Chapter 9. The Quark Model and the Standard Model

Theory, worked examples and 9 previous-year questions with full solutions — IIT JAM, GATE, JEST, TIFR and CSIR NET-JRF, 2020 onwards.

Theory

The hadrons of Chapter 8 fall into families of nearly equal mass whose charges, strangeness and reaction rates show simple regularities. This chapter explains them through isospin and SU(3) symmetry and through the quark model, and then introduces colour, the heavy quarks, the weak interaction of quarks, neutrino oscillations and the Standard Model with its Higgs boson. Natural units \(\hbar = c = 1\) are used where convenient, with \(\hbar c = 197.3\) MeV fm for numerical work.

9.1 Isospin

(a) Isospin multiplets

The proton (938.3 MeV) and neutron (939.6 MeV) have almost the same mass, and the nuclear force is charge independent. Heisenberg therefore treated them as two states of one particle, the nucleon, with isospin \(I = \tfrac12\): \(p = \lvert\tfrac12,+\tfrac12\rangle\), \(n = \lvert\tfrac12,-\tfrac12\rangle\). Isospin \(\mathbf I\) is a vector in an abstract space obeying the algebra of angular momentum, so \(\mathbf I^2\) has eigenvalues \(I(I+1)\) and \(I_3\) takes the \(2I+1\) values \(-I,\ldots,+I\). The strong interaction is invariant under rotations in isospin space, so it conserves both \(I\) and \(I_3\), and the members of an isospin multiplet have equal mass, spin and parity apart from splittings of a few MeV. Hadrons of the same \(B\), \(S\), \(J^P\) and nearly equal mass form multiplets of \(n = 2I+1\) members, so \(I\) is found by counting charge states, with \(I_3\) rising by one unit as \(Q\) rises by one. Thus \((p,n)\), \((\Xi^0,\Xi^-)\), \((K^+,K^0)\) and \((\bar K^0,K^-)\) are doublets, \(\pi\), \(\Sigma\), \(\rho\) triplets, \(\Delta\) a quartet, and \(\Lambda\), \(\Omega^-\), \(\eta\), \(\omega\) singlets. An antiparticle has the same \(I\) and opposite \(I_3\); thus \(K^+\) and \(K^-\) lie in different doublets, while the pion triplet is its own antimultiplet. The electromagnetic interaction violates \(I\) but conserves \(I_3\) (as in \(\Sigma^0\to\Lambda\gamma\)); the weak interaction violates both.

(b) The Gell-Mann–Nishijima formula

The charge of every hadron satisfies \[\boxed{\,Q = I_3 + \frac{Y}{2},\qquad Y = B + S + C + \tilde B + T\,}\] where \(Y\) is the hypercharge. Summing over a multiplet (\(\sum I_3 = 0\)) gives \(\langle Q\rangle = Y/2\): the hypercharge is twice the mean charge of the multiplet, which is how strangeness was first assigned. Each heavy flavour has its own quantum number, charm \(C\), bottomness \(\tilde B\), topness \(T\), with the sign of the quark’s charge: \(C = +1\) for \(c\), \(\tilde B = -1\) for \(b\), \(T = +1\) for \(t\), \(S = -1\) for \(s\). For the \(c\) quark, \(Q = 0 + \tfrac12(\tfrac13 + 1) = +\tfrac23\), as required; the charm enters with a plus sign. The formula fixes \(I_3\) of a single particle, not \(I\), which comes from the size of the multiplet or the quark content (only \(u\) and \(d\) carry isospin).

Example 9.1 A strongly interacting particle has \(B = 1\), \(I_3 = +\tfrac12\) and \(Q = 0\). Find \(Y\) and \(S\) and identify it.

Solution. \(Y = 2(Q - I_3) = -1\), so \(S = Y - B = -2\): a \(\Xi\), and the member with \(I_3 = +\tfrac12\) is the \(\Xi^0\) (\(uss\)). With \(Q = +1\) instead, \(Y = 1\), \(S = 0\), and the particle would be the proton.

