline{d(a')}\,\exp\Big\{t\sum_i\big(C-s_i\operatorname{tr}|J|\big)x_i(a)x_i(a')\Big\},
  \qquad\operatorname{tr}|J|=\sqrt{1+C^2},
\end{equation}
$d$ a non-vanishing diagonal factor, and the same for $T_B$. If $s_i\sqrt{1+C^2}\leq C$ for all $i$ the
exponent is $t$ times a positive semidefinite matrix; its entrywise exponential is positive semidefinite by
the Schur product theorem~\cite{HornJohnson2013}, so $T_A$ and $T_B$ act as completely positive Schur
multipliers $\mathcal T_A$, $\mathcal T_B$. Therefore
\begin{equation}
  \rho_t=\mathbb E\Big[(U_A\otimes U_B)\,(\mathcal T_A\otimes\mathcal T_B)(\rho_0)\,(U_A\otimes U_B)^\dagger\Big],
\end{equation}
a mixture of products of local completely positive maps, trace preserving in total because the diagonal of
$e^{t\Lambda}$ is one. Separable states go to separable states. Interleaved local channels and the
Lie--Trotter limit in finite dimension, with the separable set closed, give the last statement. The
construction needs only that the Gram matrices $f_A$, $f_B$ are non-singular (distinct branches); it holds
for any number of branches, any amplitudes and every $t\geq0$. If $\rmax>\sth$, then $T_A$ and $T_B$ are
indefinite and part (a) shows the state entangled. \qed

\paragraph{Dictionary for Remark~\ref{rem:wpp}.} Each $\Lambda_i$ in Eq.~\eqref{eq:pairs} is the generator
of Lemma~3 of~\cite{WangPedernalesPlenio2026} with $O_A=\sum_ax_i(a)|a\rangle\langle a|$,
$O_B=\sum_by_i(b)|b\rangle\langle b|$, coupling $\lambda=-s_i$, local rates $\Gamma_A=\Gamma_B=C$ and cross
rate $\Gamma_{AB}=-Cs_i$ (in units $Gm^2/\hbar\sigma$). Their lemma makes $e^{t\Lambda_i}$ separable when
$\lambda^2+\Gamma_{AB}^2\leq\Gamma_A\Gamma_B$, which is $s_i\sqrt{1+C^2}\leq C$.

\paragraph{Wavefunctions.} For wavefunctions only the entangling half of the criterion survives: a witness
with bounded patterns, on regions where the wavefunctions do not vanish, makes the state entangled at once.

\begin{theorem}[Wavefunctions, entangling direction]
\label{thm:continuum}
Let $\psi_A,\psi_B\in L^2(\mathbb R^3)$ and $H=0$. Suppose there are bounded neutral $q_1$, $q_2$, supported
on sets of finite measure on which $|\psi_A|,|\psi_B|\geq\varepsilon>0$, with
$C\ip{q_1}{q_1}+C\ip{q_2}{q_2}-2\,\mathrm{Re}[(C+i)\ip{q_1}{q_2}]<0$ in the smeared Coulomb form. Then
$\rho_t$ is entangled for all sufficiently small $t>0$.
\end{theorem}

\begin{proof}
The kernel $\Lambda$ is bounded ($|f|\leq f_0$). With $c=q_1\otimes|\psi_B|^2+|\psi_A|^2\otimes q_2\in L^1$ and
$\chi=c/\tilde\Psi^*\in L^2$, $\langle\chi|\rho_t^{T_1}|\chi\rangle=\int\!\!\int\bar c\,c\,e^{t\Lambda}
=t\,(\text{negative})+O(t^2\|c\|_1^2)$. A separable state has a positive partial transpose, whose quadratic
form is non-negative.
\end{proof}

Since the smeared point charges of any open set span the Coulomb space (their potentials are real-analytic),
such $q_1$, $q_2$ exist for every pair of supports with interior and every separation.