# Feynman diagram, which virtual particle?

Hi I have been asked to produce the lowest order Feynman diagram for the following scattering process:

$$a.~~~ \mu^-+\mu^-\rightarrow \mu^-+\mu^-$$

$$b.~~~ \mu^-+\mu^+\rightarrow \mu^-+\mu^+$$

The Muon can interact via the electromagnetic and weak forces (and gravity). So how do I know whether the virtual particle in each elastic scattering process is a photon or Z boson? My first guess would be the photon, but I'm unsure why.

• There is more than one diagram and I assume you should show all of them. That should also answer your question, there's a diagram for $Z$ exchange and one for $\gamma$ exchange. You can summarize them by writing $\gamma/Z$ for the propagator. – pfnuesel Jan 28 '16 at 16:51
• If you can construct a diagram for a process using different vertices (i.e. one coupling to a $Z$ boson, one to a $\gamma$ and so forth), then all of those will contribute to the S-matrix elements. – JamalS Jan 28 '16 at 21:03
• BTW -- There is a reason you've been asked to do both the same charge and the opposite charge cases. – dmckee --- ex-moderator kitten Jan 29 '16 at 3:27

## 2 Answers

You have no way to know if the virtual particle is a gamma or a Z. Actually, it does not really make sense, since no measurement can tell you what has been exchanged. You always have to envisage all the possibilities resulting from the perturbative series used to describe the theory. It's a bit as asking in the double slits young experiment, which slit has been chosen by the particle before reaching the screen.

However when the energy $\sqrt{s}$ in the center of mass of the ($\mu^+$, $\mu^-$) is close to the mass of the Z, you can assume that only the Z exchange matters (because of the propagator shape). On the contrary, when $\sqrt{s}$ is close to 0, the Z exchange can be neglected, the photon dominates.

The amplitude for low-energy muon scattering mediated by $Z$ bosons is given the the muon's "weak charge," in the same way that the amplitude for scattering mediated by photons is given by the electric charge. In units where the weak charge of the neutrino is 1, the weak charge of the charged leptons is suppressed; I think at first order it's $1-\frac14\sin^2\theta_W$, with $\theta_W$ the weak mixing angle. Interference between the EM exchange and the weak exchange will introduce a parity-violating term into the low-energy scattering. This "weak charge" has been measured for the proton and electron (which show the same suppression), but those experiments (with stable particles) are quite challenging and I'm not aware of an effort to make the same measurements for the muon. The canonical review article on the subject is here.

As Paganini says, if the momentum transfer in the scattering reaction is close the $Z$ mass, that interaction dominates.