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

## Main Question or Discussion Point

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:

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

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?

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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)

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