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Consider the 2 point function in $\phi^4$ theory which is given as something proportional to $$\int D(x-z) D(y-z) D(z-z) d^4 z,$$ where $D$ is the propagator. The corresponding Feynman diagram looks like this:

I understand where this comes from, but one thing I cannot figure out is why the vertex at $z$ is not connected to 4 propagators. Since the interaction term in the Lagrangian is $\phi^4$ shouldn't every vertex have 4 lines going into/out of it? Here there are only 3: $D(x-y)$, $D(z-y)$, and the loop $D(z-z)$. Are such loops counted twice? If so, why?

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Propagators correspond to line segments; since there are three line segments, there are three propagators. Two 'offshoots' of the vertex are covered by one propagator (the loop), but you'll notice that $z$ still appears four times in the integral, so the $\phi^4$ interaction is satisfied.

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  • $\begingroup$ Thanks for your answer. I'm still a little confused on how the $\phi^4$ interaction is satisfied. At this vertex $z$ interacts with $x$, $y$, and itself. So how do we get the fourth interaction? $\endgroup$
    – CBBAM
    Mar 19 at 17:46
  • $\begingroup$ You don't need four propagators; you just need four connections to the vertex. If you want to imagine drawing this: I go from $z$ to $x$, getting a factor of $D(x-z)$; I go from $z$ to $y$, getting a factor of $D(y-z)$; and I go from $z$ and back to $z$, getting a factor of $D(z-z)$. The interaction with itself counts twice, as it is both the interactor and interactee, but since there is only one interaction, we have one propagator. Another way of looking at it: the loop attaches two legs to the vertex (accounting for two out of four), but there is only one loop, so only one propagator. $\endgroup$ Mar 19 at 17:57
  • $\begingroup$ Thanks, I understand why there are only three propagators now. However I still don't understand why the $z$ self-interaction counts as two interactions. Would you be able to elaborate on that? $\endgroup$
    – CBBAM
    Mar 19 at 18:03
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Well, the self-loop propagator $D(z-z)$ in OP's Feynman diagram occupies 2 of the 4 legs of the 4-vertex. It is also responsible for a symmetry factor $S=2$ that the diagram should be divided with.

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  • $\begingroup$ How does the self-loop occupy 2 legs? I guess we count each loop twice when counting the total number of legs? $\endgroup$
    – CBBAM
    Mar 19 at 19:29

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