Consider a scalar field coupled to a source $$(\Box - m^2)\phi(x) = -J(x)\tag{1}.$$ Then, the response of the source is determined by the Green's function $G(x-y)$, which satisfies $$(\Box - m^2)G(x-y)=-\delta(x-y) \tag{2}.$$ In Euclidean signature the Green's function which is the solution of the previous equation, can be written as the Fourier transform $$G(x-y)= \int \frac{d^dk}{(2\pi)^d} \frac {e^{ik\cdot(x-y)}}{k^2+m^2} \tag{3}.$$ I cannot understand how given (3) the solution of (1) can be expressed as the integral $$\phi(x)=\int d^d y G(x-y)J(y)\tag{4}.$$

My guess is that one has to take (1) and act somehow (4) but right now I cannot see how to arrive to (4). I would appreciate some help.

P.S. My main problem is the fact that in the following semi-proof \begin{align} (\Box-m^2)\phi &= (\Box - m^2)\int dy \, J(y) \phi_i(x-y) \\ &= \int dy \, J(y) {\color{red}{(\Box - m^2)}}\phi_i(x-y) \\ &= \int dy \, J(y) {\color{red}{\delta(x-y)}} \\ &= J(x) \end{align} Not only I do not get the minus sign of (1) but also I do not understand why we use the homogeneous Klein-Gordon equation to get the red delta function in a problem where we begun with the inhomogeneous one!

  • 2
    $\begingroup$ I learned something with that \color{}{} command. $\endgroup$
    – BMS
    Commented Dec 15, 2014 at 17:31

1 Answer 1


First, the obvious explanation for the sign is that if $J$ has a minus sign in (1), then there should be a minus sign in (4).

For some reason your $G$ turned into $\phi_i$. Assuming that they are the same thing, then I'm not sure I understand your problem. We didn't use the homogeneous KG equation to get the delta function; we used the inhomogeneous one, with $J(x) = \delta(x-y)$. If $\phi_i$ was a solution to the homogeneous KG equation, $(\Box-m^2)\phi_i$ would be zero, not $\delta(x-y)$.

  • $\begingroup$ Hi. I am using Freedman's Supergravity textbook and both (1) and (4) (eqs 4.16 and 4.20 in the book) have these signs. As for the second part, I was a bit confused trying to understand the whole thing from various sources. But in the end, indeed this $\phi_i$ is just $G$. $\endgroup$
    – Marion
    Commented Dec 15, 2014 at 15:24
  • $\begingroup$ Sorry, the signs confused me too. (4) is correct, the problem is in the red part; the second of these (and the last line) should have a minus sign. $\endgroup$
    – Javier
    Commented Dec 15, 2014 at 15:42
  • $\begingroup$ And why is that? $\endgroup$
    – Marion
    Commented Dec 15, 2014 at 15:57
  • 1
    $\begingroup$ @Marion: Because $(\Box-m^2)\phi_i(x-y) = -\delta(x-y)$. $\endgroup$
    – Javier
    Commented Dec 15, 2014 at 16:36

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