# Pole of a free-particle propagator

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?

• This post will certainly help you. – DanielSank Aug 25 '17 at 4:31

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 $\epsilon \rightarrow 0$.