Quark model of hadrons and the SU(3) symmetry

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1 Quark moel of harons an the SU) symmetry Davi Nagy - particle physics 5) January 4, 0 Young man, if I coul remember the names of these particles, I woul have been a botanist. Enrico Fermi to his stuent Baryonic wave functions ψ = αspace) βspin) γflavour) εcolour)) Since baryons are fermions, ψ must be antisymmetric for exchange of particles. Starting from ++ = uuu with isospin I uuu = uuu an charge Q uuu = uuu ), we can get to + by stepping own I uuu = uu + uu + uu ) = + I + = u + u + u ) = 0 If we step own iagonally we get Σ + I 0 = = Q uuu = uus + usu + suu ) = Σ + I Σ + = Σ 0 an so on. Altogether 0 states which are completely symmetric in flavour for exchange of particles γf lavour) oesn t change sign). This is the baryon ecuplet.

2 Figure : Baryon ecuplet J = ) Altogether 8 states which are completely symmetric in βspin) γf lavour) for exchange of particles antisymmetric in both of them). Figure : Baryon octet J = ) Pseuoscalar mesons C u ) = u )

3 SU) symmetries q i = u s q i = U ij q i where Û is unitary with etû) =. The Gell-Mann matrices are a representation of the infinitesimal generators of the SU) group, thus any element of SU) can be written in the form Û = e iθaga where θ a R, a =,..., 8, g a = λa an the λ a are λ = λ 5 = i i 0 0 λ = λ 6 = 0 i 0 i λ = λ 7 = i 0 i 0 λ 4 = λ 8 = The matrices are traceless, hermitian an can be constructe from the commutation relations [g i, g j ] = if ijk g k where f ijk are the structure constants with sum over k implie. Then a group element is i.e. e iθλ = cosθ) sinθ) 0 sinθ) cosθ) Here, {g, g, g } constitutes a close SU) subgalgebra. correspons to Îi = g i for i =,, g q = σ ) u s = u 0 Thus, Isospin operator Hypercharge operator correspons to Ŷ = g 8 [Ŷ, Î] = 0 within SU), an we can use their common eigenvalues to characterize states, with the eigenfunctions corresponing to particles.

4 Light Quark) Mesons Mesons are combinations of a quark an an antiquark qq. Since the H space of a composite system is the tensor prouct of the state spaces of the component systems, an using the fusion rules we can ecompose the tensor prouct of two representations of a group into a irect sum of irreucible representations D i D j = k N k ijd k we can ecompose the composite system of two quarks funamental an conjugate representations corrsepon to quark an antiquark state space) as the irect sum of the trivial representation singlet) an the ajoint representation octet) = 8 Figure : an triplets Light Quark) Baryons Baryons are combinations of either three quarks qqq or three anti quarks qqq antibaryons). Their ecomposition is In orer to = resonances octet ecuplet B = Σ 0 + Λ 6 Σ + p Σ Σ0 + Λ 6 n Ξ Ξ 0 Λ 6 Gell-Mann-Okubo mass formula 6) niki p5 In case of exact SU) symmetry, all quark masses woul have to be equal. In orer to approximate the error from this, lets suppose that 4

5 . m u = m < m s, an that Then. there is no other breaking of the symmetry. H strong = H 0 + H How oes H transform uner SU)? For stationary quarks the rest energy an mass are equal q i H strong q j = q i M quark q j. Since where m u = m, so M quark = m u m m s q i M quark q j = m s + m u ) m s + m u ) m s + m u ) m u m s ) m u m s ) m u m s ) = q i m s + m u ) ˆ + m u m s ) λ 8 }{{} q j }{{} SU) invariant transforms as λ 8 Let s suppose H is small. Then we can try to use first orer perturbation H 0 ψ 0 = m 0 ψ 0 m H = ψ 0 H strong ψ 0 = m 0 + ψ 0 H ψ 0 In ψ 0 representation there is an F among F generators that transforms asλ 8 where 8ab F a F b = F af a + F + F + F ) F 8 F a F a ˆ because of Schur s lemma 5

6 F + F + F ) = II + ) thus F 8 Y H = m 0ˆ + δ m Y + δ m II + ) Y ) m H = m 0 + δ m Y + δ m II + ) Y ) 4 4 Applications For the baryon octet, there are parameters an 4 masses, which means that they are not inepenent m N + m Ξ ) = m Λ + m Σ For the ecuplet, haron mass is linear in strangeness m H = m 0 + δ m Y where we take m 0 = m 0 +δ m an δ m = δ m + δ m. The mass of Ω was preicte this way. For the pseuoscalar octet Paraoxes of the quark moel, colour symmetry. there are no free quarks. there are no qq or qqqq harons. the wave function of ++ is symmetric, because it s αspace) part is symmetric otherwise there woul be a particle with the same quantum numbers but smaller mass) niki p8 6