Timeordered derivative jump gives iδ Green source

On $t>0$, $g_{\omega }(t)=e^{-i\omega t}\mathrm{/}(2\omega )$ and $g_{\omega }^{\mathrm{\prime }}(t)=-\frac{i}{2}{e}^{-i\omega t}$, so $g_{\omega }^{\mathrm{\prime }}(0+)=-i\mathrm{/}2$. On $t<0$, $g_{\omega }(t)=e^{i\omega t}\mathrm{/}(2\omega )$ and $g_{\omega }^{\mathrm{\prime }}(t)=+\frac{i}{2}{e}^{i\omega t}$, so $g_{\omega }^{\mathrm{\prime }}(0-)=+i\mathrm{/}2$. The function is continuous at $0$, and the derivative jump is $g_{\omega }^{\mathrm{\prime }}(0+)-g_{\omega }^{\mathrm{\prime }}(0-)=-i$.

For a compactly supported smooth test $\eta$, integrate by parts twice on each half-line. The boundary terms from $g_{\omega }(0+)=g_{\omega }(0-)$ cancel, while the derivative boundary term contributes $-i\eta (0)$; away from $0$ the ordinary equation $g_{\omega }^{\mathrm{\prime }\mathrm{\prime }}=-{\omega }^2{g}_{\omega }$ holds. Hence $g_{\omega }^{\mathrm{\prime }\mathrm{\prime }}+{\omega }^2{g}_{\omega }=-i{\delta }_0$ distributionally.

In the fixed convention, $\mathrm{\square }-{\mu }^2$ acts on the spatial Fourier mode by $-({\mathrm{\partial }}_t^2+{\omega }_{\mu }(\mathbf{k}{)}^2)$. Applying it to 2.1 gives $+i{\delta }_0(t)$ for every $\mathbf{k}\mathrm{\ne }0$ when $\mu =0$, and for every $\mathbf{k}$ when $\mu >0$. The omitted massless apex has measure zero and the shell density is locally integrable there; spatial Fourier inversion therefore gives $i{\delta }_0(t,\mathbf{x})$, exactly as in [F1].