Check of the Green function for the Klein-Gordon equation

Question

In order to derive the Green function for the Klein-Gordon equation, one considers $$ \Box G(x-x')+m^2G(x-x')=\delta^4(x-x') $$ where $\delta^4(x)=\delta(x_0)\delta(x_1)\delta(x_2)\delta(x_4)$. The widely known solution (e.g. see propagator) will depend on distributions like $$ \Theta(x_0^2-x_1^2-x_2^2-x_3^2), \qquad \delta(x_0^2-x_1^2-x_2^2-x_3^2). $$ Therefore, if I substitute the solution back into the equation, it is difficult to see how to recover the more standard $\delta^4(x)$. On the other side, in momentum space, we know how to move from $d^4p$ to $d^3p/(2E)$ and keeping Lorentz invariance. What is the further condition in place for this case?

The rationale behind this question is, given an exact solution of the homogeneous equation, how I could get the Green function? Indeed, it is easy to check that the function $J_1(m x)/x$ is a solution of the homogeneous Klein-Gordon equation.

Answer 1

Suppose we have a solution to the K.G. equation $(\Box^{2}_{x}+m^{2})\phi(x)=0$. Consider the Feynman Propagator $\Delta_{F}(x-y) = \langle 0|T\{\phi(x)\phi(y)\}|0\rangle$, where the fields are time ordered. We can show that this is the Greens function for the K.G. operator by first noting that: \begin{equation} \partial_{x_{0}}\langle 0|T\{\phi(x)\phi(y)\}|0\rangle\overset{!}{=}\partial_{x_{0}}\langle 0|\theta(x_{0}-y_{0})\phi(x)\phi(y)+\theta(y_{0}-x_{0})\phi(y)\phi(x)|0\rangle = \langle 0|\delta(x_{0}-y_{0})\phi(x)\phi(y)+\theta(x_{0}-y_{0})\partial_{x_{0}}\phi(x)\phi(y)-\delta(y_{0}-x_{0})\phi(y)\phi(x)+\theta(y_{0}-x_{0})\phi(y)\partial_{x_{0}}\phi(x)|0\rangle = \langle 0|\theta(x_{0}-y_{0})\partial_{x_{0}}\phi(x)\phi(y)+\theta(y_{0}-x_{0})\phi(y)\partial_{x_{0}}\phi(x)|0\rangle = \langle 0|T\{\partial_{x_{0}}\phi(x)\phi(y)\}|0\rangle \end{equation} Repeating this again but taking note that $\Pi(x) = \partial_{x_0}\phi(x)$: \begin{equation} \partial_{x_{0}}\langle 0|T\{\partial_{x_{0}}\phi(x)\phi(y)\}|0\rangle = \langle 0|T\{\partial_{x_{0}}^{2}\phi(x)\phi(y)\}|0\rangle - [\Pi(x),\phi(y)]\delta(x_{0}-y_{0}) \\= \langle 0|T\{\partial_{x_{0}}^{2}\phi(x)\phi(y)\}|0\rangle - (2\pi)^{3}i\delta^{(3)}(x-y)\delta(x_{0}-y_{0}) = \langle 0|T\{\partial_{x_{0}}^{2}\phi(x)\phi(y)\}|0\rangle - (2\pi)^{3}i\delta^{(4)}(x^{\mu}-y^{\mu}) \end{equation} The spacial derivative part of the K.G. operator commutes with time ordering so we get that: \begin{equation} (\Box^{2}_{x}+m^{2})\langle 0|T\{\phi(x)\phi(y)\}|0\rangle \\= \langle 0|T\{\Box^{2}_{x}\phi(x)\phi(y)\}|0\rangle - ((2\pi)^{3}i\delta^{(4)}(x^{\mu}-y^{\mu})+m^{2})\langle 0|T\{\phi(x)\phi(y)\}|0\rangle \end{equation} But as $\phi(x)$ is a solution to K.G. we can then say that $\langle 0|T\{\Box^{2}_{x}\phi(x)\phi(y)\}|0\rangle = -m^{2}\langle 0|T\{\phi(x)\phi(y)\}|0\rangle$. Hence: \begin{equation} (\Box_{x}^2+m^{2})\Delta_{F}(x-y) = -i(2\pi)^{3}\delta^{(4)}(x^{\mu}-y^{\mu}) \end{equation} as required up to some normalisation factor.