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So i computed the following equal time correlation function for a one particle state. The vacuum correlations give the function $$\langle \phi(\vec x)\phi(\vec y)\rangle_0=D(\vec x-\vec y)\\ =\frac{e^{-m|\vec x-\vec y|}}{4\pi |\vec x-\vec y|}$$ while the one particle correlation function looks like: $$ \langle\vec q| \phi(\vec x)\phi(\vec y)|\vec q\rangle=2\omega_q\langle0|a_q \phi(\vec x)\phi(\vec y)a^{\dagger}_q|0\rangle\\\propto 2cos[\vec q\cdot(\vec x-\vec y)]+D(\vec x-\vec y)\\=2cos[\vec q\cdot(\vec x-\vec y)]+\frac{e^{-m|\vec x-\vec y|}}{4\pi |\vec x-\vec y|}$$

I recognize that since this is a spacelike separation that it can't be observed, the commutator of this type vanishes since the terms are invariant under $\vec x\rightarrow \vec y$. Since there is no elapsed time there is also no valid interpretation as particle propagation.

it just seems odd to me that the field has non vanishing correlations for large spatial separations. I was wondering if there is any consequence to this that people know about.

Thanks,

Selman

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