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272 November 7, 2013<br />

10.6.2 Counting one-loop diagrams<br />

The SDe approach to counting diagrams has a number of interesting or useful<br />

applications, one of which we discuss here. We can extend the treatment of the<br />

previous section as follows. For the case of purely gluonic QCD the number of<br />

one-loop diagrams including their symmetry factors can be counted by iterating<br />

the appropriate Schwinger-Dyson equation :<br />

Φ(J) = J + 1 2 Φ2 + 1 6 Φ3 + ¯h 2 (1 + Φ) Φ′ (10.77)<br />

and taking care to discard terms of order ¯h 2 or higher.<br />

gluonic 20-point function is given by<br />

As an example, the<br />

N(19) = N 0 (19) + ¯h N 1 (19) + O (¯h 2) ,<br />

N 0 (19) = 11081983532721088487500 ,<br />

N 1 (19) = 2900013601350201168582750 . (10.78)<br />

The number N 0 (19) is the actual number of diagrams since tree diagrams always<br />

have unit symmetry factor ; but the number N 1 (19) underestimates the actual<br />

number of diagrams since the symmetry factors are not trivial. We can see,<br />

however, that the only possible nonntrivial symmetry factor at the one-loop<br />

level is 1/2, as evidenced by the factor ¯h/2 in Eq.(10.77). Inspection tells us<br />

that in this theory the only elementary Feynman diagrams that have symmetry<br />

factor 1/2 are<br />

E 1 = , E 2 = , E 3 = ,<br />

E 4 = , E 5 = .<br />

All diagrams that contain one of these elementaries as a subgraph will have a<br />

symmetry factor 1/2, and it will suffice to determine their number and multiply<br />

it by two 19 . Alternatively, we may get rid of all such diagrams, and work with<br />

the difference. This is the more useful approach ; and it illustrates how we may<br />

go about using counterterms to impose constraints on the structure of Feynman<br />

diagrams. The procedure is best explained by going through it step by step.<br />

In the first place, it will become necessary to again distinguish betwee threeand<br />

four-point vertices. We therefore modify Eq.(10.77) be reinserting labels<br />

for these couplings:<br />

Φ(J) = J + g 3<br />

2 Φ2 + g 4<br />

6 Φ3 + ¯h 2 (g 3 + g 4 Φ) Φ ′ (10.79)<br />

19 This relies, of course, on the fact that there can be no diagrams containing two (or more)<br />

of the elementaries, since that would be a two-loop diagram (or even higher).

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