Theorem: Calibrated quadratic obstruction

Suppose the two hypotheses use the same interval $I$, all known dynamics and apparatus specifications match, and

at every relevant stage, setting, and history, with one constant $c\ge 0$. Put $a_G=\lambda A_{r_{\mathrm{c}}}c$.

The coefficient intervals $I$ and $I+a_G$ overlap exactly when

Whenever (eq:obstruction) holds, at least one null–alternative pair has identical complete attempted-run laws, including preparation, heralding, finite sensing, and failures. For that pair, every test satisfies

Consequently no test has both errors uniformly below $1/2$ over the entire two families.

This is family overlap, not containment of every alternative member in the null. When $a_G>\Delta b$, the coefficient intervals are disjoint, but observable separation does not follow without a detector argument.

Proof. Matching requires $b_0=b_1+a_G$. Such values in $I$ exist precisely when $[b_{-},b_{+}]\cap [b_{-}+a_G,b_{+}+a_G]\ne \varnothing$, which for $a_G\ge 0$ is (eq:obstruction). One explicit pair is $b_1=b_{-}$, $b_0=b_{-}+a_G$.

For this pair the total generators in (eq:scalar0)–(eq:scalar1) agree. Uniqueness gives identical propagation. The same apparatus coupling and instrument give identical unnormalized outcome states. Induct over stages and accessible histories to obtain identical complete laws.

For a randomized test $\phi \in [0,1]$ declaring alternative,

The remaining qualifications follow from interval overlap alone.