Domain-qualified free observables and point-field obstruction

The measure $d{\nu }_{\mu }$ is sigma-finite by [F9], and ${\omega }_{\mu }$ and $k_j$ are finite measurable multipliers by [F2]. Applying [F5] with one-particle multiplier vector $(1,{\omega }_{\mu },-k_1,-k_2,-k_3)$ gives the self-adjoint scalar operators $N_s,H_s,P_{s,j}$. On sector $n$, their multipliers are $n$, ${\sum }_{r=1}^n{\omega }_{\mu }({\mathbf{k}}_r)$, and $-{\sum }_{r=1}^n{k}_{r,j}$. Squaring these multipliers and integrating against the completed product measure yields exactly the stated domains.

The Dirac data in [F6] satisfy the measurable, finite-multiplier hypotheses of [F7], and [F16] supplies the complementary orthogonal projections and closed ranges required by that carrier. The one-particle measure is sigma-finite: each of the two species copies of R^3 is the countable union of bounded balls with finite Lebesgue measure. Let Y_m be the two species copies restricted to the ball of radius m; each Y_m has finite measure. For fixed n, the product sets Y_m^n have finite product measure and cover X^n as m grows. Their countable union over n and m is a sigma-finite cover of the Fock direct-sum measure. The particle and antiparticle sector multipliers are therefore self-adjoint on the maximal domains in [F7], which are the displayed square-integrability conditions for $H_D$ and each $P_{D,j}$. The field boundedness and equal-time CAR are exactly the inspected interfaces in [F8].

For real $f$, [F3] identifies the scalar field on ${\mathcal{D}}_{\mathrm{f}\mathrm{i}\mathrm{n}}$ with $\sqrt{2}{\mathrm{\Phi }}_{\mathrm{S}\mathrm{e}\mathrm{g}\mathrm{a}\mathrm{l}}(K_{\mu }f)$. The Hilbert space $H_{\mu }$ is separable by [F13], so the hypothesis of the Segal-domain result [F4] holds. Multiplication by the nonzero scalar $\sqrt{2}$ does not change an operator domain; hence the closure is self-adjoint on the domain in [F4] with $h=K_{\mu }f$. If $K_{\mu }f=0$, both operators are zero on all Fock space.

For a normalized $\mathrm{\Psi }\in D(A)$, Cauchy–Schwarz [F12] gives $\mathrm{\mid }⟨\mathrm{\Psi },A\mathrm{\Psi }⟩\mathrm{\mid }\le \mathrm{\parallel }A\mathrm{\Psi }\mathrm{\parallel }<\mathrm{\infty }$. Since $A$ is self-adjoint, this expectation is real. The vector $A\mathrm{\Psi }-⟨A{⟩}_{\mathrm{\Psi }}\mathrm{\Psi }$ belongs to the Hilbert space because both terms do, so its norm and the stated variance are finite without requiring $\mathrm{\Psi }\in D(A^2)$. By [F14], a product $AB$ acts only on vectors $\mathrm{\Psi }\in D(B)$ for which $B\mathrm{\Psi }\in D(A)$; it is not defined merely because both $A$ and $B$ separately exist on a common dense core.

