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

2. 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

3. 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.

4. 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.

5. T 3 |t 3, y  = t 3 |t 3, y  t 3  I 3 Y |t 3, y  = y|t 3, y  The COMMUTATION RELATIONS establish the stepping properties of these ladder (raising/lowering) operators T 3 (T + |t 3, y  ) I3I3 Y V+V+ VV TT TT UU UU Y (T + |t 3, y  ) = T + ( Y |t 3, y  ) = y(T + |t 3, y  ) = (t 3 +1)(T + |t 3, y  ) = T + (T 3 + 1)|t 3, y  = (T + T 3 + T + )|t 3, y 

6. To be applicable to quantum mechanics, the lowest dimensional representation of SU(3) – the set of 3-dimensional matrices – must act on, and their eigenvalues describe, a set of real physical states,with quantum numbers: T 3 = Y = ISO-SPIN I 3 +1/2 0 -1/2 HYPERCHARGE BARON NUMBER STRANGENESS CHARGE +1/3-2/3 +1/3 +1/3 +1/3 +1/3 Q = I 3 + ½Y Y  B+S 00 -1/3 +2/3

7. Since   U  is an assumed symmetry (mixing states within a multiplet but remaining invariant to strong interactions) consider:  *  e -i  G*  * which will also satisfy all the same equations as . Since G i =  (G i )* are obviously also traceless and hermitian (and satisfy the same algebraic Commutation Relations as the G i ) we have a completely equivalent alternate set of generators for these SU(3) transformations ~ Mathematically we say we have a dual representation (or an adjunct basis set).

8. Physically we interpret this as another possible set of fundamental states carrying the minimum quanta of isospin and hypercharge, though now the eigenvalues (the diagonal elements of 3 and 8 ) change signs. +1 11 11 downup strange +1 11 11 u d s The anti-particle quarks! I3I3 Y

9. As m strictly adds when combining | j 1 m 1 > | j 2 m 2 > the quantum numbers t 3, y must as well. +1 11 11 downup strange I3I3 Y This quark multiplet simply plots the points representing the 3 possible quark states. +1 11 11 dnup s I3I3 Y up/up A 2-quark state of up pairs would have a total t 3 =+1 and y=+2/3 +1 11 11 du s I3I3 Y uu ud us A ud state have a total t 3 =0 and y=+2/3

10. +1 11 11 du s I3I3 Y The 3  3=9 possible quark-quark states form 3  3= 3  6 mulitplets +1 11 11 11 11 Unfortunately the 3  6 mulitplets of quark-quark combinations include fractionally charged states, which do not seem to correspond to any known particles.

11. But by adding a 3rd quark (to the qq states we’ve built so far): +1 11 11 11 11 Which we can directly compare to the known spin- 1 / 2 baryons to these 6  3=18 qqq states which can be separated into 6  3= 8  10 mulitplets

12. +1 11 11 11 11 np  00 ++ 00   the spin ½ + baryon octet the spin 3 / 2 + baryon decuplet  **      ** *0*0 *+*+ 

13. Notice if you add a quark to an antiquark (or antiquark to a quark): +1 11 11 3  3=9 new states are defined, separable into 3  3= 1  8 mulitplets to be compared directly to the known 9 spin-0 and 9 spin-1 mesons +1 11 11

14. 11 11 K0K0  00 ++ 00 KK KK the spin 0  meson octet the spin 1  meson nonet +1 11 11  00 ++  0,  0 K+K+ K* 0 K*  K*  K* +

15. 1974 Accelerators 1 st breached the NEXT energy threshold and began creating NEW, HIGHER MASS particles! Endowed with another quantum number: requiring one more quark that carried it. As we will see later, by this time theorists had already extended u-d isospin symmetry into an anticipated s-c symmetry both subsets of larger SU(4) multiplets

16. c s du p n + + + + 0 0 + +   0 0        

17. L Dirac =iħc          mc 2  Look at the FREE PARTICLE Dirac Lagrangian because    e  i  and in all  pairings this added phase cancels! Dirac matrices Dirac spinors (Iso-vectors, hypercharge) This is just an SU(1) transformation, sometimes called a “GLOBAL GAUGE TRANSFORMATION.” Which is OBVIOUSLY invariant under the transformation   e i   (a simple phase change)

18. What if we GENERALIZE this? Introduce more flexibility to the transformation? Extend to: but still enforce UNITARITY? ei(x)ei(x) LOCAL GAUGE TRANSFORMATION Is the Lagrangian still invariant?   (e i  (x)  ) = L Dirac =iħc          mc 2  So: L ' Dirac =  ħc(    (x))          iħc e  i  (x)   (    )e  i  (x)   mc 2  i(    (x))  + e i  (x) (    )

19. L ' Dirac =  ħc(    (x))         iħc   (    )   mc 2  L Dirac For convenience (and to make subsequent steps obvious) define: (x)   (x)  c q L ' Dirac =  q         (   )  L Dirac then this is re-written as recognize this???? the current of the charge carrying particle described by  as it appears in our current-field interaction term

20. L =[iħc       mc 2   q      A  L ' Dirac =  q         (   )  L Dirac If we are going to demand the complete Lagrangian be invariant under even such a LOCAL gauge transformation, and that defines its transformation under the same local gauge transformation i.e., we must assume the full Lagrangian HAS TO include a current-field interaction: something that can ABSORB (account for) that extra term, it forces us to ADD to the “free” Dirac Lagrangian

21. L =[iħc       mc 2   q      A  We introduced the same interaction term 3 weeks back following electrodynamic arguments (Jackson) A  ) that “couples” to  If we chose to allow gauge invariance, it forces to introduce a vector field (here that means The search for a “new” conserved quantum number shows that for an SU(1)-invariant Lagrangian, the free Dirac Lagrangain is “INCOMPLETE.” A  ' = A  +   is exactly (check your notes!) the rule for GAUGE TRANSFORMATIONS already introduced in e&m! the transformation rule the form of the current density is correctly reproduced

22. The FULL Lagrangian also needs a term describing the free particles of the GAUGE FIELD (the photon we demand the electron interact with). Of course NOW we want the Lagrangian term that recreates that! Furthermore we now demand that now be in a form that is both Lorentz and SU(3) invariant! We’ve already introduced the Klein-Gordon equation for a massless particle, the result, the solution A = 0 was the photon field, A 

23. We will find it convenient to express this term in terms of the ANTI-SYMMETRIC electromagnetic field tensor More ELECTRODYNAMICS: The Electromagnetic Field Tensor E, B do not form 4-vectors E, B are expressible in terms of   and A  but A  =(V,A) and J  =(c ,  v x,  v y,  v z ) do! the energy of em-fields is expressed in terms of E 2, B 2 F  =   A  A  transforms as a Lorentz tensor! = E x since = B z since

24. In general =  E x =  B z =  Actually the definition you first learned: F ik =  F ki = While vectors, like J  transform as “tensors” simply transform as

25. x' =  x or x =   x' Under Lorentz transformations

26. So, simply by the chain rule: and similarly:

27. (also xyz  yzx  zxy) both can be re-written with (with the same for x  y  z) All 4 statements can be summarized in

28. The remaining 2 Maxwell Equations: are summarized by ijk = xyz, xz0, z0x, 0xy Where here I have used the “covariant form” F  = g  g  F  =