Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Cutkosky Cutting rules

  1. Aug 24, 2015 #1
    OKay, so whenever I run into explanations on the cutting rules, most of the time I see the statement to replace


    ##\frac{1}{p^2 - m^2 + i\epsilon} \rightarrow -2i\pi \delta(p^2 - m^2)## for each propagator that has been cut

    taking note that there is no factor of i in the numerator for ##\frac{1}{p^2 - m^2 + i\epsilon}##

    so for example, for ##\phi^3## theory we can have a loop amplitude given by

    ##iM(p^2) = \frac{(i\lambda)^2}{2} \int \frac{d^4k}{(2\pi)^4}\frac{i}{(k-p)^2 - m^2 + i\epsilon}\frac{i}{k^2 - m^2 + i\epsilon}##

    or
    ##iM(p^2) = -\frac{(i\lambda)^2}{2} \int \frac{d^4k}{(2\pi)^4}\frac{1}{(k-p)^2 - m^2 + i\epsilon}\frac{1}{k^2 - m^2 + i\epsilon}##


    Making the cut through the diagram, and making the above defined replacement gives

    ##-\frac{(i\lambda)^2}{2} \int \frac{d^4k}{(2\pi)^4}[-2i\pi \delta((k-p)^2 - m^2)][-2i\pi \delta(k^2 - m^2)]##

    upon simplifying we should have...

    ##-\frac{(\lambda)^2}{2} \int \frac{d^4k}{(2\pi)^2}[\delta((k-p)^2 - m^2)][\delta(k^2 - m^2)]##

    which is off, by a minus sign, from the right answer...

    I would get the right answer if I made the replacement

    ##\frac{i}{p^2 - m^2 + i\epsilon} \rightarrow -2i\pi \delta(p^2 - m^2)##
    WITH the factor of i in the numerator

    instead of ##\frac{1}{p^2 - m^2 + i\epsilon} \rightarrow -2i\pi \delta(p^2 - m^2)##
    WITHOUT the factor of i in the numerator

    What am I doing wrong?
     
  2. jcsd
  3. Aug 24, 2015 #2
    The main thing I see you're doing wrong is evaluating Feynman diagrams.
     
  4. Aug 24, 2015 #3
    Oh? Well how so?
     
  5. Aug 27, 2015 #4

    CAF123

    User Avatar
    Gold Member

    @thoughtgaze
    I have been using Cutkosky's cutting rules extensively in my summer project but I still regard myself as an amateur in QFT so please take everything I am saying with a pinch of salt as they say :) The replacement of the off shell propagator terms with delta functions when we take a 'cut' is given by, as far as I am aware, ##(k^2-m^2+i\epsilon)^{-1} \rightarrow 2 \pi \delta(k^2-m^2)##.

    The factors of ##i## come into play depending on whether we are using the normal or complex conjugated version of the Feynman rules. Given a cut diagram, there is a convention that we label the vertices of the diagram black or white, black vertices follow standard Feynman rules and are accompanied by a factor of ##i## and white ones the complex conjugated version so come with a ##-i##. In this set up, there is usually a corresponding theta function in the replacement above of the propagator terms so that energy flow from black to white vertices is counted positively but I am not sure if you are perhaps using another convention.
     
  6. Aug 27, 2015 #5
    Interesting, I have not heard of this convention to treat the vertices differently by complex conjugation. Do you have any references for this particular notion?

    Also, I have not seen ##(k^2-m^2+i\epsilon)^{-1} \rightarrow 2 \pi \delta(k^2-m^2)## in any reference I have, for example, peskin&schroeder eq. 7.56
     
  7. Sep 6, 2015 #6

    CAF123

    User Avatar
    Gold Member

    Apologies for delay in replying,
    See for example pages 9-10 of this paper http://arxiv.org/pdf/1401.3546v2.pdf.
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook




Similar Discussions: Cutkosky Cutting rules
  1. Forum Rules (Replies: 1)

Loading...