Understanding how to derive the Feynman rules out of the path integral

Click For Summary
SUMMARY

The discussion focuses on deriving the Feynman rules from the path integral formulation of interacting scalar fields, specifically using the potential \( V(\phi(x)) = \lambda \frac{\phi^3}{3!} \). The functional integral \( Z[J] \) is defined and differentiated with respect to the source \( J \) to establish correlation functions. Participants seek clarity on how perturbative expansions lead to Feynman rules and the associated Feynman diagrams, particularly for the two-point and three-point correlation functions. The conversation emphasizes the importance of correctly identifying external legs and vertices in diagrams corresponding to different orders of perturbation theory.

PREREQUISITES
  • Understanding of path integrals in quantum field theory
  • Familiarity with Feynman diagrams and their interpretation
  • Knowledge of perturbation theory in quantum mechanics
  • Basic concepts of scalar field theory and correlation functions
NEXT STEPS
  • Study the derivation of the Dyson series for \( N \)-point functions in quantum field theory
  • Learn about the Wick theorem and its application in calculating correlation functions
  • Explore the role of vertices and external legs in Feynman diagrams for various physical processes
  • Investigate the implications of Euler's formula in the context of Feynman diagrams and loop calculations
USEFUL FOR

Researchers and students in theoretical physics, particularly those focusing on quantum field theory, perturbation theory, and the application of Feynman diagrams in particle interactions.

  • #31
JD_PM said:
Well, once we include tadpoles, ##7## more contributions to ##\mathcal{O}(\lambda^4)## arise (just as one more arose for ##\mathcal{O}(\lambda^2)##). Source: Srednicki QFT, chapter 9

View attachment 279760
Good! One comment: in particle physics, people often omit the tadpoles. The reason for this is that they always disappear once one renormalizes. The reason is that the integrals over the loop momenta (that flow in the tadpoles) do not contain the external momenta. In the examples here, we then get that the tadpoles are just (divergent ) factors multiplying the two point functions, factors which are independent of the external momenta. After renormalization, these terms get canceled completely. However, when working in condensed matter systems (and at T not equal to zero), these tadpoles do not cancel and are therefore important.

Anyway, I know this is a technical point but I thought I would point it out.

[Moderator's note: An additional off topic comment has been deleted.]
 
  • Like
Likes   Reactions: JD_PM
Physics news on Phys.org
  • #32
nrqed said:
I am not sure if you are still interested in getting the rules from the path integral, but the key point is that when you square you will have to apply
\begin{align*}
\frac{(i)^2}{4}\int d z \, dw \, dx \, dy \,\left(\frac{\delta}{\delta\phi(z)}\right)\Delta_F(z-w)\left(\frac{\delta}{\delta\phi(w)}\right)
\left(\frac{\delta}{\delta\phi(x)}\right)\Delta_F(y-x)\left(\frac{\delta}{\delta\phi(y)}\right)
\end{align*} on the exponential.

Please let us discuss the up-to-second-order result

\begin{align*}
&\exp\left(\frac{i}{2}\int d^d x \int d^d y \frac{\delta}{\delta \phi(x)}\Delta_F(x-y) \frac{\delta}{\delta \phi(y)}\right)\times \exp\left(i \int d^d z \left( -\frac{\lambda}{3!}\phi^3+J\phi \right)\right)\Big|_{\phi=0} \\
&=1+\frac{i}{2}\int d^d x \int d^d y\,(i(J(x))\Delta_F(x-y)(iJ(y)) + \\
&-\lambda \int d^d z \int d^d w \Delta_F (0) \Delta_F (z-w) \\
&-\frac{1}{4}\int d^d x \int d^d y \int d^d z \int d^d w \Delta_F(x-y) \Delta_F(z-w) J_x J_y J_z J_w + \mathcal{O}(\Delta_F^4)
\end{align*}

I conceptually understood that there are three diagrams for the ##2-##function.

My question now is: how could we establish the Feynman rules out such result so that we can obtain such three diagrams algorithmically/systematically?
 
  • #33
Hi @vanhees71 , I just wanted to ping you because I thought you might be interested in #33 :smile:
 
  • #34
JD_PM said:
Hi @vanhees71 , I just wanted to ping you because I thought you might be interested in #33 :smile:
I think I've to do the calculation myself first. That needs a bit of time. I'm not sure whether this approach is in any way simpler than the standard approach though.
 
  • Like
Likes   Reactions: JD_PM

Similar threads

  • · Replies 13 ·
Replies
13
Views
2K
  • · Replies 0 ·
Replies
0
Views
1K
  • · Replies 1 ·
Replies
1
Views
2K
  • · Replies 3 ·
Replies
3
Views
2K
  • · Replies 5 ·
Replies
5
Views
2K
  • · Replies 24 ·
Replies
24
Views
2K
  • · Replies 1 ·
Replies
1
Views
2K
  • · Replies 15 ·
Replies
15
Views
3K
  • · Replies 1 ·
Replies
1
Views
2K
  • · Replies 1 ·
Replies
1
Views
1K