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

Srednicki QFT chapter 8 question

  1. Feb 11, 2010 #1

    In chapter 8 Srednicki employs the [tex] 1-i \epsilon [/tex] trick. He multiplies the Hamiltonian desity,

    [tex] H=\frac{1}{2} \Pi^2+\frac{1}{2}(\nabla\phi)^2+\frac{1}{2}m^2\phi^2 [/tex]

    by this [tex] 1-i \epsilon [/tex], and says it's equivalent to if we replaced m^2 with [tex] m^2-i \epsilon [/tex]. I cant see how this is?

  2. jcsd
  3. Feb 11, 2010 #2


    User Avatar
    Science Advisor

    He explains this in detail in the earlier chapter on the harmonic oscillator.
  4. Feb 12, 2010 #3
    In chptr 7, it's completely different,

    [tex] H(P,Q)=\frac{P^2 m^{-1}}{2}+\frac{1}{2}m\omega^2 Q^2 [/tex]

    So multiplying by [tex]1-i\epsilon[/tex], gives :

    [tex] H(P,Q)=\frac{P^2 (1-i\epsilon)m^{-1}}{2}+\frac{1}{2}(1-i\epsilon) m\omega^2 Q^2 [/tex]

    So directly you can see this is equivalent to if we used [tex] m^{-1} \rightarrow (1-i\epsilon)m^{-1} [/tex] and [tex] m\omega^2 \rightarrow (1-i\epsilon)m\omega^2 [/tex]

    I fail to see how to perform in a similar process to get Srednicki's mass substitution for the H in my first post. Thanks.
  5. Feb 12, 2010 #4

    Physics Monkey

    User Avatar
    Science Advisor
    Homework Helper

    Hi LAHLH,

    I would recommend not trying to take Srednicki's comment too literally. A useful point of view is the following. The purpose of the [tex] i \epsilon [/tex] is to get the pole structure right in the propagator. To make sense of what Srednicki is saying, try to determine if modifying only the mass by [tex] i \epsilon [/tex] is sufficient to modify the pole structure so that integrating along the real line is the "right" contour.
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook