SWIMMING WITH QUARKS M. R. Pennington Institute for Particle Physics Phenomenology, University of Durham, Durham DH1 3LE, U.K. Abstract. These six lectures, given at the XI Mexican School of Particles and Fields held at Xalapa in August 2004, are on the subject of strong coupling QCD. How this colours and shapes the hadron world in terms of (i) the hadron spectrum, (ii) chiral symmetry breaking, (iii) dynamical mass generation and (iv) confinement, are the topics discussed. 1. The quark model and beyond QCD is the theory of the strong interactions. It describes the forces between quarks and gluons that make up the hadron world. It may appear that lectures about hadrons describe ancient history, like the world of nuclei before we knew about protons and neutrons. However, there is an essential difference. Hadrons are what we observe in detectors. Though we often like to describe what is going on in terms of quarks and gluons, it is hadrons that either collide or are the result of collisions (or both) and not elementary quarks and gluons. In describing proton collisions at the LHC we imagine quarks and gluons interacting and these may in turn produce an elementary Higgs. However, though quarks and gluons may well interact perturbatively over distance scales of 10−18 m., we must not forget that as they travel over bigger distances away from the hard scattering process they have to dress themselves into the protons, pions and kaons that trigger detectors. How quarks and gluons are bound into hadrons is a long distance, strong coupling problem. It involves the regime of QCD that makes this theory and the interaction it describes far more interesting than QED. It provides the challenge for our understanding. Strong coupling QCD is responsible for the spectrum of hadrons, particularly those of light flavour. It is responsible for confinement and for the chiral symmetry breaking that shapes the light hadron world reflected in experiment. These are the topics of the present lectures. QCD has its beginnings in the spectrum of hadrons and the quark model representation of these [1]-[5]. This is our first strong coupling problem. The bound state problem we learnt about at our mother's knee is that of atoms. We know that the spectrum of light, emitted as an excited atom decays, directly reflects the fact that atoms are composed of charged objects, nuclei and electrons, with an electromagnetic force holding them together. Thus the spectrum of the system teaches us about the constituents of the system and the forces that bind them together. This is equally true of the spectrum of hadrons. Let us concentrate on mesons and recall the simple structure of some of the lightest of these [4]. Quarks have spin 1/2. Consequently a quark and antiquark can only form a system of intrinsic spin S = 0 or 1. The quark and antiquark can have relative orbital angular momentum, L , Figure 1. The multiplet of the nine lightest vector mesons on the left. The implicit horizontal axis is the z -component of isospin and the vertical axis is strangeness. ρ0; !; φ are at the origin. The main hadronic decay of each of these vector states is to pseudoscalar mesons as displayed on the right, where the length of the arrow denotes the energy released in that transition. which combines with S vectorially as usual in quantum mechanics to give the spin of the meson J , where J = L + S . This then fixes the parity and charge conjugation of the meson to be P = ( 1)L+1 ; C = ( 1)L+S : (1) − − Let us first consider the quark model states with S = 1; L = 0 . This gives the vector mesons with J P C = 1−− . With three light flavours, u; d; s we have nine vector mesons. We expect these to be divided into a flavour octet and flavour singlet, since: 3 3 = 8 1 : (2) ⊗ ⊕ The quantum numbers of the nine observed states are shown in Fig. 1. The two states in the middle of the multiplet with I = 0 , the ! and φ , are however not members of an octet and a singlet, respectively, but they magically mix so that they correspond to states of definite quark flavour. We learn this from their decays and from their masses, Fig. 1. The states decay overwhelmingly into pseudoscalar mesons by the creation of uu and dd pairs from the vacuum. This explains why the strange vector state the K ∗(890) decays to Kπ and the ρ to ππ , as shown in Fig. 2. The I = 0 octet and singlet states would both decay to 3 pions ( G -parity forbids the decay to 2 pions ). While the ! does decay to 3 pions, and has a mass very close to the ρ reflecting their composition of u and d quarks, the φ overwhelmingly decays to KK despite much greater phase space for its 3π mode, Fig. 1, and an ss composition is inferred. This is intimately related to the mass differences m(φ) m(K∗) m(K∗) m(!) 