Qm Massive Particle Kinematics — Examples

[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.

[Well-definedness and continuity] For $\phi \in \mathcal{S}$, [L3] gives $\mathrm{\mid }{e}^{ik\cdot x}\mathrm{\mid }=1$, and [L5] plus ${\prod }_j(1+x_j^2)={\sum }_{\alpha \in \{0,2{\}}^d}{x}^{\alpha }$ gives ${\int }_{{\mathbb{R}}^d}\mathrm{\mid }\phi (x)\mathrm{\mid }dx\le C_d{\sum }_{\alpha \in \{0,2{\}}^d}{p}_{\alpha ,0}(\phi )<\mathrm{\infty }\mathrm{.}$ Thus the defining integral is absolutely convergent and has the bound in (a). Since this bound is by a finite sum of defining Schwartz seminorms, $u_k$ is continuous in the Schwartz topology. Conjugate-linearity is immediate from the definition, and $u_k(\phi )=\int \overline{\phi (x)}{e}^{-2\pi ix\cdot (-k\mathrm{/}2\pi )}dx={\mathcal{F}}_2(\bar{\phi })(-k\mathrm{/}(2\pi ))$ by [L1]. This proves (a).

[Eigenrelation] Let $\phi \in \mathcal{S}$ and put $\psi :=\bar{\phi }$, a Schwartz class. Since $\overline{P_j\phi }=\overline{-i\mathrm{\hslash }{\mathrm{\partial }}_j\phi }=i\mathrm{\hslash }{\mathrm{\partial }}_j\bar{\phi }=i\mathrm{\hslash }{\mathrm{\partial }}_j\psi$, one has $u_k(P_j\phi )=\int e^{ik\cdot x}\overline{P_j\phi }dx=i\mathrm{\hslash }\int e^{ik\cdot x}{\mathrm{\partial }}_j\psi (x)dx=i\mathrm{\hslash }{\mathcal{F}}_2({\mathrm{\partial }}_j\psi )(-\frac{k}{2\pi }),$ where the last step evaluates the $L^1$ transform at $-k\mathrm{/}(2\pi )$. By [L2], ${\mathcal{F}}_2({\mathrm{\partial }}_j\psi )(\xi )=2\pi i{\xi }_j{\mathcal{F}}_2\psi (\xi )$ for every $\xi$, so at $\xi =-k\mathrm{/}(2\pi )$, $u_k(P_j\phi )=i\mathrm{\hslash }\cdot 2\pi i(-\frac{k_j}{2\pi }){\mathcal{F}}_2\psi (-\frac{k}{2\pi })=\mathrm{\hslash }{k}_j{u}_k(\phi ),$ using step 1.1. Applying the same identity to $P_j\phi$ (again a Schwartz class) gives $u_k(P_j^2\phi )=\mathrm{\hslash }{k}_j{u}_k(P_j\phi )=(\mathrm{\hslash }{k}_j{)}^2{u}_k(\phi )$. This proves (b).

