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When deriving the scattering cross section using the Lippmann-Schwinger equation we need to calculate the Green's function defined by $$G(\mathbf{r},\mathbf{r'},E)=\langle\mathbf{r}|\frac{1}{E-H+i\epsilon}|\mathbf{r'}\rangle$$ where $H=\frac{p^2}{2m}$ represents the free particle Hamiltonian. This can be calculated as $$G(\mathbf{r},\mathbf{r'},E)=-\frac{2m}{\hbar^2}\frac{e^{ik|\mathbf{r}-\mathbf{r'}|}}{4 \pi |\mathbf{r}-\mathbf{r'}|}$$ and by definition should satisfy $$(E-H)G(\mathbf{r},\mathbf{r'},E)=\delta^3(\mathbf{r}-\mathbf{r'})$$ I am trying to verify the latter expression by writing $H=-\frac{\hbar^2}{2m}\nabla^2$ and working through the derivatives using $\mathbf{R}=\mathbf{r}-\mathbf{r'}$ to simplify things a little. My method is to use the identity $$\nabla^2(fg)=f\nabla^2g+2\nabla g.\nabla f+g\nabla^2f$$ to calculate the action of $\nabla^2$ on the Green's function before adding everything together using $E=\frac{\hbar^2k^2}{2m}$ to get $$(E-H)G(\mathbf{r},\mathbf{r'},E)=e^{ik|\mathbf{r}-\mathbf{r'}|}\delta^3(\mathbf{r}-\mathbf{r'})$$ which is slightly wrong. When I think about it I see no way in which the mathematical operations performed by $E-H$ can make this extra unwanted factor disappear from the final result. Is there a flaw in anything I've said or is there an obvious place in which I have picked up this factor? Thanks for any help.

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I haven't work out the math but looks like expression you get is correct, you just need to use the following identity $$ f(x)\delta^n(x)=f(0)\delta^n(x) $$ Therefore $$ (E-H)G=e^{ik|r-r'|}\delta(r-r')=\delta(r-r') $$

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