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?

  • $\begingroup$ That's not page 60 of Srednicki in my version. It's page 72. And you're missing a $Z_0(J)$ on both of those. $\endgroup$
    – Ryan Unger
    Nov 23 '15 at 1:24
  • $\begingroup$ Hello, and welcome to Stack Exchange. It's not exactly clear what you're asking. Please consider extending the title a bit, and expanding the body of the question as well; you'll be more likely to get a quality answer if you do. $\endgroup$ Nov 23 '15 at 1:25
  • $\begingroup$ @0celo7 in the pre-publication version it is indeed 72. Print version it's 60 $\endgroup$
    – boson
    Nov 23 '15 at 1:36
  • $\begingroup$ @boson Oh, my bad. $\endgroup$
    – Ryan Unger
    Nov 23 '15 at 4:44
  • $\begingroup$ I presume in general it's $\mathcal L_I(1/i \delta /\delta J)$ $\endgroup$
    – innisfree
    Nov 29 '15 at 10:13

Yes for the $\phi^{3}$ theory the vertex has 3-lines, whereas for the $\phi^{4}$ theory this becomes 4 lines meeting at the vertex. g just refers to the number of vertices in the diagrams, so for $g^{1}$, you're summing all diagrams with one vertex, for $g^{2}$, you're summing all diagrams with two vertices, and so on.

  • $\begingroup$ I don't think so. $g$ is the coupling - the operator is $g/6 \psi^3$ in this case. $\endgroup$
    – innisfree
    Nov 29 '15 at 10:14
  • $\begingroup$ The taylor expansion of the above is $Z_{1}(J) \propto \sum_{V=0}^{\infty} \frac{1}{V!}\left[\frac{i}{6}Z_{g}g\int d^{4}x(\frac{1}{i}\frac{\delta}{\delta J})^{3}\right]^{V} ...$ From which it can be seen that for V = 2, g $\rightarrow$ $g^{2}$ and the derivative term gets squared, which creates another vertex. $\endgroup$
    – Дау
    Nov 29 '15 at 19:18
  • $\begingroup$ I think I agree with your comment, but it isn't what your answer says... $\endgroup$
    – innisfree
    Nov 30 '15 at 12:25
  • 1
    $\begingroup$ "g just refers to the number of vertices in the diagrams, so for g = 1" should read "the power of $g$... so for $g^1$ $\endgroup$
    – innisfree
    Nov 30 '15 at 12:26
  • $\begingroup$ Yes you're right, I edited it. Good catch. $\endgroup$
    – Дау
    Nov 30 '15 at 14:37

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.