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

Click For Summary
The discussion focuses on deriving Feynman rules from the path integral approach for interacting scalar fields, specifically using the functional integral Z[J]. The participants explore how to differentiate Z[J] with respect to the source J to define correlation functions and express the results in terms of Feynman diagrams, emphasizing the perturbative expansion. A specific case of the potential V(φ) = λφ³/3! is examined to illustrate the process, with the aim of understanding how to establish the Feynman rules and identify associated diagrams for correlation functions. The complexity of the calculations, particularly for second-order terms, leads to confusion about the number of propagators and external legs in the diagrams, highlighting the need for clarity in distinguishing between different physical processes. The conversation underscores the importance of correctly applying perturbation theory and understanding the relationship between external legs, vertices, and loops in Feynman diagrams.
  • #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 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 JD_PM

Similar threads

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