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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?

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  • $\begingroup$ 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. $\endgroup$ – Daniel Griscom 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
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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.

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  • $\begingroup$ 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
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    $\begingroup$ "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

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