The massive free-particle family has no strong zero-mass limit on Schwartz states

Given: $\mathrm{\hslash }>0$; the spin-zero one-particle sector of P5, with $V=0$; for $m>0$ the operator $K_m=-\frac{{\mathrm{\hslash }}^2}{2m}\mathrm{\Delta }$ with domain $W^{2,2}({\mathbb{R}}^3;\mathbb{C})$ (Kinematics and massive-particle model postulate P5: Newtonian frames and one-particle operators clause (iv), Integer-order Sobolev spaces and their norms); the free momentum operators $P_j$ of Position and momentum operators on Euclidean L2; and the explicit normalized state ${\psi }_{L_0}$ of [L3].

The free-plus-bounded-potential theorem with $V=0$ and $c=2{\pi }^2{\mathrm{\hslash }}^2\mathrm{/}m$ gives: $-\frac{{\mathrm{\hslash }}^2}{2m}\mathrm{\Delta }$ is self-adjoint on $W^{2,2}({\mathbb{R}}^3;\mathbb{C})$, and for every Schwartz function $g$ one has $-\frac{{\mathrm{\hslash }}^2}{2m}\mathrm{\Delta }g=-\frac{{\mathrm{\hslash }}^2}{2m}(\mathrm{\Delta }g)$ evaluating the Laplacian classically (Real multiplication operators and free plus bounded-potential Schrödinger operators clauses (b) and (c), Schwartz space and its seminorms).

Momentum operators: $P_j=-i\mathrm{\hslash }{D}_j$ is self-adjoint with domain $D(P_j)=\{u:D_{j}u\in L^2\}$, and $-\frac{{\mathrm{\hslash }}^2}{2m}\mathrm{\Delta }=\frac{1}{2m}{\sum }_j{P}_j^2$ as operators on the classes $\mathcal{S}\subseteq W^{2,2}$; for $u\in D(P_j)$ one has $⟨P_{j}u,u⟩=⟨u,P_{j}u⟩$ and $\mathrm{\parallel }{P}_{j}u{\mathrm{\parallel }}^2=⟨P_{j}u,P_{j}u⟩$ (Self-adjointness, Fourier representation and translation identities of position and momentum clause (b), Position and momentum operators on Euclidean L2).

An explicit length-scaled test state is obtained by putting ${\eta }_{L_0}(x)=\exp \text{\hspace{0.17em}⁣}(-1\mathrm{/}(1-\mathrm{\mid }x{\mathrm{\mid }}^2\mathrm{/}{L}_0^2))$ for $\mathrm{\mid }x\mathrm{\mid }<L_0$ and ${\eta }_{L_0}(x)=0$ for $\mathrm{\mid }x\mathrm{\mid }\ge L_0$, where $L_0>0$ has dimension $\mathrm{\ell }$. Every interior derivative is a finite sum of a polynomial in $(1-\mathrm{\mid }x{\mathrm{\mid }}^2\mathrm{/}{L}_0^2{)}^{-1}$ and the bounded coordinates times this exponential; the exponential beats every polynomial at the boundary, so all derivatives extend by zero and ${\eta }_{L_0}\in C_c^{\mathrm{\infty }}$. For $g(s)=e^{-1\mathrm{/}(1-s)}$, $g^{\mathrm{\prime }}(s)=-e^{-1\mathrm{/}(1-s)}(1-s{)}^{-2}\mathrm{\ne }0$ for $0\le s<1$, so $\mathrm{\nabla }{\eta }_{L_0}(x)=2x{L}_0^{-2}{g}^{\mathrm{\prime }}(\mathrm{\mid }x{\mathrm{\mid }}^2\mathrm{/}{L}_0^2)$ is nonzero for $0<\mathrm{\mid }x\mathrm{\mid }<L_0$. With $J:={\int }_{\mathrm{\mid }y\mathrm{\mid }<1}\exp \text{\hspace{0.17em}⁣}(-2\mathrm{/}(1-\mathrm{\mid }y{\mathrm{\mid }}^2))d^{3}y\in (0,\mathrm{\infty })$, set $N:=J^{-1\mathrm{/}2}$ and ${\psi }_{L_0}(x):=N{L}_0^{-3\mathrm{/}2}{\eta }_{L_0}(x)$. Change of variables $x=L_{0}y$ gives $\mathrm{\parallel }{\psi }_{L_0}{\mathrm{\parallel }}_2=1$ and $[{\psi }_{L_0}]={\mathrm{\ell }}^{-3\mathrm{/}2}$; direct differentiation at $x=0$ gives $\mathrm{\Delta }{\psi }_{L_0}(0)=-6N{e}^{-1}{L}_0^{-7\mathrm{/}2}\mathrm{\ne }0$.

