Local covariant physical field-strength observables

Antisymmetry gives $\frac{1}{2}\int ({\mathrm{\partial }}_a{A}_b-{\mathrm{\partial }}_b{A}_a)h^{ab}{d}^{4}x=\int {\mathrm{\partial }}_a{A}_b{h}^{ab}{d}^{4}x=-\int A_b{\mathrm{\partial }}_a{h}^{ab}{d}^{4}x,$ where integration by parts has no boundary term because $h$ is compactly supported. The current $j^b=-{\mathrm{\partial }}_a{h}^{ab}$ obeys ${\mathrm{\partial }}_b{j}^b=-{\mathrm{\partial }}_b{\mathrm{\partial }}_a{h}^{ab}=0$, since exchanging $a,b$ changes the sign of $h$. Adding ${\mathrm{\partial }}_b\lambda$ to $A_b$ changes the last integral by $\int \lambda {\mathrm{\partial }}_b{\mathrm{\partial }}_a{h}^{ab}=0$. The spatial components of $j$ are compact smooth, so [F1] supplies the vector $g_h=g_j$ and the stated finite-particle operator.

For a class $[ϵ]\in p^{\perp }\mathrm{/}\mathbb{C}p$, let $T_p[ϵ]=\mathit{ϵ}-(\mathbf{k}\mathrm{/}\omega ){ϵ}^0$ as in [F2]. The representative $\widetilde{ϵ}=(0,T_p[ϵ])$ has the same class, since $ϵ-\widetilde{ϵ}=({ϵ}^0\mathrm{/}\omega )p$. Thus we may use a transverse representative $\widetilde{ϵ}=(0,\mathit{ϵ})$. With $p_a=(-\omega ,\mathbf{k})$ and Fourier phase $e^{-i{p}_a{x}^a}$, compact support and integration by parts give $\widehat{j^b}(p)=-i{p}_a{\widehat{h}}^{ab}(p)$ for $j^b=-{\mathrm{\partial }}_a{h}^{ab}$. Its contraction with ${ϵ}_b$ is therefore $-i{\widehat{h}}^{ab}{p}_a{ϵ}_b=-\frac{i}{2}{\widehat{h}}^{ab}(p_a{ϵ}_b-p_b{ϵ}_a)$ by antisymmetry of $h$. Replacing $ϵ$ by $ϵ+cp$ leaves this expression unchanged because $p_a{p}_b-p_b{p}_a=0$. Under a Lorentz transformation both $p$ and $ϵ$ transform linearly, so their antisymmetric product transforms as a two-form. The unitary quotient action [F2] therefore intertwines the smeared tensor test with its Lorentz pullback, proving covariance without choosing a global polarization frame.

The mode formula in [F1] and the CCR give, after changing $\mathbf{k}$ to $-\mathbf{k}$ in the negative-frequency term, $[A_i(x),A_j(y)]=({\delta }_{ij}-\frac{k_i{k}_j}{\mathrm{\mid }\mathbf{k}{\mathrm{\mid }}^2})C_0(x-y)$ as the spatial Fourier multiplier distribution. Differentiating, using ${\mathrm{\partial }}_{x^0}{\mathrm{\partial }}_{y^0}=-{\mathrm{\partial }}_0^2$, gives $[E_i,E_j]=-{\mathrm{\partial }}_0^2{P}_{T,ij}{C}_0$. The massless equation ${\mathrm{\partial }}_0^2{C}_0=\mathrm{\Delta }{C}_0$ follows by differentiating the sine-over-$\mathrm{\mid }\mathbf{k}\mathrm{\mid }$ integral in [F3]. The multiplier of $-\mathrm{\Delta }{P}_{T,ij}$ is ${\delta }_{ij}\mathrm{\mid }\mathbf{k}{\mathrm{\mid }}^2-k_i{k}_j$, which is exactly the multiplier of $-{\delta }_{ij}\mathrm{\Delta }+{\mathrm{\partial }}_i{\mathrm{\partial }}_j$; this proves the stated electric commutator. For magnetic fields, each curl kills the longitudinal multiplier $k_s{k}_u\mathrm{/}\mathrm{\mid }\mathbf{k}{\mathrm{\mid }}^2$, and ${ϵ}_{irs}{ϵ}_{jts}{k}_r{k}_t={\delta }_{ij}\mathrm{\mid }\mathbf{k}{\mathrm{\mid }}^2-k_i{k}_j$ gives the same expression. For the mixed commutator, translation invariance gives ${\mathrm{\partial }}_{y^r}=-{\mathrm{\partial }}_{x^r}$, so $[E_i(x),B_j(y)]={ϵ}_{jrs}{\mathrm{\partial }}_0{\mathrm{\partial }}_r{P}_{T,is}{C}_0\mathrm{.}$ The longitudinal term is zero because ${ϵ}_{jrs}{k}_r{k}_s=0$, and relabeling the remaining indices gives ${ϵ}_{jri}{\mathrm{\partial }}_0{\mathrm{\partial }}_r{C}_0$.

If a test function is supported in the complement of $\mathrm{supp}{C}_0$, each of its derivatives has support contained in that of the test function, and the definition $⟨{\mathrm{\partial }}_a{C}_0,f⟩=-⟨C_0,{\mathrm{\partial }}_{a}f⟩$ shows the derivative distribution also vanishes there. Thus every derivative in step 2.2 remains supported in the closed causal cone from [F3]. For spacelike-separated compact test supports, every point in their difference set lies outside that cone, so pairing each commutator with their tensor product gives zero. The smeared field strengths commute on ${\mathcal{D}}_{\mathrm{f}\mathrm{i}\mathrm{n}}$.