When a process can happen via e.g. the EM and the weak-force, then to calculate the cross section of this process, we have to calculate the matrix elements of both "sub" processes, add them up and calculate the absolute value squared for the cross section: $\sigma(e^-e^+ \rightarrow q\bar{q}) \propto |M_\mathrm{fi}^\mathrm{weak} + M_\mathrm{fi}^\mathrm{em}|^2$

I've stumbled upon the (ratio) of cross sections that were used for the experimental verification for the 3 different color charges, which I don't understand.

In this case, the pairproduction via the weak process is ignored. Then the ratio of pair production of $q\bar{q}$ and $\mu^+\mu^-$ is measured. Then it says that the Quark-Antiquark pair can be produced in $N_C$ different color-anticolor configurations, so the ration $R =\frac{\sigma(e^-e^+ \rightarrow q\bar{q})}{\sigma(e^-e^+ \rightarrow \mu^+\mu^-)} = N_C \sum_f z_f^2$, where $f$ runs over all quark flavours that are accesible with the COM-Energy in the experiment, and $z_f$ is the electric charge of quark $q_f$.

My Question is, why can we here add up the cross sections of the different processes? I would have expected the ratio to be of the form $R = N_C^2 (\sum_f z_f)^2$, because I thought we have calculate the matrix element for each possible production ($M_\mathrm{fi}^{f\bar{f}} \propto N_C z_f$) and then calculate the total cross section with $\sigma(e^-e^+ \rightarrow q\bar{q}) \propto |M_\mathrm{fi}^\mathrm{u\bar{u}} + M_\mathrm{fi}^\mathrm{d\bar{d}}+\dots|^2 = N_C^2 |z_u^2 + z_d^2+\dots|^2$?

Thanks for the help!


1 Answer 1


The final states of the various quark processes $M^{u\bar u}, M^{d\bar d},\cdots$ are different so they cannot be added and then squared up. The full cross section is given by adding the different cross sections of the different processes. This is basic quantum mechanics: if the final particles are the same, then you have to add the various interactions before squaring (this is why when doing the perturbative calculation you add all the diagrams up), but if the final states are different, then they are completely different processes and need to be summed after squaring.

  • $\begingroup$ Ok I understand, thanks! $\endgroup$
    – Samuel
    Aug 19, 2021 at 15:39

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