Frequencies of the free model: for $k\in {\mathbb{R}}^3$, $\omega \in \mathbb{R}$, and $[k]={\mathrm{\ell }}^{-1}$, the smooth formal plane-wave mode ${\psi }_{k,\omega }(x,t)=e^{i(k\cdot x-\omega t)}$ satisfies $i\mathrm{\hslash }{\mathrm{\partial }}_t{\psi }_{k,\omega }=-\frac{{\mathrm{\hslash }}^2}{2m}\mathrm{\Delta }{\psi }_{k,\omega }$ at every point exactly when $\mathrm{\hslash }\omega =\frac{{\mathrm{\hslash }}^2\mathrm{\mid }k{\mathrm{\mid }}^2}{2m}$, that is $\omega =\frac{\mathrm{\hslash }\mathrm{\mid }k{\mathrm{\mid }}^2}{2m}$, because ${\mathrm{\partial }}_t{\psi }_{k,\omega }=-i\omega {\psi }_{k,\omega }$ and $\mathrm{\Delta }{\psi }_{k,\omega }=-\mathrm{\mid }k{\mathrm{\mid }}^2{\psi }_{k,\omega }$ and the exponential does not vanish (Sums, scalar multiples, products and quotients: $(f+g{)}^{\mathrm{\prime }}(c)=f^{\mathrm{\prime }}(c)+g^{\mathrm{\prime }}(c)$, $(\alpha f{)}^{\mathrm{\prime }}(c)=\alpha f^{\mathrm{\prime }}(c)$, $(fg{)}^{\mathrm{\prime }}(c)=f^{\mathrm{\prime }}(c)g(c)+f(c)g^{\mathrm{\prime }}(c)$, and $(f\mathrm{/}g{)}^{\mathrm{\prime }}(c)=(f^{\mathrm{\prime }}(c)g(c)-f(c)g^{\mathrm{\prime }}(c))\mathrm{/}g(c{)}^2$ when $g(c)\mathrm{\ne }0$, Kinematics and massive-particle model postulate P5: Newtonian frames and one-particle operators clause (iv)); the frequency has dimension ${\tau }^{-1}$ in the conventions of Dimensions, SI unit conventions and quantity types of the framework. For $k\mathrm{\ne }0$ this plane wave is not in $L^2({\mathbb{R}}^3)$, so it is a formal mode rather than a normalized state.

For $\psi \in W^{2,2}({\mathbb{R}}^3;\mathbb{C})$, with ${\mathcal{F}}_2$ the unitary Plancherel transform, the free kinetic operator has Fourier multiplier $2{\pi }^2{\mathrm{\hslash }}^2\mathrm{\mid }\xi {\mathrm{\mid }}^2\mathrm{/}m$ and $P_j$ has multiplier $2\pi \mathrm{\hslash }{\xi }_j$; hence $⟨K_m\psi ,\psi ⟩=\frac{2{\pi }^2{\mathrm{\hslash }}^2}{m}\int \mathrm{\mid }\xi {\mathrm{\mid }}^2\mathrm{\mid }{\mathcal{F}}_2\psi (\xi ){\mathrm{\mid }}^{2}d\xi =\frac{1}{2m}{\sum }_j\mathrm{\parallel }{P}_j\psi {\mathrm{\parallel }}_2^2$ (Real multiplication operators and free plus bounded-potential Schrödinger operators clauses (b)-(c), Self-adjointness, Fourier representation and translation identities of position and momentum, Plancherel theorem).