Real multiplication operators and free plus bounded-potential Schrödinger operators

[Density and symmetry of the multiplier] By [F8], the defined $L^2$ space $H$ is a complex Hilbert space. For $f\in H$ and $n\ge 1$ put $f_n:=f{\mathbf{1}}_{\{\mathrm{\mid }a\mathrm{\mid }\le n\}}$. Then $a{f}_n$ is bounded by $n\mathrm{\mid }{f}_n\mathrm{\mid }$, so $f_n\in D(M_a)$, and $f_n\to f$ in $H$ by [F1] applied to $\mathrm{\mid }{f}_n-f{\mathrm{\mid }}^2\le \mathrm{\mid }f{\mathrm{\mid }}^2$; hence $D(M_a)$ is dense. For $f,g\in D(M_a)$ the integral $\int af\bar{g}$ is defined and equals $\int f\overline{ag}$ because $a$ is real-valued, so $⟨M_{a}f,g⟩=⟨f,M_{a}g⟩$: $M_a$ is symmetric.

[Adjoint domain of the multiplier] Let $g\in D(M_a^{\ast })$ with representing vector $h=M_a^{\ast }g$, so $\int af\bar{g}=\int f\bar{h}$ for all $f\in D(M_a)$. Fix $n$ and restrict to $f\in H$ vanishing off $E_n:=\{\mathrm{\mid }a\mathrm{\mid }\le n\}$; every such $f$ lies in $D(M_a)$, and the identity reads ${\int }_{E_n}(a\bar{g}-\bar{h})f=0$ for all $f\in L^2(E_n)$, so $h=ag$ a.e. on $E_n$; as $E_n\uparrow {\mathbb{R}}^d$, $h=ag$ a.e., and since $h\in H$, $ag\in H$ and $g\in D(M_a)$. Conversely every $g\in D(M_a)$ satisfies the adjoint identity with $h=M_{a}g$. Hence $D(M_a^{\ast })=D(M_a)$ and $M_a^{\ast }=M_a$: $M_a$ is self-adjoint, giving (a).

[Unitary conjugation] Let $T:=M_{c\mathrm{\mid }\xi {\mathrm{\mid }}^2+b}$, self-adjoint on $D(T)=\{g:c\mathrm{\mid }\xi {\mathrm{\mid }}^{2}g\in H\}$ by step 2.1 applied to the real finite-valued function $c\mathrm{\mid }\xi {\mathrm{\mid }}^2+b$. Define $H_0:={\mathcal{F}}_2^{\ast }T{\mathcal{F}}_2$ on $D(H_0)={\mathcal{F}}_2^{-1}D(T)=\{f:\mathrm{\mid }\xi {\mathrm{\mid }}^2{\mathcal{F}}_{2}f\in H\}$. For $x,y\in D(H_0)$, $⟨H_{0}x,y⟩=⟨T{\mathcal{F}}_{2}x,{\mathcal{F}}_{2}y⟩$ by unitarity of ${\mathcal{F}}_2$ [F3]. If $y\in D(H_0^{\ast })$, the functional $x\mapsto ⟨H_{0}x,y⟩$ is bounded on $D(H_0)$; conjugating by the unitary ${\mathcal{F}}_2$ shows that $g\mapsto ⟨Tg,{\mathcal{F}}_{2}y⟩$ is bounded on $D(T)$, so ${\mathcal{F}}_{2}y\in D(T^{\ast })=D(T)$ and hence $y\in D(H_0)$ with $H_0^{\ast }y={\mathcal{F}}_2^{\ast }{T}^{\ast }{\mathcal{F}}_{2}y=H_{0}y$; the reverse inclusion is immediate from the symmetry identity. Thus $H_0$ is self-adjoint.

[Domain, lower bound and free action] By [F5], $D(H_0)=W^{2,2}({\mathbb{R}}^d)$ as a set with equivalent norms. For $f\in D(H_0)$ the pairing with $f$ is computed in the $\xi$-variable, where multiplication by $c\mathrm{\mid }\xi {\mathrm{\mid }}^2+b\ge b$ gives $⟨H_{0}f,f⟩=\int (c\mathrm{\mid }\xi {\mathrm{\mid }}^2+b)\mathrm{\mid }{\mathcal{F}}_{2}f(\xi ){\mathrm{\mid }}^{2}d\xi \ge b\mathrm{\parallel }{\mathcal{F}}_{2}f{\mathrm{\parallel }}^2=b\mathrm{\parallel }f{\mathrm{\parallel }}^2$, so $H_0\ge b$ in the quadratic-form sense. For a Schwartz function $f$, [F4] gives ${\mathcal{F}}_2(-(c\mathrm{/}4{\pi }^2)\mathrm{\Delta }f+bf)(\xi )=(c\mathrm{\mid }\xi {\mathrm{\mid }}^2+b){\mathcal{F}}_{2}f(\xi )$ a.e.; both sides are continuous representatives, and Plancherel inversion identifies them as elements of $H$, so $H_{0}f=-(c\mathrm{/}4{\pi }^2)\mathrm{\Delta }f+bf$. This is (b).

[Bounded perturbation by Kato-Rellich] Let $V$ be real with $\mathrm{\parallel }V{\mathrm{\parallel }}_{\mathrm{\infty }}<\mathrm{\infty }$ and let $M_V$ be the bounded multiplication operator on $H$. Then $D(H_0)\subseteq D(M_V)=H$, $M_V$ is symmetric because $V$ is real, and $\mathrm{\parallel }{M}_{V}f\mathrm{\parallel }\le \mathrm{\parallel }V{\mathrm{\parallel }}_{\mathrm{\infty }}\mathrm{\parallel }f\mathrm{\parallel }$ for every $f\in D(H_0)$, so the $H_0$-bound of $M_V$ is $0<1$ with admissible pair $(0,\mathrm{\parallel }V{\mathrm{\parallel }}_{\mathrm{\infty }})$ (Relative boundedness with respect to an operator). By [F6] the operator $H=H_0+M_V$ on $D(H_0)$ is self-adjoint and, since $H_0\ge b$, satisfies $H\ge b-\max \{0\cdot \mathrm{\mid }b\mathrm{\mid }+\mathrm{\parallel }V{\mathrm{\parallel }}_{\mathrm{\infty }},\mathrm{\parallel }V{\mathrm{\parallel }}_{\mathrm{\infty }}\}=b-\mathrm{\parallel }V{\mathrm{\parallel }}_{\mathrm{\infty }}$, hence is bounded below.

[Unitary dynamics] Applying [F7] to the self-adjoint $H$ of step 5.1 yields that $t\mapsto e^{itH}$ is a strongly continuous one-parameter unitary group, $e^{itH}D(H)=D(H)$, and the difference quotients converge exactly on $D(H)$; the physical convention $t\mapsto e^{-itH}$ is the same group at negative time. No claim in this item extends to unbounded or singular $V$, which are excluded by the boundedness and measurability hypotheses recorded in [F6].