1. 8:00am 9:00 10:00 11:00 12:00pm 1:00 2:00 3:00 4:00 5:00 6:00 7:00 Monday Tuesday Wednesday Thursday Friday Apr 25??? Make up schedule April 11 catch a 1:00pm flight April 27 leave early morning by car April 11 114 Ferguson Apr 24???

2. The Nuclear pp cycle producing energy in the sun 6 protons  4 He + 6  + 2 e + 2p 26.7 MeV Begins with the reaction 0.26 MeV neutrinos

3. 500 trillion solar neutrinos every second!

4. ALL NEUTRINOS ARE LEFT-HANDED ALL ANTI -NEUTRINOS ARE RIGHT-HANDED Helicity = m s /s =  1 Helicity = m s /s =  1

5. Dirac Equation (spin-½ particles) (   p   m    0  j  0  j  j 0 p ( ) = ( ) 0    0 0 p   p  0 where p  p x   p y   p z  0 1 1 0 0 -i i 0 1 0 0 -1 p z p x  ip y p x +ip y  p z (  0 p 0   p    m   

6. Our “Plane wave” solutions ( for FREE Dirac particles)   r,t) = a exp[i/h(Et-p r)] u (E,p)  a e (i/h)x   p  u (E,p) which gave (   p   m  u = ( )( ) E/c  mc  p  u A p  E/c  mc u B from which we note: u A = ( p   u B u B = ( p   u A c  mc  c  mc 

7. Dirac Equation (spin-½ particles) EcEc multiply from left by (-i  1      recall    i  0  1  2  3 -i      3  1 = - i  1 ) 2  2  3 = + i  2  3 = + i  2  3 ) ( )( ) = + i  i  1 ) ( ) =  1      p  )I )  =  i m      3   EcEc since    =     since (  i )   0 1 -1 0 0 1 -1 0 0 -1 so  p x  1   p x  1 I    p x  1  p y  2  p z  3 = m  -i      3  0 = +i  0  1  2  3 =  5 -i      3  2 =  2  -i      3  3 =  3 

8.     p  )I )  =  i m      3   EcEc This gives an equation that looks MORE complicated! How can this form be useful? For a ~massless particle (like the or any a relativistic Dirac particle E >> m o c 2 ) E=|p|c as m o  0 (or at least m o <<E)   p|    p  )I )  =   Which then gives: or:      p  I )  =   ^ What do you think this looks like? p  I  ^ is a HELICITY OPERATOR!  I =   0 0 

9. In Problem Set #5 we saw that if the z-axis was chosen to be the direction of a particle’s momentum were all well-defined eigenspinors of S z i.e.   p  I ) u(p)  =  u(p)  ^ “helicity states”      p  I )  =   ^       p  I )   ^  5 “measures” the helicity of  So

10. Looking specifically at  5 u(p) = = uAuBuAuB uBuAuBuA For massless Dirac particles (or in the relativistic limit)  5u(p)= 5u(p)=  p  I ) u(p)  ^

11. We’ll find a useful definition inthe “left-handed spinor” u L (p)= u(p) (1  5 ) 2 Think: “Helicity=  1” In general NOT an exact helicity state (if not massless!) Since  5 u(p) = ±u(p) for massless or relativistic Dirac particles 0 if u(p) carries helicity +1 u(p) if u(p) carries helicity  1 if neither it still measures how close this state is to being pure left-handed separates out the “helicity  1 component” Think of it as a “projection operator” that picks out the helicity  1 component of u(p)

12. Similarly, since for ANTI-particles:  5 v(p) =  (p·  I)v(p)    again for m  0 we also define: v L (p)= v(p) (1  5 ) 2 with corresponding “RIGHT-HANDED” spinors: u R (p) = u(p) (1  5 ) 2 v R (p)= v(p) (1  5 ) 2 and adjoint spinors like since  5 † =  5 since  5   = -    5

13. Chiral Spinors Particles u L = ½(1  5 ) u u R = ½(1+  5 ) u u L = u ½(1  5 ) u R = u ½(1  5 ) Anti-particles v L = ½(1  5 ) v v R = ½(1  5 ) v v L = v ½(1  5 ) v R = v ½(1  5 ) Note: u L + u R = ( ) u + ( ) u = 1   5 2 1   5 2 u and also: ( ) ( ) u = 1   5 2 1   5 2 1  2  5 +   5 ) 2 4 ( ) u 2  2  5 4 = ( ) u 1   5 2 = ( ) u

14. Chiral Spinors Particles u L = ½(1  5 ) u u R = ½(1+  5 ) u u L = u ½(1  5 ) u R = u ½(1  5 ) Anti-particles v L = ½(1  5 ) v v R = ½(1  5 ) v v L = v ½(1  5 ) v R = v ½(1  5 ) note also: ( ) ( ) u = 1   5 2 1   5 2 1  2  5 +   5 ) 2 4 ( ) u 2  2  5 4 = ( ) u 1   5 2 = ( ) u while: ( ) ( ) u = 1   5 2 1   5 2 1   5 ) 2 4 ( ) u = 0 Truly PROJECTION OPERATORS!

