Spacelike vacuum correlation without field commutator

Fix $\epsilon >0$ metres, let $L>4\epsilon$ metres, and take centers $X=(0,\mathbf{0})$ and $Y=(0,L{\mathbf{e}}_1)$ in the stated length coordinates. Apply [F3] with radii $\epsilon \mathrm{/}2$ and $\epsilon$ in ${\mathbb{R}}^4$, then translate the resulting bump to the two centers. This gives real dimensionless bumps ${\rho }_X,{\rho }_Y$ valued in $[0,1]$, each equal to $1$ on a nonempty inner ball and supported in $B_{\epsilon }(X)$ and $B_{\epsilon }(Y)$ respectively. Define $f=(1{\mathrm{m}}^{-3}){\rho }_X$ and $g=(1{\mathrm{m}}^{-3}){\rho }_Y$, so both tests have unit ${\mathrm{m}}^{-3}$. For any $x,y$ in these supports, $\mathrm{\mid }{x}^0-y^0\mathrm{\mid }\le 2\epsilon$ and $\mathrm{\mid }\mathbf{x}-\mathbf{y}\mathrm{\mid }\ge L-2\epsilon >2\epsilon$; hence every support difference is spacelike.

Put $r=\mathrm{\mid }\mathbf{z}\mathrm{\mid }$ and for $\delta >0$ metres damp the massless shell kernel by $e^{-\delta \mathrm{\mid }\mathbf{k}\mathrm{\mid }}$, so $w_{\delta }(z^0,\mathbf{z})={\int }_{{\mathbb{R}}^3}{e}^{-\delta \mathrm{\mid }\mathbf{k}\mathrm{\mid }}{e}^{-i(\mathrm{\mid }\mathbf{k}\mathrm{\mid }{z}^0-\mathbf{k}\cdot \mathbf{z})}{d}^3\mathbf{k}\mathrm{/}((2\pi {)}^{3}2\mathrm{\mid }\mathbf{k}\mathrm{\mid })$. By [F10] the spherical chart has surface element $d\sigma =\sin \theta d\theta d\phi$ and $\sigma (S^2)=4\pi$; combining this with [F7] gives $d^3\mathbf{k}=k^2\sin \theta dkd\theta d\phi$ and ${\nu }_0(\{\mathrm{\mid }\mathbf{k}\mathrm{\mid }<\kappa \})={\kappa }^2\mathrm{/}(8{\pi }^2)$ for $\kappa >0$ with units ${\mathrm{m}}^{-1}$; each ball has finite shell measure, so the shell measure is sigma-finite. For a support difference $z=x-y$, step 1.1 gives $r>0$. Rotate coordinates so the third axis is $\mathbf{z}\mathrm{/}r$; the determinant-one rotation preserves Lebesgue measure and $\mathrm{\mid }\mathbf{k}\mathrm{\mid }$ by [F8]. The angular integral is $2\pi {\int }_{-1}^1{e}^{ikru}du=4\pi \sin (kr)\mathrm{/}(kr)$ by [F10]. Thus $w_{\delta }=(4{\pi }^{2}r{)}^{-1}{\int }_0^{\mathrm{\infty }}{e}^{-(\delta +i{z}^0)k}\sin (kr)dk$. With $a=\delta +i{z}^0$ and $\mathrm{Re}a=\delta >0$, writing the sine as exponentials and using ${\int }_0^{\mathrm{\infty }}{e}^{-bk}dk=1\mathrm{/}b$ for $\mathrm{Re}b>0$ gives ${\int }_0^{\mathrm{\infty }}{e}^{-ak}\sin (kr)dk=(2i{)}^{-1}((a-ir{)}^{-1}-(a+ir{)}^{-1})=r\mathrm{/}(a^2+r^2)$. Therefore $w_{\delta }=1\mathrm{/}[4{\pi }^2(r^2+(\delta +i{z}^0{)}^2)]$.

For every $\delta >0$, the shell/spacetime integrand paired with $f(x)g(y)$ is absolutely integrable: compact support gives finite $\mathrm{\parallel }f{\mathrm{\parallel }}_1\mathrm{\parallel }g{\mathrm{\parallel }}_1$, and the radial measure calculation in step 2.1, [F2] and [F7] gives $\int e^{-\delta \mathrm{\mid }\mathbf{k}\mathrm{\mid }}d{\nu }_0=1\mathrm{/}(4{\pi }^2{\delta }^2)<\mathrm{\infty }$. Fubini [F6] therefore identifies the damped pairing with ${\int }_{{\mathbb{R}}^4}{\int }_{{\mathbb{R}}^4}f(x)g(y)w_{\delta }(x-y)d^{4}x{d}^{4}y$.

The shell vectors in [F1] lie in $L^2({\nu }_0)$, so Cauchy–Schwarz [F4] gives $\int \mathrm{\mid }{K}_{0}f{K}_{0}g\mathrm{\mid }d{\nu }_0<\mathrm{\infty }$. As $\delta \downarrow 0$, $e^{-\delta \mathrm{\mid }\mathbf{k}\mathrm{\mid }}\to 1$ almost everywhere on the shell and is bounded by $1$; dominated convergence [F5] therefore makes the damped shell pairing converge to $⟨K_{0}f,K_{0}g{⟩}_2$.

The set of support differences from step 1.1 is compact, and $\mathrm{\mid }\mathbf{x}-\mathbf{y}{\mathrm{\mid }}^2-\mathrm{\mid }{x}^0-y^0{\mathrm{\mid }}^2$ has a strictly positive lower bound there. The explicit formula in step 2.1 consequently converges uniformly on that set to $1\mathrm{/}[4{\pi }^2(\mathrm{\mid }\mathbf{x}-\mathbf{y}{\mathrm{\mid }}^2-\mathrm{\mid }{x}^0-y^0{\mathrm{\mid }}^2)]$. Taking the limit in the absolutely integrable pairing of step 3.1 and using step 4.1 and the vacuum covariance [F1] yields $⟨\mathrm{\Omega },{\mathrm{\Phi }}_0(f){\mathrm{\Phi }}_0(g)\mathrm{\Omega }⟩={\int }_{{\mathbb{R}}^4}{\int }_{{\mathbb{R}}^4}f(x)g(y)\mathrm{/}[4{\pi }^2(\mathrm{\mid }\mathbf{x}-\mathbf{y}{\mathrm{\mid }}^2-\mathrm{\mid }{x}^0-y^0{\mathrm{\mid }}^2)]d^{4}x{d}^{4}y$. The integral is finite because its denominator is bounded away from zero on the compact support product. By [F2] the kernel has units ${\mathrm{m}}^{-2}$; with tests of unit ${\mathrm{m}}^{-3}$, the double integral and vacuum correlation are dimensionless.

The kernel in step 5.1 is strictly positive throughout the support product, and each test equals $1$ on a nonempty ball, so the integral is strictly positive. By [F1] the same tests have zero field commutator and their bounded local Weyl operators commute. A nonzero vacuum two-point function is thus compatible with spacelike commutativity in this model.