Componentwise position and momentum operators on the component-valued wave-function space

[Operators as orthogonal sums] Let $f=(f_l)\in H^{(r)}$. By [F2] one has ${\mathrm{\Psi }}_0^{-1}f=(f_l)\in {⨁}_l{L}^2$, and $f\in D(Q_j)$ holds exactly when $x_j{f}_l\in L^2$ for every $l$, that is when ${\mathrm{\Psi }}_0^{-1}f\in {⨁}_{l}D(Q_j^{(1)})$; on that domain ${\mathrm{\Psi }}_0^{-1}(Q_{j}f)=(x_j{f}_l{)}_l=({⨁}_l{Q}_j^{(1)}){\mathrm{\Psi }}_0^{-1}f$. The same computation with $D_k{f}_l$ in place of $x_j{f}_l$ gives ${\mathrm{\Psi }}_0^{-1}(P_{k}f)=({⨁}_l{P}_k^{(1)}){\mathrm{\Psi }}_0^{-1}f$ on ${\mathrm{\Psi }}_0^{-1}D(P_k)={⨁}_{l}D(P_k^{(1)})$. The finite-sum condition in the domain of an orthogonal sum is automatic here: if every $\mathrm{\parallel }{Q}_j^{(1)}{f}_l\mathrm{\parallel }$ is finite then ${\sum }_l\mathrm{\parallel }{Q}_j^{(1)}{f}_l{\mathrm{\parallel }}^2=\mathrm{\parallel }{Q}_{j}f{\mathrm{\parallel }}^2<\mathrm{\infty }$, and similarly for $P_k$. This proves the identification in (a).

[A scalar difference quotient] Let $g\in L^2({\mathbb{R}}^d;\mathbb{C})$ with ${\xi }_j{\mathcal{F}}_{2}g\in L^2$, that is $g\in D(P_j^{(1)})$ by [F1], and let $t\mathrm{\ne }0$. The modulation identity ${\mathcal{F}}_2(T_{t{e}_j}g)(\xi )=e^{-2\pi it{\xi }_j}{\mathcal{F}}_{2}g(\xi )$ of [F4] gives ${\mathcal{F}}_2\left(\frac{1}{t}\right(T_{t{e}_j}g-g\left)\right)(\xi )=\frac{e^{-2\pi it{\xi }_j}-1}{t}{\mathcal{F}}_{2}g(\xi )\mathrm{.}$ Writing $\theta :=-2\pi t{\xi }_j$, the factor equals $\frac{e^{i\theta }-1}{i\theta }\cdot (-2\pi i{\xi }_j)$; by [F4] the first factor tends to $1$ and $\theta \to 0$ for every fixed $\xi$ as $t\to 0$, so the pointwise limit of the right-hand side is $-2\pi i{\xi }_j{\mathcal{F}}_{2}g(\xi )={\mathcal{F}}_2(-D_{j}g)(\xi )$ [F4]. For domination, [F4] gives $\mathrm{\mid }\frac{e^{i\theta }-1}{t}\mathrm{\mid }\le \mathrm{\mid }\theta \mathrm{/}t\mathrm{\mid }=2\pi \mathrm{\mid }{\xi }_j\mathrm{\mid }$, so the modulus of the difference $h_t(\xi ):=(\frac{e^{-2\pi it{\xi }_j}-1}{t}+2\pi i{\xi }_j){\mathcal{F}}_{2}g(\xi )$ is at most $4\pi \mathrm{\mid }{\xi }_j\mathrm{\mid }\mathrm{\mid }{\mathcal{F}}_{2}g(\xi )\mathrm{\mid }$, whose square is integrable because ${\xi }_j{\mathcal{F}}_{2}g\in L^2$. Since $h_t\to 0$ pointwise, [F6] applied to $\mathrm{\mid }{h}_t{\mathrm{\mid }}^2$ gives $\int \mathrm{\mid }{h}_t{\mathrm{\mid }}^2\to 0$, so the Fourier-space difference converges to ${\mathcal{F}}_2(-D_{j}g)$ in $L^2$. As ${\mathcal{F}}_2$ is a surjective isometry [F4], and $-D_{j}g=-\frac{i}{\mathrm{\hslash }}{P}_j^{(1)}g$, this yields ${\lim }_{t\to 0}\frac{1}{t}(T_{t{e}_j}g-g)=-\frac{i}{\mathrm{\hslash }}{P}_j^{(1)}g\text{in }{L}^2\mathrm{.}$

[Self-adjointness] Each scalar $Q_j^{(1)}$ and $P_k^{(1)}$ is self-adjoint by [F1], so [F3] makes their finite orthogonal sums self-adjoint with the displayed domains and dense domains; by step 1.1 these sums are the conjugates of $Q_j$ and $P_k$ by the unitary ${\mathrm{\Psi }}_0$, hence $Q_j$ and $P_k$ are densely defined and self-adjoint. This completes (a).

