What is the role of the i in the propagator of Feynman rules?

[NanakiXIII]

I'm probably missing something small but I haven't been able to figure this out. In the Feynman rules (for a scalar field that obeys the Klein-Gordon equation), you write a propagator for internal lines as

$$\frac{i}{k^2 - m^2 + i \epsilon}.$$

The propagator integrand is originally

$$\frac{e^{i k (x-y)}}{k^2 - m^2 + i \epsilon}.$$

Since we're dealing with an internal line, both exponentials, in $x$ and $y$, are integrated out to delta functions, leaving you with

$$\frac{1}{k^2 - m^2 + i \epsilon}.$$

That I see, but where does the $i$ in the numerator of the first expression above come from?

 

[vanhees71]

This factor i comes from the i in the path-integral formula for the generating functional for (connected Green's functions), $$W[J]=\ln Z[J]$$ with

$$Z[J]=\int \mathcal{D} \phi \exp[\mathrm{i} \int_{\mathbb{R}^4} \mathrm{d}^4 x [\mathcal{L}(\phi,\partial \phi)+J \phi]].$$

 

[NanakiXIII]

Ah, I think I got it. I had ignored the factor $i$ when I wrote things down as Wick contractions. Thanks.

 

[LAHLH]

Chapter 10 of Srednicki is a good way to see the Feynman rules emerge, including this factor of i.

 

[NanakiXIII]

He seems to adopt quite a different approach than the author of the book I'm using. I may have a look at that later. Thanks for the tip.