[No $L^2$ representative] Suppose $F\in L^2$ satisfies $\int F\bar{\phi }=u_k(\phi )$ for all $\phi \in \mathcal{S}$, and put $G:=e^{-ik\cdot x}F\in L^2$, a class of the same norm because the multiplier has modulus one [L3]. Multiplication by $e^{ik\cdot x}$ preserves the Schwartz class, since its derivatives are constant multiples of itself and Leibniz's rule expresses every derivative of the product as a finite sum of a polynomial times a derivative of $\phi$. Replacing $\phi$ by $e^{ik\cdot x}\phi$ in the assumed identity gives $\int G\bar{\phi }=\int F\overline{e^{ik\cdot x}\phi }=u_k\left(e^{ik\cdot x}\phi \right)=\int \bar{\phi },$ so $\int (G-1)\bar{\phi }=0$ for every $\phi \in \mathcal{S}$. For each $n\ge 1$ let ${\mathrm{\Omega }}_n:=(-n,n{)}^d$; the restriction $(G-1){\mathrm{\mid }}_{{\mathrm{\Omega }}_n}$ belongs to $L^2({\mathrm{\Omega }}_n)$ because $G\in L^2({\mathbb{R}}^d)$ and $1\in L^2({\mathrm{\Omega }}_n)$. Its pairing with every $v\in C_c^{\mathrm{\infty }}({\mathrm{\Omega }}_n)$ is zero, so [L4] gives $G=1$ almost everywhere on ${\mathrm{\Omega }}_n$. Since ${\bigcup }_{n\ge 1}{\mathrm{\Omega }}_n={\mathbb{R}}^d$, it follows that $G=1$ almost everywhere on ${\mathbb{R}}^d$. But then ${\int }_{{\mathbb{R}}^d}\mathrm{\mid }G{\mathrm{\mid }}^2={\int }_{{\mathbb{R}}^d}1=+\mathrm{\infty }$ by monotone convergence on expanding boxes [L4], contradicting $G\in L^2$. Hence no such $F$ exists. For $z\mathrm{\ne }0$, $\mathrm{\mid }z{e}^{ik\cdot x}\mathrm{\mid }=\mathrm{\mid }z\mathrm{\mid }$ on a set of infinite Lebesgue measure, so no nonzero scalar multiple of the plane wave belongs to $L^2$ either. This proves (c).

[Conclusion] Statement (d) records that only Schwartz classes and the functional $u_k$ were used; no normalizable class and no probability law is attached to $u_k$ by this example.

[The hypothetical operators are nonzero] Suppose $A,B\in \mathcal{B}(H)$ satisfy $AB-BA=i\mathrm{\hslash }I$. Since $H\mathrm{\ne }\{0\}$, $I\mathrm{\ne }0$ and $\mathrm{\parallel }I\mathrm{\parallel }=1$ by [F1]. Because $\mathrm{\hslash }>0$, the commutator is nonzero; hence neither $A$ nor $B$ is zero.

[Commutators with powers] For every integer $n\ge 1$, [step 1.1, algebra] $[A,B^n]=ni\mathrm{\hslash }{B}^{n-1}\mathrm{.}$ The case $n=1$ is the assumed relation. If it holds for $n$, then $[A,B^{n+1}]=[A,B^n]B+B^n[A,B]=(n+1)i\mathrm{\hslash }{B}^n$, proving it by induction. No power $B^n$ can be zero: if one were zero, this identity would force $B^{n-1}=0$, and iteration would give $I=B^0=0$, contradicting [F1].

[Factorial lower bound] By step 2.1 and [F1]-[F2], for every $n\ge 1$, [step 1.1, step 2.1, F1, F2, algebra] $n\mathrm{\hslash }\mathrm{\parallel }{B}^{n-1}\mathrm{\parallel }=\mathrm{\parallel }[A,B^n]\mathrm{\parallel }\le \mathrm{\parallel }A{B}^n\mathrm{\parallel }+\mathrm{\parallel }{B}^{n}A\mathrm{\parallel }\le 2\mathrm{\parallel }A\mathrm{\parallel }\mathrm{\parallel }{B}^n\mathrm{\parallel }\mathrm{.}$ Put $c:=\mathrm{\hslash }\mathrm{/}(2\mathrm{\parallel }A\mathrm{\parallel })>0$. Iterating the resulting inequalities $\mathrm{\parallel }{B}^n\mathrm{\parallel }\ge nc\mathrm{\parallel }{B}^{n-1}\mathrm{\parallel }$ from $\mathrm{\parallel }{B}^0\mathrm{\parallel }=\mathrm{\parallel }I\mathrm{\parallel }=1$ gives $\mathrm{\parallel }{B}^n\mathrm{\parallel }\ge c^{n}n!$ for every $n\ge 1$.

[Contradiction] By [F1], $\mathrm{\parallel }{B}^n\mathrm{\parallel }\le \mathrm{\parallel }B{\mathrm{\parallel }}^n$. Let $K:=\mathrm{\parallel }B\mathrm{\parallel }\mathrm{/}c$ and choose an integer $M>K$. For $n\ge M$, $n!\ge M!M^{n-M}$, so [step 3.1, F1, algebra] $\frac{c^{n}n!}{\mathrm{\parallel }B{\mathrm{\parallel }}^n}\ge \frac{M!}{M^M}{\left(\frac{M}{K}\right)}^n,$ which tends to $+\mathrm{\infty }$ because $M\mathrm{/}K>1$. This contradicts the simultaneous lower and upper bounds on $\mathrm{\parallel }{B}^n\mathrm{\parallel }$. Thus no bounded pair $A,B$ satisfies the commutator relation, and the displayed existential claim is refuted.

