1 Particle Physics Dr M.A. Thomson q q q n(udd) p(uud) Σ - (dds) Σ 0 (uds) Σ + (uus) Λ(uds) Ξ - (dss) Ξ 0 (uss) Part II, Lent Term 2004 HANDOUT IV
2 The Quark Model of Hadrons EVIDENCE FOR QUARKS The magnetic moments of proton and neutron are not e~ and 0 2m p ) not point-like electron-proton scattering at high q 2 deviates from Rutherford scattering ) proton has sub-structure jets are observed in e + e and pμp collisions symmetries (patterns) in masses/properties of hadron states, quarky periodic table ) sub-structure R μ Observation of cμc and bμb bound states. and much, much more... Here we will first consider the wave-functions for hadrons formed from the light quarks (up, down, strange) and deduce their static properties (mass and magnetic moments). Then we will go on to discuss the heavy quarks, charm, bottom (sometimes called beauty), and top.
3 Hadron Wave-functions in the Quark Model Due to properties of QCD quarks are always confined to hadrons (i.e. colourless states). MESONS SPIN-0,1,... q q BARYONS SPIN- 1 2, 3 2,. q q q Last year several reports of observations of pentaquark bound states qqqqq HADRON WAVE-FUNCTIONS Treat quarks as IDENTICAL fermions with states labelled with SPATIAL, SPIN, FLAVOUR, COLOUR ψ = ψ space ψ flavour ψ spin ψ colour All hadrons are COLOUR SINGLETS, i.e. net colour zero ψ qq colour = 1 p 3 (rr + gg + bb) ψ qqq colour = 1 p 6 (rgb+gbr+brg-grb-rbg-bgr) Consider ground state i.e. (L=0) mesons and baryons Determine wave-functions Determine parity Calculate masses Calculate magnetic moments
4 Parity Parity operator, ^P,! SPATIAL INVERSION ^Pψ(~r; t) = ψ( ~r; t) The Eigenvalue of ^P is called PARITY ^Pψ = Pψ; P = ±1 Particles are EIGENSTATES of PARITY and in this case P represents the INTRINSIC PARITY of a particle/anti-particle. Parity is a useful concept. IF the Hamiltonian for an interaction commutes with ^P [^P; ^H] = 0 then PARITY IS CONSERVED in the interaction: PARITY CONSERVED in STRONG interaction PARITY CONSERVED in EM interaction but NOT in the WEAK interaction Composite system of 2 particles with orbital angular momentum L: P = P 1 P 2 ( 1) L Quantum Field Theory: fermions/anti-fermions : OPPOSITE parity bosons/anti-bosons : SAME parity Choose: quarks/leptons : P = +1 anti-quarks/anti-leptons : P = 1 Gauge Bosons (fl,g,w,z) are vector fields which transform as J P = 1
5 SPIN QM Revision Quantum Mechanical LADDER operators, ^J ± = ^J x ± i^j y ^J + jj; mi = p j(j+1)-m(m+1)jj; m+1i ^J - jj; mi = p j(j+1)-m(m-1)jj; m-1i For Example - can generate all jj; mi states from jj; ji j1; 1i = j 1 2 ; 1 2 ij 1 2 ; 1 2 i = "" Since ^J x etc. formed from derivatives use product rule, d(uv) = udv+vdu, when acting on product of states: ^J - j1; 1i = (^J - j 1 2 ; 1 2 i)j 1 2 ; 1 2 i + j 1 2 ; 1 2 i(^j - j 1 2 ; 1 2 i) p 2j1; 0i = j 1 2 ; - 1 2 ij 1 2 ; 1 2 i + j 1 2 ; 1 2 ij 1 2 ; - 1 2 i j1; 0i = 1 p 2 ("# + #") For the combination of two spin half particle: j1; 1i = "" j1; 0i = p 1 2 ("# + #") j1; 1i = ## All SYMMETRIC under interchange of 1 $ 2 Also an orthogonal combination which is ANTI-SYMMETRIC under under interchange of 1 $ 2 j0; 0i = 1 p 2 ("# #")
6 Light Mesons Mesons are bound qq states. Here we consider only mesons consisting of LIGHT quarks (u,d,s). m u ο 0:3 GeV; m d ο 0:3 GeV; m s ο 0:5 GeV Ground state (L = 0) For ground states, where orbital angular momentum is zero, the meson spin (total angular momentum) is determined by the qq spin state. Two possible qq total spin states S = (0; 1) S = 0 : pseudo-scalar mesons S = 1 : vector mesons Meson Parity : (q and q have OPPOSITE parity): P = P (q)p (q)( 1) L = (+1)( 1)( 1) L = 1 (forl = 0) Flavour States: uμd, uμs, dμu, dμs, sμu, sμd (uμu, dμd, sμs) MIXTURES Expect : 9 J P = 0 MESONS : PSEUDO-SCALAR NONET 9 J P = 1 MESONS : VECTOR NONET
