Charged gauge data and SI-natural conventions

For each $\mathbf{k}\mathrm{\ne }0$, $n^{T}n=n\cdot n=1$ by [F2], so $P_T^{\ast }=P_T$ and $P_T^2=(I-n{n}^T{)}^2=I-2n{n}^T+n(n^{T}n)n^T=I-n{n}^T=P_T\mathrm{.}$ Its complex kernel is ${\mathrm{span}}_{\mathbb{C}}\{n\}$ and its range is the Hermitian perpendicular plane $n^{\perp }$, of complex dimension two. For $z\in {\mathbb{C}}^3$, put $a=n^{T}z$. The specified Hermitian pairing and realness of $n$ give $\mathrm{\parallel }{P}_{T}z{\mathrm{\parallel }}^2={\sum }_j\mathrm{\mid }{z}_j-n_{j}a{\mathrm{\mid }}^2=\mathrm{\parallel }z{\mathrm{\parallel }}^2-\mathrm{\mid }a{\mathrm{\mid }}^2\le \mathrm{\parallel }z{\mathrm{\parallel }}^2$. Thus it is contractive, with norm one because it fixes every unit vector in that nonzero perpendicular plane. The arbitrary value at the single point $\mathbf{k}=0$ does not affect its multiplication operator on $L^2$.

Under [A2], [A1] makes each complex scalar $L^2$ component complete for its first-variable-linear pairing $(u,v{)}_1=\int u\bar{v}$. The three-component direct sum is complete componentwise. Its second-variable-linear pairing is $⟨u,v{⟩}_2:=(v,u{)}_1=\int \bar{u}\cdot v$, with the same norm; [F1] fixes this convention. Thus $\mathcal{H}$ is complete. The multiplier $M_{P_T}$ is bounded and idempotent by step 1.1. If $u_r=M_{P_T}{v}_r\to u$ in $\mathcal{H}$, then $M_{P_T}u={\lim }_r{M}_{P_T}{u}_r={\lim }_r{u}_r=u$ by boundedness, so $u$ belongs to its range. Thus ${\mathcal{H}}_T$ is closed and inherits a positive Hilbert inner product. Since $P_{T}u=u$ is equivalent to $n\cdot u=0$ for unit $n$ by [F2], the pointwise range formula gives the displayed transverse condition almost everywhere.

The Clifford and adjoint conventions in the definition are the exact ones in [A1]; in particular, no change from $\eta =(-,+,+,+)$ to the slash-matrix convention $-\eta$ is made when raising the indices on $F_{ab}$. The supplier's Countable Choice hypothesis is carried explicitly in [A2].

Since ${\mu }_0$ has units kg m C${}^{-2}$ and $\mathrm{\hslash }c$ has units kg m${}^3$ s${}^{-2}$, $S_0$ has units kg m${}^2$(C s)${}^{-1}$. Dividing $A_a$, whose units are kg m(C s)${}^{-1}$, by $S_0$ gives $[a_a]=$m${}^{-1}$. Also $[\mu ]=[mc\mathrm{/}\mathrm{\hslash }]=$m${}^{-1}$ and $[{\mu }_{0}c\mathrm{/}\mathrm{\hslash }]=$C${}^{-2}$, so $[g]=1$. A derivative has units m${}^{-1}$; hence both terms in $D_a$ have units m${}^{-1}$. The two displayed Lagrangian densities therefore have units m${}^{-4}$ and their spacetime integral is dimensionless.

