\section{Simultaneous permittivity and permeability: exact corner purity}\label{sec:rectangle}
Work in fixed canonical units, or nondimensionalize once before fixing the
bounds. The Hamiltonian \eqref{eq:MaxwellHamiltonian} gives
\begin{equation}\label{eq:boxgenerator}
x'=A(a,b)x,\qquad A(a,b)=\begin{pmatrix}0&a\\-b&0\end{pmatrix},\qquad
0<a_-\le a\le a_+,\quad0<b_-\le b\le b_+.
\end{equation}
In the original canonical coordinates $a=\varepsilon^{-1}$ and
$b=k^2\mu^{-1}$; a fixed canonical rescaling changes these coefficients by
reciprocal positive factors without changing Floquet multipliers.
The admissible set is exactly the convex hull of the four material corners.
It is not a general non-Hermitian matrix family.

For a constant segment of duration $h$, put $\omega=\sqrt{ab}$. Its exact
transfer in the same fixed canonical coordinates is
\begin{equation}\label{eq:boxsegment}
E_{a,b}(h)=\begin{pmatrix}
\cos(\omega h)&a\sin(\omega h)/\omega\\
-b\sin(\omega h)/\omega&\cos(\omega h)
\end{pmatrix}.
\end{equation}
Chronological segments multiply with the latest factor on the left.

The angular and radial rates are
\begin{equation}\label{eq:boxrates}
v_{a,b}=b\cos^2\theta+a\sin^2\theta>0,\qquad
f_{a,b}=(b-a)\sin\theta\cos\theta.
\end{equation}
They obey $f_{a,b}=-\tfrac12\partial_\theta v_{a,b}$.
For $H_\eta=(f_{a,b}-\eta)/v_{a,b}$, differentiation gives
\begin{equation}\label{eq:boxderivatives}
\partial_aH_\eta=
\frac{-b\sin\theta\cos\theta+\eta\sin^2\theta}{v_{a,b}^2},\qquad
\partial_bH_\eta=
\frac{a\sin\theta\cos\theta+\eta\cos^2\theta}{v_{a,b}^2}.
\end{equation}
Assume first that both bound intervals are strict.

\begin{lemma}[The exact corner selector]\label{lem:boxselector}
For every finite $\eta$, the maximizer of $H_\eta$ on the rectangle is a
unique corner away from the following finite set of switching angles.
For $\eta>0$, let
$\theta_1=\arctan(b_+/\eta)$ and
$\theta_2=\pi-\arctan(\eta/a_+)$. In increasing angle order the arcs and
states are
\[
\begin{array}{c|c}
(0,\theta_1)&(a_-,b_+)\\
(\theta_1,\pi/2)&(a_+,b_+)\\
(\pi/2,\theta_2)&(a_+,b_-)\\
(\theta_2,\pi)&(a_+,b_+).
\end{array}
\]
For $\eta<0$, put $e=-\eta$,
$\theta_1=\arctan(e/a_-)$, and
$\theta_2=\pi-\arctan(b_-/e)$. The selector is
\[
\begin{array}{c|c}
(0,\theta_1)&(a_-,b_-)\\
(\theta_1,\pi/2)&(a_-,b_+)\\
(\pi/2,\theta_2)&(a_-,b_-)\\
(\theta_2,\pi)&(a_+,b_-).
\end{array}
\]
For $\eta=0$, only the two opposite corners are used:
$(a_-,b_+)$ on $(0,\pi/2)$ and $(a_+,b_-)$ on $(\pi/2,\pi)$.
For nonzero $\eta$ there are exactly three distinct states and four switches
per projective turn, counting the switch at the periodic cut.
\end{lemma}
\begin{proof}
On the first quadrant and for $\eta>0$, $\partial_bH>0$ for every $a,b$,
so $b=b_+$. The remaining derivative has the sign of
$\tan\theta(\eta\tan\theta-b_+)$, giving its first switch.
On the second quadrant $\partial_aH>0$ for every $a,b$, so $a=a_+$;
the other derivative has the sign of $a_+\tan\theta+\eta$, giving the
second switch. For $\eta=-e<0$, the first quadrant instead has
$\partial_aH<0$, fixing $a=a_-$, and the other derivative changes sign
at $\tan\theta=e/a_-$. The second quadrant has $\partial_bH<0$, fixing
$b=b_-$; the remaining derivative changes at
$\tan\theta=-b_-/e$. Strict inequalities away from these angles prove
uniqueness. Setting $\eta=0$ in the derivatives gives the two stated
opposite corners directly.
\end{proof}

