Part 6: Hadrons: quantum numbers and excited states

FYSH3, fall 3 Tuomas Lappi tuomas.v.v.lappi@jyu.fi Office: FL49. No fixed reception hours. fall 3 Part 6: Hadrons: quantum numbers and excited states

Introductory remarks Properties of hadrons can be understood (if not always calculated) from quark model Free quarks never seen Some of the observed systematics can also be equivalently analyzed based on observed symmetries: parity, charge conjugation, masses, spins etc. Setting (by convention) the parity of 6 hadrons one can deduce others from conservation laws Note: in this section we will neglect electroweak interaction and consider e.m. or weakly decaying partices as stable.

Observation: groups of similar mass hadrons, multiplets All within 5MeV, similar quantum numbers except electric charge Pions (pseudoscalar) π ± (39.57MeV ), π (34.98MeV ) Kaons ( strange pseudoscalar) K ± (494MeV ), K, K (497MeV ) (K +, K : S =, K, K : S = ) ρ mesons (vector) ρ,± (775MeV ) Nucleons, p(938.7mev ),n(939.57mev ), p, n baryons ++,+,, (3MeV ) Σ +,, baryons (89,9,97MeV) (S = ) Some more detached loners : pseudoscalars η(548), η (958), vectors ω(78mev ), φ(mev ), baryon Λ(6) In the quark model these are easily explained: u and d have same mass, changing u to d gives these multiplets. Before quarks were discovered the equivalent systematics was described purely in terms of symmetries and charges. 3

Hypercharge and isospin Define isospin I 3 (really 3rd component of isospin ) and hypercharge Y. Y B + S + C + e B + T I 3 = Q Y Historically: no C + e B + T known: Fixed strangeness S, trade B, Q for Y, I 3 Fixed baryon number B, trade S, Q for Y, I 3 Groups have same B, S, Y Isospin looks like angular spin, I 3 = I, I +,..., I (Originally isotopic spin ) Particles in multiplet = different isospin states of same particle. Why the funny assignment? Check B, S, Y, I 3 for hadron multiplets: B S Y I 3 π ±,,, K +, K K, K - -, ρ,±,, p,n p, n - -, ++,+,, 3 Σ +,, -,, η, η, ω, φ Λ - 4

Hypercharge and isospin of quarks Y B + S + C + e B + T I 3 Q Y Interpretation in quark model Y I 3 u 3 d 3 s 3 4 c 3 b 3 t 4 3 Isospin measures N u Nū N d + N d I 3 = for other quarks. Hadron mass independent of I 3 because: m u m d Strong interaction treats u and d equally = isospin symmetry From the assignments of isospin numbers we can guess that the mathematics of isospin is just like that of spin. i.e. SU(). One can show from the Lagrangian that this is indeed the case. With strangeness this becomes SU(3) = come back to this later. 5

SU() structure of isospin Isospin operator vector Î ˆÎi, Î j = iɛijk Î k i hî, Ĥ strong = All hadrons are eigenstates of strong interaction Hamiltonian = can be chosen as isospin eigenstates: Hadron I, I 3 Î 3 I, I 3 = I 3 I, I 3 Î I, I 3 = I(I+) I, I 3, I 3 = I, I+,... I Isospin assignments, examples π +, π, π =,,,,, Note sign in π + p, n =, fl,, fl ++, +,, = 3, 3 fl, 3, fl, 3, fl, 3, 3 fl, i (Note: hî, Îi =, so there can be a term Î in Ĥstrong = This leads to different masses for s (I = 3/) than p, n (I = /), even if other quantum numbers are the same. ) If only I 3 changes, same mass; if I different, mass different. 6

Strong interactions conserve isospin i hî, Ĥ strong = Initial and final states of a strong interaction process have same I, I 3 Scattering amplitude (derived from Hamiltonian) independent of I 3 (but can depend on I) Practical procedure. Initial/final state with several particles: tensor in isospin space.. Use Clebsch-Gordans to decompose initial state into total I, I 3 states. 3. Scattering amplitude can only depend on total I (in strong interaction process) 4. Again use CG s to find amplitudes for going from each total I, I 3 to a particular many-particle final state If initial state is just one unstable particle, step is trivial Also note that coupling a singlet I = to something is trivial... 7