Under the stated transformation, $D_a^{\mathrm{\prime }}{\psi }^{\mathrm{\prime }}=({\mathrm{\partial }}_a-ig(a_a+{\mathrm{\partial }}_a\alpha ))(e^{ig\alpha }\psi )=e^{ig\alpha }({\mathrm{\partial }}_a\psi +ig({\mathrm{\partial }}_a\alpha )\psi -ig{a}_a\psi -ig({\mathrm{\partial }}_a\alpha )\psi )=e^{ig\alpha }{D}_a\psi \mathrm{.}$ Thus ${\bar{\psi }}^{\mathrm{\prime }}(i{\gamma }^a{D}_a^{\mathrm{\prime }}-\mu ){\psi }^{\mathrm{\prime }}=\bar{\psi }(i{\gamma }^a{D}_a-\mu )\psi$, since $\overline{{\psi }^{\mathrm{\prime }}}=e^{-ig\alpha }\bar{\psi }$. Further, $f_{ab}^{\mathrm{\prime }}=f_{ab}$ because ${\mathrm{\partial }}_a{\mathrm{\partial }}_b\alpha ={\mathrm{\partial }}_b{\mathrm{\partial }}_a\alpha$. These are algebraic covariance identities for smooth fields.

If such a frame existed, then $e_1(n)\in n^{\perp }$ for every $n$ by [F1], and $\mathrm{\mid }{e}_1(n)\mathrm{\mid }=1$ by orthonormality. Therefore $e_1$ is a continuous nowhere-zero tangent vector field on $S^2$.

Since $2$ is a positive even integer, [F2] forbids the vector field produced in step 1.1. This contradiction proves the nonexistence of the global frame.

The map $Q:\mathcal{K}\to L^2({\mathbb{R}}^3)$ defined by $Qϵ=-{ϵ}^0+n\cdot \mathit{ϵ}$ is bounded because $\mathrm{\mid }Qϵ{\mathrm{\mid }}^2\le 2(\mathrm{\mid }{ϵ}^0{\mathrm{\mid }}^2+\mathrm{\mid }\mathit{ϵ}{\mathrm{\mid }}^2)$ pointwise. Thus ${\mathcal{K}}_0=\ker Q$ is closed. For $ϵ\in {\mathcal{K}}_0$, $n\cdot \mathit{ϵ}={ϵ}^0$, so $n\cdot Tϵ=0$ and $Tϵ\in {\mathcal{H}}_T$ by [F1]. Also $\mathrm{\parallel }Tϵ{\mathrm{\parallel }}_2\le \mathrm{\parallel }\mathit{ϵ}{\mathrm{\parallel }}_2+\mathrm{\parallel }{ϵ}^0{\mathrm{\parallel }}_2$, proving boundedness.

Write $\mathit{ϵ}=Tϵ+n{ϵ}^0$ and $\mathit{\delta }=T\delta +n{\delta }^0$. The cross terms in ${\mathit{ϵ}}^{†}\mathit{\delta }$ vanish because each $T$ vector is orthogonal to $n$, while the remaining $n$ term is $\overline{{ϵ}^0}{\delta }^0$ and cancels the time component $-\overline{{ϵ}^0}{\delta }^0$ of $\eta$. Hence $[ϵ,\delta {]}_{\eta }=⟨Tϵ,T\delta {⟩}_2$. Given any $u\in {\mathcal{H}}_T$, $ϵ=(0,u)$ belongs to ${\mathcal{K}}_0$ and $Tϵ=u$, so $T$ is onto.

The equation $Tϵ=0$ is equivalent to $\mathit{ϵ}=n{ϵ}^0$, so $ϵ=b(1,n)$ with $b={ϵ}^0\in L^2$; each such vector is in ${\mathcal{K}}_0$. Step 2.1 shows these vectors pair to zero with every member of ${\mathcal{K}}_0$. Conversely, if $ϵ$ pairs to zero with all of ${\mathcal{K}}_0$, choose $\delta =(0,Tϵ)$, which lies in ${\mathcal{K}}_0$ by [F1]; then $0=[ϵ,\delta {]}_{\eta }=\mathrm{\parallel }Tϵ{\mathrm{\parallel }}_2^2$, so $Tϵ=0$. Thus the radical is exactly $\mathcal{N}$.

