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If the Green's function of a second order differential operator $L$ is $G^{(0)}$, then if I add a small perturbation $\delta L$, a Green's function $G$ for the operator: $(L+ \delta L)$ should be:

$G=G^{(0)}-G^{(0)} (\delta L G^{(0)})$

correct? Now, what confuses me is: how do I write the second term?

Is it

$\int dx'( G^{(0)}(x,x') \int(dx'' \delta L(x'')G^{(0)}(x',x'')) )$?

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Ok I think I got this.

One integration is unneessary, in fact:

$(L+\delta L)(G^{(0)}(x-x')+\delta G(x-x'))= \delta(x-x')$

$(L+\delta L) G^{(0)}(x-x') +L \delta G(x-x') = \delta(x-x')$

$\delta(x-x')+ \delta L G^{(0)}(x-x') +L \delta G(x-x') =\delta(x-x')$

So $\delta G$ solves the same equation as $G^{(0)}$ with the source term $\delta L G^{(0)}(x-x')$, which means the solution is:

$\delta G(x-x')= \int dx''\left[ G^{(0)}(x'-x'') \left[\delta L(x)G^{(0)}(x-x'')\right]\right] $

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