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?
Best Answer
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.