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I'm having trouble deriving the Greens function for the Helmholtz equation. I happen to know what the answer is, but I'm struggling to actually compute it using typical tools for computing Greens functions.

In particular, I'm solving this equation:

$$ (-\nabla_x^2 + k^2) G(x,x') = \delta(x-x') \quad\quad\quad x\in\mathbb{R}^3 $$

I know that the solution is

$$ G(x,x') = \frac{e^{-i k |x-x'|}}{4\pi|x-x'|} $$

but I would like to understand how to obtain this answer (so that I can find Greens functions for other, more complicated, equations). The typical approach seems to be to solve in Fourier space and then transform back into real space, but I'm having a lot of trouble with the transformation back into real space. In particular, evaluating this integral is hard:

$$ G(x,x') = \int d^3p \frac{e^{-ip(x-x')}}{p^2+k^2}$$

(where $p$ is my Fourier space variable, and $k$ is the same $k$ as in the original equation above). Is there an easier way? Am I doing something wrong?

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1 Answer 1

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The integral is not hard! The measure $d^3p$ is equal to $|p|^2d|p|\, d(\cos \theta) d\phi$ and the exponetial $\exp\{-i{\bf p}\cdot ({\bf x}-{\bf x}')\}$ is $\exp\{-i |p| |{\bf x}-{\bf x}'|\cos\theta\}$, so everything is a straightforward integral over decoupled scalars.

We are to show that $$ G(|x|)= \int_{{\mathbb R}^3}\frac{d^3 p}{(2\pi)^3} \frac{e^{i{\bf p}\cdot {\bf x}}}{|{\bf p}|^2+m^2 } = \frac 1{4\pi|{\bf x}|} e^{-m{|\bf x}|} $$ (note that the OP's equation is not quite correct: there is no $i$ in the exponent on the RHS when his $k^2$ (my $m^2$) is positive). We use the measure I gave above to do the integrals over the angles $\theta$ and $\phi$ to see that this is equal to $$ G(|{\bf x}|)=C \frac 1{|{\bf x}|}\int_0^\infty d|p| \frac{|{\bf p}| \sin |{\bf p}||{\bf x}|}{|{\bf p}|^2+m^2} $$ for some constant $C$. We do not need to use Jordan's lemma to do the remaining integral. We observe that the elementary integral $$ \int_0^\infty e^{-mx}\sin (px) \, dx= \frac p{p^2+m^2} $$ is a half-line Fourier transform, and the desired $|{\bf p}|$ integral is its inverse Fourier transform.

When the OP's $k^2$ becomes negative we have a wave equation. This does not have a unique Green function and it is in this case that there are singularities on the integration contour of the final $|{\bf p}|$ integral. How we route the contour around them by adding $\pm i\epsilon$'s then determines whether we replace $m$ by $i|k|$ or $-i|k|$, and physical refers to outgoing waves (and a causal Green function) or incoming waves (and an anti-causal Green function). In this case some knowledge of complex variable techniques is useful, but still not necessary as we can still invert a Fourier transform.

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  • $\begingroup$ The integral is singular, with two simple poles on the real axis. The most common way to solve it is using the Residue theorem, but even in that case is not trivial $\endgroup$ Commented Mar 7, 2017 at 7:55
  • $\begingroup$ @Alessandro Zunino: I've edited by initial answer to address this issue. $\endgroup$
    – mike stone
    Commented Mar 8, 2017 at 15:25
  • $\begingroup$ Thank you! A follow up question: what parameters of this general solution can I adjust to force the Greens function to satisfy specific boundary conditions? Specifically, my Green's function lives a bound spherical domain $|x-x'|<R$, and at the domain wall I have $\frac{d}{dx}G(|x-x|) \sim \frac{G(|x-x'|}{|x-x'|}$ $\endgroup$
    – alexvas
    Commented Mar 8, 2017 at 18:19
  • $\begingroup$ @alexvas I'd try a method of images solution, but I'm not sure that it would be easy as the usual Coulomb green function has properties that do not hold when k is non-zero. The other option is a series expansion in the actual eigenfunctions for the domain. This will involve spherical Bessel functions $\endgroup$
    – mike stone
    Commented Mar 8, 2017 at 18:38
  • $\begingroup$ What do you mean when you say "eigenfunctions of the domain"? $\endgroup$
    – alexvas
    Commented Mar 10, 2017 at 0:13

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