Quark Model. Ling-Fong Li. (Institute) Note 8 1 / 26

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1 Quark Model Ling-Fong Li (Institute) Note 8 1 / 6

2 QCD Quark Model Isospin symmetry To a good approximation, nuclear force is independent of the electric charge carried by the nucleons charge independence. To implement this, assume strong interaction has SU () symmetry n $ p. These SU () generators T 1, T, T 3 satisfy commutation, Acting on n or p [T i, T j ] = i ɛ ijk T k T 3 jpi = 1 jpi, T 3jni = 1 jni, T +jni = jpi, T jpi = jni Thus n and p form a doublet under isospin p N = n and N! N 0 = e i!! τ α N, α i some real pararmeters (Institute) Note 8 / 6

3 Isospin invariance simply means that [T i, H s ] = 0 where H s is strong interation Hamiltonian. We can extend isospin assignments to other hadrons. For example we get the following isospin multiplets, (π +, π 0, π ) I = 1, (K +, K 0 ), ( K 0, K ) I = 1, η, I = 0 (Σ +, Σ 0, Σ ) I = 1, (Ξ 0, Ξ ), I = 1, Λ, I = 0 (ρ +, ρ 0, ρ ) I = 1, (K +, K 0 ), ( K 0, K ) I = 1 If isospin symmetry were exact, particles in multiplets have same masses. For example, the nucleon mass term _ m N NN = m N pp + nn is invariant under isospin transformation and gives same mass to n and p But mass di erence within isospin multiplets seems to be quite small. m n m p , m n + m p m π+ m π m π+ + m π0 Thus isospin symmetry is approximate one and maybe it is good to few %. (Institute) Note 8 3 / 6

4 In addtion to relations among masses, there are relations among couplings. For example, if we write I = 1 pion elds as! π = (π 1, π, π 3 ) then _ g N! τ! πn is invariant under isospin rotation. To see this, take in nitesmal transformation,!!! N! N 0 τ α = 1 i N, _ N! _ N 0 = _ N!!! τ α 1 + i! π!! π 0 =! π! π! α Then _ N 0!! 0 τ π N 0! _!!! τ α h!τ!π i!!!!! τ α N 1 + i + π α 1 i N ( "!τ = _ N! τ! πn + _!!! # ) α τ π!!! N i, τ π α N Using commutation relation, h τi, τ i j τ = k i ε ijk (Institute) Note 8 4 / 6

5 we get "!τ! α,!! # τ π! τ = i! π! α Write _ N 0! τ! π 0 N 0 = _ N! τ! πn! τ! π = π 0 p π + p π π 0 then _ gn! τ! p π πn = g 0 π + p n p p π π 0 n = g _ p π 0 p + p π + n + g _ p n π p π 0 n These relates various coupling constant (Institute) Note 8 5 / 6

6 SU(3) symmetry and Quark Model When Λ and k particles were discovered, they were produced in pair (associated production). It was postulated that there is a new additive quantum number, strangeness S, conserved by strong interaction but violated in decays, S (Λ 0 ) = 1, S (K 0 ) = 1 Extension to other hadrons, we get a general relation, Q = T 3 + Y where Y = B + S is called hyperchargee, and B is the baryon number. This is known as Gell-Mann-Nishijima relation. Eight-fold way : Gell-Mann, Neeman Group mesons or baryons with same spin and parity, (Institute) Note 8 6 / 6

7 (Institute) Note 8 7 / 6

8 These are the same as irreducible representations of SU (3) group. The spectra of hadrons show some pattern of SU (3) symmetry which is lots worse than isospin of SU () because the mass splitting within the SU (3) multiplets is about 0% at best. Nevertheless, it is still useful to classify hadrons in terms of SU (3) symmetry. This is known as the eight-fold way. (Institute) Note 8 8 / 6