uds MULTIPLETS 7 QUARKS Strangeness ANTIQUARKS Strangeness s d u -1/2 +1/2-1 s Isospin J P = 1/2 + u d Isospin J P = 1/2 - Strangeness ds us SPIN 0 or 1 du dd uu ud ss Isospin su sd The ideas of strangeness and isospin are historical quantum numbers assigned to different states. Essentially they count quark flavours (this was all before the formulation of the quark model) Isospin = 1 2 (n u n d n μu + n d ) Strangeness = n μs n s
8 Light Mesons ZERO ORBITAL ANGULAR MOMENTUM Strangeness K 0 ds K + us J P = 0 - PSEUDOSCALAR NONET π 0,η,η, are combinations of uu, dd, ss π - du η η, π + K - su π 0 K 0 sd ud Isospin Masses/MeV: π(140), K(495) η(550), η, (960) Strangeness K *0 ds K *+ us J P = 1 - VECTOR NONET ρ 0,φ,ω 0 are combinations of uu, dd, ss ρ - du ω 0 φ ρ + K *- su ρ 0 K *0 sd ud Isospin Masses/MeV: ρ(770), K * (890) ω(780), φ(1020)
Meson Wave-functions 9 uμd, uμs, dμu, dμs, sμu, sμd are straightforward However, (uμu, dμd, sμs) states all have zero flavour quantum numbers - therefore can MIX ß 0 (140) = 1 p 2 (uμu dμd) (550) = 1 p 6 (uμu + dμd 2sμs) 0 (960) = 1 p 3 (uμu + dμd + sμs) ρ 0 (770) = 1 p 2 (uμu dμd)! 0 (780) = 1 p 2 (uμu + dμd) ffi(1020) = sμs 9 >= >; JP = 0 9 >= >; JP = 1 Mixing coefficients determined experimentally from masses, decays. e.g. leptonic decays of vector mesons e - q Q q α Matrix element / Q α q ff V Decay rate / Q 2 qff 2 q γ e + ee / Q 2 qff 2» M fi (ρ 0! e + e ) ο e 1 1p2 q 2 (Q u e Q d e)» 1p2 2 ρ 0!e + e / 3 1 2 = 1 3 2» 1p2 2! 0!e + e / 3 + 1 2 = 1 3 18 ffi!e + e /» 1 3 2 = 1 9 PREDICT: ρ 0 :! 0 : ffi = 9 : 1 : 2 EXPERIMENT: 8:8 ± 2:6 : 1 : 1:7 ± 0:4
10 Meson Masses Meson masses partly from constituent quark masses m(k) > m(ß) hints at m s > m u ; m d But that is not the whole story m(ρ + ) > m(ß + ) (770 MeV c.f. 140 MeV) but both are uμd Only difference is in orientation of Quark SPINS "" vs. #" SPIN-SPIN INTERACTION QED: Hyperfine splitting in H 2 (L=0) Energy shift due to electron spin in magnetic field of proton E = ~μ:~b = 2 3 ~μ e:~μ p jψ(0)j 2 using ~μ = e 2m ~ S E / ff em ~S e :~S p m 1 m 2 QCD: Colour Magnetic Interaction Fundamental form of the interaction between a quark and a gluon is identical to that between an electron and a photon. Consequently, also have a COLOUR MAGNETIC INTERACTION E / ff S ~S 1 :~S 2 m 1 m 2
MESON MASS FORMULA (L=0) 11 M qq = m 1 + m 2 + A ~ S 1 : ~ S 2 m 1 m 2 where A is a constant For a state of SPIN ~S = ~S 1 + ~S 2 ~S 2 = ~S 1 2 + ~S 2 2 + 2 ~S 1 :~S 2 ~S 1 :~S 2 = 1 2 ( ~S 2 ~S 1 2 ~S 2 2 ) S 2 1 = S2 2 = S 1 (S 1 + 1) = 1 2 (1 + 1 2 ) = 3 4 giving ~S 1 :~S 2 = 1 2 ~ S 2 3 4 For J P = 0 MESONS : ~S 2 = 0 J P = 1 MESONS : ~S 2 = S(S + 1) = 2 therefore ~S 1 :~S 2 = 1 ~ 2 S 2 3 4 = 3 4 (0 mesons) ~S 1 :~S 2 = 1 2 ~ S 2 3 4 = + 1 4 (1 mesons) Giving the (L=0) Meson Mass formulae M = m 1 + m 2 M = m 1 + m 2 + 3A 4m 1 m 2 A 4m 1 m 2 (0 mesons) (1 mesons) 0 mesons lighter than 1 mesons
Can now try different values of m u=d, m s and A and try to reproduce the observed values. 12 Mass/MeV Meson Predicted Experiment ß 140 138 K 484 496 ρ 780 770! 780 782 K Λ 896 894 ffi 1032 1019 Excellent agreement using: m u = m d = 310 MeV, m s = 483 MeV, A = 0:06 GeV 3. (see Question 4 on the problem sheet) Bring on the Baryons...
