Proposition: Window integration certificate

If $V(z)\in [V_{-},V_{+}]$, $V_{-}>0$, and all $\alpha (z)$ lie within ${\theta }_W<\pi /2$ of one common phase, then the coherence after forgetting $z$ satisfies

For $H_W>0$,

If parity gives $P_{+}(z)=P_{-}(z)$ almost everywhere, then $H_W=1$.

Proof. Integrate the positive amplitude $\sqrt{P_{+}{P}_{-}}V$. Its projection along the common phase is at least the same integral times $\cos {\theta }_W$; its modulus is at most the integral. Use the visibility bounds and divide by the integrated populations. Parity makes the numerator defining $H_W$ equal either population.