Soft-current angular norm and exact cutoff coherent model

For every leg and $n\in S^2$, $1-v_i\cdot n\ge 1-\mathrm{\mid }{v}_i\mathrm{\mid }>0$. Since the leg set is finite, each summand of $K(n)$ is continuous and bounded on $S^2$. Moreover $h(K(n),(1,n))={\sum }_i{\eta }_i{Q}_i\frac{1-v_i\cdot n}{1-v_i\cdot n}={\sum }_i{\eta }_i{Q}_i=0.$ Thus $K^0(n)=n\cdot \mathbf{K}(n)$ and $h(K,K)=(K^0{)}^2-\mathrm{\mid }\mathbf{K}{\mathrm{\mid }}^2=-\mathrm{\mid }{P}_T(n)\mathbf{K}(n){\mathrm{\mid }}^2$. Consequently the angular integrand is nonnegative, bounded and measurable; the coefficient $C$ is finite and nonnegative.

By [F10] at $p=2$, the real-valued scalar space $L^2({\mathbb{R}}^3;\mathbb{R})$ has a countable dense set $D$. It is nonempty because this scalar $L^2$ space is nonempty and $D$ is dense. By [F12], $D\times D$ is at most countable; [F14] supplies a surjection from $\mathbb{N}$ onto this nonempty product. Composing with $(d,e)\mapsto d+ie$ and applying [F14] again shows that $D_{\mathbb{C}}:=\{d+ie:d,e\in D\}$ is at most countable. For every complex scalar $u\in L^2$, its real and imaginary parts $u_R,u_I$ belong to real $L^2$, and $\mathrm{\parallel }u-(d+ie){\mathrm{\parallel }}_2^2=\mathrm{\parallel }{u}_R-d{\mathrm{\parallel }}_2^2+\mathrm{\parallel }{u}_I-e{\mathrm{\parallel }}_2^2$; density of $D$ therefore makes $D_{\mathbb{C}}$ dense in complex scalar $L^2$. By two applications of [F12], $D_{\mathbb{C}}^3$ is countable and dense in $L^2({\mathbb{R}}^3;{\mathbb{C}}^3)$, since each of three components can be approximated with squared error below ${\epsilon }^2\mathrm{/}3$. For $n\in S^2$ and $z\in {\mathbb{C}}^3$, put $a={\sum }_j{n}_j{z}_j$. Using [F13] and the reality of $n_j$, direct expansion gives $\mathrm{\parallel }{P}_T(n)z{\mathrm{\parallel }}^2={\sum }_j\mathrm{\mid }{z}_j-n_{j}a{\mathrm{\mid }}^2={\sum }_j\mathrm{\mid }{z}_j{\mathrm{\mid }}^2-\mathrm{\mid }a{\mathrm{\mid }}^2\le \mathrm{\parallel }z{\mathrm{\parallel }}^2$, and $(I-n{n}^T{)}^2=I-n{n}^T$. Thus $M_{P_T}$ is a bounded idempotent on the ambient vector-valued $L^2$ space. Its image of $D_{\mathbb{C}}^3$ is countable and dense in its range: for $z=M_{P_T}z$ choose $w\in D_{\mathbb{C}}^3$ close to $z$, and $\mathrm{\parallel }{M}_{P_T}w-z{\mathrm{\parallel }}_2\le \mathrm{\parallel }w-z{\mathrm{\parallel }}_2$.

By [F6] and [F7], the $d\mathrm{\Omega }$ used in the statement is the finite Euclidean surface measure, and [F8] gives $d^{3}q\mathrm{/}((2\pi {)}^{3}2\omega )=\omega d\omega d\mathrm{\Omega }\mathrm{/}(2(2\pi {)}^3)$. Multiplying by $\mathrm{\mid }{h}_{\lambda }(\mathbf{q}){\mathrm{\mid }}^2=g^2\mathrm{\mid }{K}_{\perp }(n){\mathrm{\mid }}^2\mathrm{/}{\omega }^2$ and integrating over $\lambda <\omega <\mathrm{\Lambda }$ gives $\mathrm{\parallel }{h}_{\lambda }{\mathrm{\parallel }}^2=\frac{g^2}{2(2\pi {)}^3}{\int }_{S^2}\mathrm{\mid }{K}_{\perp }{\mathrm{\mid }}^{2}d\mathrm{\Omega }{\int }_{\lambda }^{\mathrm{\Lambda }}\frac{d\omega }{\omega }=C\log (\mathrm{\Lambda }\mathrm{/}\lambda )\mathrm{.}$ The substitution $\omega =\lambda u$ and [F15] give the displayed logarithm. The cutoffs are strictly positive and finite, so the norm is finite. The endpoint set $\{\lambda ,\mathrm{\Lambda }\}$ is null by [F11], so the strict cutoff convention gives the same integral as including either endpoint.

