A. Feynman Diagrams

Since we use several of them in this document, it is worthwhile explaining Feynman diagrams a bit more detail. Figure 5 shows an example where a neutron is decaying into a proton, an electron and an electron antineutrino. The neutron and proton are both bound states of three quarks ($udd$ for the neutron and $uud$ for the proton). As an exercise, verify that the electric charges work out right.

In this diagram, time flows from the left to the right. You can see that the decay of the neutron is really just the decay of a $d$ quark into a $u$ quark, with the other two quarks being unaffected.

Figure 5: The decay of a neutron into a proton, electron and an electron antineutrino.

We conventionally draw fermions with solid lines, photons with wavy lines, gluons with spiral lines, and all other bosons with dashed lines. Arrows on the fermion lines indicate whether it is a particle or an antiparticle, with particle lines pointing in the direction of positive time.

In addition to the purely graphical display of how particles interact, Feynman diagrams are used to calculate the cross sections (or interaction probability or rate) for particle interactions. At every vertex (point where three particles meet) there is coupling constant (the coupling constant for the electromagnetic force is $\alpha$, also known as the fine structure constant, which has a value of 1/137 (at low energies)). One multiplies all the coupling constants (and propagator terms that depend on the momenta and masses of the intermediate particles) to calculate the amplitude for the reaction. The rate of a reaction occurring is proportional to the square of the amplitude.

For a given set of initial state particles, there may be several reactions that produce identical final states. Figure 6 shows an example for the reaction $e^+e^- \to \erp\erm$, which can occur either via a photon or $Z$, or via a neutralino. When this is the case, quantum mechanics requires that one calculates the probability of the reaction by calculating the amplitude of each diagram, adding the amplitudes and then squaring the sum. It is possible for the amplitudes to cancel--depending on the masses and spins of the particles involved, the probability might end up being larger or smaller than the probability if only one diagram contributed. We use this argument in section 2.2 to explain why supersymmetry solves the Higgs divergence problem.

Figure 6: The production of pairs of right-handed selectrons via two different modes.