Call this selected state $j_\eta(\theta)$ and define
\begin{equation}\label{eq:boxTG}
\tau_\Box(\eta)=\int_0^\pi\frac{\dd\theta}{v_{j_\eta}(\theta)},\qquad
G_\Box(\eta)=\int_0^\pi\frac{f_{j_\eta}(\theta)}{v_{j_\eta}(\theta)}\dd\theta,
\qquad \Psi_\Box=G_\Box-\eta\tau_\Box.
\end{equation}
These are elementary finite formulas: on each selected arc $[x,y]$,
\begin{equation}\label{eq:boxarcs}
\tau_{[x,y]}=\int_x^y\frac{\dd\theta}{b\cos^2\theta+a\sin^2\theta},\qquad
G_{[x,y]}=\frac12\log\frac{v_{a,b}(x)}{v_{a,b}(y)}.
\end{equation}
An antiderivative for the time is the continuous lifted branch of
$\arctan(\sqrt{a/b}\tan\theta)/\sqrt{ab}$. This convention avoids
spurious jumps of the principal arctangent.

\begin{lemma}[A strictly monotone clock]\label{lem:boxclock}
The function $\tau_\Box$ is continuous and strictly decreasing, with
\begin{equation}\label{eq:boxlimits}
\tau_\Box(-\infty)=\frac\pi{\sqrt{a_-b_-}},\qquad
\tau_\Box(+\infty)=\frac\pi{\sqrt{a_+b_+}}.
\end{equation}
For every finite $\eta$, $G_\Box(\eta)>0$. At zero,
\begin{equation}\label{eq:boxzero}
G_\Box(0)=\tfrac12\log\frac{a_+b_+}{a_-b_-},\qquad
\tau_\Box(0)=\frac\pi2\left(\frac1{\sqrt{a_-b_+}}+
\frac1{\sqrt{a_+b_-}}\right).
\end{equation}
\end{lemma}
\begin{proof}
The integral $\Psi_\Box=\int\max_j(f_j-\eta)/v_j$ is convex in $\eta$.
The selector is unique almost everywhere and denominators are uniformly
positive, so difference quotients and dominated convergence give
$\Psi_\Box'=-\tau_\Box$. Hence $\tau_\Box$ is nonincreasing. Equality of
its values at two different parameters would force $\Psi_\Box$ to be
affine between them. Equality in the integral of pointwise convex functions
then forces the same uniquely maximizing corner almost everywhere at both
parameters. The explicit moving cut angles in Lemma~\ref{lem:boxselector}
contradict that conclusion. Thus the decrease is strict. The arc formulas
give continuity, including at zero. As $\eta\to+\infty$, the fastest
corner $(a_+,b_+)$ is selected except on shrinking sets; as
$\eta\to-\infty$, the slowest corner is selected. Dominated convergence
gives \eqref{eq:boxlimits}.

A constant corner has zero integrated radial gain. For $\eta>0$, the
pointwise selector strictly improves on the fastest constant corner on a
set of positive measure. Therefore
$\Psi_\Box>-\eta\pi/\sqrt{a_+b_+}$ and
\[
G_\Box=\Psi_\Box+\eta\tau_\Box>
\eta\left(\tau_\Box-\frac\pi{\sqrt{a_+b_+}}\right)>0.
\]
For $\eta<0$, comparison with the slowest corner gives
$G_\Box>\eta(\tau_\Box-\pi/\sqrt{a_-b_-})>0$.
At zero, summing the two arc gains gives \eqref{eq:boxzero}, which is
positive under strict bounds.
\end{proof}

