In general NON-ABELAIN GAUGE THEORIES :

1. In general NON-ABELAIN GAUGE THEORIES: • introduce more interactions (vertices) • for SU(2) we saw both 3 and 4 particle interaction vertices • have (still)massless gauge particles (like the photon!) • the gauge field particles posses “charge” just like the • fundamental Dirac states • not electric charge - we’re trying to think of NEW forces TheYANG-MILLS was built on the premise that there existed • 2 elementary Dirac (spin-1/2) particles of ~equal mass • serving as sources for the force fields through which they interact NO SUCH PAIRS EXIST • proton/neutron isospin states were the inspiration, but • there is NO massless vector (spin 1) iso-triplet • (isospin 1) of known particle states • r-mesons? 770 MeV/c2 • p,n,r now recognized as COMPOSITE particles • isospin of up,dn quarks generalized into SU(3) SU(4) The strong force must be independent of FLAVOR up charm top down strange bottom i.e., the strong force does not couple to flavors. SO WHERE DOES THE STRONG FORCE COME FROM?

2. Rutherford 1911 Elastic scattering of α – particles on atoms Discovery of atomic nucleus Size of nucleus 10-5 size of atom

3. 1968 SLAC-MIT Deep inelastic scattering of e- of p, d observation of ~flat q2 dependence of R= σinel/σMott R ~ const →point-like scatterers inside proton Fermi referred to these as “partons” which eventually became identified with the quarks!

4. We’ve argued for the simple energy dependence Cross section  Energy possibly punctuated by resonances Cross section  and threshold effects as higher energy opens up new production of heavier particles Energy

5. e+e- anihilation cross section Cross-section, cm2 s = cms energy, GeV

7. Dirac particlesfundamental,charged Fermions (spin- 1/2) e-e+m-m+t-t+udcst b udcstb q: -1 +1 -1 +1 -1 +1 +2/3 -1/3 +2/3 -1/3 +2/3 -1/3 -2/3 +1/3-2/3 +1/3 -2/3 +1/3 All can be described by DIRAC equations w/charge coupling to g-fields - u e- q=e q=e 2/3e ?? e- - u m105 MeV/c2 u7.5 MeV/c2 d4.2 MeV/c2 c 1100 MeV/c2 s150 MeV/c2 b 4200 MeV/c2

8. Consider e+e- annihilation in an electron-positron collider e+e-, +-,  +  -, q+q- Q-coupling any flavor e e+ e- unclear if annihilated easy signature to detect! (hadrons) (+-) N e+e-hadrons Ne+e-+- R = = total probability of all possible quarks  iqi2 Q

9. m105 MeV/c2 u7.5 MeV/c2 d4.2 MeV/c2 c 1100 MeV/c2 s150 MeV/c2 b 4200 MeV/c2 When above the u,d,s threshold 2 3 = above the J/ mass (c-production threshold) 10 9 R = = 1.11 above the upsilon mass (b-production threshold) 11 9 R = = 1.22

10. 3.67 3.33 1.22 1.11 0.66 Off by a factor of3!!!!! Could every known quark have a previously un-noticed 3-fold degeneracy????

11. Baryon States StateQuark contentMassSpin p uud 938.272MeV 1/2 n udd 939.565MeV 1/2 uds 1115.683MeV 1/2 +uus 1189.37 MeV 1/2  0uds 1192.632 MeV 1/2  -dds 1197.449 MeV 1/2  0uss 1314.9 MeV 1/2  -dss 1321.32 MeV 1/2 ++uuu1230. MeV 3/2 +uud 1231 MeV 3/2  0 udd 1233 MeV 3/2  -ddd1234 MeV 3/2 *+ uus 1382.8 MeV 3/2 *0 uds 1383.7 MeV 3/2 *- dds 1387.2 MeV 3/2 *0 uss 1531.80 MeV 3/2 *- dss 1535.0 MeV 3/2  -sss1672.45 MeV 3/2

12. U(1) introduced an interaction term:(qY g m Y )Am a charged current coupled to a vector field Yang-Mills extended this to SU(2)with an interaction term of: (qY gm tY ).Gm 2    · G each is a “vector” on “iso-space” Remember: SU(2) includes ALL POSSIBLE traceless HERMITIAN2x2 matrices the most GENERAL such matrix is a1 + a2 + a3 = a .t 1 0 0 -1 0 1 1 0 0 -i i0 · G • the G = (G1, G2, G3) are three independent • 4-vector fields, but together form an ISO-vector • transform under SU(2) the same way t does: U(· G)U† = U U†· UGU†

13.  2 Under the SU(2) Iso-spin transformations: 1 2 recall it was from that we determined the correct transformation property of G U(· G)U† G / =RTG-   Not a column vector in iso-space! Not a matrix! the 3 independent fields are isospin components which transform like under a basis transformation G must obey the same COMMUTATOR relations that  does, i.e., they form an adjoint representation of  (a set of functions isomorphic – mathematically equivalent to the  matrices).