Example, Nπ decay (π = π ± N = p, n) j pπ + (3) (uud) is I = 3/, I 3 = /, decays via + nπ + Z Now Γ( + Nπ) = K dω T ( + Nπ), K identical for N = p, n since the masses same. T only depends on I. In terms of isospin states p π z } { fi T ( + pπ ) = T (I = 3/) B @, + fiz } { z + } {, C A 3, + fl T ( + nπ + ) = fi T (I = 3/) B @, {z } n, + C A 3, + fl {z } {z } π + + fi (We ll come back to - sign in π + later) 8

Example continued, Nπ decay (N = p, n) Y Using Clebsch tables / 3/ +3/ 3/ / + +/ +/ +/ + / /3 /3 3/ / +/ /3 /3 / /, + fl, fl = / /3 /3 3/ 3, +/ /3 /3 3/ 3 / +3 3, fl r, fl = 3, Project out the + state from these: fl r + 3 3, fl fl + 3 3, fl, T ( + pπ ) = r T (I = 3/) 3 T ( + nπ + ) = 3 T (I = 3/) ( from π + =, ) And we get Γ( + pπ ) Γ( + nπ + ) = T ( + pπ ) T ( + nπ + ) = = You can also directly read the table the other way: 3, fl = 3, fl r, fl + 3, + fl, fl 9

Example : NN scattering Consider scattering processes pp pp, pn pn, nn nn Initial, final states are = 3 : only independent amplitudes T (cos θ) T (I =, cos θ) and T (cos θ) T (I =, cos θ). Note: pn pn : outgoing p to angle θ pn np : outgoing n to angle θ. p θ n p θ n p θ p n θ n Masses = kinematic factors are same in all cases: dσ(n N N N ) d cos θ = K M(N N N N )

Example : NN scattering, isospin amplitudes M(pp pp) M(pn pn) M(pn np) M(nn nn) fi, fi, fl, T fi,, fl, fl fi, fi, fl, T fi,, fl, fl fi +, fi, fl, T fi,, fl, fl fi, fi, fl, T fi,, fl, fl fi +, fi, fl, T fi,, fl, fl fi, fi, fl, T fi,, fl, fl

Example : NN scattering continued Using Clebsches these are M(pp pp) T M(pn pn) M(pn np) M(nn nn) T T + T T + «T Thus dσ(pp pp) d cos θ = dσ(nn nn) d cos θ = K T (cos θ ) dσ(pn pn) d cos θ = K 4 T (cos θ ) + T (cos θ ) dσ(pn np) d cos θ = K 4 T (cos θ ) T (cos θ ) These are nontrivial constraints. E.g. when cos θ = (9 scattering) np and pn are the same: dσ(pn pn) T () = and = dσ(pp pp) d cos θ θ =π/ 4 d cos θ θ =π/

Charge conjugation and isospin doublet «p u Consider isospin doublet such as or n d The corresponding antiparticles should also form isospin doublet. ««««p C u C Charge conjugation n p n d ū d But these cannot be isospin doublets, since: I 3 ( p) = I 3 (ū) =, = should be lower component of doublet I 3 ( n) = I 3 ( d) = = should be upper component of doublet Solution: perform isospin SU() rotation [Î, Ĥstrong] = = If ψ is strong eigenstate (hadron), also eiθ Î ψ is. Now, in this matrix notation, the representation of Î is σ/. So we can rotate by SU() matrix e iπσ / = Antifermion isospin doublets «to get the correct «. «p n «= n p ««ū d «= d ū «3

Signs of pion states in triplet We now have «u I = =, I 3 = d I =, I 3 = «d ū «I = =, I 3 = I =, I 3 = «Pions are isospin triplet, =, = `, = This is why + π + is isospin, state π has minus sign in uū d d This is of course just a convention; we could equally well choose = u d = π + = uū d d = π = dū = π d ū (Isospin rotation by, π, in stead of, π, ) Note: states π + and π + are the same physical particle. «as the antiquark doublet 4

Short- and long lived hadrons Consider 3 hadrons: I = 3/, m = 3MeV, decays strongly: Γ = 8MeV, cτ.7fm Λ π I =, S =, m = 6MeV, decays weakly: cτ = 7.89cm, Γ.5 MeV I =, m = 35MeV, decays electromagnetically: cτ = 5.nm= 5. 6 fm; Γ = 7.9 6 MeV. Compare lifetimes to typical hadronic distance: proton charge radius.88fm, or to resolution of particle detector. Λ π Weak decay, can be seen in detector E.m. decay; only decay products seen, but cτ size of hadrons R had Strong decay; cτ R had ; Γ m, Can decay even be separated from production? Q: so how do we know what the lifetime/width of the is? A: it is seen as a resonance in a Nπ cross section. N ev( s) ( s m ) + Γ 4 Breit-Wigner formula 5

