Nucleon Scattering

Nucleon Scattering p+dp+d p-dp-d p0dp0d | I,I3  | 1, 1 | 1,-1 | 1 0 If the strong interaction is I3-invariant These reactions must occur with equal strengths…equal probabilities… equal CROSS SECTIONS involve identical matrix elements

Now consider these (possible and observed) collisions p + p  d + p+ p + n  d + p0 n + n  d + p- | 1, 1> | 1,-1> < 1, 1 | < 1, 0 | < 1,-1| (|1,0 + |1,1) <pp|L|dp+> : <pn|L|dp0> : <nn|L|dp-> 1:: 1 1/2 Then the ratio of cross sections: sppdp+:spndp0: snndp-:= 2:1:2

Consider the scattering reactions: p+pp+p p-np-n p-p p-p The strong force does not discriminate between nucleon or pion charge. What can we expect for the cross section of these three reactions?

If we enforce conservation of isospin we can only connect initial and final states of the same total I, I3 p+pp+p p-np-n | 3/2, 3/2> | 3/2, -3/2> | I,I3 > 1 ½ 1 ½ -1 -½ -1 -½ But p- + p  p- + p -1 ½ -1 ½ means combining: | 1 -1 > | 1/2, 1/2 > = |3/2, -1/2> - |1/2, -1/2>) this interaction involves two matrix elements <p-p|L|p-p> = M + M

p+ + pp+ + p a. b. c. 1,1 ½ ,½ elastic scattering p- + pp- + p 1,-1 ½ ,½ but only one of the above can also participate in a charge exchange process p- + pp0 + n ? ? 1,-1 ½ ,½ 1, 0 ½ ,-½ This IS observed! So all strong interactions not SIMPLY charge independent. I3ISOSPIN independence is more general.

p+ + pp+ + p a. b. c. 1,1 ½ ,½ elastic scattering p- + pp- + p 1,-1 ½ ,½ p- + pp0 + n charge exchange process 1,-1 ½ ,½ 1, 0 ½ ,-½ These three interactions involve the ISOSPIN spaces: p+ p p- p 1 3 2 3 - p0 n 2 3 1 3 + 2 2 M½ = Mfi Recall: M3/2 Let’s denote: same by I3-indep.

p+ + pp+ + p a. b. c. elastic scattering p- + pp- + p p- + pp0 + n charge exchange process p+ p M½ p- p 1 3 2 3 - M3/2 p0 n 2 3 1 3 + 2 a. b. c. aM3/2 2 b| M3/2 + M1/2| 2 3 1 3 2 c| M3/2-M1/2| 2 3 2 3

2 p+ + pp+ + p aM3/2 a. b. c. 2 b| M3/2 + M1/2| p- + pp- + p 2 3 1 3 2 c| M3/2-M1/2| p- + pp0 + n 2 3 2 3 a: b: c = : : 2 1 9 2 9 |M3/2 +2M1/2|2 |M3/2-M1/2|2 M3/2 for the combined cross section of both processes Now ifM3/2=M1/2 then +p = -p total but also -p0n= 0

p0p p+ + p gppo p Total Cross SectionsT (10-27 cm2) p- + p  beam 2H target S1 S3 S2 S4 Measured the depletion of pion beam repeated with the tank full, empty repeated with +and  - beam for KE195 MeV (the resonance of the 3/2-spin ) 200 300 180 250 160 200 140 120 150 100 100 80 50 60 0 40 40 60 100 200 400 Photon Beam Energy (MeV) 20 0 40 100 200 400 Lab Energy of Pion Beam (MeV)

a: b: c = : : 2 1 9 2 9 |M3/2 +2M1/2|2 |M3/2-M1/2|2 M3/2 2 a: b+c = : M3/2 a: b: c = 9: 1 :2

Symmetry implies any transformation still satisfies the same Schrödinger equation, same Hamiltonian:  (U) (U) U† U U†H U= H means we must demand: [H ,U]= 0 Which means that the operator U must be associated with a CONSERVED quantity! Though U are UNITARY, not necessarily HERMITIAN, but remember: where the G is Hermitian! since you’ve already shown [H ,U]= 0 [H ,G]= 0 The GENERATOR of any SYMMETRY OPERATION is an OPERATOR of a CONSERVED OBSERVABLE (quantum number!)

Mesons Spin-0 isospinmass charge Particle I3MeV/c2 states Q pion 139.569 + +1 134.964 0 0 139.569 --1 +1 0 -1 eta 548.8 0 0 0 +1 0 -1 rho770. + +1 0 0 - -1 omega 783.0 0 0 0 Baryons Spin-1/2 +1/2 -1/2 nucleon 938.280 p +1 939.573 n 0 Spin-3/2 delta 1232. ++ +2 + +1 0 0 --1 +3/2 +1/2 -1/2 -3/2

Q = I3 + ½Y “hypercharge” or BARYON NUMBER because =1 for baryons 0 for mesons

1947 Rochester and Butler cloud chamber cosmic ray event of a neutral object decaying into two pions K0  + +

1947 Rochester and Butler cloud chamber cosmic ray event of a neutral object decaying into two pions K0  + + m = 497.72 MeV 1949 C. F. Powell photographic emulsion event K+ + + m = 493.67 MeV