The induced map ${\mathcal{K}}_0\mathrm{/}\mathcal{N}\to {\mathcal{H}}_T$, $[ϵ]\mapsto Tϵ$, is well-defined, injective by the kernel calculation, onto by step 2.1, and preserves the inner product by that same step. It is therefore an isometry onto the complete Hilbert space ${\mathcal{H}}_T$, proving positivity and completeness of the quotient.

The integral in $g_j$ is the spacetime Fourier transform of a compactly supported smooth function, hence decreases faster than every inverse power of $\mathrm{\mid }(\omega ,\mathbf{k})\mathrm{\mid }$ by repeated integration by parts. Near $\mathbf{k}=0$, it is bounded by $\int \mathrm{\mid }j\mathrm{\mid }$, while the squared prefactor is $1\mathrm{/}(2\omega )=1\mathrm{/}(2\mathrm{\mid }\mathbf{k}\mathrm{\mid })$. The radial integral of this bound is a constant times ${\int }_0^1{r}^2\mathrm{/}rdr<\mathrm{\infty }$. At infinity the rapid Fourier decay and $\mathrm{\parallel }{P}_T\mathrm{\parallel }\le 1$ make the squared norm integrable. Also $P_T{g}_j=g_j$ almost everywhere. Therefore $g_j\in {\mathcal{H}}_T$.

The one-particle space ${\mathcal{H}}_T$ is also separable, although [F2] applies to every complex Hilbert space. Let $D$ be the countable dense subset of real $L^2({\mathbb{R}}^3)$ supplied by [F4]. Its sixfold product is countable by [F5]; identifying each tuple with a three-component complex function whose real and imaginary components lie in $D$ gives a dense subset of $\mathcal{H}$. Density follows by approximating the six real $L^2$ components, since the squared complex-vector norm is the sum of their squared norms. Hence $\mathcal{H}$ is separable in its norm topology by [F6]. The image of this countable set under the bounded projection $M_{P_T}$ is countable and dense in ${\mathcal{H}}_T$: for $u\in {\mathcal{H}}_T$, $M_{P_T}u=u$, so projected approximants to $u$ converge to $u$. Thus ${\mathcal{H}}_T$ is separable in its norm topology by [F6], and [F2] applies. By [F3], a vector in ${\mathcal{D}}_{\mathrm{f}\mathrm{i}\mathrm{n}}$ has only finitely many nonzero sectors. On each such sector [F2] bounds creation and annihilation by the corresponding square-root particle number times $\mathrm{\parallel }{g}_j\mathrm{\parallel }\mathrm{\parallel }{\mathrm{\Psi }}_n\mathrm{\parallel }$. Thus both maps and their sum are defined on ${\mathcal{D}}_{\mathrm{f}\mathrm{i}\mathrm{n}}$. The same estimates bound every matrix element between fixed finite-particle vectors by a constant times $\mathrm{\parallel }{g}_j\mathrm{\parallel }$. On each fixed compact support, repeated integration by parts in step 1.1 bounds $\mathrm{\parallel }{g}_j\mathrm{\parallel }$ by finitely many smooth-test seminorms, so these matrix elements are continuous in the test-function topology. This proves the asserted distributional smearing without defining a point field.

