Propagator in Feynman rules

  • Thread starter NanakiXIII
  • Start date
  • #1
392
0
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

[tex]
\frac{i}{k^2 - m^2 + i \epsilon}.
[/tex]

The propagator integrand is originally

[tex]
\frac{e^{i k (x-y)}}{k^2 - m^2 + i \epsilon}.
[/tex]

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

[tex]
\frac{1}{k^2 - m^2 + i \epsilon}.
[/tex]

That I see, but where does the [itex]i[/itex] in the numerator of the first expression above come from?
 

Answers and Replies

  • #2
vanhees71
Science Advisor
Insights Author
Gold Member
17,949
8,919
This factor i comes from the i in the path-integral formula for the generating functional for (connected Green's functions), [tex]W[J]=\ln Z[J][/tex] with

[tex]Z[J]=\int \mathcal{D} \phi \exp[\mathrm{i} \int_{\mathbb{R}^4} \mathrm{d}^4 x [\mathcal{L}(\phi,\partial \phi)+J \phi]].[/tex]
 
  • #3
392
0
Ah, I think I got it. I had ignored the factor [itex]i[/itex] when I wrote things down as Wick contractions. Thanks.
 
  • #4
409
1
Chapter 10 of Srednicki is a good way to see the Feynman rules emerge, including this factor of i.
 
  • #5
392
0
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.
 

Related Threads on Propagator in Feynman rules

  • Last Post
Replies
11
Views
9K
  • Last Post
Replies
13
Views
1K
Replies
1
Views
823
Replies
2
Views
1K
  • Last Post
Replies
4
Views
4K
Replies
8
Views
1K
  • Last Post
Replies
11
Views
2K
  • Last Post
Replies
3
Views
4K
Replies
7
Views
6K
  • Last Post
Replies
7
Views
2K
Top