The Gaussian position and momentum packet of the massive model

[Heat-kernel slices] Comparing exponents and normalization with [F1], $\mathrm{\Gamma }(x,\frac{{\sigma }^2}{2})=(2\pi {\sigma }^2{)}^{-3\mathrm{/}2}{e}^{-\mathrm{\mid }x{\mathrm{\mid }}^2\mathrm{/}(2{\sigma }^2)}=2^{-3\mathrm{/}2}(\pi {\sigma }^2{)}^{-3\mathrm{/}4}{\psi }_{\sigma }(x),$ since $4\cdot \frac{{\sigma }^2}{2}=2{\sigma }^2$; and $\mathrm{\Gamma }(x,\frac{{\sigma }^2}{4})=(\pi {\sigma }^2{)}^{-3\mathrm{/}2}{e}^{-\mathrm{\mid }x{\mathrm{\mid }}^2\mathrm{/}{\sigma }^2}=\mathrm{\mid }{\psi }_{\sigma }(x){\mathrm{\mid }}^2\mathrm{.}$

[Schwartz class, norm and domain] By step 1.1 and [F1] the function ${\psi }_{\sigma }$ is a constant multiple of the $C^{\mathrm{\infty }}$ kernel slice $\mathrm{\Gamma }(\cdot ,{\sigma }^2\mathrm{/}2)$, hence $C^{\mathrm{\infty }}$, and every derivative obeys $\mathrm{\mid }{D}^{\beta }{\psi }_{\sigma }(x)\mathrm{\mid }\le C_{\beta }^{\mathrm{\prime }}{e}^{-\mathrm{\mid }x{\mathrm{\mid }}^2\mathrm{/}(4{\sigma }^2)}$ [F1]. For every multi-index $\alpha$, the seminorm ${\sup }_x\mathrm{\mid }{x}^{\alpha }{D}^{\beta }{\psi }_{\sigma }(x)\mathrm{\mid }\le C_{\beta }^{\mathrm{\prime }}{\sup }_x(1+\mathrm{\mid }x\mathrm{\mid }{)}^{\mathrm{\mid }\alpha \mathrm{\mid }}{e}^{-\mathrm{\mid }x{\mathrm{\mid }}^2\mathrm{/}(4{\sigma }^2)}$ is finite: the function is continuous on ${\mathbb{R}}^3$ and tends to $0$ at infinity, because $(1+r{)}^m{e}^{-r^2\mathrm{/}(4{\sigma }^2)}\to 0$ as $r\to +\mathrm{\infty }$, which follows from the substitution $x=r^2\mathrm{/}(4{\sigma }^2)$ in [F3]. Hence ${\psi }_{\sigma }\in \mathcal{S}({\mathbb{R}}^3;\mathbb{C})$ [F4]. Its norm is $\mathrm{\parallel }{\psi }_{\sigma }{\mathrm{\parallel }}^2=\int \mathrm{\mid }{\psi }_{\sigma }{\mathrm{\mid }}^2=\int \mathrm{\Gamma }(x,{\sigma }^2\mathrm{/}4)dx=1$ by step 1.1 and the normalisation of [F1]. The components $x_j{\psi }_{\sigma }$ lie in $L^2$ because $\int x_j^2\mathrm{\mid }{\psi }_{\sigma }{\mathrm{\mid }}^{2}dx=\int x_j^2\mathrm{\Gamma }(x,{\sigma }^2\mathrm{/}4)dx={\sigma }^2\mathrm{/}2$ by [F1]; and $D_j{\psi }_{\sigma }=-{\sigma }^{-2}{x}_j{\psi }_{\sigma }$, which lies in $L^2$ by the same computation, is the weak derivative by [F4]. This proves (i).

[Position density and variance] By [F5] and step 1.1, ${\mu }_{{\psi }_{\sigma }}^{Q_j}(B)={\int }_{x_j^{-1}(B)}\mathrm{\Gamma }(x,{\sigma }^2\mathrm{/}4)dx$; writing $x=(x_{\perp },t)$ and iterating by [F3] gives the density $f(t)=e^{-t^2\mathrm{/}{\sigma }^2}(\pi {\sigma }^2{)}^{-3\mathrm{/}2}{\int }_{{\mathbb{R}}^2}{e}^{-\mathrm{\mid }{x}_{\perp }{\mathrm{\mid }}^2\mathrm{/}{\sigma }^2}d{x}_{\perp }=(\pi {\sigma }^2{)}^{-1\mathrm{/}2}{e}^{-t^2\mathrm{/}{\sigma }^2}$, the marginal of $\mathrm{\Gamma }(\cdot ,{\sigma }^2\mathrm{/}4)$. Its mean vanishes and its second moment is $\int t^{2}f(t)dt=\int t^2\mathrm{\Gamma }(t,{\sigma }^2\mathrm{/}4)dt={\sigma }^2\mathrm{/}2$ by [F1]; hence the variance is ${\sigma }^2\mathrm{/}2$.