Proton-pion cross section Γ m 6

Time dependent perturbation theory Why does unstable particle show up as a peak in the spectrum? Need time-dependent perturbation theory. Hamiltonian Ĥ = Ĥ + ˆV Develop in ψ n ; eigenstates of free Hamiltonian Ĥ: i.e. Ĥ ψ n = E n ψ n. ψ(t) = X n e ient a n(t) ψ n, Initial state decaying particle ψ, i.e. a (t = ) =, a n(t = ) =, n. d Schroedinger eq. i ψ(t) = Ĥ ψ(t) dt X e iemt [E ma m(t) + iȧ m(t)] ψ m = X h e iemt Ĥ + ˆV i a m(t) ψ m m m ψ n e ient, ψ n ψ m = δ mn = iȧ n(t) = X e i(em En)t ψ n ˆV ψ m a m(t) {z } m H nm (We can assume H nn =, diagonal part in Ĥ, not ˆV.) 7

Time dependence of decay states Perturbation theory Assume ˆV is small. With the initial condition a () =, a n() =, n assume a n ˆV. Keep only H n a V, not other H nma m V. Physically: ψ is decaying particle. Only keep transitions from ψ to decay products ψ n, not transitions among ψ n. iȧ n(t) = e i(e E n)t H n a (t). If ψ decays with rate Γ, a (t) = e Γt, then iȧ n(t) = e i(e E n iγ/)t H n P n(t /Γ) = H n E E n iγ/ = H n (E E n) + (Γ/) P n(t /Γ) = π Γ H n Γ/(π) (E E n) + (Γ/) {z } P(E n E ) Separated factors to get (continuum) normalized R de P(E E ) =. 8

Add density of states We were assuming discrete spectrum of states (box normalization... ) General density of final states ρ f (E) = P δ(e E f ) statesf In other words ρ f (E) de = number of final states between E and E + de. Now probability to decay into state with energy E is P f (E) de = π Γ H f peak z } { z } { H n P(E E ) ρ f (E) de π H f P(E E )ρ f (E ) de Γ 9

Breit-Wigner Decay into channel f with Branching fraction B f = Γ f Γ = π H f ρ f (E ) Γ Probability distribution of energies P(E E ) = Γ/(π) (E E ) + (Γ/) Experimentally: lifetime of energy distribution of decay products, width. Note: f labels particle content of final state; P(E E ) the probability distribution of energy of those particles. Γ in peak is total width, not partial width to channel f. This is the unrelativistic version, similar in relativistic case.

Formation and decay of resonance Now include production of resonance in same (nonrigorous) treatment. Assume we have ψ(t) = P n an(t) ψn a c Intial state (particles ab) ψ R in out Unstable resonance state ψ b d Possible final states n. Amplitude of resonance state Initial condition a (t = ) =, a = a n = O() z } { iȧ (t) = e i(e E )t H a (t) + X O(H m ) z } { e i(em E)t H m a m(t) e i(e E )t H m This is just the creation of the resonant state. Approximate a = R also decays: add damping term by hand (instead of treating it rigorously) ȧ (t) ie i(e E )t H Γ a (t) P(R) = P (t /Γ) = a (t) t H (E E ) + Γ /4

Resonance peak in cross section We see that the transition probability from initial state ab to R is P(ab R) H R,ab (E ab E ) + Γ /4 P(R ab) Γ R ab; Peak at E ab = s = E = m R, just like the (time reversed) decay R ab. Resonance peak in cross section ab cd For s close to the resonance mass, the ab R cross section has a peak = this peak dominates the cross section σ ab cd P(ab R)Γ R cd. in a R c out σ ab cd = π (q TRF i ) Γ R abγ R cd (E ab m R ) + Γ /4 b d q TRF i is momentum of a in TRF (b rest frame); See Martin&Shaw, KJE for factors. Isospin: CG coefficients for R ab, R cd in Γ R ab, Γ R cd. Remember: scattering also without intermediate R = but no peak in σ

