As I understand it a Green's function ##G(x,y)## for a translationally invariant differential equation satisfies $$G(x+a,y+a)=G(x,y)\qquad\Rightarrow\qquad G(x,y)=G(x-y)$$ (where ##a## is an arbitrary constant shift.)(adsbygoogle = window.adsbygoogle || []).push({});

My question is, given such a translationally invariant system, how does one show that ##G(x,y)=G(x-y)##?

Is it as simple as setting ##y=-a## such that $$G(x+a,y+a)=G(x-y,0)\equiv G(x-y)$$ or is it more to it than that? (I'm slightly uncomfortable with setting ##y=-a##, it doesn't seem correct?!)

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# Green's functions for translationally invariant systems

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