[Momentum representative] By [F2] and the computation ${\psi }_{\sigma }=(\pi {\sigma }^2{)}^{-3\mathrm{/}4}{e}^{-\mathrm{\mid }x{\mathrm{\mid }}^2\mathrm{/}(2{\sigma }^2)}$, $U{\psi }_{\sigma }(p)=(\pi {\sigma }^2{)}^{-3\mathrm{/}4}(2\pi \mathrm{\hslash }{)}^{-3\mathrm{/}2}{\mathcal{F}}_2\left(e^{-\mathrm{\mid }x{\mathrm{\mid }}^2\mathrm{/}(2{\sigma }^2)}\right)\left(\frac{p}{2\pi \mathrm{\hslash }}\right);$ the Gaussian Fourier lemma with $t=(2\pi {\sigma }^2{)}^{-1}$ gives ${\mathcal{F}}_2\left(e^{-\mathrm{\mid }x{\mathrm{\mid }}^2\mathrm{/}(2{\sigma }^2)}\right)(\xi )=(2\pi {\sigma }^2{)}^{3\mathrm{/}2}{e}^{-2{\pi }^2{\sigma }^2\mathrm{\mid }\xi {\mathrm{\mid }}^2}$, and substituting $\xi =p\mathrm{/}(2\pi \mathrm{\hslash })$ yields $U{\psi }_{\sigma }(p)=(\pi {\sigma }^2{)}^{-3\mathrm{/}4}(2\pi \mathrm{\hslash }{)}^{-3\mathrm{/}2}(2\pi {\sigma }^2{)}^{3\mathrm{/}2}{e}^{-{\sigma }^2\mathrm{\mid }p{\mathrm{\mid }}^2\mathrm{/}(2{\mathrm{\hslash }}^2)}=(\pi (\mathrm{\hslash }\mathrm{/}\sigma {)}^2{)}^{-3\mathrm{/}4}{e}^{-\mathrm{\mid }p{\mathrm{\mid }}^2\mathrm{/}(2(\mathrm{\hslash }\mathrm{/}\sigma {)}^2)},$ which is (ii) with ${\sigma }_p=\mathrm{\hslash }\mathrm{/}\sigma$.

[Momentum slice] Comparing step 3.1 with [F1], $\mathrm{\mid }U{\psi }_{\sigma }(p){\mathrm{\mid }}^2=(\pi {\sigma }_p^2{)}^{-3\mathrm{/}2}{e}^{-\mathrm{\mid }p{\mathrm{\mid }}^2\mathrm{/}{\sigma }_p^2}=\mathrm{\Gamma }(p,\frac{{\sigma }_p^2}{4})\mathrm{.}$

[Momentum density and variance] By [F5] and step 4.1, ${\mu }_{{\psi }_{\sigma }}^{P_k}(B)={\int }_{\{p_k\in B\}}\mathrm{\Gamma }(p,{\sigma }_p^2\mathrm{/}4)dp$; the same marginal computation with width ${\sigma }_p$ gives density $(\pi {\sigma }_p^2{)}^{-1\mathrm{/}2}{e}^{-p^2\mathrm{/}{\sigma }_p^2}$, mean $0$ and variance ${\sigma }_p^2\mathrm{/}2={\mathrm{\hslash }}^2\mathrm{/}(2{\sigma }^2)$ by [F1]. The product of standard deviations is $\left(\sigma \mathrm{/}\sqrt{2}\right)(\mathrm{\hslash }\mathrm{/}(\sqrt{2}\sigma ))=\mathrm{\hslash }\mathrm{/}2$.

[Joint densities and scope] The joint position density is $\mathrm{\mid }{\psi }_{\sigma }{\mathrm{\mid }}^2=\mathrm{\Gamma }(\cdot ,{\sigma }^2\mathrm{/}4)$ and the momentum density is $\mathrm{\mid }U{\psi }_{\sigma }{\mathrm{\mid }}^2=\mathrm{\Gamma }(\cdot ,{\sigma }_p^2\mathrm{/}4)$ by steps 1.1 and 4.1, which is (iv). All computations used only the heat-kernel and Fourier suppliers and the sharp outcome laws of [F5]; no dynamics, detector, sampling or empirical input entered.