# Feynman diagrams in the Yukawa theory

I'm trying to find out the best method to find the Feynman diagrams which appear in the Yukawa theory $$L_{int} = -y \bar{\psi} \psi \phi$$ in the 4 point Greens function $$g^{(4)}$$ with external scalar fields up to $$O(y^4)$$.

Is there any better/faster way than to calculate the generating functional $$Z[J,\eta,\bar{\eta}]$$ and then differentiate the functional 4 times and set the sources to zero? In other words,

$$g^{(4)} = \frac{\delta Z[J,\eta,\bar{\eta}]}{i \delta J \, i \delta \eta \, i \delta \bar{\eta} } \bigg| _{J=\eta=\bar{\eta}=0}$$ with $$Z[J,\eta,\bar{\eta}]= \Omega \, exp\left( \,i \int d^4x \,(-y) \left(-\frac{\delta}{i \delta \eta(x) } \right) \left(\frac{\delta}{i \delta \bar{\eta}(x) } \right)\left(\frac{\delta}{i \delta J(x) } \right) \,\right) Z_0[J,\eta,\bar{\eta}]$$ and $$Z_0[J,\eta,\bar{\eta}] = exp\left( \int d^4z\int d^4z'\left[ \frac{i}{2}J(z)i\Delta_F(z-z')iJ(z') + i\bar{\eta}(z)iS_F(z-z')i\eta(z') \right] \right)$$ with $$\Omega$$ being the normalisation constant.

So is there a way to guess the diagrams without going through this calculation?

• Open any introductory textbook on QFT, e.g. Peskin & Schroeder, and it will have the Feynman rules for the Yukawa theory worked out. Form there it should be a pice of cake to write all (connected) diagrams of order 4. Feynman diagrams were after all invented not to go through the rigmarole you suggest. May 27, 2021 at 17:03

The moment you have the Feynman rules you don't need to consider the generating functional anymore. It's now rather a combinatorial problem. In your case, to find the diagrams of $$\phi \phi \to \phi \phi$$ up to order $$O(y^4)$$ you need to find all the possible ways to connect the four $$\phi$$ external lines with the four vertices. Ignoring the vacuum bubbles (the disconnected diagrams) you get the scattering amplitude at order $$O(y^4)$$.
• Thanks for your answer. I'm still quite lost. Why are you talking only of diagrams like $\phi \phi \rightarrow \phi \phi$? What about diagrams like $\phi \rightarrow \phi$ or $\phi \rightarrow \phi \phi \phi$ or the same configurations with dirac fields instead of scalar fields? May 27, 2021 at 16:55
• I thought you mentioned the 4-point correlator right? So $\phi \to \phi$ is 2-point. $\phi \to \phi \phi \phi$ is $\phi \phi \to \phi \phi$. What matters is the number of external legs (4). May 27, 2021 at 17:14