9.2 Addition of isospins

(a) Clebsch–Gordan coefficients

Isospins add like angular momenta: \(I = \lvert I_1 - I_2\rvert,\ldots,I_1 + I_2\) with \(I_3 = I_3^{(1)} + I_3^{(2)}\), and \(\lvert I, I_3\rangle = \sum C^{I_1I_2I}_{m_1m_2I_3}\lvert I_1,m_1\rangle\lvert I_2,m_2\rangle\) with the usual Clebsch–Gordan coefficients. The “stretched” state of largest \(I_3\) is a single product; lowering it with \(I_- = I_-^{(1)} + I_-^{(2)}\), \(I_-\lvert I,m\rangle = \sqrt{(I+m)(I-m+1)}\,\lvert I,m-1\rangle\), and orthogonality give the rest. A state at the end of a multiplet such as \(\pi^+p\) or \(pp\) is therefore always pure isospin. For two nucleons (\(\tfrac12\otimes\tfrac12 = 1\oplus0\)) \[\lvert1,1\rangle = pp,\quad \lvert1,0\rangle = \tfrac{1}{\sqrt2}(pn + np),\quad \lvert1,-1\rangle = nn,\quad \lvert0,0\rangle = \tfrac{1}{\sqrt2}(pn - np),\] so \(pn = (\lvert1,0\rangle + \lvert0,0\rangle)/\sqrt2\) is half \(I = 1\) and half \(I = 0\); the deuteron has \(I = 0\). For a pion and a nucleon (\(1\otimes\tfrac12\), pion first, \(\pi^{+,0,-} = \lvert1,+1\rangle,\lvert1,0\rangle,\lvert1,-1\rangle\)) the coefficients are:

\(1\otimes\tfrac12\)	\(\pi^+p\)	\(\pi^+n\)	\(\pi^0p\)	\(\pi^0n\)	\(\pi^-p\)	\(\pi^-n\)
\(\lvert\tfrac32,+\tfrac32\rangle\) (\(\Delta^{++}\))	\(1\)
\(\lvert\tfrac32,+\tfrac12\rangle\) (\(\Delta^{+}\))		\(\sqrt{1/3}\)	\(\sqrt{2/3}\)
\(\lvert\tfrac12,+\tfrac12\rangle\)		\(\sqrt{2/3}\)	\(-\sqrt{1/3}\)
\(\lvert\tfrac32,-\tfrac12\rangle\) (\(\Delta^{0}\))				\(\sqrt{2/3}\)	\(\sqrt{1/3}\)
\(\lvert\tfrac12,-\tfrac12\rangle\)				\(\sqrt{1/3}\)	\(-\sqrt{2/3}\)
\(\lvert\tfrac32,-\tfrac32\rangle\) (\(\Delta^{-}\))						\(1\)

Read down a column, the same numbers give the inverse, for example \(\pi^-p = \sqrt{\tfrac13}\lvert\tfrac32,-\tfrac12\rangle - \sqrt{\tfrac23}\lvert\tfrac12,-\tfrac12\rangle\). For two isovectors (\(1\otimes1 = 2\oplus1\oplus0\)) the \(I = 2\) and \(I = 0\) states are symmetric and the \(I = 1\) states antisymmetric under exchange, e.g. \(\lvert1,0\rangle = \tfrac{1}{\sqrt2}(\lvert1\rangle\lvert{-1}\rangle - \lvert{-1}\rangle\lvert1\rangle)\), so \(\langle1,0;1,0\vert1,0\rangle = 0\). For a \(\Delta\) and a nucleon (\(\tfrac32\otimes\tfrac12 = 2\oplus1\), \(\Delta\) first) the \(I = 1\) states are \[\lvert1,1\rangle = \sqrt{\tfrac34}\,\Delta^{++}n - \sqrt{\tfrac14}\,\Delta^+p,\qquad \lvert1,0\rangle = \sqrt{\tfrac12}\,\Delta^{+}n - \sqrt{\tfrac12}\,\Delta^0p .\]

(b) The scalar product \(\mathbf I^{(1)}\cdot\mathbf I^{(2)}\)

Squaring \(\mathbf I = \mathbf I^{(1)} + \mathbf I^{(2)}\) gives, in a state of total isospin \(I\), \[\boxed{\,\mathbf I^{(1)}\cdot\mathbf I^{(2)} = \tfrac12\big[I(I+1) - I_1(I_1+1) - I_2(I_2+1)\big]\,}\] which is \(+\tfrac14\) (\(I = 1\)) and \(-\tfrac34\) (\(I = 0\)) for two nucleons, and \(+\tfrac12\) (\(I = \tfrac32\)) and \(-1\) (\(I = \tfrac12\)) for a pion and a nucleon. An isospin-invariant potential \(V = F(r)\,\mathbf I^{(1)}\cdot\mathbf I^{(2)}\) is thus \(F(r)\) times a different constant in each channel; at low energy (Born approximation) the amplitude is proportional to that constant and the cross-section to its square.

(c) Isospin invariance of amplitudes

By isospin invariance (the Wigner–Eckart theorem) a strong amplitude between states of definite isospin depends only on

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Chapter 9 The Quark Model and the Standard Model