Put $a=\mathrm{\hslash }c>0$ and $T(k)=ak$. For $h\in L^2({\mathbb{R}}^3,d^{3}q\mathrm{/}[(2\pi {)}^{3}2\mathrm{\mid }q\mathrm{\mid }];{\mathbb{C}}^3)$, define $G_h(q)=\mathrm{\mid }h(q){\mathrm{\mid }}^2\mathrm{/}[(2\pi {)}^{3}2\mathrm{\mid }q\mathrm{\mid }]$ for $q\mathrm{\ne }0$ and $G_h(0)=0$; the weighted $L^2$ hypothesis gives $G_h\in L^1({\mathbb{R}}^3,d^{3}q)$. The density of the energy measure is positive and finite off $q=0$, so every energy-measure null set is Lebesgue null there; the singleton $\{0\}$ is also Lebesgue null. By [F9], $T^{-1}$ preserves Lebesgue-null sets, so composition by $T$ is well-defined on these $L^2$ equivalence classes. Define $Uh(0)=0$ and, for $k\mathrm{\ne }0$, $(Uh)(k)=\frac{a}{\sqrt{(2\pi {)}^{3}2\mathrm{\mid }k\mathrm{\mid }}}h(ak)\mathrm{.}$ The change-of-variables formula [F9] for the $L^1$ function $G_h$ gives $\mathrm{\parallel }Uh{\mathrm{\parallel }}_{L^2(d^{3}k)}^2=a^3{\int }_{{\mathbb{R}}^3}{G}_h(ak)d^{3}k={\int }_{{\mathbb{R}}^3}{G}_h(q)d^{3}q=\mathrm{\parallel }h{\mathrm{\parallel }}^2\mathrm{.}$ Conversely, for $f\in L^2(d^{3}k;{\mathbb{C}}^3)$ define $Vf(0)=0$ and $Vf(q)=\sqrt{(2\pi {)}^{3}2\mathrm{\mid }q\mathrm{\mid }}{a}^{-3\mathrm{/}2}f(q\mathrm{/}a)$ for $q\mathrm{\ne }0$. Applying [F9] to $\mathrm{\mid }f{\mathrm{\mid }}^2$ and the inverse dilation shows $\mathrm{\parallel }Vf\mathrm{\parallel }=\mathrm{\parallel }f\mathrm{\parallel }$; direct substitution gives $UVf=f$ and $VUh=h$ almost everywhere. Thus $U$ is a unitary on the ambient spaces. Since $q=ak$ and the scalar multipliers do not change the transverse condition, with null sets preserved under $T$ and $T^{-1}$ the equivalence $q\cdot h(q)=0$ a.e. iff $k\cdot (Uh)(k)=0$ a.e. follows. Therefore $U$ restricts to a unitary ${\mathcal{H}}_{E,T}\to {\mathcal{H}}_T$; in particular the energy-coordinate transverse subspace is a Hilbert space because [F2] makes ${\mathcal{H}}_T$ a closed Hilbert space. Substitution of $h_{\lambda }$ gives $(U{h}_{\lambda })(k)={\mathbf{1}}_{\{\lambda \mathrm{/}(\mathrm{\hslash }c)<\mathrm{\mid }k\mathrm{\mid }<\mathrm{\Lambda }\mathrm{/}(\mathrm{\hslash }c)\}}\mathrm{\mid }k{\mathrm{\mid }}^{-3\mathrm{/}2}\frac{g{K}_{\perp }(k\mathrm{/}\mathrm{\mid }k\mathrm{\mid })}{\sqrt{2(2\pi {)}^3}},$ whose angular coefficient has squared integral $C$ by the definition of $C$.

Put $N=N_{\lambda }$. By [F3] and [F18], the squared norm in sector $m$ of $ϵ(f_{\lambda })$ is $N^m\mathrm{/}m!$, with $N^0=1$, and the exponential series [F17] gives total squared norm $\exp (N)$. Step 1.2 establishes that ${\mathcal{H}}_T$ is separable, as required by [F4]. Setting $h=-i\sqrt{2}{f}_{\lambda }$ and $z=0$ in [F4] yields $W(h)\mathrm{\Omega }=\exp (-N\mathrm{/}2)ϵ(f_{\lambda })$, so ${\mathrm{\Psi }}_{\lambda }$ is normalized and its sector squared norms are $s_m=\exp (-N)N^m\mathrm{/}m!$. The exponential series [F17] and index shifts give ${\sum }_m{s}_m=1$ and ${\sum }_{m}m{s}_m=N$. These are norm identities only; no measurement interpretation is used.