[Fourier representation] By [F2] and [F3], ${\mathcal{F}}_2^{(r)}$ is the orthogonal sum of the unitaries ${\mathcal{F}}_2$, hence unitary. For $f\in D(P_k)$ its components satisfy ${\mathcal{F}}_2(P_k^{(1)}{f}_l)(\xi )=2\pi \mathrm{\hslash }{\xi }_k{\mathcal{F}}_2{f}_l(\xi )$ for almost every $\xi$ by [F1]; assembling components gives ${\mathcal{F}}_2^{(r)}(P_{k}f)(\xi )=2\pi \mathrm{\hslash }{\xi }_k{\mathcal{F}}_2^{(r)}f(\xi )$ for almost every $\xi$. The domain description follows because ${\xi }_k{\mathcal{F}}_2{f}_l\in L^2$ for all $l$ is exactly ${\xi }_k{\mathcal{F}}_2^{(r)}f\in L^2({\mathbb{R}}^d;{\mathbb{C}}^r)$. This proves (b).

[Translation identities] By [F2] the translation $T_a^{(r)}$ corresponds to ${⨁}_l{T}_a$, which is unitary by [F3] and [F1]; it preserves the componentwise Schwartz and Sobolev classes because each scalar $T_a$ does [F1]. The identities $D_k(T_a^{(r)}f)=T_a^{(r)}(D_{k}f)$ and $T_a^{(r)}D(Q_j)=D(Q_j)$, $T_a^{(r)}{Q}_j(T_a^{(r)}{)}^{-1}=Q_j-a_{j}I$, $T_a^{(r)}{P}_k(T_a^{(r)}{)}^{-1}=P_k$ are componentwise instances of [F1]; for instance, $(T_a^{(r)}{Q}_j(T_a^{(r)}{)}^{-1}f{)}_l=T_a(x_j{T}_{-a}{f}_l)=(x_j-a_j)f_l$ for $f_l\in D(Q_j^{(1)})$, and the domain equality follows from the scalar one. This proves (c).

[The componentwise limit] Let $f\in D(P_j)$. Applying step 1.2 to each component $f_l\in D(P_j^{(1)})$ and assembling components through the unitary ${\mathrm{\Psi }}_0$ [F2] gives $\frac{1}{t}(T_{t{e}_j}^{(r)}f-f)\to (-\frac{i}{\mathrm{\hslash }}{P}_j^{(1)}{f}_l{)}_l=-\frac{i}{\mathrm{\hslash }}{P}_{j}f$ in $H^{(r)}$ as $t\to 0$; this is the first assertion of (d).

[The group of translations and its generator] Fix $j$. The family $U_j(t):=T_{t{e}_j}^{(r)}$ is a strongly continuous one-parameter unitary group: the group law and unitarity follow from the componentwise definition of translations and [F3], and strong continuity from [F4]. By step 2.4 its generator $G$ satisfies $G\supseteq -\frac{i}{\mathrm{\hslash }}{P}_j$; by [F5] applied to the self-adjoint operator $T:=-P_j\mathrm{/}\mathrm{\hslash }$, the family $V_j(t):=\exp (-\frac{i}{\mathrm{\hslash }}t{P}_j)=\exp (itT)$ is a strongly continuous unitary group with generator $iT=-\frac{i}{\mathrm{\hslash }}{P}_j$ and domain $D(P_j)$, and it preserves $D(P_j)$ [F5]. On $D(P_j)$ both groups have the same right derivative at $0$, and both preserve $D(P_j)$; hence $w(t):=U_j(t)g-V_j(t)g$ for fixed $g\in D(P_j)$ satisfies $w^{\mathrm{\prime }}(t)=(-\frac{i}{\mathrm{\hslash }}{P}_j)w(t)$ and $\frac{d}{dt}\mathrm{\parallel }w(t){\mathrm{\parallel }}^2=2\mathrm{Re}⟨w(t),-\frac{i}{\mathrm{\hslash }}{P}_{j}w(t)⟩=0$ because $⟨P_{j}w(t),w(t)⟩$ is real; with $w(0)=0$ this gives $U_j(t)g=V_j(t)g$ for all $t$ on the dense set $D(P_j)$, hence $U_j(t)=V_j(t)$ by continuity and unitarity. This proves (d).