4
$\begingroup$

I am reading the Lancaster & Blundell's Quantum field theory for the gifted amateur, p.183, Example 19.5 (Example of symmetry factors of several Feynman diagrams) and stuck at understanding several examples. I am self-studying quantum field theory and I can't possibly understand those calculations of symmetry factors on my own. Can anyone helps?

Previously, in his book, p.182, he wrote :

"The general rule is as follows : if there are $m$ ways of arranging vertices and propagators to give identical parts of a diagram ( keeping the outer ends of external lines fixed and without cutting propagator lines) we get a factor $D_i = m$. The symmetry factor if given by the product of all symmetry factors $D = \prod_{i} D_i$."

I don't understand this rule completely. Anyway, after he wrote

"Two very useful special cases, which aren't immediately apparent from the general rule, are as follows:

  • Every propagator with two ends joined to one vertex (a loop) gives a factor $D_i=2$.

  • Every pair of vertices directly joined by $n$ propagators gives a factor $D_i =n!$. "

I also don't understand this statement. Can anyone provides some explanation?

Anyway, examples of calculation of symmetry factors in his book that I stuck is as follows :

enter image description here

I can't understand the examples (e)~ (i). I somehow managed to understand (d) roughly, by refering How to count and 'see' the symmetry factor of Feynman diagrams?. But even if I try to find symmetry factors by imitating answer in the linked question, I won't be able to understand it and it will only add to confusion.

Can anyone explain the examples (e)~(i) by more step by step argument, if possible, using pictures ? I want to understand the 'all' examples completely.

For example, for (g), why its symmetry factor is $2$ , not $4$? And for (i), why is the symmetry factor only 2?

Understanding this issue will enhance understanding feynman diagram considerably. Can anyone helps?

EDIT : As a first attempt, in my opinion, I rougly guessed that

(e) Symmetric factor is $4$, since there are two loops. But in the Blundell's book, for the (e), he wrote that " (e) has a factor of $2$ from the bubble. It also has two vertices joined by two lines, contributing a factor $2!$" This statement gives me the impression that I guessed the value -by noting existence of two loops-the wrong way.

(f) Symmetric factor is $8$, since there are two loops and the two loops can be swapped ?

(g) Symmetric factor is $4$, since it seems that there are two loops

(h) I don't know why symmetry factor is $8$. It seems that there are three loops. Why the loop formed by the two external vertices is neglected?

(i) Similarly to (h), since it seems that there are two loops, symmetry factor is $4$.

Why the discrepancy occurs? Note that I am very begginer for symmetry factor. And please understanding my messy explanation and bad english.

Dreaming of complete understanding !

$\endgroup$
0

2 Answers 2

0
$\begingroup$

Rather than trying to give an encyclopedic account on symmetry factors of Feynman diagrams, let's focus on 2 issues, which seems to be relevant here:

  1. Depending on circumstances, the external legs (of same species) can be

    • distinguishable. (This is the case in Lancaster & Blundell example 19.5. This happens e.g. if the external legs are amputated.)

    • non-distinguishable. (This happens e.g. if each external leg is attached to an external $J$-source.)

    The 2 cases may lead to different symmetry factors.

  2. A self-loop (of a real field, inside a diagram) always has a symmetry of order 2. A loop consisting of more than 1 propagator (inside a diagram) may or may not have symmetry.

$\endgroup$
0
$\begingroup$

The linked answer How to count and 'see' the symmetry factor of Feynman diagrams? is all right for computing symmetry factors of particular diagrams of small sizes, but not quite for a more systematic definition and computation.

The symmetry factor $D$ is the size of the group of permutations of the set of half-edges which preserve the grouping (set partition) of half-edges into edges, and the grouping into vertices, and also fix the external half-edges, pointwise.

The underlying mathematical theory behind this is called the Joyal theory of combinatorial species. See my previous answer Problem understanding the symmetry factor in a Feynman diagram for an example of how this works. See this other answer Symmetry factor in Srednicki? for references.

For the graph (h) above the set of half-edges $H$ has 20 elements. The group of permutations to be counted is generated by the transpositions which exchanges the two half-edges of the top tadpole, by the transposition which exchanges the two half-edges of the bottom tadpole, and by the more complicated involution $\tau$ which exchanges the (a) subgraph on top (6 half-edges) with the one one the bottom (another 6 half-edges). The permutation $\tau$ is a product of 6 transpositions with disjoint supports.

$\endgroup$

Not the answer you're looking for? Browse other questions tagged or ask your own question.