Resonance particle properties Measure hadronic reaction vs. invariant mass; peak = resonance Obtain resonance particle mass, lifetime from peak center, width. Deduce quantum numbers (isospin etc.) from decay products (or initial particles). Information about decay channels Γ R X from height of peak. πn scattering, baryonic resonances π + p, in isospin basis, + /, +/ = 3/, +3/ = ++ π p, in isospin basis, /, +/ = c 3/, / + c /, / = c + c N = more peaks, also N resonance (quantum numbers of neutron). Production of resonances in a + b R + X c + d + X = other particles X involved Formation of resonances: a + b R c + d = cleaner signature 3

πn cross section plot Cross section (mb) π + ptotal π + pelastic - P lab GeV/c πp. 3 4 5 6 7 8 9 3 4 s GeV πd. 3 4 5 6 7 8 9 3 4 5 6 π dtotal Cross section (mb) π ptotal π pelastic P lab GeV/c - 4

Example: mesonic resonances Consider π + p n + π + π +. Plot cross section vs. invariant mass of X = π + π Note: now not s, but m of subset of final particles. Peaks have J PC =, ++, 3 ρ (77), J PC =, I =, I 3 = f (7) J PC = ++, I =, I 3 = ρ 3 (69) J PC = 3, I =, I 3 = X π + π quantum numbers:. J X = J ππ = L, since J π =.. C X = C ππ = ( ) L 3. P X = P ππ = ( ) L 4. B =, Q = I 3 = 5. I π = I ππ =,, B = = X meson I X =, or and I 3,X = π + π angular distribution = measure L = Identify 5

Why resonance, classical mechanics analogue Classical Harmonic oscillator with friction and external driving force spring friction external force z } { z } { z } { ẍ(t) = ωx(t) γẋ(t) + F sin(ωt) If γ =, F =, solution x(t) = A sin(ω t + ϕ ) Resonance: driving frequency ω = characteristic frequency ω At t /γ modes with sin(ω ot) die away with friction, remains: x(t) A sin(ωt + ϕ) tan ϕ = γω t ω A F = ω (ω ω ) + γ ω This is Breit-Wigner functional form, analogy F Driving force matrix elements ab R, ω ω R cd Driving frequency energy of ab ( energy = frequency!) Characteristic frequency mass of resonance γ Friction decay width (Note at ω ω (ω ω ) = (ω ω) (ω + ω) 4ω (ω ω), so A (F /ω ) /((ω ω) + γ /4)) ) 6

Strong interaction process, quark diagram Recall: gluon interactions in hadrons not calculable in perturbation theory. Example: + π + + n u π + But: strong inteactions conserve quark flavor number: just draw quark lines: + u d u π + d u u u d n d d d + u u n Feynman diagram: cannot calculate all these gluons. d d Quark diagram Quark diagram Feynman diagram In quark diagram: only draw quark lines keep track of quark flavor conservation Gluon can create quark-antiquark pair = in quark diagram can create q q pairs from nothing (same flavor!) In Feynman diagram vertices are important part in quark diagram the lines 7

Zweig rule φ s s B φ K K + 49.% φ s s B φ K K 34.% φ s s B φ pions 5% s K ū u s K+ s K d d s K d π ū u d ρ+ Experimental observation: Zweig rule Meson decays involving q q annihilation are suppressed. This is particularly important for heavy quark physics. (Note: decay is φ πρ πππ) 8

Baryons, mesons All observed resonances so far are baryons or mesons. Some exotic bound quark states, such as qqqq q are possible by symmetries, but not seen in nature. Quark diagrams are a convenient way to see whether there could be resonance peak in cross section. π + u u π d d + u u K+ K+ s s p d u u ++ Resonance peak seen in π + p π + p d u u p p d u u Z ++ d u No peaks in K + p K + p = No Z ++ resonance found, u p 9

Recent history: pentaquark craze Starting in fall of, claims of evidence for pentaquark = qqqq q. E.g. uudd s state Θ + (53) Γ = 5MeV resonance peak. (From talk by Schumacher, PANIC5) 3

But then the evidence went away... (From talk by Schumacher, PANIC5) By 5 the pentaquark was dead. Lesson: finding resonances is a tricky and complicated business. 3

SU() symmetry of isospin Two light quarks and their antiquarks form SU() doublets. ««u ū and d d Isospin SU() invariance: we can rotate these doublets by U SU() ««««u u ū ū d = U and d d = U, d and the new states u, d, ū, d behave just like the old ones under the strong interaction. What about other quarks? Strong interaction treats all flavors similarly, so for N flavors we should have SU(N) symmetry. Next step: add s quark, quark triplet = flavor SU(3) symmetry @ u d s A @ u d s A = U @ u d s A, with U U =, det U =. 3

