\subsection{An explicit gap for every measurement}
\label{app:familyquantitative}

The population mismatch also bounds how closely a measurement can
approach NH. Write \(a=\ell_1\), \(b=\ell_2\), and, for an arbitrary
complex report \(w\) and vector \(\psi\), put
\[
\begin{aligned}
 e_1&=w\psi_1-a\psi_2,&e_2&=w\psi_2-b\psi_3,\\
 e_3&=w\psi_3-\psi_4,&e_4&=\bar w\psi_4-\psi_3.
\end{aligned}
\]
Let \(E^2=\sum_i|e_i|^2\), \(R=\psi^\dagger F(w)\psi\), and
\(d=|\psi_2|^2-b^2|\psi_3|^2\). Equation~\eqref{eq:familyallcomplex}
gives \(R\ge pE^2\). We shall prove the one-sided estimate
\begin{equation}
 d\le\sqrt{19}\,\|\psi\|E
 \le\sqrt{19/p}\,\|\psi\|\sqrt R .
 \label{eq:familydefectbound}
\end{equation}
No bound on \(|d|\) is required.

Set \(\rho=|w|^2\) and \(\rho_0=7/25\). The residuals obey
\[
 \bar w e_3+e_4=(\rho-1)\psi_3,\qquad
 a|\psi_2|^2=\Re[(w\psi_1-e_1)\bar\psi_2].
\]
For \(\rho\le\rho_0\), first take absolute values in the second identity:
\(a|\psi_2|^2\le|e_1||\psi_2|+|\psi_1||w\psi_2|\).
Then use \(|w\psi_2|\le b|\psi_3|+|e_2|\) and the first identity to obtain
\[
 d\le\frac1a\left[|\psi_2||e_1|+
 |\psi_1|\left(|e_2|+
 \frac{b\sqrt\rho|e_3|+b|e_4|}{1-\rho}\right)\right].
\]
The corresponding two-row feature matrix has disjoint row supports.
Its squared operator norm is
\[
 \frac{1+b^2(1+\rho)/(1-\rho)^2}{a^2}
 \le9+\frac{4(1+\rho_0)}{(1-\rho_0)^2}
 =\frac{1529}{81}<19.
\]
Here \(a^{-2}\le9\), \(b^2/a^2\le4\), and
\((1+\rho)/(1-\rho)^2\) increases on \([0,\rho_0]\).

For \(\rho\ge\rho_0\), use instead the exact identity
\[
 \rho d=-b^2\Re[(\bar w e_3+e_4)\bar\psi_3]
       +\Re[e_2(b\bar\psi_3+\bar w\bar\psi_2)].
\]
It implies
\[
 d\le (|\psi_2|,|\psi_3|)
 \begin{pmatrix}
 \rho^{-1/2}&0&0\\
 b/\rho&b^2/\sqrt\rho&b^2/\rho
 \end{pmatrix}
 \begin{pmatrix}|e_2|\\|e_3|\\|e_4|\end{pmatrix}.
\]
Call this nonnegative matrix \(K(\rho,b)\). Its entries decrease
with \(\rho\) and increase with \(b\), so its operator norm is at
most that of \(K_0=K(7/25,\sqrt{2/3})\). This comparison follows by
applying the matrices to the componentwise absolute value of a vector;
entrywise order is not being used as a positive-semidefinite order.
The leading principal minor and determinant of
\(19I_2-K_0K_0^T\) are, respectively,
\[
 \frac{108}{7}>0,\qquad \frac{20194}{1029}>0.
\]
Thus \(\|K_0\|<\sqrt{19}\), completing
Eq.~\eqref{eq:familydefectbound} on the entire complex plane.

Put \(Q_b=\operatorname{diag}(0,1,-b^2,0)\). Spectral decomposition
and Cauchy--Schwarz extend the one-sided bound to every positive matrix
\(A\):
\[
 \Tr(Q_bA)\le\sqrt{19/p}
       \sqrt{\Tr A\,\Tr[F(w)A]}.
\]
For any normalized POVM \(M\), rescale reports by \(z=gw\) and
use its density with respect to \(\mu=\Tr M\). Normalization gives
\(\int\Tr(Q_bA)\dd\mu=1-b^2>0\) and \(\mu(\C)=4\).
Integrating first and only then squaring yields
\[
 (1-b^2)^2\le\frac{76}{p}
          \int\Tr[F(w)M(\dd w)].
\]
Since \(F_g(z)=g^2F(z/g)\) and \(g=\eta/2\), every measurement obeys
\begin{equation}
 \mathcal L-\mathcal L_{\mathrm{NH}}
 \ge\frac{p(1-b^2)^2}{304}\eta^2
 \ge\frac{\eta^2}{16416}.
 \label{eq:familyexplicitgap}
\end{equation}
The argument applies to unbounded finite-risk reports and continuous or
singular POVMs; infinite-risk procedures satisfy the bound trivially.
It does not require phase covariance or finite support.

There is also a contrast-independent relative statement. With
\(T=\Tr T_0=1-2p+p(a^2+b^2)\), positivity is immediate from
\(T=p(a^2+b^2)+q+t\). Also \(a^2+b^2=1-4u/3\le1\), so
\(T\le1-p\le5/6\).
Since \(1-\mathcal L_{\mathrm{NH}}=g^2T\),
\begin{equation}
 \frac{\mathcal L_{\min}-\mathcal L_{\mathrm{NH}}}
      {1-\mathcal L_{\mathrm{NH}}}
 \ge\frac{p(1-b^2)^2}{76T}\ge\frac1{3420}.
 \label{eq:familyrelativegap}
\end{equation}
Thus a fixed fraction of the risk reduction predicted by NH is
unattainable throughout this family. The coefficients are conservative
lower bounds, not optimal constants or exact minimum risks.
The unscaled gap vanishes at \(\eta=0\), which is excluded.

The ancillary scripts \texttt{verify\_gap.py} and
\texttt{independent\_gap.py} check the residual identities and rational
endpoint inequalities by symbolic algebra and, independently, sparse
Gaussian-rational polynomial arithmetic. A wrong conjugation is
rejected; replacing 19 by 18 fails the small-radius endpoint test.