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The free-particle propagator is given by $$\Delta_F(x-y) = \frac{1}{(2\pi)^4} \int \frac{e^{-ik\cdot(x-y)}}{k^2-m^2+i\epsilon} \, d^4k.$$

In the book Quantum Field Theory, Ryder says that $\Delta_F(x-y)$ has a pole at $k^2=m^2$ (page 203).

However, the pole is actually at $k^2=m^2 - i\epsilon$. Why is the $-i\epsilon$ term not mentioned?

DanielSank
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rainman
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1 Answers1

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Well I don't have the book, but usually the $i \epsilon$ is added intentionally (and artificially) to address the pole, and bring it off the real axis. Somewhere later in the calculation, you take the limit $\epsilon \rightarrow 0$.

Gilbert
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    Very true. There's a discussion of this issue here: https://physics.stackexchange.com/questions/138217/complex-integration-by-shifting-the-contour – DanielSank Aug 25 '17 at 04:31
  • Ah yes, thanks for the link! It's a very nice and clear discussion. – Gilbert Aug 25 '17 at 04:46