\begin{theorem}[Complete fixed-period Maxwell corner-purity theorem]\label{thm:boxfixed}
Let $S>0$ and both positive coefficient intervals be strict. Define
\[
\mathcal N_\Box=
\left\{n\in\N:\frac{n\pi}{\sqrt{a_+b_+}}<S<
\frac{n\pi}{\sqrt{a_-b_-}}\right\}.
\]
For each $n\in\mathcal N_\Box$, there is a unique $\eta_n$ with
$\tau_\Box(\eta_n)=S/n$. Then
\begin{equation}\label{eq:boxoptimum}
\boxed{\sup_{a(\cdot),b(\cdot)}\log\rho(M)
=F_\Box(S):=\max\bigl(\{0\}\cup
\{nG_\Box(\eta_n):n\in\mathcal N_\Box\}\bigr).}
\end{equation}
Every positive-growth optimizer is, up to translation and null sets, $n$
repetitions of the corner program in Lemma~\ref{lem:boxselector}, with its
arc times from \eqref{eq:boxarcs}, for a winning $n$. There are no additional
positive equality cases. In physical parameters it takes values only at
corners of $\{\varepsilon_-,\varepsilon_+\}\times\{\mu_-,\mu_+\}$.
At $\eta_n=0$, two opposite material corners suffice; at $\eta_n\ne0$,
three are necessary for that canonical branch. All four are never needed
for this bounds-only rectangular problem.
\end{theorem}
\begin{proof}
For an expanding eigenline, $\theta'=v_{a,b}>0$ and its lift advances $n\pi$.
Pointwise angular bounds by the slowest and fastest corners give the
stated winding interval. Equality at an endpoint would force its constant
corner almost everywhere and contradict positive growth. The clock lemma
gives a unique $\eta_n$.
For the actual control set
\[
E(\theta)=\max_{(a,b)\text{ in rectangle}}H_{\eta_n}(\theta,a,b)
-H_{\eta_n}(\theta,a(\theta),b(\theta))\ge0.
\]
Time cancellation gives the exact identity
\begin{equation}\label{eq:boxdefect}
nG_\Box(\eta_n)-\log\rho(M)
=\int_{\theta_0}^{\theta_0+n\pi}E(\theta)\dd\theta.
\end{equation}
The selector realizes one full angular turn with gain $G_\Box>0$, so its
one-turn multipliers are $-e^{\pm G_\Box}$. Repetition attains each branch.
If equality holds, the nonnegative defect vanishes and the unique corner
selector holds almost everywhere. The clock fixes all switch times.
A nonhyperbolic profile has zero growth and cannot improve a positive
branch. If the winding set is empty, no growing profile exists.
\end{proof}

\begin{corollary}[Free-period rectangle rate and quantum optimum]\label{cor:boxfree}
There is a unique $\gamma_\Box>0$ with
$G_\Box(\gamma_\Box)=\gamma_\Box\tau_\Box(\gamma_\Box)$.
It is the largest amplitude growth per unit time over all periodic and
nonperiodic measurable rectangle controls. Its periodic equality profiles
are precisely repetitions of the canonical $\eta=\gamma_\Box$ cell.
For strict intervals this free-rate optimizer uses three material corners
and four switches per turn. Its asymptotic squeezing and logarithmic
pair-count optimizers coincide with the classical ones, with rates
$\gamma_\Box$ and $2\gamma_\Box$ per unit time.
\end{corollary}
\begin{proof}
$\Psi_\Box$ is strictly decreasing with derivative $-\tau_\Box<0$,
positive at zero and tending to $-\infty$. At its unique positive root,
the zero-mean angular calibration has a bounded periodic primitive. The
gauge proof of Theorem~\ref{thm:free} applies word for word to the rates
\eqref{eq:boxrates}: its derivative is bounded by $\gamma_\Box$ along
every measurable trajectory, and equality follows from the exact defect.
The quantum rates follow from Theorem~\ref{thm:quantum} for this same
canonical monodromy.
\end{proof}