15. Why do we always speak of beta decay as a process “governed by the WEAK FORCE”? What do DECAYS have to do with FORCE ? Where’s the FORCE FIELD? What IS the FORCE FIELD? What VECTOR PARTICLE is exchanged? n e-e- p e _ What’s been “seen” We’ve identified complicated 4-branch vertices, but only for the mediating BOSONS… Not the FERMIONS! ++ ++   ee  e _ semi-leptonic decays leptonic decay

16. p e+e+ n e _ We’ve also “seen” the inverse of some of these processes:  ee e 

17. The semi-leptonic decays (with participating hadrons) must Internally involve the transmutation of individual quarks: dduddu u d ee e _ u ?? udud _ ++  duuduu e+e+ u d d e _

18. Protons, quarks, pions and muons are all electrically charged so do participate in:  ee ee p p Can we use QED as a prototype by comparing or to ??? ee e n p leptonbaryon Charge-carrying currents imply a charged vector boson exchange! (we’ve already seen gluons carry color)

19. d e   u e Then might explain  -decay! +2/3  1/3 11  quark-flavor coupling ℓ- ℓ coupling e e  u d _ To explain  + decay: requires a +1 charge carrier What about   decays? and    e e   e _ explains  decays but coupling only to the left-handed particle states coupling strength modulated by left-handed components

20.  e e   W    d e  W    W   u e u e e  W   d We’ve seen the observed weak interactions:    e  + e +  n  p + e  + e p + e  n + e  could all be explained in terms of the interaction picture of vector boson exchanges if we imagine a the existence of a W 

21.  W   We’ve identified two fundamental vertices to describe the observed “weak” interactions. e W  e  or u W  d Quark coupling Flips isospin! Changes mass! Changes electric charge! Lepton coupling Changes electric charge! Changes mass! Some new “weak charge” that couples to an energy/momentum carrying W ±

22. e  e Continuing the analogy to qJ  lepton A  e ee e In general for a Quantum Mechanical charge carrier, the expression for “current” is of the form but these newest currents would have to allow  e O   e coupling to a “weak-field” W  Which must carry electric charge (why?) but not couple to it (why?)

23. If this interaction reflects a symmetry, how many weak fields must there be? U(1) SU(2) SU(3) one field (the photon) YANG MILLS: (gluons) 3 fields COLOR: 8 fields U(1) is clearly inadequate U(2) would mean 3 weak fields we know we need W +, W  Could there be a neutral W 0 ? But YANG-MILLS assumes we have “ISO”DOUBLET states!

24. Left-handed weak iso-doublets (in a new weak “iso”-space) u d e e  Right-handed weak iso-singlets u R d R e R L L NOT part of a doublet… NOT linked by the weak force to neutrinos NOTE: there is NO ( e ) R We’ve discovered we do have:

25. Left-handed weak iso-doublets (in a new weak “iso”-space) u +½ d  ½ e +½ e   ½ Right-handed weak iso-singlets u R 0 d R 0 e R 0 L L With ISO-SPIN we identified a complimentary “hypercharge” representing another quantum value that could be simultaneously diagonalized with ISO-SPIN operators. We generalize that concept into a NEW HYPERCHARGE in this “weak” space.

26. Y L = 2Q – 2I 3 weak Left-handed I 3 weak u L +½ d L  ½ ( e ) L +½ e  L   ½ Right-handed u R 0 d R 0 e R 0 Y R = 2(Q) – 2(0)= 2Q Y L = 2(-1) – 2(-1/2) Y L = 2(0) – 2(+1/2) Y L = 2(-1/3) – 2(-1/2) Y L = 2(2/3) – 2(1/2) =  1 = 1/3

27. Not all weak participants have ELECTRIC CHARGE Its NOT electric charge providing the coupling All weak participants (by definition) carry weak iso-spin u +½ d  ½ e +½ e   ½ u +½ Y L = 1/3 d  ½ Y L = 1/3 e +½ Y L =  1 e   ½ Y L =  1 L L e W  e  u W  d But interactions are only well-defined by the theory if the fermion legs to a vertex have equal coupling strengths L L Y L = 2Q – 2I 3 weak

28. With DOUBLET STATES and an associated “charge” defined we can attempt a Yang-Mills gauge-field model to explain the weak force but with some warnings...

29. e  W 0  e  The Yang-Mills theory requires introducing a 3 rd field: Could this be the photon? How do we distinguish this process from  exchange? Maybe the noted U(1) symmetry is part of a much larger symmetry: U(1)  S U(2) ? U EM (1)  U Y (1) ×SU L (2) U(1)  U(1) × S U(2)

30. Straight from U(1) and the SU(2) Yang-Mills extension, consider: charge-like coupling to a photon-like field some new Yang- Mills coupling T i =  i /2 for left-handed doublets = 0 for right-handed singlets This looks like it could be U(1) with = q and B   A  Yg12Yg12 This all means we now work from a BIG comprehensive Lagrangian summed over all possible fermions f to include terms for u, d, c, s, t, b, e, , , e, , 

31. which contains, for example: plus similar terms for , , c, s, , , t, b,

32. Straight from U(1) and the SU(2) Yang-Mills extension, consider: charge-like coupling to a photon-like field some new Yang- Mills coupling T i =  i /2 for left-handed doublets = 0 for right-handed singlets This looks like it could be U(1) with = q and B   A  Yg12Yg12 This all means we now work from a BIG comprehensive Lagrangian summed over all possible fermions f to include terms for u, d, c, s, t, b, e, , , e, , 

33. which contains, for example: plus similar terms for , , c, s, , , t, b,