Tuesday, 14 April 2015

quantum field theory - Formula for Symmetry Factor


In $\phi^3$ theory, are there any formula for determining the Symmetry factor as that is found for the $\phi^4$ theory in any standard book of Quantum Field Theory?



Answer



Peskin and Schroeder provide a sufficiently detailed explanation regarding the computation of symmetry factors for Feynman diagrams. The paper by Palmer et al. present a general formula,


$$S=\frac{1}{R}\left( \frac{1}{2}\right)^{D_1}\left( \frac{1}{2!}\right)^{D_2}\left( \frac{1}{3!}\right)^{D_3}\left( \frac{1}{4!}\right)^{D_4}$$


where the constants are defined in their paper, which require an understanding of the derivation and its notation. The expression is applicable to QED, QCD and $\phi^{3,4}$ theory but generalizable to others. For a diagram such as (considered in their paper as figure 1),



enter image description here


For this case, $D_1=D_3=D_4=0$, $R=1$ and $D_2$ = 1 which yield, $S=1/2$ as expected. In Dong's paper for real and complex scalar field theories, he presents the general formula,


$$S=g2^\beta 2^d \prod_n \left( n!\right)^{\alpha_n}$$


where (quoting from the paper): $g$ is the number of of interchanges of vertices leaving the diagram topologically unchanged, $\beta$ is the number of lines connecting a vertex to itself, $d$ is the number of double bubbles, and $\alpha_n$ is the number of vertex pairs connected by $n$ identical lines.


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