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PhysicsSymmetry factor of Feynman diagram in $\phi^4$ theory at second order
[+4] [2] TheQuantumMan
[2021-01-10 07:44:21]
[ quantum-field-theory conventions feynman-diagrams symmetry ]
[ https://physics.stackexchange.com/questions/606583/symmetry-factor-of-feynman-diagram-in-phi4-theory-at-second-order ]

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I am trying to figure out the symmetry factor of the above diagram in the $\phi^4$ theory. The correct answer is $S=4$ (in that the Feynman diagram gets a multiplicative factor of $\frac{1}{4}$) but I get $S=8$ with the following reasoning.

For each closed loop, we can exchange the lines that start/end in the loop and belong to it. So, each loop gives a factor of $2$. Moreover, we can exchange the two vertices and simultaneously exchange the propagators of each loop. This gives an extra factor of $2$. I think it's the last bit that I'm getting wrong. Peskin and Schroeder write (p.93) that "A third possible type of symmetry is the equivalence of two vertices" but I don't think I understand precisely what this statement means. For example, why would the two vertices of this particular Feynman diagram not be equivalent? We can simply rename the integration variables and interchange them.

(1) I think, exchanging lines of second loop should not be count as an independent symmetry, because you can replace it with exchanging two loops, exchanging lines in first loop, and then exchanging loops back — the result of such combination will be exchanging lines of second loop. - warlock
Oh I see. So, generally, the echange of vertices could be consider a symmetry as long it does not produce this kind of "degeneracy" - TheQuantumMan
I am not 100% sure, but I think — yes, it is true. - warlock
(1) When you expand the exponential of the interaction to second order to get the two vertices, it comes with a factor if 1/2. This cancels the 2 from the possibility of exchanging the two vertices. Is that the factor you need? - kaylimekay
Well, the symmetry factor is the inverse of the number you take when you include the $1/n!$ from the exponential expandion, the $1/N!$ that's in front of the interaction parameter and the number of equivalent diagrams. But practically, it's the number of ways you can change stuff on the diagram while keeping it the same. - TheQuantumMan
physics.stackexchange.com/q/121402 - TheQuantumMan
There's a related question there. With that formula, I do get the desired $S=4$ but apparently, we cannot exchange vertices that are connected to external lines (or this is just how I understood the sentence "number of permutations of vertices that leave the diagram unchanged with fixed external lines" that's found in the answer in that link - TheQuantumMan
@TheQuantumMan The reason you do not count interchanges of vertices of the same type (here, all vertices are the same, but you could have different interaction terms) is because the symmetry factor resulting from such permutations always exactly cancels the $1/N!$ from the Taylor expansion. Nothing more. In particular, this is not restricted to vertices connected to an external line. - Richard Myers
But in the link I gave above, the formula does take into account permutation of vertices (the factor $g$). - TheQuantumMan
[+7] [2021-01-10 11:50:35] Prahar

Personally, I've always found the concept of symmetry factors more confusing than they're worth. I find it much easier and faster to simply count the number of contractions. The diagram you are looking at includes the fields $$\phi_1 \phi_2 \phi_x\phi_x\phi_x\phi_x \phi_y\phi_y\phi_y\phi_y$$ $\phi_1$ can contract with 8 possible fields (any of the $\phi_x$ or $\phi_y$). Without loss of generality, we can call the field it contracts with $\phi_x$. $\phi_2$ must then contract with $\phi_y$ in 4 different ways. One of the remaining 3 $\phi_x$ (chosen in 3 ways) contracts with one of the remaining 3 $\phi_y$ (chosen in 3 different ways). All other contractions are then uniquely determined. The number of contractions is then $$8\times 4\times 3\times 3$$ Putting in the factors of $\frac{1}{4!}$ for each vertex and a factor of $\frac{1}{2!}$ from the exponentiation, the overall factor in the Feynman diagram is $$ \frac{1}{2!} \frac{1}{4! }\frac{1}{4!} 8\times 4\times 3\times 3 = \frac{1}{4} . $$


Thanks for the answer. I struggle a bit with the specifics of symmetry factors and while I would rather use contractions to figure it out, simply calculating the symmetry factor directly seems to be much more efficient. - TheQuantumMan
@TheQuantumMan - To each his own. The symmetry factor in the diagram you have is not that hard to work out. It's just that once the diagrams become more complicated like 4 or 5 loop diagrams in theories with 3-pt and 4-pt vertices (eg. QCD), I get completely lost, and even if I have the right answer for the symmetry factor, I'm never confident in what I have unless I verify it by doing the contractions. - Prahar
BTW, to add to the @Qmechanic's answer, if $\phi_1$ and $\phi_2$ were indistinguishable, then I would have to divide by another factor of 2 so the overall factor would be $\frac{1}{8}$. - Prahar
I see. I suppose I'm also in the same situation as you where I feel much more confident with contractions rather than directly calculating symmetry factors. Again, thanks for the answer. - TheQuantumMan
@Prahar you contracted one of the 3 remaining \phi_x with one of the remaining \phi_y, but wouldn’t that correspond to a diagram without bubbles? Bubbled have contractions at the same point, so shouldn’t you contract them \phi_x with \phi_x - user310742
@Obama2020 - Bubbles have ALL contractions at the same point. In the contraction I did here, $\phi_1$ was already contracted with $\phi_x$ so already this cannot possibly be a bubble diagram. You can see this explicitly by literally drawing the diagram. - Prahar
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[+5] [2021-01-10 11:26:16] Qmechanic [ACCEPTED]

OP's main issue seems to be that the symmetry factor [1] $S$ depends on whether the external legs are distinguishable ($S=4$) or indistinguishable ($S=8$), respectively.

  1. On one hand, P&S on p. 93 consider the distinguishable situation$^1$ where there are not integrated over the external positions/momenta.

  2. On the other hand, OP apparently considers the indistinguishable situation where there are attached sources $J$ to each external legs and integrated over the external positions/momenta.

P&S sentence "A third possible type of symmetry is the equivalence of two vertices" seems to refer to the $\frac{1}{n!}$ in Taylor expansion of the exponential function of an interaction term.

--

$^1$ Although P&S on p. 93 are not considering amputated diagrams it might be helpful to mention that the external legs are treated as distinguishable in an amputated diagram.

[1] https://en.wikipedia.org/wiki/Feynman_diagram#Symmetry_factors

So, when the momentum attached to a vertex is fixed (so the corresponding leg is distinguished), we do not consider the vertex as part of the possible permutations of vertices that might constitue symmetries of the diagram? - TheQuantumMan
I updated the answer. - Qmechanic
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