Self-adjointness, Fourier representation and translation identities of position and momentum

[Position operators] The function $x\mapsto x_j$ is real, measurable and finite-valued, so [F4] applied to $a(x)=x_j$ gives that $Q_j=M_a$ is self-adjoint with $D(Q_j)=\{f:x_{j}f\in L^2\}$. This is (a).

[Unitary conjugation of a self-adjoint operator] Let $U$ be unitary and let $S$ be self-adjoint. Put $D(A):=UD(S)$ and $Au:=US{U}^{\ast }u$ for $u\in D(A)$. The domain is dense because $D(S)$ is dense and $U$ is a surjective isometry. Then $A$ is symmetric: for $u=Uz$, $v=Uw$ with $z,w\in D(S)$, $⟨Au,v⟩=⟨USz,Uw⟩=⟨Sz,w⟩=⟨z,Sw⟩=⟨Uz,USw⟩=⟨u,Av⟩$ by unitarity. If now $v\in D(A^{\ast })$ with $A^{\ast }v=w^{\mathrm{\prime }}$, then for every $z\in D(S)$, writing $y:=U^{\ast }v$, $⟨Sz,y⟩=⟨USz,v⟩=⟨AUz,v⟩=⟨Uz,w^{\mathrm{\prime }}⟩=⟨z,U^{\ast }{w}^{\mathrm{\prime }}⟩$; hence $y\in D(S^{\ast })=D(S)$ and $Sy=U^{\ast }{w}^{\mathrm{\prime }}$. Therefore $v=Uy$ with $y\in D(S)$, that is $v\in D(A)$, and $Av=USy=U{U}^{\ast }{w}^{\mathrm{\prime }}=w^{\mathrm{\prime }}$. So $D(A^{\ast })=D(A)$ and $A^{\ast }=A$.

[The two momentum domains agree] Fix $j$, define $b_j(\xi ):=2\pi \mathrm{\hslash }{\xi }_j$ and let ${\tilde{P}}_j:={\mathcal{F}}_2^{\ast }{M}_{b_j}{\mathcal{F}}_2$ on $D({\tilde{P}}_j):=\{f\in L^2:b_j{\mathcal{F}}_{2}f\in L^2\}$, where $M_{b_j}$ is maximal multiplication as in [F4]. Let $f\in D(P_j)$. Then $D_{j}f\in L^2$ and [F2] gives ${\mathcal{F}}_2(D_{j}f)=2\pi i{\xi }_j{\mathcal{F}}_{2}f$, so $b_j{\mathcal{F}}_{2}f=-i\mathrm{\hslash }{\mathcal{F}}_2(D_{j}f)\in L^2$; thus $f\in D({\tilde{P}}_j)$ and ${\tilde{P}}_{j}f={\mathcal{F}}_2^{\ast }(b_j{\mathcal{F}}_{2}f)=-i\mathrm{\hslash }{\mathcal{F}}_2^{\ast }{\mathcal{F}}_2(D_{j}f)=-i\mathrm{\hslash }{D}_{j}f=P_{j}f$.

[Translations commute with weak derivatives] Let $a\in {\mathbb{R}}^d$ and assume only $f\in L^2$ and $D_{j}f\in L^2$ for one fixed $j$; by [F5] $T_a$ is an isometry of $L^2$. For every $\phi \in C_c^{\mathrm{\infty }}$ the function $\phi (\cdot +a)$ is again a test function. Its partial difference quotient at $y$ is exactly that of $\phi$ at $y+a$, so ${\mathrm{\partial }}_j(\phi (\cdot +a))=(D_j\phi )(\cdot +a)$. Two uses of [F6], applied to positive and negative parts of real and imaginary components of the integrable functions, give $\int (T_{a}f)D_j\phi =\int f(y)D_j\phi (y+a)dy=\int f(y)D_j(\phi (\cdot +a))(y)dy=-\int (D_{j}f)(y)\phi (y+a)dy=-\int (T_a{D}_{j}f)(x)\phi (x)dx$. Thus $D_j(T_{a}f)=T_a(D_{j}f)\in L^2$, without assuming any other weak derivative of $f$. Applying this separately for every $j$ when $f\in W^{1,2}$, and then to each $D_{k}f$ when $f\in W^{2,2}$, proves preservation of both Sobolev spaces. The inverse $T_{-a}$ gives equality of these spaces under translation. Moreover $T_a$ is complex-linear and [F6] gives $⟨T_{a}f,T_{a}g⟩=\int f(y)\overline{g(y)}dy=⟨f,g⟩$; with its inverse this proves unitarity.

