Gaussian preparation spread and an independent additive reading error

[Normalisation and moments of the Gaussian] Substituting $x=\sigma t$ in ${\rho }_{\sigma }$ gives $\int {\rho }_{\sigma }=(2\pi {\sigma }^2{)}^{-1\mathrm{/}2}\sigma \int e^{-t^2\mathrm{/}2}dt=1$ by [F1] with $a=1$. Hence $\mathrm{\parallel }{\psi }_{\sigma }\mathrm{\parallel }=1$. The mean $\int x{\rho }_{\sigma }(x)dx$ vanishes because the integrand is odd; the second moment is $\int x^2{\rho }_{\sigma }(x)dx=(2\pi {\sigma }^2{)}^{-1\mathrm{/}2}{\sigma }^3\int t^2{e}^{-t^2\mathrm{/}2}dt=(2\pi {\sigma }^2{)}^{-1\mathrm{/}2}{\sigma }^3\sqrt{2\pi }={\sigma }^2$ by [F1].

[The additive-error law] Let $\rho$ be a density as in (b). Since $\mathrm{\mid }y\mathrm{\mid }\le 1+y^2$, its finite second moment implies $\int \mathrm{\mid }y\mathrm{\mid }\rho (y)dy<\mathrm{\infty }$, as required for the first-moment identities in [F3]. The same nonnegative change of variables gives $\int x^2\rho (x\pm e)dx=\int (y\mp e{)}^2\rho (y)dy\le 2\int y^2\rho (y)dy+2e^2<\mathrm{\infty }$; thus all means and variances below are finite. By [F3], $\int {\rho }_e=\frac{1}{2}\int \rho +\frac{1}{2}\int \rho =1$, so ${\rho }_e$ is a probability density; likewise $m({\rho }_e)=\frac{1}{2}\int x\rho (x+e)dx+\frac{1}{2}\int x\rho (x-e)dx=\frac{1}{2}\int (y-e)\rho (y)dy+\frac{1}{2}\int (y+e)\rho (y)dy=m(\rho )$, and the cross terms cancel: $V({\rho }_e)=\frac{1}{2}\int (y-e-m{)}^2\rho (y)dy+\frac{1}{2}\int (y+e-m{)}^2\rho (y)dy=\int (y-m{)}^2\rho (y)dy+e^2=V(\rho )+e^2$. This is (b).

[Momentum moments] By [F2], $\int {\psi }_{\sigma }^{\mathrm{\prime }}\overline{{\psi }_{\sigma }}dx=-\frac{1}{2{\sigma }^2}\int x{\rho }_{\sigma }(x)dx=0$, so $b=⟨P{\psi }_{\sigma },{\psi }_{\sigma }⟩=0$; and $\mathrm{\parallel }D{\psi }_{\sigma }{\mathrm{\parallel }}^2=\frac{1}{4{\sigma }^4}\int x^2{\rho }_{\sigma }(x)dx=\frac{1}{4{\sigma }^2}$, so $\mathrm{\parallel }P{\psi }_{\sigma }{\mathrm{\parallel }}^2={\mathrm{\hslash }}^2\mathrm{/}(4{\sigma }^2)$, i.e. $\mathrm{\Delta }P=\mathrm{\hslash }\mathrm{/}(2\sigma )$ and $\mathrm{\Delta }Q=\sigma$ by step 1.1. These are the maximal-domain quantities of Position-momentum uncertainty on the maximal domains of the line and give $\mathrm{\Delta }Q\mathrm{\Delta }P=\mathrm{\hslash }\mathrm{/}2$: the equality case of Covariance-refined preparation uncertainty without operator products clause (c), since $x_Q$ and $y_P$ are then linearly dependent (both proportional to $x{\psi }_{\sigma }(x)$). This is (a).

[Gaussian case] By step 1.1, $V({\rho }_{\sigma })={\sigma }^2$, and step 1.2 gives $V(({\rho }_{\sigma }{)}_e)={\sigma }^2+e^2$. The mixture is centered, and translation in each component gives its fourth central moment ${\int }_{\mathbb{R}}{x}^4({\rho }_{\sigma }{)}_e(x)dx=\frac{1}{2}{\int }_{\mathbb{R}}(y-e{)}^4{\rho }_{\sigma }(y)dy+\frac{1}{2}{\int }_{\mathbb{R}}(y+e{)}^4{\rho }_{\sigma }(y)dy=3{\sigma }^4+6{\sigma }^2{e}^2+e^4,$ using evenness, $\int y^2{\rho }_{\sigma }={\sigma }^2$, and $\int y^4{\rho }_{\sigma }=3{\sigma }^4$ from [F1]. Any $g_{\mu ,\tau }$ has mean $\mu$, variance ${\tau }^2$, and fourth central moment $3{\tau }^4$ by the same Gaussian moments and translation. Thus if $e>0$ and $({\rho }_{\sigma }{)}_e$ were a Gaussian density, its mean and variance would force $\mu =0$ and ${\tau }^2={\sigma }^2+e^2$, whose Gaussian fourth moment $3({\sigma }^2+e^2{)}^2$ differs from the displayed mixture fourth moment by $2e^4>0$, a contradiction. At $e=0$ the mixture is ${\rho }_{\sigma }$. Hence the model variance does not determine $\sigma$ when $e$ is unknown, although it implies the stated upper bound; this proves (c).