13 BARYON WAVE-FUNCTIONS Baryons made from 3 indistinguishable quarks (flavour treated as another quantum number in the wave-function) ψ baryon = ψ space ψ flavour ψ spin ψ colour ψ baryon must be ANTI-SYMMETRIC under interchange of any 2 quarks. Ground State (L=0) Here we will only consider the baryon ground states. The ground states have zero ORBITAL ANGULAR momentum, ) ψ space is symmetric All hadrons are COLOUR SINGLETS ψ colour = 1 p 6 (rgb+gbr+brg-grb-rbg-bgr) i:e: ψ colour is anti-symmetric Therefore ψ space ψ colour is anti-symmetric ) ψ spin ψ flavour must be SYMMETRIC
Start with the combination of three spin-half particles. Trivial to write down the spin wave-function for the j 3 2 ; 3 2 i state : j 3 2 ; 3 2 i = """ 14 Generate the other states using ^J - q 3 2 5 2-3 2 ^J - j 3 2 ; 3 2 i = (^J - ") "" + " (^J - ") " + "" (^J - ") 1 2 2 ; 1 2i = #"" + "#" + ""# j 3 2 ; 1i 2 = p 1 3 (#"" + "#" + ""#) Giving the spin 3 2 states : j 3 2 ; 3i 2 = """ j 3 2 ; 1i 2 = p 1 3 (#"" + "#" + ""#) j 3 2 ; - 1 2 i = 1 p 3 ("## + #"# + ##") j 3 2 ; - 3 2 i = ### ALL SYMMETRIC under interchange of any two spins For the spin-half states, first consider the case where the first two quarks are in a j0; 0i state. j0; 0i (12) = 1 p 2 ("# #") j 1 2 ; 1 2 i (123) = j0; 0i (12) j 1 2 ; 1 2 i = 1 p 2 ("#" #"") j 1 2 ; - 1 2 i (123) = j0; 0i (12) j 1 2 ; - 1 2 i = 1 p 2 ("## #"#) ANTI-SYMMETRIC under interchange of 1 $ 2.
3-quark spin-half states can ALSO be formed from the state with the first two quarks in a SYMMETRIC spin wave-function. Can construct a 3-particle j 1 2 ; 1 2 i (123) state from 15 Taking a linear combination: j1; 0i (12) j 1 2 ; 1 2 i (3) and j1; 1i (12) j 1 2 ; - 1 2 i (3) j 1 2 ; 1 2 i = aj1; 1ij 1 2 ; - 1 2 i + bj1; 0ij 1 2 ; 1 2 i with a 2 + b 2 = 1. Act upon both sides with ^J + : ^J + j 1 2 ; 1 2 i = a(^j + j1; 1i)j 1 2 ; - 1 2 i + aj1; 1i(^J + j 1 2 ; - 1 2 i) +b(^j + j1; 0i)j 1 2 ; 1 2 i + bj1; 0i(^J + j 1 2 ; 1 2 i) 0 = aj1; 1ij 1 2 ; 1 2 i + p 2bj1; 1ij 1 2 ; 1 2 i a = p 2b which with a 2 + b 2 = 1 implies: q q 2 a = 3 ; b = 1 3 Giving j 1 2 ; 1i 2 = q q 2 3 j1; 1ij 1 2 ; - 1i 1 2 3 j1; 0ij 1 2 ; 1i 2 j 1 2 ; 1i 2 = p 1 6 (2 ""# "#" #"") similarly j 1 2 ; - 1 2 i = 1 p 6 (2 ##" #"# "##)
3 QUARK SPIN WAVE-FUNCTIONS 3 ❶ 2 j 3 2 ; 3i 2 = """ j 3 2 ; 1 2 i = 1 p 3 (#"" + "#" + ""#) j 3 2 ; - 1 2 i = 1 p 3 ("## + #"# + ##") j 3 2 ; - 3 2 i = ### 16 SYMMETRIC under interchange of any two quarks ❷ 1 2 j 1 2 ; 1i 2 = p 1 6 (2 ""# "#" #"") j 1 2 ; - 1 2 i = 1 p 6 (2 ##" #"# "##) SYMMETRIC under interchange of 1 $ 2 ❸ 1 2 j 1 2 ; 1 2 i = 1 p 2 ("# #") " j 1 2 ; - 1 2 i = 1 p 2 ("# #") # ANTI-SYMMETRIC under interchange of 1 $ 2 ψ spin ψ flavour must be symmetric under interchange of any two quarks.
