Explicit formal QED vertex/free-line convention — Examples

For $\mathbf{n}\in S^2$, set $b(\mathbf{n})=\lambda (I_3-\mathbf{n}{\mathbf{n}}^T)\mathbf{e}$. Then $\mathbf{n}\cdot b(\mathbf{n})=0$. Since $\mathbf{e}$ is unit, $\mathrm{\mid }b(\mathbf{n}){\mathrm{\mid }}^2={\lambda }^2(1-(\mathbf{n}\cdot \mathbf{e}{)}^2)$. The function is continuous on $S^2$ and hence measurable.

By the orthogonal invariance in [F4], take $\mathbf{e}$ as the third coordinate axis and put $u=\mathbf{n}\cdot \mathbf{e}$. The chart $X(u,\varphi )=(\sqrt{1-u^2}\cos \varphi ,\sqrt{1-u^2}\sin \varphi ,u)$ for $-1<u<1$, $0<\varphi <2\pi$ has orthogonal tangent vectors with squared lengths $(1-u^2{)}^{-1}$ and $1-u^2$, so [F5] gives area density $1$. The omitted seam is a coordinate line in sphere graph charts and the poles are singletons; [F7] makes their chart integrals zero. Thus spherical integration gives $C_b={\int }_{S^2}\mathrm{\mid }b(\mathbf{n}){\mathrm{\mid }}^{2}d\mathrm{\Omega }=2\pi {\lambda }^2{\int }_{-1}^1(1-u^2)du=\frac{8\pi {\lambda }^2}{3}>0.$ Using $d^3\mathbf{k}=r^{2}drd\mathrm{\Omega }$ and $\mathrm{\mid }{f}_{ϵ}(r\mathbf{n}){\mathrm{\mid }}^2=r^{-3}\mathrm{\mid }b(\mathbf{n}){\mathrm{\mid }}^2$ gives $\mathrm{\parallel }{f}_{ϵ}{\mathrm{\parallel }}^2=C_b{\int }_{ϵ}^{\kappa }\frac{dr}{r}=C_b\log (\kappa \mathrm{/}ϵ),{\int }_{{\mathbb{R}}^3}\mathrm{\hslash }c\mathrm{\mid }\mathbf{k}\mathrm{\mid }\mathrm{\mid }{f}_{ϵ}(\mathbf{k}){\mathrm{\mid }}^2{d}^3\mathbf{k}=\mathrm{\hslash }c{C}_b{\int }_{ϵ}^{\kappa }dr=\mathrm{\hslash }c{C}_b(\kappa -ϵ)\mathrm{.}$ Both integrals are finite because $0<ϵ<\kappa$. The profile has units ${\mathrm{m}}^{3\mathrm{/}2}$ because $\mathrm{\mid }\mathbf{k}{\mathrm{\mid }}^{-3\mathrm{/}2}$ has those units and $b$ is dimensionless; the energy integral has units joules because $\mathrm{\hslash }c$ has units joule-metres. Thus $b$ satisfies the nonzero transverse-profile hypothesis of [L2] with the conventions [F1].

The definition of $b$ gives $f_{ϵ}(\mathbf{k})={\mathbf{1}}_{ϵ<\mathrm{\mid }\mathbf{k}\mathrm{\mid }<\kappa }\mathrm{\mid }\mathbf{k}{\mathrm{\mid }}^{-3\mathrm{/}2}b(\mathbf{k}\mathrm{/}\mathrm{\mid }\mathbf{k}\mathrm{\mid })$. Applying [L2] with $C_b=8\pi {\lambda }^2\mathrm{/}3$ and the Hamiltonian of [L1] gives the normalized coherent vector, its energy-domain membership and energy expectation, weak convergence to zero, and failure of strong Fock-space convergence. Put $r=\mathrm{\parallel }{f}_{ϵ}{\mathrm{\parallel }}^2$. By [F3], its squared sector norms are $e^{-r}{r}^n\mathrm{/}n!$. The Fock sector norm formula [F3] and normalization from [L2] give $1=\mathrm{\parallel }C(f_{ϵ}){\mathrm{\parallel }}^2=e^{-r}{\sum }_{m\ge 0}{r}^m\mathrm{/}m!$. The maximal number multiplier from [F6] has domain condition ${\sum }_{n\ge 1}{n}^2\mathrm{\parallel }C(f_{ϵ}{)}_n{\mathrm{\parallel }}^2<\mathrm{\infty }$. Since $n^2=n(n-1)+n$, reindexing the two nonnegative sums gives $e^{-r}{\sum }_{n\ge 1}{n}^2\frac{r^n}{n!}=(r^2+r)e^{-r}{\sum }_{m\ge 0}\frac{r^m}{m!}=r^2+r<\mathrm{\infty }\mathrm{.}$ Thus $C(f_{ϵ})$ belongs to the number-operator domain. The transverse restriction preserves this domain because each sector multiplier $nI$ preserves every transverse symmetric sector. On the same sectors, the number expectation is $⟨C(f_{ϵ}),NC(f_{ϵ})⟩=e^{-r}{\sum }_{n\ge 1}n\frac{r^n}{n!}=r{e}^{-r}{\sum }_{m\ge 0}\frac{r^m}{m!}=r\mathrm{\parallel }C(f_{ϵ}){\mathrm{\parallel }}^2=r,$ by reindexing and the Fock sector norm formula [F3]; [L2] gives $\mathrm{\parallel }C(f_{ϵ})\mathrm{\parallel }=1$. By step 2.1, $r=C_{\lambda }\log (\kappa \mathrm{/}ϵ)$, which is the claimed photon-number expectation. The item-level suppliers are present in the active manifests, but [L1] has no Step-1 record and [L2] is escalated, so this application remains conditional rather than accepted.