Write $\omega =\mathrm{\mid }\mathbf{k}\mathrm{\mid }$, and use the Fourier convention $\widehat{f}(\mathbf{k})={\int }_{{\mathbb{R}}^3}f(\mathbf{x})e^{-i\mathbf{k}\cdot \mathbf{x}}{d}^3\mathbf{x}$. The expansion implied by the definition of $g_j$ is $A_i(x^0,\mathbf{x})={\int }_{{\mathbb{R}}^3}\frac{d^3\mathbf{k}}{(2\pi {)}^{3\mathrm{/}2}(2\omega {)}^{1\mathrm{/}2}}(a_i(\mathbf{k})e^{-i\omega x^0+i\mathbf{k}\cdot \mathbf{x}}+a_i^{†}(\mathbf{k})e^{i\omega x^0-i\mathbf{k}\cdot \mathbf{x}}),$ $E_i(x^0,\mathbf{x})=-{\mathrm{\partial }}_0{A}_i(x^0,\mathbf{x})={\int }_{{\mathbb{R}}^3}\frac{d^3\mathbf{k}}{(2\pi {)}^{3\mathrm{/}2}}{\left(\frac{\omega }{2}\right)}^{1\mathrm{/}2}(i{a}_i(\mathbf{k})e^{-i\omega x^0+i\mathbf{k}\cdot \mathbf{x}}-i{a}_i^{†}(\mathbf{k})e^{i\omega x^0-i\mathbf{k}\cdot \mathbf{x}})\mathrm{.}$ Smearing the first line with $j\in C_c^{\mathrm{\infty }}({\mathbb{R}}^4;{\mathbb{R}}^3)$ gives exactly $a(g_j)+a^{†}(g_j)$: the annihilation term uses conjugate-linearity of $a$, while the creation term is linear in its argument. Here the formula is for $\mathbf{k}\mathrm{\ne }0$; values at $\mathbf{k}=0$ are irrelevant to the integrals. The symbols $a_i(\mathbf{k})$ and $a_i^{†}(\mathbf{k})$ are distributional notation for the Fock maps: for Schwartz vectors $u,v:{\mathbb{R}}^3\to {\mathbb{C}}^3$ set $a(u):=a(P_{T}u)$ and $a^{†}(v):=a^{†}(P_{T}v)$. Then [F2] gives $[a(u),a^{†}(v)]=\int \overline{u_i(\mathbf{k})}{P}_{T,ij}(\mathbf{k})v_j(\mathbf{k})d^3\mathbf{k}I$, equivalently the kernel notation $[a_i(\mathbf{k}),a_j^{†}(\mathbf{q})]=P_{T,ij}(\mathbf{k}){\delta }^3(\mathbf{k}-\mathbf{q})I$. The mode integrals are understood after test smearing; near $\mathbf{k}=0$ their squared one-particle weights behave respectively as $1\mathrm{/}\omega$ and $\omega$, both locally integrable against $d^3\mathbf{k}$, and at infinity the Fourier transforms of the spatial tests decrease rapidly.