Consider 3 cases: ❶ Quarks all SAME Flavour: uuu,ddd,sss 17 ψ flavour is SYMMETRIC under interchange of any two quarks. REQUIRE ψ spin to be SYMMETRIC under interchange of any two quarks. ONLY satisfied by SPIN- 3 2 states. no uuu, ddd, sss, SPIN- 1 2 3 SPIN- 3 2 states : uuu,ddd,sss. baryons with L = 0 ❷ Two quarks have same Flavour: ddu,uud,.. For the like quarks, ψ flavour is SYMMETRIC. REQUIRE ψ spin to be SYMMETRIC under interchange of LIKE quarks 1 $ 2. satisfied by SPIN- 3 2 and SPIN- 1 2 6 SPIN- 3 2 and 6 SPIN- 1 2 ssd states: uud, uus, ddu, dds, ssu, For example: PROTON wave-function (s z = 1 2 ): p "= 1 p 6 (2u " u " d # u " u # d " u # u " d ") The above form for the proton wavefunction is sufficient to derive masses and magnetic moments, however, note that the fully symmetrized wavefunction includes cyclic permutations: 1 p 18 ( 2u " u " d # u " u # d " u # u " d " + 2d # u " u " d " u " u # d " u # u " + 2u " d # u " u " d " u # u # d " u ")
18 ❸ All Quarks have DIFFERENT Flavour: uds Two possibilities for (ud) part: i) FLAVOUR SYMMETRIC 1 p 2 (ud + du) require spin wave-function to be SYMMETRIC under interchange of ud satisfied by SPIN- 3 2 and SPIN- 1 2 states! ONE SPIN- 3 2! ONE SPIN- 1 2 uds state uds state ii) FLAVOUR ANTI-SYMMETRIC 1 p 2 (ud du) require spin wave-function to be ANTI-SYMMETRIC under interchange of ud only satisfied by SPIN- 1 2 state! ONE SPIN- 1 2 uds state In total TEN (3+6+1) SPIN- 3 states with the 2 required overall symmetry and EIGHT (0+6+2) SPIN- 1 2 states Quark Model predicts Baryons appear in DECUPLETS of SPIN- 3 2 and OCTETS of SPIN- 1 2
19 Strangeness SPIN 1/2 n(udd) p(uud) Σ - (dds) Σ 0 (uds) Σ + (uus) Λ(uds) Isospin 940 MeV 1190 MeV 1115 MeV Ξ - (dss) Ξ 0 (uss) 1320 MeV SPIN 3/2 - (ddd) 0 (ddu) + (uud) ++ (uuu) Σ - (dds) Σ 0 (uds) Σ + (uus) 1230 MeV 1385 MeV Ξ - (dss) Ξ 0 (uss) 1530 MeV Ω - (sss) 1670 MeV
BARYON MASS FORMULA (L=0) M qqq =m 1 +m 2 +m 3 +A 0 S1 ~ : S ~ 2 S + ~ 1 : S ~ 3 S + ~ 2 : S ~ 3 m 1 m 2 m 1 m 3 m 2 m 3 20 where A 0 is a constant EXAMPLE m 1 = m 2 = m 3 = m q here M qqq = 3m q + A 0 X i<j ~S i :~S j m 2 q ~S 2 = (~S 1 + ~S 2 + ~S 3 ) 2 ~S 2 = ~S 1 2 + ~S 2 2 + ~S 3 2 + 2 X 2 X i<j ~S i :~S j = S(S + 1) 3 1 2 ( 1 2 + 1) i<j ~S i :~S j 2 X i<j ~S i :~S j = S(S + 1) 9 4 X i<j X i<j ~S i :~S j = 3 4 J = 1 2 ~S i :~S j = + 3 4 J = 3 2 e.g. proton(uud) versus (uud) m p = 3m u 3 4 m = 3m u + 3 4 A 0 m 2 u A 0 m 2 u
Again try different values of m u=d, m s and A 0 and try to reproduce the observed values. Mass/MeV Baryon Predicted Experiment p/n 939 939 Λ 1116 1114 ± 1193 1179 Ξ 1318 1327 1232 1239 ± Λ 1384 1381 Ξ Λ 1533 1529 Ω 1672 1682 21 Excellent agreement using: m u = m d = 363 MeV, m s = 538 MeV, A 0 = 0:026 GeV 3. QCD predicts A 0 = A=2 where A is the corresponding constant in the meson mass formula. Recall A = 0:06 GeV 3 provided a good description of meson masses.
