I am interested to learn about the Green function for the D'Alembert operator in arbitrary dimensions. While searching through the web I came across the following document:


There the fundamental solution for the general Green function is given in eq. (14). Special cases for 3, 2 and 1 dimensions are given further below. Now, I was trying to start with eq. (14) and reduce it to the 3 dimensional result eq. (15) by setting $d=3$ and taking the appropriate limit - but I do not get the same result. In fact, restricting to the imaginary part I actually get zero in the limit $\epsilon\rightarrow 0^+$. It would be great if someone could demonstrate how to do this. Thanks for any suggestion!


The general expression given in the paper as eq. (14) for the wave propagator $G_n(x)$ in $n$ spacetime dimensions is, denoting $x=(x^0,\mathbf x)$ and $x^2=(x^0)^2-|\mathbf x|^2$, $$\begin{align} G_n(x) &= \frac{\Gamma(n/2)}{(n-2)\pi^{n/2}}\lim_{\epsilon\to0^+} \mathrm{Im}[-(x^0-i\epsilon)^2+|\mathbf x|^2]^{1-n/2}\\ &=\frac{\Gamma(n/2)}{(n-2)\pi^{n/2}}\lim_{\epsilon\to0^+} \mathrm{Im}(-x^2+i\epsilon \,\mathrm{sign}(x^0))^{1-n/2} \end{align}$$ For n=4 (i.e. three space dimensions), $$ G_4(x)=\frac{1}{2\pi^2}\lim_{\epsilon\to 0^+}\mathrm{Im}\frac{1}{-x^2+i\epsilon \,\mathrm{sign}(x^0)}\,. $$ Now, recall the distributional identity $$ \lim_{\epsilon\to 0^+}\frac{1}{t\pm i\epsilon} =\mathrm{PV}\,\frac{1}{t}\mp i\pi\delta(t)\,, $$ where $\mathrm{PV}$ is the Cauchy principal value. Thus, $$ G_4(x)=-\frac{\delta(x^2)}{2\pi}\mathrm{sign}(x^0)\,. $$ Using also $$ \delta(f(t))=\frac{\delta(t-f^{-1}(0))}{|f'(t)|}, $$ we arrive at $$ G_4(x)=-\frac{\delta(x^0-|\mathbf x|)}{4\pi |\mathbf x|}+\frac{\delta(x^0+|\mathbf x|)}{4\pi |\mathbf x|}\,. $$ This is, up to a sign, the next to last equation eq. (15).

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