Restricted momentum and energy multipliers of the transverse one-photon space

[Restriction of the spectral measure] Let $m$ be one of the two displayed functions; both are real, measurable and finite-valued. By [F4] the multiplier $M_m$ commutes with $M_{\mathrm{\Pi }}$, so $E(B)=M_{{\mathbf{1}}_{m^{-1}(B)}}$ (componentwise) commutes with $M_{\mathrm{\Pi }}$ for every Borel set $B$ and therefore maps $\mathfrak{h}$ into itself: for $f\in \mathfrak{h}$ one has $E(B)f={\mathbf{1}}_{m^{-1}(B)}f\in \mathfrak{h}$ because transversality holds pointwise. Thus $E_{\mathfrak{h}}(B):=E(B){\mathrm{\mid }}_{\mathfrak{h}}$ is a family of projections on $\mathfrak{h}$ with $E_{\mathfrak{h}}(B{)}^2=E_{\mathfrak{h}}(B)=E_{\mathfrak{h}}(B{)}^{\ast }$, $E_{\mathfrak{h}}(\mathrm{\varnothing })=0$, $E_{\mathfrak{h}}(\mathbb{R})={\mathrm{i}\mathrm{d}}_{\mathfrak{h}}$, and $\sigma$-additivity and strong convergence inherited from $E$; hence $E_{\mathfrak{h}}$ is a projection-valued measure on $\mathfrak{h}$.

[No eigenvalues] Let $\lambda \in \mathbb{R}$ and let $f\in D(M_m^{\mathfrak{h}})$ satisfy $M_m^{\mathfrak{h}}f=\lambda f$, that is $m(k)f(k)=\lambda f(k)$ for almost every $k$. Then $f=0$ almost everywhere on $\{k:m(k)\mathrm{\ne }\lambda \}$, so $f$ is supported in $N_{\lambda }:=\{k:m(k)=\lambda \}$ up to a null set. For $m=\mathrm{\hslash }{k}_j$ the set $N_{\lambda }$ is null by [F5]; for $m=\mathrm{\hslash }c\mathrm{\mid }k\mathrm{\mid }$ it is null by [F6]. In both cases $f=0$ almost everywhere, hence $f=0$ in $L^2$. This proves (c).

[The restricted integral is the restricted multiplier] By [F3] applied to $E_{\mathfrak{h}}$ and the real function $\lambda \mapsto \lambda$ the operator $A:=\int \lambda d{E}_{\mathfrak{h}}(\lambda )$ is self-adjoint with domain $\{f\in \mathfrak{h}:\int {\lambda }^{2}d{E}_{\mathfrak{h},f}(\lambda )<\mathrm{\infty }\}$. For every $g\in \mathfrak{h}$ and Borel $B$ we have $⟨E_{\mathfrak{h}}(B)g,g⟩={\int }_{m^{-1}(B)}\mathrm{\parallel }g(k){\mathrm{\parallel }}^2{d}^{3}k$: the scalar statements of [F2] applied to the three components of $g$ combine by sesquilinearity, and the unit case extends by scaling in $g$, so the scalar measure $E_{\mathfrak{h},g}$ is the push-forward of $\mathrm{\parallel }g{\mathrm{\parallel }}^2{d}^{3}k$ under $k\mapsto m(k)$; hence $\int {\lambda }^{2}d{E}_{\mathfrak{h},f}(\lambda )=\int m(k{)}^2\mathrm{\parallel }f(k){\mathrm{\parallel }}^2{d}^{3}k$ by the change-of-variables formula for a push-forward measure against the Borel function $\lambda \mapsto {\lambda }^2$, the two sides agreeing on indicators of Borel sets and extending by linearity and monotone convergence. Therefore $D(A)=\{f\in \mathfrak{h}:\int m^2\mathrm{\parallel }f{\mathrm{\parallel }}^2{d}^{3}k<\mathrm{\infty }\}=D(M_m^{\mathfrak{h}})$.

[Identification on the domain] Let $f\in D(A)$ and put $m_n:=m{\mathbf{1}}_{\{\mathrm{\mid }m\mathrm{\mid }\le n\}}$ for $n\ge 1$; then $\mathrm{\mid }{m}_n\mathrm{\mid }\le \mathrm{\mid }m\mathrm{\mid }$ and $m_n\to m$ pointwise, so [F3] gives $m_n(E_{\mathfrak{h}})f\to Af$. For $g\in \mathfrak{h}$ the bounded operator $m_n(E_{\mathfrak{h}})$ satisfies $⟨m_n(E_{\mathfrak{h}})g,g⟩=\int m_{n}d{E}_{\mathfrak{h},g}=\int m_n(k)\mathrm{\parallel }g(k){\mathrm{\parallel }}^2{d}^{3}k=⟨m_{n}g,g⟩$ by the push-forward description of step 2.1; both operators are bounded and self-adjoint and their quadratic forms agree, so $m_n(E_{\mathfrak{h}})=M_{m_n}{\mathrm{\mid }}_{\mathfrak{h}}$: the bounded functional calculus of the restricted multiplier is multiplication by $m_n$. The convergence $m_{n}f\to mf$ in $L^2({\mathbb{R}}^3;{\mathbb{C}}^3)$ holds by dominated convergence, since $\mathrm{\parallel }{m}_{n}f-mf{\mathrm{\parallel }}^2\le 4m^2\mathrm{\parallel }f{\mathrm{\parallel }}^2$ is integrable under $f\in D(A)$; its limit $mf$ lies in $\mathfrak{h}$ because transversality is preserved pointwise, with no assertion that $mf$ belongs to $D(A)$; that stronger assertion would require $m^{2}f\in L^2$. Hence $Af=mf=M_m^{\mathfrak{h}}f$ for every $f\in D(A)$; conversely if $f\in D(M_m^{\mathfrak{h}})$ then $\int m^2\mathrm{\parallel }f{\mathrm{\parallel }}^2=\mathrm{\parallel }mf{\mathrm{\parallel }}^2<\mathrm{\infty }$, so $f\in D(A)$. Therefore $A=M_m^{\mathfrak{h}}$ with the displayed domain, and (a) and (b) follow.

[Positive variance] Let $f\in \mathfrak{h}$ be a unit vector in $D(M_m^{\mathfrak{h}})$ and put $\mu :=⟨M_m^{\mathfrak{h}}f,f⟩$, a real number because $M_m^{\mathfrak{h}}$ is self-adjoint. Then $\mathrm{\parallel }{M}_m^{\mathfrak{h}}f{\mathrm{\parallel }}^2-⟨M_m^{\mathfrak{h}}f,f{⟩}^2=\mathrm{\parallel }(M_m^{\mathfrak{h}}-\mu I)f{\mathrm{\parallel }}^2\ge 0,$ and equality holds exactly when $(M_m^{\mathfrak{h}}-\mu I)f=0$, that is when $f$ is an eigenvector of $M_m^{\mathfrak{h}}$ with eigenvalue $\mu$. By step 1.2 no nonzero eigenvector exists, so the left-hand side is strictly positive. This proves (d) and completes the proof.