Consider the elastic scattering of particles by a potential $V$ in Quantum Mechanics. In the zone of influence of the potential the Hamiltonian may be written as
$$H = H_0 + V,$$
being $H_0$ the free particle Hamiltonian. The eigenvalue equation for the Hamiltonian $H$ is thus
$$H|\psi\rangle = E|\psi\rangle,$$
and if $E = \hbar^2 k^2/2\mu$, we can rewrite this in the position representation as:
$$\left(\nabla^2+k^2\right)\psi=U\psi,$$
being $U = 2\mu V/\hbar^2$. One can solve this equation with the method of Green's functions. The general solution for a certain $k$ is
$$\psi_k(\mathbf{r})=\psi_{k}^0(\mathbf{r})-\dfrac{1}{4\pi}\int G(\mathbf{r},\mathbf{r}')U(\mathbf{r}')\psi_k(\mathbf{r}')d^3\mathbf{r}',$$
being $\psi_k^0$ the solution for the free equation, that is, $H_0|\psi_k^0\rangle = E|\psi_k^0\rangle$ and where $G$ is the Green's function satisfying:
$$(\nabla^2+k^2)G(\mathbf{r},\mathbf{r}')=-4\pi\delta(\mathbf{r}-\mathbf{r}').$$
Now if we find $G$ we reduce the problem to an integral equation. We need thus to find $G$.
My problem here is the following: to find $G$ we need boundary conditions of the problem. I can't understand, though, what boundary conditions we should impose here.
So to solve scattering problems using Green's functions like that, what are the boundary conditions we need to impose to compute the Green's function?
I've seem this method in some notes and as things are written it seems the only imposed condition is that $G(\mathbf{r},\mathbf{r}')=G(\mathbf{r}-\mathbf{r}')$. This supposition gives two Green's functions:
$$G^{\pm}(\mathbf{r},\mathbf{r}')=\dfrac{e^{\pm ik |\mathbf{r}-\mathbf{r}'|}}{|\mathbf{r}-\mathbf{r}'|}$$
This then seems to be necessary to get all the solutions.
Is this the only condition necessary or there are any boundary conditions on $G$? What is the physical meaning of the condition $G(\mathbf{r},\mathbf{r}')=G(\mathbf{r}-\mathbf{r}')$ and why should we impose it?