Order of scalar interaction impact Feynman diagrams

[Higgsy]

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?

 

[king vitamin]

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.