For real $f,h\in \mathcal{S}({\mathbb{R}}^3;{\mathbb{R}}^3)$, define the equal-time spatial smearings by pairing the mode distributions with the spatial tests, $A(f):={\sum }_i{\int }_{{\mathbb{R}}^3}{f}_i(\mathbf{x})A_i(0,\mathbf{x})d^3\mathbf{x}$ and $E(h):={\sum }_j{\int }_{{\mathbb{R}}^3}{h}_j(\mathbf{y})E_j(0,\mathbf{y})d^3\mathbf{y}$. The one-particle vectors in these definitions are $g_f(\mathbf{k})=\frac{P_T(\mathbf{k})\widehat{f}(\mathbf{k})}{(2\pi {)}^{3\mathrm{/}2}(2\omega {)}^{1\mathrm{/}2}},e_h(\mathbf{k})=-i\frac{{\omega }^{1\mathrm{/}2}{P}_T(\mathbf{k})\widehat{h}(\mathbf{k})}{(2\pi {)}^{3\mathrm{/}2}{2}^{1\mathrm{/}2}}\mathrm{.}$ These belong to ${\mathcal{H}}_T$ by the preceding bounds, and $A(f)=a(g_f)+a^{†}(g_f)$, $E(h)=a(e_h)+a^{†}(e_h)$. The CCR [F2] gives $[A(f),E(h)]=-i{\int }_{{\mathbb{R}}^3}\frac{d^3\mathbf{k}}{(2\pi {)}^3}\overline{{\widehat{f}}_i(\mathbf{k})}{P}_{T,ij}(\mathbf{k}){\widehat{h}}_j(\mathbf{k})I\mathrm{.}$ For real $f,h$, this Fourier integral is real because $P_T$ is real and even and $\widehat{f}(-\mathbf{k})=\overline{\widehat{f}(\mathbf{k})}$, $\widehat{h}(-\mathbf{k})=\overline{\widehat{h}(\mathbf{k})}$. Equivalently, as a tempered distribution in $\mathbf{z}=\mathbf{x}-\mathbf{y}$, $[A_i(0,\mathbf{x}),E_j(0,\mathbf{y})]=-i{K}_{ij}(\mathbf{x}-\mathbf{y}),K_{ij}(\mathbf{z})=(2\pi {)}^{-3}{\int }_{{\mathbb{R}}^3}{P}_{T,ij}(\mathbf{k})e^{i\mathbf{k}\cdot \mathbf{z}}{d}^3\mathbf{k}\mathrm{.}$ The last integral defines a tempered distribution because $P_T$ is measurable and bounded: for $\phi \in \mathcal{S}({\mathbb{R}}^3)$ its pairing is $(2\pi {)}^{-3}\int P_{T,ij}(\mathbf{k})\widehat{\phi }(-\mathbf{k})d^3\mathbf{k}$, which is absolutely convergent. Since the Fourier symbol of $P_{T,ij}(-i\mathrm{\nabla })$ is $P_{T,ij}(\mathbf{k})$, this distribution is exactly $P_{T,ij}(-i{\mathrm{\nabla }}_{\mathbf{x}}){\delta }^3(\mathbf{x}-\mathbf{y})$. The sign is negative because $E=-{\mathrm{\partial }}_{0}A$: the annihilation/creation cross terms contribute $-i\mathrm{/}2$ each at equal time. Thus the stated equal-time commutator holds distributionally after smearing, with its stated sign.

For $\epsilon >0$, $m_{\epsilon }(\mathbf{k})=e^{-\epsilon \mathrm{\mid }\mathbf{k}\mathrm{\mid }}\mathrm{/}\mathrm{\mid }\mathbf{k}{\mathrm{\mid }}^2$ is integrable: [F3] and $\sigma (S^2)=4\pi$ give $\int \mathrm{\mid }{m}_{\epsilon }\mathrm{\mid }{d}^3\mathbf{k}=4\pi {\int }_0^{\mathrm{\infty }}{e}^{-\epsilon s}ds<\mathrm{\infty }$. Its inverse Fourier integral is radial. Rotate the polar axis to $\mathbf{x}$ using [F4], and apply the zonal identity separately to the real and imaginary parts of $u\mapsto e^{is\mathrm{\mid }\mathbf{x}\mathrm{\mid }u}$. The polar formula [F3] is stated for nonnegative Borel functions; since this complex integrand is absolutely integrable, apply it separately to the positive and negative parts of its real and imaginary parts and recombine by [F5]. The angular integral is $4\pi \sin (sr)\mathrm{/}(sr)$, where $r=\mathrm{\mid }\mathbf{x}\mathrm{\mid }$. Thus $G_{\epsilon }(r)=(2\pi {)}^{-3}\int e^{i\mathbf{k}\cdot \mathbf{x}}{m}_{\epsilon }(\mathbf{k})d^3\mathbf{k}=\frac{1}{2{\pi }^{2}r}{\int }_0^{\mathrm{\infty }}{e}^{-\epsilon s}\frac{\sin (sr)}{s}ds\mathrm{.}$ Since $\sin (sr)\mathrm{/}s={\int }_0^r\cos (st)dt$ and ${\int }_0^r{\int }_0^{\mathrm{\infty }}{e}^{-\epsilon s}\mathrm{\mid }\cos (st)\mathrm{\mid }dsdt\le r\mathrm{/}\epsilon$, Fubini [F7] applies. The real part of ${\int }_0^{\mathrm{\infty }}{e}^{-(\epsilon -it)s}ds=(\epsilon -it{)}^{-1}$ is $\epsilon \mathrm{/}({\epsilon }^2+t^2)$; integrating it over $t\in [0,r]$ gives $\arctan (r\mathrm{/}\epsilon )$. Therefore $G_{\epsilon }(r)=\frac{\arctan (r\mathrm{/}\epsilon )}{2{\pi }^{2}r}\mathrm{.}$