14. Generalizing YANG-MILLS L ~SFi mnFimn= ( ) . ( ) GAUGE FIELDS Fimn = mGin - n Gim - 2ge ijk GjmGkn c The exact form depends on the non-abelian nature of the generators (their commutator rules) because[ti,tj ] = ieijktk More generally Fimn = mGin - n Gim - 2gCijk GjmGkn c “structure constant” of SU(n)

15. So now, in the face of evidence that quarks carry some 3-fold degeneracy that needs to be explained LET’S TRY: SU(3) This time  must be not only a Dirac spinoru(s)(p) but a 3-column vector in some NEW SPACE as well:

16. SU(3) will require introducing an interaction term: (g Y g m Y ) .Gm l1, l2, … l8 span the 8-dim space (the 8-independent parameters) of the 3x3 traceless, Hermitian generators of SU(3) Gmthe 8-independent VECTOR fields behave like the lioperators: 0 1 0 0 -i 0 1 0 0 l1= 1 0 0 l2=i0 0 l3= 0 -1 0 0 0 0 0 0 0 0 0 0 0 0 1 0 0 -i 0 0 0 l4= 0 0 0 l5=0 0 0 l6=0 0 1 -1 0 0 i 0 0 0 1 0 0 0 0 1 0 0 l7=0 0 -il8=1/ /30 1 0 0 i0 0 0 -2

17. Since we have the explicit matrix form of the i, its easy to check any of these! [i, j] = 2iC ijkk 8×8×8 different C ijk but only = 56 distinct combinations 8·7·6 3·2·1 (since exchanges/rearrangements are already defined) C jik = C ikj=-C ijk only 9 of these are non-zero (with only 3 different values!) C 123 = 1 C 147=C 246=C 257 =C 345= C 516= C 637=1/2 C 458 = C 678 = 3/2

18. SU(3) States If we write the 3COLOR states as 0 b = 0 1 1 g = 0 0 0 r = 1 0 Look at 1r= g changes r g charge 1g= r HEY!!!!! Electrons don’t change their charge when emitting/absorbing photons! but remember our “gluons” carry “charge” 0 1 0 0 -i 0 1 0 0 l1= 1 0 0 l2=i0 0 l3= 0-1 0 0 0 0 0 0 0 0 0 0 0 0 1 0 0 -i 0 0 0 l4= 0 0 0 l5=0 0 0 l6=0 0 1 -1 0 0 i 0 0 0 1 0 0 0 0 1 0 0 l7=0 0 -il8=1/ /30 1 0 0 i0 0 0-2

19. SU(3) States If we write the 3COLOR states as 0 b = 0 1 1 g = 0 0 0 r = 1 0 1r= g 2r creates the same color exchange but with different coefficients: a change of phase 1g= r 3, 8: preserve/retain color but may change phase (still exchange/transfer energy & momentum - like s) 4b= 6b= g r 4, 5: 6, 7: 4g= 6r= b b 0 1 0 0 -i 0 1 0 0 l1= 1 0 0 l2=i0 0 l3= 0-1 0 0 0 0 0 0 0 0 0 0 0 0 1 0 0 -i 0 0 0 l4= 0 0 0 l5=0 0 0 l6=0 0 1 -1 0 0 i 0 0 0 1 0 0 0 0 1 0 0 l7=0 0 -il8=1/ /30 1 0 0 i0 0 0-2

20. or 6b= r 6, 7: 6r= b How does this work? The gluons form a color OCTET (a multiplet of 8). How can you get an OCTET from a basis of 3?

21. In SU(3) the 3 fundamental COLOR “charges” (the basis set in the lowest order representation) can be combined in pairs of COLOR/antiCOLOR to give either a SINGLET or an OCTET x x = x xx x x 3 3 = 8 1 x x Must be color/anticolor states like rbrgbrbggrgb While the fundamental process of the “strong” force is q  gq with a primitive vertex: q b r q Gm6or Gm7 Gm1 = (1/2 ) (rg + gr ) Gm2 = (-i/2 )(rg-gr ) rg or gr or inverting

22. 6of the gluon fields areindependent linear combinationsof the simple gluon fields we enumerated Gm1 = (rg + gr)/2 Gm4 = (bg + gb)/2Gm6 = (rb + br)/2 Gm2=-i(rg - gr)/2 Gm5=-i(bg - gb)/2Gm7=-i(rb + br)/2 TheCOLOR SINGLETwould be1/3 ( rr + gg + bb ) However, just as we argued that (isospin) part of the deuteron’s wavefunction had to be anti-symmetric If the color singlet gluon existed it would be exchanged between color singlet (color-less?) states which as you’ll see momentarily would suggest a long range strong force between neutron and proton the (color) part of any baryon’s wavefunction has to be antisymmetric and that characteristic must not change when color is exchanged! …which simply does not exist …and as a color singlet it would be directly observable as a free particle.