On the support, the one-photon energy ${\omega }_{\gamma }(k)=\mathrm{\hslash }c\mathrm{\mid }k\mathrm{\mid }$ satisfies $0<{\omega }_{\gamma }<\mathrm{\Lambda }$. By [F5] and [L1], the energy on sector $m$ is the sum of its $m$ one-photon energies, bounded above by $m\mathrm{\Lambda }$. The exponential series [F17] gives ${\sum }_m{m}^2{s}_m=N^2+N<\mathrm{\infty }$, so ${\sum }_m\mathrm{\parallel }(H_{\gamma }{\mathrm{\Psi }}_{\lambda }{)}_m{\mathrm{\parallel }}^2\le {\mathrm{\Lambda }}^2(N^2+N)$ and the vector lies in the operator domain. For each $m\ge 1$, the sector quadratic form of the unnormalized tensor $f_{\lambda }^{\otimes m}\mathrm{/}\sqrt{m!}$ is $m⟨f_{\lambda },{\omega }_{\gamma }{f}_{\lambda }⟩N^{m-1}\mathrm{/}m!$. Summing with the factor $\exp (-N)$ gives $⟨{\mathrm{\Psi }}_{\lambda },H_{\gamma }{\mathrm{\Psi }}_{\lambda }⟩=⟨f_{\lambda },{\omega }_{\gamma }{f}_{\lambda }⟩$. In energy coordinates the latter integral is $\frac{g^2}{2(2\pi {)}^3}{\int }_{S^2}\mathrm{\mid }{K}_{\perp }{\mathrm{\mid }}^{2}d\mathrm{\Omega }{\int }_{\lambda }^{\mathrm{\Lambda }}d\omega =C(\mathrm{\Lambda }-\lambda )\mathrm{.}$ This is a free-model Hilbert-space expectation in joules, not an empirical measurement.

If $C=0$, [F16] applied to the nonnegative integrand defining $C$ gives $K_{\perp }=0$ almost everywhere, so $h_{\lambda }=f_{\lambda }=0$ as Hilbert-space vectors and ${\mathrm{\Psi }}_{\lambda }=\mathrm{\Omega }$. If $C>0$, the function $b=g{K}_{\perp }\mathrm{/}\sqrt{2(2\pi {)}^3}$ is a nonzero transverse member of $L^2(S^2;{\mathbb{C}}^3)$, with squared norm integral $C$. Its cutoffs $ϵ=\lambda \mathrm{/}(\mathrm{\hslash }c)$ and $\kappa =\mathrm{\Lambda }\mathrm{/}(\mathrm{\hslash }c)$ satisfy $0<ϵ<\kappa$, and step 3.1 identifies its exact wavefunction with [L1]'s current family. Conditional on [L1], its weak-zero and no-strong-limit claims apply, and its energy formula agrees with step 5.1 because $\mathrm{\hslash }cC(\kappa -ϵ)=C(\mathrm{\Lambda }-\lambda )$.

For each integer $j\ge 1$ put ${\lambda }_j=\mathrm{\Lambda }{2}^{-j}$, ${ϵ}_j={\lambda }_j\mathrm{/}(\mathrm{\hslash }c)=\kappa 2^{-j}$, $f_j=f_{{\lambda }_j}$ and ${\mathrm{\Psi }}_j={\mathrm{\Psi }}_{{\lambda }_j}$. The increment $f_{j+1}-f_j$ is supported on the disjoint shell ${ϵ}_{j+1}<\mathrm{\mid }k\mathrm{\mid }<{ϵ}_j$, so its squared norm is $C\log 2>0$, since [F15] gives $\log 2={\int }_1^{2}du\mathrm{/}u\ge 1\mathrm{/}2>0$. Also $⟨f_j,f_{j+1}⟩=\mathrm{\parallel }{f}_j{\mathrm{\parallel }}^2$, because the increment is orthogonal to $f_j$. By [F4] and the normalization in step 4.1, $\mathrm{\mid }⟨{\mathrm{\Psi }}_j,{\mathrm{\Psi }}_{j+1}⟩\mathrm{\mid }=\exp (-C\log 2\mathrm{/}2)<1$ by [F19]. For every $z\in \mathbb{C}$ with $\mathrm{\mid }z\mathrm{\mid }=1$, $\mathrm{\parallel }{\mathrm{\Psi }}_{j+1}-z{\mathrm{\Psi }}_j{\mathrm{\parallel }}^2=2-2\mathrm{Re}(z⟨{\mathrm{\Psi }}_{j+1},{\mathrm{\Psi }}_j⟩)\ge 2(1-\exp (-C\log 2\mathrm{/}2))>0.$ Taking the infimum over $z$ gives $d([{\mathrm{\Psi }}_j],[{\mathrm{\Psi }}_{j+1}])\ge \sqrt{2(1-\exp (-C\log 2\mathrm{/}2))}>0$. If the ray sequence converged, the triangle inequality for $d$ would force these adjacent distances to tend to zero, a contradiction. This fills the interface difference in [L1] without asserting that its current statement contains a projective-ray result.