# Twisting internal/external lines in Feynman diagrams. How does it result in different diagrams?

I'm following Griffiths' Introduction to Elementary Particles. To introduce the Feynman rules, he proposes ABC theory, where these three spin-$$0$$ particles interact through a fundamental vertex:

One of the exercises (6.12a of the 2008 edition) is to draw all lowest order diagrams for the interaction $$A + A \rightarrow A + A$$. This is the proposed solution:

What I'm having trouble is understanding in which way these are different diagrams, apart from the obvious difference between $$2$$ and $$5$$. For example, I can obtain diagram $$1$$ from $$2$$ by swapping the two vertexes on the right. Or I can get $$4$$ from $$2$$ by swapping the two vertexes on top. I can also draw this diagram:

which is isomorphic to $$2$$ by swapping the two right vertexes and the two left. So how does swapping one pair of vertexes produces different diagrams but swapping two pairs doesn't? What's the 'physical' difference between $$1$$ and $$2$$ in the proposed solution, and why isn't my diagram any 'physically' different from $$2$$?

• @Qmechanic it's exercise 6.12a, page 223 of the 2008 edition. The solution is in the solution manual. Oct 20, 2023 at 12:54

In the context of Feynman diagrams, you want to think of the 4 external lines as being distinct, while the internal lines are only labeled by their momenta (which is suppressed) and particle type. Later on, when computing an actual observable (like a scattering amplitude), you will need to sum the Feynman diagrams for different permutations of the external states to account for identical particles, but this will happen at a later step.

Therefore, you can label the four external $$A$$ lines as

• $$A_1$$ (upper left)
• $$A_2$$ (lower left)
• $$A_3$$ (upper right)
• $$A_4$$ (lower right)

Then we can trace the following line through diagram 1: $$A_1 \rightarrow C \rightarrow A_4$$ The only other place this line occurs is in diagram 6. However, diagram 6 is different from diagram 1, because in diagram 1 the loop connects the external lines in the order $$A_1 \rightarrow A_4 \rightarrow A_3 \rightarrow A_2$$ (using edges $$CBCB$$), whereas in diagram 6 the loop connects the external lines in the order $$A_1 \rightarrow A_4 \rightarrow A_2 \rightarrow A_3$$ (using edges $$CBCB$$). (I'm not conveying this idea perfectly since the loop doesn't have a direction, so you could also reverse the arrows, but the main idea is that the two loops are topologically different based on how they connect to the external lines).

Your "new" diagram is actually equivalent to diagram 1. If you trace out the loop, it connects the external lines in the order $$A_1\rightarrow A_4 \rightarrow A_3 \rightarrow A_2$$, using the edges $$CBCB$$.

In this notation it's maybe obvious why there are 6 options. You can always follow the edges in the order $$CBCB$$ and start with external line $$A_1$$. Then there are 3 options for the second line in the sequence, 2 options for the third, and one for the fourth, leading to $$3\times 2\times 1 = 6$$ possible diagrams.

• I think I understand now: we think about how the loop connects to the external lines to differentiate diagrams because if two diagrams have the same loop, then the conservation of momentum in the vertexes (expressed by the delta functions in the Feynman rules) will be the same. So in general two diagrams are the same iff the total conservation of momentum is the same, at least in this case were we have 3 different particles. Correct? Oct 20, 2023 at 21:45
• @blundered_bishop I would say the diagrams are different if they lead to different loop integrations. We don't actually know the Feynman rules for this theory, so we can't definitively say what the loop integrals are, but "generically" the topologically different diagrams (in the sense of my answer) will lead to different loop integrals. For example, say the mass of the $B$ particle is $m_B$ and the mass of the $C$ particle is $m_C$, and let $g$ be the coupling constant associated with the vertex, $k$ be the loop momentum, and $p_a$ with $a=\{1,2,3,4\}$ be the momenta of the external lines... Oct 21, 2023 at 1:02
• ... Then the loop integral for diagram 1 would be $$\int \frac{d^4 k}{(2\pi)^4}\frac{g^4}{(k^2 + m_C^2)((p_4+k)^2+m_B^2)((p_4+p_2+k)^2+m_C^2)((p_4+p_2+p_3)^2+m_B^2)}$$ which is different from what you'd get from diagram 6 (for example) $$\int \frac{d^4 k}{(2\pi)^4} \frac{g^4}{(k^2 + m_C^2)((p_4+k)^2+m_B^2)((p_4+p_3+k)^2+m_C^2)((p_4+p_2+p_3)^2+m_B^2)}$$ because of the third factor in the denominator of the two integrands. Oct 21, 2023 at 1:03