Explicit form of SU(3) generators t a = λ a / Gell-Mann matrices: λ = @ A, λ = @ i i A, λ 3 = @ A, λ 4 = @ A, λ 5 = @ i A, λ 6 = @ A, i λ 7 = @ i i A, λ 8 = @ 3 «An SU() doublet is the two eigenstates of σ 3 / = the only diagonal generator = enough to know one eigenvalue (±/) SU(3) states are characterized by the eigenvalues of the two diagonal generators, t 3, t 8 isospin and hypercharge A. 33

SU() and SU(3) ladder operators Recall for SU(): One diagonal generator «σ 3 / «Eigenstates, for spin / representation from other two can form ladder operators σ + = (σ + iσ )/ = «σ = (σ iσ )/ = that raise or lower eigenvalue. «For SU(3) diagonal generators: t 3 (eigenvalue isospin I 3 ) and t 8 (eigenvalue 3 Y ; Y = hypercharge). Eigenstates characterized by values of I 3 and Y Other generators combined into ladder operators t ± it ; t 4 ± it 5 ; t 6 ± it 7 that raise/lower I 3, Y Fundametal representation (like spin / for SU()) generated by t a s: flavor u, d, s (or color of QCD quarks) higher representations (like spin, 3/ etc for SU()) with higher I, Y ; but ladder operators raise/lower quantum numbers the same way. 34

SU(3) multiplets Hadrons can indeed be organized into representations of SU(3): Y t it Y t it t 4 + it 5 t 4 it 5 t + it t 4 + it 5 t 4 it 5 t + it t 6 it 7 t 6 + it 7 I 3 t 6 it 7 t 6 + it 7 I 3 These are not the simplest representation = but effect of ladder operators is the same. But different hypercharge/strangeness states do not have the same masses. We do not go into representations of SU(3) here, subject for a full group theory course. 35

Mass matrix in SU(N) rotation The mass part of the Ĥ (for quarks) is mu u u + m d d d + m s s s. How does this transform in a SU(3) rotation? ( u d s ) @ m u m d m s A @ u d s ( u d s )U @ If masses were equal m u @ m d A = m @ m s A m u m d m s Quark masses break flavor SU(N) symmetry A U T @ u d s A = U mu T = m(uu ) = m A SU() of isospin is a very good symmetry becuse m u m d m s m u,d = strange particles heavier, flavor SU(3) not as good. 36

Neutral meson mixing and mass matrix Neutral mesons states uū, d d, s s have same charge, parity, C-parity etc. = they can mix. Physical mesons Pseudoscalar J PC = + : π, η, η (L =, spin singlet S = ) Vector J PC = : ρ, ω φ (L =, spin triplet S = ) are linear combinations of uū, d d, s s. What linear combinations? The eigenvectors of the mass matrix. Components of the mass matrix H ij, i, j = uū, d d, s s have contributions from Quark masses m qi δ ij = only important for s quark Gluon exchange between quark and antiquark (binding potential from QCD); only diagonal component δ ij, do not change flavor q i q i g q i q i Quark-antiquark annihilation and creation of a new pair: same for all component combinations H ij q i q i g g q j q j 37

Simple model for neutral meson mass matrix Choose basis uū = @ A, d d = @ Simplest SU(3) symmetric mass matrix would be m + ɛ ɛ ɛ H m = @ ɛ m + ɛ ɛ A. ɛ ɛ m + ɛ Strange quark mass is heavier, add parameter: m + ɛ ɛ ɛ H m = @ ɛ m + ɛ ɛ A. ɛ ɛ m + ɛ A, s s = @ This can be used to describe vector mesons ρ, ω, φ (exercise) Pseudoscalar mesons π, η, η are more complicated. A. 38

Pseudoscalar mixings Two natural assumptions for physical particle states: Flavor SU(3) symmetry conserved SU(3) octet states π = ( uū d d ), η 8 = 6 ( uū + d d s s ), SU(3) singlet state η = 3 ( uū + d d + s s ), Flavor SU(3) completely broken: strange quark very heavy. Eigenstates: Isospin triplet π = ( uū d d ), and singlet ( uū + d d ), Strange quark state = s s For neutral pseudoscalar mesons neither! Physical states are a mixture η(547) = cos θ η 8 sin θ η =.58( uū + d d ).5698 s s η (958) = sin θ η 8 + cos θ η =.49( uū + d d ) +.879 s s θ = (Either KJE or I has a typo here in numerical values... ) 39