9 Quark Model One peculiar feature of eight fold way is that octet and decuplet are not smallest representation of SU (3) group. In 1964, Gell-mann and Zweig independently propose the quark model: all hadrons are built out of spin 1 quarks which transform as the fundamental representation of SU (3), with the quantum numbers q i q 1 q q 3 A u d s Q T T 3 Y S B u /3 1/ +1/ 1/3 0 1/3 d 1/3 1/ 1/ 1/3 0 1/3 s 1/3 0 0 /3 1 1/3 In this scheme, mesons are q q bound states. For examples, A π + d u π 0 1 p (ūu d d ). π ūd K + su K 0 sd, K ūs. η 0 1 p 6 (ūu + dd ss) (Institute) Note 8 9 / 6

10 and baryons are qqq bound states, p uud, n ddu Σ + ud + du suu, Σ 0 s p, Σ sdd Ξ 0 ssu, Ξ ssd, Λ 0 s(ud du) p. Quantum numbers of hadrons are all carried by the quarks. But we do not know the dynamics which bound the quarks into hadrons. Since quarks are the fundamental constituent of hadrons it is important to nd these particles. But over the years none have been found. Paradoxes of simple quark model 1 Quarks have fractional charges while all observed particles have integer charges. At least one of the quarks is stable. None has been found. Hadrons are exclusively made out q q, qqq bound states. In other word, qq, qqqq states are absent. 3 The quark content of the baryon N ++ is uuu. If we chose the spin state to be 3, 3 then all quarks are in spin-up state~ α 1 α α 3 is totally symmetric. If we assume that the ground state has l = 0, then spatical wave function is also symmetric. This will leads to violation of Pauli exclusion principle. (Institute) Note 8 10 / 6

11 Color degree of freedom To solve these problems, introduce color degrees of freedom for each quark and postulates that only color singlets are physical observables. 3 colors are needed to get antisymmetric wave function for N ++ and remains a color singlet state. In other words each quark comes in 3 colors, u α = (u 1, u, u 3 ), d α = (d 1, d, d 3 ) All hadrons form singlets under SU (3) color symmetry, e.g. N ++ u α (x 1 )α β (x )u γ (x 3 )ε αβγ Futhermore, color singlets can not be formed from the combination qq, qqqq and they are not in the observed specrum. Needless to say that a single quark is not observable. (Institute) Note 8 11 / 6

12 Gell-Mann Okubo mass formula Since SU (3) is not an exact symmetry, we want to understand the pattern of the SU (3) breaking. Experimentally, SU () seems to be a good symmetry, we will assume isospin symmetry to set m u = m d. We will assume that we can write the hadron masses as linear combinations of quark masses. 1 o mesons m π = m(m o + m u ) mk = m (m o + m u + m s ) mη = m m o + 3 (m u + m s ) m some constant mass Eliminate the quark masses we get 4m k = m π + 3m η This is Gell-Mann Okubo mass formula. Experimentally, we hav LHS = 4m k ' 0.98(Gev ) while RHS = m π + 3m η ' 0.9(Gev ) This formula works quite well. (Institute) Note 8 1 / 6

13 1 + baryon m N = m 0 + 3m u m Σ = m o + m u + m s m Ξ = m o + m u + m s m Λ = m o + m u + m s We get the mass relation m Σ + 3m Λ = m N + m Ξ Expermentally, m Σ + 3m Λ '.3 Gev, and m N + m Ξ '.5 Gev baryon m Ω m Ξ = m Ξ m Σ = m Σ m N This is the equal spacing rule. In fact when this relation is derived the particle Ω has not yet been found and this relation is used to predicted the mass of Ω and subsequent discovery gives a very strong support to SU (3) symmetry. (Institute) Note 8 13 / 6