[Density and symmetry] $D(T)$ contains $C_c^{\mathrm{\infty }}(0,L)$, which is dense in $L^2$ [F1]; hence $T$ is densely defined, and it is linear on the linear subspace $D(T)$ [F1]. For $u,v\in D(T)$, the integration-by-parts identity [F1] gives $⟨Tu,v⟩-⟨u,Tv⟩=-i\mathrm{\hslash }(u^{\ast }(L)\overline{v^{\ast }(L)}-u^{\ast }(0)\overline{v^{\ast }(0)})=0,$ because both traces of both functions vanish. Hence $T$ is symmetric.

[The adjoint] Let $v\in H^1(0,L)$. For every $u\in D(T)$, [F1] gives $⟨Tu,v⟩=-i\mathrm{\hslash }{\int }_0^{L}Du\bar{v}=-i\mathrm{\hslash }[u^{\ast }\overline{v^{\ast }}{]}_0^L+{\int }_0^{L}u\overline{-i\mathrm{\hslash }Dv}=⟨u,-i\mathrm{\hslash }Dv⟩$, since $u^{\ast }$ vanishes at both endpoints; so $u\mapsto ⟨Tu,v⟩$ is the restriction of a bounded functional and $v\in D(T^{\ast })$ with $T^{\ast }v=-i\mathrm{\hslash }Dv$ [F2]. Conversely, if $v\in D(T^{\ast })$, testing the identity $⟨Tu,v⟩=⟨u,T^{\ast }v⟩$ against $u\in C_c^{\mathrm{\infty }}(0,L)\subseteq D(T)$ shows that the distribution $-i\mathrm{\hslash }Dv$ equals the $L^2$ class $T^{\ast }v$; hence $v$ has an $L^2$ weak derivative, that is $v\in H^1(0,L)$, and $T^{\ast }v=-i\mathrm{\hslash }Dv$. Thus $D(T^{\ast })=H^1(0,L)$. The constant class $1$ lies in $H^1(0,L)$ with $1^{\ast }(0)=1\mathrm{\ne }0$, so $D(T^{\ast })\mathrm{\ne }D(T)$ and $T$ is not self-adjoint. This proves (a).

[Dirichlet functions are not momentum eigenvectors] Fix $n\ge 1$ and put $u_n(x):=\sqrt{2\mathrm{/}L}\sin (n\pi x\mathrm{/}L)$, a unit class in $H^1(0,L)$ with $u_n^{\ast }(0)=u_n^{\ast }(L)=0$, so $u_n\in D(T)$ and its weak derivative is the classical derivative $u_n^{\mathrm{\prime }}(x)=\sqrt{2\mathrm{/}L}(n\pi \mathrm{/}L)\cos (n\pi x\mathrm{/}L)$ [F1]. Let $S\supseteq T$ be any self-adjoint extension. Then $u_n\in D(S)$ and $S{u}_n=T{u}_n=-i\mathrm{\hslash }{u}_n^{\mathrm{\prime }}=-i\mathrm{\hslash }\sqrt{2\mathrm{/}L}\frac{n\pi }{L}\cos (n\pi x\mathrm{/}L)\mathrm{.}$ Suppose $S{u}_n=\lambda u_n$ for some $\lambda \in \mathbb{R}$; then the continuous functions $\cos (n\pi x\mathrm{/}L)$ and $\sin (n\pi x\mathrm{/}L)$ would be proportional on $[0,L]$. At $x=0$ one has $\sin 0=0$ and $\cos 0=1$, so proportionality would force the sine multiple to vanish identically, a contradiction. Hence $u_n$ is not an eigenvector of $S$, for any self-adjoint extension $S$. This proves (c).