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?
Best Answer
Why do you want to set a boundary condition? Clearly here your problem has no boundary since you can detect some particles very far away from the place where the interaction with the potential takes place. However, surely you will need an intitial condition provided by $\psi^0_\textbf{k}(\textbf{r})$.
I feel (tell me if I'm wrong) that what you mean by "boundary condition" refers actually to what kind of solution you are interested in. As you pointed out, the equation $$ (\nabla^2+k^2)G(\textbf{r},\textbf{r}')=-4\pi\delta(\textbf{r}-\textbf{r}') $$ has two solution called retarded $G^+$ and advanced $G^-$ Green functions. This gives rise to two different kind of solutions to your differential equation :
The causal solution : $$ \psi_\textbf{k}(\textbf{r})=\psi^{\text{(in)}}_\textbf{k}(\textbf{r})-\frac{1}{4\pi}\int\mathrm{d}\textbf{r}'\,G^+(\textbf{r}-\textbf{r}')\,U(\textbf{r}')\,\psi_\textbf{k}(\textbf{r}') $$ where $\psi^{\text{(in)}}_\textbf{k}\equiv \psi^{\text{0}}_\textbf{k}$ is interpreted as the incoming field, i.e. the asymptotic field you obtain when taking the limit $t\rightarrow -\infty$ and which evolves freely with time. This is usually the solution people are interested in because it describes the scattering event as a result of the interaction with the potential.
The anti-causal solution : $$ \psi_\textbf{k}(\textbf{r})=\psi^{\text{(out)}}_\textbf{k}(\textbf{r})-\frac{1}{4\pi}\int\mathrm{d}\textbf{r}'\,G^-(\textbf{r}-\textbf{r}')\,U(\textbf{r}')\,\psi_\textbf{k}(\textbf{r}') $$ where $\psi^{\text{(out)}}_\textbf{k}$ is interpreted as the outcoming field, i.e. the asymptotic field you obtain when taking the limit $t\rightarrow +\infty$ and which had evolved freely with time from the past.
In both of these cases, the limits $t\rightarrow\pm\infty$ allow you to get rid of the integral terms and give rise to different interpretations of the solutions.
It seems to me that this is not directly related to your initial problem. One can show independently that the property $$ G(\textbf{r},\textbf{r}')=G(\textbf{r}-\textbf{r}') $$ is just a consequence of the fact that the dispersion relation $\langle\textbf{k}| H|\textbf{k}\rangle =E(\textbf{k})$ only depends on the magnitude of the impulsion, i.e. $E(\textbf{k})\equiv E(k)$, which is a consequence of the fact that your system is invariant under translation.