Baryon Magnetic Moments 22 Assume the bound quarks within baryons behave as DIRAC point-like SPIN-1/2 particles with fractional charge, q q. Then quarks will have magnetic dipole moments: ^μ q = q q m q ^S where m q is the quark mass. Magnitude of the magnetic dipole moment: μ u = 2 3 μ q = hq " j q q m q ^Sjq "i with ^Sjq "i = 1 ~jq "i 2 giving μ q = q q~ 2m q e~ 2m u ; μ d = 1 3 e~ 2m d ; μ s = 1 3 e~ 2m s For quarks bound within a L=0 baryon, X, the baryon magnetic moment is the expectation value of the sum of the individual quark magnetic moments. ^μ B = q 1 ^S 1 + q 2 ^S 2 + q 3 ^S 2 m 1 m 2 m 3 μ X = hx "j^μ B jx "i where jx "i is the baryon wave-function for the spin up state.
p " = 1 p 6 (2u " u " d # (u " u # +u # u ")d ") ) μp = 1 6 h(2 ""# ("# + #") ")j^μ 1 + ^μ2 + ^μ3j(2 ""# ("# + #") ")i 1 h(2" "# (" # +# ") ")j^μ 1j(2" "# (" # +# ") ")i 6 = 1 6 h(2" "# (" # +# ") ")j(2μ 1" "# (μ1" # μ1# ") ")i = 2 3 μ 1 = 2 3 μ u = 4 9 e~ 2mu μp = 4 3 μ u 1 3 μ d = 4 9 e~ 2mu + 4 9 where μn is the NUCLEAR MAGNETON μn = e~=2mp = e~ 2mu=d e~ 2mu + 1 9 e~ 2md For a spin-up proton: Consider the contribution from quark 1 (an up-quark): Summing over the other contributions gives: m p = μn mu=d 23
Repeat for other (L=0) Baryons, PREDICT 24 μ n μ p = 2 3 compared to the experimentally measured value of 0:685 Baryon μ B in Predicted Experiment Quark Model [μ N ] [μ N ] 4 p 3 μ u 1 3 μ d +2:79 +2:793 4 n 3 μ d 1 3 μ u 1:86 1:913 Λ μ s 0:61 0:614 ± 0:005 ± + 4 Ξ 0 4 3 μ u 1 3 μ s +2:68 +2:46 ± 0:01 3 μ s 1 3 μ u 1:44 1:25 ± 0:014 Ξ 4 3 μ s 1 3 μ d 0:51 0:65 ± 0:01 Ω 3μ s 1:84 2:02 ± 0:05 Impressive agreement with data using m u = m d = 336 MeV and m s = 509 MeV. (see Question 5 on the problem sheet)
SUMMARY 25 Baryons and mesons are complicated objects. However, the Quark model can be used to make predictions for masses/magnetic moments. The predictions give reasonably consistent values for the constituent quark masses. m u=d m s Meson Masses 310 MeV 483 MeV Baryon Masses 363 MeV 538 MeV Baryon mag. moms. 336 MeV 510 MeV m u ß 335 MeV m d ß 335 MeV m s ß 510 MeV What about the HEAVY quarks (charm, bottom, top)...
Discovery of the J=ψ 26 1974 : Discovery of a NARROW RESONANCE in e + e collisions collisions at p s ß 3:1 GeV Observe resonances R μ at low p s - many bumps e + e! hadrons J=ψ(3097) e + e! μ + μ e + e! e + e Observed width, ο 3 MeV, all due to experimental resolution! Actual WIDTH, J=ψ ο 87 kev.
