# Green's function of D'Alembertian

I want to find the Feynman Green's function of the D'Alembertian operator but I get stuck at one point.

$$G(x-y)$$ satisfies

$$\square_xG(x-y)=\delta^{(4)}(x-y),\tag{1}$$ where $$\square_x=\eta_{\mu\nu}\frac{\partial}{\partial x^\mu}\frac{\partial}{\partial x^\nu}$$ and $$\eta_{\mu\nu}=\text{diag}(1,-1,-1,-1)$$.

Fourier-transforming the equation, using that

$$G(x-y)=\int\frac{d^4k}{(2\pi)^4}\tilde{G}(k)e^{ik_\mu(x^\mu-y^\mu)}\tag{2}$$

I get

$$-\int\frac{d^4k}{(2\pi)^4}k_\mu k^\mu\tilde{G}(k)e^{ik_\mu(x^\mu-y^\mu)}=\int\frac{d^4k}{(2\pi)^4}e^{ik_\mu(x^\mu-y^\mu)},\tag{3}$$

so (calling $$k^0\equiv\omega$$ and $$|\vec{k}|\equiv k$$)

$$\tilde{G}(k)=\frac{-1}{k_\mu k^\mu}=\frac{-1}{\omega^2-k^2}=\lim_{\epsilon\to0}\frac{-1}{[\omega-(k-i\epsilon)][\omega+(k-i\epsilon)]}.\tag{4}$$

Now I replace this in $$(2)$$

$$G(x-y)=\lim_{\epsilon\to0}\frac{-1}{(2\pi)^4}\int d^3\vec{k}~e^{-i\vec{k}\cdot(\vec{x}-\vec{y})}\int\limits_{-\infty}^\infty d\omega\frac{e^{i\omega(x^0-y^0)}}{[\omega-(k-i\epsilon)][\omega+(k-i\epsilon)]}\tag{5}.$$

Doing the integral in $$\omega$$ using residues and setting $$\epsilon\to0$$ I get

\begin{align} G(x-y)=&\frac{i}{(2\pi)^3}\int d^3\vec{k} \frac{e^{-ik|\vec{x}-\vec{y}|\cos\theta}}{2k}e^{ik|x^0-y^0|}=...= \\\\ =&\frac{-1}{4\pi|\vec{x}-\vec{y}|}\int\limits_0^\infty \frac{dk}{2\pi}\Big(e^{-ik|\vec{x}-\vec{y}|}-e^{ik|\vec{x}-\vec{y}|}\Big)e^{ik|x^0-y^0|}\tag{6} \end{align}

I know the answer involves Dirac deltas, but I can't see how to extend this integral from $$-\infty$$ to $$\infty$$, maybe I'm doing something wrong. Any help would be appreciated.

• Which Green's function? Which boundary conditions? The Feynman Green's function? Oct 20, 2021 at 19:13
• yes, Feynman Green's function
– AFG
Apr 15, 2023 at 16:25
• Apr 17, 2023 at 10:13