14 ω φ mixing For 1 mesons, situation is di erent. If we make an analogy with o mesons, we get Gell-Mann Okubo mass relation as, 3mω = 4mK m ρ Using m K = 890 Mev, m ρ = 770 Mev we get m ω = 96.5 Mev from this. But experimentally, m ω = 783Mev But there is a φ meson with mass m φ = 100 Mev and has same SU () quatntum number as ω. In principle, when SU (3) symmetry is broken, ω φ mixing is possible. Suppose there is a signi cant ω φ mixing we want to see whether this can save the mass relation. Denote the SU (3) octet state by V 8 and singlet state by V 1 V 8 = 1 p 6 (ūu + d d ss), V 1 = 1 p 3 (ūu + d d + ss) Write the mass matrix as M = m 88 m 18 m 18 m 11 Assume that the octet mass is that predicted by Gell-Mann Okubo mass relation, i.e. 3m 88 = 4m K m ρ After diagonalizing the mass matrix M, we get R + m MR = M d = ω 0 0 mφ cos θ sin θ, with R = sin θ cos θ (Institute) Note 8 14 / 6

15 Thus ω = cos θv 8 sin θv 1 φ = sin θv 8 + cos θv 1 and sin θ = s (m 88 mω) (mφ mω) Using m 88 = 96.5 Mev from Gell-Mann Okubo mass formula, we get sin θ = 0.76 q This is very close to the ideal mixing sin θ = 3 = 0.81 where mass eigenstates have a simple form, ω = 1 p (ūu + d d ) φ = ss This means that the physical φ meson is mostly made out of s quarks in this scheme. (Institute) Note 8 15 / 6

16 1 Zweig rule Since ω and φ have same quantum numbers under SU (), one expects they have similar decay widths. Experimentally, ω! 3π mostly, but φ! 3π is very suppressed relative to φ! KK channel even though φ! KK has very small phase space since m φ = 100 Mev and m k 494 Mev. B (φ! KK ) 85%, B (φ! πππ) 8% Quark diagrams In term quarks contents, the decays of φ meson proceed as following diagrams (Institute) Note 8 16 / 6

17 Zweig rule postulates that processes involving quark-antiquark annihilation are highly suppressed for some reason. This explains why φ has a width Γ φ 4.6 Mev smaaler than Γ ω 8.5 Mev. J/ψ and charm quark In 1974 the ψ/j(3100) particle was discovered with unusually narrow width, Γ 70 kev, compared to Γ ρ 150 Mev, Γ ω 10Mev. Simple explanation, ψ/j cc and is below the threshold of decaying into mesons containing charm quark. It can only decay by c c annihilation in the initial state. By Zweig rule, these decays are highly suppressed and have very narrow width. (Institute) Note 8 17 / 6

18 Asymptotic freedom 1 λφ 4 theory E ective coupling constant λ satis es L = 1 [( µφ) m φ ] λ 4! φ4 d λ dt = β( λ), β(λ) 3λ 16π + 0(λ3 ) It is not asymptotically free. Generalization: λφ 4! λ ijkl φ i φ j φ k φ l, λ ijkl is totally symmetric. It turns out that β ijkl = d λ ijkl dt = 1 16π [λ ijmnλ mnkl + λ ikmn λ mnjl + λ ilmn λ mnjk ] Take the special case, i = j = k = l β 1111 = 3 16π λ 11mnλ mn11 > 0 is not asymptotically free. (Institute) Note 8 18 / 6

19 Yukawa interaction L = ψ(i γ µ µ m)ψ + 1 [( µφ) µ φ ] λφ 4 + f ψψφ The equations for e ective coupling constant are β λ = d λ dt = Aλ + B λf + Cf 4, A > 0 β f = df dt = Df 3 + E λ f, D > 0 To get β λ < 0, with A > 0, we need f λ. This means we can drop E term in β f. With D > 0, Yukawa coupling f is not asymptotically free. Generalization to more than one fermion elds or more scalar elds conclusion remains the same. 3 Abelian gauge theory(qed) L = ψi γ µ ( µ iea µ )ψ m ψψ scalar QED : Both are not asymptotically free. de dt = β e = e 3 1π + O (e 5 ) 1 4 F µνf µν de dt = β e = e 3 48π + O (e 5 ) (Institute) Note 8 19 / 6