Choose $b_d$ from [F18] for $d=3,4$. Put $Z_d=\int b_d(x)d^{d}x$ and ${\rho }_d=Z_d^{-1}{b}_d$. The integral is finite because $0\le b_d\le 1$ and its support lies in a bounded ball, and it is positive because $b_d=1$ on a ball of positive volume. Every derivative of $b_d$ is continuous with compact support, so all its coordinate-weighted derivatives are bounded. Thus ${\rho }_d$ is Schwartz, nonnegative, nonzero, and has integral one. In particular $0<\mathrm{\parallel }{\rho }_d{\mathrm{\parallel }}_2<\mathrm{\infty }$. Define ${\widehat{\rho }}_d(\xi )=\int e^{-i\xi \cdot x}{\rho }_d(x)d^{d}x$. The integral triangle inequality gives $\mathrm{\mid }{\widehat{\rho }}_d(\xi )\mathrm{\mid }\le \int {\rho }_d=1$; dominated convergence [F10], with majorant ${\rho }_d$, gives ${\widehat{\rho }}_d(\xi )\to 1$ as $\xi \to 0$. Set $\rho ={\rho }_4$ and, for $\epsilon >0$, define ${\rho }_{\epsilon ,x_0}(x)={\epsilon }^{-4}{\rho }_4((x-x_0)\mathrm{/}\epsilon )$. For every Schwartz test $\varphi$, change of variables and dominated convergence [F10], with majorant $\mathrm{\parallel }\varphi {\mathrm{\parallel }}_{\mathrm{\infty }}{\rho }_4(y)$, give $\int {\rho }_{\epsilon ,x_0}(x)\varphi (x)d^{4}x\to \varphi (x_0)$, so this is a point-mass approximation. For $p=(p^0,\mathbf{p})\in H_{\mu }^{+}$ the same change of variables gives $K_{\mu }{\rho }_{\epsilon ,x_0}(p)=e^{ip\circ x_0}{\widehat{\rho }}_4(\epsilon (-p^0,\mathbf{p}))\mathrm{.}$ Thus $K_{\mu }{\rho }_{\epsilon ,x_0}(p)\to e^{ip\circ x_0}$ for every fixed shell momentum and $\mathrm{\mid }{K}_{\mu }{\rho }_{\epsilon ,x_0}(p)\mathrm{\mid }\le 1$. The shell measure has infinite total mass: in radial coordinates, for $r\ge \max (1,\mu )$, ${\omega }_{\mu }(r)\le \sqrt{2}r$, so its radial density is bounded below by a positive constant times $rdr$, whose integral diverges. Fatou's lemma [F11] therefore gives ${\mathrm{lim\hspace{0.17em}inf}}_{{\epsilon }_m\downarrow 0}\mathrm{\parallel }{K}_{\mu }{\rho }_{{\epsilon }_m,x_0}{\mathrm{\parallel }}^2\ge {\int }_{H_{\mu }^{+}}1d{\nu }_{\mu }=\mathrm{\infty }$. By [F1, F3] and the creation formula on the vacuum, $\mathrm{\parallel }{\mathrm{\Phi }}_{\mu }({\rho }_{{\epsilon }_m,x_0})\mathrm{\Omega }\mathrm{\parallel }=\mathrm{\parallel }{K}_{\mu }{\rho }_{{\epsilon }_m,x_0}\mathrm{\parallel }$, so these vacuum vectors have no finite-norm point-field limit. More generally, suppose $f_{\epsilon }$ satisfies the scalar hypotheses in the Statement. For any sequence ${\epsilon }_j\to 0$, Fatou [F11] applied to $\mathrm{\mid }{K}_{\mu }{f}_{{\epsilon }_j}{\mathrm{\mid }}^2$ gives ${\mathrm{lim\hspace{0.17em}inf}}_j\mathrm{\parallel }{K}_{\mu }{f}_{{\epsilon }_j}{\mathrm{\parallel }}^2\ge \int 1d{\nu }_{\mu }=\mathrm{\infty }$. Thus the norms diverge as $\epsilon \to 0$: otherwise a bounded sequence with ${\epsilon }_j<1\mathrm{/}j$ could be selected using [A1], contradicting this conclusion. The vacuum creation identity gives the asserted field-vector norm. To verify the sufficient concentration condition, let $M={\sup }_{\epsilon }\mathrm{\parallel }{f}_{\epsilon }{\mathrm{\parallel }}_1$ and let $g$ be any bounded continuous scalar function. Since $\int f_{\epsilon }=1$, splitting the integral at radius $\delta$ gives $\mathrm{\mid }\int f_{\epsilon }(x)g(x)d^{4}x-g(x_0)\mathrm{\mid }\le M{\sup }_{\mathrm{\mid }x-x_0\mathrm{\mid }<\delta }\mathrm{\mid }g(x)-g(x_0)\mathrm{\mid }+2\mathrm{\parallel }g{\mathrm{\parallel }}_{\mathrm{\infty }}{\int }_{\mathrm{\mid }x-x_0\mathrm{\mid }\ge \delta }\mathrm{\mid }{f}_{\epsilon }(x)\mathrm{\mid }{d}^{4}x$. First choose $\delta$ using continuity at $x_0$, then take $\epsilon \to 0$. Applying this to Schwartz $g$ and to $g(x)=e^{ip\circ x}$ proves both required limits. Every rescaling of a nonnegative normalized Schwartz profile has $M=1$ and satisfies concentration by its integrable tail; hence the general conclusion includes every such profile, not only the chosen bumps.

The scalar one-particle space contains a unit vector $e$, for example a normalized indicator of an annulus $1<\mathrm{\mid }\mathbf{k}\mathrm{\mid }<2$, whose shell measure is finite and positive by [F2]. The symmetric vectors $e^{\otimes n}$ are unit vectors in mutually orthogonal sectors. Put $C=({\sum }_{n\ge 1}{n}^{-4}{)}^{-1\mathrm{/}2}$ and ${\mathrm{\Psi }}_n=C{n}^{-2}{e}^{\otimes n}$. By [F15], $\sum n^{-4}$ converges and is positive, so $C$ is well-defined and $\mathrm{\Psi }$ is normalized. Also by [F15], $\sum n^{-2}<\mathrm{\infty }$, so $\mathrm{\Psi }\in D(N_s)$ and $N_s\mathrm{\Psi }=(C{n}^{-1}{e}^{\otimes n}{)}_{n\ge 1}$ is a Fock vector. Its image under one more application of $N_s$ would have squared norm $C^2{\sum }_{n\ge 1}1=\mathrm{\infty }$, so $N_s\mathrm{\Psi }\mathrm{\notin }D(N_s)$. By the composition-domain rule [F14] with $A=B=N_s$, $\mathrm{\Psi }\mathrm{\notin }D(N_s^2)$. Its mean and variance are finite by 2.2.