120 MeV ; − ' − ' which reflect the mass difference between the s quark and the u and d . For a state made of ss quarks to decay to 3π requires the initial strange quarks to annihilate and for all the quarks Figure 2. Quark line diagrams illustrating how the creation of uu and dd pairs allows the vector mesons, K∗; ρ, φ to decay into two pseudoscalar mesons. of the final state to be created from the vacuum. We learn that such a process is suppressed. This is known as the OZI rule. The fact that the ρ0 , ! and φ have composition 1 uu dd ; ss ; (3) p2 is confirmed by their decay rate to e+e− . As illustrated in Fig. 3, in this decay the quarks annihilate to form a virtual photon that then couples to an e+e− pair. Since the coupling of the photon to quarks is proportional to their charge, these decay rates are a measure of the square of the mean charge of the constituent quarks. If all other factors are equal for each Figure 3. Feynman graph of the neutral vector meson, V , decaying into e+e− , illustrating how the matrix element for this process is proportional to the average charge of the constituent quarks in the meson. meson, then Γ(ρ0 e+e−) : Γ(φ e+e−) : Γ(! e+e−) = ! ! ! 1 2 1 2 1 2 1 2 1 2 : : + = 9 : 2 : 1 :(4) p 3 3 3 p 3 3 2 − − − 2 − Experiment agrees with this very well. Exactly, the same structure of an ideally mixed nonet shown in Fig. 1 applies to the tensor 0 mesons where the quarks combine with S = 1; L = 1 . There the heaviest, the f2(1525) (the spin-2 analogue of the φ ) overwhelmingly decays to KK . The f2(1270) composed of u and d quarks can decay to KK too by the creation of an ss pair, but strange quarks being Figure 4. Feynman graph for a qq tensor meson, T , to decay into two photons. The square of its modulus gives the decay rate. heavier than u and d quarks, this has reduced probability. These assignments can be tested this time in their decay rate to two photons. A spin-2 meson naturally couples to two spin-1 photons. This rate is related to the square of the mean charge squared of the constituent quarks as illustrated in Fig. 4 by Γ(T γγ) α2 e2 2 Π ; (5) ! ∼ h q i T where ΠT is the probability that the quarks in a tensor meson, T , annihilate. In the non- relativistic quark model this is proportional to (0) 2 , where (0) is the wavefunction at the origin. If all these factors are the same for eachjneutralj tensor meson, then for an ideally mixed multiplet 0 Γ(f2(1270) γγ) : Γ(a2(1320) γγ) : Γ(f2(1525) γγ) = 25 : 9 : 2 : ! ! ! Experiment very approximately reproduces this with 25 : (10 1) : (1 1) . That this is not more exact is a combination of imperfect experiments and the fact that the masses of the states are only roughly equal so there are inevitably corrections to the naive non-relativistic quark model. This is highlighted by the application of the same formula, Eq. (5), to pseudoscalar mesons: π; η; η0 , for which experiment gives the ratio 1 : 50 : 600 . But Hayne and Isgur [6] 3 proposed that the naive result of Eq. (5) should be corrected by a factor of MP , where MP is the mass of the pseudoscalar meson. For the π; η; η0 this contributes factors of 1 : 67 : 357 multiplying the quark charges, bringing far better agreement with experiment. What does QCD add to this picture of qq mesons? We believe confinement requires that hadronic states should be colour singlets. This means their wavefunctions must be symmetric under permutations of the colours. Then the ρ+ instead of being simply ud has a colour wavefunction: + 1 ρ = uR dR + uG dG + uB dB + ; (6) j i pNc j · · · i where R (red), G (green), B (blue), label the Nc colours. While the number of colours matters little for mesons, for baryons ·it· ·is a different matter. A colour singlet wavefunction for a state of three quarks demands that each comes in a different colour. So a proton, made of uud quarks, has a singlet wavefunction 1 p = uR uG dB + uG uB dR + uB uR dG uR uB dG uB uG dR uG uR dB : (7) j i p6 j − − − i While the wavefunction is symmetric under permutations of the colours, it is odd under the interchange of any pair.