\section{A scalar-average illustration} \label{app:scalarillustration} We begin with a maximally mixed average state, for which weak commutativity holds automatically. This simpler model introduces the comparison matrix used to solve the nonscalar model in the main text. Let the only nonzero entries of a weighted ladder \(C\) be \begin{equation} C_{12}=1,\qquad C_{23}=2,\qquad C_{34}=3. \label{eq:ladder} \end{equation} Its Hermitian components \(X=(C+C^\dagger)/2\) and \(Y=(C-C^\dagger)/(2i)\) are finite-level quadratures, not canonical oscillator position and momentum. Put \begin{equation} \begin{split} U&=\operatorname{diag}(-1,-1,-1,1),\\ K_\pm&=(\1_4\pm U)/8,\qquad B=CK_++K_-C,\\ B_1&=(B+B^\dagger)/2,\qquad B_2=(B-B^\dagger)/(2i). \end{split} \label{eq:shiftmoments} \end{equation} Thus \(B\) has superdiagonal entries \(1/4,1/2,3/2\). For \(s=\pm1\) and \(j=1,2\), prepare one copy of \begin{equation} p_{s,j}=\tfrac14,\qquad \theta_{s,j}=10s e_j,\qquad \rho_{s,j}=\1_4/4+s B_j/5. \label{eq:shiftprior} \end{equation} The classical label is not given to the receiver. Both \(B_j\) are traceless and Hermitian; diagonal dominance gives \(\rho_{s,j}\succeq\1_4/20\). The average and first moments are exactly \(S=\1_4/4,B_j\), with \(\lambda=100\). The unique posterior means are \(S_j=4B_j\). Weak commutativity follows from \(\Tr[S_1,S_2]=0\), although the operators themselves need not commute. Indeed, a scalar average state makes this trace test automatic: the obstruction concerns what that test fails to detect. \begin{theorem}[Exact Bayesian risk and strict gap] For the ensemble \eqref{eq:shiftprior} and the loss \eqref{eq:risk}, \begin{equation} \mathcal L_{\mathrm{NH}}=\frac{377}{4},\qquad \mathcal L_{\min}=100-4f(u_*,v_*)=\frac{377}{4}+\Gamma_*. \label{eq:gap} \end{equation} The algebraic pair and rational function \(f\) are defined below and in Appendix~\ref{app:optimal}. The exact value satisfies \(0.03592634498992<\Gamma_*<0.03592634498993\). An explicit eight-effect POVM attains the minimum. The matching lower bound holds for every POVM and measurable estimator, with no bound on outcome number or estimate range. \end{theorem} The terminating decimals bound an algebraic value; they are not its exact specification. For comparison, the Fourier projectors onto \((1,w,w^2,w^3)^T/2\), \(w\in\{1,i,-1,-i\}\), with reports \(9w/4\), have risk \(1519/16\) and are therefore not optimal. \subsection{Minimizing the NH expression} For a Hermitian contraction \(-\1\preceq R\preceq\1\), define \(K_\pm(R)=S^{1/2}(\1\pm R)S^{1/2}/2\) and \[ q_R(Z)=\Tr[ZK_+(R)Z^\dagger+Z^\dagger K_-(R)Z]. \] The quadratic plus trace norm in \eqref{eq:nh} is \(\sup_R q_R(Z)\), where \(Z=V_1+iV_2\). For \eqref{eq:ladder}, \[ iS^{1/2}[X,Y]S^{1/2} =\operatorname{diag}(-1,-3,-5,9)/8, \] so \(U\) is the active sign and \eqref{eq:shiftmoments} is its stationarity condition. For every perturbation \(E\), the active branch minus the linear term changes by \(q_U(E)\ge0\). The supremum is no smaller than that branch and agrees with it at \(C\). This proves a \emph{global} NH minimum, without uniqueness. In particular, \begin{equation} T_0=CK_+C^\dagger+C^\dagger K_-C =\operatorname{diag}(0,1/4,13/4,9/4), \label{eq:t0} \end{equation} and \(\mathcal L_{\mathrm{NH}}=\lambda-\Tr T_0=377/4\). \subsection{A lower bound for every measurement} Group the outcomes of \(P\) by their reported estimates \(a\). This gives a POVM \(M\) on the complex estimate plane \(z=a_1+ia_2\). The positive comparison matrix is \begin{equation} F(z)=|z|^2S-\bar zB-zB^\dagger+T_0, \label{eq:globalf} \end{equation} and direct risk expansion yields \begin{equation} \mathcal L(P,a)-\mathcal L_{\mathrm{NH}} =\int\Tr[M(\dd z)F(z)]. \label{eq:globaldual} \end{equation} There is no requirement that the measurement's first moments equal \(X,Y\). Finite risk suffices for integrability; infinite-risk procedures obey the bound trivially. Operator-dual formulations go back to Ref.~\cite{Holevo1973} and are used for the Bayesian NH bound in Ref.~\cite{Albarelli2026}, alongside distinct conic formulations of local bounds \cite{Hayashi2023}. A simple rational witness is the real matrix \begin{equation} G=\begin{pmatrix} 0&-1/16&0&2/47\\ -3/16&0&1/94&0\\ 0&-24/47&0&-3/4\\ -2/47&0&-3/4&0 \end{pmatrix}. \label{eq:g} \end{equation} It satisfies \(G+G^T=-(B+B^\dagger)\), and \(GG^T\) is diagonal. Consequently \begin{equation} \begin{aligned} A&=T_0-4GG^T\\ &=\frac1{141376}\operatorname{diag}(-3233,15399,-6080,-1024),\\ \Tr A&=\gamma=\frac{2531}{70688}. \end{aligned} \label{eq:a} \end{equation} For \(z=rw\), \(r\ge0\), \(|w|=1\), put \(D(w)=\operatorname{diag}(1,w,w^2,w^3)\) and \(Q(r)=r\1_4/2+2G\). The following identity holds over the entire estimate plane: \begin{equation} F(rw)-A=D(w)Q(r)Q(r)^TD(w)^\dagger\succeq0. \label{eq:factor} \end{equation} At the origin take \(r=0,w=1\). This is a polynomial coefficient identity, not a sampled positivity test. Although \(A\) is indefinite, POVM normalization gives \[ \int\Tr[M(\dd z)A]=\Tr A. \] Integrating \eqref{eq:factor} in \eqref{eq:globaldual} proves the simple rational lower bound \(\mathcal L_{\min}-\mathcal L_{\mathrm{NH}}\ge\gamma\). The number \(\gamma\) is a certified lower bound, not the exact excess risk. Its scale can be compared with the NH-allowed gain \(\lambda-\mathcal L_{\mathrm{NH}}=23/4\), giving a missed fraction of at least \(\gamma/(23/4)\). This is a relaxation error, not a sensing advantage or an incompatibility measure invariant under arbitrary changes of loss. Parameter-independent ancillas, classical randomization, and sequential instruments on the same received copy induce effective POVMs and are included. Every abort outcome must remain part of the normalized experiment; either conditional postselection or the use of extra parameter-encoded copies changes the task. \subsection{A matching optimal measurement} The rational comparison can be improved until it is met by a measurement. Write \((w_0,m_0,t_0)=(1/4,1/2,3/2)\), and define \begin{equation} \begin{split} d&=v(t_0+v)-u(w_0+u),\\ a&=u m_0(t_0+v)/d,\qquad b=u m_0(w_0+u)/d,\\ G(u,v)&=\begin{pmatrix} 0&u&0&a\\-w_0-u&0&b&0\\ 0&-m_0-b&0&v\\-a&0&-t_0-v&0 \end{pmatrix}. \end{split} \label{eq:gramfamily} \end{equation} Put \(f(u,v)=\|G(u,v)\|_F^2\). The selected pair \((u_*,v_*)\) is the unique stationary point of \(f\) in the rational rectangle \[ \begin{gathered} -.06554650