\subsection{A quantitative rectangle consequence}
For a fixed finite $\eta$, let $Z_\eta$ be the selected switch set and
$z_\eta=|Z_\eta|$ on the projective circle (two or four). Write
$h=\max_j H_j$ over the four distinct corners and, away from switches,
$g(\theta)=\min_{j\ne j_\eta(\theta)}(h-H_j)$. Define
\begin{equation}\label{eq:boxmargin}
c_\eta=\inf_{\theta\notin Z_\eta}
\frac{g(\theta)}{\dist(\theta,Z_\eta)}>0.
\end{equation}
This positivity is explicit from the sector proof: at each non-axis switch,
the sign-changing coefficient is a nonzero multiple of
$\eta\tan\theta-b_+$, $a_+\tan\theta+\eta$, or its displayed negative-$\eta$
analogue. It has a simple zero, while the other two corners have a positive
gap. At an axis the exchanged corner has a nonzero linear slope, and for
$\eta=0$ the two selected opposite corners have strictly distinct ratios
$a/b$; all other gaps there also grow linearly. The finite one-sided
limits of the ratio in \eqref{eq:boxmargin} are positive. Compactness away
from switches proves the stated positive infimum.

\begin{proposition}[Rectangle structural stability]\label{prop:boxstability}
For strict intervals and fixed $S$ with $F_\Box(S)>0$, the distance in
mean $|a-a_*|+|b-b_*|$ to the union of winning corner-program translates
is bounded by $C_\Box\sqrt{F_\Box-\log\rho(M)}$, with a finite constant
computable from the arc formulas, the margins \eqref{eq:boxmargin}, and
the finite winning-branch gaps. The finite-cycle quantum certificate of
Theorem~\ref{thm:quantumstability} holds with this distance and constant.
\end{proposition}
\begin{proof}
Put $v_0=\min(a_-,b_-)$, $v_1=\max(a_+,b_+)$ and
$d_\Box=(a_+-a_-)+(b_+-b_-)$ in the fixed canonical units.
Represent the actual point by convex weights $p_j$ of corners. Fractional
convexity gives weights $q_j=p_jv_j/v_A$ and
$E=\sum_jq_j(h-H_j)$. If
$e=|a-a_*|+|b-b_*|$, then
$e\le d_\Box\sum_{j\ne *}p_j$ and
\[
E\ge\frac{v_0 c_\eta}{v_1d_\Box}\dist(\theta,Z_\eta)e
=:\kappa\dist(\theta,Z_\eta)e.
\]
Over $n$ turns, a distance-$t$ neighborhood of the switch set has length
at most $2nz_\eta t$. The layer-cake argument gives
$\int\dist(\theta,Z_\eta)e\ge Q^2/(4nz_\eta d_\Box)$, where
$Q=\int e$. Thus
$Q\le\sqrt{4nz_\eta d_\Box E_{\rm tot}/\kappa}$.
The clocks differ by at most $Q/v_0^2$; direct temporal error is at most
$Q/v_0$. Moving the $nz_\eta$ switches adds at most
$nz_\eta d_\Box Q/v_0^2$. Divide this bound by $S$ on winning branches.
The maximum distance $d_\Box$ and the minimum positive loser-branch gap
handle all other profiles exactly as in Theorem~\ref{thm:globalstability}.
Finally $\|A\|_2\le v_1$ gives $K_0\le\max(\Omega,\Omega^{-1})e^{v_1S}$
for the quantum certificate.
\end{proof}

\paragraph{Collapsed intervals and limitations.}
If only $a$ is fixed, a fixed canonical rescaling and the time change
$\sqrt{a b_-}\,t$ reduce the problem to the original scalar oscillator
with $R^2=b_+/b_-$. If only $b$ is fixed, exchange the canonical coordinates
by a fixed symplectic rotation first and use $R^2=a_+/a_-$.
Adjacent identical sectors then merge. If both intervals collapse, all
coefficients are constant and growth is zero. Thus the original scalar
bounds-only theorem is also a literal face of this rectangle theorem.
The universal square-root exponent cannot be improved across this class
because it contains the scalar sharpness example. No sharper strict-rectangle
exponent is asserted separately.