The free Klein-Gordon propagator in momentum space $\sim (p^2-m^2+i\epsilon)^{-1}$ has just a single pole at $p^2=m^2$. The passage to Fourier space is difficult but possible. The result is very illuminating in terms of how disturbances of a free scalar field propagate.

In an interacting theory, with a dressed Klein-Gordon propagator in momentum space,


is there a picture of what disturbances look like in position space?


just use the Kallen-Lehmann spectral representation: once Fourier Transformed in real space you have an integral representation of the (time-ordered) 2-point functions $$ \langle T\Phi(x_1)\Phi(x_2)\rangle=\int_0^\infty d\mu^2 \rho(\mu^2)\Delta(x_1-x_2;\mu^2) $$ where $\Delta(x;\mu^2)$ is the free propagator for a scalar of mass (squared) $\mu^2$, and $\rho(\mu^2)$ is positive distribution that sum up to $1=\int_0^\infty d\mu^2 \rho(\mu^2)$ and has a delta function at $\mu^2=m^2$ if $\Phi$ is associated to some particle state of mass $m$.

  • $\begingroup$ This is exactly what I was looking for! So, the pole will lead to a propagation like a typical massive particle, and the continuum part will lead to propagation at speeds less than that pole part because I'm integrating over larger masses. Right? $\endgroup$ – QuantumDot Nov 10 '12 at 0:37

The propagator looks:

$\Delta_{F}(t,\vec{x})=\frac{im}{4\pi^{2}\sqrt{|\vec{x}|^{2}-t^{2}}}K_{1}\Big(m\sqrt{|\vec{x}|^{2}-t^{2}} \Big)$

Where K_{1} is a Bessel function. I found this result in the book Quantum Fiel Theory of Gerald Folland, there you can other propagator in position space too.

  • $\begingroup$ Thanks, but this is the free propagator in position space; I already new this result. My question is about the dressed propagator. $\endgroup$ – QuantumDot Aug 5 '16 at 11:01

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