Smeared Coulomb photon operator distribution

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.