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?

  • 1
    $\begingroup$ 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! $\endgroup$ Feb 5, 2020 at 15:55
  • $\begingroup$ 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. $\endgroup$
    – scaphys
    Feb 5, 2020 at 18:06
  • $\begingroup$ Update: I missed the Jacobian. The question is therefore solved. I will write it up as an answer. $\endgroup$
    – scaphys
    Feb 5, 2020 at 18:23

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


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