For every test $\phi \in C_c^{\mathrm{\infty }}({\mathbb{R}}^3)$ [F8], the bound $\mathrm{\mid }{G}_{\epsilon }(\mathbf{x})\mathrm{\mid }\le (4\pi \mathrm{\mid }\mathbf{x}\mathrm{\mid }{)}^{-1}$ and [F3] make $G_{\epsilon }\phi$ dominated by an integrable function; [F5] gives the distributional limit $G_{\epsilon }\to G$. To identify the same limit in Fourier variables, repeated integration by parts in a coordinate with $\mathrm{\mid }{k}_j\mathrm{\mid }\ge \mathrm{\mid }\mathbf{k}\mathrm{\mid }\mathrm{/}\sqrt{3}$ gives $\mathrm{\mid }\widehat{\phi }(\mathbf{k})\mathrm{\mid }\le C_N(1+\mathrm{\mid }\mathbf{k}\mathrm{\mid }{)}^{-N}$ for every integer $N$, while $\mathrm{\mid }\widehat{\phi }\mathrm{\mid }\le \mathrm{\parallel }\phi {\mathrm{\parallel }}_1$ near zero. Hence $\mathrm{\mid }\mathbf{k}{\mathrm{\mid }}^{-2}\widehat{\phi }(-\mathbf{k})$ is integrable by [F3]. Fubini [F7] identifies $\int G_{\epsilon }\phi$ with $(2\pi {)}^{-3}\int m_{\epsilon }(\mathbf{k})\widehat{\phi }(-\mathbf{k})d^3\mathbf{k}$. Since $\mathrm{\mid }{m}_{\epsilon }\mathrm{\mid }\le \mathrm{\mid }\mathbf{k}{\mathrm{\mid }}^{-2}$, [F5] shows this tends to the inverse Fourier pairing with $\mathrm{\mid }\mathbf{k}{\mathrm{\mid }}^{-2}$. Both limits are therefore the same distribution $G$, proving ${\mathcal{F}}^{-1}(\mathrm{\mid }\mathbf{k}{\mathrm{\mid }}^{-2})=G$. By [F1] this is the published normalized fundamental solution.

Integration by parts against compact smooth tests gives Fourier multiplier $-k_i{k}_j$ for ${\mathrm{\partial }}_i{\mathrm{\partial }}_{j}G$. Step 2.1 therefore gives ${\mathcal{F}}^{-1}\text{\hspace{0.17em}⁣}({\delta }_{ij}-\frac{k_i{k}_j}{\mathrm{\mid }\mathbf{k}{\mathrm{\mid }}^2})={\delta }_{ij}{\delta }^3+{\mathrm{\partial }}_i{\mathrm{\partial }}_{j}G\mathrm{.}$ For $\mathbf{x}\mathrm{\ne }0$, put $s=\mathrm{\mid }\mathbf{x}{\mathrm{\mid }}^2$ by [F6], so $G=(4\pi {)}^{-1}{s}^{-1\mathrm{/}2}$. The power, chain and sum rules [F7] give ${\mathrm{\partial }}_{i}G=-\frac{x_i}{4\pi s^{3\mathrm{/}2}},{\mathrm{\partial }}_i{\mathrm{\partial }}_{j}G=-\frac{{\delta }_{ij}}{4\pi r^3}+\frac{3x_i{x}_j}{4\pi r^5}\mathrm{.}$ Substitution into [F2] proves the stated equal-time commutator kernel.