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

Order of scalar interaction impact Feynman diagrams

  1. Nov 22, 2015 #1
    On page 60 of srednicki (72 for online version) for the $$\phi^{3}$$ interaction for scalar fields he defines

    $$Z_{1}(J) \propto exp\left[\frac{i}{6}Z_{g}g\int d^{4}x(\frac{1}{i}\frac{\delta}{\delta J})^{3}\right]Z_0(J)$$

    Where does this come from? I.e for the quartic interaction does this just become

    $$Z_{1}(J) \propto exp\left[\frac{i}{6}Z_{g}g\int d^{4}x(\frac{1}{i}\frac{\delta}{\delta J})^{4}\right]Z_0(J)$$

    and for the feynman diagrams the $$\phi ^{3}$$ theory has 3-line vertices whereas the $$\phi^{4}$$ has 4-line vertices? Then how do the feynman diagrams change as we change the order of g?
     
  2. jcsd
  3. Nov 28, 2015 #2
    Thanks for the post! This is an automated courtesy bump. Sorry you aren't generating responses at the moment. Do you have any further information, come to any new conclusions or is it possible to reword the post?
     
  4. Nov 28, 2015 #3

    king vitamin

    User Avatar
    Gold Member

    We define the generating functional,

    [tex]
    Z[J] = \int \mathcal{D} \phi \exp \left[i \int d^dx \left(\mathcal{L}_0(\phi) + \mathcal{L}_1(\phi) + J(x)\phi(x) \right)\right]
    [/tex]

    where [itex]\mathcal{L}_0[/itex] is solvable, be which I mean I can write down the "free" generating functional

    [tex]
    Z_0[J] = \int \mathcal{D} \phi \exp \left[i \int d^dx \left(\mathcal{L}_0(\phi) + J(x)\phi(x) \right)\right]
    [/tex]

    exactly as an analytic functional of [itex]J(x)[/itex]. In particular, I can take functional derivatives with respect to [itex]J[/itex]. Then by taking derivatives, we can evaluate the following functional integrals:
    [tex]
    \int \mathcal{D}\phi (\phi(x))^n (\phi(y))^m \cdots \exp \left[i \int d^dx \left(\mathcal{L}_0(\phi) + J(x)\phi(x) \right)\right] = \frac{1}{i}\frac{\delta^n}{\delta \phi(x)^n}\frac{1}{i}\frac{\delta^m}{\delta \phi(y)^m} \cdots Z_0[J].
    [/tex]

    This is basically already the content of your expression. We assume [itex]e^{\int d^dx\mathcal{L}_0(\phi)}[/itex] is just an analytic function of [itex]\phi[/itex] so that it can be defined by a polynomial power series like the above, and we can formally write
    [tex]
    Z[J] = \exp\left( \int d^dx \mathcal{L}_1\left( \frac{1}{i}\frac{\delta}{\delta \phi} \right) \right) Z_0[J].
    [/tex]

    So for ANY interaction, you just replace the interaction lagrangian with [itex]\phi(x) \rightarrow \frac{1}{i}\frac{\delta}{\delta \phi(x)}[/itex] acting on the free generating function. The easiest way to deal with [itex]\phi^4[/itex] theory is with the Lagrangian
    [tex]
    \mathcal{L}_1 = \frac{g}{24} \phi(x)^4
    [/tex]
    (the factor of 24 will help you later for the same reason the factor of 6 helps you in phi^3 theory). So the generating functional is
    [tex]
    Z[J] = \exp\left( \frac{g}{24}\int d^dx \left( \frac{1}{i}\frac{\delta}{\delta \phi} \right)^4 \right) Z_0[J].
    [/tex]
    Then expanding the exponential in powers of [itex]g[/itex] gives you the Feynman expansion.
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook




Similar Discussions: Order of scalar interaction impact Feynman diagrams
  1. Feynman Diagrams (Replies: 2)

Loading...