Section 11.2 Feynman Diagrams and Particle Interactions
We now have all the pieces that we need to produce diagrams corresponding to the particle interactions that we discussed in the previous class. All of the conservation laws that were used to analyze those interactions will be incorporated into these interaction diagrams.
In the previous section, we reviewed the key features of Feynman diagrams as they pertain to the various interactions. A quark or lepton goes along, then emits a messenger particle, changing the properties of the original particle. Another particle receives and absorbs the messenger, and the interaction is complete. Or perhaps the messenger decays into a quark-antiquark or lepton-antilepton pair.
We can expand the Feynman diagram idea to composite particles like hadrons by just “carrying along” the non-interacting constituents. Here's the key point: in every diagram, you should consider very carefully each vertex where a messenger is emitted or absorbed. Every vertex must conserve both charge and color. The net charge before the emission/absorption process is the same as the net charge after the process. Similarly, (net color before) = (net color after). If you are careful to conserve both charge and color, then your diagrams will most likely be correct.
A few examples will best illustrate this. Consider the beta-decay of a free neutron, a process governed by the weak interaction:
To show the internal structure, simply expand the composite particle into quarks and antiquarks and let the reaction go. See Figure 11.1. Note the two types of \(W\)-boson vertices in the diagram: one changes the flavor of a quark and the other creates a lepton-antilepton pair.
Take a look at the two vertices in this diagram. The first is the vertex corresponding to the emission of the \(W^-\) from the down quark, which is changed to an up quark by the emission. Ignore the up and down quark at the left of the diagram and consider only the particles involved at that vertex. If you draw a horizontal dotted line, then everything below the line is what was present before the emission and everything above the line is what was present after the emission. Now, consider charge conservation. Charge before: \(Q_{ before} = Q_d = -1/3\) (recall that \(Q\) refers to the charge of the particle). Charge after: \(Q_{ after} = Q_u + Q_{W^-} = +2/3 - 1 = -1/3\text{,}\) so this vertex conserves charge. Note: we won't draw the horizontal dotted lines anymore.
Checkpoint 1: convince yourself that charge is also conserved at the second vertex: the decay of the \(W^-\) into the \(e^-\) and \(\overline\nu_e\text{.}\)
Note that color is automatically conserved at both of these vertices, since the \(W\) particles never have any net color, nor do the \(e^-\) and \(\overline\nu_e\text{.}\)
For the strong interaction there are several possibilities. The simplest type is an exchange reaction, an example of which is shown in Figure 11.2. The quarks just get jumbled up in the collision and then rearrange themselves to make up the end products. Note that there is still the exchange of a gluon here. Also, note that color is conserved at the vertices at both ends of the gluon. At the first (lower) vertex, the color is initially \(R\) (for the \(d\) quark in the proton). After the vertex, the color is \(B\) (for the \(d\) quark) + \(R\overline B\) (for the gluon that has just been emitted).
Checkpoint 2: convince yourself that color is also conserved at the second vertex.
Another possibility is to annihilate and create quark-antiquark pairs, mediated by a gluon. Consider the bombardment of a stationary proton with a high-energy pions, as shown in [cross-reference to target(s) "fig_pi_p_reaction" missing or not unique]. There are several ways in which the colors could be labeled in this diagram. Let's assume that the two quarks in the \(\pi^-\) before the reaction have colors \(G\) and \(\overline G\text{.}\) The three quarks in the \(p\) must always have colors \(R\text{,}\) \(G\text{,}\) and \(B\text{,}\) according to the colorless rule. The gluon will then have the color combination \(R\overline G\text{.}\)
A couple of additional points to note about the reaction in [cross-reference to target(s) "fig_pi_p_reaction" missing or not unique]. First, since strange quarks are more massive than up quarks, considerable kinetic energy (of the \(\pi^-\)) must be converted to mass-energy for this reaction to occur. In fact, a \(\pi^-\) hitting a stationary proton must have a kinetic energy of at least \(770\Xunits{MeV}\) to produce a \(\Lambda\) and \(K^0\text{.}\) Second, the gluons that mediate the strong interaction cannot change the net flavor of the quarks. Up and antiup are destroyed together; strange and antistrange are created together. As a result, the additive properties like strangeness, baryon number and charge are automatically conserved.
A third possibility for the strong interaction involves the decay of excited states of quark configurations. This can occur whenever the mass of the parent particle is high enough to have energy available for creating a quark-antiquark pair. See Figure 11.3 for an example.
Consider the decays of the particles labeled \(\Sigma^-(1197)\) and \(\Sigma^{*-}(1387)\) (the numbers in parenthesis are the masses in MeV/\(c^2\)). A proposed decay scheme might be like that shown in Figure 11.3. Note that we've shown the colors on this diagram since it involves emission and decay of a gluon. However, when we check conservation of energy, we find that the mass of the products is greater than \(1197\Xunits{MeV/ c^2 }\) but less than \(1387\Xunits{MeV/ c^2 }\text{.}\) This means the \(\Sigma^{*-}(1387)\) has enough rest energy to decay this way but the \(\Sigma^-(1197)\) does not. It must find another way.