Green's function for the screened Poisson equation

Question

Assuming we are given a Lagrangian \begin{equation} \mathcal{L}(\phi(r),\partial^i\phi(r)) = \frac{1}{2} \partial_i\phi \partial^i\phi + \frac{m^2}{2} \phi^2 + \lambda \phi, \end{equation} the equations of motion obtained from the Euler-Lagrange equations are \begin{equation} (\Delta-m^2) \phi = \lambda. \end{equation} In order to find the Green's function for this system, the standard procedure would be to impose $(\Delta-m^2)G(r) = \delta(r)$, going to Fourier space in order to get an algrbraic equation ($\Delta \rightarrow k^2$) and get $G(r)$ by performing the inverse Fourier transform.

However, assuming we know that $G \propto r^{-1}e^{-mr}$, is there a way to find the proper normalization for the Green's function directly from the equations of motion, without using the above procedure?

Answer 1

Integrating $(\Delta-m)G(\mathbf{r}) = \delta(\mathbf{r})$ with the ansatz $G(r) = A r^{-1} \exp(-mr)$ over a sphere of radius R yields \begin{align} 1 &= A\int_V \Delta\Big(\frac{e^{-mr}}{r}\Big)-m^2\frac{e^{-mr}}{r} \;\mathrm{d}V\\ &= A\int_V \partial_r\left[r^2\partial_r\Big(\frac{e^{-mr}}{r}\Big)\right]-m^2re^{-mr} \;\mathrm{d}r\mathrm{d}\Omega\\ &= 4\pi A\int_0^R \partial_r\left[r^2\partial_r\Big(\frac{e^{-mr}}{r}\Big)\right]-m^2re^{-mr} \;\mathrm{d}r\\ &= -4\pi A \left[(1+mR)e^{-mR}\right]+\left[1-(1+mR)e^{-mr}\right] \\ &= -4 \pi A. \end{align} Therefore, $A = -\frac{1}{4\pi}$.