 - 1950 Carl Anderson(Cal Tech) p   p +  - m=1115.6 MeV mp=938.27 MeV

1952 Brookhaven Cosmotron 1st modern accelerator artificially creating these particles for study 1954 6.2-GeV p synchrotronLawrence,Berkeley 1960 28-GeV p synchrotronCERN, Geneva 33-GeV p synchrotronBrookhaven Lab 1962 6-GeV e synchrotron Cambridge 1963 12.5-GeV p synchrotronArgonne Lab 1964 6.5-GeV p synchrotron DESY,Germany 1966 21-GeV e LinacSLAC (Standford)

Spin-0 Pseudoscalar Mesons isospinmass charge Particle I3MeV/c2 states Q +1 0 -1 pion 139.569 + +1 134.964 0 0 139.569 --1 kaon 493.67 K+ +1 497.72 K0 0 +1/2 -1/2 kaon 497.72 K00 493.67 K--1 +1/2 -1/2 eta 548.8 0 0 0 rho 770. + +1 0 0 - -1 +1 0 -1 omega 783.0 0 0 0 Spin-1/2 Baryons nucleon 938.280 p +1 939.573 n 0 +1/2 -1/2 lambda 1115.6  0 0 Sigma 1385. + +1 0 0 --1 +1 0 -1 Cascade 1533. + +1 --1 +1/2 -1/2

Spin-3/2 Baryons isospinmass charge Particle I3MeV/c2 states Q Delta 1232. ++ +2 + +1 0 0 --1 +3/2 +1/2 -1/2 -3/2 Sigma-star 1385. + +1 0 0 --1 +1 0 -1 Cascade-star 1533. *+ +1 *--1 +1/2 -1/2

pdg.lbl.gov/pdgmail

FRANK & EARNEST

These new heavier particle states were produced as copiously asps in nuclearcollisions (and in fact decay into ps) all evidence ofSTRONGINTERACTIONS but unlike STRONGproduction/decay phenomena like nuclear resonances (all with p final decay products, like the D) which decay “instantly”, i.e., as readily as they are produced or ELECTROMAGNETICproduction/decay phenomena atomic (electron) resonances (all with g decay) these new states decayed slowly like the weak decays p n + e- +  p m +  which decay via neutrinos (accepted as the “signature” of a weak decay) ...Strange!

...strange... What else was about them? Observed p- + p+ K+ + S-  K0+ S0  K0+ L NEVER Observed p- + p+ p+ + S-  K0 + n p0+ L all still conserve mass, charge, isospin “Associated production” p- + p+ K- + S+ Also NEVER observe: but DO see: p+ + p+ K+ + S+

1952-53 (Pais, Gell-man) “Strangeness” K+ K0 K- K0 L S+ S0 S- +1 +1 -1 -1 -1 -1 -1 -1 Q = I3 + ½Y YB+S

Spin-0 Pseudoscalar Mesons isospinmass charge Strangeness Particle I3MeV/c2 states Q S +1 0 -1 pion 139.569 + +1 134.964 0 0 139.569 --1 0 0 0 kaon 493.67 K+ +1 497.72 K0 0 +1/2 -1/2 +1 +1 -1 -1 kaon 497.72 K00 493.67 K--1 +1/2 -1/2 eta 548.8 0 0 0 0 rho 770. + +1 0 0 - -1 +1 0 -1 0 0 0 omega 783.0 0 0 0 0 Spin-1/2 Baryons nucleon 938.280 p +1 939.573 n 0 +1/2 -1/2 0 0 lambda 1115.6  0 -1 0 -1 -1 -1 Sigma 1385. + +1 0 0 --1 +1 0 -1 -2 -2 Cascade 1533. + +1 --1 +1/2 -1/2

SU(2) Combining SPIN or ISOSPIN ½ objects gives new states described by the DIRECT PRODUCT REPRESENTATION built from two 2-dim irreducible representations: one 2(½)+1 and another 2(½)+1 yielding a 4-dim space.  ispin=1 triplet +½ the isospin 0 singlet state  =  = isospin space 1 2 (  -  ) -½  which we noted reduces to2  2 = 1  3

SU(2)- Spin added a new variable to the parameter space defining all state functions - it introduced a degeneracy to the states already identified; each eigenstate became associated with a 2+1multiplet of additional states - the new eigenvalues were integers, restricted to a range (- to + ) and separated in integral steps - only one of its 3 operators, J3, was diagonal, giving distinct eigenvalues. The remaining operators, J1 and J2, actually mixed states. - however, a pair of ladder operators could constructed: J+= J1 + iJ2 and J-= J1 - iJ2 which stepped between eigenstates of a given multiplet. n -1/2 +1/2 -1 0 1

The SU(3) Generators are Gi = ½i just like theGi = ½iare forSU(2) The ½ distinguishes UNITARY from ORTHOGONAL operators. iappear in the SU(2) subspaces in block diagonal form. 3’sdiagonal entries are just the eigenvalues of the isospin projection. 8is ALSO diagonal! It’s eigenvalues must represent a NEW QUANTUM number! Notice, like hypercharge (a linear combination of conserved quantities), 8 is a linear combinations of 2 diagonal matrices: 2 SU(2) subspaces.

The remaining matrices MIX states. In exactly the same way you found the complete multiplets representing angular momentum/spin, we can define T± G1± iG2 U± G6± iG7 V± G4± iG5 T±, T3are isospin operators By slightly redefining our variables we can associate the eigenvalues of 8 with HYPERCHARGE.

Nucleon Scattering