# Green's function for the screened Poisson equation

Assuming we are given a Lagrangian $$$$\mathcal{L}(\phi(r),\partial^i\phi(r)) = \frac{1}{2} \partial_i\phi \partial^i\phi + \frac{m^2}{2} \phi^2 + \lambda \phi,$$$$ the equations of motion obtained from the Euler-Lagrange equations are $$$$(\Delta-m^2) \phi = \lambda.$$$$ In order to find the Green's function for this system, the standard procedure would be to impose $$(\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?

• The normalization constant, $1/4\pi$ is the same as for the unscreened Poisson equation, for the very same reason. Do you remember how to get that by integrating both sides (perhaps from EM)? The $\delta$ on the r.h.s. is a 3 dimensional one! – Cosmas Zachos Feb 5 at 15:55
• In the unscreened case I can integrate over both sides and apply the divergence theorem, but I am having trouble to do the same thing here, as $\int_0^R m^2 r^{-1}\exp(-mr) \mathrm{d}r$ diverges. – scaphys Feb 5 at 18:06
• Update: I missed the Jacobian. The question is therefore solved. I will write it up as an answer. – scaphys Feb 5 at 18:23

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, $$A = -\frac{1}{4\pi}$$.