Let ${ϵ}_j=\kappa 2^{-j}$ for $j\ge 1$. By step 3.1 the difference $f_{{ϵ}_{j+1}}-f_{{ϵ}_j}$ is supported on ${ϵ}_{j+1}<\mathrm{\mid }\mathbf{k}\mathrm{\mid }<{ϵ}_j$, so spherical integration gives $\mathrm{\parallel }{f}_{{ϵ}_{j+1}}-f_{{ϵ}_j}{\mathrm{\parallel }}^2=C_b{\int }_{{ϵ}_{j+1}}^{{ϵ}_j}\frac{dr}{r}=C_b\log 2>0.$ From the Fock direct-sum pairing in [F3], the tensor product pairing on sector $n$ gives $⟨C(f),C(g)⟩=e^{-(\mathrm{\parallel }f{\mathrm{\parallel }}^2+\mathrm{\parallel }g{\mathrm{\parallel }}^2)\mathrm{/}2}{\sum }_{n=0}^{\mathrm{\infty }}\frac{⟨f,g{⟩}^n}{n!}=e^{-(\mathrm{\parallel }f{\mathrm{\parallel }}^2+\mathrm{\parallel }g{\mathrm{\parallel }}^2)\mathrm{/}2+⟨f,g⟩},$ where the series is absolutely convergent by $\mathrm{\mid }⟨f,g⟩\mathrm{\mid }\le \mathrm{\parallel }f\mathrm{\parallel }\mathrm{\parallel }g\mathrm{\parallel }$. Hence $\mathrm{\mid }⟨C(f),C(g)⟩\mathrm{\mid }=e^{-\mathrm{\parallel }f-g{\mathrm{\parallel }}^2\mathrm{/}2}$. By [L2], $\mathrm{\parallel }C(f_{{ϵ}_j})\mathrm{\parallel }=1$ for each $j$. For any choices of phases ${\alpha }_j\in \mathbb{R}$, the unit vectors $v_j=e^{i{\alpha }_j}C(f_{{ϵ}_j})$ satisfy $\mathrm{\parallel }{v}_{j+1}-v_j{\mathrm{\parallel }}^2=2-2\mathrm{Re}⟨v_j,v_{j+1}⟩\ge 2(1-e^{-C_b\log 2\mathrm{/}2})>0.$ This bound holds for every phase sequence, so none is Cauchy. Any phase-adjusted norm limit as $ϵ\downarrow 0$ would restrict to a Cauchy sequence along ${ϵ}_j$; therefore no such limit exists.

By [L2], each vector has norm one and the weak limit is zero. Since the zero vector has norm zero, it is not a normalized state; the nonexistence of a norm limit follows from steps 3.1 and 4.1.

By [F1], the physical sector and the regional charge $Q_R$ are those of the exact finite regulator. The supplier [F2] gives the coefficient identity $Q_R\psi ={\sum }_{e\in \mathrm{\partial }R}ϵ(e,R)E_e\psi$ on ${\mathcal{D}}_R$: summing the Gauss multipliers over $x\in R$ cancels every link with both endpoints in $R$, leaving only the signed boundary links.

The hypotheses on $O$ match [F2], so $O\psi$ remains physical and in all boundary electric domains. Therefore $Q_{R}O\psi ={\sum }_{e\in \mathrm{\partial }R}ϵ(e,R)E_{e}O\psi =O{\sum }_{e\in \mathrm{\partial }R}ϵ(e,R)E_e\psi =O{Q}_R\psi ,$ which proves $[Q_R,O]\psi =0$ on ${\mathcal{D}}_R$.

If also $[Q_R,O]\psi =qO\psi$ with fixed $q\mathrm{\ne }0$, step 2.1 gives $qO\psi =0$ for every $\psi \in {\mathcal{D}}_R$. Division by the nonzero scalar $q$ yields $O\psi =0$ there; no conclusion about vectors outside this domain is needed.