# Green's function for Klein-Gordon equation in x-space

1. Oct 21, 2012

### RedSonja

I'm trying to derive the x-space result for the Green's function for the Klein-Gordon equation, but my complex analysis skills seems to be insufficient. The result should be:

\begin{eqnarray}
G_F(x,x') = \lim_{\epsilon \rightarrow 0} \frac{1}{(2 \pi)^4} \int d^4k\frac{e^{-ik\cdot(x-x')}}{k^2-\frac{m^2c^2}{\hbar^2}+i\epsilon}
= \left(-\frac{1}{4\pi} \delta (s^2) + \frac{mc^2}{8 \pi \hbar s} H_1^{(1)} \left(\frac{mc^2}{\hbar} s \right) \right) \:\Theta (s)
- \frac{m}{4\pi^2 s} K_1 \left(i\frac{mc^2}{\hbar} s \right) \:\Theta(-s))\\
\end{eqnarray}

with $s^2 = c^2(t-t')^2-(\vec{x}-\vec{x}')^2$, the source point $x'$, $\Theta(s)$ the Heaviside stepfunction, $H_1^{(1)}(x)$ the Hankel function of the first kind, and $K_1(x)$ the modified Bessel function of the second kind.

I changed to spherical polar coordinates and did the $\phi$ and $\theta$ integrals first. From there I've tried several approaches, but I always end up with a complex integral of a multivalued function:

\int \frac{f(z)}{\sqrt{z^2 \pm a^2}} dz

From residues I get zero when $f(z)$ is an exponential function, but that can't be right?

Both my quantum field theory and many-particle books work in k-space and I haven't found the derivation here.

Would someone be kind enough to guide me throught the correct procedure for the different steps of the Fourier transform of the Feynman propagator?

2. Oct 21, 2012

### andrien

Have you tried greiner quantum electrodynamics.I think the x-space green function is derived there(but perhaps for spin 1/2 but that will lead you the way)

3. Oct 21, 2012

### Bill_K

It ain't easy!

G(x,x') = (2π)-4∫eip·(x-x')/(p2 + m2-i0) d4p

Making use of the integral identity

0 e-i(ξ-i0)s ds ≡ -i/(ξ-i0)

we obtain

G(x,x') = i/(2π)40 ds ∫exp{-i[(p2+m2-i0)s - p·(x-x')]} d4p

Now complete the square in the exponent and use the Gaussian integral,

-∞ eiax2 dx ≡ √(π/a) exp{(i a/|a|)(π/4)}

G(x,x') = (4π)-20 s-2 exp{-i[m2s - (x-x')2/4s]}ds

From there, in a few more steps it reduces to Schlafli's integral representation of the Bessel function.

Last edited: Oct 21, 2012
4. Oct 21, 2012

### RedSonja

Cheers, Bill_K!
I just knew there had to be a neat trick for this!

5. Oct 22, 2012