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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,

$$G\sim\frac{1}{p^2-m^2-\Sigma(p^2)+i\epsilon}$$

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

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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 $m$.

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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? –  QuantumDot Nov 10 '12 at 0:37
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