Zoom in to the CHARMONIUM (cμc) region, i.e. p s ο 2mc mass of charm quark, m c = 1:5 GeV 27 Resonances due to formation of BOUND unstable cμc states. The lowest energy of these is the narrow J=ψ state. e - e + γ J/ψ The particle physics of decaying particles: Particle Lifetimes Decay Widths Partial Widths Resonances i.e. the physics of e + e! Z 0. c c γ µ - µ + Much of this will be familiar from Nuclear Physics
28 Particle Decay Most Particles are transient beings - only the privileged few live forever (e ; u; d; fl; ::). Transition rate Undecayed! decayed given by Fermi s Golden Rule: T fi = 2ßjM fi j 2 ρ(e f ) T fi is the transition probability per unit time. Note, previously used fi for transition probability. Start with N particles, in time dt, NT fi dt will decay: dn = NT fi dt Z 1 N dn = Z T fi dt N = N 0 e T fit = N 0 e t=fi where fi is the mean lifetime fi = 1 T fi Finite lifetime ) UNCERTAIN energy E E t ο ~ Decaying states do not correspond to a single energy - they have a width, E E fi ο ~ E ο ~=fi = ~T fi The WIDTH, E, of a particle state is: ffl inversely proportional to the lifetime fi ffl EQUAL to the transition rate T fi using Natural Units
Decaying States!Resonances 29 QM description of decaying states Consider a state with energy E 0 and lifetime fi, need to modify the time dependent part of the wave-function (E 0 = ~!): ψ(t) = ψ 0 e ( ie 0t)! ψ 0 e ( ie 0t) e ( t 2fi ) ψ Λ ψ = ψ Λ 0 ψ 0 e( t fi ) i.e. probability density decays exponentially as required. The frequencies present in the wave-function are given by the Fourier transform of ψ(t) Z 1 f(!) = f(e) = = 0 Z 1 0 ψ 0 e t(ie 0+ 1 2fi ) e (iet) dt ψ 0 e t(i(e 0 E)+ 1 2fi ) dt ψ = 0 (E 0 E) i=(2fi) Probability of finding state with energy E = f(e) Λ f(e) ψ Λ 0 ψ 0 (E 0 E) 2 + 1 4fi 2 Finite lifetime, uncertain energy
Cross-section for producing the decaying state has this energy dependence - i.e. RESONANT when E = E 0 30 σ 0 σ 0 /2 Γ BREIT-WIGNER distribution E 0 -Γ/2 E 0 E 0 +Γ/2 Consider full-width at half-maximum amplitude (FWHM) ff(e) / 1 (E 0 E) 2 + 4fi 1 ff(e = E 0 ) / 4fi 2 ff(e) = ff 0 1 4fi 2 (E 0 E) 2 + 4fi 1 2 now ff(e 0 ± 1 2fi ) = ff 0 1 4fi 2 (E 0 E 0 ± 2fi 1 )2 + 4fi 1 2 = ff 0 2 So FWHM corresponds to ± 1 2fi (using Natural units) = 1 fi = T fi,ortotal WIDTH
Partial Decay Widths 31 Particles with more than one DECAY MODE, e.g. J=ψ! hadrons (87:7 ± 0:5 %) J=ψ! e + e (5:9 ± 0:1 %) J=ψ! μ + μ (5:9 ± 0:1 %) The numbers in brackets are the decay BRANCHING FRACTIONS i.e. 87:7 % of the time J=ψ! hadrons Transition rate for DECAY MODE k: T k fi = 2ßjM J=ψ!k j2 ρ(e) Total decay rate J=ψ!anything: X Tfi total = k THIS determines the lifetime, fi = 1 Tfi total T k fi The TOTAL WIDTH of a particle state, : = ~T total fi = ~ Define the PARTIAL WIDTHS, k k = ~T k fi X k T k fi = X k k Finally, BRANCHING FRACTION f k : f k = T k fi T total fi = k
Partial Widths : Example J=ψ 32 LIFETIME: J=ψ has lifetime fi J=ψ = 7:6 10 21 s immeasurably small! TOTAL WIDTH: J=ψ = fi J=ψ = 87 ± 5 kev ~ BRANCHING FRACTIONS: J=ψ! hadrons (87:7 ± 0:5 %) J=ψ! e + e (5:9 ± 0:1 %) J=ψ! μ + μ (5:9 ± 0:1 %) PARTIAL WIDTHS e:g: J=ψ!e + e = J=ψ 0:059 = 5 kev J=ψ!μ +μ = 5 kev J=ψ!hadrons = 77 kev COMMON MISCONCEPTIONS: Different partial widths for the decay modes DOES NOT mean different widths for the resonance curve. This is determined by the lifetime (i.e. the TOTAL width). Different partial widths do not imply a different constant in the exponential lifetime expression for J=ψ! e + e - only the total decay rate matters
Resonant Production 33 e + Z 0 q e - e + γ e - J/ψ c c q γ µ - µ + Cross section for particle scattering/annihilation via an intermediate DECAYING state, i! j! f. Start from ff = 2ßjMj 2 ρ(e) Previously (handout II) used time-ordered PT and looked at the terms in the perturbation expansion: M fj 1 E i E j M ji Sum over all time-orderings to give propagator (Feynman diagrams). M fj 1 q 2 m 2 M ji For a DECAYING intermediate state ψ(t) = ψ 0 e ( iet)! ψ 0 e ( iet) e ( t i:e: E! E i = E i 2fi 2 2fi ) Make this substitution for E j in the perturbation expansion.
