I am troubled for if we need to divide the factor of $2$ for two identical output states in decay and scattering process in quantum field theory.

Consider Peskin and Schroeder's QFT book, on page 127, problem 4.2, the book consider the process of a scalar particle decay: $$\Phi \rightarrow \phi \ \phi $$ and in Zhong-Zhi Xianyu's solution: $$\Gamma = \frac{1}{2} F_0 $$ where $F_0$ is given in eq.(4.86), and divide $2$ is for two identical output bosons.

While in problem 4.3 (a), the book considers: $$\Phi^1 \Phi^1 \rightarrow \Phi^2 \Phi^2 $$ process, and $\Phi^1$, $\Phi^2$ have identical mass, if we consider the scattering differential cross section in center of mass frame, which is given in eq.(4.85).

However, Xianyu's solution don't divide by $2$ in this case. I am thinking since output particles are still two identical bosons, why in this case we don't divide by $2$.


1 Answer 1


Let's me finish this post myself.

First, we need to figure out where causes the symmetry factor $S$, it comes from integration in momentum space: (for example, Peskin and Schroeder's book, page 106, eq.(4.80)) $$\underset{f}\Pi\int \frac{d^3 p_f}{(2\pi)^3}\frac{1}{2E_f} $$ normally, when we do this integration, we regard the output particle are distinguishable, so $d^3p_1 d^3p_2$ and $d^3p_2 d^3p_1$ represent different final states. However, for identical output particles. Physically, we lose the order effect of $d^3p_1$ and $d^3p_2$, (we even cannot label which particle is $1$ or $2$) and we need to divide by $2$.

For a more general case, if their have $n_i^{\prime}$ identical outgoing particles of type $i$, we can define a symmetry factor (See Srednicki's QFT book, page 84, eq.(11.35) and (11.36)): $$S=\underset{i}\Pi\ n_i^{\prime}!$$ and $$\sigma = \frac{1}{S}\int d\sigma $$

Last, back to my post, for problem 4.3(a) in Peskin's book, maybe now they don't divide the symmetry factor when calculating differential cross section.


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