# Third-order Feynman diagrams of 2-point function in $\phi^4$-theory [closed]

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

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.

• Think about what they represent and that will give you the answer. Alternatively think about the contractions you need to take. – Oбжорoв Jan 19 at 20:41
• 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. – Moeman Jan 19 at 20:58
• 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. – Moeman Jan 19 at 21:16

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.

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