Fix a unit spinor $z\in {\mathbb{C}}^4$ and, for $\epsilon >0$, define $u_{\epsilon }(\mathbf{x})={\epsilon }^{-3}{\rho }_3((\mathbf{x}-{\mathbf{x}}_0)\mathrm{/}\epsilon )z$. For every spatial Schwartz test, change of variables and [F10] show that its pairing with $u_{\epsilon }$ tends to the test's value at ${\mathbf{x}}_0$ times $z$, while $\mathrm{\parallel }{u}_{\epsilon }{\mathrm{\parallel }}_2={\epsilon }^{-3\mathrm{/}2}\mathrm{\parallel }{\rho }_3{\mathrm{\parallel }}_2\to \mathrm{\infty }$. For a general family in the Dirac Statement, set $e=z_0\mathrm{/}\mathrm{\parallel }{z}_0\mathrm{\parallel }$ and $h_a(\mathbf{x})=b_3((\mathbf{x}-{\mathbf{x}}_0)\mathrm{/}a)e$ for $a>0$. By [F18], this is a spinor Schwartz test, $h_a({\mathbf{x}}_0{)}^{†}{z}_0=\mathrm{\parallel }{z}_0\mathrm{\parallel }$, and $\mathrm{\parallel }{h}_a{\mathrm{\parallel }}_2=a^{3\mathrm{/}2}\mathrm{\parallel }{b}_3{\mathrm{\parallel }}_2$. Given $L>0$, choose $a$ with $\mathrm{\parallel }{h}_a{\mathrm{\parallel }}_2<\mathrm{\parallel }{z}_0\mathrm{\parallel }\mathrm{/}(2L)$. The assumed pairing limit then gives $\mathrm{\mid }⟨h_a,u_{\epsilon }{⟩}_2\mathrm{\mid }>\mathrm{\parallel }{z}_0\mathrm{\parallel }\mathrm{/}2$ for all sufficiently small $\epsilon$. Cauchy–Schwarz [F12] implies $\mathrm{\parallel }{u}_{\epsilon }{\mathrm{\parallel }}_2>L$. Thus every such family has $\mathrm{\parallel }{u}_{\epsilon }{\mathrm{\parallel }}_2\to \mathrm{\infty }$. The following operator estimate applies to either the explicit bump family or this arbitrary family. By [F16], choose countable dense sets in the two Dirac energy ranges. Conjugating the negative-range dense set preserves density, and the Cartesian product of the two dense sets is countable and dense in the Hilbert direct sum. Therefore the one-particle space for the CAR Fock representation is separable and [F17] applies. For $A_{\epsilon }={\mathrm{\Psi }}_t(u_{\epsilon })$, [F8] gives $A_{\epsilon }{A}_{\epsilon }^{†}+A_{\epsilon }^{†}{A}_{\epsilon }=\mathrm{\parallel }{u}_{\epsilon }{\mathrm{\parallel }}_2^{2}I$. By [F17], the creation and annihilation terms are bounded, hence so is their sum. Cauchy–Schwarz [F12] gives ${\sup }_{\mathrm{\parallel }x\mathrm{\parallel }=\mathrm{\parallel }y\mathrm{\parallel }=1}\mathrm{\mid }⟨A_{\epsilon }x,y⟩\mathrm{\mid }\le \mathrm{\parallel }{A}_{\epsilon }\mathrm{\parallel }$, while choosing $y=A_{\epsilon }x\mathrm{/}\mathrm{\parallel }{A}_{\epsilon }x\mathrm{\parallel }$ for each unit $x$ with $A_{\epsilon }x\mathrm{\ne }0$ gives equality after taking the supremum over $x$. The adjoint identity swaps $x$ and $y$, so the same two-vector supremum equals $\mathrm{\parallel }{A}_{\epsilon }^{†}\mathrm{\parallel }$. For bounded operators $X,Y$, each unit $\xi$ satisfies $\mathrm{\parallel }XY\xi \mathrm{\parallel }\le \mathrm{\parallel }X\mathrm{\parallel }\mathrm{\parallel }Y\xi \mathrm{\parallel }\le \mathrm{\parallel }X\mathrm{\parallel }\mathrm{\parallel }Y\mathrm{\parallel }$ and $\mathrm{\parallel }(X+Y)\xi \mathrm{\parallel }\le \mathrm{\parallel }X\xi \mathrm{\parallel }+\mathrm{\parallel }Y\xi \mathrm{\parallel }\le \mathrm{\parallel }X\mathrm{\parallel }+\mathrm{\parallel }Y\mathrm{\parallel }$; taking suprema gives product submultiplicativity and the triangle inequality. Applying these inequalities to the CAR identity yields $\mathrm{\parallel }{u}_{\epsilon }{\mathrm{\parallel }}_2^2=\mathrm{\parallel }{A}_{\epsilon }{A}_{\epsilon }^{†}+A_{\epsilon }^{†}{A}_{\epsilon }\mathrm{\parallel }\le 2\mathrm{\parallel }{A}_{\epsilon }{\mathrm{\parallel }}^2$. Thus $\mathrm{\parallel }{A}_{\epsilon }\mathrm{\parallel }\ge \mathrm{\parallel }{u}_{\epsilon }{\mathrm{\parallel }}_2\mathrm{/}\sqrt{2}\to \mathrm{\infty }$, excluding a bounded equal-time point-field norm limit.