# How to determine the squared average amplitude for $\nu_e(p_1)+d(p_2)\rightarrow e^-(p_3)+u(p_4)$?

I have the following charged current interaction, at quark level, by the process: $$\nu_e(p_1)+d(p_2)\rightarrow e^-(p_3)+u(p_4)$$ By assuming that the energy is such that I can neglect the lepton and quark masses, I obtained a certain amplitude $$M$$. I want to determine $$\langle|M|^2\rangle$$ and since in charged current interactions only left-handed chiral particle states participate, I have that $$M\equiv M(\downarrow\downarrow,\downarrow\downarrow)$$

My question is the following: the solution to the problem gives that $$\langle|M|^2\rangle=M\equiv \frac{1}{2}|M(\downarrow\downarrow,\downarrow\downarrow)|^2$$. Why the factor $$\frac{1}{2}$$? Don't I only have one possible combination for the helicities, which is the one in the image? Why not only $$\langle|M|^2\rangle=M\equiv |M(\downarrow\downarrow,\downarrow\downarrow)|^2$$?