[Converse inclusion] Let $f\in D({\tilde{P}}_j)$ and set $g:={\mathcal{F}}_2^{\ast }(2\pi i{\xi }_j{\mathcal{F}}_{2}f)\in L^2$. We show that $g$ is the weak $j$-derivative of $f$. Let $\phi \in C_c^{\mathrm{\infty }}({\mathbb{R}}^d)$ be complex-valued; it is enough to verify the bilinear identity for the real and imaginary parts, so assume first that $\phi$ is real-valued. Applying [F2] to $\phi$ (whose classical derivative is its weak derivative as specified in the defining supplier) and [F1], $\int f{D}_j\phi =⟨f,D_j\phi ⟩=⟨{\mathcal{F}}_{2}f,{\mathcal{F}}_2(D_j\phi )⟩=⟨{\mathcal{F}}_{2}f,2\pi i{\xi }_j{\mathcal{F}}_2\phi ⟩=\int {\mathcal{F}}_{2}f\overline{2\pi i{\xi }_j{\mathcal{F}}_2\phi }=-\int (2\pi i{\xi }_j{\mathcal{F}}_{2}f)\overline{{\mathcal{F}}_2\phi }=-⟨g,\phi ⟩=-\int g\phi$. The Fourier pairings remain sesquilinear; real-valuedness of $\phi$ identifies only the spatial inner products with the bilinear test integrals. By linearity the identity holds for all complex $\phi \in C_c^{\mathrm{\infty }}$. Hence $g$ is the $L^2$ weak $j$-derivative of $f$, that is $f\in D(P_j)$ and $P_{j}f=-i\mathrm{\hslash }g={\mathcal{F}}_2^{\ast }(2\pi \mathrm{\hslash }{\xi }_j{\mathcal{F}}_{2}f)={\tilde{P}}_{j}f$.

[Momentum as a Fourier multiplier] The function $b_j$ of step 1.3 is real, measurable and finite-valued. By [F4], $M_{b_j}$ is self-adjoint with domain $D(M_{b_j})=\{h\in L^2:b_{j}h\in L^2\}$. The operator ${\tilde{P}}_j$ of step 1.3 has domain ${\mathcal{F}}_2^{-1}D(M_{b_j})$, so step 1.2 with $U={\mathcal{F}}_2^{\ast }$ and $S=M_{b_j}$ proves its self-adjointness.

[Translation identities] Since $T_a^{-1}=T_{-a}$ and $T_a$ is unitary by [F5], step 1.4 gives $T_a{P}_j{T}_a^{-1}h=-i\mathrm{\hslash }{T}_a{D}_j(T_{-a}h)=-i\mathrm{\hslash }{T}_a{T}_{-a}{D}_{j}h=-i\mathrm{\hslash }{D}_{j}h=P_{j}h$ for $h\in D(P_j)$, and $T_{a}D(P_j)=D(P_j)$ because the same computation applies to $T_{-a}$. For $Q_j$, $Q_j{T}_{-a}h$ is the class $x\mapsto x_{j}h(x+a)$ and applying $T_a$ gives $x\mapsto (x-a{)}_{j}h(x)$; hence $T_a{Q}_j{T}_a^{-1}=Q_j-a_j$ as class-valued maps, and $D(Q_j-a_j)=\{h:(x_j-a_j)h\in L^2\}=D(Q_j)$ because $\mathrm{\mid }(x_j-a_j)h\mathrm{\mid }\le \mathrm{\mid }{x}_{j}h\mathrm{\mid }+\mathrm{\mid }{a}_j\mathrm{\mid }\mathrm{\mid }h\mathrm{\mid }$ and conversely $\mathrm{\mid }{x}_{j}h\mathrm{\mid }\le \mathrm{\mid }(x_j-a_j)h\mathrm{\mid }+\mathrm{\mid }{a}_j\mathrm{\mid }\mathrm{\mid }h\mathrm{\mid }$.

[Conclusion of (b)] Steps 1.3 and 2.1 give $D(P_j)=D({\tilde{P}}_j)=\{f:{\xi }_j{\mathcal{F}}_{2}f\in L^2\}$ and $P_j={\tilde{P}}_j$, which is self-adjoint by step 2.2; and for $f\in D(P_j)$, ${\mathcal{F}}_2(P_{j}f)={\mathcal{F}}_2({\tilde{P}}_{j}f)=2\pi \mathrm{\hslash }{\xi }_j{\mathcal{F}}_{2}f$ a.e. If $d=1$ then $\{f:\xi {\mathcal{F}}_{2}f\in L^2\}=\{f:⟨\xi ⟩{\mathcal{F}}_{2}f\in L^2\}=W^{1,2}(\mathbb{R};\mathbb{C})$ by [F3], because ${\xi }^2\le ⟨\xi {⟩}^2=1+{\xi }^2$ and $⟨\xi {⟩}^2\le 1+2{\xi }^2$ for $\mathrm{\mid }\xi \mathrm{\mid }\ge 1$ while both weights are bounded by constants on $\mathrm{\mid }\xi \mathrm{\mid }\le 1$. This is (b).

[Conclusion of (c)] Steps 1.4 and 2.3 give all assertions of (c): unitarity of $T_a$ is [F5], the momentum and position conjugation identities and the preservation of $W^{1,2}$ and $W^{2,2}$ are step 1.4 and its iteration.