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The Green's function of $(\nabla^2 -m^2)$ is

$$\Delta(x,y)=\int d^4k \frac{e^{ik\cdot(x-y)}}{k^2+m^2+i\varepsilon}$$

so that $(\nabla^2 -m^2)\Delta(x,y)=\delta^{(4)}(x-y)$

Is it possible to find the Green's function of:

$$f(x)(\nabla^2 -m^2)$$

where $f(x)$ is an arbitrary function? Such a Green's function $G(x,y)$ satisfies:

$$f(x)(\nabla^2 -m^2)G(x,y)=\delta^{(4)}(x-y)$$

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  • $\begingroup$ Would that mean that $(\square-m^2)G(x, y) =f(x)\delta(x-y)$ or $(\square-m^2)f(x)G(x, y) =\delta(x-y)$? $\endgroup$
    – Nihar Karve
    Feb 15, 2021 at 3:00
  • $\begingroup$ Possibly related: physics.stackexchange.com/q/598679/278763 $\endgroup$
    – Richard Myers
    Feb 15, 2021 at 3:08
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    $\begingroup$ @Nihar Karve. I added a clarification $\endgroup$
    – user84158
    Feb 15, 2021 at 3:11

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