20 4 Non-Abelian gauge theories are asymptotically free. L = 1 T r (F µν F µν ) where F µν = µ A ν ν A µ ig [A µ, A ν ], A µ = T a A a µ with The equation for e ective coupling constant [T a, T b ] = if abc T c, T r (T a, T b ) = 1 δ ab dg dt = β(g ) = g 3 16π ( 11 3 )t (V ) < 0 where t (V )δ ab = t r [T a (V )T b (V )], t (V ) = n for SU (n) The graphs which contribute to β funciton are given below. (Institute) Note 8 0 / 6

21 Thus this is asymptotically free. If in addition,there are fermions and scalars with representation, T a (F ) and T a (s), then β g = g π 3 t (V ) t (F ) t (s) where t (F )δ ab = t r (T a (F )T b (F )) (Institute) Note 8 1 / 6

22 t (S )δ ab = t r (T a (S )T b (S )) Thus as long as t (F ),and t (s) are not too large, we still have β g < 0. (Institute) Note 8 / 6

23 QCD Quark model needs colors degrees of freedom to overcome paradoxes of simple quark model. On the other-hands, Bjroken scaling in deep inelastic scattering seems to require asymptotically free theory. These suggest gauging the color degrees of freedom of quarks ) Quantum chromodynamics L QCD = 1 t r (G µν G µν ) + q k (i γ µ D µ k m k )q k where G µν = µ A ν ν A µ ig [A µ, A ν ] D µ q k = ( µ iga µ )q k, A µ = A a λ a µ β g = 1 16π (11 3 n f ) = bg 3 where n f :number of avors. The equation for e ective coupling constant d ḡ dt = bḡ 3 t = ln λ and the solution is ḡ (t) = g 1 + bg t where g = ḡ (g, 0) (Institute) Note 8 3 / 6

24 For large momenta, λp i, ḡ (t) decreases like ln λ Conveient to de ne α s (Q ) = ḡ (t) 4π then we can write α s (Q ) = α s (µ ) 1 + 4πbα s (µ ) ln(q /µ ) Introduce Λ by the relation, ln Λ = ln µ 1 4πb α s (µ ) then α s (Q ) = 4π (11 3 n f ) ln Q /Λ Thus the e ective coupling constant α s (Q ) decreases slowly 1 ln Q. QCD can make prediction about scaling violation (small) in the forms of integral over structure functions. (Institute) Note 8 4 / 6

25 Quark con nements Since α s (Q ) is small for large Q, it is reasonable to believe that α s (Q ) is large for small Q. If α s (Q ) is large enough between quarks so that quarks will never get out of the hadrons. This is called quark con nement. It is most attractive way to "explain" why quarks cannot be detected as free particles. QCD and Flavor symmetry QCD Lagrangian is of the form L QCD = 1 t r (G µν G µν ) + k q k i γ µ D µ q k + k q k m k q k Consider the simple case of 3 avors. G µν = µ A ν ν A µ ig [A µ, A ν ], A µ = A a λ a µ D µ q k = ( µ iga µ )q k, q k = (u, d, s ) q k = (u, d, s) L QCD = 1 t r (G µν G µν ) + (ūi γ µ D µ u + d i γ µ D µ d + si γ µ D µ s) + m u ūu + m d dd + m s ss In the limit m u = m d = m s = 0, $ QCD, is invariant under SU (3) L SU (3) R transformation u d s 1 A L! U L u d s 1 A L u d s 1 A R! U R u d s (Institute) Note 8 5 / 6 1 A R

26 However, hadron spectra shows only approximate SU (3) symmetry, not SU (3) SU (3) symmetry. We can reconcile this by the scheme SU (3) SU (3) is broken spontaneously to SU (3) so that particles group into SU (3) multiplet. This would require 8 Goldstone bosons, which are massless. However in real world quark masses are not zero, these Goldstone bosons are not exactly massless. But if this symmetry breaking makes sense at all, these Goldstone bosons should be light. Thus we can identify them as pseudoscalar mesons. In other worlds, pseudoscalar mesons are almost Goldstone bosons. (Institute) Note 8 6 / 6