Srednicki Ch. 11 (p.84) provides an argument for introducing by hand a symmetry factor $S$ in the final integral for the total cross-section.

$$ \sigma = \frac1{S} \int d\sigma. \tag{11.36} $$

The argument claims that the final state is an unordered list of outgoing momenta and since the integral treats it otherwise, we should account for the miscalculation by introducing the symmetry factor by hand.

I don't get it. Could you please illustrate what he means?

Besides, from kindergarten mathematics, I know that $\int df = f$ for any integrable function $f$. Therefore, the argument sounds apologetic and wrong to me.

So, why is the symmetry factor really introduced?


Peskin & Schroeder Exercise 4.2 : Decay of a scalar particle

$$ \mathcal L = \frac12(\partial_\mu \Phi)^2 - \frac12 M^2\Phi^2 + \frac12(\partial_\mu \phi)^2 - \frac12 m^2\phi^2 - \mu \Phi \phi^2$$

The decay $\Phi \to \phi\phi$ is characterized by the amplitude $\mathcal M = -\color{red}2\mu$. Note the symmetry factor of $\color{red}2$ here in the amplitude already accounts for the multiple identical contractions. The differential decay rate would now be given as

$$ \frac{d\Gamma}{d\Omega} = \frac{\mu^2}{16\pi^2M}\sqrt{1-(\frac{2m}M)^2}\,.$$

Now, to get $\Gamma$ from this, why does everyone divide by a factor of $2$ after integrating over the solid angle?


It is a result of integrating over $d^3p_1d^3p_2$ (momenta of the final-state particles). Such integration counts both $$\left(\vec p_1,\vec p_2\right)=\left(\vec a, \vec b\right)$$ and $$\left(\vec p_1,\vec p_2\right)=\left(\vec b, \vec a\right)\;,$$ while these two situations are physically identical if particles $1$ and $2$ are indistinguishable. Hence, to avoid counting the same physical situation twice, you have to divide by 2.

The same works for any set of $n$ identical particles in the final state: you have to divide by $n!$, that is, the number of possible assignments of $n$ different momenta to $n$ indistinguishable particles.

  • $\begingroup$ @Nanashi: are you satisfied with this answer? $\endgroup$ – mdi Dec 9 '20 at 5:53

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