Order of scalar interaction impact Feynman diagrams

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SUMMARY

The discussion focuses on the derivation of the generating functional for scalar field theories, specifically the $$\phi^{3}$$ and $$\phi^{4}$$ interactions as outlined in Srednicki's work. The expression for the generating functional is given by $$Z_{1}(J) \propto exp\left[\frac{i}{6}Z_{g}g\int d^{4}x(\frac{1}{i}\frac{\delta}{\delta J})^{3}\right]Z_0(J)$$ for the cubic interaction and similarly for the quartic interaction. The transition from cubic to quartic interactions involves changing the order of the coupling constant $$g$$ and the corresponding Feynman diagrams reflect this change through the number of line vertices. The discussion emphasizes the importance of functional derivatives in evaluating integrals related to these generating functionals.

PREREQUISITES
  • Understanding of quantum field theory concepts, particularly scalar fields.
  • Familiarity with generating functionals in quantum mechanics.
  • Knowledge of functional derivatives and their applications in field theory.
  • Experience with Feynman diagrams and their representation of interactions.
NEXT STEPS
  • Study the derivation of the generating functional for $$\phi^{4}$$ theory in detail.
  • Learn about the role of functional derivatives in quantum field theory.
  • Explore the implications of different interaction terms on Feynman diagrams.
  • Investigate the analytic properties of generating functionals in quantum mechanics.
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The discussion is beneficial for theoretical physicists, particularly those specializing in quantum field theory, as well as graduate students seeking to deepen their understanding of scalar interactions and generating functionals.

Higgsy
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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?
 
We define the generating functional,

<br /> 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]<br />

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

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

exactly as an analytic functional of J(x). In particular, I can take functional derivatives with respect to J. Then by taking derivatives, we can evaluate the following functional integrals:
<br /> \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].<br />

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

So for ANY interaction, you just replace the interaction lagrangian with \phi(x) \rightarrow \frac{1}{i}\frac{\delta}{\delta \phi(x)} acting on the free generating function. The easiest way to deal with \phi^4 theory is with the Lagrangian
<br /> \mathcal{L}_1 = \frac{g}{24} \phi(x)^4<br />
(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
<br /> Z[J] = \exp\left( \frac{g}{24}\int d^dx \left( \frac{1}{i}\frac{\delta}{\delta \phi} \right)^4 \right) Z_0[J].<br />
Then expanding the exponential in powers of g gives you the Feynman expansion.
 

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