M fj 1 E i E j M ji! M fj 1 E i E j + i =2 M ji 34 In the limit 2 fi m 2 the propagator for a DECAYING state becomes: 1 q 2 m! 1 2 q 2 m 2 + im EXAMPLE: ( NON-EXAMINABLE) e - e + e + e! J=ψ! μ + μ γ J/ψ In centre-of-mass system (q 2 = E 2 ) would expect cross section to be of form: c c ff ο jm e + e!ψ j 2 1 (E 2 -m 2 ψ )2 + 2 jm ψ!μ +μ j 2 ρ(e μ ) ψ m2 ψ First, note partial width for J=ψ! μ + μ : γ µ - µ + μμ ο jm ψ!μ +μ j 2 ρ(e μ ) giving ff ο jm e + e!ψ j 2 1 (E 2 -m 2 ψ )2 + 2 μμ ψ m2 ψ Partial width for J=ψ! e + e : ee ο jm ψ!e + e j 2 ρ(e e ) with ρ(e e ) = 4ß E e 2 (2ß) 3
35 giving jm ψ!e + e j 2 ο ß ee E 2 e BUT jm ψ!e + e j 2 = jm e + e!ψ j 2 ff = ß E 2 e (E 2 m 2 ψ )2 + m 2 ψ 2 ψ 4m 2 ψ ee μμ (The factor 4m ψ comes from the relativistic wave-function normalization) Need to sum over possible SPIN states of intermediate particle and average over initial electron states. Introduces SPIN-FACTOR g = 2J + 1 (2S 1 + 1)(2S 2 + 1) where J is the ANGULAR MOMENTUM of the resonance, S 1, S 2 are the SPINS of the initial particles. ff = g 4ßm2 ψ ee μμ Ee 2 (E 2 m 2 ψ )2 + m 2 ψ 2 ψ Which for 2 fi m 2 ψ conventional form: may be written in the more ff = g ß E 2 e ee μμ (E m ψ ) 2 + 2 ψ =4 THE BREIT-WIGNER RESONANCE FORMULA NOTE: E e is the energy of a SINGLE initial state particle. (see Question 6 on the problem sheet)
Charmonium 36 In e + e collisions observe cμc bound states as resonances, e.g. e + e! J=ψ! μ + μ. Only produce cμc bound states with same J PC as photon 1. Only directly produce 1 states in e + e collisions Other states observed via DECAY e.g.
Photonic decays e.g. ψ(3685)! χ + fl χ! J=ψ(3097) + fl 37 Crystal Ball Expt. 1982 e + e! ψ(3686) First 3 peaks: ψ(3686)! χ(3556) fl ψ(3686)! χ(3510) fl ψ(3686)! χ(3415) fl Fourth peak: χ(3556)! J=ψ(3097) fl χ(3510)! J=ψ(3097) fl Knowing the cμc levels provides a probe of the QCD potential: Because QCD is a theory of a strong confining force (self-interacting gluons) it is VERY difficult to calculate the exact form of the QCD potential from first principles. However, it is possible to experimentally determine the QCD potential by finding an appropriate form which gives the observed charmonium states. In practice one finds that V QCD = 4 ff S 3 r + kr with ff S = 0:2 and k = 1 GeV fm 1 provides a good description of the EXPERIMENTALLY OBSERVED levels in the charmonium system. yet to explain why low energy states are so narrow...
