Factor of 1/2 in the cross section of Møller scattering I have seen everywhere (like Wikipedia) that in
Møller scattering the cross section is calculated with the formula:
$$\frac{\mathrm d \sigma}{\mathrm d \Omega} = \frac{1}{64 \pi^2 E_{CM}^2 } \frac{|\pmb{p}_i|}{|\pmb{p}_f|} \overline{|M_{fi}|}^2 $$
But shouldn't there be an extra factor of $\frac{1}{2} $ due to indistinguishableness of the final two electrons? This should change the formula to:
$$\frac{\mathrm d \sigma}{\mathrm d \Omega} = \frac{1}{128 \pi^2 E_{CM}^2 } \frac{|\pmb{p}_i|}{|\pmb{p}_f|} \overline{|M_{fi}|}^2 $$
Why isn't this accounted for?
 A: The indistinguishable particle factor is handled in $\overline{|M_{fi}|^2}$. (Note that the square goes under the over-bar).
There are two separate diagrams at tree level. The $t$-channel process:
$$ e^-_1e^-_2 \rightarrow e^-_3e^-_4$$

and the $u$-channel, describing the crossed final state:
$$ e^-_1e^-_2 \rightarrow e^-_4e^-_3$$

They add coherently, with a minus sign in-front of the crossed-channel:
$$iM_{fi} = i(M_t - M_u)$$
$$M_{fi}=-i(-ie)^2\big[
\frac 1 t\bar u(p_3)\gamma^{\mu}u(p_1)\bar u(p_4)\gamma_{\mu}u(p_2)-
\frac 1 u\bar u(p_3)\gamma^{\mu}u(p_2)\bar u(p_4)\gamma_{\mu}u(p_1)
\big]$$
You then square that and take the appropriate traces, which sums over the initial and final spin states. Hence, you need to divide by 4 in order to average over the initial spin states (unless your beam/target is/are polarized):
$$\overline{|M_{fi}|^2}=\frac 1 4 \sum_{\rm spins}|M_{fi}|^2 $$
$$=2e^4\big[\frac 1 {t^2}(s^2+u^2-8m^2(s+u)+24m^4)+
\frac 1{u^2}(s^2+t^2-8m^2(s+t)+24m^4)
+\frac 2{tu}(s^2-8m^2s+12m^4)\big]$$
So the $1/t^2$ term is from $t$-channel scatterings, and the $1/u^2$ term is exchanged final state. The $1/{tu}$ term is from interference between the two amplitudes. Clearly, it's not just a factor of $1/2$.
From there, you transform to the lab frame:
$$\frac{\mathrm d \sigma}{\mathrm d \Omega} = \frac{1}{64 \pi^2 E_{CM}^2 } \frac{|\pmb{p}_i|}{|\pmb{p}_f|} \overline{|M_{fi}|^2} $$
