Nucleon Scattering d dd dd dd dd dd d | I,I 3  | 1, 1  | 1,  1  | 1 0  If the strong interaction is I 3 -invariant These reactions must occur with equal strengths…equal probabilities… equal CROSS SECTIONS involve identical matrix elements

p + p  d +   p + n  d +    n + n  d +   Now consider these (possible and observed) collisions | 1, 1 > | 1,-1 > < 1, 1 | < 1, 0 | < 1,-1 | : : (|1,0  + |1,1  ) 1 : : 1  pp  d  + :  pn  d  0 :  nn  d   : = 2 : 1 : 2 1/  2 Then the ratio of cross sections:

p pn np pn n p pp p Consider the scattering reactions: 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, I 3 | I,I 3 > | 3/2, 3/2 > | 3/2, -3/2 > But = M + M this interaction involves two matrix elements p pn np pn n   + p    + p means combining: 1 ½ -1 -½ -1 ½ | 1 -1 > | 1/2, 1/2 > = |3/2, -1/2 >  |1/2, -1/2 > )

  + p    + p   + p    + n   + p    + p a. b. c. elastic scattering but only one of the above can also participate in a charge exchange process 1,1 1,-1 ½,½ 1, 0½,  ½ This IS observed! So all strong interactions not SIMPLY charge independent. I 3 ISOSPIN independence is more general. ? ?

  + p    + p   + p    + n   + p    + p a. b. c. elastic scattering charge exchange process These three interactions involve the ISOSPIN spaces:   p   p   n   Recall: 2 = M fi 2 M½M½ M 3/2 same by I 3 -indep. Let’s denote: 1,1 1,-1 ½,½ 1, 0½,  ½

  + p    + p   + p    + n   + p    + p a. b. c. elastic scattering charge exchange process   p   p   n   M½M½ M 3/2 a. b. c.  a  M 3/2 2  b  | M 3/2 + M 1/2 |  c  | M 3/2  M 1/2 | 2 2323 2323

  + p    + p   + p    + n   + p    + p a. b. c.  a  M 3/2 2  b  | M 3/2 + M 1/2 |  c  | M 3/2  M 1/2 | 2 2323 2323  a :  b :  c = : : M 3/2 |M 3/2 +2 M 1/2 | |M 3/2  M 1/2 | for the combined cross section of both processes Now if M 3/2 = M 1/2 then   + p =   - p total but also   - p  0 n = 0

2 H target S1 S2 S3 S4  beam Total Cross Section  T ( cm 2 )   + p   + p  p    p  p  0 p Lab Energy of Pion Beam (MeV) Photon Beam Energy (MeV) 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  )

 a :  b :  c = : : M 3/2 |M 3/2 +2 M 1/2 | |M 3/2  M 1/2 |  a :  b +  c = : M 3/2 2  a :  b :  c = 9 : 1 : 2

Symmetry implies any transformation  still satisfies the same Schrödinger equation, same Hamiltonian: (U)(U)(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! sinceyou’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 isospin mass charge Particle I 3 MeV/c 2 states Q nucleon p n 0 pion      1 delta   + +1  0 0    1 rho 770.  + +1  0 0    1 eta  /2  1/   1 +3/2 +1/2  1/2  3/2 Spin-0 Baryons Spin-1/2 Spin-3/2 omega  0 0 0

Q = I 3 + 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 K 0    +  

1947 Rochester and Butler cloud chamber cosmic ray event of a neutral object decaying into two pions K 0    +   1949 C. F. Powell photographic emulsion event K +        m = MeV m = MeV

p      p +   m  = MeV m p = MeV 1950 Carl Anderson (Cal Tech)

1952 Brookhaven Cosmotron 1st modern accelerator artificially creating these particles for study GeV p synchrotron Lawrence,Berkeley GeV p synchrotron CERN, Geneva 33-GeV p synchrotron Brookhaven Lab GeV e synchrotron Cambridge GeV p synchrotron Argonne Lab GeV p synchrotron DESY,Germany GeV e Linac SLAC (Standford)

Spin-0 Pseudoscalar Mesons nucleon p n 0 pion      1 rho 770.  + +1  0 0    1 eta  /2  1/   1 Spin-1/2 Baryons omega  isospin mass charge Particle I 3 MeV/c 2 states Q lambda  0 Sigma  + +1  0 0     1 Cascade  + +1    1 +1/2  1/2 0 kaon K K /2  1/2 kaon K K   1 +1/2  1/2

Delta   + +1  0 0    1 +3/2 +1/2  1/2  3/2 Spin-3/2 Baryons isospin mass charge Particle I 3 MeV/c 2 states Q Sigma-star   + +1   0 0      1 Cascade-star  * + +1  *   1 +1/2  1/2

pdg.lbl.gov/pdgmail

FRANK & EARNEST

These new heavier particle states were produced as copiously as  s in nuclear collisions (and in fact decay into  s) all evidence of STRONG INTERACTIONS these new states decayed slowly like the weak decays p  n + e  +    + which decay via neutrinos (accepted as the “signature” of a weak decay) but unlike STRONG production/decay phenomena like nuclear resonances (all with  final decay products, like the  ) which decay “instantly”, i.e., as readily as they are produced ELECTROMAGNETIC production/decay phenomena atomic (electron) resonances (all with  decay) or

What else wasabout them? Observed   + p +  K + +      K  +      K  +  NEVER Observed   + p +   + +      K  + n   +  all still conserve mass, charge, isospin “Associated production” Also NEVER observe:   + p +    +   but DO see:   + p +   + +  

K+KKKK+KKK +1  1  Q = I 3 + ½Y Y  B+S (Pais, Gell-man) “Strangeness”

Spin-0 Pseudoscalar Mesons nucleon p n 0 pion      1 rho 770.  + +1  0 0    1 eta  /2  1/   1 Spin-1/2 Baryons omega  isospin mass charge Strangeness Particle I 3 MeV/c 2 states Q S lambda  0 Sigma  + +1  0 0     1 Cascade  + +1    1 +1/2  1/2 0 kaon K K /2  1/2 kaon K K   1 +1/2  1/  1 1111 111111 2222 11

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. isospin space +½ ½½  =   =  which we noted reduces to 2  2 = 1  3 the isospin 0 singlet state (   1212 ispin=1 triplet

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 +1 multiplet of additional states - the new eigenvalues were integers, restricted to a range (- to + ) and separated in integral steps - only one of its 3 operators, J 3, was diagonal, giving distinct eigenvalues. The remaining operators, J 1 and J 2, actually mixed states. - however, a pair of ladder operators could constructed: J + = J 1 + iJ 2 and J  = J 1 - iJ 2 which stepped between eigenstates of a given multiplet. nn -1/2+1/201

The SU(3) Generators are G i = ½ i just like the G i = ½  i are for SU(2) The ½ distinguishes UNITARY from ORTHOGONAL operators.  i appear in the SU(2) subspaces in block diagonal form. 3 ’s diagonal entries are just the eigenvalues of the isospin projection. 8 is 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.

In exactly the same way you found the complete multiplets representing angular momentum/spin, we can define T ±  G 1 ± iG 2 U ±  G 6 ± iG 7 V ±  G 4 ± iG 5 The remaining matrices MIX states. T ±, T 3 are isospin operators By slightly redefining our variables we can associate the eigenvalues of 8 with HYPERCHARGE.