Why is the J=ψ so narrow? 1 3 S 1 ψ(3097) ß 0:09 MeV 2 3 S 1 ψ(3685) ß 0:28 MeV 3 3 S 1 ψ(3770) ß 25 MeV 4 3 S 1 ψ(4040) ß 52 MeV 38 Width depends on whether the decay to lightest mesons containing c quarks is kinematically possible: D (dμc), D + (cμd) m D ± = 1869:4 ± 0:5 MeV IF m(ψ) > 2m(D) c ψ c d c d c IF m(ψ) < 2m(D) ψ c c D + D - π + π 0 π - ψ! D + D ALLOWED ordinary STRONG DECAY ) large width ZWEIG RULE Unconnected lines in the quark diagram lead to SUPPRESSION of the decay amplitude ) narrow width Requires at least 3 Gluons / ff 6 s, i.e. suppressed ) NARROW
Bottomonium 39 In 1977 (9460) state discovered lowest energy 3 S 1 bound b μ b state ) m b ο 4.7 GeV Similar properties to the ψ Bottomonium spectrum well described by same QCD potential as used for charmonium Evidence that the QCD potential does not depend on Quark type Very similar properties to charmonium
Summary 40 m ß u 335 MeV m ß d 335 MeV m ß s 510 MeV m ß c 1:5 GeV m ß b 4:5 GeV m ß t 175 GeV..and have discussed decaying intermediate states. e - e + γ J/ψ c c γ µ - µ + and hopefully have a better understanding of the Breit-Wigner formula. ff = g ß E 2 e ee μμ (E m ψ ) 2 + 2 ψ =4 We ll see this again when it comes to e + e! Z 0
NON-EXAMINABLE 41 APPENDIX: J=ψ!e+ e Calculation of the partial width for J=ψ! e + e J/ψ c c Notation : J=ψ mass = m ψ, Q c is charge of charm quark, electron/positron energy E e = m ψ =2 (neglecting m e ) γ e - e + e + e = ~T fi = 2ßjMj 2 ρ(e e ) neglecting m e dρ(e e ) = E2 e dω (2ß) 3 Assume isotropic decay and integrate over solid angle: E e = m ψ 2 Matrix element: ρ(e e ) = 4ßE2 e (2ß) 3 ) ρ(e e ) = ßm2 ψ (2ß) 3 M = he + ee j 1 q 2 jμcq ceci M = Q c e 2 1 q 2 ψ cμc(0) jmj 2 = Q 2 c (4ßff)2 1 q 4 jψ cμc(0)j 2
q = (E ψ ; ~0) = (m ψ ; ~0) 42 jmj 2 = Q2 c (4ßff)2 m 4 ψ e + e = 2ß Q2 c (4ßff)2 m 4 ψ = 4ß Q2 c ff2 m 2 ψ jψcμc(0)j 2 jψcμc(0)j 2 ßm2 ψ (2ß) 3 jψcμc(0)j 2 Need the wave-function of the cμc system ) probability that the cμc are in the same place. For arguments sake, take the wave-function of the J=ψ to correspond to that of ground state in the strong version of the Coulomb potential (i.e. Hydrogen atom ): ψcμc(r; ; ffi) = 1 p ß 1 a 0 3 2 exp jψcμc(r = 0)j 2 = 1 ß 1 a0 3 r a 0 e + e = 4 Q2 c ff2 a 3 0 m 2 ψ Take a 0 = 1:5 GeV 1 = 0:3 fm which is a reasonable value for actual potential. 2 2 1 2 1 3 1 2 e + e = 4 GeV 3 137 1:5 3:1 e + e = 2:9 10 6 GeV = 2:9 kev EXPERIMENTAL VALUE = 5:1 kev
It is somewhat fortuitous that we obtain roughly the correct answer. Many approximations, e.g. the wave-function will not correspond to that in a Coulomb potential,... The calculation illustrates a number of points: Decay rates are hard to calculate, we are no longer dealing with fundamental particles. Need to know the wave-function of the decaying state. Need to know the QCD potential However calculations of ratios more reliable since they use same assumptions. 43 ANOTHER EXAMPLE: The (9460) is the bμb state corresponding to the J=ψ. For a Coulomb potential a 0 = EXPT.!e + e = 4 Q2 b ff2 a 0 3 m 2 J=ψ!e + e!e + e J=ψ!e + e!e + e 3 2m q ff s = Q2 c Q 2 b m 3 c m 3 b = 4 1:53 0:3 3 4:75 3 = 4:0 = 5:1 1:25 ff 3 s (3:1) m 2 ff 3 s (9:5) m 2 ψ 9:5 2 0:2 3 3:1 2 = 4:1 ± 0:2 NOTE the use of a running value of ff S
APPENDIX : Charmonium 44 NON-EXAMINABLE Compare bound cμc states (CHARMONIUM), with bound e + e states POSITRONIUM For Positronium use non-relativistic Schrödinger Equation with V em = ff em r E n = ff2 emm e 4n 2 Levels split by spin-spin and spin-orbit interactions ) FINE STRUCTURE E n ο ff4 emm e n 3 For lowest energy bound cμc states can still use non-relativistic Schrödinger Equation V QCD = 4 3 ff S r for QCD potential ) solve numerically + kr Relative magnitude of FINE STRUCTURE E n E n ο ff2 n ff S ) larger splitting.