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$\newcommand{\Braket}[1]{\left<\Omega|#1|\Omega\right>}$ Hello,

I am currently studying QFT and have a problem concerning the 2-point correlation function in $\phi^4$-theory. When I draw all the Feynman diagrams contributing to $\left<\Omega|\phi(x)\phi(y)|\Omega\right>$ (so without the vacuum bubbles) up to $O(\lambda^3)$, I get the following:

enter image description here

However, I am not sure about the ones in the curly brackets. Are those the same diagrams? If so, how does the symmetry factor account for them?

I tried to find an answer to this, but nobody draws the third order diagrams.

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  • $\begingroup$ Think about what they represent and that will give you the answer. Alternatively think about the contractions you need to take. $\endgroup$ – Oбжорoв Jan 19 at 20:41
  • $\begingroup$ So find the solution to your problem and you have the solution to the problem? No offense, but if I didn't already think about that I wouldn't have asked the question here. $\endgroup$ – Moeman Jan 19 at 20:58
  • $\begingroup$ When thinking about the contractions (and also intuitively) I would say they should be the same. However, the rules to get the symmetry factor I know don't seem to apply here in order to take both cases into account. $\endgroup$ – Moeman Jan 19 at 21:16
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If you denote the external points by $x,y$ and the internal points by $1,2,3$ than the diagrams of the last line in your picture are $ x1, 11,12,23,23,23,3y$ and $x1,12,12,12,23,33,3y$ with integration over the internal points and what is between brackets is contracted. Yes, they are the same.

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  • $\begingroup$ Thank you very much! This clears it up for me. $\endgroup$ – Moeman Jan 20 at 11:18

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